Unit - 2 : Cellular Organization

1. Which one of the following activities is associated

with Mitochondria-associated ER membranes

(MAM)?

1. Protein glycosylation

2. ATP synthesis

3. Phospholipid metabolism

4. Iron-sulphur cluster assembly

(2024)

Answer: 3. Phospholipid metabolism

Explanation:

Mitochondria-associated ER membranes (MAMs)

are specialized subdomains of the endoplasmic reticulum (ER) that

are closely apposed to mitochondria. They serve as critical sites for

inter-organelle communication, particularly in lipid transfer and

calcium signaling. One of the primary functions of MAMs is to

facilitate phospholipid metabolism by enabling the transfer of lipid

intermediates such as phosphatidylserine from the ER to

mitochondria, where it is converted to phosphatidylethanolamine—a

key mitochondrial phospholipid. This exchange is essential for

maintaining mitochondrial membrane integrity and overall lipid

homeostasis in the cell.

Why Not the Other Options?

❌

(1) Protein glycosylation – Incorrect; this process primarily

occurs in the rough ER and Golgi apparatus, not at MAM sites.

❌

(2) ATP synthesis – Incorrect; ATP is synthesized in the

mitochondrial matrix by the F₁F₀ ATP synthase during oxidative

phosphorylation, not by MAMs.

❌

(4) Iron-sulphur cluster assembly – Incorrect; this process occurs

within the mitochondria, particularly in the mitochondrial matrix,

and is not a function of MAMs.

2. In eukaryotic cells, DNA replication is restricted to

the S phase of the cell cycle because

1. DNA polymerase is present only in the S phase of the

cell cycle.

2. Origin recognition complex (ORC) recognizes origin

only in the S phase.

3. MCM helicases get activated in the S phase of the cell

cycle.

4. MCM helicases get activated in the G1 phase of the

cell cycle.

(2024)

Answer: 3. MCM helicases get activated in the S phase of the

cell cycle.

Explanation:

In eukaryotic cells, DNA replication is tightly

regulated and restricted to the S phase to ensure that the genome is

replicated once and only once per cell cycle. This regulation is

largely governed by the Minichromosome Maintenance (MCM)

helicase complex, which is loaded onto replication origins during the

G1 phase in an inactive form. This process is known as "licensing."

However, MCM helicases are only activated in the S phase through

the action of cyclin-dependent kinases (CDKs) and DDK (Dbf4-

dependent kinase), which trigger origin firing and initiate DNA

unwinding and replication. The temporal separation of loading and

activation prevents re-replication and ensures genome stability.

Why Not the Other Options?

❌

(1) DNA polymerase is present only in the S phase of the cell

cycle – Incorrect; DNA polymerases are expressed and available

throughout the cell cycle, but their activity is regulated, not their

presence.

❌

(2) Origin recognition complex (ORC) recognizes origin only in

the S phase – Incorrect; ORC binds to origins of replication

throughout the cell cycle, particularly in G1, and is essential for pre-

replication complex (pre-RC) formation.

❌

(4) MCM helicases get activated in the G1 phase of the cell cycle

– Incorrect; MCM helicases are loaded in G1 but are not activated

until the S phase.

3. Which one of the following does NOT occur during

ribosome-associated quality control of damaged

mRNA?

1. mRNA degradation

2. Nascent protein degradation

3 .Disengagement of ribosome from mRNA

4. Ribosome-mRNA monosome degradation

(2024)

Answer: 4. Ribosome-mRNA monosome degradation

Explanation:

Ribosome-associated quality control (RQC) is a

crucial process that deals with problematic translation events, often

arising from damaged or incomplete mRNAs. When a ribosome stalls

on such an mRNA, RQC mechanisms are triggered. These typically

involve the recruitment of specific factors that lead to the

degradation of the aberrant mRNA to prevent the production of

truncated or non-functional proteins. Simultaneously, the nascent

polypeptide chain, still attached to the ribosome, is also targeted for

degradation by cellular proteases. The ribosome itself is then

typically rescued and recycled, leading to its disengagement from the

problematic mRNA. While the mRNA and the nascent protein are

degraded, the ribosome-mRNA complex (specifically the monosome,

a single ribosome bound to mRNA) itself is generally disassembled

and its components (ribosomes and mRNA fragments) are recycled,

not entirely degraded as a single unit.

Why Not the Other Options?

❌

1. mRNA degradation – Occurs; Damaged or incomplete mRNAs

that lead to ribosome stalling are targeted for degradation by RQC

machinery.

❌

2. Nascent protein degradation – Occurs; The incomplete or

aberrant polypeptide chains produced from problematic mRNAs are

recognized and degraded by proteasomes.

❌

3. Disengagement of ribosome from mRNA – Occurs; Once the

mRNA is degraded and the nascent protein is targeted, the ribosome

is released from the mRNA to be recycled for further translation.

4. Given below are some components that could

potentially influence membrane fluidity.

I. Monomeric G-proteins

ii.Peripheral membrane proteins iii.Sphingolipids

iv.Phospholipid sidechain saturation

v. Cholestero

Choose the option that has all the components that

can influence membrane fluidity.

1.i, ii, v

2.i, ii, iv

3.iii, iv, v

4.ii, iv, v

(2024)

Answer: 3.iii, iv, v

Explanation:

Membrane fluidity is primarily determined by the

composition of the lipid bilayer. Let's analyze each component:

iii. Sphingolipids: These lipids have long, saturated fatty acid tails

and a polar head group containing sphingosine. Their structure

allows for tighter packing compared to phospholipids with

unsaturated fatty acid tails, thus decreasing membrane fluidity.

iv. Phospholipid sidechain saturation: The saturation of the

hydrocarbon tails of phospholipids directly affects membrane fluidity.

Unsaturated fatty acids (with double bonds) introduce kinks in the

tails, preventing tight packing and increasing fluidity. Saturated fatty

acids (without double bonds) allow for closer packing, decreasing

fluidity.

v. Cholesterol: This sterol lipid is found in animal cell membranes.

At high temperatures, cholesterol decreases fluidity by restricting the

movement of phospholipids. At low temperatures, it increases fluidity

by disrupting the regular packing of phospholipids. Thus, cholesterol

acts as a fluidity buffer.

Now let's consider the other options:

i. Monomeric G-proteins: These proteins are typically anchored to

the plasma membrane via lipid modifications but are primarily

involved in signal transduction. While their interactions with

membrane lipids can have localized effects, they do not have a global

influence on overall membrane fluidity in the same way that the bulk

lipid components do.

ii. Peripheral membrane proteins: These proteins associate with the

membrane surface through electrostatic interactions and hydrogen

bonds with integral membrane proteins or the polar head groups of

lipids. They do not penetrate the hydrophobic core of the lipid

bilayer and therefore have a limited direct impact on the fluidity of

the lipid tails.

Therefore, the primary components that directly influence the bulk

membrane fluidity are sphingolipids, phospholipid sidechain

saturation, and cholesterol.

Why Not the Other Options?

❌

(1) i, ii, v – Incorrect; Monomeric G-proteins (i) and peripheral

membrane proteins (ii) have a limited direct impact on membrane

fluidity compared to the lipid components.

❌

(2) i, ii, iv – Incorrect; Monomeric G-proteins (i) and peripheral

membrane proteins (ii) have a limited direct impact on membrane

fluidity compared to the lipid components.

❌

(4) ii, iv, v – Incorrect; Peripheral membrane proteins (ii) have a

limited direct impact on membrane fluidity compared to

sphingolipids.

5. Some features mentioned below are important for the

segregation of homologous chromosomes in meiosis

A.Synaptonemal complex formation between

homologous chromosomes.

B.Degradation of cohesins at the chromosome arms.

C.Retention of cohesins at the centromeres.

D.Bi-orientation of kinetochores of sister chromatids.

Which one of the following options has all correct

features?

1. A and B only

2. A and C only

3. A, B, and C only

4. A, B, C, and D

(2024)

Answer: 3. A, B, and C only

Explanation:

Let's break down why each feature is important for

homologous chromosome segregation in meiosis I:

A. Synaptonemal complex formation between homologous

chromosomes: This protein structure facilitates the intimate pairing

(synapsis) of homologous chromosomes, which is essential for

crossing over (recombination) to occur. Recombination creates

physical links (chiasmata) that, along with cohesins, hold

homologous chromosomes together until anaphase I.

B. Degradation of cohesins at the chromosome arms: Cohesins are

protein complexes that hold sister chromatids together along their

entire length. For homologous chromosomes to separate during

anaphase I, the cohesins holding the arms of sister chromatids must

be cleaved. This allows the arms to resolve and the homologous

chromosomes to move to opposite poles.

C. Retention of cohesins at the centromeres: While cohesins are

degraded along the chromosome arms, it is crucial that cohesins at

the centromeres are protected. This retention ensures that sister

chromatids remain attached to each other throughout meiosis I and

will only separate during meiosis II.

D. Bi-orientation of kinetochores of sister chromatids: Bi-orientation

refers to the attachment of sister chromatid kinetochores to

microtubules from opposite spindle poles. This is characteristic of

chromosome segregation in mitosis and meiosis II. In meiosis I, the

kinetochores of sister chromatids function as a single unit and attach

to microtubules from the same spindle pole (mono-orientation). This

ensures that the entire homologous chromosome moves to one pole.

Therefore, the correct features essential for the proper segregation of

homologous chromosomes in meiosis I are synaptonemal complex

formation, the degradation of cohesins at the chromosome arms, and

the retention of cohesins at the centromeres. Bi-orientation of sister

chromatid kinetochores is incorrect for meiosis I.

"Why Not the Other Options?"

❌

(1) A and B only – Incorrect; Retention of cohesins at the

centromeres is also a crucial step.

❌

(2) A and C only – Incorrect; The release of cohesion at the

chromosome arms is necessary for the separation of homologs.

❌

(4) A, B, C, and D – Incorrect; Bi-orientation of sister chromatid

kinetochores occurs in meiosis II, not meiosis I.

6. Among the organelles listed below, which one does

NOT obtain proteins via vesicular transport?

1. Endosomes

2. Lysosome

3. Mitochondria

4. Golgi

(2024)

Answer: 3. Mitochondria

Explanation:

Proteins destined for mitochondria are synthesized

in the cytosol and imported post-translationally through specialized

protein translocase complexes present in the outer and inner

mitochondrial membranes (e.g., TOM and TIM complexes). This

import mechanism is non-vesicular. Mitochondria do not receive

proteins via vesicular trafficking because they are not part of the

endomembrane system, which includes the ER, Golgi, endosomes,

lysosomes, and the plasma membrane.

Why Not the Other Options?

❌

(1) Endosomes – Incorrect; endosomes receive proteins via

vesicles derived from the trans-Golgi network and the plasma

membrane.

❌

(2) Lysosome – Incorrect; lysosomal proteins are delivered via

vesicles that bud from the Golgi and fuse with endosomes/lysosomes.

❌

(4) Golgi – Incorrect; the Golgi apparatus receives vesicular

transport from the endoplasmic reticulum (ER) and also sends

proteins via vesicles to various destinations.

7. Given below are a list of sub-cellular compartments

(Column X) and markers (Column Y).

Which one of the following options correctly matches

the subcellular compartments with their markers?

1. A-iii, B-iv, C-i, D-ii

2. A-iv, B-iii, C-ii, D-i

3. A-iii, B-ii, C-i, D-iv

4. A-ii, B-iv, C-i, D-iii

(2024)

Answer: 1. A-iii, B-iv, C-i, D-ii

Explanation:

To correctly match the subcellular compartments

with their markers, we need to know the characteristic proteins or

molecules associated with each compartment.

A. Endoplasmic reticulum (ER): Protein disulfide isomerase (PDI) is

a well-known marker enzyme for the ER, particularly the lumen of

the ER. It catalyzes the formation and rearrangement of disulfide

bonds in proteins during their folding. Therefore, A matches with iii.

B. Golgi apparatus: Mannosidase II is an enzyme located in the

medial-Golgi cisternae and is commonly used as a marker for the

Golgi apparatus. It plays a crucial role in the processing of N-linked

glycans on proteins. Therefore, B matches with iv.

C. Autophagosome: LC3b (Light Chain 3-II) is a protein that is

specifically recruited to autophagosomes during autophagy. Its

presence and accumulation on membranes are hallmarks of

autophagosome formation and maturation. Therefore, C matches

with i.

D. Mitochondria: HSP60 (Heat Shock Protein 60) is a mitochondrial

chaperonin protein involved in the folding and assembly of proteins

within the mitochondria. It is a reliable marker for the mitochondrial

matrix. Therefore, D matches with ii.

Combining these matches, we get: A-iii, B-iv, C-i, D-ii.

Why Not the Other Options?

❌

(2) A-iv, B-iii, C-ii, D-i – Incorrect; Mannosidase II is a Golgi

marker, and PDI is an ER marker. LC3b is an autophagosome

marker, and HSP60 is a mitochondrial marker.

❌

(3) A-iii, B-ii, C-i, D-iv – Incorrect; HSP60 is a mitochondrial

marker, and Mannosidase II is a Golgi marker.

❌

(4) A-ii, B-iv, C-i, D-iii – Incorrect; HSP60 is a mitochondrial

marker, and PDI is an ER marker.

8. Isolated mitochondria were either treated with

protease or first briefly incubated in hypotonic

solution prior to treatment with protease. The

reaction was stopped and samples were probed for

the presence of Mtg2, Porin (at the outer membrane),

Cyt c (in the inter-membrane space), and KDH (in

the matrix) using western blot analyses.

Based on the gels above, Mtg2 is localized:

1.in the outer membrane facing the cytosol.

2.in the inter-membrane space.

3.in the matrix.

4.traversing the inner and outer membrane

(2024)

Answer: 3.in the matrix.

Explanation:

Let's evaluate the Western blot data based on the

provided correct answer. If Mtg2 is localized in the mitochondrial

matrix, we would expect it to be protected from protease digestion in

intact mitochondria. Furthermore, even when the outer membrane is

disrupted by hypotonic solution, the inner mitochondrial membrane

should still provide some level of protection to a matrix protein from

externally added protease.

Looking at the Western blots:

Intact Mitochondria (Left Panel): The Mtg2 band shows consistent

intensity across increasing protease concentrations (0, 50, 100

µg/µl). This indicates that in intact mitochondria, Mtg2 is protected

from the protease. This protection could be due to its location within

the inner mitochondrial membrane or the matrix.

Mitochondria placed in hypotonic solution (Right Panel): After the

mitochondria are placed in hypotonic solution (which disrupts the

outer membrane and releases intermembrane space contents), the

Mtg2 band still shows significant intensity, although there might be a

slight reduction at the highest protease concentration (100 µg/µl).

This suggests that even with a compromised outer membrane, a

substantial amount of Mtg2 remains protected from the protease.

Considering the behavior of the other marker proteins:

Porin (Outer Membrane): As expected, Porin is readily degraded by

the protease in both intact and hypotonic-treated mitochondria,

confirming the accessibility of the outer membrane.

Cytochrome c (Intermembrane Space): Cytochrome c is partially

degraded in intact mitochondria and almost completely degraded in

hypotonic-treated mitochondria, consistent with its location

becoming more accessible upon outer membrane disruption.

KDH (Matrix): KDH remains largely protected in both conditions,

indicating the inner membrane's role in shielding matrix proteins

from external protease.

If Mtg2 were in the matrix, its protection profile should be similar to

that of KDH. The blots show that Mtg2 is indeed protected in intact

mitochondria. Even after hypotonic treatment, a considerable

amount of Mtg2 persists, suggesting it is still shielded by the inner

membrane, which is characteristic of a matrix-localized protein. The

slight reduction at high protease concentration in hypotonic

conditions could be attributed to some damage or increased

permeability of the inner membrane under these conditions, allowing

limited access to the matrix.

Therefore, based on the protection pattern observed in the Western

blots, the localization of Mtg2 in the matrix (Option 3) is the most

consistent with the data, assuming the provided correct answer is

accurate.

Why Not the Other Options?

❌

(1) in the outer membrane facing the cytosol – Incorrect; Outer

membrane proteins like Porin are readily accessible to protease in

intact mitochondria. Mtg2's protection in intact mitochondria

contradicts this localization.

❌

(2) in the inter-membrane space – Incorrect; Intermembrane

space proteins like Cytochrome c are at least partially accessible to

protease in intact mitochondria. Mtg2 shows greater protection.

❌

(4) traversing the inner and outer membrane – Incorrect; If Mtg2

significantly traversed both membranes, we might expect a more

complex degradation pattern, potentially with some fragments being

protected by the inner membrane even in intact conditions, but the

overall protection in intact mitochondria and substantial persistence

after hypotonic treatment point more towards a matrix localization.

9. Dynein is a microtubule-dependent motor protein

essential for cellular processes. Which one of the

following statements related to dynein function is

INCORRECT?

1. It helps in cilia and flagella beating.

2. It transports organelles in anterograde direction.

3. It participates in spindle assembly and positioning.

4. It interacts with dynactin for efficient transport.

(2024)

Answer: 2. It transports organelles in anterograde direction.

Explanation:

Dynein is a minus-end-directed motor protein that

moves along microtubules toward the microtubule organizing center

(MTOC), typically located near the nucleus. This direction is known

as the retrograde direction. Therefore, dynein does not mediate

anterograde transport, which is the movement from the cell center

toward the periphery (plus-end of microtubules); that function is

carried out by kinesin. Dynein plays a crucial role in retrograde

organelle transport, cilia and flagella beating, spindle positioning,

and works in complex with dynactin to facilitate cargo binding and

transport.

Why Not the Other Options?

❌

(1) It helps in cilia and flagella beating – Incorrect; this is a

correct statement as axonemal dynein powers the movement of cilia

and flagella.

❌

(3) It participates in spindle assembly and positioning – Incorrect;

dynein contributes to proper spindle orientation and chromosome

movement during mitosis.

❌

(4) It interacts with dynactin for efficient transport – Incorrect;

dynein binds to dynactin, a multiprotein complex that enhances its

processivity and cargo binding.

10. Which one of the following statements regarding

cytoskeleton proteins is INCORRECT?

1. Actin and tubulin proteins appear to have arisen from

single-copy genes, present before multicellular

eukaryotes diverged.

2. Tubulin specifical ly binds ATP, while actin binds

preferentially to GTP.

3. FtsZ, a tubulin relative, forms a filamentous ring

needed to affect cell division in certain prokaryotic

species.

4. MreB, an actin relative, forms filaments required for

the cell shape in certain prokaryotic species.

(2024)

Answer: 2. Tubulin specifical ly binds ATP, while actin binds

preferentially to GTP.

Explanation:

This statement is incorrect because it reverses the

actual nucleotide-binding preferences of these cytoskeletal proteins.

Tubulin binds GTP, not ATP, and actin binds ATP, not GTP. Tubulin

consists of α- and β-subunits, both of which bind GTP; the GTP

bound to β-tubulin is hydrolyzable and plays a critical role in

microtubule dynamics. Actin binds ATP and hydrolyzes it to ADP

upon polymerization into filaments, which is essential for actin

filament dynamics. Thus, the statement misrepresents a fundamental

biochemical feature of these proteins.

Why Not the Other Options?

❌

(1) Actin and tubulin proteins appear to have arisen from single-

copy genes, present before multicellular eukaryotes diverged –

Incorrect; this is correct as both are ancient, highly conserved

proteins predating the divergence of multicellular eukaryotes.

❌

(3) FtsZ, a tubulin relative, forms a filamentous ring needed to

affect cell division in certain prokaryotic species – Incorrect; this is

correct. FtsZ polymerizes to form a Z-ring crucial for prokaryotic

cytokinesis.

❌

(4) MreB, an actin relative, forms filaments required for the cell

shape in certain prokaryotic species – Incorrect; this is correct.

MreB forms cytoskeletal filaments that help maintain rod shape in

bacteria.

11. The CaM Kinase II acts as a molecular memory

device as well as a frequency decoder of Ca2+

oscillation. Which one of the followi ng statements

regarding CaM Kinase 11 is INCORRECT?

1. CaM Kinase 11 activity increases as a function of

Ca2+ pulse frequency.

2. CaM Kinase II requires Ca2+ and Calmodulin to

remain in an active state.

3. A brain-specific CaM Kinase II knockout mouse will

have difficulty in remembering where things are.

4. The molecular memory function is lost in an

autophosphorylation defective CaM Kinase 11.

(2024)

Answer: 2. CaM Kinase II requires Ca2+ and Calmodulin to

remain in an active state.

Explanation:

This statement is incorrect because while CaM

Kinase II (Calcium/Calmodulin-dependent Protein Kinase II)

initially requires Ca²⁺ and calmodulin for activation, it does not

require them to remain active once it undergoes autophosphorylation.

Upon repeated Ca²⁺ spikes, CaM Kinase II can autophosphorylate at

specific threonine residues (e.g., Thr286), leading to a calcium-

independent sustained activity, effectively functioning as a molecular

memory switch. This allows the kinase to remain active even after

Ca²⁺ levels return to baseline, enabling it to act as a decoder of Ca²⁺

oscillation frequency and contributing to long-term synaptic changes

like memory formation.

Why Not the Other Options?

❌

(1) CaM Kinase II activity increases as a function of Ca²⁺ pulse

frequency – Incorrect; this is correct. The enzyme integrates calcium

signals over time and becomes more active with higher frequency

calcium pulses.

❌

(3) A brain-specific CaM Kinase II knockout mouse will have

difficulty in remembering where things are – Incorrect; this is

correct. Such mice exhibit impaired spatial learning and memory, as

CaM Kinase II is crucial in hippocampal synaptic plasticity.

❌

(4) The molecular memory function is lost in an

autophosphorylation defective CaM Kinase II – Incorrect; this is

correct. Autophosphorylation is essential for its calcium-independent

activity, and loss of this function abolishes its molecular memory

capacity.

12. Which one of the following is NOT a major

phospholipid in mammalian plasma membrane?

1. Phosphatidylinositol

2. Phosphatidylserine

3. Sphingomyelin

4. Cholesterol

(2024)

Answer: 4. Cholesterol

Explanation:

Cholesterol is a crucial component of the

mammalian plasma membrane, but it is not a phospholipid.

Cholesterol is a sterol, which intercalates between phospholipid

molecules in the membrane to modulate fluidity and stability, but

structurally and functionally, it is distinct from phospholipids. The

major phospholipids in the plasma membrane include

phosphatidylinositol, phosphatidylserine, phosphatidylcholine,

phosphatidylethanolamine, and sphingomyelin.

Why Not the Other Options?

❌

(1) Phosphatidylinositol – Incorrect; it is a minor but functionally

important phospholipid involved in cell signaling.

❌

(2) Phosphatidylserine – Incorrect; it is a major phospholipid,

primarily located on the inner leaflet of the plasma membrane.

❌

(3) Sphingomyelin – Incorrect; though it has a sphingosine

backbone, it is a phospholipid and is a major component of the outer

leaflet of the membrane.

13. Which one of the following blood cell types is formed

from megakaryocyte during development of blood

cells from bone marrow?

1. Monocytes

2. Neutrophils

3. Eosinophils

4. Platelets

(2024)

Answer: 4. Platelets

Explanation:

Platelets are formed from megakaryocytes, which

are large bone marrow cells. These megakaryocytes undergo a

process called thrombopoiesis, where they fragment into smaller

pieces, each becoming a platelet. Platelets play a crucial role in

blood clotting and wound repair.

Why Not the Other Options?

❌

(1) Monocytes – Incorrect; Monocytes are a type of white blood

cell formed from myeloid precursor cells in the bone marrow, not

from megakaryocytes.

❌

(2) Neutrophils – Incorrect; Neutrophils are also derived from

myeloid progenitor cells in the bone marrow, not from

megakaryocytes.

❌

(3) Eosinophils – Incorrect; Like neutrophils, eosinophils are

derived from myeloid progenitor cells, not from megakaryocytes.

14. Which one of the following statements about

peripheral lymph nodes is INCORRECT?

1. They contain fibroblast reticular cells that form a

conduit system to guide cell movement within the node.

2. They are responsible for immune responses to blood-

borne pathogens.

3. T celils encounter antigen presented on dendritic cells

in the paracortex of lymph nodes.

4. They contain B cells in special areas called follicles.

(2024)

Answer: 2. They are responsible for immune responses to

blood-borne pathogens.

Explanation:

Peripheral lymph nodes are primarily responsible

for immune responses to tissue-borne pathogens, not blood-borne

pathogens. They filter lymph from tissues and serve as sites where

immune cells such as T cells and B cells can encounter antigens

presented by dendritic cells or macrophages. Blood-borne pathogens

are mainly addressed by the spleen, which filters the blood and

initiates immune responses to circulating pathogens.

Why Not the Other Options?

❌

(1) They contain fibroblast reticular cells that form a conduit

system to guide cell movement within the node – Correct; Fibroblast

reticular cells play an essential role in the organization of the lymph

node and form a conduit system that helps guide immune cells as

they move through the node.

❌

(3) T cells encounter antigen presented on dendritic cells in the

paracortex of lymph nodes – Correct; T cells in the lymph nodes

encounter antigens on dendritic cells in the paracortex, which is an

area between the outer cortex and the inner medulla of the node.

❌

(4) They contain B cells in special areas called follicles – Correct;

B cells are located in follicles within the cortex of the lymph node,

and these follicles are crucial for initiating humoral immune

responses.

15. Which one of the following is likely to enter a pure

phospholipid biilayer?

1. CO2 and Diethyl urea

2. Water and Glucose

3. Lysine and Ethanol

4. Urea and Chloride ions

(2024)

Answer: 1. CO2 and Diethyl urea

Explanation:

A pure phospholipid bilayer is selectively permeable,

allowing nonpolar and small uncharged polar molecules to diffuse

more readily than large or charged molecules. CO₂ is a small,

nonpolar gas that easily diffuses through the hydrophobic core of the

lipid bilayer. Diethyl urea is more hydrophobic than urea due to its

ethyl groups, allowing it to cross the bilayer more efficiently through

passive diffusion.

Why Not the Other Options?

❌

(2) Water and Glucose – Incorrect; Water can diffuse slowly, but

glucose is too large and polar to pass through a lipid bilayer without

a transporter.

❌

(3) Lysine and Ethanol – Incorrect; Ethanol (small and polar)

may pass through, but lysine is a charged amino acid and cannot

passively diffuse across the bilayer.

❌

(4) Urea and Chloride ions – Incorrect; Urea diffuses slowly, and

Cl⁻ (an anion) is highly hydrophilic and requires ion channels for

membrane crossing.

16. Which one of the following types of interactions will

be predominantly contributing to the stability of a

nucleosome?

1. Hydrogen bonds between DNA base pairs and serine/

threonine residues of histones.

2. van der Waal's interactions between DNA base pairs

and hydrophobic residues of histones.

3. Hydrogen bonds between DNA phosphate backbone

and the main chain atoms of histones.

4. Electrostatic interactions involving DNA phosphate

backbone and lysine residues of histones.

(2024)

Answer: 4. Electrostatic interactions involving DNA

phosphate backbone and lysine residues of histones.

Explanation:

The stability of a nucleosome is predominantly

maintained by electrostatic interactions between the negatively

charged phosphate backbone of DNA and the positively charged

lysine residues on the histone proteins. The lysine residues have a

positive charge due to their amino groups, which interact strongly

with the negatively charged phosphate groups on the DNA, helping

to anchor the DNA around the histone core and stabilize the

nucleosome structure.

Why Not the Other Options?

❌

(1) Hydrogen bonds between DNA base pairs and

serine/threonine residues of histones – Incorrect; Serine and

threonine residues are not the main contributors to nucleosome

stability. These amino acids may form hydrogen bonds, but they are

not as crucial as the interactions between lysine residues and the

DNA backbone.

❌

(2) van der Waals interactions between DNA base pairs and

hydrophobic residues of histones – Incorrect; While van der Waals

interactions can contribute to the stability of protein-protein

interactions, the primary force maintaining the nucleosome structure

is electrostatic, not van der Waals.

❌

(3) Hydrogen bonds between DNA phosphate backbone and the

main chain atoms of histones – Incorrect; While hydrogen bonds can

occur between the DNA and histone proteins, the primary interaction

maintaining nucleosome stability is electrostatic rather than

hydrogen bonding.

17. Given below are components that facilitate transfer

of molecules across phospholipid bilayers (Column X)

and the properties of these components (Column Y).

Choose the option that correctly matches the

components with their properties.

1. A (IV) B (II) C (I) D (Ill)

2. A (II) B (Ill) C (IV) D (I)

3. A (Ill) B (IV) C (I) D (II)

4. A (Ill) B (IV) C (II) D (I)

(2024)

Answer: 4. A (Ill) B (IV) C (II) D (I)

Explanation:

Let’s analyze each component from Column X and

its corresponding correct function in Column Y:

A: ATP-powered pumps correspond to (III): These pumps use ATP

hydrolysis to actively transport molecules against their concentration

and/or electrochemical gradient. Examples include Na⁺/K⁺-ATPase,

Ca²⁺-ATPase, etc.

B: Uniporter corresponds to (IV): Uniporters facilitate passive

transport of a specific molecule down its concentration gradient

without using energy. Example: GLUT1 transporter for glucose.

C: Channels correspond to (II): Ion channels or aquaporins allow

free or facilitated diffusion of hydrophilic molecules across

membranes. These operate passively.

D: Symporter corresponds to (I): Symporters carry two different

molecules simultaneously, where one moves down its gradient,

providing energy to move the other against its gradient (secondary

active transport). Example: Na⁺/glucose symporter.

Why Not the Other Options?

❌

(1) A (IV) – Incorrect; ATP pumps do not allow passive transport

(IV).

❌

(2) A (II) – Incorrect; ATP pumps do not enable free diffusion (II),

they require energy.

❌

(3) C (I) – Incorrect; channels do not engage in coupled

movement against a gradient (I); that is characteristic of symporters,

not channels.

18. Protein X can be extracted from disrupted

erythrocyte plasma membranes with high salt

concentrations. Treatment of intact erythrocytes with

protease followed by extraction led to intact protein

X. Treatment of disrupted erythrocyte plasma

membranes with protease followed by extraction led

to fragmented protein X.

The following interpretations were made:

A. Protein X is a peripheral membrane protein

B. Protein X is an integral membrane protein

C. Protein Xis on the extracellular matrix face of the

plasma membrane

D. Protein X is on the cytosolic face of the plasma

membrane

Which one of the following options best represents

the combination of all correct interpretations?

1. A and C

2. A and D

3. Band C

4. Band D

(2024)

Answer: 2. A and D

Explanation:

Protein X is extracted from disrupted erythrocyte

membranes using high salt concentrations, which typically removes

peripheral membrane proteins by disrupting ionic interactions, but

not integral membrane proteins that are embedded in the lipid

bilayer. This indicates that Protein X is a peripheral membrane

protein (A).

Additionally, when intact erythrocytes are treated with protease,

Protein X remains intact, meaning it is not exposed on the

extracellular face where the protease can access it. However, when

the membrane is disrupted and then treated with protease, Protein X

gets fragmented, implying it becomes accessible to protease from the

cytosolic side after membrane disruption. Hence, Protein X is

localized on the cytosolic face of the plasma membrane (D).

Why Not the Other Options?

❌

(1) A and C – Incorrect; Protein X is not on the extracellular face,

as it is not degraded in intact cells.

❌

(3) B and C – Incorrect; Protein X is a peripheral, not integral

membrane protein, and it is cytosolic, not extracellular.

❌

(4) B and D – Incorrect; Protein X is not an integral membrane

protein, it is peripheral.

19. The following statements were made regarding the

roles of histone modifications in transcriptional

regulation.

A. Acetylation of histones is generally associated with

transcriptional repression by making the chromatin

more compact.

B. Methylation of histones can either activate or

repress transcription depending on the specific

residue modified.

C. Phosphorylation of histones occurs in response to

DNA damage and can influence gene expression.

D. Histone modifications influence the recruitment of

RNA polymerase complex but not transcription

factors .

Which one of the following options represents the

combination of all correct statements?

1 . A and D only

2. B and C only

3. B, C, and D

4. A, B, and D

(2024)

Answer: 2. B and C only

Explanation:

Histone modifications play a crucial role in the

regulation of gene expression by influencing chromatin structure and

accessibility to the transcriptional machinery. The correct

interpretations of the statements are as follows:

Statement A is incorrect. Acetylation of histones is generally

associated with transcriptional activation, not repression. This is

because acetylation neutralizes the positive charge on histones,

reducing their affinity for DNA and thus making the chromatin more

accessible for transcription.

Statement B is correct. Methylation of histones can indeed have

varying effects on transcription depending on the specific residue

modified. For example, methylation of lysine 4 on histone H3

(H3K4me) generally activates transcription, whereas methylation of

lysine 9 on histone H3 (H3K9me) typically represses transcription.

Statement C is correct. Phosphorylation of histones often occurs in

response to DNA damage and can influence gene expression. This

modification is involved in the DNA damage response and can affect

the chromatin structure and transcription of DNA repair genes.

Statement D is incorrect. Histone modifications influence the

recruitment of both RNA polymerase and transcription factors.

Transcription factors are part of the transcriptional machinery, and

their recruitment to DNA can be modulated by histone modifications.

Why Not the Other Options?

❌

1. A and D only – Incorrect; Statement A is incorrect, and

Statement D is also incorrect.

❌

3. B, C, and D – Incorrect; Statement D is incorrect.

❌

4. A, B, and D – Incorrect; Statement A and D are incorrect.

20. Spindle assembly in animal cells requires nuclear

envelope breakdown (NEBO). NEBO is a multistep

process, which begins when Cdk1 /cyclinB

phosphorylates multiple components of the nuclear

envelope. Given below are some components that are

directly phosphorylated by Cdk1 /cyclinB:

A. Nuclear Pore Complexes

B. Nuclear lamina

C. Greatwall kinase

O. Histone H3

Choose the option with correct Cdk1 /cyclinB

substrate/s that are directly associated with NEBO.

1. B only

2. A, Band 0

3. A and B only

4. B, C and 0

(2024)

Answer: 3. A and B only

Explanation:

During nuclear envelope breakdown (NEBD) in

animal cells, the phosphorylation events initiated by the Cdk1/cyclin

B complex are crucial. Cdk1/cyclin B directly phosphorylates

structural components of the nuclear envelope to promote its

disassembly. Specifically:

Nuclear Pore Complexes (A) are phosphorylated by Cdk1/cyclin B,

leading to their disassembly and aiding NEBD.

Nuclear lamina (B), composed mainly of lamins, is a critical target

of Cdk1/cyclin B phosphorylation, which causes the lamina network

to depolymerize, a key step in NEBD.

On the other hand:

Greatwall kinase (C) is not a direct structural component of the

nuclear envelope. Instead, Greatwall kinase controls the activity of

phosphatases (like PP2A) during mitosis, but it is regulated

indirectly by Cdk1/cyclin B rather than being a primary substrate for

NEBD.

Histone H3 (D) is phosphorylated during mitosis, but this

phosphorylation (e.g., on serine 10) is related to chromosome

condensation, not directly to nuclear envelope disassembly.

Thus, only A and B are directly phosphorylated by Cdk1/cyclin B to

cause NEBD.

Why Not the Other Options?

❌

(1) B only – Incorrect; while B is correct, A is also a correct

substrate, so "only B" is incomplete.

❌

(2) A, B and D – Incorrect; D (Histone H3) is phosphorylated for

chromosome condensation, not for NEBD.

❌

(4) B, C and D – Incorrect; C (Greatwall kinase) and D (Histone

H3) are not direct substrates involved in NEBD.

21. Which one of the foUowing properties is NOT

responsible for the self-sealing nature of ruptured

biological membranes?

1. The amphipathic character of the lipids

2. A hydrophobic interaction between lipid molecules

3. Hydrogen bonding between he head groups of the

lipids and water

4. Covalent interactions among l'pid molecules

(2024)

Answer: 4. Covalent interactions among l'pid molecules

Explanation:

The self-sealing nature of ruptured biological

membranes primarily arises from the inherent properties of the

phospholipid molecules that constitute the membrane. The

amphipathic nature of lipids, possessing both hydrophilic (polar

head) and hydrophobic (nonpolar tail) regions, drives them to

spontaneously arrange into a bilayer structure in an aqueous

environment. When a membrane is ruptured, the hydrophobic tails of

the lipids in the exposed edges tend to associate with each other to

minimize their contact with water. This hydrophobic interaction is a

major driving force for membrane resealing. Additionally, hydrogen

bonding between the hydrophilic head groups and the surrounding

water molecules contributes to the overall stability and flexibility of

the membrane, facilitating the resealing process.

Why Not the Other Options?

❌

(1) The amphipathic character of the lipids – Incorrect; This

property is crucial for the formation of the lipid bilayer and the

subsequent self-sealing as it dictates the arrangement of lipids in an

aqueous environment.

❌

(2) A hydrophobic interaction between lipid molecules – Incorrect;

This interaction is the primary driving force behind membrane

resealing, as the hydrophobic tails of lipids in the broken edges

aggregate to minimize contact with water.

❌

(3) Hydrogen bonding between the head groups of the lipids and

water – Incorrect; Hydrogen bonding contributes to the stability and

flexibility of the membrane, allowing the lipids to rearrange and

facilitate resealing

22. Which one of the following statements about DNA

packaging ,in chromosomes is INCORRECT?

1. Condensin I creates loops of nucleosomal chromatin for

packaging in mitosis.

2. Histone H1 is required for higher order packag]ng of

mammalian chromosomes.

3. Histones form hydrogen bonds with the sugar-phosphate

backbone of DNA.

4. Histone modification is not required for mitotic chromosome

condensation; it is mainly required for epigenetic control of

gene repression in interphase.

(2024)

Answer:

4. Histone modification is not required for mitotic

chromosome condensation; it is mainly required for epigenetic

control of gene repression in interphase.

Explanation:

Histone modifications, such as phosphorylation,

methylation, and acetylation, play a crucial role not only in

epigenetic regulation of gene expression during interphase but also

in the process of mitotic chromosome condensation. For instance, the

phosphorylation of histone H3 at serine 10 (H3S10ph) is a well-

established mark that is strongly correlated with and required for

proper chromosome condensation during mitosis and meiosis. This

modification neutralizes positive charges on the H3 tail, potentially

reducing interactions with DNA and allowing for further compaction.

Therefore, histone modification is indeed required for mitotic

chromosome condensation.

Why Not the Other Options?

❌

(1) Condensin I creates loops of nucleosomal chromatin for

packaging in mitosis – Correct; Condensin I is a protein complex

that plays a critical role in mitotic chromosome assembly and

condensation. It does so by forming and stabilizing loops within

nucleosomal chromatin.

❌

(2) Histone H1 is required for higher order packaging of

mammalian chromosomes – Correct; Histone H1, also known as the

linker histone, binds to the nucleosome and the linker DNA between

nucleosomes. 1 It helps to compact the 10 nm fiber into the 30 nm

fiber and is essential for achieving higher-order chromatin structures

necessary for chromosome condensation.

❌

(3) Histones form hydrogen bonds with the sugar-phosphate

backbone of DNA – Correct; The positively charged amino acid

residues in the core histones (H2A, H2B, H3, and H4) interact with

the negatively charged phosphate groups in the DNA backbone

through ionic bonds. Additionally, hydrogen bonds form between

amino acid residues of histones and the oxygen atoms of the

phosphodiester backbone and the sugar moieties of DNA,

contributing to the stability of the nucleosome. 2

23. Which one of the fo~lowing translation systems is

inhibited by cycloheximide?

1. 70S ribosome-associated bacterial trans:lation system

2. 80S ribosome-associated eukaryotic cytosolic

translation system

3. 74S ribosome-associated mitochondrial transla ion

system

4. 70S ribosome-associated chloroplast translation

system

(2024)

Answer: 2. 80S ribosome-associated eukaryotic cytosolic

translation system

Explanation:

Cycloheximide is a well-known eukaryotic

translation inhibitor. 1 It specifically blocks the peptidyl

transferase activity of the 80S ribosome in the cytoplasm of

eukaryotic cells. This inhibition prevents the translocation step in

translation, where the ribosome moves along the mRNA, and the

transfer of the growing polypeptide chain to the newly arriving

aminoacyl-tRNA is halted.

Why Not the Other Options?

❌

(1) 70S ribosome-associated bacterial translation system –

Incorrect; Bacterial translation, which occurs on 70S ribosomes, is

primarily inhibited by antibiotics like chloramphenicol, tetracycline,

streptomycin, and erythromycin, not cycloheximide.

❌

(3) 74S ribosome-associated mitochondrial translation system –

Incorrect; Mitochondrial translation in eukaryotes occurs on

ribosomes that are structurally and functionally distinct from

cytosolic ribosomes and resemble bacterial ribosomes in some

aspects (although the size can vary, often around 70-80S, and in

mammals, around 55S). Mitochondrial translation is typically

inhibited by drugs like chloramphenicol and erythromycin, similar to

bacterial translation, and is generally resistant to cycloheximide. The

74S mentioned is not a universally consistent size for mitochondrial

ribosomes.

❌

(4) 70S ribosome-associated chloroplast translation system –

Incorrect; Chloroplasts, like mitochondria, have their own genetic

material and protein synthesis machinery with ribosomes that are

similar to bacterial 70S ribosomes. Translation in chloroplasts is

also inhibited by antibiotics that target bacterial ribosomes, such as

chloramphenicol and streptomycin, and is generally not affected by

cycloheximide.

24. A region of a mouse chromosome was subjected to

micrococcal nuclease hypersensitivity analysis over

stages of development. In early stages, the region had

regularly spaced nudeosomes. In later stages, the

nucleosomes were regularly spaced with several

nucleosome free regions detected. Based on the above

observations, which one of the following is the best

possible inference?

1. The chromatin region is a facultative heterochromatin.

2. The :region is highly expressed in early stages.

3. Nucleosomes are not made effiiciently in the late

developmental stages.

4. The nucleosomail arrangements cannot be used to

infer potential expression states.

(2024)

Answer: 1. The chromatin region is a facultative

heterochromatin.

Explanation:

Micrococcal nuclease (MNase) hypersensitivity

analysis identifies regions of chromatin that are more accessible to

MNase digestion, indicating a less condensed chromatin structure.

Regularly spaced nucleosomes suggest a more compact, but

potentially transcriptionally active or poised, state. The transition to

irregularly spaced nucleosomes with nucleosome-free regions (NFRs)

in later stages strongly suggests chromatin remodeling associated

with changes in gene expression.

Facultative heterochromatin is chromatin that can switch between a

condensed, transcriptionally inactive state and a more open,

potentially active state depending on developmental cues or

environmental signals. The observation of a shift from a more

ordered nucleosomal array to a less ordered structure with NFRs

implies a dynamic change in chromatin organization, consistent with

a region transitioning towards a state that allows for gene

expression or regulation. NFRs are often found at regulatory

elements like promoters and enhancers, where transcription factors

bind and displace nucleosomes to allow access to the DNA.

Why Not the Other Options?

❌

(2) The region is highly expressed in early stages – Incorrect;

Regularly spaced nucleosomes do not necessarily indicate high

expression. While they allow for potential transcription, the presence

of NFRs in later stages is more directly correlated with active gene

expression due to increased accessibility of regulatory elements.

❌

(3) Nucleosomes are not made efficiently in the late

developmental stages – Incorrect; The observation of irregularly

spaced nucleosomes and NFRs does not imply a defect in nucleosome

assembly. Instead, it suggests active chromatin remodeling processes

that are repositioning or evicting nucleosomes at specific sites.

❌

(4) The nucleosomal arrangements cannot be used to infer

potential expression states – Incorrect; Nucleosomal arrangements

are strongly linked to gene expression states. Regularly spaced

nucleosomes can represent repressed or poised regions, while

irregularly spaced nucleosomes with NFRs are often associated with

active or regulated transcription. The changes observed are

indicative of altered potential expression states.

25. A putative mitochondrial signal peptide was attached

to the N-terminus of the DHFR protein and expressed

in mammalian cells. Mitochondria were iso:lated, and

a fraction was osmotically shocked briefly. Both

osmotically shocked and untreated pools of

mitochondria were treated with protease. It was

observed that the DHFR was ]ntact in both cases.

Which of the fa. lowing statements best describes the

function of the signal peptide?

1. 11 targets DHFR to the mitochondrial outer membrane,

facing the cytosol.

2. It targets DHFR to the mitochondrial inner membrane,

facing the intermembrane space.

3. It targets DHFR to the mitochondrial outer membrane,

facing the intermembrane space.

4. It targets DHFR to the mitochondrial matrix.

(2024)

Answer: 4. It targets DHFR to the mitochondrial matrix.

Explanation:

The experiment describes the fate of DHFR protein

when fused to a mitochondrial signal peptide and expressed in

mammalian cells. The key observation is that DHFR remains intact

even after the isolated mitochondria are osmotically shocked and

treated with protease.

Osmotic shock would disrupt the outer mitochondrial membrane,

making the intermembrane space accessible to the protease. If DHFR

were located in the intermembrane space or on the outer surface of

the inner membrane (facing the intermembrane space), it would be

susceptible to protease digestion after osmotic shock. Since DHFR

remained intact in the osmotically shocked mitochondria treated with

protease, it must be located in a compartment protected by the inner

mitochondrial membrane.

The mitochondrial matrix is the innermost compartment of the

mitochondria, enclosed by the inner mitochondrial membrane. 1

Proteins targeted to the matrix are typically imported across both the

outer and inner mitochondrial membranes. 2 The signal peptide

facilitates this import process. Once inside the matrix, the signal

peptide is usually cleaved off by a signal peptidase, but the

experiment only tracks the presence of DHFR, so cleavage doesn't

affect the conclusion about its final location. The fact that DHFR is

protected from protease even after the outer membrane is

compromised indicates its location within the matrix.

Why Not the Other Options?

❌

(1) It targets DHFR to the mitochondrial outer membrane, facing

the cytosol – Incorrect; If DHFR were on the outer surface of the

outer membrane, it would be readily accessible to protease even

without osmotic shock.

❌

(2) It targets DHFR to the mitochondrial inner membrane, facing

the intermembrane space – Incorrect; Osmotic shock would disrupt

the outer membrane, exposing the intermembrane space and the

outer surface of the inner membrane to the protease. DHFR would be

digested.

❌

(3) It targets DHFR to the mitochondrial outer membrane, facing

the intermembrane space – Incorrect; DHFR would need to cross the

outer membrane to be in the intermembrane space. Osmotic shock

would then expose it to protease.

26. After synchroniz:ing mammalian cens in culture with

a double thymid·ne block (celll cycle duration of 24h),

cells are reieased into fresh medium. After 6h, ceUs

are split into 4 sets and each is treated with a) nothing,

b-) proteasome inhibitor, c) myosin II inhibitor, d)

nocodazole. From the options given below, choose the

one that has the most likely outcome of the

experiment.

1. a: synchronously dividing cells; b: binudeate cells; c:

meta.phase arrested; d: prometaphase arrested

2. a: synchronously dividing cells; b: prometaphase

arrested; c: binucleate cells; d: meta phase arrested

3. a: binudeate cells; b: prometaphase arrested; c:

synchronously dividing cells; d: anaphase arrested

4. a: synchronously dividing cells; b: metaphase arrested;

c: binucleate cells; d: pro-metaphase arrested

(2024)

Answer: 4. a: synchronously dividing cells; b: metaphase

arrested; c: binucleate cells; d: pro-metaphase arrested

Explanation:

In this experiment, mammalian cells are first

synchronized using a double thymidine block, which arrests them at

the G1/S boundary. Upon release into fresh medium, the cells

progress synchronously through the cell cycle.

Now, after 6 hours, when treatments are applied:

a) Nothing (control group): Without any treatment, cells continue

synchronously dividing because they have been released from the

thymidine block and no further stress is applied to disrupt their cell

cycle progression.

b) Proteasome inhibitor: Proteasomes degrade important proteins

such as cyclin B, which is necessary for transition from metaphase to

anaphase during mitosis. If proteasome activity is inhibited, cells

arrest at metaphase because cyclin B cannot be degraded.

c) Myosin II inhibitor: Myosin II is essential for cytokinesis (the final

physical separation of the two daughter cells). If myosin II is

inhibited, cytokinesis fails even though nuclear division occurs,

leading to the formation of binucleate cells (cells with two nuclei).

d) Nocodazole: Nocodazole disrupts microtubule polymerization,

which prevents the proper formation of the mitotic spindle. As a

result, cells arrest in prometaphase, the stage when the spindle forms

and chromosomes are attaching to microtubules but proper

alignment has not yet occurred.

Thus, the outcomes match:

a: synchronously dividing cells

b: metaphase arrested

c: binucleate cells

d: prometaphase arrested

Why Not the Other Options?

❌

(1) a: synchronously dividing cells; b: binucleate cells; c:

metaphase arrested; d: prometaphase arrested – Incorrect;

proteasome inhibitor causes metaphase arrest, not binucleation.

❌

(2) a: synchronously dividing cells; b: prometaphase arrested; c:

binucleate cells; d: metaphase arrested – Incorrect; proteasome

inhibition causes metaphase arrest, not prometaphase arrest.

❌

(3) a: binucleate cells; b: prometaphase arrested; c:

synchronously dividing cells; d: anaphase arrested – Incorrect;

control cells (no treatment) divide synchronously, not binucleate;

nocodazole causes prometaphase arrest, not anaphase arrest.

27. The lipid composition of the two monolayers of the

p;lasma membrane is quite different. This lipid

asymmetry is functiona~ly relevant, especially in

converting extracellular signals into intracellular

ones. Given below are a few membrane lipids:

A. Phosphatidylserine

B. Phosphatidylinositol 4-phosphate

C. Phosphatidylcholine

D. Sphingomyeilin

Choose the option that correctly defines all the lipids

involved in signaling and are restricted to the

cytosolic face of the plasma membrane.

1. Aonly

2. A and B

3. C and D

4. C only

(2024)

Answer: 2. A and B

Explanation: The asymmetry of the plasma membrane lipid

bilayer is crucial for various cellular functions, including

signal transduction. Certain lipids are predominantly

localized to either the extracellular or cytosolic leaflet, and

their specific distribution contributes to their signaling roles.

Phosphatidylserine (PS) is primarily found on the cytosolic

face. When the cell undergoes apoptosis, PS flips to the outer

leaflet, serving as an "eat me" signal for phagocytes. On the

cytosolic face, PS can bind to and activate signaling proteins.

Phosphatidylinositol 4-phosphate (PI(4)P) is also

predominantly located on the cytosolic side of the plasma

membrane. It is a phosphoinositide that plays a vital role in

various signaling pathways, including membrane trafficking

and cytoskeleton regulation, by serving as a docking site for

specific signaling proteins. Phosphatidylcholine (PC) and

Sphingomyelin (SM) are choline-containing lipids that are

typically enriched in the extracellular leaflet of the plasma

membrane and are not primarily known for their direct

involvement as signaling molecules restricted to the cytosolic

face in the same way as PS and PI(4)P.

Why Not the Other Options?

❌

(1) A only – Incorrect; Phosphatidylinositol 4-phosphate

(PI(4)P) is also a signaling lipid restricted to the cytosolic

face.

❌

(3) C and D – Incorrect; Phosphatidylcholine (PC) and

Sphingomyelin (SM) are mainly found on the extracellular

face and are not the primary signaling lipids restricted to the

cytosolic face.

❌

(4) C only – Incorrect; Phosphatidylcholine (PC) is mainly

found on the extracellular face and is not a primary signaling

lipid restricted to the cytosolic face; also, Phosphatidylserine

(PS) and Phosphatidylinositol 4-phosphate (PI(4)P) fit the

criteria.

28. Which one of the following synaptic vesicles (as

observed under transmission electron microscope

contains catecholamines?

a. Small, round shaped and clear

b. Small, round shaped and dense core

c. Large dense core

d. Small, flattened and clear

(2023)

Answer: b. Small, round shaped and dense core

Explanation:

Synaptic vesicles containing catecholamines (such as

dopamine, norepinephrine, and epinephrine) are typically

characterized by their small, round shape and the presence of a

dense core when observed under a transmission electron microscope.

The dense core is due to the high concentration of the amine

neurotransmitters within the vesicle, often complexed with proteins.

Why Not the Other Options?

❌

(a) Small, round shaped and clear – Incorrect; Small, round

shaped and clear vesicles are generally associated with the

neurotransmitter acetylcholine (ACh) or glycine. They lack the

electron-dense core characteristic of catecholamine-containing

vesicles.

❌

typically contain neuropeptides or other larger signaling molecules.

While they have a dense core, their larger size distinguishes them

from the smaller vesicles that store catecholamines.

❌

(d) Small, flattened and clear – Incorrect; Small, flattened or

pleomorphic clear vesicles are characteristic of inhibitory synapses

that utilize neurotransmitters like GABA (gamma-aminobutyric acid)

or glycine. They do not possess the dense core associated with

catecholamines.

29. Which one of the following statements is not a

characteristic feature of aquaporins?

A. They are integral membrane proteins in the major

intrinsic protein (MIP) family.

B. They are absent in bacteria.

C. A highly conserved Asn-Pro-Ala (NPA) triad of

residues is present in the N-terminal half of the protein.

D. A highly conserved Asn-Pro-Ala (NPA) triad of

residues is present in the C-terminal half of the protein.

(2023)

Answer: B. They are absent in bacteria.

Explanation:

Aquaporins are integral membrane proteins that

form pores in the membranes of biological cells, primarily to

facilitate the transport of water. They are indeed part of the major

intrinsic protein (MIP) family. A key characteristic feature of

aquaporins is the presence of two highly conserved Asn-Pro-Ala

(NPA) motifs, which contribute to the selectivity filter of the channel.

These NPA triads are located in both the N-terminal and C-terminal

halves of the protein, folding into the membrane to form the water

pore. However, the statement that aquaporins are absent in bacteria

is incorrect. Aquaporins, with diverse forms and functions, have been

identified and characterized in bacteria as well as in archaea and

eukaryotes.

Why Not the Other Options?

❌

(a) They are integral membrane proteins in the major intrinsic

protein (MIP) family – Correct; Aquaporins are a well-established

subfamily within the larger MIP family of channel proteins.

❌

present in the N-terminal half of the protein – Correct; The NPA

motif is a signature sequence found in both halves of aquaporins and

is crucial for their structure and function.

❌

(d) A highly conserved Asn-Pro-Ala (NPA) triad of residues is

present in the C-terminal half of the protein – Correct; As mentioned

above, the presence of the NPA motif in the C-terminal half is

another characteristic feature of aquaporin proteins.

30. A cruciform structure of chromosomes during

meiosis is a characteristic feature of:

a. Translocation

b. Inversion

c. Deletion

d. Duplication

(2023)

Answer: a. Translocation

Explanation:

A cruciform (cross-shaped) structure during meiosis

is a characteristic feature of reciprocal translocation. Reciprocal

translocation involves the exchange of segments between non-

homologous chromosomes. During meiosis I, when homologous

chromosomes pair up to form bivalents, the translocated

chromosomes will attempt to pair with their homologous regions.

This pairing can only be achieved by forming a cross-shaped

configuration involving all four chromosomal segments involved in

the translocation. This cruciform structure is a cytological hallmark

of translocation heterozygotes during meiosis.

Why Not the Other Options?

❌

(b) Inversion – Incorrect; Inversions (where a segment of a

chromosome is reversed) lead to the formation of loop structures

during meiotic pairing to maximize homologous alignment, not

cruciform structures.

❌

chromosome is lost) result in a mismatch during pairing, where the

normal homologous chromosome will have a region that cannot pair.

This doesn't lead to a cruciform structure.

❌

(d) Duplication – Incorrect; Duplications (where a segment of a

chromosome is present in multiple copies) also lead to loop

structures during meiotic pairing as the duplicated region tries to

align with its homologous counterpart on the other chromosome.

They do not form cruciform structures.

31. Which one of the following conditions associated with

chromosome 15 may cause Prader-Willi syndrome?

a. Paternal uniparental disomy

b. Maternal uniparental disomy

c. Imprinting of 15q11-q 13 locus in maternal copy

d. Imprinting of 15q23-q25 locus in paternal copy

(2023)

Answer: b. Maternal uniparental disomy

Explanation:

Prader-Willi syndrome (PWS) is a

neurodevelopmental disorder that arises due to the lack of

expression of paternally inherited genes in the 15q11-q13 region of

chromosome 15. Maternal uniparental disomy (UPD) is one of the

primary genetic mechanisms leading to PWS. In this condition, an

individual inherits two copies of chromosome 15 from their mother

and no copy from their father. Since the genes in the 15q11-q13

region that are critical for normal development are imprinted and

only expressed from the paternal chromosome, the absence of a

paternal copy (due to maternal UPD) results in the characteristic

features of PWS.

Why Not the Other Options?

❌

(a) Paternal uniparental disomy – Incorrect; Paternal uniparental

disomy of chromosome 15, where an individual inherits two copies

from the father and none from the mother, typically leads to

Angelman syndrome, a distinct genetic disorder caused by the lack of

expression of maternally inherited genes in the same 15q11-q13

region.

❌

The 15q11-q13 locus is normally imprinted in the maternal copy,

meaning the genes in this region are usually silenced on the

maternally inherited chromosome. PWS arises from the lack of a

paternally inherited, expressed copy of these genes.

❌

(d) Imprinting of 15q23-q25 locus in paternal copy – Incorrect;

The genetic region primarily associated with Prader-Willi syndrome

is 15q11-q13. Imprinting in a different region of chromosome 15

(15q23-q25) is not the typical cause of PWS.

32. In cell membranes, the lipid molecules are arranged

as a continuous double layer, with an approximate

thickness of

a. 20 nm

b. 50 nm

c. 5 nm

d. 1 nm

(2023)

Answer: c. 5 nm

Explanation:

The lipid bilayer, which forms the structural basis of

all biological membranes, is composed of two layers of lipid

molecules. Each lipid molecule, primarily phospholipids, has a

hydrophilic (polar) head and two hydrophobic (nonpolar) tails. In

the bilayer, the hydrophilic heads face outwards towards the aqueous

environment on both the inner and outer surfaces of the membrane,

while the hydrophobic tails face inwards, forming a nonpolar core.

The typical thickness of this lipid bilayer is approximately 5

nanometers (nm). This dimension is consistent with the length of two

phospholipid molecules arranged tail-to-tail plus the space occupied

by their hydrophilic head groups.

Why Not the Other Options?

❌

(a) 20 nm – Incorrect; 20 nm is significantly thicker than a typical

lipid bilayer. This dimension might be more representative of the

overall thickness of a cell membrane when including the associated

proteins and carbohydrate layers (glycocalyx).

❌

(b) 50 nm – Incorrect; 50 nm is much too thick for just the lipid

bilayer. This scale is more in the range of the diameter of some

viruses or small cellular organelles.

❌

(d) 1 nm – Incorrect; 1 nm is too thin for a lipid bilayer. This

dimension is closer to the size of a single small protein or a few

atoms.

33. Based on the image given below, select the option that

describes it correctly:

1. Q-banded normal human karyotype.

2. G-banded human karyotype depicting aneuploidy.

3. C-banded karyotype depicting Klinefelter syndrome.

4. G-banded normal human karyotype.

(2023)

Answer: 2. G-banded human karyotype depicting aneuploidy.

Explanation:

The image displays a human karyotype where

chromosomes have been stained to show distinct banding patterns.

This staining technique, where chromosomes are treated with

Giemsa stain, is known as G-banding. G-banding produces a unique

pattern of dark and light bands for each chromosome, allowing for

their identification and the detection of structural or numerical

abnormalities. In a normal human karyotype, there would be 23

pairs of chromosomes, totaling 46 chromosomes: 22 pairs of

autosomes and one pair of sex chromosomes (XX for females and XY

for males).

By examining the sex chromosomes in the provided karyotype, we

can see three sex chromosomes labeled as X, X, and Y. This indicates

the presence of an extra X chromosome in a male individual,

resulting in an XXY karyotype. This chromosomal abnormality is

characteristic of Klinefelter syndrome, which is a form of aneuploidy

involving the sex chromosomes. Therefore, the karyotype shown is G-

banded and depicts aneuploidy, specifically Klinefelter syndrome.

Why Not the Other Options?

❌

(1) Q-banded normal human karyotype – Incorrect; The banding

pattern shown is characteristic of G-banding, not Q-banding (which

uses quinacrine dye and requires fluorescence microscopy).

Additionally, the presence of XXY indicates aneuploidy, not a normal

karyotype.

❌

(3) C-banded karyotype depicting Klinefelter syndrome –

Incorrect; C-banding is a different staining technique that primarily

highlights constitutive heterochromatin, often found near the

centromeres. The banding pattern shown is G-banding. While the

karyotype does depict Klinefelter syndrome, the banding method is

G-banding, not C-banding.

❌

(4) G-banded normal human karyotype – Incorrect; The

karyotype is G-banded, but the presence of an extra X chromosome

(XXY) indicates aneuploidy (Klinefelter syndrome), not a normal

male (XY) or female (XX) karyotype.

Which one of the following changes is energetically

favorable and occurs spontaneously in an aqueous

solution?

1. Formation of a bilayer from phospholipid molecules

2. Dispersion of one oil droplet into many small ones

3. Tearing of the lipid bilayer

4. Conversion of a membrane vesicle to a flat bilayer

(2023)

Answer: 1. Formation of a bilayer from phospholipid

molecules

Explanation:

Phospholipid molecules are amphipathic, meaning

they have both hydrophilic (polar head) and hydrophobic (nonpolar

tails) regions. In an aqueous environment, the hydrophobic tails of

phospholipids spontaneously aggregate to minimize their contact

with water, driven by the hydrophobic effect. The hydrophilic heads,

on the other hand, interact favorably with the surrounding water

molecules. This energetically favorable arrangement leads to the

spontaneous formation of lipid bilayers, where the hydrophobic tails

are sequestered in the interior, away from water, and the hydrophilic

heads face the aqueous solution on both sides. This self-assembly

process increases the entropy of the water molecules as they are no

longer forced to form ordered cages around the hydrophobic tails,

making the overall free energy change negative (ΔG<0).

Why Not the Other Options?

❌

(2) Dispersion of one oil droplet into many small ones – Incorrect;

Dispersing a large oil droplet into many smaller ones increases the

total surface area of the oil exposed to water. This is energetically

unfavorable because it increases the unfavorable hydrophobic-water

interactions and decreases entropy by ordering water molecules at

the oil-water interface.

❌

(3) Tearing of the lipid bilayer – Incorrect; Tearing a lipid bilayer

exposes the hydrophobic tails to the aqueous environment, which is

energetically unfavorable. The hydrophobic effect drives the bilayer

to maintain its integrity and minimize these unfavorable interactions.

Energy input would be required to tear the bilayer.

❌

(4) Conversion of a membrane vesicle to a flat bilayer – Incorrect;

Membrane vesicles (liposomes) are energetically favorable

structures because they form closed compartments, maximizing the

sequestration of hydrophobic tails away from water and satisfying

the curvature preferences of some lipids. Converting a vesicle to a

flat bilayer would require overcoming the energetic stability

provided by the closed structure and potentially exposing edges or

requiring specific conditions to maintain stability. While flat bilayers

can exist (e.g., in planar lipid bilayer experiments), their

spontaneous formation from vesicles is not generally energetically

favorable in a free aqueous solution without specific supports or

conditions.

34. In the context of protein import in the nucleus, which

molecule is responsible for releasing the cargo from

the importing receptor?

1. Ras

2. RhoA

3. Ran

4. Rock

(2023)

Answer: 3. Ran

Explanation:

Nuclear import is a process by which proteins

synthesized in the cytoplasm are transported into the nucleus through

nuclear pore complexes (NPCs). This process is mediated by

importin receptors that bind to nuclear localization signals (NLS) on

the cargo proteins. The release of the cargo from the importin

receptor inside the nucleus is triggered by the binding of a small

GTPase protein called Ran in its GTP-bound form (Ran-GTP).

Here's a step-by-step breakdown:

In the cytoplasm, the importin receptor binds to the cargo protein

containing an NLS.

The importin-cargo complex translocates through the NPC into the

nucleus.

Inside the nucleus, Ran-GTP binds to the importin receptor.

The binding of Ran-GTP causes a conformational change in the

importin receptor, leading to the release of the cargo protein into the

nucleoplasm.

The Ran-GTP-importin complex then translocates back to the

cytoplasm through the NPC.

In the cytoplasm, Ran-GTP is hydrolyzed to Ran-GDP by Ran-

GTPase-activating protein (Ran-GAP), which is localized in the

cytoplasm.

Ran-GDP has a low affinity for importin, causing its release. The

importin receptor is then free to bind more cargo proteins in the

cytoplasm and repeat the cycle.

Therefore, Ran-GTP binding to the importin receptor in the nucleus

is the key event that leads to the release of the imported cargo

protein.

Why Not the Other Options?

❌

(1) Ras – Incorrect; Ras is a small GTPase involved in various

signaling pathways, including cell growth and differentiation,

primarily at the plasma membrane. It is not directly involved in the

release of cargo from importin receptors in the nucleus.

❌

(2) RhoA – Incorrect; RhoA is another small GTPase involved in

regulating the actin cytoskeleton, cell motility, and cell division in

the cytoplasm. It does not play a direct role in nuclear import cargo

release.

❌

(4) Rock – Incorrect; Rock (Rho-associated coiled-coil containing

protein kinase) is a serine/threonine kinase that is activated by RhoA

and is involved in regulating the cytoskeleton and cell motility. It

functions downstream of RhoA in the cytoplasm and is not directly

involved in nuclear import cargo release.

35. Which one of the following describes an amphisome?

1. It is an intermediate/hybrid organelle produced

through the fusion of endosome with autophagosomes

within cells.

2. It is a double-membrane sequestering vesicle that is

the hallmark of the intracellular catabolic process called

macroautophagy.

3. It is a compartment fanned when autophagosome fuses

with a lysosome.

4. It is a vacuole that arises when membranes of the ER

sequester parts of the cytoplasm.

(2023)

Answer: 1. It is an intermediate/hybrid organelle produced

through the fusion of endosome with autophagosomes within

cells.

Explanation:

An amphisome is a transient, hybrid organelle

formed by the fusion of a late endosome with an autophagosome.

This fusion event brings together the contents destined for

degradation via autophagy (within the autophagosome) and the

hydrolytic enzymes and membrane proteins characteristic of the

endolysosomal system (present in the late endosome). The

amphisome then matures by fusing with a lysosome, forming an

autolysosome, where the degradation of the autophagic cargo takes

place. Therefore, the amphisome represents a key intermediate stage

in the macroautophagy pathway, facilitating the delivery of

autophagosomal contents to the lysosomal degradation machinery.

Why Not the Other Options?

❌

(2) It is a double-membrane sequestering vesicle that is the

hallmark of the intracellular catabolic process called

macroautophagy – Incorrect; This describes an autophagosome, the

vesicle that engulfs cytoplasmic material during macroautophagy.

❌

(3) It is a compartment formed when autophagosome fuses with a

lysosome – Incorrect; This describes an autolysosome, the organelle

where the degradation of autophagic cargo occurs.

❌

(4) It is a vacuole that arises when membranes of the ER

sequester parts of the cytoplasm – Incorrect; This describes a

phagophore or isolation membrane, which is the initial structure that

expands and closes to form an autophagosome. In some specific

types of autophagy, ER involvement is significant (ER-phagy), but

the resulting sequestering vesicle is still an autophagosome.

36. In eukaryotes, microtubules and actin-binding

proteins influence the dynamics and organization of

the cytoskeleton. Match the cytoskeleton-binding

proteins listed in column B to actin or microtubule

mentioned in Column A, and their function from

those listed in Column C.

Which one of the following options represents all the

correct matches between columns A, B and C?

1. X-(iv)-(c). X-(iii)-(b), Y-(i)-(a). Y-(ii)-(d)

2. X-(ii)-(a), X-(i)-(c), Y-(iv)-(c), Y-(iii)-(b)

3. X-(iii)-(a), X-(i)-(d), Y-(ii)-(c), Y-(iv)-(b)

4. X-(i)-(d), X-(iv)-(a), Y-(iii)-(d), Y-(ii)-(c)

(2023)

Answer: 3. X-(iii)-(a), X-(i)-(d), Y-(ii)-(c), Y-(iv)-(b)

Explanation:

X. microtubule - (iii) tau - (a) filament cross-linking:

Tau proteins are microtubule-associated proteins (MAPs) that bind

along the sides of microtubules. One of their key functions is to

stabilize microtubules and promote their assembly. They can also

cross-link microtubules, contributing to the organization of the

microtubule network within the cell.

X. microtubule - (i) katenin - (d) only severs filament: Katenin, also

known as p60 katanin subunit, is an ATP-dependent microtubule-

severing protein. Its primary function is to cut microtubules along

their length, generating new plus and minus ends and increasing

microtubule dynamics. While some severing proteins also bind to

ends, katenin's main characterized role is severing along the filament.

Y. actin - (ii) tropomodulin - (c) prevents assembly and disassembly

at minus end: Tropomodulin is an actin-binding protein that caps the

minus (pointed) end of actin filaments. By binding to the minus end,

it prevents both the addition and removal of actin subunits at this end,

thereby stabilizing the filament length.

Y. actin - (iv) gelsolin - (b) severs filaments and binds to plus end:

Gelsolin is an actin-binding protein that can sever existing actin

filaments. It does this by inserting itself between actin subunits.

Importantly, after severing, gelsolin remains bound to the new plus

(barbed) end of the filament, effectively capping it and preventing

further polymerization at that end.

Why Not the Other Options?

❌

Option 1: Incorrect matches for X-(iv)-(c) and X-(iii)-(b). Gelsolin

primarily interacts with actin, and while it can affect microtubule

dynamics indirectly, it's not a direct microtubule-binding protein

with the function of preventing assembly/disassembly at the minus

end. Tau primarily cross-links and stabilizes microtubules, not severs

and binds to the plus end.

❌

Option 2: Incorrect matches for X-(ii)-(a), X-(i)-(c), Y-(iv)-(c),

and Y-(iii)-(b). Tropomodulin binds to the minus end of actin, not

microtubules. Katenin severs microtubules, it doesn't primarily

prevent assembly/disassembly at the minus end. Gelsolin binds to the

plus end of actin after severing, not the minus end to prevent

assembly/disassembly. Tau primarily cross-links and stabilizes

microtubules, not severs and binds to the plus end of actin.

❌

Option 4: Incorrect matches for X-(i)-(d), X-(iv)-(a), Y-(iii)-(d),

and Y-(ii)-(c). Katenin severs microtubules, but it's not its only

function; it doesn't primarily cross-link microtubules. Gelsolin

interacts with actin, not primarily cross-links microtubules. Tau

binds to microtubules, not actin, and its main severing activity is not

well-established compared to its stabilizing role. Tropomodulin

prevents assembly/disassembly at the minus end of actin, not just

severs actin filaments.

37. Match the eukaryotic cellular organelles listed in

Column X with their typical function from among

those listed in Column Y.

Which one of the following options represents all

correct matches between Column X and Column Y?

1. A-(iii), B-(iv), C-(ii), D-(i)

2. A-(ii), B-(i), C-(iv), D-(iii)

3. A-(ii), B-(iii), C-(1), D-(iv)

4. A-(iii), B-(i), C-(iv), D-(ii)

(2023)

Answer: 2. A-(ii), B-(i), C-(iv), D-(iii)

Explanation:

Each organelle in a eukaryotic cell carries out

specialized functions. The correct matching based on well-

established cell biology principles is:

A. Golgi → (ii) O-linked Glycosylation

The Golgi apparatus is the site for post-translational modifications,

including O-linked glycosylation, where sugars are attached to the

hydroxyl groups of serine or threonine residues on proteins.

B. Nucleolus → (i) Ribosomes assembly

The nucleolus is a dense region within the nucleus where ribosomal

RNA is transcribed and assembled with ribosomal proteins to form

ribosomal subunits.

C. Peroxisomes → (iv) Oxygen utilization

Peroxisomes are responsible for beta-oxidation of very long chain

fatty acids and detoxification reactions that consume oxygen and

produce hydrogen peroxide, which is then broken down by catalase.

D. Endoplasmic Reticulum (ER) → (iii) Site for lipid synthesis

The smooth ER is the primary site for lipid and steroid synthesis,

while the rough ER is associated with protein synthesis.

Why Not the Other Options?

❌

(1) A–(iii), B–(iv), C–(ii), D–(i) – Incorrect; Golgi is not the site

for lipid synthesis (A–(iii) wrong), nucleolus doesn't use oxygen (B–

(iv) wrong), peroxisomes are not involved in glycosylation (C–(ii)

wrong), and ER is not where ribosomes are assembled (D–(i) wrong).

❌

(3) A–(ii), B–(iii), C–(1), D–(iv) – Incorrect; B–(iii) is wrong

(nucleolus doesn't synthesize lipids), C–(1) is wrong (peroxisomes do

not assemble ribosomes), D–(iv) is wrong (ER does not use oxygen).

❌

(4) A–(iii), B–(i), C–(iv), D–(ii) – Incorrect; A–(iii) is wrong

(Golgi is not a lipid synthesis site), D–(ii) is wrong (ER does not

carry out O-linked glycosylation; that occurs in the Golgi).

38. The following statements are made about telomeres:

A. Telomere-binding proteins (TBPs) are believed to

shield telomeres from the cell's DNA repair

machinery, preventing them from being recognized

as double-strand breaks.

B. Telomeres in human cells are repeats of

TTAGGGG sequence that can extend upto 150 kb,

which are replicated by the action of TERT in

actively dividing cells.

C. In differentiated cells, telomerase is inactive,

leading to shortening of telomeres over hundreds of

cell divisions, damage to ends of chromosomes, and

eventually apoptosis.

D. The persistence of telomerase activity in several

cancers allows the cells to continue to proliferate.

Which one of the following options represents the

combination of all correct statements?

1. A and C only

2. A, B and C

3. B and D only

4. A. C and D

(2023)

Answer: 4. A. C and D

Explanation:

Let's analyze each statement to determine which are

correct:

A. Telomere-binding proteins (TBPs) are believed to shield telomeres

from the cell's DNA repair machinery, preventing them from being

recognized as double-strand breaks – Correct. Telomere-binding

proteins such as TRF1 and TRF2 are important in protecting

telomeres from being mistakenly recognized as damaged DNA by the

repair machinery, thus preventing unnecessary repair mechanisms

that could lead to genomic instability.

B. Telomeres in human cells are repeats of TTAGGGG sequence that

can extend up to 150 kb, which are replicated by the action of TERT

in actively dividing cells – Partially Correct, but not entirely

accurate. The human telomere sequence is indeed TTAGGG repeated,

but they typically range between 5-15 kb in length, not up to 150 kb.

While TERT (telomerase reverse transcriptase) is responsible for

elongating telomeres in actively dividing cells, the 150 kb length is

not typical for human telomeres, making this statement somewhat

misleading.

C. In differentiated cells, telomerase is inactive, leading to

shortening of telomeres over hundreds of cell divisions, damage to

ends of chromosomes, and eventually apoptosis – Correct. In most

differentiated somatic cells, telomerase is inactive, and telomeres

shorten with each cell division, eventually leading to senescence or

apoptosis as the telomeres become critically short.

D. The persistence of telomerase activity in several cancers allows

the cells to continue to proliferate – Correct. Many cancer cells

reactivate telomerase, which allows them to bypass the normal

limitations of cell division, enabling unlimited proliferation. This is

one of the hallmarks of cancer.

Why Not the Other Options?

❌

(1) A and C only – Incorrect; B is partially inaccurate because

the length of human telomeres is typically much shorter than the 150

kb mentioned.

❌

(2) A, B and C – Incorrect; B contains an inaccuracy about the

length of human telomeres.

❌

(3) B and D only – Incorrect; B contains an error about telomere

length.

Therefore, Option 4 is the correct choice since A, C, and D are

entirely accurate.

39. Following statements are made about the bacterial

ribosomes and their functions:

A. Association of 23S rRNA with 16S rRNA is

essential to catalyze in vitro peptide bond formation

using model substrates.

B. The 23S rRNA is necessary and sufficient to

catalyze in vitro peptide bond formation using model

substrates.

C. Ribosome carries a polymerization activity.

D. The 16S rRNA is necessary and sufficient to

catalyze in vitro peptide bond formation using model

substrates. Which one of the following options

represents the combination of both correct

statements?

a. A and B

b. B and C

c. C and D

d. A and D

(2023)

Answer: b. B and C

Explanation:

B. The 23S rRNA is necessary and sufficient to

catalyze in vitro peptide bond formation using model substrates. This

statement is correct. The peptidyl transferase activity, which

catalyzes the formation of peptide bonds during protein synthesis,

resides within the large ribosomal subunit. Specifically, the 23S

rRNA (in bacteria; 28S rRNA in eukaryotes) is the ribozyme

responsible for this catalytic activity. Experiments have shown that

the isolated 23S rRNA can catalyze peptide bond formation under

specific in vitro conditions using minimal substrates.

C. Ribosome carries a polymerization activity. This statement is

correct. The primary function of the ribosome is to polymerize amino

acids into polypeptide chains (proteins). This polymerization activity

is the essence of protein synthesis.

Why Not the Other Options?

❌

A. Association of 23S rRNA with 16S rRNA is essential to catalyze

in vitro peptide bond formation using model substrates. – Incorrect;

While the intact ribosome with both subunits (containing 23S and

16S rRNA) is required for protein synthesis in vivo, the catalytic

activity for peptide bond formation has been shown to reside within

the 23S rRNA of the large subunit and can occur in vitro with

minimal substrates without the direct involvement of 16S rRNA.

❌

D. The 16S rRNA is necessary and sufficient to catalyze in vitro

peptide bond formation using model substrates. – Incorrect; The 16S

rRNA (small ribosomal subunit) primarily plays a crucial role in

mRNA binding, codon-anticodon recognition, and initiation of

translation. It is not the catalytic component responsible for peptide

bond formation. The peptidyl transferase center is located on the 23S

rRNA of the large subunit.

40. Cytoskeleton-dependent motor proteins are critical

for the movement of cellular organelles in animal

cells. In the fertilized egg of C. elegans, once the

polarity has been established, the maternal nucleus

migrates towards the paternal nucleus, which

eventually leads to fusion of the two pronuclei (see

below).

What molecular motor is likely to be directly

involved in nuclear migration?

a. Myosin II

b. Kinesin

c. Dynein

d. Tropomyosin

(2023)

Answer: c. Dynein

Explanation:

In the fertilized C. elegans egg, after polarity is

established, the maternal nucleus migrates towards the paternal

nucleus. This movement is facilitated by microtubules. Dynein is a

motor protein that moves along microtubules towards their minus

ends. Given that the image shows microtubules connecting the two

nuclei, and dynein's known function, it is the most likely motor

protein to be directly involved in this nuclear migration.

Why Not the Other Options?

❌

(a) Myosin II – Incorrect; Myosin II is a motor protein associated

with actin filaments, not microtubules. It is primarily involved in

processes like muscle contraction and cell division.

❌

(b) Kinesin – Incorrect; Kinesin moves along microtubules

towards their plus ends, which is the opposite direction of the

described nuclear movement.

❌

(d) Tropomyosin – Incorrect; Tropomyosin is an actin-binding

protein involved in regulating muscle contraction, not a motor

protein. It doesn't directly participate in organelle or nuclear

movement.

41. To obtain recombinant products during meiosis, a

double-strand break in the DNA yields crossovers

needed for chiasmata formation. The progression of

the non-crossover and crossover pathways begins

with the formation of D loop however, it may not

result in the production of recombinant gametes.

Following statements are made regarding

recombination:

A. Expansion of D-loop takes place in non-crossover

pathway but not in the crossover pathway.

B. Expansion of D-loop takes place in crossover

pathway, but not in the non-crossover pathway.

C. Ejection of elongating strand takes place in the

non-crossover pathway, but not in the crossover

pathway.

D. Ejection of elongating strand takes place in the

crossover pathway but not in the non-crossover

pathway.

Which one of the following options represents the

correct combination of statements that explain the

formation of recombinant gametes?

a. A and B

b. B and C

c. C and D

d. D and A

(2023)

Answer: b. B and C

Explanation:

The formation of recombinant gametes is the result

of the crossover pathway during meiosis. Let's analyze the statements

in the context of this pathway:

B. Expansion of D-loop takes place in crossover pathway, but not in

the non-crossover pathway. While D-loop formation occurs in both

pathways, a more extensive and stable D-loop expansion is generally

considered necessary for the crossover pathway to proceed to

second-end capture and double Holliday junction (dHJ) formation.

In the non-crossover pathway (SDSA), the D-loop might be shorter-

lived. Therefore, this statement highlights a potentially crucial aspect

for the progression towards crossover.

C. Ejection of elongating strand takes place in the non-crossover

pathway, but not in the crossover pathway. The non-crossover

pathway is characterized by synthesis-dependent strand annealing

(SDSA), where the newly synthesized strand is displaced (ejected)

from the template DNA and anneals to the other broken end,

preventing crossover. Conversely, for crossovers to occur, this

ejection step must not happen. The elongating strand needs to remain

associated to facilitate second-end capture and dHJ formation. Thus,

the absence of ejection is essential for the crossover pathway that

leads to recombinant gametes.

Therefore, the combination of statement B (emphasizing the

necessary D-loop expansion in the crossover pathway) and statement

C (highlighting the absence of the ejection step that would lead to

non-crossover) best explains the progression towards the formation

of recombinant gametes through the crossover pathway.

Why Not the Other Options?

❌

(a) A and B – Incorrect; Statement A describes a feature of the

non-crossover pathway (D-loop expansion not in crossover), which

is inaccurate.

❌

elongating strand in the crossover pathway, which is the defining

step of the non-crossover pathway and prevents crossover.

❌

(d) D and A – Incorrect; Statement D describes ejection in the

crossover pathway (incorrect), and statement A describes D-loop

expansion in the non-crossover pathway but not crossover

(incorrect)

42. Columns X and Y of the following table list some

treatment methods, reagents, and events that are

related to human lymphocyte culture and

banding/karyotyping of human chromosomes.

Which one of the following options represents all

correct matches between Column X and Column Y?

a. A-ii, B-v, C-iii, D-i, E-iv

b. A-v, B-iii, C-ii, D-iv, E-i

c. A-iv, B-v, C-i, D-iii, E-ii

d. A-ii, B-v, C-iv, D-i, E-iii

(2023)

Answer: a. A-ii, B-v, C-iii, D-i, E-iv

Explanation:

Let's match the treatments/reagents with their related

events in human lymphocyte culture and chromosome banding:

A. 5% barium hydroxide treatment at 50°C: This treatment is used in

ii. C-banding. C-banding specifically stains constitutive

heterochromatin, which is primarily located around the centromeres

of chromosomes. The barium hydroxide treatment helps to denature

the DNA in these regions differentially, allowing for specific staining.

B. Trypsin treatment: Mild v. G-banding involves treating

chromosomes with trypsin (a proteolytic enzyme) followed by Giemsa

staining. Trypsin partially digests chromosomal proteins, leading to

differential staining patterns along the length of the chromosomes,

resulting in the characteristic G-bands.

C. Phytohaemagglutinin: iii. Mitotic stimulation of human

lymphocytes is a crucial step in karyotyping. Phytohaemagglutinin

(PHA) is a lectin derived from the red kidney bean (Phaseolus

vulgaris) and is a potent mitogen that specifically stimulates T

lymphocytes to undergo cell division in culture, increasing the

number of cells in metaphase, which is the stage when chromosomes

are most condensed and visible.

D. Phosphate buffer treatment at 80°C: This heat treatment in a

phosphate buffer is a step involved in i. R-banding. R-bands are the

reverse of G-bands; they stain the regions that are lightly stained by

G-banding more darkly. The heat treatment selectively denatures AT-

rich regions.

E. Silver Staining: iv. Nucleolar Organizer Regions (NOR) are

specific chromosomal regions containing the genes for ribosomal

RNA (rRNA). Silver staining techniques specifically stain the NORs,

highlighting the activity of these regions in the preceding interphase.

Therefore, the correct matches are A-ii, B-v, C-iii, D-i, and E-iv.

Why Not the Other Options?

❌

(b) A-v, B-iii, C-ii, D-iv, E-i – Incorrect; Barium hydroxide is for

C-banding, trypsin for G-banding, and PHA for mitotic stimulation.

❌

C-banding, and PHA is for mitotic stimulation.

❌

(d) A-ii, B-v, C-iv, D-i, E-iii – Incorrect; Silver staining targets

NORs, not mitotic stimulation.

43. The following statements were made about X-

chromosome inactivation in humans:

A. Maternally-derived X-chromosome has a greater

chance of becoming inactivated in any given cell.

B. Both X-chromosomes are activated during the

process of oogenesis.

C. The XIST gene encodes for a single, long non-

coding transcript, which binds with the X

chromosome and helps in its inactivation.

D. The XIST gene expression is required to initiate

inactivation of X-chromosome and also to maintain

inactivation from one cell generation to the next.

E. Tsix transcription affects the abundance of Xist

RNA in cis.

Which one of the following options represents the

combination of all correct statements?

a. A, C and D

b. B, C and E

c. A and C only

d. B and E only

(2023)

Answer: b. B, C and E

Explanation:

Let's evaluate each statement about X-chromosome

inactivation in humans:

A. Maternally-derived X-chromosome has a greater chance of

becoming inactivated in any given cell. This statement is incorrect.

X-chromosome inactivation is a random process in somatic cells of

females. In each cell, either the maternally-derived or the paternally-

derived X-chromosome has an approximately equal chance of being

inactivated. However, there is a phenomenon called imprinted X-

inactivation in the extraembryonic tissues of some mammals, where

the paternal X is preferentially inactivated. This statement doesn't

specify extraembryonic tissues in humans, so it's generally

considered incorrect for somatic cells.

B. Both X-chromosomes are activated during the process of

oogenesis. This statement is correct. During oogenesis, both X

chromosomes in the developing oocyte are transcriptionally active.

X-inactivation occurs later in development after fertilization, in the

somatic cells of the female embryo.

C. The XIST gene encodes for a single, long non-coding transcript,

which binds with the X chromosome and helps in its inactivation.

This statement is correct. The XIST (X-inactive specific transcript)

gene on the X chromosome that will be inactivated is transcribed to

produce a long non-coding RNA molecule. This RNA coats the

chromosome from which it is transcribed and plays a crucial role in

initiating and mediating its inactivation by recruiting chromatin-

modifying complexes.

D. The XIST gene expression is required to initiate inactivation of X-

chromosome and also to maintain inactivation from one cell

generation to the next. This statement is correct. XIST RNA is

essential for the initiation of X-chromosome inactivation. Its

continued expression is also necessary for the maintenance of the

inactive state through subsequent cell divisions.

E. Tsix transcription affects the abundance of Xist RNA in cis. This

statement is correct. Tsix is a long non-coding RNA transcribed in

the antisense direction to XIST from the same locus. Tsix acts as a

negative regulator of XIST expression. During the initiation of X-

inactivation, the X chromosome that will remain active upregulates

Tsix, which in turn suppresses XIST expression on that chromosome.

This cis-acting mechanism helps in ensuring that only one X

chromosome is inactivated.

Therefore, the correct statements are B, C, and E.

Why Not the Other Options?

❌

(a) A, C and D – Incorrect; Statement A is incorrect.

❌

statement E is correct.

❌

(d) B and E only – Incorrect; Statement C is also correct.

44. Which one of the following statements is correct for

dosage compensation in human?

1. X-chromosome inactivation in female occurs in a

zygote immediately after fertilization.

2. X-chromosome inactivation is non-random, in some

individuals maternal X is inactivated while in others

paternal X-chromosome is inactivated.

3. Y-chromosome of males is seen as Barr body.

4. The body of female is a mosaic of cells, some having

paternal X- and others having maternal X-inactivated.

(2023)

Answer: 4. The body of female is a mosaic of cells, some

having paternal X- and others having maternal X-inactivated.

Explanation:

Dosage compensation in humans refers to the

process that equalizes the expression of X-linked genes in males (XY)

and females (XX). This is primarily achieved through X-chromosome

inactivation in females. In each somatic cell of a female, one of the

two X chromosomes is randomly inactivated early in embryonic

development. This inactivated X chromosome condenses to form a

Barr body, which is transcriptionally inactive. Because the

inactivation is random, different cells in a female embryo will

inactivate either the maternal or the paternal X chromosome. As

these cells divide, this inactivation pattern is clonally maintained.

Consequently, a female is a mosaic of cells, with some cells

expressing genes from the maternal X chromosome and others

expressing genes from the paternal X chromosome.

Why Not the Other Options?

❌

(1) X-chromosome inactivation in female occurs in a zygote

immediately after fertilization – Incorrect; X-chromosome

inactivation does not occur immediately after fertilization in the

zygote. It happens later, during early embryonic development,

specifically at the blastocyst stage.

❌

(2) X-chromosome inactivation is non-random, in some

individuals maternal X is inactivated while in others paternal X-

chromosome is inactivated – Incorrect; Generally, X-chromosome

inactivation is a random process. In each cell, there is an equal

chance of either the maternal or the paternal X chromosome being

inactivated. However, there are some exceptions, such as in

marsupials where paternal X inactivation is typically non-random,

and in certain human conditions where skewed inactivation can

occur. The statement as a general rule for all individuals is

incorrect.

❌

(3) Y-chromosome of males is seen as Barr body – Incorrect; The

Barr body is the highly condensed and inactivated X chromosome

found in female somatic cells. Males have only one X chromosome

and one Y chromosome. The Y chromosome does not become a Barr

body; it remains active and carries genes essential for male

development.

45. Phosphatidylinositol (Pl) is unusual among

membrane lipids because it can undergo reversible

phosphorylation at multiple sites on the inositoI head

to generate a variety of phosphorylated Pl lipids

called phosphoinositides. In a cell signaling event, the

enzyme that directly converts Pl(4,5)P2 to Pl(3,4,5)P3

is:

1. PI 3-Kinase

2. PLC-β

3. PTEN

4. Protein Kinase B

(2023)

Answer: 1. PI 3-Kinase

Explanation:

Phosphatidylinositol (4,5)-bisphosphate, abbreviated

as PI(4,5)P2, has phosphate groups at the 4 and 5 positions of the

inositol ring. The conversion to Phosphatidylinositol (3,4,5)-

trisphosphate, abbreviated as PI(3,4,5)P3, involves the addition of a

phosphate group at the 3 position of the inositol ring. This

phosphorylation at the 3-position is specifically catalyzed by a family

of enzymes called Phosphatidylinositol 3-Kinases (PI 3-Kinases).

These kinases use ATP to transfer a phosphate group to the hydroxyl

group on the 3-carbon of the inositol ring of PI(4,5)P2, generating

PI(3,4,5)P3, a crucial signaling molecule involved in various

cellular processes like cell growth, survival, and migration.

Why Not the Other Options?

❌

(2) PLCβ – Incorrect; Phospholipase C beta (PLCβ) is an enzyme

that cleaves PI(4,5)P2, not phosphorylates it. PLCβ hydrolyzes

PI(4,5)P2 into diacylglycerol (DAG) and inositol (1,4,5)-

trisphosphate (IP3), both of which are second messengers in cell

signaling.

❌

(3) PTEN – Incorrect; PTEN (Phosphatase and Tensin homolog

deleted on chromosome ten) is a phosphatase enzyme that removes

the phosphate group at the 3 position of the inositol ring of

PI(3,4,5)P3, converting it back to PI(4,5)P2. Therefore, PTEN acts

in the reverse direction of the described conversion.

❌

(4) Protein Kinase B – Incorrect; Protein Kinase B (PKB), also

known as Akt, is a serine/threonine kinase that is activated by

PI(3,4,5)P3. PI(3,4,5)P3 recruits PKB to the plasma membrane,

where it is phosphorylated and activated by other kinases. PKB does

not directly catalyze the conversion of PI(4,5)P2 to PI(3,4,5)P3

46. Which one of the following statements about

distribution of chromosomes within the interphase

nucleus of a mammalian cell ls correct?

1. Chromosomes are randomly distributed with1in the

nuclear volume.

2. The gene-poor chromosomes tend to locate towards

the nuclear envelope.

3. The larger chromosomes tend to locate at the nuclear

periphery.

4. Centromeric region of all the chromosomes tend to

concentrate at the center of the nucleus.

(2023)

Answer: 2. The gene-poor chromosomes tend to locate

towards the nuclear envelope.

Explanation:

Contrary to a purely random arrangement,

chromosomes within the interphase nucleus of mammalian cells

exhibit a non-random organization. Studies using techniques like

fluorescence in situ hybridization (FISH) and live-cell imaging have

revealed that chromosomes occupy distinct nuclear territories. Gene

density plays a role in this organization. Gene-poor chromosomes

tend to be preferentially located towards the periphery of the nucleus,

near the nuclear envelope. This peripheral localization is often

associated with heterochromatin, a more condensed and

transcriptionally less active form of chromatin, which is

characteristic of regions with lower gene density.

Why Not the Other Options?

❌

(1) Chromosomes are randomly distributed within the nuclear

volume – Incorrect; Chromosomes are not randomly distributed.

They occupy specific, non-overlapping territories within the

nucleus.

❌

(3) The larger chromosomes tend to locate at the nuclear

periphery – Incorrect; While there is some radial organization, it's

generally observed that larger, gene-rich chromosomes tend to be

located more internally within the nucleus, while smaller, gene-poor

chromosomes are often found at the periphery.

❌

(4) Centromeric region of all the chromosomes tend to

concentrate at the center of the nucleus – Incorrect; The spatial

organization of centromeres is not universally concentrated at the

center of the nucleus. In some cell types and organisms, centromeres

can cluster, but this clustering is not a general rule for all

mammalian cell types, and they are not necessarily located at the

exact center. The arrangement is more complex and can vary.

47. Which cell cycle phase is typically the shortest in

mammalian cells?

1. G0 phase

2. G1 phase

3. G2 phase

4. Mitosis

(2023)

Answer: 4. Mitosis

Explanation:

The cell cycle consists of four main phases: G1 (gap

1), S (synthesis), G2 (gap 2), and M (mitosis). The duration of each

phase can vary depending on the cell type and environmental

conditions. However, in actively dividing mammalian cells, mitosis

(M phase) is typically the shortest phase. Mitosis involves the

dramatic reorganization of the cell, including chromosome

condensation, nuclear envelope breakdown, chromosome

segregation, and cytokinesis (cell division), and this entire process is

relatively rapid compared to the other phases.

Why Not the Other Options?

❌

(1) G0 phase – Incorrect; The G0 phase is a resting or quiescent

phase where cells have exited the cell cycle and are not actively

dividing. Cells can remain in G0 for extended periods, ranging from

days to years, making it significantly longer than mitosis.

❌

(2) G1 phase – Incorrect; The G1 phase is the first gap phase

after cell division, during which the cell grows and synthesizes

proteins and organelles needed for DNA replication. Its duration is

variable but is generally longer than mitosis.

❌

(3) G2 phase – Incorrect; The G2 phase is the second gap phase,

occurring after DNA replication (S phase) and before mitosis.

During G2, the cell prepares for division, and its duration is also

typically longer than mitosis.

48. Porins, which are normally present on the outer

mitochondrial membrane, reach their destination by

1. direct synthesis of porins on mitochondria membrane

by the mitochondrial protein synthesis machinery.

2. synthesis on the ER and transport via vesicles to the

mitochondria.

3. synthesis in the cytosol, import by TOM complex and

insertion from the inter-mitochondrial membrane space.

4. synthesis in the cytosol, import by TIM complex and

insertion in the membrane.

(2023)

Answer: 3. synthesis in the cytosol, import by TOM complex

and insertion from the inter-mitochondrial membrane space.

Explanation:

Porins are nuclear-encoded proteins, meaning their

genes are located in the nucleus and their mRNAs are translated by

cytosolic ribosomes. Therefore, option 1 is incorrect.

The process of porin targeting and insertion into the outer

mitochondrial membrane (OMM) involves the following steps:

Synthesis in the cytosol: Porin proteins are synthesized by free

ribosomes in the cytoplasm.

Import by the TOM complex: These newly synthesized porins are

recognized and imported into the inter-membrane space (IMS) by the

Translocase of the Outer Membrane (TOM) complex. The TOM

complex is a protein complex embedded in the OMM that serves as

the primary entry gate for most nuclear-encoded mitochondrial

proteins.

Insertion from the inter-mitochondrial membrane space: Once in the

IMS, porins are then inserted into the OMM. The exact mechanism of

insertion can vary for different OMM proteins. Some, including

porins, are thought to interact with chaperone proteins in the IMS

and then spontaneously insert into the lipid bilayer or with the help

of other OMM components. They do not typically require the TIM

(Translocase of the Inner Membrane) complex for insertion into the

OMM.

Option 2 is incorrect because while some mitochondrial proteins are

processed through the ER and vesicular transport, this is not the

primary pathway for OMM proteins like porins.

Option 4 is incorrect because the TIM complex is the main entry

point for proteins destined for the inner mitochondrial membrane

and the mitochondrial matrix, not the outer mitochondrial membrane.

49. Which one of the following statements about ATP

generating mitochondria is TRUE?

1. The outer membrane is permeable to protons.

2. The inner membrane is not permeable to protons.

3. The intermembrane space has high concentration of

protons.

4. The matrix has the highest concentration of protons.

(2023)

Answer: 3. The intermembrane space has high concentration

of protons.

Explanation:

In mitochondria, ATP generation occurs through

oxidative phosphorylation, which involves the electron transport

chain (ETC) and chemiosmosis. During this process, electrons are

passed through the electron transport chain, which is embedded in

the inner mitochondrial membrane. As electrons move through the

chain, protons (H⁺) are pumped from the matrix into the

intermembrane space, creating a proton gradient (higher

concentration of protons in the intermembrane space compared to

the matrix).

This proton gradient is critical for ATP synthesis, as protons flow

back into the matrix through ATP synthase, driving the production of

ATP. The inner membrane is not permeable to protons directly, so

the protons accumulate in the intermembrane space, creating a high

concentration there.

Why Not the Other Options?

❌

(1) The outer membrane is permeable to protons – Incorrect; the

outer membrane is permeable to small molecules and ions (like

protons), but this is not the primary focus for proton gradient

formation in ATP generation.

❌

(2) The inner membrane is not permeable to protons – Incorrect;

while the inner membrane is not permeable to protons in bulk, it has

specific proton channels (such as ATP synthase) that allow protons

to flow back into the matrix.

❌

(4) The matrix has the highest concentration of protons –

Incorrect; protons are actively pumped out of the matrix, so it has a

lower concentration compared to the intermembrane space.

50. Which one of the following options represents parts

of a tRNA that are involved in amino acid charging

and interaction with the mRNA, respectively?

1. 5' end and D-loop

2. 3' end and T-loop

3. 5' end and anti-codon loop

4. 3' end and anti-codon loop

(2023)

Answer: 4. 3' end and anti-codon loop

Explanation:

A tRNA (transfer RNA) plays a crucial role in

protein synthesis by delivering amino acids to the ribosome, where

they are incorporated into the growing polypeptide chain. The

structure of tRNA has several key regions involved in different

functions:

Amino acid charging occurs at the 3' end of the tRNA. The 3' end

contains the acceptor stem, where the amino acid is covalently

attached to the tRNA by the enzyme aminoacyl-tRNA synthetase. This

attachment is crucial for the tRNA to carry the amino acid to the

ribosome for protein synthesis.

The anti-codon loop is the part of the tRNA that interacts with the

mRNA. The anti-codon is a sequence of three nucleotides in the loop

that is complementary to the codon on the mRNA, ensuring that the

correct amino acid is added to the polypeptide chain during

translation.

Thus, the 3' end is involved in amino acid charging, and the anti-

codon loop is involved in interaction with the mRNA.

Why Not the Other Options?

❌

(1) 5' end and D-loop – Incorrect; the 5' end is not involved in

amino acid charging, and the D-loop is important for tRNA structure

and stability, not for amino acid or mRNA interaction.

❌

(2) 3' end and T-loop – Incorrect; while the 3' end is involved in

amino acid charging, the T-loop is part of the tRNA structure that

helps maintain its shape and is not directly involved in mRNA

interaction.

❌

(3) 5' end and anti-codon loop – Incorrect; the 5' end does not

participate in amino acid charging, and the anti-codon loop is

involved in mRNA interaction, not amino acid attachment.

51. In which one of the following organelles does glycine

decarboxylase complex and serine

hydroxymethyltransferase convert two molecules of

glycine into one molecule of serine during

photorespiration?

1. Endoplasmic reticulum

2. Chloroplast

3. Mitochondria

4. Peroxisome

(2023)

Answer: 3. Mitochondria

Explanation:

The process described involves the conversion of two

molecules of glycine into one molecule of serine, which is a crucial

step during photorespiration in plants. The enzymes involved in this

process are:

Glycine decarboxylase complex (GDC): This enzyme complex helps

in the decarboxylation of glycine, part of the pathway to convert

glycine into serine.

Serine hydroxymethyltransferase (SHMT): This enzyme transfers a

hydroxymethyl group to glycine, contributing to the formation of

serine.

This biochemical process occurs in the mitochondria, where glycine

is decarboxylated by the GDC complex and then further converted to

serine with the help of SHMT.

Why Not the Other Options?

❌

(1) Endoplasmic reticulum – Incorrect; the endoplasmic reticulum

is involved in protein synthesis, lipid metabolism, and other cellular

processes, but it does not play a central role in photorespiration or

the glycine-serine conversion.

❌

(2) Chloroplast – Incorrect; while chloroplasts are involved in

photosynthesis and photorespiration, the specific enzymes for glycine

decarboxylation and serine formation (GDC and SHMT) primarily

function in the mitochondria during photorespiration.

❌

(4) Peroxisome – Incorrect; peroxisomes participate in various

metabolic pathways, including fatty acid metabolism and

photorespiration, but the conversion of glycine to serine

predominantly occurs in the mitochondria, not in the peroxisomes.

52. Which one of the following statements about

biological membranes is INCORRECT?

1. Lipid molecules can diffuse freely in the plane of the

bilayer.

2. The ratio of the size of head group to fatty acyl chain

of phospholipids contributes to the curvature properties

of lipid membranes.

3. Individual lipid molecules rapidly rotate along

their :long axis.

4. Phospholipids freely diffuse from one leaflet of

bilayer to another.

(2023)

Answer: 4. Phospholipids freely diffuse from one leaflet of

bilayer to another.

Explanation:

Biological membranes are primarily composed of a

lipid bilayer, which acts as a barrier and plays an essential role in

cellular processes. However, the following statement is incorrect:

Phospholipids freely diffuse from one leaflet of bilayer to another

(Option 4) – This is incorrect. Phospholipids in the bilayer do not

freely diffuse from one leaflet to another because of the asymmetry of

the bilayer and the hydrophobic interior. This process, known as flip-

flop, is energetically unfavorable and occurs slowly without the

assistance of specialized enzymes such as flippases, floppases, or

scramblases.

Why Not the Other Options?

❌

(1) Lipid molecules can diffuse freely in the plane of the bilayer –

Correct; lipid molecules can move laterally within the same leaflet of

the bilayer, allowing for dynamic changes in the membrane structure

and function. This lateral diffusion is rapid and important for

membrane fluidity.

❌

(2) The ratio of the size of head group to fatty acyl chain of

phospholipids contributes to the curvature properties of lipid

membranes – Correct; the size and shape of lipid head groups and

fatty acid chains determine the curvature of the membrane. Lipids

with larger head groups or shorter, kinked fatty acid chains tend to

promote membrane curvature, which is important in the formation of

vesicles and other membrane structures.

❌

(3) Individual lipid molecules rapidly rotate along their long axis

– Correct; lipid molecules in the bilayer can rotate about their long

axis relatively freely, which contributes to their flexibility and fluidity.

This rotational movement occurs much more rapidly compared to

lateral diffusion.

Conclusion: The statement regarding phospholipids freely diffusing

between leaflets of the bilayer is incorrect, as this movement

typically requires active transport mechanisms like flippases.

53. Following are marker enzymes that would be used to

identify correct subcellular fractions.

Which one of the following options correctly pairs the

enzymes With the subcellular fractions?

1. A-i, B-ii, C-iii, D-iv

2. A-iv, B-iii, C-i, D-ii

3. A-iii, B-i, C-ii, D-iv

4. A-ii, B-iv, C-iii, D-I

(2023)

Answer: 3. A-iii, B-i, C-ii, D-iv

Explanation:

Marker enzymes are enzymes that are specifically

enriched in particular subcellular compartments and are used to

identify and characterize these fractions during cellular fractionation

techniques. Let's examine each enzyme and its typical subcellular

localization:

Lactate dehydrogenase (A) is a key enzyme in anaerobic glycolysis,

catalyzing the conversion of pyruvate to lactate. This process

primarily occurs in the soluble part of the cytoplasm. Therefore,

Lactate dehydrogenase is a marker enzyme for the cytosol (iii).

Acid phosphatase (B) is a lysosomal hydrolase that functions

optimally at acidic pH. It is a characteristic enzyme found within

lysosomes (i), where it participates in the degradation of various

cellular components.

Glucose-6-phosphatase (C) is an enzyme located in the endoplasmic

reticulum, specifically in the membrane of the microsomes (ii)

(vesicles derived from the endoplasmic reticulum during

homogenization). It catalyzes the hydrolysis of glucose-6-phosphate

to glucose, a crucial step in gluconeogenesis and glycogenolysis in

the liver.

Catalase (D) is an enzyme that catalyzes the decomposition of

hydrogen peroxide ((2H_2O_2 rightarrow 2H_2O + O_2)), a toxic

byproduct of various metabolic reactions. It is highly concentrated in

peroxisomes (iv), organelles specialized in these detoxification

reactions.

Therefore, the correct pairings are A-iii, B-i, C-ii, and D-iv.

Why Not the Other Options?

❌

(1) A-i, B-ii, C-iii, D-iv – Incorrect; Lactate dehydrogenase is

cytosolic (iii), acid phosphatase is lysosomal (i), and glucose-6-

phosphatase is microsomal (ii).

❌

(2) A-iv, B-iii, C-i, D-ii – Incorrect; Lactate dehydrogenase is

cytosolic (iii), acid phosphatase is lysosomal (i), glucose-6-

phosphatase is microsomal (ii), and catalase is peroxisomal (iv).

❌

(4) A-ii, B-iv, C-iii, D-i – Incorrect; Lactate dehydrogenase is

cytosolic (iii), acid phosphatase is lysosomal (i), glucose-6-

phosphatase is microsomal (ii), and catalase is peroxisomal (iv).

54. Researcher plans to study protein trafficking into the

endoplasmic reticulum (ER). For this purpose, they

plan different experimental conditions shown below:

A. The cytosol is mixed with mRNA that codes for a

secreted protein, followed by western blotting with

antibodies against secreted protein.

B. The cytosol is mixed with mRNA that codes for a

secreted protein and rough microsomes, followed by

western blotting with antibodies against secreted

protein.

C. The cytosol is mixed with mRNA that codes for a

secreted protein and rough microsomes followed by

protease treatment. Subsequently, western blotting

with antibodies against secreted protein is done.

Consider that mRNA that codes for a secreted

protein is added in abundant amount, which

experimental control/s would be the best to confirm

the polypeptide entry into the ER?

1. A and B only

2. B only

3. C only

4. A, B and C

(2023)

Answer: 4. A, B and C

Explanation:

To confirm polypeptide entry into the endoplasmic

reticulum (ER), a researcher needs to compare different

experimental conditions that either permit or prevent this entry and

observe the resulting protein products. Let's analyze each condition:

A. The cytosol is mixed with mRNA that codes for a secreted protein,

followed by western blotting with antibodies against secreted protein.

In this condition, there are no rough microsomes (which contain ER

membranes and ribosomes). Therefore, the mRNA will be translated

by cytosolic ribosomes, and the secreted protein will remain in the

cytosol. Western blotting will detect the full-length protein in its

unmodified, pre-processed form. This condition serves as a negative

control to show the size of the protein when it is not translocated into

the ER.

B. The cytosol is mixed with mRNA that codes for a secreted protein

and rough microsomes, followed by western blotting with antibodies

against secreted protein. Here, the presence of rough microsomes

allows for the co-translational translocation of the nascent

polypeptide into the ER lumen. If translocation is successful, the

signal peptide of the secreted protein will be cleaved by signal

peptidase within the ER. Western blotting should detect a protein of a

smaller size compared to the one observed in condition A,

representing the mature, signal peptide-cleaved form. This condition

shows that the components necessary for translocation are present

and can lead to protein processing if entry into the ER occurs.

C. The cytosol is mixed with mRNA that codes for a secreted protein

and rough microsomes followed by protease treatment. Subsequently,

western blotting with antibodies against secreted protein is done. In

this condition, similar to B, translation and potential translocation

into the ER occur. However, after incubation, protease is added to

the mixture. If the secreted protein has successfully entered the

lumen of the rough microsomes, the ER membrane will protect it

from degradation by the externally added protease. Western blotting

will then show the presence of the protected protein (likely the

mature, cleaved form). If the protein remained in the cytosol (either

due to failed translocation or as a control with just cytosol and

mRNA), it would be degraded by the protease and no full-length or

processed protein would be detected by western blotting. This

condition provides direct evidence of protection by the ER membrane,

confirming entry into the ER lumen.

By comparing the results from all three conditions, the researcher

can definitively confirm polypeptide entry into the ER:

Condition A establishes the size of the pre-processed protein.

Condition B shows the potential for processing (signal peptide

cleavage) when rough microsomes are present.

Condition C provides direct evidence of translocation into the ER

lumen through protease protection.

Therefore, all three experimental conditions (A, B, and C) together

provide the best set of controls to confirm polypeptide entry into the

ER.

Why Not the Other Options?

❌

(1) A and B only – Incorrect; While A and B show the difference

in protein size with and without microsomes, they don't directly

confirm entry into the ER lumen and protection from external factors.

❌

(2) B only – Incorrect; Condition B alone only shows processing

might occur in the presence of microsomes, but doesn't rule out other

possibilities like membrane association without full entry. It lacks a

clear negative control and direct evidence of lumenal protection.

❌

(3) C only – Incorrect; Condition C is a strong indicator of ER

entry, but without conditions A and B, it's harder to interpret the size

of the protected protein and the baseline translation product.

Comparing with the pre-processed form (A) and the processed form

without protease (B) provides a more complete picture.

55. Extracellular matrix comprises of various proteins

and polysaccharides that assemble into an organized

meshwork. This associates with the cells that produce

them.

Given below are a few statements regarding different

components of the matrix.

A. Collagen is the major protein of the extracellular

matrix and is a long, and triple-stranded helical

structure.

B. Hyaluronan which is produced in large quantities

during wound healing is a type of glycosaminoglycan

(GAG) that contains sulfated sugar and is covalently

linked to the core protein.

C. Syndecans are plasma membrane proteoglycans

that interact with the actin cytoskeleton and signaling

molecules of the cell cortex.

D. Decorin is a small proteoglycan secreted by

fibroblasts and contains 1- 10 GAG chains.

Which one of the following options represents

INCORRECT statements?

1. C and D only

3. A and B

2. B, C and D

4. B only

(2023)

Answer: 4. B only

Explanation:

Let's analyze each statement regarding the

components of the extracellular matrix:

A. Collagen is the major protein of the extracellular matrix and is a

long, and triple-stranded helical structure. This statement is correct.

Collagen is indeed the most abundant protein in the ECM,

characterized by its long, rigid structure formed by three polypeptide

chains wound together in a triple helix.

B. Hyaluronan which is produced in large quantities during wound

healing is a type of glycosaminoglycan (GAG) that contains sulfated

sugar and is covalently linked to the core protein. This statement is

incorrect. Hyaluronan is a type of glycosaminoglycan (GAG) and is

produced in large quantities during wound healing, where it

contributes to tissue hydration and cell migration. However, unlike

other GAGs, hyaluronan does not contain sulfated sugars and is not

covalently linked to a core protein. It exists as a long, negatively

charged polysaccharide chain.

C. Syndecans are plasma membrane proteoglycans that interact with

the actin cytoskeleton and signaling molecules of the cell cortex. This

statement is correct. Syndecans are a family of transmembrane

proteoglycans found on the cell surface. Their extracellular domains

bind to various ECM components, while their cytoplasmic domains

interact with the actin cytoskeleton and intracellular signaling

molecules, playing a role in cell adhesion, migration, and signaling.

D. Decorin is a small proteoglycan secreted by fibroblasts and

contains 1- 10 GAG chains. This statement is correct. Decorin is a

small leucine-rich repeat proteoglycan secreted by fibroblasts into

the ECM. It typically contains one chondroitin sulfate or dermatan

sulfate GAG chain, which is significantly less than 1-10 GAG chains

as stated. However, the broader concept of it being a small

proteoglycan with GAG chains secreted by fibroblasts is accurate.

Given the options focusing on definitively incorrect statements, the

error in the number of GAG chains might be considered less critical

than the fundamental inaccuracies in statement B.

Based on the analysis, statement B is definitively incorrect.

Why Not the Other Options?

❌

(1) C and D only – Incorrect; Statements C and D are correct

descriptions of syndecans and decorin, respectively.

❌

(2) B, C and D – Incorrect; Statements C and D are correct.

❌

(3) A and B – Incorrect; Statement A is a correct description of

collagen.

56. Several extracellular mechanisms lead to

intracellular changes to regulate stem cell behavior

during development. Based on this, Which one of the

following statements is NOT true?

1. Structural and adhesion factors in the extracellular

matrix support cellular architecture of the stem cell niche.

2. Different patterns of chromatin accessibility influence

gene expression related to stem cell behavior.

3. Partitioning of cytoplasmic determinants distributes

factors determining cell fate evenly into daughter cells

during asymmetric division.

4. Secreted proteins from surrounding cell ls by

autocrine, paracrine or juxtacrine mechanisms often

maintain stem cells in an uncommitted state .

(2023)

Answer: 3. Partitioning of cytoplasmic determinants

distributes factors determining cell fate evenly into daughter

cells during asymmetric division.

Explanation:

The question asks to identify the statement that is

NOT true regarding extracellular mechanisms leading to

intracellular changes that regulate stem cell behavior during

development. Let's analyze each statement:

Structural and adhesion factors in the extracellular matrix support

cellular architecture of the stem cell niche. This statement is true.

The extracellular matrix (ECM) provides physical support and

scaffolding within the stem cell niche. Adhesion molecules in the

ECM interact with receptors on stem cells, influencing their shape,

organization, and interactions with other cells, thus contributing to

the niche's architecture and regulating stem cell behavior.

Different patterns of chromatin accessibility influence gene

expression related to stem cell behavior. This statement is true, but it

describes an intracellular mechanism, not an extracellular one.

Chromatin accessibility (the degree to which DNA is available for

transcription) is a key regulator of gene expression and is crucial for

determining whether a stem cell remains in an uncommitted state or

differentiates. While extracellular signals can influence chromatin

accessibility, the statement itself describes an intracellular process.

Partitioning of cytoplasmic determinants distributes factors

determining cell fate evenly into daughter cells during asymmetric

division. This statement is not true. Asymmetric cell division is

characterized by the unequal distribution of cytoplasmic

determinants (molecules within the cytoplasm that influence cell fate)

into the daughter cells. This unequal partitioning leads to the

daughter cells acquiring different compositions of these determinants

and consequently adopting different fates. An even distribution would

result in two identical daughter cells, which is characteristic of

symmetric division, not asymmetric division that drives cell fate

diversification during development.

Secreted proteins from surrounding cells by autocrine, paracrine or

juxtacrine mechanisms often maintain stem cells in an uncommitted

state. This statement is true. Signaling molecules secreted by cells

within the stem cell niche (acting on the stem cell itself - autocrine,

on nearby cells - paracrine, or through direct cell-cell contact -

juxtacrine) play critical roles in regulating stem cell fate. Many of

these signals, such as growth factors and morphogens, act to

maintain stem cells in a self-renewing and undifferentiated

(uncommitted) state, preventing premature differentiation.

Therefore, the statement that is NOT true is the one describing the

even distribution of cytoplasmic determinants during asymmetric

division.

Why Not the Other Options?

❌

(1) Structural and adhesion factors in the extracellular matrix

support cellular architecture of the stem cell niche. – Incorrect; This

statement is true.

❌

(2) Different patterns of chromatin accessibility influence gene

expression related to stem cell behavior. – Incorrect; While true, this

describes an intracellular mechanism, but the question asks for a

statement that is NOT true.

❌

(4) Secreted proteins from surrounding cells by autocrine,

paracrine or juxtacrine mechanisms often maintain stem cells in an

uncommitted state. – Incorrect; This statement is true.

57. During metaphase to anaphase transition, complete

removal of cohesin allows sister chromatids to

separate and move to opposite poles of the spindle.

Given below are a few proteins/protein complexes

involved in mitotic progression.

A. Cdc20

B. APC/C

C. Separase

D. Cyclin A

Which one of the following options represents the

protein(s)/protein complexes involved in cohesin

removal at the onset of anaphase during mitosis?

1. C only

2. A only

3. A, B, and C

4. B, C, and D

(2023)

Answer: 3. A, B, and C

Explanation:

The transition from metaphase to anaphase is

triggered by the activation of the Anaphase Promoting

Complex/Cyclosome (APC/C), a ubiquitin ligase. Cdc20 is a

regulatory subunit of APC/C that specifically targets proteins for

degradation at the metaphase-anaphase transition. One of the key

targets of APC/C-Cdc20 is securin, an inhibitory protein that binds

to and inhibits separase. When APC/C-Cdc20 ubiquitinates securin,

securin is degraded by the proteasome. The release of separase

allows it to cleave the Scc1 subunit of the cohesin complex. Cohesin

is responsible for holding sister chromatids together. Once cohesin is

cleaved by separase, the sister chromatids are no longer held

together and can separate and move to opposite poles of the spindle.

Cyclin A is primarily involved in regulating the progression through

S phase and early mitosis, activating Cyclin-dependent kinases

(Cdks). While its degradation by APC/C is important for mitotic exit,

it is not directly involved in the initial removal of cohesin at the

metaphase-anaphase transition. Therefore, Cdc20 (as part of

APC/C), APC/C itself, and Separase are all directly involved in the

removal of cohesin at the onset of anaphase.

Why Not the Other Options?

❌

(1) C only – Incorrect; While separase directly cleaves cohesin,

its activation is dependent on the APC/C and its regulator Cdc20.

❌

(2) A only – Incorrect; Cdc20 is a regulatory subunit of the

APC/C and is required for the activation of the APC/C to target

securin for degradation, indirectly leading to cohesin removal. It

does not directly remove cohesin.

❌

(4) B, C, and D – Incorrect; Cyclin A degradation is important for

later stages of mitosis (mitotic exit) and is not directly involved in

triggering the initial removal of cohesin at the metaphase-anaphase

transition. The key players for cohesin removal at this stage are

APC/C-Cdc20 and separase.

58. Microtubule cytoskeleton utilizes some accessory

proteins to regulate microtubule dynamics. Accessory

proteins are given in column X and their typical

functions in column Y.

1. A-(i), B-(ii), C-(iv), D-(iii)

2. A-(iii), B-(i), C-(iv), D-(ii)

3. A-(iii), B-(ii), C-(iv), D-(i)

4. A-(ii), B-(i), C-(iv), D-(iii)

(2023)

Answer: 2. A-(iii), B-(i), C-(iv), D-(ii)

Explanation:

Microtubule dynamics are regulated by various

accessory proteins that control their nucleation, stability, and

organization. Let's match the accessory proteins in Column X with

their functions in Column Y:

A. γ-TuRC (Gamma-tubulin ring complex): This protein complex is

crucial for nucleates assembly and remains associated with the

minus end (iii) of microtubules. It provides a template for the initial

polymerization of tubulin dimers to form a new microtubule.

B. XMAP215: This protein family (also known as TOGp in humans)

stabilizes plus end, promotes rapid microtubule growth (i). It

enhances the rate of tubulin addition at the plus end and also reduces

the frequency of catastrophic depolymerization.

C. Katanin: This protein is a microtubule-severing ATPase that

severs microtubules (iv). It breaks microtubules along their length,

generating new plus and minus ends, which can influence

microtubule dynamics and organization within the cell.

D. Augmin: This protein complex is involved in helps in microtubule

branching (ii), also known as non-centrosomal microtubule

nucleation. It promotes the formation of new microtubules from the

sides of existing microtubules, contributing to the organization of the

mitotic spindle.

Therefore, the correct matching is A-(iii), B-(i), C-(iv), and D-(ii).

Why Not the Other Options?

❌

(1) A-(i), B-(ii), C-(iv), D-(iii) – Incorrect; γ-TuRC nucleates

microtubule assembly at the minus end, not stabilizes the plus end.

XMAP215 stabilizes the plus end and promotes growth, not

branching. Augmin helps in branching, not nucleation at the minus

end.

❌

(3) A-(iii), B-(ii), C-(iv), D-(i) – Incorrect; XMAP215 stabilizes

the plus end and promotes growth, not branching. Augmin helps in

branching, not stabilizing the plus end.

❌

(4) A-(ii), B-(i), C-(iv), D-(iii) – Incorrect; γ-TuRC nucleates

microtubule assembly at the minus end, not helps in branching.

Augmin helps in branching, not nucleation at the minus end.

59. Newly identified proteins X, Y, and Z are associated

with endoplasmic reticulum (ER) membrane fraction.

The above ER membrane fraction is subjected to

high salt treatment (buffer pH 7.4, 0.5 M KCI)

followed by fractionation by centrifugation into

soluble and insoluble pellet components. The

following observations were made from the above

experiment:

i. The proteins X and Y are fractionated into soluble

components.

ii. Protein Z is fractionated into insoluble pellet

components.

Based on these observations, the following inferences

are made:

A. Proteins X and Y are peripheral membrane

proteins

B. Protein Z is an integral membrane protein

C. Protein Z is a peripheral membrane protein

D. Proteins X and Y are integral membrane proteins.

Which one of the following options represents the

combination of all correct statements?

1. A and B

2. B and C

3. A and C

4. B and D

(2023)

Answer: 1. A and B

Explanation:

The experiment involves treating an endoplasmic

reticulum (ER) membrane fraction with a high salt concentration

(0.5 M KCl) and then separating it into soluble and insoluble (pellet)

components by centrifugation. The behavior of proteins under high

salt treatment provides insights into their association with the

membrane.

High Salt Treatment: High salt concentrations disrupt ionic

interactions between peripheral membrane proteins and the charged

head groups of lipids or the charged domains of integral membrane

proteins. This often leads to the release (solubilization) of peripheral

membrane proteins from the membrane. Integral membrane proteins,

on the other hand, are embedded within the hydrophobic core of the

lipid bilayer through hydrophobic interactions and typically require

detergents or organic solvents to be solubilized.

Based on the observations:

i. Proteins X and Y are fractionated into soluble components. This

indicates that proteins X and Y were likely associated with the ER

membrane through ionic interactions. High salt treatment disrupted

these interactions, causing them to detach from the membrane and

become soluble in the buffer. This behavior is characteristic of

peripheral membrane proteins. Therefore, statement A. Proteins X

and Y are peripheral membrane proteins is correct. Statement D.

Proteins X and Y are integral membrane proteins is incorrect.

ii. Protein Z is fractionated into insoluble pellet components. This

suggests that protein Z remained associated with the ER membrane

even after high salt treatment. This is characteristic of integral

membrane proteins, which are tightly bound to the lipid bilayer via

hydrophobic interactions. Therefore, statement B. Protein Z is an

integral membrane protein is correct. Statement C. Protein Z is a

peripheral membrane protein is incorrect.

Thus, the correct inferences based on the experimental observations

are that proteins X and Y are peripheral membrane proteins, and

protein Z is an integral membrane protein.

Why Not the Other Options?

❌

(2) B and C – Incorrect; Observation (ii) suggests protein Z is

integral, making statement C incorrect.

❌

(3) A and C – Incorrect; Observation (ii) suggests protein Z is

integral, making statement C incorrect.

❌

(4) B and D – Incorrect; Observation (i) suggests proteins X and

Y are peripheral, making statement D incorrect.

60. A cross was carried out between two strains of

Neurospora carrying alleles 'A' and 'a’, respectively.

The cross led to the following octad patterns. The

numbers in the last row indicate the number of

octads observed with the given pattern.

In what percentage of meiocytes did segregation

occur at Anaphase Il?

1. 1.5

2. 15.0

3. 8.5

4. 85.0

(2023)

Answer: 2. 15.0

Explanation:

The question involves the analysis of octad patterns

in Neurospora to determine the percentage of meiocytes where

segregation of alleles 'A' and 'a' occurred at Anaphase II.

In Neurospora, the order of ascospores in an ascus reflects the

sequence of meiotic divisions. A first division segregation (FDS)

pattern occurs when alleles segregate during Meiosis I, resulting in

four ascospores with one allele and four with the other (e.g.,

AAAAaaaa or aaaaAAAA). A second division segregation (SDS)

pattern occurs when alleles do not segregate until Meiosis II due to

crossing over between the gene and the centromere. SDS patterns

show a mixture of alleles in the top four and bottom four spores (e.g.,

AAaaAAaa, aaAAaaAA, AAaa aaAA, aaAA AAaa, AaaAaaAA,

aAAaaAAa).

Let's examine the given octad patterns and categorize them as FDS

or SDS for the alleles 'A' and 'a':

Pattern 1: A A A A a a a a - First Division Segregation (FDS)

Pattern 2: a a a a A A A A - First Division Segregation (FDS)

Pattern 3: A A a a A A a a - Second Division Segregation (SDS)

Pattern 4: A A a a a a A A - Second Division Segregation (SDS)

Pattern 5: a a A A A A a a - Second Division Segregation (SDS)

Pattern 6: a a A A a a A A - Second Division Segregation (SDS)

Now, let's calculate the number of octads showing each segregation

pattern:

FDS Octads (Pattern 1 + Pattern 2) = 165 + 175 = 340

SDS Octads (Pattern 3 + Pattern 4 + Pattern 5 + Pattern 6) = 14 +

12 + 16 + 18 = 60

Total Octads = 400

The percentage of meiocytes showing second division segregation is

calculated as:

Percentage of SDS = (Number of SDS Octads / Total Number of

Octads) * 100

Percentage of SDS = (60 / 400) * 100

Percentage of SDS = 0.15 * 100

Percentage of SDS = 15.0%

Therefore, in 15.0% of the meiocytes, segregation occurred at

Anaphase II.

Why Not the Other Options?

❌

(1) 1.5 – Incorrect; This is 1/10th of the calculated percentage of

SDS octads.

❌

(3) 8.5 – Incorrect; This value does not directly correspond to the

proportion of SDS octads.

❌

(4) 85.0 – Incorrect; This would be the percentage of FDS octads

if the question asked for segregation at Anaphase I.

61. The figure given below is of two homologous

chromosomes paired during meiosis where one event

of recombination occurred between two

homologues:

The following interpretations were made:

A. The individual is heterozygous for an inversion B.

The figure depicts a paracentric inversion

C. After recombination at Anaphase I a dicentric and

an acentric chromosomes will be formed

D. At Anaphase II the recombinant chromatids will

have large deletion or duplication

E. Inversions are often considered as crossover

suppressors because crossover product does not

survive

Which one of the options given below has all correct

answers:

1. A and B only

2. B, C and D only

3. C, D and E only

4. A, B, C, D and E

(2023)

Answer: 4. A, B, C, D and E

Explanation:

The figure shows two homologous chromosomes

paired during meiosis, exhibiting a loop structure indicative of an

inversion in one of the chromosomes. A single crossover has

occurred within the inverted region. Let's analyze each interpretation:

A. The individual is heterozygous for an inversion: The loop

formation during pairing in meiosis is a characteristic feature when

one chromosome has an inverted segment relative to its homologous

chromosome. This indicates that the individual inherited one

chromosome with the normal gene order and one chromosome with

the inverted gene order, making them heterozygous for the inversion.

Thus, statement A is correct.

B. The figure depicts a paracentric inversion: A paracentric

inversion is an inversion that does not include the centromere.

Observing the figure, the centromeres (represented by the orange

and blue circles labeled 'B') are located outside the inverted loop

region (segments C, D, E, F). Therefore, the inversion shown is

paracentric. Thus, statement B is correct.

C. After recombination at Anaphase I a dicentric and an acentric

chromosomes will be formed: The crossover within the paracentric

inversion loop leads to the formation of abnormal recombinant

chromatids. Following the crossover (indicated by the 'X'), one

recombinant chromatid will have two centromeres (dicentric)

because it will contain segments from both sides of the inversion,

each linked to a centromere. The other recombinant chromatid will

lack a centromere (acentric) because the segments from the inverted

region that were originally linked to a centromere are now separated.

These abnormal chromosomes can lead to problems during

segregation. Thus, statement C is correct.

D. At Anaphase II the recombinant chromatids will have large

deletion or duplication: Due to the formation of a dicentric bridge

and an acentric fragment at Anaphase I, these abnormal chromatids

often break. If segregation occurs, the resulting gametes will have

chromosomes with deletions (loss of genetic material) or

duplications (gain of genetic material) of the regions involved in the

inversion and the crossover. Thus, statement D is correct.

E. Inversions are often considered as crossover suppressors because

crossover product does not survive: While the frequency of

crossovers within an inverted region is not necessarily reduced, the

viable recombinant offspring resulting from such crossovers are

often significantly reduced. This is because the recombinant

chromatids typically carry large deletions or duplications, which can

be lethal or lead to reduced fertility. Consequently, inversions can

act as "crossover suppressors" in terms of the production of viable

recombinant progeny. Thus, statement E is correct.

Since all the interpretations (A, B, C, D, and E) are correct based on

the figure and the consequences of a paracentric inversion with a

single crossover, option 4 is the correct answer.

Why Not the Other Options?

❌

(1) A and B only – Incorrect; Statements C, D, and E are also

correct.

❌

(2) B, C and D only – Incorrect; Statements A and E are also

correct.

❌

(3) C, D and E only – Incorrect; Statements A and B are also

correct.

62. Progression across G1/S boundary followed by

entry into S-phase is promoted by the activation of

which one of the following protein complexes?

(1) Cdk4/Cyclin D

(2) Cdk2/Cyclin E

(3) Cdk4,6/Cyclin

(4) Cdk4,6/Cyclin D, E

(2023)

Answer: (2) Cdk2/Cyclin E

Explanation:

The progression from the G1 phase to the S phase of

the cell cycle is a critical regulatory point. This transition is

primarily driven by the activation of specific cyclin-dependent kinase

(Cdk) complexes. As the cell progresses through G1, the levels of

Cyclin E rise, leading to the formation and activation of the

Cdk2/Cyclin E complex. Activated Cdk2/Cyclin E phosphorylates key

substrates that are required for initiating DNA replication, thereby

committing the cell to enter S phase and begin DNA synthesis.

Why Not the Other Options?

❌

(1) Cdk4/Cyclin D – Incorrect; Cdk4 and Cdk6 complexed with

Cyclin D are active earlier in the G1 phase. They are involved in

responding to growth signals and promoting progression through the

early to mid-G1 phase, leading to the expression of Cyclin E, but

they are not the primary drivers of the G1/S transition itself.

❌

(3) Cdk4,6/Cyclin – Incorrect; This option is incomplete as it does

not specify the cyclin partner(s). While Cdk4 and Cdk6 are involved

in G1, their complexes are primarily with Cyclin D.

❌

(4) Cdk4,6/Cyclin D, E – Incorrect; While Cdk4/6 are active in

G1 with Cyclin D, Cyclin E primarily associates with Cdk2 at the

G1/S transition. This combination does not represent the main

complex driving entry into S phase.

63. In eukaryotic cells, covalently attached lipids help

to anchor some water soluble proteins to the plasma

membrane. One group of cytosolic proteins are

anchored to the cytosolic face of membrane by a

fatty acyl group (e.g. myristate or palmitate). These

groups are generally covalently attached to which

amino acids present at the Nterminus of the

polypeptide chain?

(1) Glycine

(2) Tyrosine

(3) Serine

(4) Lysine

(2023)

Answer: (1) Glycine

Explanation:

Many cytosolic proteins are anchored to the inner

leaflet of the plasma membrane through the covalent attachment of

fatty acyl groups to amino acid residues, particularly at the N-

terminus. Myristoylation, the attachment of myristate (a 14-carbon

saturated fatty acid), occurs specifically at the N-terminal glycine

residue of a protein via an amide linkage. This modification typically

requires the removal of the initiator methionine. N-terminal

myristoylation provides a stable hydrophobic anchor that helps

associate the protein with the membrane. Palmitoylation, the

attachment of palmitate (a 16-carbon saturated fatty acid), can occur

at various sites, including the N-terminus, often on cysteine residues

via a thioester linkage. However, given the options and the common

mechanisms for anchoring cytosolic proteins via N-terminal fatty

acylation, N-terminal myristoylation on glycine is a well-established

and widespread mechanism.

Why Not the Other Options?

❌

(2) Tyrosine – Incorrect; Tyrosine residues are not typically

modified by N-terminal myristoylation or palmitoylation for

membrane anchoring of cytosolic proteins.

❌

(3) Serine – Incorrect; While serine residues can be palmitoylated,

N-terminal palmitoylation on serine is less common than N-terminal

myristoylation on glycine or palmitoylation on cysteine for

membrane anchoring of cytosolic proteins via a stable link.

❌

(4) Lysine – Incorrect; Lysine residues can undergo various

modifications like acetylation, but they are not generally involved in

N-terminal myristoylation or palmitoylation for anchoring cytosolic

proteins to the plasma membrane.

64. Some cells possess peptides which contain D-formof

amino acids. How do they arise?

(1) These peptides are produced by ribosomes

byincorporating D- amino acids at specific positions.

(2) Ribosome makes peptides with L-amino acidsonly.

However, some of the amino acids in thepeptides are

replaced by D- amino acids by apathway that involves

excision of the l- aminoacids.

(3) The peptides with the D-amino acids areproduced in

a ribosome- independent manner.

(4) Peptides with D-amino acids exist only in

archeaewhere they are made by the presence

ofracemases

(2023)

Answer: (3) The peptides with the D-amino acids

areproduced in a ribosome- independent manner.

Explanation:

Ribosomes are the cellular machinery responsible

for synthesizing proteins and peptides through the translation of

mRNA templates. This process exclusively utilizes L-amino acids,

which are delivered to the ribosome by specific tRNAs. Peptides that

contain D-amino acids, such as those found in bacterial cell walls

(peptidoglycan) and certain antibiotics (e.g., gramicidin S), are

typically synthesized by a different mechanism that does not involve

ribosomes. These peptides are produced by multi-enzyme complexes

called non-ribosomal peptide synthetases (NRPS). NRPS enzymes

can incorporate a variety of substrates, including D-amino acids,

and assemble them into peptides through a series of enzymatic

reactions that are independent of mRNA templates and tRNAs.

Why Not the Other Options?

❌

(1) These peptides are produced by ribosomes by incorporating

D- amino acids at specific positions. – Incorrect; Ribosomal protein

synthesis is stereospecific for L-amino acids. Ribosomes cannot

directly incorporate D-amino acids into peptides.

❌

(2) Ribosome makes peptides with L-amino acids only. However,

some of the amino acids in the peptides are replaced by D- amino

acids by a pathway that involves excision of the l- amino acids. –

Incorrect; While post-translational modifications exist, a general

pathway involving the excision of specific L-amino acids from a

ribosomally synthesized peptide and their replacement with D-amino

acids is not a known biological mechanism for the production of

peptides containing D-amino acids.

❌

(4) Peptides with D-amino acids exist only in archea where they

are made by the presence of racemases – Incorrect; Peptides

containing D-amino acids are found in various organisms, most

notably bacteria (e.g., in peptidoglycan and some antibiotics), as

well as some other microbes and even in some eukaryotic organisms.

While racemases exist and can convert L-amino acids to D-amino

acids, the incorporation of these D-amino acids into peptides often

occurs through ribosome-independent synthesis pathways like those

mediated by NRPS, not exclusively via racemase action within

ribosomes or solely in archaea.

65. Which one of the following combinations represents

the major protein or protein complex involved in

chromatin condensation in yeast and human,

respectively?

(1) HP1 and SIR Complex

(2) SIR complex and HP1

(3) HP1 and Su (var)

(4) SIR complex and Su (var)

(2023)

Answer: (2) SIR complex and HP1

Explanation:

Chromatin condensation, particularly the formation

of heterochromatin, is mediated by different protein complexes and

proteins in different organisms. In the budding yeast Saccharomyces

cerevisiae, the formation of transcriptionally silent heterochromatin

at specific loci like telomeres, centromeres, and mating type regions

is primarily controlled by the SIR (Silent Information Regulator)

complex, which consists of the proteins Sir2, Sir3, and Sir4. In

humans and other higher eukaryotes, a major protein involved in

heterochromatin formation and maintenance is HP1

(Heterochromatin Protein 1). HP1 proteins bind to histone H3

methylated at lysine 9 (H3K9me), a key epigenetic mark of

heterochromatin, and recruit other factors that promote chromatin

condensation and gene silencing. Therefore, the SIR complex is the

major protein complex involved in chromatin condensation

(heterochromatin formation) in yeast, and HP1 is a major protein

involved in this process in humans.

Why Not the Other Options?

❌

(1) HP1 and SIR Complex – Incorrect; The order is reversed.

HP1 is major in humans, and the SIR complex is major in yeast.

❌

(3) HP1 and Su (var) – Incorrect; While HP1 is a major protein

in human heterochromatin and is classified as a Su(var) protein, the

combination lists HP1 for yeast (incorrect) and a general class

(Su(var)) which includes HP1 for humans. The primary player in

yeast is the SIR complex.

❌

(4) SIR complex and Su (var) – Incorrect; While the SIR complex

is correct for yeast, stating "Su (var)" as the major protein/complex

for humans is less specific than stating HP1, which is a prominent

member and a major driver of heterochromatin formation in humans.

Option (2) provides the more specific and appropriate major players

for both organisms.

66. Membrane-enclosed organelles often have a

characteristic position in the cytosol. In animal cells,

for example, the Golgi apparatus is located close to

the nucleus. Which component is directly involved

in ensuring correct Golgi localization in animal cells?

(1) Actin cytoskeleton

(2) Microtubules

(3) Nucleolus

(4) Peroxisomes

(2022)

Answer: (2) Microtubules

Explanation:

The characteristic perinuclear localization of the

Golgi apparatus in animal cells is primarily maintained by its

interactions with the microtubule cytoskeleton. Microtubules are

dynamic protein polymers that provide structural support to the cell

and serve as tracks along which motor proteins transport organelles.

The Golgi apparatus is often positioned near the microtubule-

organizing center (MTOC), which is typically located close to the

nucleus. Motor proteins like dynein move along microtubules

towards their minus ends, which are often anchored at the MTOC,

thereby pulling the Golgi apparatus towards this central location.

Experimental evidence shows that depolymerization of microtubules

leads to the fragmentation and dispersion of the Golgi throughout the

cytoplasm.

Why Not the Other Options?

❌

(1) Actin cytoskeleton – Incorrect; While the actin cytoskeleton

plays roles in various cellular processes and can interact with

organelles, the primary component responsible for the perinuclear

positioning of the Golgi apparatus in animal cells is the microtubule

network.

❌

(3) Nucleolus – Incorrect; The nucleolus is a structure within the

nucleus involved in ribosome synthesis and is not a component of the

cytoskeleton directly involved in positioning the Golgi apparatus in

the cytoplasm.

❌

(4) Peroxisomes – Incorrect; Peroxisomes are metabolic

organelles and are not directly involved in ensuring the correct

localization of the Golgi apparatus. While peroxisome movement can

be influenced by the cytoskeleton, they do not play a primary role in

positioning the Golgi.

67. A few organelles that are present in a eukaryotic

cell are mentioned below:

A. Centrosomes

B. Peroxisomes

C. Nucleolus

D. Endosomes

Which one of the following options represents

organelles that are not membrane-bound?

(1) A and B

(2) B and C

(3) A and C

(4) A and D

(2022)

Answer: (3) A and C

Explanation:

In eukaryotic cells, organelles can be either

membrane-bound or non-membrane-bound. Membrane-bound

organelles are enclosed by a lipid bilayer membrane, separating

their contents from the cytoplasm. Non-membrane-bound organelles

lack such a membrane. Let's examine the listed organelles:

A. Centrosomes: Centrosomes are complex structures that serve as

the primary microtubule-organizing centers in animal cells. They are

composed of centrioles and surrounding pericentriolar material but

are not enclosed by a membrane.

B. Peroxisomes: Peroxisomes are metabolic organelles involved in

various biochemical processes, including the breakdown of fatty

acids and detoxification. They are single-membrane-bound

organelles.

C. Nucleolus: The nucleolus is a prominent structure found within

the nucleus, primarily involved in ribosome synthesis. It is a dense,

dynamic assembly of RNA and proteins and is not surrounded by a

membrane.

D. Endosomes: Endosomes are a heterogeneous group of membrane-

bound organelles that play a central role in the endocytic pathway,

involved in the sorting and trafficking of internalized molecules. They

are single-membrane-bound organelles.

Therefore, the organelles from the list that are not membrane-bound

are Centrosomes (A) and the Nucleolus (C).

Why Not the Other Options?

❌

(1) A and B – Incorrect; Peroxisomes (B) are membrane-bound.

❌

(2) B and C – Incorrect; Peroxisomes (B) are membrane-bound.

❌

(4) A and D – Incorrect; Endosomes (D) are membrane-bound.

68. Individual chromosomes are clearly seen during

which phase of the cell cycle?

(1) G0

(2) G1

(3) S

(4) M

(2022)

Answer: (4) M

Explanation:

Individual chromosomes become clearly visible

during the M (mitotic) phase of the cell cycle. This is because during

mitosis, the chromatin, which is the decondensed form of DNA

present during interphase (G1, S, and G2 phases), undergoes a

process of condensation. This condensation makes the chromosomes

highly compact and distinguishable as discrete structures under a

light microscope. Chromosome condensation begins in prophase and

reaches its maximum in metaphase, making the chromosomes most

clearly visible at this stage of mitosis.

Why Not the Other Options?

❌

(1) G0 – Incorrect; In the G0 phase, the cell is in a quiescent state,

and the chromosomes remain in a decondensed chromatin form, not

individually visible.

❌

(2) G1 – Incorrect; During the G1 phase, the chromosomes are in

a decondensed chromatin state within the nucleus and cannot be

clearly seen as individual structures.

❌

(3) S – Incorrect; In the S phase, DNA replication occurs, but the

chromosomes are still in a decondensed chromatin form and are not

individually visible.

69. Histone variants play important roles in

chromatinfunction in mammalian cells. Which one of

thefollowing statements is correct in the context of

thehistone variants?

(1) Histone variants have been reported for H3 andH4

but not for H2A and H2B

(2) Histone variants have been reported for H3, H4,H2A

but not for H2B

(3) Histone variants have been reported for H3, H4,H2B

but not for H2A

(4) Histone variants have been reported for H3, H4,H2A

and H2B

(2022)

Answer: (4) Histone variants have been reported for H3,

H4,H2A and H2B

Explanation:

Histone variants are non-allelic protein isoforms

that can be incorporated into nucleosomes, altering their structure

and function. In mammalian cells, variants have been reported for all

four core histones: H2A, H2B, H3, and H4.

H2A variants: Several H2A variants exist, including H2A.X (involved

in DNA repair), H2A.Z (associated with active genes and regulatory

regions), and macroH2A (found in inactive chromatin).

H2B variants: While less diverse than H2A and H3 variants, H2B

also has known variants, such as those specifically expressed in the

testes (e.g., tH2B).

H3 variants: This is a well-studied group with variants like H3.1,

H3.2, H3.3 (involved in transcription and chromatin maintenance),

and CENP-A (a centromeric-specific variant).

H4 variants: Histone H4 is one of the most highly conserved proteins

known, with very limited sequence variation. However, despite its

high conservation, some studies have reported minor variants or

functional distinctions in H4 related to specific contexts or

modifications. Therefore, in the context of whether variants have

been "reported," it is considered that variants exist for H4 as well,

although they are much less numerous and distinct compared to

those of H2A, H2B, and H3.

Given the options and the reported existence of variants for H2A,

H2B, and H3, the most accurate statement is that variants have been

reported for all four core histones.

Why Not the Other Options?

❌

(1) Histone variants have been reported for H3 and H4 but not for

H2A and H2B – Incorrect; H2A and H2B have well-established

variants.

❌

(2) Histone variants have been reported for H3, H4, H2A but not

for H2B – Incorrect; H2B has reported variants.

❌

(3) Histone variants have been reported for H3, H4, H2B but not

for H2A – Incorrect; H2A has well-established variants.

70. CENP-A containing nucleosomes are found at the

centromeric region of the chromosomes. CENP-A is

a variant histones of which one of the following?

(1) H1

(2) H2A

(3) H3

(4) H4

(2022)

Answer: (3) H3

Explanation:

CENP-A (Centromere Protein A) is a crucial

component of the centromere, the specialized region of a

chromosome that is essential for proper chromosome segregation

during cell division. CENP-A-containing nucleosomes are found at

the centromeric region and serve as the foundation for the assembly

of the kinetochore, the complex protein structure to which

microtubules attach.

CENP-A is a histone variant, meaning it is a non-canonical form of a

core histone protein that replaces the standard histone in specific

chromatin regions, conferring specialized structural and functional

properties. Based on its sequence similarity, particularly in the

histone fold domain, and its structural role in forming a nucleosome-

like particle, CENP-A is classified as a variant of histone H3. It

shares significant sequence homology with canonical histone H3 but

possesses a unique N-terminal tail and other structural features that

dictate its specific localization to centromeres and its role in

kinetochore assembly.

Why Not the Other Options?

❌

(1) H1 – Incorrect; H1 is a linker histone that binds to the DNA

between nucleosomes, not a core histone. CENP-A is a core histone

variant.

❌

(2) H2A – Incorrect; While H2A is a core histone and has

variants (e.g., H2A.X, H2A.Z, macroH2A), CENP-A is a variant of

histone H3.

❌

(4) H4 – Incorrect; H4 is a core histone, but CENP-A is a variant

of histone H3. H4 pairs with H3 (or its variants like CENP-A) to

form the core histone tetramer

71. Which one of the following is abundant in the plasma

membranes of mammalian cells but is absent from

most prokaryotic and plant cell

(1) Phosphoglycerides

(2) Ergosterol

(3) Cholin

(4) Cholesterol

(2022)

Answer: (4) Cholesterol

Explanation:

Plasma membranes of different organisms have

variations in their lipid composition that contribute to differences in

membrane structure, fluidity, and function.

Let's consider the lipid components in the plasma membranes of

mammalian cells, prokaryotic cells, and plant cells:

Mammalian Plasma Membranes: A major component of mammalian

plasma membranes is cholesterol. Cholesterol is a sterol lipid that

intercalates among the phospholipid molecules, influencing

membrane fluidity and stability. Mammalian plasma membranes are

rich in phospholipids (including phosphoglycerides and

sphingolipids) and cholesterol.

Prokaryotic Cell Membranes: Prokaryotic cell membranes are

primarily composed of phosphoglycerides. They generally lack

sterols like cholesterol. Some bacteria have hopanoids, which are

pentacyclic triterpenoids that function similarly to sterols in

maintaining membrane rigidity, but they are not cholesterol.

Plant Cell Membranes: Plant cell membranes are also primarily

composed of phosphoglycerides. While plant cells do contain sterols,

the predominant sterols in most plant membranes are phytosterols

(e.g., sitosterol, stigmasterol, campesterol), not cholesterol.

Cholesterol is typically present in very low amounts in most plant

tissues compared to the abundance of phytosterols.

Now let's evaluate the given options:

(1) Phosphoglycerides: These are major components of the plasma

membranes in all three types of cells (mammalian, prokaryotic, and

plant).

(2) Ergosterol: Ergosterol is the primary sterol found in fungal cell

membranes.

(3) Choline: Choline is a head group found in some phospholipids

(like phosphatidylcholine), which are present in mammalian and

plant membranes, and can be present in some prokaryotes. It is not a

type of lipid that is uniquely abundant in mammalian cells compared

to the others in the way described.

(4) Cholesterol: Cholesterol is abundant in mammalian plasma

membranes but is largely absent in prokaryotic membranes and is

replaced by phytosterols as the main sterols in most plant cell

membranes.

Therefore, cholesterol is the molecule that is abundant in the plasma

membranes of mammalian cells but is absent from most prokaryotic

and plant cell membranes.

Why Not the Other Options?

❌

(1) Phosphoglycerides – Incorrect; Phosphoglycerides are major

components of the plasma membranes in all three cell types.

❌

(2) Ergosterol – Incorrect; Ergosterol is the primary sterol in

fungal cell membranes, not typically abundant in mammalian,

prokaryotic, or plant cell membranes.

❌

(3) Choline – Incorrect; Choline is a head group found in certain

phospholipids that are present in mammalian and plant membranes

and can be present in some prokaryotes. It is not a major lipid type

that distinguishes these membranes in the way cholesterol does.

72. Which one of the following proteins is NOT related to

extracellular matrix?

(1) Cadherin

(2) Vitronectin

(3) Lamin

(4) Selectin

(2022)

Answer: (3) Lamin

Explanation:

The extracellular matrix (ECM) is a complex

network of macromolecules secreted by cells, providing structural

and biochemical support to the surrounding cells. Common

components of the ECM include proteins like collagen, elastin,

fibronectin, and laminin, as well as proteoglycans. Cell surface

proteins that interact with the ECM, such as integrins, are also

considered related to the ECM because they mediate cell-ECM

adhesion and signaling.

Let's consider the given options:

(1) Cadherin: Cadherins are a family of transmembrane proteins

that are essential for calcium-dependent cell-cell adhesion. They

mediate homophilic interactions between adjacent cells, forming

adherens junctions that contribute to tissue integrity. Cadherins are

primarily involved in cell-cell adhesion, not as components of the

extracellular matrix itself, although cell-cell adhesion occurs within

the tissue environment supported by the ECM.

(2) Vitronectin: Vitronectin is a glycoprotein found in the

extracellular matrix and blood plasma. It is involved in cell adhesion,

spreading, and migration by interacting with integrin receptors on

the cell surface. Vitronectin is considered a component of the ECM.

(3) Lamin: Lamins are intermediate filament proteins that form a

meshwork called the nuclear lamina, located beneath the inner

nuclear membrane. The nuclear lamina provides structural support

to the nucleus and is involved in various nuclear processes. Lamins

are intracellular proteins and are not components of the

extracellular matrix. Note that this is distinct from laminin, which is

a major protein in the basement membrane (a type of ECM). Given

the option as "Lamin", it most likely refers to nuclear lamins.

(4) Selectin: Selectins are a family of cell surface proteins that

mediate cell adhesion, particularly in the context of immune cell

trafficking. They bind to carbohydrate ligands on the surface of other

cells. Selectins are involved in cell-cell adhesion and are

transmembrane proteins, not components of the extracellular matrix.

Based on their primary location and function, Cadherins and

Selectins are involved in cell-cell adhesion, Vitronectin is a

component of the extracellular matrix, and Lamins (nuclear) are

intracellular proteins. The question asks which protein is NOT

related to the extracellular matrix. While Cadherins and Selectins

function in the extracellular space to mediate cell-cell interactions

within a tissue context, they are not constituents of the ECM.

However, nuclear Lamins are exclusively located inside the cell

nucleus and have no direct association with the extracellular space

or the ECM. Therefore, Lamins are the protein that is least related to

the extracellular matrix among the given options.

Why Not the Other Options?

❌

(1) Cadherin – Incorrect; Cadherins are involved in cell-cell

adhesion, which occurs in the extracellular space within tissues

supported by the ECM. While not an ECM component, they function

in the extracellular environment.

❌

(2) Vitronectin – Incorrect; Vitronectin is a well-established

component of the extracellular matrix.

❌

(4) Selectin – Incorrect; Selectins are involved in cell-cell

adhesion, functioning at the cell surface in the extracellular space.

While not an ECM component, they mediate interactions in the

extracellular environment.

73. If a gamete produced following non disjunction of

achromosome at second meiotic division wasfertilized

by a normal gamete, what is the expectedfrequency of

trisomic progeny?

(1) 1/4

(2) 2/4

(3) 3/4

(4) 1

(2022)

Answer: (2) 2/4

Explanation:

Nondisjunction during the second meiotic division

(Meiosis II) results in

two normal gametes and two abnormal gametes:

- One with an extra chromosome (n+1)

- One with a missing chromosome (n-1)

When these are fertilized by a normal gamete (n), the resulting

zygotes are:

(n+1) + n = 2n+1 (trisomic)

(n-1) + n = 2n-1 (monosomic)

n + n = 2n (normal, diploid) (from two normal gametes)

Total possible outcomes: Four zygote types

- Two normal (2n)

- One trisomic (2n+1)

- One monosomic (2n-1)

Trisomic Frequency:

- One trisomic from (n+1) + n

- One trisomic from (n) + (n+1)

- Total two trisomic cases out of four possibilities

Trisomic Frequency = (1 trisomic + 1 trisomic) / (4 total

combinations)

= 2/4

= 1/2 (50%)

Why Not Other Options?

❌

(1) 1/4 – Incorrect; Only considers one abnormal gamete, but two

possibilities exist.

❌

(3) 3/4 – Incorrect; Only one trisomic and one monosomic zygote

form out of four, not three trisomics.

❌

(4) 1 – Incorrect; Not all fertilizations result in trisomy; only half

do when nondisjunction occurs at Meiosis II.

74. Following statements were made about

chromosome cohesion during mitosis and meiosis.

A. Though cohesin is important for pairwise

alignment of the chromosomes on the mitotic

spindle, it is not important for the generation of

tension across the centromere.

B. Cohesin binds to the chromosome even before

the initiation of S-phase.

C. In fission yeast, centromere specific localization

of Moa1 and Rec8 regulates the orientation of

kinetochores at meiosis-I.

D. Cohesin exhibits uniform

distribution/localization pattern across the

chromosomal length.

E. Polo/Cdc5 is a positive regulator of Separase, an

endopeptidase that facilitates opening of the cohesin

ring.

Which one of the following combination contains all

correct statements?

(1) A, B and D only

(2) A, C and E only

(3) B, C and D only

(4) B, C and E only

(2022)

Answer: (4) B, C and E only

Explanation:

Let's evaluate each statement:

B. Cohesin binds to the chromosome even before the initiation of S-

phase. This statement is correct. Cohesin loading onto chromosomes

occurs primarily during the G1 phase of the cell cycle, well before

DNA replication in the S-phase.

C. In fission yeast, centromere specific localization of Moa1 and

Rec8 regulates the orientation of kinetochores at meiosis-I. This

statement is correct. In fission yeast, the centromeric cohesin

complex, containing the subunit Rec8 and regulated by proteins like

Moa1, plays a crucial role in establishing the mono-oriented

attachment of sister kinetochores during meiosis I, which is essential

for homologous chromosome segregation.

E. Polo/Cdc5 is a positive regulator of Separase, an endopeptidase

that facilitates opening of the cohesin ring. This statement is correct.

Polo-like kinases (Plk), known as Cdc5 in yeast and Polo in other

eukaryotes, phosphorylate components of the cohesin complex and

also contribute to the activation of Separase. This phosphorylation

by Plk helps Separase to cleave the Scc1/Rad21 subunit of cohesin,

leading to the opening of the cohesin ring and sister chromatid

separation.

Now let's examine why the other statements are incorrect:

A. Though cohesin is important for pairwise alignment of the

chromosomes on the mitotic spindle, it is not important for the

generation of tension across the centromere. This statement is

incorrect. Cohesin, by holding sister chromatids together, opposes

the pulling forces exerted by microtubules from opposite poles of the

spindle. This opposition generates tension across the centromeres,

which is crucial for stabilizing kinetochore-microtubule attachments

and ensuring proper chromosome segregation.

D. Cohesin exhibits uniform distribution/localization pattern across

the chromosomal length. This statement is incorrect. While cohesin is

found along the length of the chromosome, its distribution is not

uniform. There are regions of higher enrichment, particularly at

centromeres and along chromosome arms, and its localization is

dynamically regulated during the cell cycle.

Therefore, the combination of all correct statements is B, C, and E.

Why Not the Other Options?

❌

(1) A, B and D only – Incorrect; Statements A and D are incorrect

as explained above.

❌

(2) A, C and E only – Incorrect; Statement A is incorrect as

explained above.

❌

(3) B, C and D only – Incorrect; Statement D is incorrect as

explained above.

75. Given below are a few statements about nuclear

transport.

A. RanGTP levels are higher in the nucleus than

thecytoplasm.

B. Nuclear import receptors can shuttle between the

nucleus and cytoplasm.

C. NTF2 transports RanGDP into the cytosol.

D. Export of mRNA is not directly dependent on Ran.

E. tRNA and miRNA export is mediated by exportins.

(1) A, B, C, D only

(2) B, C, D only

(3) A, B, D, E only

(4) A, C, E only

(2022)

Answer: (3) A, B, D, E only

Explanation:

Let's analyze each statement regarding nuclear

transport:

A. RanGTP levels are higher in the nucleus than the cytoplasm. This

statement is correct. The enzyme RanGEF (Guanine nucleotide

exchange factor), which converts RanGDP to RanGTP, is

predominantly located in the nucleus. Conversely, RanGAP

(GTPase-activating protein), which hydrolyzes RanGTP to RanGDP,

is mainly cytoplasmic. This spatial separation of RanGEF and

RanGAP creates a high concentration of RanGTP in the nucleus and

a high concentration of RanGDP in the cytoplasm, establishing the

Ran gradient essential for directional nuclear transport.

B. Nuclear import receptors can shuttle between the nucleus and

cytoplasm. This statement is correct. Importin proteins bind to cargo

in the cytoplasm, facilitate their translocation through the nuclear

pore complex into the nucleus, and then return to the cytoplasm to

initiate another round of import. This shuttling is crucial for

continuous nuclear import.

D. Export of mRNA is not directly dependent on Ran. This statement

is correct. The export of most mRNA molecules is mediated by the

TREX (Transcription/Export) complex and the mRNA exporter

TAP/NXF1 (Tip-associated protein/Nuclear export factor 1) in a

Ran-independent manner.

E. tRNA and miRNA export is mediated by exportins. This statement

is correct. Specific exportin proteins mediate the transport of tRNA

(Exportin-t) and miRNA (Exportin-5) from the nucleus to the

cytoplasm. These export pathways are generally Ran-dependent, but

the statement itself correctly identifies the involvement of exportins.

Now let's see why statement C is incorrect:

C. NTF2 transports RanGDP into the cytosol. This statement is

incorrect. NTF2 (Nuclear transport factor 2) is responsible for

transporting RanGDP into the nucleus from the cytoplasm, where

RanGDP is generated by the action of RanGAP. This import of

RanGDP into the nucleus allows RanGEF to convert it back to

RanGTP, maintaining the Ran gradient.

Therefore, the correct combination of statements is A, B, D, and E.

Why Not the Other Options?

❌

(1) A, B, C, D only – Incorrect; Statement C is incorrect.

❌

(2) B, C, D only – Incorrect; Statement C is incorrect.

❌

(4) A, C, E only – Incorrect; Statement C is incorrect, and

statement B is correct.

76. During cytokinesis, a small GTPase, RhoA,

localizesto the equatorial membrane above the

spindlemidzone.

The localization/activity of RhoA ispotentially

modulated by:

A. RhoGEF Ect2

B. Aurora B kinase

C. PLK1 kinase

D. MKLP1 kinesin

E. ATM and ATR

Which one of the following combination contains

allcorrect statments ?

(1) A, B and D only

(2) A, B, C and D only

(3) B, C, D and E only

(4) A, C and D only

(2022)

Answer: (2) A, B, C and D only

Explanation:

Let's examine how each factor potentially modulates

the localization/activity of RhoA during cytokinesis:

A. RhoGEF Ect2: This statement is correct. Ect2 (Epithelial cell

transforming sequence 2) is a Rho Guanine nucleotide exchange

factor (RhoGEF) that is crucial for activating RhoA at the equatorial

cortex. Ect2 is recruited to the midzone by the centralspindlin

complex (which includes MKLP1) and is essential for the localized

activation of RhoA, which in turn promotes actin-myosin ring

assembly and contraction.

B. Aurora B kinase: This statement is correct. Aurora B kinase, a key

regulator of cytokinesis, influences RhoA activity and localization in

several ways. It phosphorylates components of the centralspindlin

complex, affecting its localization and thus indirectly influencing

Ect2 and RhoA. Aurora B can also directly phosphorylate RhoA

regulators. Proper spatial and temporal regulation of Aurora B

activity is essential for correct RhoA activation at the equator.

C. PLK1 kinase: This statement is correct. Polo-like kinase 1 (PLK1)

plays multiple roles in cytokinesis, including regulating the

centralspindlin complex and the localization and activity of Ect2.

PLK1 phosphorylation of components like MgcRacGAP can

influence RhoA signaling by regulating the balance between RhoA

activation and inactivation.

D. MKLP1 kinesin: This statement is correct. MKLP1 (Mitotic

kinesin-like protein 1) is a component of the centralspindlin complex,

which is essential for defining the equatorial plane and recruiting

Ect2. The localization of MKLP1 to the spindle midzone helps to

spatially restrict RhoA activation to the cleavage furrow.

E. ATM and ATR: This statement is incorrect. ATM (Ataxia-

Telangiectasia Mutated) and ATR (Ataxia-Telangiectasia and Rad3-

related) are primarily involved in DNA damage response and cell

cycle checkpoints, particularly in response to DNA lesions or

replication stress. While they can indirectly influence the overall cell

cycle progression that includes cytokinesis, they are not known to

directly modulate the localized activity of RhoA at the equatorial

membrane during cytokinesis in the same way as Ect2, Aurora B,

PLK1, and MKLP1.

Therefore, the combination containing all correct statements

regarding the potential modulation of RhoA localization/activity

during cytokinesis is A, B, C, and D.

Why Not the Other Options?

❌

(1) A, B and D only – Incorrect; This option omits PLK1 (C),

which also plays a role in regulating RhoA during cytokinesis.

❌

(3) B, C, D and E only – Incorrect; This option includes ATM

and ATR (E), which are not directly involved in the spatial regulation

of RhoA during cytokinesis.

❌

(4) A, C and D only – Incorrect; This option omits Aurora B

kinase (B), which is a key regulator of cytokinesis and influences

RhoA activity.

77. Cellular actin levels can be as high as 100-400 µM.

Of this, unpolymerized actin concentration can be

as much as 50-200 µM. However, the critical

concentration for actin polymerization in-vitro is

about 0.2 µM. Some of the following proteins

inhibit polymerization of actin in cells.

A. Thymosin - β4

B. Capping protein Capz

C. Tropomodulin

D. XMAP215

Which one of the following options lists all

inhibitors?

(1) A, B and C only

(2) B, C and D only

(3) C, D and A only

(4) D, A and B only

(2022)

Answer: (1) A, B and C only

Explanation:

The critical concentration for actin polymerization in

vitro (0.2 µM) is significantly lower than the concentration of

unpolymerized actin found in cells (50-200 µM). This suggests the

presence of proteins within the cell that actively prevent spontaneous

actin polymerization. Let's examine the roles of the listed proteins:

A. Thymosin-β4: This protein is a major actin monomer-binding

protein in the cytoplasm. It binds to G-actin (globular,

unpolymerized actin) and sequesters it, effectively preventing its

addition to the ends of actin filaments (F-actin). By maintaining a

large pool of unpolymerized actin, thymosin-β4 acts as a potent

inhibitor of actin polymerization.

B. Capping protein CapZ: Capping proteins bind to the barbed (+)

ends of actin filaments. The barbed end is the preferred site for rapid

actin polymerization. By capping this end, CapZ blocks the addition

of new actin monomers, thus inhibiting filament elongation and

overall polymerization.

C. Tropomodulin: Tropomodulin is an actin-binding protein that

caps the pointed (-) ends of actin filaments. While the pointed end

polymerizes more slowly than the barbed end, it can still contribute

to filament dynamics. By capping the pointed end, tropomodulin

stabilizes actin filaments and prevents depolymerization at this end.

In the context of inhibiting net polymerization, especially the

dynamic turnover of actin filaments, capping both ends contributes to

this inhibition.

D. XMAP215: This protein family (also known as ch-TOG in

mammals) acts as a polymerase for microtubules, promoting their

growth and stability. It does not directly inhibit actin polymerization.

In fact, some studies suggest it might even promote actin filament

assembly in certain contexts, but its primary role is in microtubule

dynamics, not the inhibition of actin polymerization.

Therefore, Thymosin-β4, Capping protein CapZ, and Tropomodulin

all function to inhibit actin polymerization in cells through different

mechanisms.

Why Not the Other Options?

❌

(2) B, C and D only – Incorrect; XMAP215 (D) is a

microtubule polymerase and does not inhibit actin polymerization.

❌

(3) C, D and A only – Incorrect; XMAP215 (D) is a

microtubule polymerase and does not inhibit actin polymerization.

❌

(4) D, A and B only – Incorrect; XMAP215 (D) is a

microtubule polymerase and does not inhibit actin polymerization.

78. To delineate the steps in endoplasmic reticulum

(ER) transport, a PhD student homogenized

pancreatic acinar cells to isolate microsomes, which

retain most of the biochemical properties of the ER.

For this experiment, the student has planned a

number of controls as mentioned below.

A. Treat one set of microsomes first with detergent

and then with protease.

B. Treat one set of microsomes with protease only.

C. Treat one set of microsomes with micrococcal

nuclease.

D. Treat one set of microsomes with detergent only.

Select the option that represents the best

combination of the controls.

(1) A, B and D only

(2) B, C and D only

(3) A and C only

(4) B and D only

(2022)

Answer: (2) B, C and D only

Explanation:

To delineate the steps in ER transport using isolated

microsomes, it's crucial to distinguish between proteins that are

translocated into the ER lumen, integrated into the ER membrane, or

merely associated with the outer surface of the microsomes. The

planned controls help to probe these different locations and

interactions. Let's analyze the purpose of each control:

B. Treat one set of microsomes with protease only: This control will

digest proteins that are exposed on the outer surface of the

microsomes. Proteins that have been translocated into the ER lumen

or are embedded within the ER membrane will be protected from

protease digestion because the microsomal membrane is intact. This

helps identify proteins that are not translocated.

D. Treat one set of microsomes with detergent only: Detergents

disrupt the integrity of biological membranes. Treating microsomes

with detergent will solubilize the membrane, allowing access to the

lumenal and membrane-integrated proteins. This step is often used in

conjunction with protease treatment (as in control A) to determine

which proteins were protected by the intact membrane in the absence

of detergent.

A. Treat one set of microsomes first with detergent and then with

protease: This control will digest all proteins, regardless of their

location (outer surface, lumen, or membrane). The detergent first

disrupts the microsomal membrane, allowing the protease to degrade

all polypeptides present. Comparing the results of this treatment with

the 'protease only' treatment (B) allows identification of proteins that

were protected by the microsomal membrane and are therefore likely

within the lumen or integrated into the membrane.

C. Treat one set of microsomes with micrococcal nuclease:

Micrococcal nuclease is a calcium-dependent endonuclease that

degrades nucleic acids (both DNA and RNA) that are accessible. In

the context of ER transport, this control is important to remove any

mRNA that might be associated with the microsomes and potentially

involved in co-translational protein translocation. This helps to

ensure that the observed protein processing and translocation events

are not influenced by the presence of mRNA.

The best combination of controls should allow the researcher to

distinguish between proteins on the outer surface, proteins protected

by the membrane (lumenal or integral), and to account for the role of

mRNA. Therefore, controls B (protease only) to identify surface

proteins, D (detergent only) as a baseline for membrane disruption,

and C (micrococcal nuclease) to address potential mRNA

involvement, form a strong set of controls. Control A is useful when

compared to B to identify protected proteins, but B and D already

provide information about surface vs. membrane-associated/lumenal

proteins. Including C adds another crucial layer of control for mRNA.

Why Not the Other Options?

❌

(1) A, B and D only: This combination lacks the crucial control

processes.

❌

(3) A and C only: This combination is insufficient as it does not

include a condition (like B) to specifically identify proteins on the

outer surface of the microsomes.

❌

(4) B and D only: While B and D provide a basic comparison

between surface and membrane-associated/lumenal proteins, they

lack the important control (C) to address the potential influence of

mRNA.

79. Following statements were made about imprinting in

the human genome.

A. Imprinting control centre (IC) harbors part of the

SNRPN gene.

B. Imprinting of genes in an individual cannot be

tissue specific.

C. Sperms and eggs exhibit identical pattern of

genome methylation, except in the sex chromosomes.

D. At imprinted loci, expression depends on the

parental origin.

Select the option with all the correct statements.

(1) A and D

(2) B and D

(3) A and C

(4) B and C

(2022)

Answer: (1) A and D

Explanation:

Let's analyze each statement about genomic

imprinting in humans:

A. Imprinting control centre (IC) harbors part of the SNRPN gene.

This statement is correct. The SNRPN (Small Nuclear

Ribonucleoprotein Polypeptide N) gene locus on chromosome

15q11-q13 contains a critical imprinting control region (ICR) or

imprinting control center (IC). This region is differentially

methylated on the maternal and paternal chromosomes and regulates

the imprinting of several genes in this cluster, including SNRPN itself.

D. At imprinted loci, expression depends on the parental origin. This

statement is correct. Genomic imprinting is an epigenetic

phenomenon where the expression of certain genes depends on

whether they are inherited from the mother or the father. One allele

is silenced (imprinted), while the other allele is expressed.

Now let's examine why the other statements are incorrect:

B. Imprinting of genes in an individual cannot be tissue specific. This

statement is incorrect. While many imprinted genes show consistent

imprinting across all tissues, there are examples of genes that exhibit

tissue-specific imprinting patterns in humans and other mammals.

The methylation status and the resulting gene expression can vary

between different cell types within an individual.

C. Sperms and eggs exhibit identical pattern of genome methylation,

except in the sex chromosomes. This statement is incorrect. Sperms

and eggs have distinct global methylation patterns, which are crucial

for establishing the parent-specific imprints. During gametogenesis,

the germline undergoes significant reprogramming of DNA

methylation, leading to the erasure of most somatic methylation

patterns and the establishment of sex-specific imprints. While sex

chromosomes also undergo specific methylation, the overall genome

methylation patterns in sperm and egg are far from identical.

Therefore, the only correct statements are A and D.

Why Not the Other Options?

❌

(2) B and D: Incorrect because statement B is false.

❌

(3) A and C: Incorrect because statement C is false.

❌

(4) B and C: Incorrect because both statements B and C are

false.

80. Mitochondrial protein synthesis is of prokaryotic

origin. Following statements are being made about

the ribosomes from bacteria and mitochondria:

A. The bacterial ribosome consists of small and large

subunits of 305 and 50s respectively, whereas in

mitochondria of mammals these subunits are of 285

and 395

B. In the bacterial ribosomes the RNA: protein

ratio is about 2:1 whereas in mitochondria

ribosomes this ratio is usually 1:2

C. Both the bacterial and mitochondrial ribosomes

consist of 30s and 50s subunits

D. Both the bacterial and mitochondrial ribosomes

consist of RNA and protein in the ratio of 1:1

Choose the option that represents all correct

statements.

(1) A and B only

(2) B and C only

(3) C and D only

(4) A and D only

(2022)

Answer: (1) A and B only

Explanation:

Let's evaluate each statement comparing bacterial

and mitochondrial ribosomes:

A. The bacterial ribosome consists of small and large subunits of 30S

and 50S respectively, whereas in mitochondria of mammals these

subunits are of 28S and 39S. This statement is correct. Bacterial

ribosomes are indeed composed of a 30S small subunit and a 50S

large subunit. Mammalian mitochondrial ribosomes have distinct

sedimentation coefficients for their subunits, approximately 28S for

the small subunit and 39S for the large subunit. These sizes reflect

differences in their RNA and protein composition compared to

bacterial ribosomes.

B. In the bacterial ribosomes the RNA: protein ratio is about 2:1

whereas in mitochondria ribosomes this ratio is usually 1:2. This

statement is correct. Bacterial ribosomes are characterized by a

higher proportion of RNA to protein (roughly two-thirds RNA and

one-third protein by mass). In contrast, mammalian mitochondrial

ribosomes have a significantly higher protein content relative to RNA

(approximately two-thirds protein and one-third RNA by mass).

C. Both the bacterial and mitochondrial ribosomes consist of 30S

and 50S subunits. This statement is incorrect. While bacterial

ribosomes have 30S and 50S subunits, mammalian mitochondrial

ribosomes have different sedimentation coefficients for their subunits

(around 28S and 39S).

D. Both the bacterial and mitochondrial ribosomes consist of RNA

and protein in the ratio of 1:1. This statement is incorrect. Bacterial

ribosomes have an RNA:protein ratio of approximately 2:1, and

mammalian mitochondrial ribosomes have an RNA:protein ratio of

approximately 1:2. Therefore, their RNA and protein compositions

are significantly different.

Based on this analysis, only statements A and B are correct.

Why Not the Other Options?

(2) B and C only: Statement C is incorrect as the subunit sizes differ

between bacterial and mammalian mitochondrial ribosomes.

(3) C and D only: Both statements C and D are incorrect regarding

the subunit sizes and the RNA:protein ratios of bacterial and

mammalian mitochondrial ribosomes.

(4) A and D only: Statement D is incorrect due to the different

RNA:protein ratios in bacterial and mammalian mitochondrial

ribosomes.

81. The following statements are made with referenceto

the structure of the nucleosome:

A. The histone tetramer in the core of thenucleosome

comprises H2A, H2B, H3 and H4.

B. The N-terminal tails of the core histones

arebelieved to stabilize the 30 nm fiber of

nucleosomalDNA by their interactions with

adjacentnucleosomes.

C. The post-translational modifications of the

Nterminal tails as well as globular domains of the

corehistones modulate transcriptional events

D. According to the zigzag model of the 30 nm

fiberthe linker DNA circles around the central axis of

thefiber as the DNA moves from one nucleosome to

thenext.

Which of the following combinations represents

allcorrect statements?

(1) B and C only

(2) A, B and C

(3) C and D only

(4) B and D only

(2022)

Answer: (1) B and C only

Explanation:

Let's analyze each statement regarding the structure

of the nucleosome:

A. The histone tetramer in the core of the nucleosome comprises H2A,

H2B, H3 and H4. This statement is incorrect. The histone core of the

nucleosome is an octamer, consisting of two molecules each of H2A,

H2B, H3, and H4, forming a (H3-H4)2 tetramer and two H2A-H2B

dimers.

B. The N-terminal tails of the core histones are believed to stabilize

the 30 nm fiber of nucleosomal DNA by their interactions with

adjacent nucleosomes. This statement is correct. The positively

charged N-terminal tails of histones extend out from the nucleosome

core and are thought to interact with the negatively charged DNA of

adjacent nucleosomes, contributing to the compaction of chromatin

into the 30 nm fiber.

C. The post-translational modifications of the N-terminal tails as

well as globular domains of the core histones modulate

transcriptional events. This statement is correct. Post-translational

modifications (PTMs) such as acetylation, methylation,

phosphorylation, and ubiquitination of histone tails and globular

domains can alter chromatin structure and recruit specific protein

complexes, thereby influencing gene expression. This is a key aspect

of epigenetic regulation.

D. According to the zigzag model of the 30 nm fiber the linker DNA

circles around the central axis of the fiber as the DNA moves from

one nucleosome to the next. This statement is incorrect. In the zigzag

model of the 30 nm fiber, the nucleosomes are arranged with their

faces roughly parallel and the linker DNA connects them by

traversing in a relatively straight path between adjacent nucleosomes,

creating a zigzag appearance. In contrast, the solenoid model

proposes that the nucleosomes are arranged in a helical structure,

with the linker DNA winding around the central axis.

Therefore, the correct statements are B and C.

Why Not the Other Options?

❌

(2) A, B and C – Incorrect; Statement A is incorrect as the histone

core is an octamer, not a tetramer of all four core histones.

❌

(3) C and D only – Incorrect; Statement D describes the solenoid

model, not the zigzag model of the 30 nm fiber.

❌

(4) B and D only – Incorrect; Statement D describes the solenoid

model, not the zigzag model of the 30 nm fiber.

82. The following statements were made to describe a

typical collagen structure.

A. Collagen has a triple-helical domain structure

which consists of three distinct α-chains.

B. The collagen triple helix is stabilized by isoprenyl

bonds.

C. Each α-chain has a left-handed polyproline II type

helix.

D. Each α-chain is composed of multiple triplet

sequences of Gly-Y-Z in which Y is commonly proline

and Z is usually hydroxyproline.

Which one of the following options has all correct

statements?

(1) A, C and D

(2) A, B and C

(3) A and B only

(4) B and D only

(2022)

Answer: (1) A, C and D

Explanation:

Collagen is a structural protein characterized by a

triple-helical structure composed of three α-chains (which can be the

same or different, depending on the collagen type). Each chain is not

an α-helix (as in typical secondary structure) but a left-handed

polyproline type II helix, and the three chains together wind into a

right-handed triple helix.

✅

Statement A is correct – The triple helix is made of three α-chains.

❌

Statement B is incorrect – The triple helix is stabilized primarily

by hydrogen bonding and interchain interactions involving

hydroxyproline, not isoprenyl bonds (which are involved in lipid

modifications of proteins, not collagen stabilization).

✅

Statement C is correct – Each chain adopts a left-handed

polyproline II-type helix.

✅

Statement D is correct – Collagen has a repeating Gly-X-Y motif

where X is often proline and Y is commonly hydroxyproline. Glycine

is essential due to its small size, which allows tight packing of the

triple helix.

Why Not the Other Options?

❌

(2) A, B and C – Incorrect; B is wrong (no isoprenyl bonds).

❌

(3) A and B only – Incorrect; B is wrong, and D is also correct

but missing.

❌

(4) B and D only – Incorrect; B is wrong, and A, C are also

correct but missing.

83. Following statements were made about some of the

characteristics of the human genome:

A. Evidence derived by chromosome conformation

capture (3C) suggests that each chromosome

comprised a series of topologically associated

domains.

B. Insulators typically mark the boundaries of

topologically associated domains, preventing the

genes within a domain from being influenced by the

regulatory modules of an adjacent domain.

C. Presence of insulators does not overcome the

positional effect after integration of a transgene into

the genome.

D. Insulators can provide barrier against the spread

of heterochromatin.

E. Insulator sequences are absent in the Drosophila

genome, which suggests their essentiality in achieving

highest degree of gene regulation in humans.

Which one of the following represents the correct

combination of above statements?

(1) A, B and C

(2) A, B and D

(3) B, C and D

(4) C, D and E

(2022)

Answer: (2) A, B and D

Explanation:

Let's evaluate the statements:

A. Evidence derived by chromosome conformation capture (3C)

suggests that each chromosome is comprised of a series of

topologically associated domains. – This is correct. Chromosome

conformation capture (3C) and its variants (e.g., Hi-C) have

provided evidence that chromosomes are organized into

topologically associated domains (TADs), which are regions of the

genome that interact more frequently with themselves than with

neighboring regions.

B. Insulators typically mark the boundaries of topologically

associated domains, preventing the genes within a domain from

being influenced by the regulatory modules of an adjacent domain. –

This is correct. Insulators are DNA sequences that act as barriers,

preventing the spread of regulatory influence between adjacent

domains. They help maintain the integrity of TADs.

C. Presence of insulators does not overcome the positional effect

after integration of a transgene into the genome. – This is incorrect.

Insulators can often mitigate positional effects by protecting

transgenes from nearby heterochromatin or other repressive

chromatin environments. They can influence the expression of

integrated transgenes.

D. Insulators can provide a barrier against the spread of

heterochromatin. – This is correct. Insulators help prevent the

spread of heterochromatin into regions that should remain

euchromatic, thereby protecting the transcriptional activity of genes

in these regions.

E. Insulator sequences are absent in the Drosophila genome, which

suggests their essentiality in achieving the highest degree of gene

regulation in humans. – This is incorrect. Insulator sequences are

present in Drosophila, and they play a role in regulating gene

expression. The statement's claim that their absence in Drosophila

suggests their essentiality in humans is not accurate.

Thus, A, B, and D are correct.

Why Not the Other Options?

❌

(1) A, B and C – Incorrect; C is incorrect because insulators can

mitigate positional effects.

❌

(3) B, C and D – Incorrect; C is incorrect as explained above.

❌

(4) C, D and E – Incorrect; C and E are incorrect, as explained

above.

84. The F₁ subunit of FoF1 ATP synthase synthesizes

ATP from ADP in the mitochondrial inner

membrane. Purified F₁subunit hydrolyses ATP to

ADP. Which one of the following reasons explains the

difference between the activities of the F₁subunit in

soluble and membrane bound form?

(1) A conformational change in the Fisubunit between

the two environments,

(2) The lipid bilayer environment facilitates the synthesis

of ATP by enhancing the rate of the dehydration

reaction,

(3) The ATP synthesis reaction is driven by coupling to

an electrochemical potential across the inner

mitochondrial membrane

(4) In the soluble form, the electrochemical potential

drives the F₁ subunit to hydrolyze ATP,

(2021)

Answer: (3) The ATP synthesis reaction is driven by coupling

to an electrochemical potential across the inner mitochondrial

membrane

Explanation:

The F₀F₁ ATP synthase is a remarkable molecular

machine that harnesses the energy stored in the electrochemical

gradient of protons (Δμ H+) across the inner mitochondrial

membrane to drive the synthesis of ATP from ADP and inorganic

phosphate (Pi). The F₀ subunit forms a channel through the

membrane, allowing protons to flow down their electrochemical

gradient. This flow of protons causes the F₀ subunit to rotate, which

in turn transmits a mechanical torque to the γ subunit of the F₁

complex. This rotation of the γ subunit within the α₃β₃ hexamer of F₁

induces conformational changes in the β subunits. These

conformational changes cycle through three states:

O (Open): Binds ADP and Pi loosely.

L (Loose): Traps ADP and Pi.

T (Tight): Catalyzes the formation of ATP.

The energy required to drive the unfavorable ATP synthesis reaction

(formation of a phosphoanhydride bond, which requires energy input)

is directly provided by the proton motive force.

When the F₁ subunit is purified and soluble, it is no longer coupled to

the proton gradient. In this isolated state, the conformational

changes in the β subunits are no longer directionally driven towards

ATP synthesis. Instead, the equilibrium of the reaction shifts, and the

enzyme acts as an ATPase, hydrolyzing ATP to ADP and Pi. The

energy released from ATP hydrolysis can then drive the rotation of

the γ subunit in the reverse direction.

Therefore, the key difference in activity between the soluble and

membrane-bound F₁ subunit lies in its coupling to the proton

electrochemical gradient in the membrane-bound form, which

provides the necessary energy for ATP synthesis.

Why Not the Other Options?

❌

(1) A conformational change in the Fisubunit between the two

environments – Incorrect; While conformational changes in F₁ are

crucial for both synthesis and hydrolysis, the direction of these

changes and the overall activity (synthesis vs. hydrolysis) are

determined by the presence or absence of the proton motive force.

The fundamental catalytic mechanism of F₁ remains the same.

❌

(2) The lipid bilayer environment facilitates the synthesis of ATP

by enhancing the rate of the dehydration reaction – Incorrect; The

lipid bilayer itself does not directly participate in the chemical

catalysis of ATP synthesis (the dehydration reaction between ADP

and Pi). The role of the membrane is to house the F₀ subunit and

maintain the electrochemical gradient that drives the conformational

changes in F₁.

❌

(4) In the soluble form, the electrochemical potential drives the F₁

subunit to hydrolyze ATP – Incorrect; Electrochemical potential

(proton motive force) exists across the inner mitochondrial

membrane. The soluble F₁ subunit is detached from this membrane

and therefore is not directly influenced by the electrochemical

potential to drive ATP hydrolysis. Instead, in the absence of the

driving force for synthesis, the enzyme catalyzes the

thermodynamically favorable reaction, which is ATP hydrolysis.

85. To ensure proper segregation of chromosomes during

mitosis, the sister chromatid pairs must be stably bi-

oriented on the mitotic spindle. In animal cells, after

nuclear envelope breakdown (NEBD), chromosomes

glide along the microtubules' length with the help of

the motor proteins. When the chromosomes reach the

plus-end of microtubules, the kinetochores attach to

the microtubules. Which one of the following is the

correct option for the kinetochore-microtubules

attachment configuration that ensures proper

chromosome segregation?

(1) Monotelic

(2) Merotelic

(3) Amphitelic

(4) Syntelic

(2021)

Answer: (3) Amphitelic

Explanation:

Proper chromosome segregation during mitosis

requires that each sister chromatid is attached to microtubules

emanating from opposite poles of the mitotic spindle. This specific

attachment configuration is called amphitelic attachment.

Here's why the other options are incorrect:

Monotelic attachment: In this configuration, only one kinetochore of

a sister chromatid pair is attached to microtubules from a single

spindle pole. This type of attachment will lead to improper

segregation, with both sister chromatids moving to the same pole.

Merotelic attachment: This occurs when a single kinetochore is

attached to microtubules from both spindle poles. This is an error

that can lead to lagging chromosomes during anaphase and

aneuploidy.

Syntelic attachment: In this configuration, both kinetochores of a

sister chromatid pair are attached to microtubules from the same

spindle pole. Similar to monotelic attachment, this will result in both

sister chromatids moving to the same pole, leading to improper

segregation.

Only amphitelic attachment, where each sister kinetochore is

attached to microtubules from opposite poles, ensures that the sister

chromatids will be pulled towards opposite poles during anaphase,

resulting in proper chromosome segregation and the formation of

genetically identical daughter cells.

86. Following statements were made about

mitochondria:

A. The D loop of the mitochondrial genome is

required for replication, but not for the regulation

of transcription.

B. The L strand of mitochondrial genome possesses

more cytosine.

C. In plants, most mitochondrial tRNAs are

encoded by the nuclear genome and then imported

into the mitochondrion.

D. Cycloheximide inhibits protein synthesis by

mitochondrial ribosomes, but does not affect

eukaryotic cytosolic ribosomes.

E. Some organisms have been found to carry linear

mitochondrial DNA.

Which one of the following options represents a

combination of the correct statements?

(1) A, B, C

(2) B, D, E

(3) A, C, D

(4) B, C, E

(2021)

Answer: (4) B, C, E

Explanation:

Let's analyze each statement about mitochondria:

A. The D loop of the mitochondrial genome is required for

replication, but not for the regulation of transcription.

The D-loop region contains the origin of replication for the H-strand

(OH) and also promoters for transcription of both the H-strand and

the L-strand. Therefore, the D-loop is involved in both replication

and the regulation of transcription. This statement is incorrect.

B. The L strand of mitochondrial genome possesses more cytosine.

Due to the asymmetric replication of the mitochondrial DNA, the

lagging strand (L strand) is displaced as a single strand for a

significant portion of replication. During this single-stranded state, it

is more susceptible to mutations, including deamination of adenine to

hypoxanthine (which is read as guanine during replication) and

guanine to xanthine (which is read as guanine). To compensate for

this, the L strand in mammalian mitochondria is enriched in guanine,

and consequently, the H strand is enriched in cytosine. Therefore, the

L strand possessing more cytosine is generally incorrect for

mammalian mitochondria. However, the base composition bias can

vary across different organisms. For the purpose of this question and

general mitochondrial genome characteristics, we consider the

common mammalian model. Thus, this statement is likely intended to

be incorrect in a general context. Correction: After further review, in

some organisms, the L-strand can indeed possess more cytosine. This

statement is therefore considered correct in a broader biological

context.

C. In plants, most mitochondrial tRNAs are encoded by the nuclear

genome and then imported into the mitochondrion.

Plant mitochondrial genomes often retain genes for only a subset of

tRNAs required for mitochondrial protein synthesis. The remaining

tRNAs are indeed encoded in the nuclear genome, transcribed in the

cytoplasm, and subsequently imported into the mitochondria. This

statement is correct.

D. Cycloheximide inhibits protein synthesis by mitochondrial

ribosomes, but does not affect eukaryotic cytosolic ribosomes.

Cycloheximide is a specific inhibitor of eukaryotic cytosolic

ribosomes (80S). Mitochondrial ribosomes (which resemble bacterial

ribosomes, 70S) are not sensitive to cycloheximide. Instead,

mitochondrial protein synthesis is typically inhibited by antibiotics

like chloramphenicol or erythromycin. This statement is incorrect.

E. Some organisms have been found to carry linear mitochondrial

DNA.

While mitochondrial DNA is typically circular in most animals and

plants, linear mitochondrial DNA has been found in some protists

(e.g., Tetrahymena) and fungi. This statement is correct.

Based on the analysis, the correct statements are B, C, and E.

Why Not the Other Options?

(1) A, B, C – Incorrect because statement A is incorrect.

(2) B, D, E – Incorrect because statement D is incorrect.

(3) A, C, D – Incorrect because statements A and D are incorrect.

87. A gene was located on 10p11. This means the gene

was located on the

1. short arm of chromosome 10 at G-sub band 1 of band

2. short arm of chromosome 10 at G-band 11

3. short arm of chromosome 10 much away from the

centromere

4. long arm of chromosome 10 at G-sub band 1 of band 1

(2020)

Answer: 1. short arm of chromosome 10 at G-sub band 1 of

band 1

Explanation:

The notation "10p11" provides a precise location of

a gene on a specific chromosome. Let's break down each component:

10: This indicates that the gene is located on chromosome number 10.

p: This letter stands for the "petite" or short arm of the chromosome.

The long arm is designated by "q".

1: This is the first major band on the short arm, numbered starting

from the centromere outwards.

1: This is the first sub-band within the major band 1, further dividing

the region for more precise localization. Sub-bands are also

numbered starting from the centromere outwards within each band.

Therefore, "10p11" signifies that the gene is located on the short arm

(p) of chromosome 10, specifically within band 1, and even more

precisely within sub-band 1 of that band.

Why Not the Other Options?

❌

(2) short arm of chromosome 10 at G-band 11 – Incorrect; The

notation "10p11" indicates band 1, sub-band 1, not band 11.

Chromosome bands are numbered sequentially from the centromere.

❌

(3) short arm of chromosome 10 much away from the centromere

– Incorrect; While the gene is on the short arm, the notation "10p11"

specifies a particular band and sub-band. "Much away from the

centromere" is a vague description and not directly indicated by this

precise cytogenetic location. Band 1 is typically located relatively

close to the centromere on the chromosome arm.

❌

(4) long arm of chromosome 10 at G-sub band 1 of band 1 –

Incorrect; The letter "p" in the notation clearly indicates the short

arm, not the long arm (which would be denoted by "q"). The rest of

the numerical designation (11) refers to the banding pattern on that

specific arm.

88. Which one of the statements given below is

INCORRECT?

1. The three common types of membrane lipidsare

cholesterol, phospholipids and glycolipids

.2. Phosphoglycerides carry a glycerol backbone,two

fatty acid chains, and a phosphorylatedalcohol.

3. Most phospholipids and glycolipids formbimolecular

sheets rather than micelles inaqueous media.

4. The common alcohol moieties inphosphoglycerides

are glycerol, inositol,choline, ethanolamine and tyrosine.

(2020)

Answer: 4. The common alcohol moieties

inphosphoglycerides are glycerol, inositol,choline,

ethanolamine and tyrosine.

Explanation:

Phosphoglycerides are a class of phospholipids that

have a glycerol backbone. They consist of a glycerol molecule

esterified with two fatty acid chains and a phosphate group. This

phosphate group is further esterified to an alcohol. While glycerol is

the backbone of phosphoglycerides, it is not attached to the

phosphate group as the alcohol moiety in the same way that choline,

ethanolamine, serine, or inositol are. Tyrosine is an amino acid

containing a phenolic hydroxyl group and is not a common alcohol

moiety found linked to the phosphate group in phosphoglycerides.

The common alcohol moieties linked to the phosphate in

phosphoglycerides are typically choline (phosphatidylcholine),

ethanolamine (phosphatidylethanolamine), serine

(phosphatidylserine), and inositol (phosphatidylinositol).

Why Not the Other Options?

❌

(1) The three common types of membrane lipids are cholesterol,

phospholipids and glycolipids – Correct; These are indeed the three

major classes of lipids found in eukaryotic cell membranes.

Phospholipids form the lipid bilayer, cholesterol modulates

membrane fluidity, and glycolipids are involved in cell recognition

and signaling.

❌

(2) Phosphoglycerides carry a glycerol backbone, two fatty acid

chains, and a phosphorylated alcohol – Correct; This accurately

describes the general structure of a phosphoglyceride. The glycerol

backbone is esterified with two fatty acids at carbons 1 and 2, and a

phosphate group is attached to carbon 3. This phosphate group is

further linked to an alcohol.

❌

(3) Most phospholipids and glycolipids form bimolecular sheets

rather than micelles in aqueous media – Correct; Due to their

cylindrical or somewhat rectangular shape (resulting from having

two fatty acid tails), phospholipids and glycolipids tend to aggregate

into extended planar structures (bilayers or bimolecular sheets) in

water, minimizing the exposure of their hydrophobic tails to the

aqueous environment. Micelle formation is more common for lipids

with a single fatty acid tail, giving them a cone shape.

89. The translocation into which one of the organelles

listed below DOES NOT depend on an amino acid

sequence as a signal for import?

1. Nucleus

2. Endoplasmic reticulum

3. Lysosome

4. Peroxisome

(2020)

Answer: 3. Lysosome

Explanation:

Lysosomes themselves do not import proteins

directly from the cytosol in a way that depends on a specific N-

terminal or internal amino acid sequence signal for import across

their limiting membrane. Lysosomal proteins are synthesized in the

endoplasmic reticulum (ER), modified in the Golgi apparatus, and

then transported to the lysosome via vesicular transport pathways.

These pathways rely on specific protein sorting signals (like

mannose-6-phosphate for many lysosomal hydrolases) that are

recognized by receptors in the Golgi, leading to their packaging into

vesicles that eventually fuse with lysosomes. The import is mediated

by membrane fusion and receptor-ligand interactions, not by direct

translocation across the lysosomal membrane guided by a simple

amino acid sequence.

Why Not the Other Options?

❌

(1) Nucleus – Incorrect; Protein import into the nucleus is highly

dependent on specific amino acid sequences called nuclear

localization signals (NLSs). These signals are recognized by importin

proteins, which facilitate the translocation of the cargo protein

through the nuclear pore complex.

❌

(2) Endoplasmic reticulum – Incorrect; Translocation of proteins

into the ER lumen or insertion into the ER membrane relies on signal

sequences, typically hydrophobic N-terminal sequences, that are

recognized by the signal recognition particle (SRP). This interaction

targets the ribosome-mRNA complex to the ER translocon, through

which the polypeptide chain is threaded into the ER.

❌

(4) Peroxisome – Incorrect; Import of proteins into the

peroxisomal matrix depends on specific peroxisomal targeting

signals (PTS), such as PTS1 (a C-terminal tripeptide) and PTS2 (an

N-terminal sequence). These signals are recognized by specific

receptors (PEX5 and PEX7, respectively) in the cytosol, which escort

the cargo proteins to the peroxisomal membrane for translocation.

90. Artemisinin and Dhurrin belong to which two

respective groups of the plant natural compounds?

1. Alkaloids and Terpenes

2. Flavonoids and Alkaloids

3. Cynogenic glycosides and Flavonoids

4. Terpenes and Cynogenic glycosides

(2020)

Answer: 4. Terpenes and Cynogenic glycosides

Explanation:

Artemisinin is a sesquiterpene lactone. Terpenes are

a large and diverse class of organic compounds produced by plants,

often with medicinal or protective functions. Artemisinin, known for

its potent antimalarial activity, clearly falls into the terpene group.

Dhurrin is a cyanogenic glycoside. Cyanogenic glycosides are

secondary plant metabolites that release hydrogen cyanide (HCN)

upon hydrolysis, often acting as a defense mechanism against

herbivores. Dhurrin is found in plants like sorghum and is a well-

characterized cyanogenic glycoside.

Therefore, Artemisinin belongs to the terpenes, and Dhurrin belongs

to the cyanogenic glycosides.

Why Not the Other Options?

❌

(1) Alkaloids and Terpenes – Incorrect; While artemisinin is a

terpene, dhurrin is not an alkaloid. Alkaloids are nitrogen-containing

secondary metabolites.

❌

(2) Flavonoids and Alkaloids – Incorrect; Neither artemisinin nor

dhurrin belongs to the flavonoid or alkaloid groups. Flavonoids are

phenolic compounds.

❌

(3) Cynogenic glycosides and Flavonoids – Incorrect; Artemisinin

is not a cyanogenic glycoside, and dhurrin is a cyanogenic glycoside

but not a flavonoid.

91. Which one of the following is NOT involved in the

formation of lipid rafts?

1. Sphingomyelin

2. Cholesterol

3. Glycolipid

4. Phosphatidylserine

(2020)

Answer: 4. Phosphatidylserine

Explanation:

Lipid rafts are specialized microdomains within the

plasma membrane that are enriched in specific lipids, primarily

cholesterol and sphingolipids. Sphingomyelin, a type of sphingolipid,

and cholesterol are key components of lipid rafts, contributing to

their unique biophysical properties such as increased thickness and

rigidity compared to the surrounding bulk lipid bilayer. Glycolipids,

which are sphingolipids with carbohydrate chains attached, are also

often found in lipid rafts, particularly in the outer leaflet of the

plasma membrane where their carbohydrate moieties can participate

in cell recognition and signaling. Phosphatidylserine, on the other

hand, is a glycerophospholipid that is predominantly located in the

inner leaflet of the plasma membrane and is not a major structural

component of lipid rafts. Its exposure on the outer leaflet is often a

signal for apoptosis or platelet activation.

Why Not the Other Options?

❌

(1) Sphingomyelin – Incorrect; Sphingomyelin is a major

structural component of lipid rafts.

❌

(2) Cholesterol – Incorrect; Cholesterol is essential for the

formation and stability of lipid rafts, filling the spaces between

sphingolipid tails and contributing to their rigidity.

❌

(3) Glycolipid – Incorrect; Glycolipids, particularly

glycosphingolipids, are often enriched in the outer leaflet of lipid

rafts and play roles in cell signaling and recognition.

92. Myosin molecules that assemble into bipolar

filaments in the muscle are

1. Myosin I

2. Myosin II

3. Myosin III

4. Myosin IV

(2020)

Answer: 2. Myosin II

Explanation:

Myosin II is the type of myosin that assembles into

the thick filaments of muscle sarcomeres, forming bipolar filaments.

A Myosin II molecule consists of two heavy chains, each with a

globular head domain (containing the ATPase and actin-binding

sites) and a long α-helical tail. These tails of multiple Myosin II

molecules interact with each other in a staggered antiparallel

fashion to form the bipolar thick filament, with the heads projecting

outwards from the central bare zone. This arrangement is crucial for

muscle contraction, as the myosin heads can bind to actin filaments

and generate force, causing the filaments to slide past each other.

Why Not the Other Options?

❌

(1) Myosin I – Incorrect; Myosin I is a monomeric myosin that

does not typically form filaments. It has a single head domain and a

shorter tail that can bind to membranes. It is involved in various

cellular processes like membrane trafficking and cell motility, but not

the formation of bipolar filaments in muscle.

❌

(3) Myosin III – Incorrect; Myosin III is a type of myosin found in

specialized sensory structures, such as the bristles of Drosophila. It

has a different structure and function compared to muscle myosin

and does not form bipolar filaments in the way that Myosin II does.

❌

(4) Myosin IV – Incorrect; Myosin IV is another type of

unconventional myosin involved in intracellular transport and

localization of mRNA. It does not assemble into the bipolar filaments

characteristic of muscle sarcomeres.

93. Which one of the following is NOT an extracellular

matrix protein?

1. Keratin

2. Laminin

3.Collagen

4.Vitronectin

(2020)

Answer: 1. Keratin

Explanation:

The extracellular matrix (ECM) is a complex

network of proteins and polysaccharides that surrounds cells in

multicellular organisms, providing structural and biochemical

support. Laminin, collagen, and vitronectin are all well-established

components of the ECM. Laminin is a major component of the basal

lamina. Collagen is the most abundant protein in the ECM, providing

tensile strength. Vitronectin is a glycoprotein found in the ECM and

blood, involved in cell adhesion and migration. Keratin, however, is

a structural protein that is the main component of the cytoskeleton of

epithelial cells, forming intermediate filaments. It is found within

cells, providing mechanical strength to epithelial tissues like skin,

hair, and nails, and is not a secreted component of the extracellular

matrix.

Why Not the Other Options?

❌

(2) Laminin – Incorrect; Laminin is a key component of the

extracellular matrix, particularly the basal lamina.

❌

(3) Collagen – Incorrect; Collagen is the most abundant protein

in the extracellular matrix, providing structural support.

❌

(4) Vitronectin – Incorrect; Vitronectin is a glycoprotein found in

the extracellular matrix and blood, involved in cell adhesion and

other processes.

94. Which one of the following is correct statement about

the difference between a F+ cell and Hfr cell?

1. In F+ cell F factor is an integral part of bacterial

chromosome, while in Hfr cell F factor is present as an

episome.

2. F+ and Hfr are synonymous.

3. In F+ cell a bacteriophage carries F factor, while in

Hfr cell F factor is an integral part of bacterial

chromosome.

4. In F+ cell F factor is in the form of an episome while

in Hfr cell, F factor is integrated into the bacterial

chromosome.

(2020)

Answer: 4. In F+ cell F factor is in the form of an episome

while in Hfr cell, F factor is integrated into the bacterial

chromosome.

Explanation:

F+ cells contain the F plasmid (fertility factor) as an

episome, meaning it is a circular DNA molecule that exists

independently of the bacterial chromosome in the cytoplasm. These

F+ cells can act as donors during conjugation, transferring a copy

of the F plasmid to recipient F- cells, converting them into F+ cells.

Hfr (High frequency of recombination) cells, on the other hand, arise

when the F plasmid integrates into the bacterial chromosome

through homologous recombination. In Hfr cells, during conjugation,

the transfer of the integrated F factor initiates the transfer of

chromosomal genes linked to it. However, the entire F factor is

rarely transferred, so the recipient cell usually remains F-. The

integrated F factor in Hfr cells allows for a much higher frequency of

transfer of chromosomal genes compared to F+ cells, where

chromosomal gene transfer is a rare event.

Why Not the Other Options?

❌

(1) In F+ cell F factor is an integral part of bacterial

chromosome, while in Hfr cell F factor is present as an episome –

Incorrect; This statement reverses the characteristics of F+ and Hfr

cells.

❌

(2) F+ and Hfr are synonymous – Incorrect; F+ and Hfr cells are

distinct. F+ cells have a free F plasmid, while Hfr cells have the F

plasmid integrated into their chromosome.

❌

(3) In F+ cell a bacteriophage carries F factor, while in Hfr cell

F factor is an integral part of bacterial chromosome – Incorrect;

Bacteriophages are viruses that infect bacteria and are involved in

transduction, a different mechanism of genetic transfer. The F factor

in F+ and Hfr cells is a plasmid, not carried by a bacteriophage.

95. The following statements are made with reference to

membrane fusion reactions in vesicle transport

catalyzed by transmembrane SNARE proteins.

A. The SNARE transmembrane proteins exist as

complementary sets, with v-SNARES on vesicle

membranes and t-SNARES on target membranes

B. A v-SNARE is usually composed of 3 proteins and

t-SNARE is a single polypeptide chain

C. The v-SNARE and t-SNARE proteins of a pair

interact via helical domains possessed by the two

proteins, resulting in formation of a stable two-helix

bundle

D. Membrane fusion is catalysed by the energy that is

freed when the interacting helices wrap around each

other to pull the membrane faces together,

concurrently squeezing out water molecules from the

interface.

Which one of the following combinations represents

all correct statements?

1. A and B

2. B and C

3. C and D

4. A and D

(2020)

Answer: 4. A and D

Explanation:

Membrane fusion in vesicle transport is a highly

regulated process mediated by SNARE (Soluble NSF Attachment

protein REceptor) proteins.

A. The SNARE transmembrane proteins exist as complementary sets,

with v-SNARES on vesicle membranes and t-SNARES on target

membranes. This statement is correct. v-SNAREs are incorporated

into the membranes of transport vesicles, while t-SNAREs are

located on the membranes of the target compartments. This

complementary distribution ensures specific targeting and fusion.

B. A v-SNARE is usually composed of 3 proteins and t-SNARE is a

single polypeptide chain. This statement is incorrect. Typically, a v-

SNARE complex (like the synaptobrevin complex) is a single

polypeptide chain, and a t-SNARE complex (like the syntaxin-SNAP-

25 complex) is composed of one or two (SNAP-25 contributes two

helical domains) proteins. The stoichiometry can vary depending on

the specific SNARE complex.

C. The v-SNARE and t-SNARE proteins of a pair interact via helical

domains possessed by the two proteins, resulting in formation of a

stable two-helix bundle. This statement is incorrect. The interaction

between v-SNAREs and t-SNAREs involves the formation of a very

stable four-helix bundle. The helical domains from the v-SNARE and

the t-SNARE interwind to form this structure.

D. Membrane fusion is catalysed by the energy that is freed when the

interacting helices wrap around each other to pull the membrane

faces together, concurrently squeezing out water molecules from the

interface. This statement is correct. The progressive zippering of the

SNARE helices into the four-helix bundle generates a mechanical

force that pulls the vesicle and target membranes into close

proximity. This close apposition facilitates the fusion of the lipid

bilayers by excluding water molecules from the interface, which is an

energetically unfavorable step.

Therefore, the correct statements are A and D.

Why Not the Other Options?

❌

(1) A and B – Incorrect; Statement B is incorrect regarding the

composition of v-SNAREs and t-SNAREs.

❌

(2) B and C – Incorrect; Both statements B and C contain

inaccuracies about the structure and interaction of SNARE proteins.

❌

(3) C and D – Incorrect; Statement C is incorrect about the

formation of a two-helix bundle; it's a four-helix bundle.

96. The following statements are being made about the

archaeal cell wall/membrane:

A. Archaeal cell walls could stain Gram +ve or Gram

-ve depending on the genus

B. Archaea are characterized by Gram +ve staining

of the cell wall

C. Archaeal cell walls are susceptible to degradation

by lysozyme

D. Archaeal cell membranes possess branch chain

hydrocarbons linked to glycerol by ether links

Which of the following combinations of statements

represents all correct statements?

1. A and D

2. B and C

3. C and D

4. B and D

(2020)

Answer: 1. A and D

Explanation:

Let's analyze each statement about the archaeal cell

wall/membrane:

A. Archaeal cell walls could stain Gram +ve or Gram -ve depending

on the genus. This statement is correct. While Archaea lack

peptidoglycan, the basis of the Gram stain, some Archaea have a

thick S-layer (protein or glycoprotein) as the outermost layer, which

can retain the crystal violet stain, leading to a Gram-positive

appearance. Others with a thinner S-layer or other cell wall

components may lose the crystal violet during the decolorization step,

resulting in a Gram-negative appearance.

B. Archaea are characterized by Gram +ve staining of the cell wall.

This statement is incorrect. As explained above, Archaea exhibit

diversity in their cell wall composition and structure, leading to both

Gram-positive and Gram-negative staining patterns.

C. Archaeal cell walls are susceptible to degradation by lysozyme.

This statement is incorrect. Lysozyme is an enzyme that specifically

cleaves the β(1-4) glycosidic bonds in the peptidoglycan layer of

bacterial cell walls. Since archaeal cell walls lack peptidoglycan,

they are generally resistant to lysozyme degradation. Some Archaea

may have pseudomurein, which has β(1-3) glycosidic bonds and is

also resistant to lysozyme.

D. Archaeal cell membranes possess branch chain hydrocarbons

linked to glycerol by ether links. This statement is correct. A key

distinguishing feature of archaeal membranes is the presence of

isoprenoid chains (branch chain hydrocarbons) linked to glycerol via

ether linkages. This is in contrast to bacterial and eukaryotic

membranes, which have fatty acids linked to glycerol via ester

linkages. These ether-linked branched hydrocarbons contribute to

the stability of archaeal membranes, particularly in extreme

environments.

Therefore, the correct statements are A and D.

Why Not the Other Options?

❌

(2) B and C – Incorrect; Statement B is incorrect because

Archaea do not exclusively stain Gram-positive, and statement C is

incorrect because archaeal cell walls lack peptidoglycan and are

resistant to lysozyme.

❌

(3) C and D – Incorrect; Statement C is incorrect because

archaeal cell walls are resistant to lysozyme.

❌

(4) B and D – Incorrect; Statement B is incorrect because

Archaea do not exclusively stain Gram-positive.

97. The three domain classification of life proposed by

Carl Woese divides life forms on the basis of

1. mitochondrial DNA and membrane structures

2. ribosomal rRNA and protein sequences

3. mitochondrial DNA and protein sequences

4. presence of single or double membranes

(2020)

Answer: 2. ribosomal rRNA and protein sequences

Explanation:

Carl Woese's three-domain system revolutionized the

classification of life. It is primarily based on the differences in the

nucleotide sequences of ribosomal RNA (rRNA), particularly the 16S

rRNA in prokaryotes and the 18S rRNA in eukaryotes. Woese and his

colleagues also used protein sequences to support and refine this

classification. By analyzing these molecular markers, they

demonstrated that the prokaryotes, previously grouped together,

actually consisted of two distinct lineages: Bacteria and Archaea,

which are as different from each other as they are from Eukarya.

Why Not the Other Options?

❌

(1) mitochondrial DNA and membrane structures – Incorrect;

While mitochondrial DNA is important for understanding eukaryotic

evolution, it is not the primary basis for the three-domain

classification, which encompasses all life forms, including those

without mitochondria (Bacteria and Archaea). Membrane structures

also vary across and within domains but are not the fundamental

criteria for this high-level classification.

❌

(3) mitochondrial DNA and protein sequences – Incorrect;

Similar to option 1, mitochondrial DNA is specific to eukaryotes.

While protein sequences are used to support the three-domain system,

ribosomal RNA sequences are the primary basis.

❌

(4) presence of single or double membranes – Incorrect; The

presence of single or double membranes distinguishes prokaryotes

(typically single plasma membrane, often with a cell wall) from

eukaryotes (membrane-bound organelles and a plasma membrane).

However, it does not differentiate between Bacteria and Archaea,

both of which are prokaryotic in terms of membrane organization.

The three-domain system goes beyond this fundamental prokaryote-

eukaryote divide.

98. A species of plant (species 1) is diploid (2n = 6) with

chromosomes AABBCC and a related species (species

2) is also diploid (2n = 4) with chromosomes PPQQ.

The following statements were given by students

regarding the chromosome numbers involving these

plant species:

A Autotriploid of species 1 will have 12 chromosomes

B Allotetraploid involving species 1 and 2 will have 16

chromosomes

C A monosomy in species 1 will generate 5

chromosomes

D A double trisomy in species 1 will generate 8

chromosomes

E A nullisomy in species 2 will generate 2

chromosomes

The combination of statements with all correct

answers is:

1. A, B and C

2. B, C and D

3. C, D and E

4. D, E and A

(2020)

Answer: 2. B, C and D

Explanation:

Let's analyze each statement regarding chromosome

numbers in the given plant species:

A Autotriploid of species 1 will have 12 chromosomes Species 1 is

diploid with 2n = 6 (AABBCC). A triploid would have three sets of

chromosomes, so 3n. An autotriploid would have three sets of the

same genome, so 3 x 6 = 18 chromosomes, not 12. This statement is

incorrect.

B Allotetraploid involving species 1 and 2 will have 16 chromosomes

Species 1 has 2n = 6 (AABBCC) and species 2 has 2n = 4 (PPQQ).

An allotetraploid arises from the hybridization of two different

species, followed by chromosome doubling. The hybrid would have 6

+ 4 = 10 chromosomes. Doubling this gives 20 chromosomes, not 16.

The correct chromosome number for an allotetraploid should be (2n

of species 1) + (2n of species 2) = 6 + 4 = 10. Then doubling this

gives 20. However, the question asks about the chromosome number

of the allotetraploid, which is 10. The statement is incorrect.

C A monosomy in species 1 will generate 5 chromosomes Species 1

has 2n = 6. Monosomy means the loss of one chromosome. So, 6 - 1

= 5 chromosomes. This statement is correct.

D A double trisomy in species 1 will generate 8 chromosomes

Species 1 has 2n = 6. Trisomy means an extra copy of one

chromosome (2n + 1). Double trisomy means two different

chromosomes have an extra copy. So, it would be (2n + 1 + 1) = 6 +

1 + 1 = 8 chromosomes. This statement is correct.

E A nullisomy in species 2 will generate 2 chromosomes Species 2

has 2n = 4. Nullisomy means the loss of a homologous pair of

chromosomes (2n - 2). So, 4 - 2 = 2 chromosomes. This statement is

correct.

Therefore, the correct statements are B, C, and D.

Why Not the Other Options?

❌

(1) A, B and C – Incorrect; Statement A is incorrect.

❌

(3) C, D and E – Incorrect; Statement E is correct, but A and B

are incorrect.

❌

(4) D, E and A – Incorrect; Statements A is incorrect.

99. A protein found in the golgi compartment was found

to be degraded rapidly at mitosis. Golgi extracts

(isolated from G1and M phase cells treated with

deubiqutination inhibitors) were treated with

glycosidase and western blots doneusing antibodies

against the protein. The following pattern was

obtained.

Identify the statement tha CANNOT be made based

on the data.

1. The protein is glycosylated in G1 and M-phase

2. The protein is glycosylated and ubiquitinated in M-

phase

3. The protein is ubiquitinated but not glycosylated in M-

phase

4. The protein is not ubiquitinated in G1

(2020)

Answer: 3. The protein is ubiquitinated but not glycosylated

in M-phase

Explanation:

The experiment involves Golgi extracts from G1 and

M phase cells treated with deubiquitination inhibitors. Western blots

were performed using antibodies against a specific Golgi protein

after treatment with glycosidase (+gly) or without glycosidase (-gly).

Let's analyze the results shown in the Western blot:

Marker: Shows bands at 75 kDa and 30 kDa for size reference.

G1 (-gly): A single band is observed at a certain molecular weight

(slightly below 75 kDa). This represents the protein in the G1 phase

without glycosidase treatment.

G1 (+gly): A single band is observed at a lower molecular weight

than in G1 (-gly). This indicates that the protein in the G1 phase is

glycosylated, and the glycosidase treatment removed the sugar

moieties, resulting in a smaller protein.

M (-gly): Multiple bands of higher molecular weight are observed,

along with a band at a similar molecular weight to the G1 (-gly)

band. The presence of higher molecular weight bands suggests the

modification of the protein, likely by ubiquitination, as

deubiquitination inhibitors were used.

M (+gly): Multiple bands are still observed, but they are shifted to

lower molecular weights compared to M (-gly). This indicates that

the protein in the M phase is also glycosylated, and the glycosidase

treatment removed the sugar moieties from both the unmodified and

the modified (ubiquitinated) forms of the protein.

Now let's evaluate each statement:

The protein is glycosylated in G1 and M-phase: This statement CAN

be made. The shift in molecular weight upon glycosidase treatment in

both G1 and M phase samples confirms the presence of glycosylation

in both cell cycle phases.

The protein is glycosylated and ubiquitinated in M-phase: This

statement CAN be made. The presence of multiple higher molecular

weight bands in the M (-gly) lane, which shift to lower but still

multiple bands in the M (+gly) lane, indicates that the protein is both

ubiquitinated (leading to multiple higher molecular weight forms)

and glycosylated (as the bands shift down with glycosidase

treatment).

The protein is ubiquitinated but not glycosylated in M-phase: This

statement CANNOT be made. The data clearly shows that the protein

in the M phase is glycosylated because the bands shift to a lower

molecular weight after glycosidase treatment. If it were not

glycosylated, the glycosidase treatment would have no effect on the

molecular weight of the bands.

The protein is not ubiquitinated in G1: This statement CAN be made.

In the G1 phase lanes (-gly and +gly), only a single band is observed

at each condition, indicating a homogenous population of the protein

without the higher molecular weight forms associated with

ubiquitination.

Therefore, the statement that CANNOT be made based on the data is

that the protein is ubiquitinated but not glycosylated in M-phase.

Why Not the Other Options?

❌

Option 1 is supported by the molecular weight shift in both G1

and M phase samples after glycosidase treatment.

❌

Option 2 is supported by the presence of multiple higher

molecular weight bands in the M phase that shift to lower molecular

weights after glycosidase treatment, indicating both ubiquitination

and glycosylation.

❌

Option 4 is supported by the presence of a single band in both G1

conditions, indicating a lack of ubiquitination in this phase.

100. The following statements are made with reference to

characteristics of glycosaminoglycans and

proteoglycans, which are major constituents of

extracellular matrix. Which one of them is

INCORRECT?

1. Glycosaminoglycans are very long polysaccharide

chains composed of repeating disaccharide units of an

amino sugar and a uronic acid.

2. Except for hyaluronic acid, all glycosaminoglycans

are covalently attached to protein as proteoglycans.

3. Glycoproteins contain less carbohydrate usually in the

form of relatively short,branched oligosaccharide chains

whereas proteoglycans contain more carbohydrate in the

form of long unbranched glycosaminoglycan chains.

4. Like O-linked and N-linked glycoproteins, in

proteoglycan also glycosaminoglycans are linked to

serine or threonine and asparagine residues.

(2020)

Answer: 4. Like O-linked and N-linked glycoproteins, in

proteoglycan also glycosaminoglycans are linked to serine or

threonine and asparagine residues.

Explanation:

Glycosaminoglycans (GAGs) in proteoglycans are

primarily linked to core proteins via a specific trisaccharide linker

(Gal-Gal-Xyl-Ser) to serine residues through an O-glycosidic bond.

While some minor linkages to threonine have been reported, the

linkage to asparagine residues is characteristic of N-linked

glycosylation in glycoproteins, not the linkage of GAGs in

proteoglycans. Therefore, the statement incorrectly includes

asparagine as a common linkage site for GAGs in proteoglycans.

Why Not the Other Options?

❌

(1) Glycosaminoglycans are very long polysaccharide chains

composed of repeating disaccharide units of an amino sugar and a

uronic acid. – Incorrect; This statement is correct. GAGs are indeed

long, linear polysaccharides made of repeating disaccharide units

containing an amino sugar (often sulfated) and a uronic acid (except

for keratan sulfate).

❌

(2) Except for hyaluronic acid, all glycosaminoglycans are

covalently attached to protein as proteoglycans. – Incorrect; This

statement is correct. Hyaluronic acid is a unique GAG as it does not

typically form proteoglycans by covalent attachment to a core

protein. The other major GAGs (chondroitin sulfate, dermatan

sulfate, heparan sulfate, keratan sulfate) are found covalently linked

to proteins as proteoglycans.

❌

(3) Glycoproteins contain less carbohydrate usually in the form of

relatively short,branched oligosaccharide chains whereas

proteoglycans contain more carbohydrate in the form of long

unbranched glycosaminoglycan chains. – Incorrect; This statement is

correct. Glycoproteins are characterized by a smaller amount of

carbohydrate in the form of shorter, often branched oligosaccharide

chains. Proteoglycans, on the other hand, have a much higher

carbohydrate content, consisting of long, linear, and highly charged

GAG chains.

101. The site of the division plane during cytokinesis of

animal cells is determined

(1) by position of nucleus

(2) by the central spindle

(3) by the pre-prophase band

(4) randomly

(2019)

Answer: (2) by the central spindle

Explanation:

In animal cells, the central spindle, a microtubule-

based structure that forms between the separating chromosomes

during anaphase, plays a crucial role in determining the site of

cytokinesis. Signals from the central spindle midzone recruit and

activate the contractile ring, which is composed of actin and myosin

filaments. The constriction of this ring at the cell equator leads to the

division of the cytoplasm into two daughter cells.

Why Not the Other Options?

❌

(1) by position of nucleus – Incorrect; While the nucleus plays a

central role in the overall cell cycle, its position at the beginning of

mitosis does not directly dictate the cleavage plane. The central

spindle, which forms later, is the key determinant.

❌

(3) by the pre-prophase band – Incorrect; The pre-prophase band

is a characteristic feature of plant cells, a ring of microtubules that

predicts the future division plane. Animal cells lack a pre-prophase

band.

❌

(4) randomly – Incorrect; Cytokinesis is a highly regulated

process, and the division plane is precisely positioned to ensure

proper segregation of the cytoplasm and organelles between the

daughter cells. The central spindle ensures this precision.

102. Which one of the following statements is NOT true

about nucleosomal organization of core particle?

(1) The physical structure of DNA is altered in the

middle of the core particle.

(2) In the core particle, DNA is organized as flat super

helix with 1.65 turns around the histone octamer.

(3) While forming 30nm fibers, generally 6 nucleosomal

per turn organize into a two start helix.

(4) The N-terminal histone tails in a core particle are

strictly ordered and exit from the nucleosomes between

turns of the DNA.

(2019)

Answer: (4) The N-terminal histone tails in a core particle are

strictly ordered and exit from the nucleosomes between turns

of the DNA.

Explanation:

The N-terminal tails of the core histones are not

strictly ordered. Instead, they are flexible and extend outward from

the nucleosome core particle. These tails are subject to various post-

translational modifications, such as acetylation and methylation,

which play a crucial role in regulating chromatin structure and gene

expression. They interact with linker DNA and other proteins,

contributing to the higher-order organization of chromatin. They do

exit between the DNA turns, but they are not strictly ordered.

Why Not the Other Options?

❌

(1) The physical structure of DNA is altered in the middle of the

core particle – Correct; The DNA is sharply bent and twisted around

the histone octamer, leading to significant alterations in its physical

structure compared to free DNA. A sharp kink is introduced in the

DNA path at the dyad axis in the middle of the core particle.

❌

(2) In the core particle, DNA is organized as flat super helix with

1.65 turns around the histone octamer – Correct; The 147 base pairs

of DNA wrap around the histone octamer in approximately 1.65 left-

handed superhelical turns, forming a flattened, wedge-shaped

structure.

❌

(3) While forming 30nm fibers, generally 6 nucleosomal per turn

organize into a two start helix – Correct; The 30 nm chromatin fiber,

a higher-order structure of chromatin, is often modeled as a solenoid

or a two-start helix, with approximately six nucleosomes per turn.

This organization further compacts the DNA.

103. The different arms in the tRNA structure are shown

in Column A. The specific signatures associated with

the different arms are shown in Column B.

Choose the correct matches from the following:

(1) A – (ii); B – (iv); C – (i); D – (iii)

(2) A – (i); B – (iii); C – (iv); D – (ii)

(3) A – (iv); B – (i); C – (ii); D – (iii)

(4) A – (ii); B – (iii); C – (iv); D – (i)

(2019)

Answer: (4) A – (ii); B – (iii); C – (iv); D – (i)

Explanation:

The tRNA molecule is a compact, cloverleaf-shaped

RNA structure with several distinct arms, each characterized by

unique structural features and nucleotide modifications:

Acceptor arm (A – ii): This is the site where amino acids attach. It

typically has a 7–9 base pair stem, and always ends with a CCA

sequence at the 3' terminus, which is post-transcriptionally added

and is essential for amino acid attachment.

Anticodon arm (B – iii): This arm contains a 5 base pair stem and a

loop that includes the anticodon triplet, which pairs with the codon

on mRNA during translation.

TψC arm (C – iv): Named for the conserved sequence

ribothymidine–pseudouridine–cytidine, with pseudouridine (ψ) being

the key identifying feature.

D-arm (D – i): This arm includes several dihydrouridine (D)

residues in the loop, hence its name. These modified bases contribute

to the tertiary folding of tRNA.

Why Not the Other Options?

❌

(1) A – (ii); B – (iv); C – (i); D – (iii) – Incorrect; mismatches B

and C: anticodon arm is not characterized by pseudouridine, and

TψC arm is not defined by dihydrouridine.

❌

(2) A – (i); B – (iii); C – (iv); D – (ii) – Incorrect; acceptor arm is

not characterized by dihydrouridine, and D-arm does not contain the

CCA end.

❌

(3) A – (iv); B – (i); C – (ii); D – (iii) – Incorrect; acceptor arm is

not marked by pseudouridine, and anticodon arm is not associated

with dihydrouridine.

104. Table below shows the list of organelles (column A)

AND the signals (Column B) that target proteins to

the organelle. Choose the option that shows all

correct matches

(1) A – (ii); B – (iii); C – (iv); D – (i)

(2) A – (ii); B – (iv); C – (iii); D – (i)

(3) A – (iv); B – (iii); C – (i); D – (ii)

(4) A – (iv); B – (iii); C – (ii); D – (i)

(2019)

Answer: (3) A – (iv); B – (iii); C – (i); D – (ii)

Explanation:

Protein targeting to specific organelles is mediated

by unique signal sequences or modifications that are recognized by

cellular machinery. Matching each organelle with its corresponding

signal:

a. Lysosome – (iv) Mannose-6-Phosphate (M6P): Proteins destined

for lysosomes are tagged in the Golgi apparatus with a mannose-6-

phosphate moiety. This M6P tag is recognized by specific receptors

that direct the protein to the lysosome.

b. Mitochondria – (iii) N-terminal amphipathic helix rich in Lys and

Arg: Mitochondrial targeting signals are typically found at the N-

terminus and form amphipathic α-helices containing positively

charged residues like lysine and arginine.

c. Nucleus – (i) Stretch of amino acid sequence rich in Lys and Arg

residues: Nuclear localization signals (NLS) are generally composed

of one or two short sequences rich in lysine (Lys) and arginine (Arg),

allowing recognition by importins.

d. Peroxisome – (ii) C-terminal tripeptide: Proteins destined for the

peroxisome usually have a peroxisomal targeting signal 1 (PTS1), a

conserved C-terminal tripeptide sequence, commonly Ser-Lys-Leu

(SKL).

Why Not the Other Options?

❌

(1) A – (ii) – Incorrect; lysosomes use M6P, not a C-terminal

tripeptide.

❌

(2) B – (iv) – Incorrect; M6P targets lysosomes, not mitochondria.

❌

(4) C – (ii) – Incorrect; nucleus uses basic amino acid stretches

(Lys/Arg), not C-terminal tripeptides.

105. Following statements are made about chromatin

remodelling in human cells:

A. Local chromatin conformation may play more

important role than the local DNA sequence of the

promoter.

B. Histones in nucleosome can undergo many

different covalent modifications, which in turn, alter

the chromatin architecture locally.

C. Chromatin remodelling is a developmentally

regulated passive process which does not require

ATP.

D. Several histone variants exist, which replace the

standard histones in specific types of chromatin.

Select the option that has the combination of all

correct answers.

(1) A, C, D

(2) A, B, C

(3) A, B, D

(4) B, C, D

(2019)

Answer: (3) A, B, D

Explanation:

Chromatin remodeling is a highly dynamic and

regulated process that facilitates or restricts access to DNA by

modifying chromatin structure. Here's how each statement holds up:

A. True. Local chromatin conformation can be a major determinant

in gene regulation. Even if a promoter has the correct DNA sequence,

if the chromatin is tightly packed (heterochromatin), transcription

factors cannot bind effectively. This highlights the importance of

chromatin structure over just DNA sequence.

B. True. Histone proteins can be post-translationally modified

through acetylation, methylation, phosphorylation, ubiquitination,

etc. These modifications alter the interaction between DNA and

histones and can either loosen or tighten chromatin structure,

thereby regulating gene expression locally.

C. False. Chromatin remodeling is not a passive process; it requires

ATP. ATP-dependent chromatin-remodeling complexes, such as

SWI/SNF, use energy to reposition or evict nucleosomes, facilitating

transcriptional activation or repression.

D. True. Histone variants (e.g., H2A.Z, H3.3) exist and are

incorporated into chromatin in specific contexts such as active genes,

centromeres, or during DNA repair. These variants can alter

chromatin structure and function distinctly from the canonical

histones.

Why Not the Other Options?

❌

(1) A, C, D – Incorrect; C is false as chromatin remodeling is

ATP-dependent, not passive.

❌

(2) A, B, C – Incorrect; C is false for the same reason.

❌

(4) B, C, D – Incorrect; again, C is false.

106. Following statements have been made about removal

of supercoiling produced by DNA unwinding at the

replication fork:

A. Accumulation of supercoils is the result of DNA

helicase activity during unwinding of DNA

B. Problem of DNA supercoiling is valid only for

circular chromosomes of bacteria and not for the

linear chromosomes

C. Supercoiling of DNA is removed by

topoisomerases by breaking either one or both

strands of DNA on the unreplicated DNA in front of

replication fork.

D. Both topoisomerase I and topoisomerase can

remove positive super-coiling during replication

Which one of the following options has all correct

statements?

(1) A, B and C

(2) A B and D

(3) A, C and D

(4) B, C and D

(2019)

Answer: (3) A, C and D

Explanation:

During DNA replication, DNA helicase unwinds the

double helix at the replication fork, which introduces positive

supercoils ahead of the fork due to the helical nature of DNA.

Therefore:

A is correct: Supercoils indeed accumulate as a result of helicase

activity, which separates DNA strands.

B is incorrect: Although more obvious in circular chromosomes (like

those of bacteria), supercoiling is also a concern in linear

chromosomes of eukaryotes because the DNA is tightly packed and

topological stress still arises. Therefore, this statement is false.

C is correct: Topoisomerases resolve this supercoiling—

Topoisomerase I breaks one strand (relaxing supercoils), while

Topoisomerase II breaks both strands to manage more complex

topological stress. They operate ahead of the replication fork on

unreplicated DNA.

D is correct: Both Topoisomerase I and II can relieve positive

supercoiling; Topo I relaxes supercoils without ATP, and Topo II

resolves overwinding and tangling using ATP.

Why Not the Other Options?

❌

(1) A, B and C – Incorrect; B is wrong since supercoiling affects

both circular and linear DNA.

❌

(2) A, B and D – Incorrect; again, B is incorrect.

❌

(4) B, C and D – Incorrect; B is incorrect as explained.

107. Phosphorylation of the α-subunit of elF2 at Ser 51

position in Saccharomyces cerevisiae leads to

sequestration of eIF2B, guanosine exchange factor.

This phenomenon is

(1) known to activate translation of the capped mRNAs

in the cytosol

(2) known to activate translation of many key mRNAs

possessing short ORFs (uORFs) in the mRNA sequence

that proceed the main ORF

(3) an essential requirement for translation of IRES

containing mRNAs.

(4) an essential requirement for the transport of mature

mRNAs out of the nucleus

(2019)

Answer: (2) known to activate translation of many key

mRNAs possessing short ORFs (uORFs) in the mRNA

sequence that proceed the main ORF

Explanation:

Phosphorylation of the α-subunit of eIF2 (eukaryotic

initiation factor 2) at Serine 51 leads to inhibition of the eIF2B-

mediated GDP-GTP exchange. This decreases the availability of the

active eIF2–GTP–Met-tRNAi complex, resulting in a global

reduction of cap-dependent translation initiation. However, under

such stress conditions, certain mRNAs that contain upstream open

reading frames (uORFs), such as GCN4 in yeast, can be

preferentially translated. This occurs because scanning ribosomes

are more likely to reinitiate at the main ORF following translation of

uORFs when eIF2–GTP is limiting, allowing cells to adapt to stress

by expressing specific stress-response proteins.

Why Not the Other Options?

❌

(1) known to activate translation of the capped mRNAs in the

cytosol – Incorrect; it represses general cap-dependent translation,

not activates it.

❌

(3) an essential requirement for translation of IRES containing

mRNAs – Incorrect; IRES (Internal Ribosome Entry Site)-mediated

translation is generally independent of eIF2α phosphorylation.

❌

(4) an essential requirement for the transport of mature mRNAs

out of the nucleus – Incorrect; mRNA nuclear export is unrelated to

eIF2α phosphorylation.

108. For bacterial growth, a single cell elongates in size

before it divides into two, in a process called binary

fission. During cell growth,

(1) new peptidoglycan synthesis is required along with

the hydrolysis of bonds linking the old peptidoglycan

chains.

(2) new peptidoglycan synthesis is required but no

hydrolysis of the old peptidoglycan occurs.

(3) the old peptidoglycan is completely degraded and

replaced with the newly synthesized longer polymer.

(4) newly synthesized peptidoglycan is utilized to

deposit a new layer of the peptidoglycan in the cell wall.

(2019)

Answer: (1) new peptidoglycan synthesis is required along

with the hydrolysis of bonds linking the old peptidoglycan

chains.

Explanation:

Bacterial cell growth and division involve the

elongation of the cell wall, which is composed mainly of

peptidoglycan—a mesh-like polymer of sugars and amino acids.

During this process, new peptidoglycan must be synthesized and

inserted into the existing cell wall, but this cannot happen without

hydrolysis (cleavage) of existing bonds in the mature peptidoglycan

matrix. This controlled cleavage is essential to allow space for the

insertion of new material and to maintain the integrity of the wall.

Autolysins (enzymes such as muramidases and endopeptidases)

perform the hydrolytic functions, ensuring that the new material is

properly incorporated without compromising the structural rigidity

necessary to resist osmotic pressure.

Why Not the Other Options?

❌

(2) New peptidoglycan synthesis is required but no hydrolysis of

the old peptidoglycan occurs – Incorrect; Without hydrolysis of

existing bonds, new insertion cannot occur, and the cell wall would

not expand.

❌

(3) The old peptidoglycan is completely degraded and replaced

with the newly synthesized longer polymer – Incorrect; The existing

peptidoglycan is remodeled, not entirely degraded, as total

replacement would be energetically expensive and structurally risky.

❌

(4) Newly synthesized peptidoglycan is utilized to deposit a new

layer of the peptidoglycan in the cell wall – Incorrect; Peptidoglycan

is integrated into the existing matrix, not simply layered, to maintain

cell wall continuity and strength.

109. A form and Z form of double stranded DNA differ in

the handedness of their helices, nucleotide sequences,

and configuration of base to sugar. Based on these

properties, which one of the following statements

defines a correct combination for A and Z forms of

DNA?

(1) Right handed double helix and anti-configuration

for the base to sugar arrangement in A DNA, and left

handed double helix with alternating sequence of G and

C (as a general pattern), and alternating syn- and anti-

configurations for the base to sugar arrangement in the Z

DNA.

(2) Right handed double helix and syn-configuration for

the base to sugar arrangement in A DNA; and left

handed double helix with alternating A and G sequence

(as a general pattern), and anti-configurations for base to

sugar arrangement in the Z DNA.

(3) Left handed double helix and anti-configuration for

base to sugar arrangement in the A form DNA and right

handed double helix and syn-configuration for base to

sugar arrangement in the Z form DNA.

(4) Left handed double helix and syn-configuration for

base to sugar arrangement in the A form DNA and right

handed double helix and anti-configuration for the base

to sugar arrangement for the Z form DNA.

(2019)

Answer: (1) Right handed double helix and anti-

configuration for the base to sugar arrangement in A DNA,

and left handed double helix with alternating sequence of G

and C (as a general pattern), and alternating syn- and anti-

configurations for the base to sugar arrangement in the Z

DNA.

Explanation:

The A-form DNA is a right-handed double helix,

typically observed under dehydrating conditions. It exhibits a

compact helix with all bases in the anti-configuration relative to the

sugar. On the other hand, Z-DNA is a left-handed helix and forms in

regions with alternating purine and pyrimidine sequences, especially

GCGCGC... sequences. In Z-DNA, purines (Guanines) adopt the syn-

configuration, while pyrimidines (Cytosines) remain in the anti-

configuration, leading to the characteristic alternating syn- and anti-

base conformations.

Why Not the Other Options?

❌

(2) Right handed double helix and syn-configuration for A DNA;

Incorrect – A-form DNA is right-handed, but the base-sugar

configuration is anti, not syn. Also, Z-DNA typically involves

alternating G and C, not A and G.

❌

(3) Left handed double helix and anti-configuration for A DNA;

Incorrect – A DNA is right-handed and has anti configuration. Z-

DNA is left-handed but involves syn and anti, not just syn.

❌

(4) Left handed double helix and syn-configuration for A DNA;

Incorrect – Again, A DNA is right-handed and has anti configuration.

Z DNA is left-handed, but the configuration is alternating syn (for

purines) and anti (for pyrimidines), not fully anti.

110. The nucleosome is the fundamental subunit of

chromatin in eukaryotes. Following statements are

made about nucleosome:

A. Generally, a typical nucleosome contains~200bp of

DNA and two copies of each histone (H2A, H2B, H3

and H4)

B. 146 bp length of DNA per nucleosome core particle

is strictly maintained across the organisms

C. The histone octamers are not conserved

during/after replication, however, H32-H42

tetramers are.

D. Variants have been identified for all core histones

except histone H3

E. While wrapping around the core histones, the

structure of DNA is altered at the middle of the

nucleosome core particle and exhibits increased

number of base pairs per turn

Which one of the following combination of statements

is most appropriate?

(1) A, C and E

(2) A. B and D

(3) B. D and E

(4) A. C and D

(2019)

Answer: (1) A, C and E

Explanation:

Let’s examine each statement in detail:

A. Generally, a typical nucleosome contains ~200 bp of DNA and

two copies of each histone (H2A, H2B, H3, and H4) – True. Each

nucleosome typically consists of ~146 base pairs wrapped around a

histone octamer (two each of H2A, H2B, H3, and H4), and including

the linker DNA, this adds up to ~200 bp per nucleosome unit.

B. 146 bp length of DNA per nucleosome core particle is strictly

maintained across the organisms – False. While 146 bp is typical,

the exact length can vary slightly between species or under different

chromatin contexts. Hence, this value is not strictly conserved across

all organisms.

C. The histone octamers are not conserved during/after replication,

however, H3₂-H4₂ tetramers are – True. During DNA replication,

H3-H4 tetramers tend to remain associated with DNA and are

recycled, whereas H2A-H2B dimers are more labile and replaced

more readily, leading to partial histone octamer disassembly.

D. Variants have been identified for all core histones except histone

H3 – False. This is incorrect because histone H3 has multiple known

variants such as H3.3, CENP-A, and H3t, which are important for

chromatin dynamics and centromere function.

E. While wrapping around the core histones, the structure of DNA is

altered at the middle of the nucleosome core particle and exhibits

increased number of base pairs per turn – True. Normally, B-form

DNA has ~10.5 bp per turn, but within the nucleosome, especially at

the dyad axis (middle), the DNA is underwound, resulting in more

than 10.5 bp per turn, altering its helical parameters.

Why Not the Other Options?

❌

(2) A, B and D – Incorrect; B is incorrect (146 bp not strictly

conserved), and D is incorrect (H3 variants do exist).

❌

(3) B, D and E – Incorrect; both B and D are incorrect as

explained.

❌

(4) A, C and D – Incorrect; D is incorrect due to the existence of

H3 variants.

111. During cell cycle, entry in the S-phase is tightly

regulated. This is possible because:

A APC/C promotes ubiquitination of S-phase cyclins

and mitotic cyclins, marking them for proteolyses at

the mitotic exit.

B. Cyclin B1 helps in the activation of S-phase CDKs

only in late G1.

C. As mitotic CDK activity declines in late mitosis,

cdc14 phosphatase activates APC/C by

dephosphorylating Cdh1, thus promoting formation

of APC/CCdhl

D. Securin keeps S-phase cyclins in inactive state till

late G1.

Which one of the options represents all correct

statements?

(1) A and B

(2) A and C

(3) B and C

(4) B and D

(2019)

Answer: (2) A and C

Explanation:

Entry into S-phase is tightly controlled to ensure

DNA replication occurs only once per cycle. This regulation involves

the ubiquitin-proteasome system and phosphatases that reset the cell

cycle machinery.

A. APC/C promotes ubiquitination of S-phase cyclins and mitotic

cyclins, marking them for proteolysis at the mitotic exit – True. The

Anaphase Promoting Complex/Cyclosome (APC/C), particularly in

association with the co-activator Cdh1, plays a key role in targeting

S-phase cyclins (like Cyclin A) and mitotic cyclins (like Cyclin B) for

degradation, preventing premature entry into the next cell cycle

phase and allowing proper resetting after mitosis.

B. Cyclin B1 helps in the activation of S-phase CDKs only in late G1

– False. Cyclin B1 is primarily associated with CDK1 and mitosis,

not S-phase entry. Activation of S-phase CDKs (like CDK2) involves

Cyclin E (late G1) and Cyclin A (S-phase), not Cyclin B1.

C. As mitotic CDK activity declines in late mitosis, Cdc14

phosphatase activates APC/C by dephosphorylating Cdh1, thus

promoting formation of APC/CCdh1 – True. In late mitosis,

declining CDK1 activity allows Cdc14 to dephosphorylate Cdh1,

enabling formation of the APC/CCdh1 complex, which targets

remaining cyclins and other substrates for degradation, ensuring exit

from mitosis and proper G1 entry.

D. Securin keeps S-phase cyclins in inactive state till late G1 – False.

Securin is an inhibitor of separase, involved in anaphase regulation

(preventing sister chromatid separation), and has no role in

regulating S-phase cyclins.

Why Not the Other Options?

❌

(1) A and B – Incorrect; B is incorrect because Cyclin B1 is

involved in mitosis, not S-phase.

❌

(3) B and C – Incorrect; B is incorrect as explained.

❌

(4) B and D – Incorrect; both B and D are incorrect.

112. Thousands of proteins that are synthesized in the

cytoplasm are imported into the nucleus through the

nuclear pore complex (NPC) every minute. These

proteins contain Nuclear Localization Signal (NLS)

that direct their selective transport into the nucleus.

This nuclear import requires.

A. A small monomeric G-protein Ran

B. A nuclear transport receptor that interacts with

the NLS on a cargo protein.

C. A GTPase activating protein (GAP) bound to the

chromatin tethered to the nuclear membrane.

D. A Guanine Exchange Factor (GEF) bound to the

chromatin inside the nucleus.

Which one of the following options represents all

correct statements?

(1) A, C, D

(2) A, B, D

(3) B, C only

(4) A, C only

(2019)

Answer: (2) A, B, D

Explanation:

Nuclear import of proteins via the Nuclear Pore

Complex (NPC) is a tightly regulated and energy-dependent process

that ensures only proteins with a proper Nuclear Localization Signal

(NLS) are imported into the nucleus. This process depends on several

essential components:

A. A small monomeric G-protein Ran – Correct. Ran is a key

regulator of nucleocytoplasmic transport. It exists in two forms: Ran-

GTP (predominantly in the nucleus) and Ran-GDP (in the

cytoplasm), and its gradient drives the directionality of transport.

B. A nuclear transport receptor that interacts with the NLS on a

cargo protein – Correct. Importins, the nuclear transport receptors,

bind to NLS-containing cargo proteins in the cytoplasm and guide

them through the NPC into the nucleus.

C. A GTPase activating protein (GAP) bound to the chromatin

tethered to the nuclear membrane – Incorrect. RanGAP is localized

in the cytoplasm, not bound to chromatin or tethered to the nuclear

membrane. It promotes the conversion of Ran-GTP to Ran-GDP in

the cytoplasm, thus maintaining the Ran-GTP/Ran-GDP gradient.

D. A Guanine Exchange Factor (GEF) bound to the chromatin inside

the nucleus – Correct. RanGEF (also known as RCC1) is chromatin-

associated and located inside the nucleus, where it promotes the

exchange of GDP for GTP on Ran, ensuring a high concentration of

Ran-GTP in the nucleus.

Why Not the Other Options?

❌

(1) A, C, D – Incorrect; C is incorrect because RanGAP is

cytoplasmic, not chromatin-tethered.

❌

(3) B, C only – Incorrect; A and D are essential components; C is

incorrect.

❌

(4) A, C only – Incorrect; B and D are also required, and C is

incorrect.

113. The membrane phospholipid structures in bacteria

and archaea differ. Which one of the following

correctly states the differences between the two?

(1) The bacterial membrane phospholipid consist of D-

glycerol linked to hydrophobic chains (tails) with ester

bonds whereas those of the archaeal membranes consist

of L-glycerol linked to hydrophobic tails through ether

bonds.

(2) The bacterial membrane phospholipid consist of L-

glycerol linked to hydrophobic chains (tails) with ester

bonds whereas those of the archaeal membranes consist

of D-glycerol linked to hydrophobic tails through ether

bonds.

(3) The bacterial membrane phospholipid consist of D-

glycerol linked to hydrophobic chains (tails) with ether

bonds whereas those of the archaeal membranes consist

of L-glycerol linked to hydrophobic tails through ester

bonds.

(4) The bacterial membrane phospholipid consist of L-

glycerol linked to hydrophobic chains (tails) with ether

bonds whereas those of the archaeal membranes consists

of D-glycerol linked to hydrophobic tails through ester

bonds.

(2019)

Answer: (1) The bacterial membrane phospholipid consist of

D-glycerol linked to hydrophobic chains (tails) with ester

bonds whereas those of the archaeal membranes consist of L-

glycerol linked to hydrophobic tails through ether bonds.

Explanation:

Bacterial and archaeal membranes differ

fundamentally in the chirality of the glycerol backbone and the

linkage type connecting the glycerol to hydrophobic chains. In

bacteria, the phospholipids are composed of D-glycerol-3-phosphate,

and the fatty acid chains are attached through ester bonds. In

contrast, archaea use L-glycerol-1-phosphate, and the hydrophobic

isoprenoid chains are connected via more chemically stable ether

bonds. These structural differences reflect their distinct evolutionary

lineages and provide archaea with greater stability under extreme

environmental conditions.

Why Not the Other Options?

❌

(2) The bacterial membrane uses D-glycerol, not L-glycerol;

archaeal membranes use L-glycerol and ether bonds, not D-glycerol.

❌

(3) The linkage in bacterial membranes is ester, not ether;

archaeal membranes use ether linkages, not ester.

❌

(4) Bacterial membranes use D-glycerol, not L-glycerol, and

archaea do not use ester bonds—they use ether bonds.

114. Ribosomes prepared from a bacterium were

fractionated by sucrose density gradient

centrifugation (panel i) to separate the 30S, 50S and

70S populations. When the ribosome preparation was

incubated individually with either elongation factor-

G (EF-G), or a newly identified protein X, or GTP,

the profile remained unchanged. Likewise, no

changes were seen in the profile when the ribosomal

preparation was incubated with EF-G + GTP or

protein X+ GTP. However, when the ribosomal

preparation was incubated with protein X, EF-G and

GTP together, it resulted in a change of profile which

showed a decrease of the 70S peak area and increase

in the peak areas for 30S and 50S (panel ii).

Choose the option that defines a correct conclusion

from the observations.

(1) Protein X is an anti-association factor which

functions in the presence of EF-G and GTP

(2) Protein X is a dissociation factor which functions in

the presence of EF-G and GTP

(3) Protein X binds GTP

(4) EF-G is known to bind GTP, hence it can be

concluded that the effect of GTP is through EF-G and

protein X does not bind GTP.

(2019)

Answer: (2) Protein X is a dissociation factor which functions

in the presence of EF-G and GTP

Explanation:

The sucrose density gradient centrifugation profiles

in panels (i) and (ii) show sedimentation of bacterial ribosomal

subunits. In panel (i), we observe distinct peaks for 30S, 50S, and

70S ribosomal particles, with a strong 70S peak indicating intact

ribosomes. In panel (ii), when protein X, EF-G, and GTP are

combined, the 70S peak decreases, while 30S and 50S peaks increase,

indicating that the 70S ribosomes are dissociating into subunits. This

dissociation only occurs when all three components are present,

implying that protein X functions as a dissociation factor, requiring

both EF-G and GTP to act. EF-G is a GTPase involved in

translocation during translation and is known to assist in ribosome

disassembly under certain conditions, especially during ribosome

recycling. Protein X, therefore, is not acting alone but cooperates

with EF-G and GTP hydrolysis to dissociate the ribosome.

Why Not the Other Options?

❌

(1) Protein X is an anti-association factor... – Incorrect; an anti-

association factor prevents ribosomal subunits from assembling into

70S, but here the effect is on already assembled 70S ribosomes,

indicating dissociation, not prevention.

❌

(3) Protein X binds GTP – Incorrect; the experiment does not

demonstrate GTP binding by protein X, only that it requires GTP to

function in concert with EF-G.

❌

(4) EF-G is known to bind GTP... protein X does not bind GTP –

Incorrect; this conclusion is not supported by the data, as the binding

capability of protein X to GTP is not directly assessed; the inference

is not justifiable from the given evidence.

115. The individuals considered in this question are

having two haploid sets of autosomes and no Y-

chromosome. The X:A ratio of the individuals, the

type of organisms chosen, their primary sex and

number of Barr bodies expected in their cells are

shown in the table below:

Three Select the option below with all correct

matches:

(1) i-A-II-a; ii-A-II-d; i-B-I-a; ii-B-III-a

(2) i-A-I-a; ii-A-II-c; i-B-II-a; ii-B-I-a

(3) i-A-II-c; ii-A-I-d; i-B-I-c; ii-B-II-b

(4) i-A-II-a; ii-A-II-d; i-B-III-a; ii-B-I-a

(2019)

Answer: (1) i-A-II-a; ii-A-II-d; i-B-I-a; ii-B-III-a

Explanation:

This question involves interpreting the X:A ratio (X

chromosome to autosome sets) and its effects on sex determination in

humans and Drosophila, along with the expected number of Barr

bodies in human cells.

Let’s analyze each case:

Case i: X:A = 0.5

This means one X chromosome and two sets of autosomes (X:2A).

In humans, sex is determined by presence or absence of Y

chromosome (not X:A ratio). A human with one X and no Y (i.e., XO)

has Turner syndrome, phenotypically female, but the question labels

this as male, so this is likely referring to Drosophila, where:

In Drosophila, sex is determined by X:A ratio.

X:A = 0.5 → Male (even without Y chromosome).

Since it's a single X, no Barr body is formed.

So: i-B-I-a

Case ii: X:A = 2

Two X chromosomes and one set of autosomes (2X:1A) → X:A = 2

In Drosophila, X:A > 1 → Metafemale (non-viable or sterile female).

In Humans, having two X chromosomes and two sets of autosomes

(normally X:A = 1), but here if we assume 4 X chromosomes (X:A =

2), this suggests a female with multiple X chromosomes.

Each additional X beyond the first is inactivated (forms Barr body).

4 X chromosomes → 3 Barr bodies.

So:

ii-A-II-d

ii-B-III-a

Back to i-A

i-A (X:A = 0.5 in Human): likely XO → Turner syndrome

(phenotypic female).

But if we go by the question key, it’s matched to II (female) with a

(zero Barr bodies) → this fits XO.

So:

i-A-II-a

Final Matching from Option (1):

i-A-II-a

✅

ii-A-II-d

✅

(4 Xs → 3 Barr bodies)

i-B-I-a

✅

(0.5 X:A in Drosophila → male, no Barr body)

ii-B-III-a

✅

(X:A = 2 → metafemale, still only one X, no

inactivation needed)

Why Not the Other Options?

❌

(2) Assigns ii-A to c (only 2 Barr bodies), incorrect for 4 Xs.

❌

(3) Mixes up sex and Barr body count.

❌

(4) Assigns i-B to III (metafemale), wrong since X:A = 0.5 gives

male in Drosophila.

116. Match the following tumor cell origin with their

nomenclature.

(1) A - a, B - c, C - d, D - b

(2) A - d, B - a, C - b, D - c

(3) A - c, B - b, C – d, D - a

(4) A - b, B - d, C - a, D – c

(2018)

Answer: (4) A - b, B - d, C - a, D – c

Explanation:

Tumor nomenclature is largely based on the tissue of

origin of the cancerous cells. Sarcomas (b) originate from

mesenchymal tissues like muscle. Teratocarcinomas (d) are derived

from germ cells, which are pluripotent and can differentiate into

various tissue types. Carcinomas (a) arise from epithelial cells that

line surfaces and cavities in the body. Leukemias (c) are cancers of

the white blood cells, which are part of the hematopoietic system.

Why Not the Other Options?

❌

(1) A - a, B - c, C - d, D - b – Incorrect; Muscle cell tumors are

sarcomas, germ cell tumors can be teratocarcinomas, epithelial cell

tumors are carcinomas, and white blood cell tumors are leukemias.

❌

(2) A - d, B - a, C - b, D - c – Incorrect; Muscle cell tumors are

sarcomas, germ cell tumors can be teratocarcinomas, epithelial cell

tumors are carcinomas.

❌

(3) A - c, B - b, C – d, D - a – Incorrect; Muscle cell tumors are

sarcomas, germ cell tumors can be teratocarcinomas, epithelial cell

tumors are carcinomas, and white blood cell tumors are leukemias.

117. Following are a list of extracellular matrix proteins

(Column A) along with their functional

characteristics (Column B):

Which one of the following is the correct match?

(1) A - i; B-iv; C-iii; D-ii

(2) A - ii; B-iii; C-iv; D-i

(3) A-iii; B-iv; C-i; D-ii

(4) A- iv; B-i; C-ii; D-iii

(2018)

Answer: (3) A-iii; B-iv; C-i; D-ii

Explanation:

Let's analyze each extracellular matrix protein and

its functional characteristic:

A. Connexin: Connexins are a family of transmembrane proteins that

assemble to form gap junction channels between adjacent cells.

These channels allow the direct passage of ions and small molecules

between the cytoplasm of connected cells. Connexins are typically

four-pass transmembrane proteins and are the major constituents of

gap junctions. Therefore, A matches with (iii).

B. Plasmodesmata: Plasmodesmata are unique intercellular

junctions found in plant cells. They are channels that traverse the

cell walls of adjacent plant cells, directly connecting their cytoplasm

and allowing for the transport of various substances. Therefore, B

matches with (iv).

C. ICAM (Intercellular Adhesion Molecule): ICAMs are cell surface

proteins belonging to the immunoglobulin (Ig) superfamily. They are

expressed on various cell types, including endothelial cells, and

serve as ligands for integrins expressed on white blood cells. These

interactions are crucial for leukocyte adhesion to the endothelium

during inflammation and immune responses. Therefore, C matches

with (i).

D. Selectin: Selectins are a family of cell surface carbohydrate-

binding proteins (lectins) expressed on endothelial cells (E-selectin),

platelets (P-selectin), and leukocytes (L-selectin). They mediate the

initial weak adhesion of leukocytes to the endothelium during

leukocyte rolling in the bloodstream, which is a crucial step in

leukocyte extravasation to sites of inflammation. Therefore, D

matches with (ii).

Based on these matches:

A - (iii)

B - (iv)

C - (i)

D - (ii)

This corresponds to option (3).

Why Not the Other Options?

❌

(1) A - i; B-iv; C-iii; D-ii – Incorrect; Connexins are not

primarily recognized by white blood cell integrins, and ICAMs are

not the main constituents of gap junctions.

❌

(2) A - ii; B-iii; C-iv; D-i – Incorrect; Connexins are not

carbohydrate-binding proteins, plasmodesmata are not the main

constituents of gap junctions, and ICAMs are recognized by integrins.

❌

(4) A- iv; B-i; C-ii; D-iii – Incorrect; Connexins are not unique to

plants, plasmodesmata do not interact with white blood cell integrins,

and ICAMs are not carbohydrate-binding proteins.

118. In metazoan cell cycle, metaphase to anaphase

transition is regulated by the activity of:

(1) Cdkl/cyclinB

(2) APC/C

(3) Cdc25

(4) Wee1

(2018)

Answer: (2) APC/C

Explanation:

The transition from metaphase to anaphase is

triggered by the activation of the Anaphase Promoting

Complex/Cyclosome (APC/C), which is a ubiquitin ligase. The

APC/C targets securin for degradation. Securin is an inhibitor of

separase, an enzyme that cleaves cohesins, the protein complexes

holding sister chromatids together. Once securin is degraded,

separase becomes active, leading to the separation of sister

chromatids and the onset of anaphase.

Why Not the Other Options?

❌

(1) Cdk1/cyclin B – Incorrect; Cdk1/cyclin B (also known as MPF

or M-phase promoting factor) is crucial for driving entry into mitosis

and progression through the early stages of mitosis up to metaphase.

Its activity needs to be downregulated for the metaphase-to-anaphase

transition.

❌

(3) Cdc25 – Incorrect; Cdc25 is a phosphatase that activates

Cdk1/cyclin B by removing inhibitory phosphorylation. It is

important for entry into mitosis, not the metaphase-to-anaphase

transition.

❌

(4) Wee1 – Incorrect; Wee1 is a kinase that inhibits Cdk1/cyclin B

by adding inhibitory phosphorylation. Its activity needs to be

suppressed for entry into mitosis.

119. Which one of the following statements is true about

human chromosomes?

(1) The chromosomes that have highest gene density

generally localize towards the centre of the nucleus.

(2) The chromosomes that have highest gene· density

generally localize near the nuclear periphery to facilitate

rapid transport of the nascent transcripts.

(3) The centromeres of different chromosomes tend to

cluster together at the centre of the nucleus.

(4) Chromosomal positioning in the nucleus is absolutely

random.

(2018)

Answer: (1) The chromosomes that have highest gene density

generally localize towards the centre of the nucleus.

Explanation:

Studies on interphase chromosome organization in

human cells have revealed non-random positioning of chromosomes

within the nucleus. Generally, chromosomes with higher gene density

tend to be located more towards the interior (center) of the nucleus.

This positioning is thought to be associated with a more

transcriptionally permissive environment within the nucleus, where

factors and machinery required for active gene expression are more

concentrated.

Why Not the Other Options?

❌

(2) The chromosomes that have highest gene density generally

localize near the nuclear periphery to facilitate rapid transport of the

nascent transcripts. – Incorrect; While the nuclear periphery is

involved in some aspects of RNA processing and transport, gene-rich

chromosomes are typically found in the nuclear interior, where

transcription is more active.

❌

(3) The centromeres of different chromosomes tend to cluster

together at the centre of the nucleus. – Incorrect; Centromeres and

telomeres exhibit specific spatial organization but do not generally

cluster at the center of the nucleus. Centromeres often associate with

heterochromatic regions and can be found clustered near the

nucleolus or the nuclear periphery, depending on the cell type and

stage.

❌

(4) Chromosomal positioning in the nucleus is absolutely random.

– Incorrect; Chromosomes occupy preferred spatial territories

within the nucleus, and their positioning is influenced by factors such

as gene density, transcriptional activity, and interaction with nuclear

structures. This organization is not random.

120. Following observation was made when a mammalian

cell in one phase of cell cycle was fused with a cell in

another phase of cell cycle:

A. Fusion of a cell in G1 phase with S phase caused

the G1 nucleus to enter S phase.

B. Upon fusion of a G2 cell with an S phase cell, G2

cell does not enter S phase.

C. Upon fusion of a G1 cell with G2 cell, G1 nucleus

enters G2 phase.

D. Fusion of an S phase cell with a M phase causes

the S phase cell to immediately enter mitosis.

Choose the combination with all correct statements.

(1) A, B, C

(2) A, C, D

(3) B, C, D

(4) A, B, D

(2018)

Answer: (4) A, B, D

Explanation:

Cell fusion experiments have provided valuable

insights into the regulation of the cell cycle. Let's analyze each

statement based on the known principles of cell cycle control:

A. Fusion of a cell in G1 phase with S phase caused the G1 nucleus

to enter S phase. This statement is correct. The cytoplasm of an S

phase cell contains factors that promote DNA replication. When a

G1 cell is fused with an S phase cell, these S phase-promoting factors

can diffuse into the G1 nucleus and trigger the initiation of DNA

synthesis, causing the G1 nucleus to prematurely enter S phase.

B. Upon fusion of a G2 cell with an S phase cell, G2 cell does not

enter S phase. This statement is correct. G2 phase cells have already

completed DNA replication and contain factors that prevent re-

replication. The fusion of a G2 cell with an S phase cell does not

cause the G2 nucleus to undergo another round of DNA synthesis.

The S phase nucleus continues its replication, but the G2 nucleus

remains in G2.

C. Upon fusion of a G1 cell with G2 cell, G1 nucleus enters G2

phase. This statement is incorrect. The G2 phase is characterized by

the presence of factors that prepare the cell for mitosis. While the G1

nucleus might be influenced by the G2 cytoplasm, it does not directly

"enter" G2 phase in the sense of skipping S phase and acquiring G2

characteristics. The G1 nucleus would still need to undergo DNA

replication (S phase) before it can enter G2 phase. The fusion might,

however, influence the timing of subsequent S phase entry for the G1

nucleus.

D. Fusion of an S phase cell with a M phase causes the S phase cell

to immediately enter mitosis. This statement is correct. M phase cells

contain factors, such as M-phase promoting factor (MPF), that

induce the events of mitosis, including chromosome condensation

and nuclear envelope breakdown. When an S phase cell is fused with

an M phase cell, these mitotic factors can override the normal S

phase progression and cause the S phase nucleus to undergo

premature chromosome condensation (PCC) and enter a state

resembling mitosis, even though DNA replication might not be

complete.

Therefore, the correct statements are A, B, and D.

Why Not the Other Options?

❌

(1) A, B, C – Incorrect; Statement C is incorrect.

❌

(2) A, C, D – Incorrect; Statement C is incorrect.

❌

(3) B, C, D – Incorrect; Statement C is incorrect.

121. In an experiment, intact chromatin was isolated and

digested with micrococcal nuclease in independent

tubes for 30 min, 1 h, 2h, and 4h. Further, the DNA

was purified from each tube, separated on agarose gel

and Southern hybridization was performed with

rRNA gene probe and a centromeric DNA probe.

Which one of the following patterns of signal

intensity from both of the probes is likely to be

obtained following Southern hybridization?

(1) With increasing time, compared to centromeric probe,

a rapid increase in signal intensity of rRNA gene probe

was observed.

(2) With increasing time, compared to centromeric probe,

a rapid decrease in signal intensity of rRNA gene probe

was observed.

(3) Irrespective of incubation period, both probes

produced identical band intensities.

(4) Treatment with micrococcal nuclease would instantly

degrade the DNA, hence, no hybridi:zation signal would

be obtained· in any of the samples.

(2018)

Answer: (2) With increasing time, compared to centromeric

probe, a rapid decrease in signal intensity of rRNA gene probe

was observed.

Explanation:

Micrococcal nuclease is an enzyme that

preferentially digests linker DNA, the regions of DNA between

nucleosomes in chromatin. Euchromatin, which generally contains

actively transcribed genes like rRNA genes, is typically less

condensed and more accessible to nucleases compared to

heterochromatin, which often contains repetitive sequences like

centromeric DNA and is more tightly packaged.

In this experiment:

rRNA genes are actively transcribed and thus likely located in more

accessible euchromatic regions. This means the linker DNA around

rRNA gene-containing nucleosomes will be more readily digested by

micrococcal nuclease. As the incubation time with the nuclease

increases, the rRNA gene sequences will be progressively fragmented

into smaller pieces that may not be efficiently retained during DNA

purification or may produce a smear on the gel, leading to a

decrease in the intensity of specific bands hybridizing to the rRNA

gene probe.

Centromeric DNA is typically found in heterochromatin, which is

more condensed and less accessible to micrococcal nuclease. The

linker DNA in these regions will be digested more slowly. Therefore,

the centromeric DNA will be relatively protected from extensive

digestion compared to the rRNA genes, and the signal intensity from

the centromeric probe will likely decrease more gradually with

increasing incubation time.

Consequently, with increasing digestion time, we would expect to see

a more rapid decrease in the signal intensity of the rRNA gene probe

compared to the centromeric DNA probe.

Why Not the Other Options?

❌

(1) With increasing time, compared to centromeric probe, a rapid

increase in signal intensity of rRNA gene probe was observed. –

Incorrect; Nuclease digestion leads to fragmentation of DNA, which

would decrease the signal intensity over time as the probe's target

sequences are broken down.

❌

(3) Irrespective of incubation period, both probes produced

identical band intensities. – Incorrect; The differential accessibility

of euchromatin (containing rRNA genes) and heterochromatin

(containing centromeric DNA) to micrococcal nuclease would result

in different rates of DNA fragmentation and thus different signal

intensities over time.

❌

(4) Treatment with micrococcal nuclease would instantly degrade

the DNA, hence, no hybridization signal would be obtained in any of

the samples. – Incorrect; Micrococcal nuclease digestion is time-

dependent. At shorter incubation times, the DNA would be partially

digested, allowing for hybridization signals to be detected. Complete

degradation would require much longer incubation times or higher

enzyme concentrations.

122. Which one of the following proteins is most likely to

be found in the inter-membrane space of

mitochondria? A protein containing

(1) an N-terminal matrix targeting sequence followed by

hydrophobic stop-transfer anchor sequence

(2) an N-terminal matrix targeting sequence followed by

a cleavable hydrophobic sequence that blocks complete

translocation

(3) a protein with multiple internal sequences that are

recognized by Tim 22 complex

(4) a protein with an outer membrane localization

sequence followed by a matrix targeting signal

(2018)

Answer: (2) an N-terminal matrix targeting sequence

followed by a cleavable hydrophobic sequence that blocks

complete translocation

Explanation:

Proteins targeted to the inter-membrane space (IMS)

of mitochondria typically follow a specific import pathway that

involves a two-step process:

Import into the matrix: The protein usually contains an N-terminal

matrix targeting sequence (MTS) that directs it across the outer and

inner mitochondrial membranes via the TOM (Translocase of the

Outer Membrane) and TIM23 (Translocase of the Inner Membrane)

complexes.

Retention in the IMS: To prevent complete translocation into the

matrix, these IMS proteins often possess a second signal, such as a

hydrophobic stop-transfer anchor sequence or a cleavable

hydrophobic sequence.

In the case of a cleavable hydrophobic sequence that blocks complete

translocation, the MTS directs the protein through the TOM complex

and partially into the TIM23 channel. The hydrophobic sequence

then gets stuck in the inner membrane, preventing further

translocation into the matrix. The MTS is subsequently cleaved by a

matrix protease, and the remaining protein, anchored in the inner

membrane, has its soluble domain exposed to the IMS. This anchored

protein can then be processed further or released into the IMS via

other mechanisms.

Let's examine why the other options are less likely:

(1) an N-terminal matrix targeting sequence followed by

hydrophobic stop-transfer anchor sequence: This type of protein is

typically destined for the inner mitochondrial membrane. The MTS

initiates import, and the stop-transfer sequence halts translocation

through the TIM23 complex, anchoring the protein in the inner

membrane.

(3) a protein with multiple internal sequences that are recognized by

Tim 22 complex: The TIM22 complex is primarily involved in the

insertion of multi-pass transmembrane proteins into the inner

mitochondrial membrane. These proteins lack cleavable N-terminal

MTS and instead have internal targeting signals.

(4) a protein with an outer membrane localization sequence followed

by a matrix targeting signal: If a protein has a strong outer

membrane localization sequence, it is likely to be inserted into the

outer membrane. A subsequent matrix targeting signal might be

cryptic or not efficiently recognized once the protein is anchored in

the outer membrane. While some proteins might have dual

localization signals, this specific arrangement is less common for

IMS proteins compared to the mechanism described in option (2).

Therefore, a protein with an N-terminal matrix targeting sequence

followed by a cleavable hydrophobic sequence that blocks complete

translocation is most likely to be found in the inter-membrane space

of mitochondria.

Why Not the Other Options?

❌

(1) an N-terminal matrix targeting sequence followed by

hydrophobic stop-transfer anchor sequence – Incorrect; These

proteins are usually embedded in the inner mitochondrial membrane.

❌

(3) a protein with multiple internal sequences that are recognized

by Tim 22 complex – Incorrect; These proteins are typically inserted

into the inner mitochondrial membrane as multi-pass transmembrane

proteins.

❌

(4) a protein with an outer membrane localization sequence

followed by a matrix targeting signal – Incorrect; The outer

membrane localization signal would likely dominate, targeting the

protein to the outer membrane.

123. A single protofilament of microtubule grows at the

speed of 2 μm/min. Considering that there is no

catastrophe in the microtubule nucleation and the

size of the tubulin unit is of the order of 5 nm, how

many tubulin units are added to the growing

microtubule per minute?

(1) 400

(2) 1600

(3) 3200

(4) 5200

(2018)

Answer: (1) 400

Explanation:

We are given the growth speed of a single

protofilament of a microtubule and the size of a tubulin unit. We need

to calculate how many tubulin units are added per minute.

Given:

Growth speed = 2 μm/min

Size of one tubulin unit = 5 nm

First, we need to convert the growth speed and the size of the tubulin

unit to the same units. Let's convert everything to nanometers (nm)

and minutes.

Growth speed = 2 μm/min

Since 1 μm = 1000 nm,

Growth speed = 2 × 1000 nm/min = 2000 nm/min

Size of one tubulin unit = 5 nm

Now, to find the number of tubulin units added per minute, we divide

the growth speed by the size of one tubulin unit:

Number of tubulin units added per minute = (Growth speed) / (Size

of one tubulin unit)

Number of tubulin units added per minute = (2000 nm/min) / (5

nm/unit)

Number of tubulin units added per minute = 400 units/min

Therefore, 400 tubulin units are added to the growing microtubule

per minute.

Why Not the Other Options?

❌

(2) 1600 – Incorrect; This value is four times the calculated value,

suggesting a potential error in calculation or understanding the

relationship between growth speed and unit size.

❌

(3) 3200 – Incorrect; This value is eight times the calculated

value, indicating a larger discrepancy.

❌

(4) 5200 – Incorrect; This value is significantly higher than the

calculated number of tubulin units added per minute. The calculation

directly based on the given speed and unit size yields 400.

124. mRNA of a gene was depleted in human cells using

siRNA that arrest cells in the G2 phase of the cell

cycle. In order to test whether the G2 arrest is due to

an off-target or all on-target effect of siRNA

mediated mRNA depletion, an investigator can:

A. re-introduce an ectopic copy of the gene coding for

the wild-type mRNA and protein

B. re-introduce an ectopic copy of the gene that is

different in its mRNA sequence at the siRNA target

site but encodes for the same protein

C. re-introduce an ectopic copy of the gene that codes

for-different mRNA and protein

D. utilize few more siRNAs targeting different

regions of the mRNA in question

Choose the combination with correct statements:

(1) A, B, C only

(2) C and D only

(3) B and D only

(4) B, C, D only

(2018)

Answer: (3) B and D only

Explanation:

To determine whether the G2 arrest observed upon

siRNA-mediated mRNA depletion is due to an on-target effect

(specific to the intended gene) or an off-target effect (due to the

siRNA binding to and affecting other unintended mRNAs), an

investigator can employ several strategies. Let's analyze each option:

A. re-introduce an ectopic copy of the gene coding for the wild-type

mRNA and protein: If the G2 arrest is an on-target effect, re-

introducing the wild-type gene and restoring the protein levels

should rescue the phenotype (i.e., cells should no longer be arrested

in G2). This helps confirm the role of the targeted gene but doesn't

definitively rule out off-target effects being the primary cause of the

arrest observed with the initial siRNA.

B. re-introduce an ectopic copy of the gene that is different in its

mRNA sequence at the siRNA target site but encodes for the same

protein: If the G2 arrest is due to the siRNA specifically targeting the

intended mRNA (on-target effect), then introducing a modified mRNA

that cannot be recognized by the same siRNA but still produces the

same protein should rescue the phenotype. If the arrest persists, it

would suggest an off-target effect of the siRNA. This is a strong test

for on-target specificity.

C. re-introduce an ectopic copy of the gene that codes for different

mRNA and protein: Introducing a completely different gene and

protein is unlikely to rescue a G2 arrest caused by the depletion of a

specific gene. This would primarily serve as a negative control and

wouldn't help distinguish between on-target and off-target effects of

the original siRNA.

D. utilize few more siRNAs targeting different regions of the mRNA

in question: If the G2 arrest is an on-target effect, multiple different

siRNAs targeting the same mRNA but at different sequences should

also lead to a similar phenotype (G2 arrest). If different siRNAs

targeting the same gene consistently cause the same effect, it

strengthens the argument for an on-target mechanism. If different

siRNAs lead to different or no effects, it might suggest off-target

effects are playing a significant role.

Therefore, the most effective strategies to test for the specificity of the

siRNA-mediated G2 arrest are to introduce a modified version of the

target gene that is resistant to the siRNA (option B) and to use

multiple independent siRNAs targeting the same gene (option D).

Why Not the Other Options?

❌

(1) A, B, C only – Incorrect; Option C does not directly help in

distinguishing between on-target and off-target effects.

❌

(2) C and D only – Incorrect; While option D is helpful, option C

is not. Option B provides a more direct test for on-target specificity.

❌

(4) B, C, D only – Incorrect; Option C does not directly address

the question of on-target versus off-target effects of the initial siRNA.

125. The cell maintains a high concentration of protons

inside the lysosome because of

(1) antiporter in the lysosomal membrane

(2) ATP-powered proton pump in the lysosomal

membrane

(3) facilitated diffusion proton channel in the lysosomal

membrane

(4) facilitated diffusion proton uniporter in the lysosomal

membrane

(2018)

Answer: (2) ATP-powered proton pump in the lysosomal

membrane

Explanation:

Lysosomes are acidic organelles with a pH of

around 4.5-5.0, which is significantly lower than the neutral pH

(around 7.4) of the cytoplasm. This acidic environment is crucial for

the function of the hydrolytic enzymes within the lysosome that

degrade various cellular components. Maintaining this proton

gradient requires energy, as protons are being moved against their

concentration gradient (from a lower concentration in the cytoplasm

to a higher concentration in the lysosome). This energy is supplied by

the hydrolysis of ATP by a specific type of P-type ATPase located in

the lysosomal membrane. This ATP-powered proton pump actively

transports protons (H+) into the lysosomal lumen, thus establishing

and maintaining the acidic pH.

Why Not the Other Options?

❌

(1) antiporter in the lysosomal membrane – Incorrect; Antiporters

facilitate the movement of one ion or molecule across a membrane

coupled with the movement of another in the opposite direction.

While there might be antiporters in the lysosomal membrane

involved in other transport processes, they wouldn't directly cause

the accumulation of protons without an initial energy input to create

the gradient.

❌

(3) facilitated diffusion proton channel in the lysosomal

membrane – Incorrect; Facilitated diffusion allows ions or molecules

to move across a membrane down their concentration gradient

through a channel protein. This process is passive and does not

require energy input. A proton channel would allow protons to flow

out of the lysosome down their concentration gradient, dissipating

the high proton concentration inside.

❌

(4) facilitated diffusion proton uniporter in the lysosomal

membrane – Incorrect; A uniporter facilitates the movement of a

single type of ion or molecule across a membrane down its

concentration gradient. Similar to a channel, a proton uniporter

would allow protons to move out of the lysosome, not accumulate

inside. Facilitated diffusion is a passive process and cannot establish

or maintain a concentration gradient against the direction of

thermodynamic favorability.

126. Which one of the following permits the rapid

diffusion of small, water-soluble molecules between

the cytoplasm of adjacent cells?

(1) Tight junctions

(2) Anchoring junctions

(3) Adherens junctions

(4) Gap junctions

(2018)

Answer: (4) Gap junctions

Explanation:

Gap junctions are specialized intercellular

connections between animal cells that directly link the cytoplasm of

adjacent cells. They are formed by clusters of transmembrane protein

complexes called connexons, which align to create channels

spanning the intercellular space. These channels have a relatively

small diameter, allowing the passage of small, water-soluble

molecules such as ions, sugars (although some sources suggest

larger sugars might not pass), amino acids, nucleotides, and

signaling molecules (like cAMP and IP3) directly from the cytoplasm

of one cell to the cytoplasm of its neighbor. This direct

communication facilitates rapid coordination of cellular activities

and signaling between connected cells.

Why Not the Other Options?

❌

(1) Tight junctions – Incorrect; Tight junctions (also known as oc

They are crucial for maintaining tissue barriers and

compartmentalization.

❌

(2) Anchoring junctions – Incorrect; Anchoring junctions

(including adherens junctions and desmosomes) primarily function to

provide mechanical strength and link the cytoskeletons of adjacent

cells or a cell to the extracellular matrix. They do not directly permit

the passage of small molecules between cells.

❌

(3) Adherens junctions – Incorrect; Adherens junctions are a type

of anchoring junction that uses cadherin proteins to link the actin

cytoskeletons of adjacent cells. Their main role is in cell adhesion

and tissue stability, not in facilitating the direct cytoplasmic

exchange of small molecules.

127. Match the enzymes in Column A with their respective

biological function in Column B:

Choose the correct combinations of answers from the

options given below:

(1) A - iii, B - i, C - ii, D - iv

(2) A - i, B - iii, C - iv, D - ii

(3) A - iv, B- ii, C - i, D- iii

(4) A- ii, B- iv, C - iii, D – i

(2018)

Answer: (1) A - iii, B - i, C - ii, D - iv

Explanation:

Lipases (A) are enzymes that catalyze the hydrolysis

of lipids, specifically breaking down triacylglycerols into glycerol

and fatty acids. Therefore, A matches with iii.

Flippases (B) are ATP-dependent enzymes that facilitate the

movement of specific phospholipids, namely aminophospholipids like

phosphatidylserine (PS) and phosphatidylethanolamine (PE), from

the extracellular (exoplasmic) leaflet to the cytosolic leaflet of a

biological membrane. This translocation is crucial for maintaining

the asymmetric distribution of phospholipids in the plasma

membrane. Therefore, B matches with i.

Floppases (C) are also ATP-dependent enzymes that catalyze the

movement of phospholipids across the plasma membrane bilayer, but

in the opposite direction to flippases. They typically translocate

phospholipids from the cytosolic leaflet to the extracellular leaflet.

Therefore, C matches with ii.

Scramblases (D) are enzymes that facilitate the bidirectional

movement of various phospholipids across the membrane bilayer

down their concentration gradient. This process is ATP-independent

and leads to a more randomized distribution of phospholipids

between the two leaflets, often occurring in response to cellular

signals like increased calcium levels during apoptosis. Therefore, D

matches with iv.

Why Not the Other Options?

❌

(2) A - i, B - iii, C - iv, D - ii – Incorrect; Lipases hydrolyze

triacylglycerols (iii), Flippases move aminophospholipids from

extracellular to cytosolic (i), Floppases move phospholipids from

cytosolic to extracellular (ii), and Scramblases facilitate movement

down the concentration gradient (iv).

❌

(3) A - iv, B- ii, C - i, D- iii – Incorrect; Lipases hydrolyze

triacylglycerols (iii), Flippases move aminophospholipids from

extracellular to cytosolic (i), Floppases move phospholipids from

cytosolic to extracellular (ii), and Scramblases facilitate movement

down the concentration gradient (iv).

❌

(4) A- ii, B- iv, C - iii, D – i – Incorrect; Lipases hydrolyze

triacylglycerols (iii), Flippases move aminophospholipids from

extracellular to cytosolic (i), Floppases move phospholipids from

cytosolic to extracellular (ii), and Scramblases facilitate movement

down the concentration gradient (iv).

128. An integral membrane protein (P) has been identified

as a cell surface protein of hepatocytes and assigned

to bind to hepatitis B virus (HBV) and promote its

entry into cytosol. Upon binding to HBV particles,

the C-terminal of P interacts with F-actin in the

cytosol and in turn, helps in the entry of the HBV

particles. P was successfully cloned and expressed in

animal cells in culture wherein its N- terminal is

exposed on the surface while the C-terminal resides

in the cytosol. The recombinant protein P so

expressed retains its complete structure and function.

From the list of experiments given below, which one

of the experiments will you perform to show that C-

terminal of the protein P via interacting with F-actin

helps in HBV entry?

(1) Incubating radiolabelled HBV with hepatocytes in

culture and follow up its association with F-actin by

immuno- precipitation analysis using anti-F-actin

antibody.

(2) Incubating radiolabelled HBV with hepatocytes over-

expressing the C- terminal mutant of P and repeat the

rest of the experiment as in "1"

(3) Incubating radiolabelled HBV with· hepatocytes

over-expressing the N- terminal mutant of P and repeat

the rest of the experiment as in "1".

(4) Using wild type P as well as C-terminal mutant of P

and their individual over- expression in a heterologous

cell line (completely devoid of endogenous P protein)

and then repeat experiment as in "1".

(2018)

Answer: (4) Using wild type P as well as C-terminal mutant

of P and their individual over- expression in a heterologous

cell line (completely devoid of endogenous P protein) and

then repeat experiment as in "1".

Explanation:

This experiment directly addresses the role of the C-

terminal of protein P and its interaction with F-actin in HBV entry by

isolating the system. By using a heterologous cell line that does not

naturally express protein P, you eliminate any potential confounding

factors from the endogenous protein. Overexpressing both the wild-

type P and a mutant of P specifically lacking or having a non-

functional C-terminal (that cannot interact with F-actin) in this cell

line allows for a clear comparison. If HBV entry is indeed dependent

on the C-terminal interaction with F-actin, you would observe:

In cells overexpressing wild-type P: Radiolabeled HBV would bind

to the cell surface via the N-terminal of P, and the C-terminal of P

would interact with F-actin, leading to HBV entry and association

with F-actin, which can be detected by immunoprecipitation with

anti-F-actin antibody followed by detection of radiolabeled HBV.

In cells overexpressing the C-terminal mutant of P: Radiolabeled

HBV would still bind to the cell surface via the N-terminal, but the

mutated C-terminal would be unable to interact with F-actin. This

would result in significantly reduced or absent association of HBV

with F-actin in the immunoprecipitation analysis, indicating that this

interaction is crucial for HBV entry.

This controlled system in a heterologous cell line provides strong

evidence for the specific role of the C-terminal of protein P and its

interaction with F-actin in HBV entry.

Why Not the Other Options?

❌

(1) Incubating radiolabelled HBV with hepatocytes in culture and

follow up its association with F-actin by immuno-precipitation

analysis using anti-F-actin antibody – Incorrect; While this

experiment can show an association of HBV with F-actin in

hepatocytes, it doesn't specifically prove that this interaction is

mediated by the C-terminal of protein P. The observed association

could be due to other cellular factors.

❌

(2) Incubating radiolabelled HBV with hepatocytes over-

expressing the C- terminal mutant of P and repeat the rest of the

experiment as in "1" – Incorrect; Overexpressing a mutant in the

presence of the endogenous wild-type protein P in hepatocytes might

lead to complex and difficult-to-interpret results. The endogenous P

could still mediate some level of HBV entry via F-actin interaction,

masking the effect of the mutant.

❌

(3) Incubating radiolabelled HBV with hepatocytes over-

expressing the N- terminal mutant of P and repeat the rest of the

experiment as in "1" – Incorrect; A mutation in the N-terminal of P

would likely affect HBV binding to the cell surface, which is the

initial step. This experiment would primarily assess the role of the N-

terminal in binding, not the C-terminal in downstream entry

mechanisms involving F-actin.

129. The cell membrane of neuron maintains intracellular

conditions that differ from those of the extracellular

environment. Such difference in intraand

extracellular conditions are critical to the function of

the nerve cell as the nerve cell membrane resembles a

charged capacitor. Assuming the electric field (E)

across a parallel-plate capacitor is uniform and if

membrane thickness is 7 nm and potential difference

across the membrane is -60 mV, calculate E of the

membrane.

(1) 6 × 10

V m

-1

(2) 7 × 10

V m

-1

(3) 8.6 × 10

V m

-1

(4) 6.6 × 10

V m

-1

(2018)

Answer: (3) 8.6 × 106 V m-1

Explanation:

The electric field (E) across a parallel-plate

capacitor with a uniform electric field is related to the potential

difference (V) across the plates and the distance (d) between the

plates by the formula:

E=d

∣

In this case, the nerve cell membrane acts as the parallel-plate

capacitor. We are given:

Membrane thickness (d) = 7 nm = 7×10−9 m

Potential difference across the membrane (V) = -60 mV =

−60×10−3 V

We need to calculate the magnitude of the electric field, so we take

the absolute value of the potential difference:

∣

−60×10−3V

∣

=60×10−3V

Now, we can plug the values into the formula:

E=7×10−9m60×10−3V

E=760 × 10−910−3 Vm−1

E≈8.57×10−3−(−9)Vm−1

E≈8.57×10−3+9Vm−1

E≈8.57×106Vm−1

Rounding to one decimal place as in the options, we get

8.6×106Vm−1.

Why Not the Other Options?

❌

(1) 6 × 10⁵ V m⁻¹ – Incorrect; This value is obtained by an

incorrect calculation involving the given values.

❌

(2) 7 × 10⁵ V m⁻¹ – Incorrect; This value is obtained by an

incorrect calculation involving the given values.

❌

(4) 6.6 × 10⁶ V m⁻¹ – Incorrect; This value is obtained by an

incorrect calculation involving the given values.

130. Two liposome preparations ("X" and "Y") are made

using basic lipid composition as phosphatidylcholine

(PC) and cholesterol (Chol). In "Y" a ganglioside

(asialo-GM1) is incorporated during the preparation

besides PC and Chol. In an attempt to find out the

localization of asialo-GM1, in the membrane bilayer

of "Y" (taking "X" as a negative control) and

considering liposome as a true depiction of lipid

bilayer structure of cellular membrane, following

reagents are provided as probes: A. Phospholipase A

B. Galactose binding lectin C. Exoglycosidase D.

Cyclodextrin Choose the most appropriate reagent(s)

from the above list to ascertain the localization of

asialo-GM1-

(1) Only A

(2) Both C and D

(3) Both B and C

(4) Both A and Da

(2018)

Answer: (3) Both B and C

Explanation:

Asialo-GM1 is a ganglioside, which is a

glycosphingolipid containing one or more sialic acid residues (in the

case of asialo-GM1, the sialic acid is absent, hence "asialo"). The

carbohydrate portion of gangliosides extends into the aqueous

environment. To determine the localization of asialo-GM1 in the

liposome bilayer, we need probes that can interact specifically with

its carbohydrate moiety.

B. Galactose binding lectin: Asialo-GM1 contains galactose residues

in its carbohydrate chain. Lectins are carbohydrate-binding proteins

with high specificity. A galactose-binding lectin can bind to the

galactose residues exposed on the surface of the liposomes. If the

lectin binds to liposome "Y" (containing asialo-GM1) but not

significantly to liposome "X" (lacking asialo-GM1), it indicates that

the galactose residues of asialo-GM1 are accessible on the outer

surface of the liposomes.

C. Exoglycosidase: Exoglycosidases are enzymes that specifically

cleave terminal carbohydrate residues from glycoconjugates. If an

exoglycosidase that specifically removes terminal galactose residues

is incubated with liposome "Y" and subsequently analyzed for the

presence of asialo-GM1 (e.g., by thin-layer chromatography or mass

spectrometry), a reduction or absence of asialo-GM1 would indicate

that the galactose residues are exposed and accessible to the enzyme.

This would further confirm the localization of asialo-GM1 with its

carbohydrate portion facing outwards.

Why Not the Other Options?

❌

(1) Only A – Incorrect; Phospholipase A enzymes hydrolyze

phospholipids at specific ester bonds, releasing fatty acids and

lysophospholipids. While it can probe the accessibility of

phospholipids in the bilayer, it does not specifically target the

carbohydrate moiety of asialo-GM1.

❌

(2) Both C and D – Incorrect; While exoglycosidase (C) can

probe the carbohydrate part of asialo-GM1, cyclodextrin (D) is a

cyclic oligosaccharide that can encapsulate hydrophobic molecules,

particularly cholesterol. It can be used to study cholesterol's role in

membrane structure but does not directly provide information about

the localization of the carbohydrate headgroup of asialo-GM1.

❌

(4) Both A and D – Incorrect; Neither phospholipase A (A) nor

cyclodextrin (D) specifically targets the carbohydrate portion of

asialo-GM1 to determine its localization.

131. Eg5 is a well-studied protein in Xenopus. To

understand the function of Eg5 in mammalian cells, a

team of researchers treated mammalian cells during

late G2 phase with Eg5-inhibitor. The following

diagrams represent images of mitotic cells.

Based on the above observation, what function might

be attributed to Eg5 during mitosis?

(1) Eg5 inhibits actin dynamics.

(2) Eg5 can activate GPCR signalling.

(3) Eg5 has motor activity.

(4) Eg5 can impact mitochondrial dynamics.

(2018)

Answer: (3) Eg5 has motor activity.

Explanation:

The image shows two diagrams representing mitotic

cells. The "Control" cell exhibits chromosomes aligned at the

metaphase plate, indicating a normal progression through mitosis.

The "Eg5 inhibitor" treated cell shows chromosomes that are not

properly aligned; they appear scattered and unable to form a clear

metaphase plate.

Eg5 is a kinesin-5 motor protein. Kinesins are a family of motor

proteins that use the energy from ATP hydrolysis to move along

microtubules. Kinesin-5 proteins are unique in that they have motor

domains at both ends and can cross-link antiparallel microtubules.

During mitosis, Eg5 plays a crucial role in establishing and

maintaining the bipolar spindle. It pushes the spindle poles apart by

sliding antiparallel microtubules past each other at the spindle

midzone. This outward force generated by Eg5 is essential for proper

chromosome alignment at the metaphase plate.

When Eg5 is inhibited, the outward pushing force on the spindle

poles is reduced, leading to a collapse or shortening of the spindle.

Consequently, the chromosomes, which rely on the bipolar spindle

for proper alignment and segregation, fail to congress to the

metaphase plate and remain disorganized, as observed in the "Eg5

inhibitor" diagram. This phenotype directly points to a motor activity

of Eg5 that is essential for spindle organization and chromosome

alignment during mitosis.

Why Not the Other Options?

❌

(1) Eg5 inhibits actin dynamics. – Incorrect; The observed

phenotype primarily involves chromosome alignment and spindle

organization, which are microtubule-dependent processes. There is

no direct evidence from this experiment to suggest that Eg5 inhibits

actin dynamics.

❌

(2) Eg5 can activate GPCR signalling. – Incorrect; G protein-

coupled receptors (GPCRs) are involved in cell signaling pathways.

While cell signaling is crucial for regulating mitosis, the direct effect

of Eg5 inhibition on chromosome alignment strongly suggests a

mechanical role for Eg5 in spindle function, not a signaling role via

GPCRs.

❌

(4) Eg5 can impact mitochondrial dynamics. – Incorrect;

Mitochondrial dynamics (fusion and fission) are important for

cellular function, but the observed mitotic defect of misaligned

chromosomes directly implicates a role for Eg5 in the mitotic spindle,

which is composed of microtubules, not mitochondria. There is no

direct link established by this experiment between Eg5 and

mitochondrial dynamics.

132. The extracellular matrix (ECM) is a complex

combination of secreted proteins that is involved in

holding cells and tissues together. The components of

ECM form a network by binding to each other and

communicate with cells by binding to adhesion

receptors on the cell surface. ECM comprises mainly

two classes of macromolecules, proteoglycans and

very high molecular weight large proteins. Which one

of the following statements regarding ECM

constituents is INCORRECT?

(1) Proteoglycans are a subset of secreted or cell surface-

attached glycoproteins containing covalently linked

specialized polysaccharide chains called glycosamino-

glycans (GAGs).

(2) GAGs are long branched polymers of specific

repeating disaccharides of sialic acid and glucose or

galactose.

(3) Major types of GAGs present in ECM are heparan

sulphate, chondroitin sulphate, dermatan sulphate,

keratan sulphate and hyaluronan.

(4) Major types of large proteins present in ECM are

collagen, laminin, elastin and fibronectin.

(2018)

Answer: (2) GAGs are long branched polymers of specific

repeating disaccharides of sialic acid and glucose or galactose.

Explanation:

(1) Proteoglycans are a subset of secreted or cell

surface-attached glycoproteins containing covalently linked

specialized polysaccharide chains called glycosamino- glycans

(GAGs). This statement is CORRECT. Proteoglycans are indeed

defined by the presence of one or more covalently attached GAG

chains to a core protein.

(2) GAGs are long branched polymers of specific repeating

disaccharides of sialic acid and glucose or galactose. This statement

is INCORRECT. Glycosaminoglycans (GAGs) are long, unbranched

polysaccharides composed of repeating disaccharide units. While

they do contain derivatives of glucose (like N-acetylglucosamine)

and galactose (like N-acetylgalactosamine), a key characteristic is

that one of the sugars in the repeating disaccharide is usually an

amino sugar (N-acetylglucosamine or N-acetylgalactosamine), and

the other is typically a uronic acid (like glucuronic acid or iduronic

acid) or sulfate group. Sialic acid is a terminal sugar found on some

glycoproteins and glycolipids, but it is not a core repeating unit in

the major GAGs.

(3) Major types of GAGs present in ECM are heparan sulphate,

chondroitin sulphate, dermatan sulphate, keratan sulphate and

hyaluronan. This statement is CORRECT. These are the five major

classes of GAGs found in the extracellular matrix.

(4) Major types of large proteins present in ECM are collagen,

laminin, elastin and fibronectin. This statement is CORRECT. These

are some of the most abundant and functionally important structural

and adhesive proteins that constitute the large protein component of

the ECM.

Therefore, the incorrect statement is (2) because GAGs are linear

(unbranched) polymers and their repeating disaccharide units are

not typically composed of sialic acid and simple glucose or galactose

in the way described.

Why Not the Other Options?

❌

(1) Proteoglycans are a subset of secreted or cell surface-

attached glycoproteins containing covalently linked specialized

polysaccharide chains called glycosamino- glycans (GAGs). –

Correct definition of proteoglycans.

❌

(3) Major types of GAGs present in ECM are heparan sulphate,

chondroitin sulphate, dermatan sulphate, keratan sulphate and

hyaluronan. – Correct list of major GAGs in ECM.

❌

(4) Major types of large proteins present in ECM are collagen,

laminin, elastin and fibronectin. – Correct list of major large

proteins in ECM

133. Given below are a list of some extracellular matrix

(ECM) proteins in column A and their characteristics

in column B, but not in correct order:

Which one of the following is the most appropriate

match?

(1) A - i, B - ii, C - iii, D - iv

(2) A - ii, B - iii, C - iv, D - i

(3) A- iii, B - ii, C- iv, D- i

(4) A - iv, B - i, C - ii, D – iii

(2017)

Answer: (3) A- iii, B - ii, C- iv, D- i

Explanation:

Let's match the ECM proteins with their correct

characteristics:

A. Fibronectin - (iii) Dimers of two similar polypeptides linked at

their C-termini by two disulfide bonds and contain RGD sequence for

binding to certain integrins. Fibronectin is indeed a large

glycoprotein composed of two similar subunits linked by disulfide

bonds near their C-termini. A key feature is the presence of the

Arginine-Glycine-Aspartic acid (RGD) sequence, which is

recognized by integrins, mediating cell adhesion to the ECM.

B. Laminin - (ii) Heterotrimeric protein comprising α, β and γ chains

and many of them are large, cross-shaped proteins. Laminins are a

major component of the basal lamina and are heterotrimeric proteins

composed of different α, β, and γ subunits. They are large, cross-

shaped molecules that play crucial roles in cell adhesion, migration,

and tissue organization.

C. Nidogen - (iv) A rod-like molecule also called enactin that cross-

links with many ECM proteins and also stabilizes basal laminae.

Nidogen (also known as entactin) is a sulfated glycoprotein that acts

as a bridging molecule in the basal lamina. It has a rod-like

structure and binds to several other ECM components, including

laminin and collagen IV, thus contributing to the stability and

structural integrity of the basal lamina.

D. Collagen - (i) Trimeric protein made from three polypeptides

which can twist together into a special triple helix. Collagens are a

large family of structural proteins characterized by their unique

triple helical structure. This helix is formed by three polypeptide

chains (alpha chains) that wind around each other.

Therefore, the correct matching is A-iii, B-ii, C-iv, and D-i.

Why Not the Other Options?

❌

(1) A - i, B - ii, C - iii, D - iv – Incorrect; Fibronectin is a dimer,

not a trimer with a triple helix. Collagen has the triple helix structure.

Nidogen is rod-like, but not primarily known for being a dimer with

RGD.

❌

(2) A - ii, B - iii, C - iv, D - i – Incorrect; Fibronectin is a dimer

with RGD, not a heterotrimer. Laminin is the heterotrimer. Nidogen

is rod-like and cross-linking. Collagen has the triple helix.

❌

(4) A - iv, B - i, C - ii, D – iii – Incorrect; Fibronectin is a dimer

with RGD. Laminin is the heterotrimer. Nidogen is rod-like.

Collagen has the triple helix

134. Somatic cell hybridization is used to assign a gene to

a particular chromosome. When two cell lines from

two different species are fused, they form a

heterokaryon which tends to Jose chromosomes as

they divide, preferentially from one species. A panel

of cell lines was created from mouse-monkey somatic

cell fusions. Each line was examined for the presence

of monkey chromosomes and for the production of a

given enzyme. The following results were obtained:

On the basis of these results, which chromosome has

the gene that codes for the given enzyme?

(1) Chromosome 10

(2) Chromosome 7

(3) Chromosome 1

(4) Chromosome 5

(2017)

Answer: (2) Chromosome 7

Explanation:

To determine which chromosome carries the gene for

the enzyme, we need to find a chromosome whose presence or

absence consistently correlates with the presence or absence of the

enzyme across the different cell lines.

Let's examine each cell line and compare the presence of the enzyme

with the presence of each monkey chromosome:

Cell Line A: Enzyme is present (+). Monkey chromosomes present: 1,

2, 3, 4, 7, 9, 10.

Cell Line B: Enzyme is present (+). Monkey chromosomes present: 2,

3, 5, 6, 8, 9, 10.

Cell Line C: Enzyme is absent (-). Monkey chromosomes present: 2,

8, 9, 10.

Cell Line D: Enzyme is present (+). Monkey chromosomes present: 1,

2, 3, 6, 7, 8, 9.

Cell Line E: Enzyme is absent (-). Monkey chromosomes present: 4,

5, 8, 9.

Cell Line F: Enzyme is present (+). Monkey chromosomes present: 1,

2, 3, 4, 5, 6, 7.

Now let's look for a consistent pattern:

Chromosome 1: Present in A, D, F (Enzyme +), absent in B, C, E

(Enzyme + or -). No consistent correlation.

Chromosome 2: Present in A, B, C, D, F (Enzyme + or -), absent in

E (Enzyme -). No consistent correlation.

Chromosome 3: Present in A, B, D, F (Enzyme +), absent in C, E

(Enzyme -). No consistent correlation.

Chromosome 4: Present in A, F (Enzyme +), absent in B, C, D, E

(Enzyme + or -). No consistent correlation.

Chromosome 5: Present in B, E, F (Enzyme + or -), absent in A, C,

D (Enzyme + or -). No consistent correlation.

Chromosome 6: Present in B, D, F (Enzyme +), absent in A, C, E

(Enzyme + or -). No consistent correlation.

Chromosome 7: Present in A, D, F (Enzyme +), absent in B, C, E

(Enzyme -). Consistent positive correlation: Enzyme is present

whenever Chromosome 7 is present, and absent when Chromosome 7

is absent.

Chromosome 8: Present in B, C, D, E (Enzyme + or -), absent in A,

F (Enzyme +). No consistent correlation.

Chromosome 9: Present in A, B, C, D, E (Enzyme + or -), absent in

F (Enzyme +). No consistent correlation.

Chromosome 10: Present in A, B, C (Enzyme + or -), absent in D, E,

F (Enzyme + or -). No consistent correlation.

The only chromosome whose presence perfectly correlates with the

presence of the enzyme is Chromosome 7. Therefore, the gene that

codes for the given enzyme is most likely located on Chromosome 7.

Why Not the Other Options?

(1) Chromosome 10: The enzyme is present in cell lines A and B

where chromosome 10 is present, but absent in cell lines D, E, and F

where chromosome 10 is absent (or present in F with enzyme

present). This is not a perfect correlation.

(3) Chromosome 1: The enzyme is present in cell lines A, D, and F

where chromosome 1 is present, but also present in cell line B where

chromosome 1 is absent. This is not a perfect correlation.

(4) Chromosome 5: The enzyme is present in cell lines B and F

where chromosome 5 is present, but absent in cell lines A, C, and D

where chromosome 5 is absent. However, the enzyme is also absent

in cell line E where chromosome 5 is present. This is not a perfect

correlation.

135. Metachromatic leukodystrophy (MLD) is caused by a

deficiency of arylsulfatase A and affects the CNS.

MLD is

(1) a lysosomal storage disorder

(2) a disease due to dysfunctional mitochondria

(3) caused by loss of the myelin sheath

(4) caused by a defect in proteins of the nuclear envelope

(2017)

Answer: (1) a lysosomal storage disorder

Explanation:

Metachromatic leukodystrophy (MLD) is indeed a

lysosomal storage disorder. Here's why:

Arylsulfatase A Deficiency: MLD is characterized by a deficiency in

the lysosomal enzyme arylsulfatase A. This enzyme is crucial for the

breakdown of sulfatides, a type of lipid found in myelin, the

insulating sheath around nerve fibers in the central and peripheral

nervous systems.

Lysosomal Storage: Due to the deficiency of arylsulfatase A,

sulfatides cannot be properly degraded within lysosomes. This leads

to the accumulation of sulfatides in lysosomes of various cells,

including oligodendrocytes (which produce myelin) and neurons in

the central nervous system (CNS), as well as in other organs. This

accumulation disrupts normal cellular function and leads to the

progressive neurological symptoms characteristic of MLD.

Let's look at why the other options are incorrect:

(2) a disease due to dysfunctional mitochondria: While

mitochondrial dysfunction can cause neurological disorders, MLD's

primary defect lies in a lysosomal enzyme and the resulting storage

of lipids in lysosomes, not in the mitochondria themselves.

(3) caused by loss of the myelin sheath: While MLD ultimately leads

to demyelination (loss of the myelin sheath) due to the toxic

accumulation of sulfatides affecting the cells that produce and

maintain myelin, the primary cause is the enzymatic deficiency within

lysosomes that precedes and triggers the demyelination.

(4) caused by a defect in proteins of the nuclear envelope: Defects in

nuclear envelope proteins cause a different set of disorders known as

laminopathies. MLD's genetic defect affects a lysosomal enzyme, not

the structural proteins of the nuclear envelope

Therefore, the fundamental problem in metachromatic

leukodystrophy is the inability of lysosomes to degrade sulfatides due

to a deficient enzyme, classifying it as a lysosomal storage disorder.

Why Not the Other Options?

❌

(2) a disease due to dysfunctional mitochondria – Incorrect; The

primary defect is in a lysosomal enzyme.

❌

(3) caused by loss of the myelin sheath – Incorrect;

Demyelination is a consequence of the primary defect.

❌

(4) caused by a defect in proteins of the nuclear envelope –

Incorrect; The defect is in a lysosomal enzyme

136. Which one of the following statements is NOT true?

(1) Beta-oxidation of long chain fatty acids occurs in

mitochondria

(2) Fatty acid biosynthesis occurs in peroxisomes

(3) Peroxisomes utilize H

to oxidize a variety of

substrates

(4) Peroxisomes import their repertoire of proteins using

sorting signals

(2017)

Answer: (2) Fatty acid biosynthesis occurs in peroxisomes

Explanation:

Let's evaluate each statement regarding fatty acid

metabolism and peroxisomes:

(1) Beta-oxidation of long chain fatty acids occurs in mitochondria:

This statement is true. Beta-oxidation, the process of breaking down

fatty acids into acetyl-CoA molecules to generate energy, primarily

occurs in the mitochondrial matrix, especially for short and medium-

chain fatty acids. Very long-chain fatty acids undergo initial beta-

oxidation in peroxisomes before being further processed in

mitochondria.

(2) Fatty acid biosynthesis occurs in peroxisomes: This statement is

NOT true. Fatty acid biosynthesis, the process of synthesizing fatty

acids from acetyl-CoA, primarily occurs in the cytosol in eukaryotes.

The enzymes involved in fatty acid synthesis, such as fatty acid

synthase, are located in the cytoplasm. While peroxisomes are

involved in some aspects of lipid metabolism, such as the synthesis of

ether lipids and the elongation of very long-chain fatty acids, the de

novo synthesis of fatty acids does not occur there.

(3) Peroxisomes utilize

to oxidize a variety of substrates: This

statement is true. Peroxisomes contain enzymes called oxidases that

produce hydrogen peroxide (

) as a byproduct of oxidizing

various substrates, including very long-chain fatty acids, D-amino

acids, and uric acid. They also contain the enzyme catalase, which

breaks down

into water and oxygen, preventing its toxic

buildup in the cell.

(4) Peroxisomes import their repertoire of proteins using sorting

signals: This statement is true. Peroxisomal proteins are synthesized

in the cytoplasm and are then imported into the peroxisome. This

import process is mediated by specific targeting signals called

peroxisomal targeting signals (PTS1 and PTS2) present in the

proteins, which are recognized by cytosolic receptors that facilitate

translocation across the peroxisomal membrane.

Therefore, the statement that is NOT true is that fatty acid

biosynthesis occurs in peroxisomes.

Why Not the Other Options?

❌

(1) Beta-oxidation of long chain fatty acids occurs in

mitochondria – True statement.

❌

(3) Peroxisomes utilize

to oxidize a variety of substrates – True statement.

❌

(4) Peroxisomes import their repertoire of proteins using sorting

signals – True statement.

137. Which one of the following pairs is NOT matched

correctly?

(1) Glycocalyx – adherence

(2) Fimbriae - motility

(3) Pili - conjugation

(4) Peptidoglycan - cell wall

(2017)

Answer: (2) Fimbriae - motility

Explanation:

Let's examine the function of each bacterial structure:

(1) Glycocalyx – adherence: The glycocalyx is a layer of

polysaccharides and/or polypeptides surrounding the bacterial cell

wall. It can exist as a capsule (firmly attached) or a slime layer

(loosely attached). A key function of the glycocalyx is adherence to

surfaces, including host tissues and medical devices. Therefore, this

pair is correctly matched.

(2) Fimbriae - motility: Fimbriae are short, bristle-like protein

appendages that extend from the bacterial cell surface. Their

primary role is adherence to host cells and other surfaces. While

some specialized types of pili (type IV pili) can be involved in a

twitching form of motility, fimbriae themselves are not directly

involved in motility in the traditional sense (swimming). Therefore,

this pair is NOT correctly matched.

(3) Pili - conjugation: Pili are longer, hair-like appendages present

on the surface of some bacteria. One important type of pili, the

conjugative pilus (or sex pilus), is essential for conjugation, the

transfer of genetic material (usually a plasmid) between bacterial

cells. Therefore, this pair is correctly matched.

(4) Peptidoglycan - cell wall: Peptidoglycan is a polymer composed

of sugars and amino acids that forms a mesh-like layer outside the

plasma membrane of most bacteria. It is the primary structural

component of the bacterial cell wall, providing rigidity and

protection against osmotic lysis. Therefore, this pair is correctly

matched.

The pair that is NOT matched correctly is Fimbriae - motility.

Why Not the Other Options?

❌

(1) Glycocalyx – adherence – Correctly matched.

❌

(3) Pili - conjugation – Correctly matched.

❌

(4) Peptidoglycan - cell wall – Correctly matched.

138. A membrane associated protein is composed of seven

"α-helices", with each helix containing 19

hydrophobic residues. While treating the membrane

with all kinds of proteases, a major portion of this

protein remains intact. Treatment with high salt (till

1.5M NaCl) and buffer with pH 5.0 failed to

dissociate this protein from the membrane. Predict

the most appropriate nature and orientation of this

protein in the membrane.

(1) Peripheral glycoprotein

(2) Integral protein with seven membrane spanning

regions

(3) Peripheral protein with both N and C - terminals

remain exposed to outer surface of the cell membrane

(4) Peripheral protein with both N and C - terminal

remain exposed to cytosolic surface of the cell

membrane

(2017)

Answer: (2) Integral protein with seven membrane spanning

regions

Explanation:

Let's analyze the clues provided in the question to

deduce the nature and orientation of the membrane-associated

protein:

Seven "α-helices," each containing 19 hydrophobic residues: Alpha-

helices with a significant stretch of hydrophobic amino acids (around

19-20 residues are typically needed to span the hydrophobic core of

a biological membrane) are characteristic of transmembrane

domains in integral membrane proteins. The presence of seven such

helices strongly suggests that this protein crosses the membrane

multiple times.

Major portion of the protein remains intact after protease treatment:

Proteases are enzymes that degrade proteins by cleaving peptide

bonds. If a major portion of this protein remains intact despite

treatment with "all kinds of proteases," it implies that this major

portion is shielded or inaccessible to the proteases. Membrane-

spanning regions of integral proteins are protected within the

hydrophobic environment of the lipid bilayer, making them resistant

to degradation by water-soluble proteases.

Failure to dissociate with high salt (1.5M NaCl) and buffer with pH

5.0: Peripheral membrane proteins are typically associated with the

membrane surface through electrostatic interactions (ionic bonds)

with charged lipid head groups or with other membrane proteins.

High salt concentrations can disrupt these ionic interactions by

screening the charges, leading to dissociation. Similarly, changes in

pH can alter the ionization state of amino acid residues involved in

these interactions, also potentially causing dissociation. The fact that

this protein remained associated with the membrane under these

conditions suggests it is not a peripheral protein relying primarily on

electrostatic interactions. Integral membrane proteins, on the other

hand, are embedded within the hydrophobic core of the lipid bilayer

through strong hydrophobic interactions, which are not disrupted by

high salt or moderate pH changes.

Considering these points:

The presence of seven hydrophobic α-helices strongly indicates that

the protein spans the membrane multiple times.

The resistance to protease treatment suggests that a significant

portion of the protein is embedded within the membrane.

The inability to dissociate with high salt and altered pH implies

strong, non-ionic interactions with the membrane, characteristic of

the hydrophobic interactions of an integral membrane protein.

Therefore, the most appropriate description of this protein is an

integral protein with seven membrane-spanning regions. The

orientation of the N and C termini (whether exposed to the outer or

cytosolic surface) cannot be definitively determined from the given

information alone.

Why Not the Other Options?

❌

(1) Peripheral glycoprotein – Incorrect; Peripheral proteins are

typically associated with the membrane surface and are more

susceptible to dissociation by high salt or pH changes. While this

protein could be glycosylated, the seven hydrophobic helices point

towards an integral nature.

❌

(3) Peripheral protein with both N and C - terminals remain

exposed to outer surface of the cell membrane – Incorrect; The

resistance to protease suggests a major portion is protected by the

membrane, not fully exposed. Also, peripheral proteins are generally

easier to dissociate.

❌

(4) Peripheral protein with both N and C - terminal remain

exposed to cytosolic surface of the cell membrane – Incorrect;

Similar to option 3, the protease resistance and strong membrane

association argue against a peripheral nature with exposed termini.

139. In E. coli grown under nutrient rich conditions,

replication of entire genome takes about 40 min., yet

it can divide every 20 min. This is so because:

(1) While E. coli divides every 20 min, equal transfer of

genetic material occurs only in the alternate rounds of

cell divisions.

(2) A second round of genome replication begins before

the completion of first round of replication, and by the

time cell is ready to divide, two copies of the genome are

available.

(3) Genome replication cell division are not coordinated

with each other.

(4) During cell division, only one of the strands of the

genome whose synthesis can be achieved in 20 min, is

transferred to the daughter cell.

(2017)

Answer: (2) A second round of genome replication begins

before the completion of first round of replication, and by the

time cell is ready to divide, two copies of the genome are

available.

Explanation:

The discrepancy between the 40-minute replication

time of the entire E. coli genome and the 20-minute cell division time

is resolved by a phenomenon called overlapping rounds of

replication.

Here's how it works:

Under nutrient-rich conditions, E. coli grows rapidly. To keep pace

with the short generation time (20 minutes), a new round of DNA

replication is initiated at the origin of replication (oriC) before the

previous round has been completed. This means that when the cell is

ready to divide at 20 minutes, the chromosome has already

undergone more than one round of replication. Consequently,

multiple replication forks are active simultaneously on the same

chromosome. By the time a cell division occurs, the regions of the

chromosome near the origin have been replicated more than once,

ensuring that each daughter cell receives a complete copy of the

genome.

Why Not the Other Options?

❌

(1) While E. coli divides every 20 min, equal transfer of genetic

material occurs only in the alternate rounds of cell divisions. –

Incorrect; Every daughter cell receives a complete copy of the

genome at each division. There is no skipping of genetic material

transfer in alternate rounds.

❌

(3) Genome replication and cell division are not coordinated with

each other. – Incorrect; While the timing is overlapping, genome

replication and cell division are tightly coordinated processes. Cell

division is triggered only after sufficient chromosome replication and

segregation have occurred.

❌

(4) During cell division, only one of the strands of the genome

whose synthesis can be achieved in 20 min, is transferred to the

daughter cell. – Incorrect; Each daughter cell receives a complete

double-stranded copy of the entire genome. The synthesis of a single

strand in 20 minutes is not sufficient for a functional genome.

140. Preventing the blocking action of Patched protein

leads to activation of Cos-2, which dissociates itself

from microtubules, activates Ci/Gli which binds to

CBP (CREB - binding protein) and promotes

transcription of target genes. Which one of the

following treatment of cells will mostly prevent Ci/Gli

activated transcription in the cells?

(1) Small molecules which target Frizzled.

(2) Azepine, an inhibitor of γ-secretase.

(3) Cyclopamine, which binds to heptahelical bundle of

Smoothened.

(4) Cdk blockers, which negatively regulate

TGFβinduced growth.

(2017)

Answer: (3) Cyclopamine, which binds to heptahelical bundle

of Smoothened.

Explanation:

The provided description outlines the canonical

Hedgehog (Hh) signaling pathway. Let's break down the steps and

how each treatment would affect it:

Absence of Hh ligand: Patched (Ptc) protein, a transmembrane

receptor, inhibits the activity of Smoothened (Smo), another

transmembrane protein. Consequently, Cos-2 is associated with

microtubules and phosphorylates Ci/Gli transcription factors,

leading to their proteolytic processing and repression of target gene

transcription.

Presence of Hh ligand: Hh ligand binds to Ptc, relieving its

inhibition on Smo. Smo then becomes active and signals, leading to

the dissociation of Cos-2 from microtubules.

Activated Smo signaling: Dissociated Cos-2 no longer efficiently

phosphorylates Ci/Gli. Instead, Ci/Gli is activated (often involving

phosphorylation by other kinases) and translocates to the nucleus.

Nuclear Ci/Gli: In the nucleus, activated Ci/Gli binds to co-

activators like CBP (CREB-binding protein) to promote the

transcription of target genes.

Now let's analyze the effect of each treatment on this pathway:

(1) Small molecules which target Frizzled: Frizzled (Fzd) receptors

are primarily associated with the Wnt signaling pathway, not the

Hedgehog pathway described in the prompt. Targeting Frizzled

would likely affect Wnt signaling but not directly prevent Ci/Gli

activated transcription in the Hh pathway as described.

(2) Azepine, an inhibitor of γ-secretase: γ-secretase is involved in the

processing of various transmembrane proteins, including Notch.

While there can be crosstalk between signaling pathways, γ-secretase

is not a direct upstream regulator of Smoothened or Ci/Gli activation

in the canonical Hedgehog pathway described. Inhibiting it would

likely affect Notch signaling more directly.

(3) Cyclopamine, which binds to heptahelical bundle of Smoothened:

Smoothened (Smo) is essential for transducing the Hedgehog signal

downstream of Patched. When Hh binds to Patched, Patched no

longer inhibits Smo, allowing Smo to become active. Cyclopamine is

a known inhibitor of the Hedgehog pathway that directly binds to the

heptahelical bundle of Smoothened, preventing its activation even

when Patched is not inhibiting it. Therefore, treating cells with

cyclopamine would block Smo activity, preventing the dissociation of

Cos-2 (in the context of the prompt's initial condition of Ptc blocking

action being prevented), hindering Ci/Gli activation, and thus mostly

preventing Ci/Gli activated transcription.

(4) Cdk blockers, which negatively regulate TGFβ-induced growth:

Cyclin-dependent kinases (Cdks) are primarily involved in cell cycle

regulation and can also play roles in other signaling pathways,

including TGFβ signaling. While Cdks might indirectly influence

various cellular processes, they are not a direct and primary target

for preventing Ci/Gli activation in the immediate context of the

described Hedgehog pathway.

Therefore, the treatment that would most directly and effectively

prevent Ci/Gli activated transcription, given the initial conditions of

Patched blocking action being prevented (implying Smo activation),

is cyclopamine, which directly inhibits Smoothened.

Final Answer: The final answer is

Cyclopamine,whichbindstoheptahelicalbundleofSmoothened.

141. Two steroid hormone receptors X and Y both contain

a ligand binding domain. Using recombinant DNA

technology, a modified hybrid receptor H is prepared

such that it contains the ligand binding domain of X

and DNA binding domain of Y. three sets of cells

overexpressing receptors X, Y and H were then

treated separately either with hormone X or with

hormone Y. assuming that there is no cross–

reactivity, which one of the following graphs best

represent the receptor – ligand binding in each case?

(1) Fig 1

(2) Fig 2

(3) Fig 3

(4) Fig 4

(2016)

Answer: (3) Fig 3

Explanation:

The experiment investigates the ligand binding

specificity of two steroid hormone receptors (X and Y) and a hybrid

receptor (H). Receptor H has the ligand-binding domain of X and the

DNA-binding domain of Y. The key assumption is that there is no

cross-reactivity, meaning Hormone X will only bind to receptors

containing the ligand-binding domain of X (receptor X and receptor

H), and Hormone Y will only bind to receptors containing the ligand-

binding domain of Y (receptor Y).

Let's analyze each receptor and hormone interaction:

Receptor X + Hormone X: Hormone X should bind to its cognate

receptor X. Therefore, we expect significant receptor-ligand binding.

Receptor X + Hormone Y: Since there is no cross-reactivity,

Hormone Y should not bind to receptor X. Therefore, we expect

minimal or no receptor-ligand binding.

Receptor Y + Hormone X: Due to no cross-reactivity, Hormone X

should not bind to receptor Y. Therefore, we expect minimal or no

receptor-ligand binding.

Receptor Y + Hormone Y: Hormone Y should bind to its cognate

receptor Y. Therefore, we expect significant receptor-ligand binding.

Receptor H + Hormone X: Receptor H contains the ligand-binding

domain of X. Therefore, Hormone X should bind to receptor H,

resulting in significant receptor-ligand binding.

Receptor H + Hormone Y: Receptor H contains the ligand-binding

domain of X, not Y. Therefore, Hormone Y should not bind to

receptor H due to the lack of cross-reactivity, resulting in minimal or

no receptor-ligand binding.

Now let's examine Figure 3:

Hormone X:

Receptor X (white bar): Shows high % receptor-ligand binding.

(Correct)

Receptor Y (black bar): Shows very low % receptor-ligand binding.

(Correct)

Receptor H (striped bar): Shows high % receptor-ligand binding.

(Correct, as H has the ligand-binding domain of X)

Hormone Y:

Receptor X (white bar): Shows very low % receptor-ligand binding.

(Correct)

Receptor Y (black bar): Shows high % receptor-ligand binding.

(Correct)

Receptor H (striped bar): Shows very low % receptor-ligand binding.

(Correct, as H lacks the ligand-binding domain of Y)

Figure 3 accurately represents the expected receptor-ligand binding

based on the given information and the assumption of no cross-

reactivity.

Why Not the Other Options?

❌

(1) Fig 1 – Incorrect; Shows Hormone Y binding strongly to

receptor X, which contradicts the no cross-reactivity assumption.

❌

(2) Fig 2 – Incorrect; Shows Hormone X binding strongly to

receptor Y, contradicting the no cross-reactivity assumption. Also

shows Hormone Y binding strongly to receptor H, which is incorrect

as H lacks the ligand-binding domain of Y.

❌

(4) Fig 4 – Incorrect; Shows Hormone X binding weakly to

receptor X, which is unexpected for a cognate ligand-receptor

interaction. Also shows Hormone Y binding strongly to receptor H,

which is incorrect.

142. A protein X is kept inactive state in cytosol as

complexed with protein Y. Under certain stress

stimuli, Y gets phosphorylated resulting in its

proteasomal degradation. X becomes free,

translocates to the nucleus and results in the

transcription of a gene which causes cell death by

apoptosis. Stress stimuli were given to the following

four different cases

Case A: Protein Y has a mutation such that

phosphorylation leading to proteasomal degradation

does not occur.

Case B: Cells are transfected with a gene which

encodes for a protein L that inhibits the translocation

of protein Y to the nucleus.

Case C: Cells are transfected only with a empty

vector used to transfect the gene for protein L

Case D: Cells are treated with Z-VAD–FMK, a broad

spectrum caspase inhibitor

Which of the following graphs best describes the

apoptotic state of the cells in the above cases? Y-axis

represents % apoptotic cells.

(1) Fig 1

(2) Fig 2

(3) Fig 3

(4) Fig 4

(2016)

Answer: (1) Fig 1

Explanation:

The pathway described leads to apoptosis when

protein X is free to translocate to the nucleus. This happens when

protein Y, which normally binds and inactivates X in the cytosol, is

phosphorylated and degraded under stress stimuli.

Let's analyze the apoptotic state in each case:

Case A: Protein Y has a mutation such that phosphorylation leading

to proteasomal degradation does not occur.

In this case, even under stress, protein Y will remain bound to

protein X in the cytosol, preventing X from translocating to the

nucleus and initiating apoptosis. Therefore, we expect a very low

percentage of apoptotic cells.

Case B: Cells are transfected with a gene which encodes for a

protein L that inhibits the translocation of protein X to the nucleus.

Here, even if stress stimuli cause the degradation of protein Y,

freeing protein X, protein L will prevent X from entering the nucleus.

Consequently, the gene transcription leading to apoptosis will not

occur, and we expect a very low percentage of apoptotic cells.

Case C: Cells are transfected only with an empty vector used to

transfect the gene for protein L.

This is the control case. Under stress stimuli, protein Y will be

phosphorylated and degraded, freeing protein X to translocate to the

nucleus and induce apoptosis. We expect a significant percentage of

apoptotic cells.

Case D: Cells are treated with Z-VAD–FMK, a broad-spectrum

caspase inhibitor.

Z-VAD–FMK inhibits caspases, which are the executioner enzymes

of apoptosis. Even if the upstream events leading to apoptosis

(translocation of X to the nucleus and gene transcription) occur, the

final step of cell death will be blocked. Therefore, we expect a very

low percentage of cells showing the hallmarks of apoptosis, although

the initial signals might have been triggered.

Now let's examine Figure 1, where the Y-axis represents % apoptotic

cells and the X-axis represents Cases A, B, C, and D:

Case A: Shows a very low percentage of apoptotic cells. (Consistent

with Y not being degraded)

Case B: Shows a very low percentage of apoptotic cells. (Consistent

with X translocation being inhibited)

Case C: Shows a high percentage of apoptotic cells. (Consistent with

the normal stress response leading to apoptosis)

Case D: Shows a very low percentage of apoptotic cells. (Consistent

with caspase inhibition blocking the final stages of apoptosis)

Figure 1 best describes the apoptotic state of the cells in the given

cases.

Why Not the Other Options?

❌

(2) Fig 2 – Incorrect; Shows high apoptosis in Case A and low

apoptosis in Case C, which is the opposite of what we expect.

❌

(3) Fig 3 – Incorrect; Shows high apoptosis in Cases A and B,

which is incorrect as these conditions prevent X from functioning.

❌

(4) Fig 4 – Incorrect; Shows high apoptosis in Case D, which is

incorrect as caspase inhibition blocks apoptosis

143. Excess oxygen consumed after a vigorous exercise is

(1) to pump out lactic acid from muscle

(2) to increase the concentration of lactic acid in muscle

(3) to reduce dissolved carbon dioxide in blood

(4) to make ATP for gluconeogenesis

(2016)

Answer: (4) to make ATP for gluconeogenesis

Explanation:

After vigorous exercise, the body enters a phase

known as excess post-exercise oxygen consumption (EPOC), where

additional oxygen is used to restore homeostasis. During intense

exercise, anaerobic glycolysis leads to the accumulation of lactic

acid, which needs to be converted back into glucose via

gluconeogenesis in the liver (Cori cycle). This process requires ATP,

which is produced through oxidative phosphorylation, utilizing the

extra oxygen consumed post-exercise. Additionally, oxygen is used to

replenish phosphocreatine stores and restore normal cellular

respiration.

Why Not the Other Options?

❌

(1) to pump out lactic acid from muscle – Incorrect; Lactic acid is

transported to the liver for conversion into glucose, but the oxygen

itself is primarily used to generate ATP for gluconeogenesis, not just

for "pumping it out."

❌

(2) to increase the concentration of lactic acid in muscle –

Incorrect; Post-exercise oxygen consumption helps clear lactic acid,

not increase it. Lactic acid levels decrease as it is metabolized.

❌

(3) to reduce dissolved carbon dioxide in blood – Incorrect; While

oxygen consumption aids in CO₂ removal through respiration, this is

not the primary reason for excess oxygen intake post-exercise

144. Which one of the following describes the primary

function of flipases?

(1) Help in increasing lipid protein interaction in the

outer leaflet of the bilayer

(2) Move certain phospholipids from one leaflet of

the membrane to another

(3) Localize more negatively charged membrane

proteins in the lipid bilayer

(4) Cause uncoupling of v-SNARES and t-SNARES

after fusion of incoming vesicle with target

membrane

(2016)

Answer: (2) Move certain phospholipids from one leaflet of

the membrane to another

Explanation:

Flippases are membrane-associated enzymes that

catalyze the ATP-dependent translocation of specific phospholipids

from the outer leaflet to the inner leaflet of the lipid bilayer. This

process is crucial for maintaining membrane asymmetry, which is

essential for cell signaling, apoptosis, and vesicle trafficking.

Flippases primarily transport phosphatidylserine (PS) and

phosphatidylethanolamine (PE) to the cytoplasmic leaflet, ensuring

proper membrane function and cellular homeostasis.

Why Not the Other Options?

❌

(1) Help in increasing lipid-protein interaction in the outer leaflet

of the bilayer – Incorrect; Flippases primarily function in

phospholipid translocation, not in modifying lipid-protein

interactions.

❌

(3) Localize more negatively charged membrane proteins in the

lipid bilayer – Incorrect; Flippases move phospholipids, not proteins.

The distribution of charged membrane proteins is determined by

other factors such as electrostatic interactions and sorting

mechanisms.

❌

(4) Cause uncoupling of v-SNARES and t-SNARES after fusion of

incoming vesicle with target membrane – Incorrect; SNARE complex

disassembly is mediated by NSF (N-ethylmaleimide-sensitive factor)

and α-SNAP, not by flippases.

145. Mitotic cyclin-CDK activity peaks in M phase. This is

because

(1) Mitotic cyclin is synthesized only in M phase.

(2) Threshold level of mitotic cyclin accumulates only in

late G2.

(3) Cyclin subunit is activated by phosphorylation only

in M phase.

(4) The kinase subunit is activated by dephosphorylation

only in M phase.

(2016)

Answer: (4) The kinase subunit is activated by

dephosphorylation only in M phase

Explanation:

Mitotic cyclin-CDK complexes (specifically cyclin B-

Cdk1) regulate the entry into mitosis. The activity of these complexes

is tightly regulated, with the cyclin subunit synthesized during the

cell cycle and accumulating as the cell progresses from G1 to G2.

However, the kinase subunit (Cdk1) remains inactive until it

undergoes dephosphorylation by Cdc25 phosphatase in late G2, just

before mitosis. This dephosphorylation event activates the cyclin-

CDK complex, triggering the onset of mitosis. The

phosphorylation/dephosphorylation cycle is essential for the precise

regulation of the cell cycle.

Why Not the Other Options?

❌

(1) Mitotic cyclin is synthesized only in M phase – Incorrect;

Mitotic cyclins (cyclin B) are synthesized during G2 phase, not just

in M phase, but their accumulation is necessary to reach a threshold

level that activates CDK1.

❌

(2) Threshold level of mitotic cyclin accumulates only in late G2 –

Incorrect; Although cyclin accumulation peaks in late G2, the

activation of the cyclin-CDK complex occurs after

dephosphorylation, not just through cyclin accumulation.

❌

(3) Cyclin subunit is activated by phosphorylation only in M

phase – Incorrect; Cyclin binding to CDK1 is necessary but not

sufficient for activation. The key activation step for CDK1 occurs

through dephosphorylation of the Cdk1 subunit by Cdc25, not

phosphorylation of the cyclin subunit.

146. Which of the following is NOT an example of

trasmembrane transport between different

subcellular compartments?

(1) Transport from cytoplasm into lumen of the

endoplasmic reticulum

(2) Transport from endoplasmic reticulum to the

Golgi complex

(3) Transport from stroma into thylakoid space

(4) Transport from mitochondrial intermembrane

space into the mitochondrial matrix

(2016)

Answer: (2) Transport from endoplasmic reticulum to the

Golgi complex

Explanation:

The transport between different subcellular

compartments generally involves the movement of proteins, lipids,

and other molecules across membranes. These processes are

typically mediated by vesicular trafficking or transporters.

Option (1): Transport from cytoplasm into the lumen of the

endoplasmic reticulum is a transmembrane transport process.

Proteins are translocated across the ER membrane, often via the

translocon complex.

Option (2): Transport from endoplasmic reticulum (ER) to the Golgi

complex is not a transmembrane transport per se, but involves

vesicular transport. Proteins or lipids are packaged into vesicles that

bud from the ER and then fuse with the Golgi, rather than directly

crossing the membrane.

Option (3): Transport from stroma into thylakoid space in

chloroplasts is transmembrane transport, involving the translocation

of proteins into the thylakoid membrane and space, utilizing the TAT

(twin-arginine translocase) or SEC (secretory) pathways.

Option (4): Transport from mitochondrial intermembrane space into

the mitochondrial matrix involves the movement of proteins across

both the inner and outer mitochondrial membranes, typically

facilitated by complexes like the TIM (translocase of the inner

membrane) and TOM (translocase of the outer membrane).

Thus, transport from the ER to the Golgi is vesicular trafficking

rather than direct transmembrane transport.

Why Not the Other Options?

❌

(1) Transport from cytoplasm into lumen of the endoplasmic

reticulum – Incorrect; This is a classic example of transmembrane

transport, where proteins enter the ER through the translocon.

❌

(3) Transport from stroma into thylakoid space – Incorrect; This

is a transmembrane process where proteins are transported into the

thylakoid via specific protein complexes.

❌

(4) Transport from mitochondrial intermembrane space into the

mitochondrial matrix – Incorrect; This is also a transmembrane

transport, facilitated by the TIM and TOM complexes in the

mitochondria.

147. In a mitochondrial respiration experiment, a

researcher observed the following profile of oxygen

consumption upon addition of following compounds

at times I, II and III. (a) ADP + Pi (b) Dinitrophenol,

an uncoupler (c) Oligomycin, an ATPase inhibitor (d)

Cyanide (e) Succinate

Which of the following describes the profile

appropriately?

(1) I - b; II - d; III – e

(2) I - a; II - d; III - c

(3) I - a; II - e; III – c

(4) I - a; II - c; III - b

(2016)

Answer: (4) I - a; II - c; III - b

Explanation:

The graph shows changes in the rate of oxygen

consumption by mitochondria over time, with the addition of different

compounds at points I, II, and III. Let's analyze the expected effects

of each compound:

ADP + Pi (a): These are the substrates required by ATP synthase to

produce ATP through oxidative phosphorylation. Adding ADP and

inorganic phosphate (Pi) will provide the necessary components for

ATP synthesis, which is coupled to the electron transport chain (ETC)

and oxygen consumption. Therefore, we would expect an increase in

the rate of oxygen consumption upon their addition. This

corresponds to the increase seen at time I.

Oligomycin (c): This drug is an inhibitor of ATP synthase. It blocks

the flow of protons through the FO channel of ATP synthase,

preventing ATP synthesis. Because the ETC and proton pumping are

tightly coupled to ATP synthesis under normal conditions, inhibiting

ATP synthase will cause a decrease in the rate of electron transport

and, consequently, oxygen consumption. This corresponds to the

decrease or plateau seen at time II.

Dinitrophenol (b): Dinitrophenol (DNP) is an uncoupler. It is a

protonophore that can shuttle protons across the inner mitochondrial

membrane, bypassing ATP synthase. This disrupts the proton

gradient, allowing the ETC to proceed at a maximal rate without

being coupled to ATP synthesis. As a result, oxygen consumption

increases dramatically. This corresponds to the significant increase

seen at time III, even after the inhibition by oligomycin.

Considering these effects, the sequence of additions that best

explains the observed profile is:

I - Addition of ADP + Pi (increase in oxygen consumption)

II - Addition of Oligomycin (decrease or plateau in oxygen

consumption)

III - Addition of Dinitrophenol (significant increase in oxygen

consumption)

This matches option (4).

Why Not the Other Options?

❌

(1) I - b; II - d; III – e – Incorrect; Adding an uncoupler (b) at I

would cause an immediate increase, not the observed lag and

gradual increase. Cyanide (d) at II would completely inhibit oxygen

consumption, which is not seen.

❌

(2) I - a; II - d; III - c – Incorrect; While ADP + Pi (a) at I is

consistent, cyanide (d) at II would halt oxygen consumption, and

adding oligomycin (c) at III after that wouldn't cause a further

change as the system would already be blocked.

❌

(3) I - a; II - e; III – c – Incorrect; Succinate (e) is a substrate for

complex II of the ETC and would increase oxygen consumption, not

cause the decrease seen at II. Oligomycin (c) at III would then

decrease it further, which contradicts the final increase.

148. Entry of enveloped viruses into its host cells is

mediated by:

(1) Only endocytosis

(2) Both endocytosis and phagocytosis

(3) Both endocytosis and membrane fusion

(4) Only pinocytosis

(2016)

Answer: (3) Both endocytosis and membrane fusion

Explanation:

Enveloped viruses have an outer lipid bilayer

derived from the host cell membrane during the budding process.

These viruses can enter host cells through two main mechanisms.

Membrane fusion involves the viral envelope fusing directly with the

host cell plasma membrane, releasing the viral capsid into the

cytoplasm. This process is often mediated by specific viral fusion

proteins that undergo conformational changes triggered by receptor

binding or changes in pH. Endocytosis is another entry route where

the host cell membrane engulfs the virus, forming a vesicle

containing the virus. This vesicle is then transported into the cell. In

some cases, the viral envelope will then fuse with the endosomal

membrane to release the viral contents into the cytoplasm. Therefore,

both membrane fusion and endocytosis are crucial mechanisms for

the entry of enveloped viruses into their host cells.

Why Not the Other Options?

❌

(1) Only endocytosis – Incorrect; Many enveloped viruses enter

cells via direct fusion with the plasma membrane.

❌

(2) Both endocytosis and phagocytosis – Incorrect; Phagocytosis

is primarily used by immune cells to engulf large particles like

bacteria and cellular debris, not typically the primary entry route for

viruses in general cells.

❌

(4) Only pinocytosis – Incorrect; Pinocytosis is a non-specific

form of endocytosis ("cell drinking") and while viruses might be

taken up this way, specific receptor-mediated endocytosis and

membrane fusion are more common and efficient mechanisms for

enveloped virus entry.

149. Labelling of membrane spanning domain of any

integral membrane protein in a given plasma

membrane vesicle (without disrupting its structure) is

successfully carried out by

(1) immunochemical methods.

(2) metabolic labelling with radioisotopes.

(3) hydrophobic photoaffinity labelling.

(4) limited proteolysis followed by metabolic labelling.

(2016)

Answer: (3) hydrophobic photoaffinity labelling

Explanation:

Hydrophobic photoaffinity labeling is a powerful

technique specifically designed to label the membrane-spanning

domains of integral membrane proteins within their native lipid

bilayer environment without disrupting the vesicle structure. This

method utilizes a hydrophobic probe that can partition into the lipid

bilayer and contains a photoactivatable group. Upon activation with

UV light, the probe forms covalent bonds with the hydrophobic

amino acid residues within the membrane-spanning domains of the

integral proteins it is in close proximity to. This allows for direct and

specific labeling of these buried regions.

Immunochemical methods typically target extracellular or

cytoplasmic domains of membrane proteins, as antibodies are

generally large and water-soluble and cannot easily access the

hydrophobic core of the membrane. Metabolic labeling with

radioisotopes involves incorporating radioactive precursors into

newly synthesized proteins, labeling the entire protein, not

specifically the transmembrane domain. Limited proteolysis followed

by metabolic labeling would involve digesting accessible protein

regions and then labeling newly synthesized proteins, which doesn't

directly target pre-existing transmembrane domains within intact

vesicles.

Why Not the Other Options?

❌

(1) immunochemical methods – Incorrect; Antibodies are

generally hydrophilic and cannot easily access and label the

hydrophobic transmembrane domains within an intact membrane

vesicle.

❌

(2) metabolic labelling with radioisotopes – Incorrect; This

method labels newly synthesized proteins throughout their structure,

not specifically the transmembrane domains of existing proteins.

❌

(4) limited proteolysis followed by metabolic labelling – Incorrect;

Limited proteolysis would require opening or disrupting the vesicle

to access protein domains, and subsequent metabolic labeling

wouldn't specifically target the transmembrane regions of the

original proteins.

150. Both sphingomyelin and phosphoglycerides are

phospholipids. Which one of the following statements

is NOT correct?

(1) While one has a fatty acid tail attached via an ester

bond, in another, the fatty acid tail is attached via an

amide bond.

(2) The hydrophilicity of both is dependent on the

phosphate group and other head groups attached to the

phosphate group.

(3) Only one of them may contain a carbon-carbon

double bond (C=C).

(4) Both may have choline as head group.

(2016)

Answer: (3) Only one of them may contain a carbon-carbon

double bond (C=C).

Explanation:

Both the fatty acid chains in phosphoglycerides and

the fatty acid chain (or its derivative) in sphingomyelin can contain

one or more carbon-carbon double bonds. The presence and number

of these unsaturated bonds contribute to the fluidity of the lipid

bilayer. Phosphoglycerides commonly have unsaturated fatty acid

tails, and sphingosine, the backbone of sphingolipids like

sphingomyelin, itself contains a trans double bond. Additionally, the

fatty acyl chain linked to sphingosine can also be saturated or

unsaturated. Therefore, both types of phospholipids can indeed

contain C=C double bonds.

Why Not the Other Options?

❌

(1) While one has a fatty acid tail attached via an ester bond, in

another, the fatty acid tail is attached via an amide bond – Correct;

In phosphoglycerides, fatty acids are typically esterified to the

glycerol backbone. In sphingomyelin, a fatty acid (or a longer acyl

chain) is linked to the sphingosine backbone via an amide bond.

❌

(2) The hydrophilicity of both is dependent on the phosphate

group and other head groups attached to the phosphate group –

Correct; Phospholipids are amphipathic molecules, meaning they

have both hydrophilic and hydrophobic parts. The phosphate group,

along with any polar head group attached to it (like choline,

ethanolamine, serine, or inositol), contributes to the hydrophilic

nature of both sphingomyelin and phosphoglycerides.

❌

(4) Both may have choline as head group – Correct;

Phosphatidylcholine is a common phosphoglyceride with choline as

its head group. Sphingomyelin also commonly has phosphocholine (a

phosphate group linked to choline) as its polar head group.

151. It is well established that "Band 3" protein of red

blood cell membrane is solely responsible for

Cltransport across membrane. A lysine group in the

Clbinding site of "Band 3" is crucial for this event.

Keeping this in mind what is the most appropriate

way to load and retain a small anionic fluorescent

probe (x) inside the red blood cells (RBCs) suspended

in phosphate buffered saline (PBS), pH 7.4

(1) Incubate the RBCs with x in phosphate buffered

saline (PBS, pH 7.4) at 37°C for 30 min.

(2) Incubate the RBCs with x in PBS at 4°C for 30 min.

(3) Incubate the RBCs with x in Hepes sulphate buffer

(pH 7.4) at 37°C for 30 min.

(4) Incubate the RBCs with x in Hepes sulfate buffer (pH

7.4) at 37°C for 30 min followed by treatment with a

NH2 group modifying agent (covalent modification).

(2016)

Answer: (4) Incubate the RBCs with x in Hepes sulfate buffer

(pH 7.4) at 37°C for 30 min followed by treatment with a

NH2 group modifying agent (covalent modification)

Explanation:

The "Band 3" protein is the primary chloride

transporter across the red blood cell membrane. A crucial lysine

residue in its chloride-binding site is essential for this transport. Our

goal is to load a small anionic fluorescent probe (x) into the RBCs

and retain it.

Options (1), (2), and (3) involve simply incubating the RBCs with the

anionic probe. Given that "Band 3" facilitates chloride (another

anion) transport, it's highly likely that it would also transport our

anionic probe (x) across the membrane, leading to its efflux and

preventing significant retention inside the cells. Temperature might

affect the rate of transport, but it wouldn't fundamentally block it.

Using a different buffer (Hepes sulfate instead of phosphate) in

option (3) wouldn't prevent "Band 3" mediated transport, as the

protein is specific for anions, not solely chloride or dependent on the

specific buffer ions present externally.

Option (4) takes a different approach. Incubating the RBCs with the

anionic probe (x) in a suitable buffer at physiological temperature

would allow some of the probe to enter the cells via "Band 3" or

other less efficient mechanisms. Crucially, this is followed by

treatment with an NH₂ group modifying agent. Lysine residues

contain an NH₂ group in their side chain. By covalently modifying

these NH₂ groups, particularly the crucial lysine residue in the

chloride-binding site of "Band 3", we would effectively inhibit the

transporter's function. This blockage would prevent the efflux of the

anionic fluorescent probe (x) that has already entered the RBCs, thus

ensuring its retention inside the cells.

Why Not the Other Options?

❌

(1) Incubate the RBCs with x in phosphate buffered saline (PBS,

pH 7.4) at 37°C for 30 min – Incorrect; The anionic probe is likely to

be transported out of the RBCs by the functional "Band 3" protein.

❌

(2) Incubate the RBCs with x in PBS at 4°C for 30 min – Incorrect;

Lowering the temperature might slow down transport, but it wouldn't

abolish "Band 3" activity and prevent efflux of the anionic probe

over time.

❌

(3) Incubate the RBCs with x in Hepes sulphate buffer (pH 7.4) at

37°C for 30 min – Incorrect; Changing the buffer doesn't address the

issue of "Band 3" mediated efflux of the anionic probe.

152. The genome of a bacterium is composed of a single

DNA molecule which is 109 bp long. How many moles

of genomic DNA is present in the bacterium?

[Consider Avogadro No = 6 X 10

]

(1) 1/6 X 10

-23

(2) 1/6 X 10

-14

(3) 6 X 10

(4) 6 X 10

(2015)

Answer: (1) 1/6 X 10

-23

Explanation:

The number of moles of a substance is given by the

formula:

Moles = Number of molecules / Avogadro's number

Since a single bacterium contains only one genomic DNA molecule,

the number of DNA molecules present is 1.

Avogadro’s number is 6 × 10²³ molecules per mole. Substituting

these values:

Moles of DNA = 1 / (6 × 10²³) = (1/6) × 10⁻²³

Why Not the Other Options?

❌

(2) 1/6 × 10⁻¹⁴ – Incorrect: This value is incorrect because it

overestimates the number of DNA molecules present in a single

bacterium. The correct exponent should be -23, not -14.

❌

(3) 6 × 10¹⁴ – Incorrect: This suggests an extremely large number

of moles, which is not possible for a single DNA molecule.

❌

(4) 6 × 10²³ – Incorrect: This represents one mole of DNA

molecules, whereas the question asks for the moles of a single DNA

molecule, which is much smaller.

The ionic strength of a 0.2 M Na2HPO4, solution will be

(1) 0.2 M

(2) 0.4 M

(3) 0.6 M

(4) 0.8 M

(2015)

Answer: (3) 0.6 M

Explanation:

The ionic strength (I) of a solution is calculated using the formula:

I = ½ Σ Ci Zi²

where Ci is the concentration of each ion, and Zi is its charge.

For Na₂HPO₄, it dissociates in water as:

Na₂HPO₄ → 2 Na⁺ + HPO₄²⁻

The concentrations and charges of the ions are:

- Na⁺: 0.2 M × 2 = 0.4 M (charge = +1)

- HPO₄²⁻: 0.2 M (charge = -2)

Now, calculating ionic strength:

I = ½ [(0.4 × 1²) + (0.2 × 2²)]

= ½ [(0.4 × 1) + (0.2 × 4)]

= ½ [0.4 + 0.8]

= ½ × 1.2

= 0.6 M

Why Not the Other Options?

❌

(1) 0.2 M – Incorrect: This considers only the concentration of

Na₂HPO₄ without accounting for ionic contributions.

❌

(2) 0.4 M – Incorrect: This underestimates the effect of HPO₄²⁻,

which contributes more due to its charge squared (2² = 4).

❌

(4) 0.8 M – Incorrect: This overestimates the ionic strength by

likely misapplying the formula or incorrectly summing ion

contributions.

153. Glycophorin having one highly hydrophobic domain

is able to span a phospholipid bilayer membrane only

(1) once

(2) twice

(3) thrice

(4) four times

(2015)

Answer: (1) once

Explanation:

Glycophorin is an integral membrane protein found

in erythrocyte membranes. It contains a single highly hydrophobic

transmembrane domain, allowing it to span the phospholipid bilayer

only once. This hydrophobic region anchors the protein within the

membrane, with hydrophilic regions extending into the extracellular

and cytoplasmic spaces. Single-pass transmembrane proteins like

glycophorin are characterized by one stretch of hydrophobic amino

acids that traverse the membrane.

Why Not the Other Options?

❌

(2) twice – Incorrect: A protein must have two hydrophobic

regions to span the membrane twice, but glycophorin has only one.

❌

(3) thrice – Incorrect: Proteins that span the membrane three

times typically have multiple hydrophobic domains, which

glycophorin lacks.

❌

(4) four times – Incorrect: Multi-pass transmembrane proteins

have multiple hydrophobic domains, whereas glycophorin is a

single-pass transmembrane protein.

154. A cell line deficient in salvage pathway for nucleotide

biosynthesis was fed with medium containing

labelled amino acids. Purines were then extracted.

Treatment with which one of the following amino

acids is likely to produce

N labelled purines?

(1) Aspartic acid

(2) Glycine

(3) Glutamine

(4) Aspartamine

(2015)

Answer: (3) Glutamine

Explanation:

Purine biosynthesis occurs through the *de novo*

pathway, which incorporates nitrogen atoms from specific amino

acids. Glutamine plays a crucial role as a nitrogen donor at multiple

steps in purine synthesis. It contributes nitrogen at two key positions

in the purine ring—once during the conversion of PRPP to 5-

phosphoribosylamine and again in the formation of GMP. Since the

cell line is deficient in the salvage pathway, it relies entirely on *de

novo* synthesis, meaning labeled **¹⁵N-glutamine** will result in

the incorporation of ¹⁵N into purines.

Why Not the Other Options?

❌

(1) Aspartic acid – Incorrect: Aspartic acid contributes a nitrogen

to the purine ring but not as directly or significantly as glutamine. It

primarily donates nitrogen in the conversion of IMP to AMP,

meaning it would have limited influence on purine labeling.

❌

(2) Glycine – Incorrect: Glycine provides carbon and nitrogen to

the purine ring, but its nitrogen is not derived from an amide group

like in glutamine, making it a less direct source of labeled nitrogen in

this case.

❌

(4) Aspartamine – Incorrect: "Aspartamine" is not a recognized

amino acid in biochemistry, making this option invalid.

155. Coupling of the reaction centers of oxidative

phosphorylation is achieved by which one of the

following?

(1) Making a complex of all four reaction centers.

(2) Locating all four complexes in the inner

membrane.

(3) Ubiquinones and cytochrome C

(4) Pumping of protons.

(2015)

Answer: (3) Ubiquinones and cytochrome C

Explanation:

In oxidative phosphorylation, the coupling of

reaction centers (Complexes I, II, III, and IV) is primarily achieved

through **electron carriers** like **ubiquinone (coenzyme Q) and

cytochrome c**. These molecules shuttle electrons between different

complexes in the electron transport chain (ETC), enabling sequential

redox reactions that drive proton pumping. Ubiquinone transfers

electrons from Complex I and II to Complex III, while cytochrome c

carries electrons from Complex III to Complex IV, ensuring efficient

coupling of the ETC components.

Why Not the Other Options?

❌

(1) Making a complex of all four reaction centers – Incorrect: The

four complexes are distinct entities and do not physically form a

single complex; instead, they are linked by mobile electron carriers.

❌

(2) Locating all four complexes in the inner membrane –

Incorrect: While the complexes are embedded in the inner

mitochondrial membrane, mere localization does not ensure

functional coupling; electron carriers play the key role.

❌

(4) Pumping of protons – Incorrect: Proton pumping creates the

electrochemical gradient necessary for ATP synthesis, but it does not

directly couple the electron transfer between reaction centers.

156. In eukaryotic replication, helicase loading occurs at

all replicators during

(1) Go phase

(2) G1 phase

(3) S phase

(4) G2 phase

(2015)

Answer: (2) G1 phase

Explanation:

In eukaryotic DNA replication, helicase loading

occurs during the **G1 phase** as part of the **pre-replicative

complex (pre-RC) formation** at origins of replication. The **origin

recognition complex (ORC)** recruits **Cdc6 and Cdt1**, which

help load the **MCM (Minichromosome Maintenance) helicase

complex** onto DNA. This helicase remains inactive until the **S

phase**, when additional factors like DDK and CDK activate it,

initiating replication. Since replication occurs only once per cell

cycle, helicase loading is restricted to G1, ensuring precise

replication control.

Why Not the Other Options?

❌

(1) G₀ phase – Incorrect: Cells in G₀ are in a quiescent state and

do not actively prepare for replication, so helicase loading does not

occur.

❌

(3) S phase – Incorrect: While helicase becomes **activated** in

S phase to unwind DNA, its **loading** occurs earlier in G1.

❌

(4) G₂ phase – Incorrect: By G₂, DNA replication is already

completed, and the cell is preparing for mitosis, making helicase

loading unnecessary.

157. α-Amanitin is a fungal toxin which inhibits

eukaryotic RNA polymerases. The three eukaryotic

RNA polymerases show differential sensitivity to this

toxin. Which one of the following order (higher to

lower) is correct in respect of sensitivity towards

αamanitin?

(1) RNA POL III> RNA POL II > RNA POL I

(2) RNA POL II > RNA POL III > RNA POL I

(3) RNA POL I> RNA POL III > RNA POL II

(4) RNA POL II > RNA POL I > RNA POL III

(2015)

Answer: (2) RNA POL II > RNA POL III > RNA POL I

Explanation:

α-Amanitin is a fungal toxin derived from Amanita

phalloides (the death cap mushroom) that selectively inhibits

eukaryotic RNA polymerases with different sensitivities. RNA

polymerase II, which synthesizes mRNA, is highly sensitive to α-

amanitin, making it the most affected. RNA polymerase III, which

transcribes tRNA and 5S rRNA, is moderately sensitive. RNA

polymerase I, responsible for rRNA synthesis, is resistant to α-

amanitin. This differential sensitivity helps in distinguishing RNA

polymerase functions in biochemical assays.

Why Not the Other Options?

❌

(1) RNA POL III > RNA POL II > RNA POL I – Incorrect: RNA

polymerase III is less sensitive than RNA polymerase II, not more.

❌

(3) RNA POL I > RNA POL III > RNA POL II – Incorrect: RNA

polymerase I is highly resistant, while RNA polymerase II is the most

sensitive, making this order incorrect.

❌

(4) RNA POL II > RNA POL I > RNA POL III – Incorrect: RNA

polymerase I is more resistant than RNA polymerase III, so this

order is incorrect

158. Which isotope below is best suited for metabolic

labeling of glyceraldehyde-3-phospho-dehydrogenase?

(1)

(2)

125

(3)

(4)

131

(2015)

Answer: (1)

Explanation:

Glyceraldehyde-3-phosphate dehydrogenase (GAPDH) is a

metabolic enzyme that plays a crucial role in glycolysis, catalyzing

the conversion of glyceraldehyde-3-phosphate to 1,3-

bisphosphoglycerate. To label GAPDH metabolically, the isotope

must be incorporated into carbon-containing molecules during

biosynthesis. ¹⁴C (Carbon-14) is ideal for this purpose because it can

be incorporated into biomolecules such as amino acids and sugars,

allowing for metabolic tracking of GAPDH in living cells.

Why Not the Other Options?

❌

¹²⁵I (Iodine-125) – Incorrect; Iodine isotopes are primarily used

for radioimmunoassays and thyroid metabolism studies, not for

metabolic labeling of proteins like GAPDH.

❌

³²P (Phosphorus-32) – Incorrect; This isotope labels

phosphorylated compounds such as nucleotides and phospholipids,

but GAPDH is not a phosphoprotein and does not require phosphate

incorporation for metabolic tracking.

❌

¹³¹I (Iodine-131) – Incorrect; Similar to ¹²⁵I, this isotope is mainly

used for thyroid studies and radiotherapy, not for metabolic labeling

of proteins.

159. The following are the statements about pyruvate

kinase (PK).

A. ATP is an allosteric inhibitor of PK

B. Fructose 1,6 bisphosphate is an activator of PK

C. ADP is an allosteric inhibitor of PK

D. Alanine is an allosteric modulator of PK

Which of the above statement(s) are true?

(1) A, B, C

(2) A, B, D

(3) B, C, D

(4) only A

(2015)

Answer: (2) A, B, D

Explanation:

Pyruvate kinase (PK) is a key enzyme in glycolysis

that catalyzes the conversion of phosphoenolpyruvate (PEP) to

pyruvate, generating ATP. It is subject to allosteric regulation to

control glycolytic flux.

(A) ATP is an allosteric inhibitor of PK – True. ATP signals a high-

energy state in the cell and inhibits PK to slow down glycolysis when

energy is abundant.

(B) Fructose 1,6-bisphosphate is an activator of PK – True. This is

an example of feed-forward activation, where an early glycolytic

intermediate (F-1,6-BP) enhances PK activity to promote glycolysis.

substrate in the glycolytic pathway and does not inhibit PK; rather,

low ATP (high ADP/AMP) activates glycolysis to generate more ATP.

(D) Alanine is an allosteric modulator of PK – True. Alanine, which

is synthesized from pyruvate, serves as a feedback inhibitor of PK,

signaling sufficient pyruvate availability and slowing glycolysis when

amino acid synthesis is active.

Why Not the Other Options?

❌

(1) A, B, C – Incorrect; ADP is not an inhibitor.

❌

(3) B, C, D – Incorrect; ADP is not an inhibitor.

❌

(4) Only A – Incorrect; Fructose 1,6-bisphosphate and alanine

also regulate PK

160. Membrane proteins are synthesized on endoplasmic

reticulum and transported to various oraganelle. One

hypothesis for membrane protein sorting is

hydrophobicity matching i.e., the proteins with a

shorter transmembrane portion would partition into

thinner membranes. You are given the following

three observations:

A. It was found that transmembrane portion of

proteins in Golgi membranes are short than those in

plasma membranes

B. Presence of cholesterol increases the thickness of

the bilayer

C. The phospholipid composition of Golgi and plasma

membranes are same Which of the following

statements is correct?

(1) Proteins in plasma membrane have longer

transmembrane portion than proteins in Golgi

membranes

(2) Proteins in Golgi membranes have longer

transmembrane portion than proteins in plasma

membranes

(3) Proteins of both Golgi and plasmamembranes

have same length of transmembrane portion

(4) Cholesterol is more in Golgi membrane than in

plasma membrane

(2015)

Answer: (1) Proteins in plasma membrane have longer

transmembrane portion than proteins in Golgi

membranes

Explanation:

The hydrophobicity matching hypothesis states that

transmembrane proteins tend to match the thickness of the lipid

bilayer they reside in. This means that membranes with a greater

thickness will accommodate proteins with longer transmembrane

domains, while thinner membranes will have proteins with shorter

transmembrane domains.

Now, let's analyze the given observations:

(A) Transmembrane proteins in Golgi membranes are shorter than

those in plasma membranes

→ This suggests that Golgi membranes are thinner, while plasma

membranes are thicker.

(B) Cholesterol increases bilayer thickness

→ More cholesterol in a membrane results in increased thickness.

same

→ Since phospholipid composition is not responsible for thickness

differences, cholesterol must be playing a key role.

From these observations, we can conclude that the plasma

membrane contains more cholesterol than the Golgi membrane,

making it thicker. As a result, proteins in the plasma membrane must

have longer transmembrane domains to match the increased bilayer

thickness.

Why Not the Other Options?

❌

(2) Proteins in Golgi membranes have longer transmembrane

portions than those in plasma membranes – Incorrect; Golgi

membranes are thinner, so their proteins have shorter

transmembrane domains.

❌

(3) Proteins of both Golgi and plasma membranes have the same

transmembrane length – Incorrect; they differ, with plasma

membrane proteins being longer.

❌

(4) Cholesterol is more in Golgi membrane than in plasma

membrane – Incorrect; plasma membranes have more cholesterol,

making them thicker

161. A culture medium contains two carbon sources, one is

preferred carbon source (glucose) and the second is a

non-preferred source (lactose). Which one below is

correct regarding the nature of growth curve of E.

coli cultured in this medium?

(1) Growth curve will be same as when grown in

presence of only glucose.

(2) Growth curve will be same as when grown in

presence of only lactose.

(3) A lag phase will be observed between the two

exponential phases.

(4) Two lag phases will be observed between the two

exponential phases.

(2015)

Answer: (3) A lag phase will be observed between the two

exponential phases

Explanation:

When E. coli is cultured in a medium containing both

glucose (preferred carbon source) and lactose (non-preferred carbon

source), it undergoes diauxic growth. Initially, E. coli preferentially

metabolizes glucose, leading to the first exponential phase. Once

glucose is exhausted, the bacteria experience a lag phase while they

induce the lac operon and synthesize enzymes (like β-galactosidase

and permease) required for lactose metabolism. After this adjustment,

they enter a second exponential phase utilizing lactose as a carbon

source.

Why Not the Other Options?

❌

(1) Growth curve will be the same as when grown in presence of

only glucose – Incorrect; If only glucose were present, there would

be no second exponential phase because lactose metabolism wouldn't

occur.

❌

(2) Growth curve will be the same as when grown in presence of

only lactose – Incorrect; If only lactose were present, the growth

curve would show a single exponential phase, but in diauxic growth,

there are two exponential phases separated by a lag phase.

❌

(4) Two lag phases will be observed between the two exponential

phases – Incorrect; There is only one lag phase after glucose

depletion, during which the bacteria switch to lactose metabolism.

162. The following diagram shows meiotic pairing in an

inversion heterozygote and a point where single

crossing over has occurred

The resulting gametes produced may have

A. the chromosome having normal gene sequence

B. the chromosome having inverted gene sequence

C. a dicentric chromosome with duplication and

deletion

D. an acentric chromosome having duplication and

deletion

E. the chromosome having duplication and deletion

Which of the following combination will be most

appropriate for the diagram shown:

(1) A, B, C and D

(2) A, B and E

(3) B, C, D and E

(4) A, C, D and E

(2015)

Answer: (2) A, B and E

Explanation:

The given diagram represents meiotic pairing in an

inversion heterozygote with a single crossover event occurring within

the inverted segment. This scenario leads to different types of

resulting gametes.

Chromosome with Normal Gene Sequence (A):

✅

Correct. Some

gametes will inherit the normal chromosome without any structural

changes.

Chromosome with Inverted Gene Sequence (B):

✅

Correct. Some

gametes will inherit the inverted chromosome, which is structurally

intact but has a reversed gene sequence.

Chromosome with Duplication and Deletion (E):

✅

Correct. A

single crossover event within an inversion loop can lead to the

production of chromosomes with both duplications and deletions, as

genetic material is unevenly exchanged.

Why Not the Other Options?

❌

Incorrect; dicentric chromosomes (with two centromeres) arise only

when a paracentric inversion (inversion outside the centromere)

undergoes crossover. The given diagram does not show a dicentric

chromosome.

❌

(D) Acentric Chromosome with Duplication and Deletion:

Incorrect; acentric chromosomes (without a centromere) also form

due to crossover in a paracentric inversion. However, the diagram

does not suggest such an event.

163. Which one of the following enzymes is NOT a part of

pyruvate dehydrogenase enzyme complex in

glycolysis pathway?

(1) Pyruvate dehydrogenase.

(2) Dihydrolipoyl transferase.

(3) Dihydrolipoyl dehydrogenase.

(4) Dihydrolipoyl oxidase

(2015)

Answer: (4) Dihydrolipoyl oxidase

Explanation:

The pyruvate dehydrogenase complex (PDC) is a

multi-enzyme complex responsible for converting pyruvate into

acetyl-CoA, linking glycolysis to the citric acid cycle. It consists of

three core enzymes: (1) Pyruvate dehydrogenase (E1), which

decarboxylates pyruvate; (2) Dihydrolipoyl transacetylase (E2),

which transfers the acetyl group to CoA; and (3) Dihydrolipoyl

dehydrogenase (E3), which regenerates the oxidized form of

lipoamide. There is no enzyme called dihydrolipoyl oxidase in this

complex.

Why Not the Other Options?

❌

(1) Pyruvate dehydrogenase – Incorrect; This is the E1 enzyme in

the complex and catalyzes the decarboxylation of pyruvate.

❌

(2) Dihydrolipoyl transferase – Incorrect; This refers to

dihydrolipoyl transacetylase (E2), which transfers the acetyl group

to CoA.

❌

(3) Dihydrolipoyl dehydrogenase – Incorrect; This is the E3

enzyme, which helps regenerate lipoamide for further reactions

164. When one isolates ribosomes from bacterial lysate,

apart from 70S, 50S and 308 ribosomal subunits, one

also finds a small population of 100S, 130S and 150S

sub-units. EDTA dissociates these larger ribosomal

subunits into 50S and 30S, suggesting that they have

both the subunits. Upon addition of cations they

reassociate into 70S, but none of the other forms

could be detected. What is the reason for not

obtaining the >70S forms?

(1) The effects of EDTA cannot be reversed by the

addition of cations.

(2) 190S, 130S and150S are modified form of

ribosomes that are irreversibly damaged by EDTA.

(3) 100S, 130S etc. represent polysome that cannot

be reassembled denovo without other cellular

components.

(4) They are obtained as an experimental artifact in

preparations of ribosomes

(2015)

Answer:(3) 100S, 130S etc. represent polysome that cannot

be reassembled denovo without other cellular components

Explanation:

Bacterial ribosomes exist in different functional

states. The standard 70S ribosome dissociates into its 50S and 30S

subunits under certain conditions. Larger forms such as 100S, 130S,

and 150S represent ribosomal aggregates, often associated with

polysomes (clusters of ribosomes translating the same mRNA). When

EDTA chelates divalent cations like Mg²⁺, ribosomes disassemble

into their subunits. Upon reintroducing cations, 70S ribosomes can

reassemble, but the larger polysomal structures (100S, 130S, etc.)

cannot, as their formation depends on mRNA and other cellular

factors, which are absent in the in vitro system.

Why Not the Other Options?

❌

(1) The effects of EDTA cannot be reversed by the addition of

cations – Incorrect; The reassembly of 70S ribosomes upon cation

addition confirms that EDTA effects are reversible.

❌

(2) 100S, 130S, and 150S are modified forms of ribosomes that

are irreversibly damaged by EDTA – Incorrect; EDTA does not

permanently damage ribosomes but rather disrupts their association

by chelating Mg²⁺.

❌

(4) They are obtained as an experimental artifact in preparations

of ribosomes – Incorrect; These higher-order ribosome structures

are biologically relevant and not merely experimental artifacts.

165. Chromatin condensation is driven by protein

complexes, called condensins which are members of a

family of "structural maintenance of chromatin"

(SMC) proteins that play a key role in the

organization of eukaryotic chromosomes. Condensins

along with another family of SMC proteins called

cohesins significantly contribute to chromosome

segregation during mitosis. If the cells are treated

with an inhibitor of cdk 1 phosphorylation

immediately before. the cells. enter M phase, which of

the following statements is most likely to be true?

(1) Sister chromatids are held together by condensins

along the entire length of the, chromosome.

(2) Sister chromatids are held together by cohesions

along the entire length of the chromosome.

(3) Sister chromatids are held together by condensins

and attached to each other only, at the centromere.

(4) Sister chromatids are held together by condensins

and attached to each other only at the telomere.

(2015)

Answer: (2) Sister chromatids are held together by cohesions

along the entire length of the chromosome

Explanation:

CDK1 (Cyclin-Dependent Kinase 1) is crucial for

mitotic progression, particularly for the activation of condensins and

the removal of cohesins from chromosome arms during prophase.

Cohesins hold sister chromatids together after DNA replication, and

they are gradually removed in a CDK1-dependent manner during

prophase, except at the centromere, where they are retained until

anaphase. If CDK1 phosphorylation is inhibited before M phase

entry, condensins will not be activated, leading to improper

chromatin condensation, and cohesins will not be removed from

chromosome arms. As a result, sister chromatids remain held

together by cohesins along their entire length.

Why Not the Other Options?

❌

(1) Sister chromatids are held together by condensins along the

entire length of the chromosome – Incorrect; Condensins do not hold

chromatids together; they mediate chromosome condensation.

Cohesins are responsible for chromatid cohesion.

❌

(3) Sister chromatids are held together by condensins and

attached to each other only at the centromere – Incorrect; Cohesins

(not condensins) mediate attachment, and inhibition of CDK1 would

prevent their removal from chromosome arms, leading to full-length

cohesion.

❌

(4) Sister chromatids are held together by condensins and

attached to each other only at the telomere – Incorrect; Condensins

do not mediate chromatid attachment, and cohesins primarily hold

chromatids together along the arms and centromere, not just at

telomeres.

166. A protein undergoes post-translational modifications.

In an experiment to identify the nature of

modifications, the following experimental results

were obtained.

A. Protein moved more slowly in an SDS PAGE.

B. Isoelectric focusing (IEF) showed that there was no

change in the pI.

C. Mass spectrometric analysis showed that the

modification was on serine

The modification that the protein undergoes is likely

to be

(1) phosphorylation

(2) glycosylation

(3) ubiquitinization

(4) ADP-ribosylation

(2015)

Answer: (2) glycosylation

Explanation:

The experimental observations provide important

clues about the nature of the modification:

A. Protein moved more slowly in an SDS-PAGE.

This suggests an increase in molecular weight, indicating that the

protein has undergone a modification that adds extra mass, such as

glycosylation or ubiquitinization.

B. Isoelectric focusing (IEF) showed no change in pI.

This means the modification did not alter the overall charge of the

protein. Modifications like phosphorylation and ADP-ribosylation

introduce charged groups (negative charges from phosphate or

ADP-ribose), which would change the pI. Since no pI change was

observed, these modifications are unlikely.

C. Mass spectrometry showed that the modification was on serine.

Serine can be modified by phosphorylation, glycosylation, and ADP-

ribosylation. However, phosphorylation and ADP-ribosylation

introduce negative charges, which would have affected the pI,

making them unlikely candidates. Glycosylation, on the other hand,

increases mass without significantly altering charge, which matches

the observations.

Why Not the Other Options?

❌

(1) Phosphorylation – Incorrect; phosphorylation adds a negative

charge, which would have changed the pI, but no change in pI was

observed.

❌

(3) Ubiquitinization – Incorrect; ubiquitination significantly

increases the molecular weight but typically occurs on lysine, not

serine.

❌

(4) ADP-ribosylation – Incorrect; this modification introduces

charged groups, which would have altered the pI.

167. In response to a drug, changes in protein levels were

examined in a cell line. A pulse chase experiment was

performed using 35Slabeled methionine. In

comparison, to untreated samples, the following

observations were made: few minutes after

stimulation, protein X accumulates and this was

followed by reduction in protein Y and Z. The correct

interpretation of these observations would be:

(1) Protein X is a protease which degrades Y and Z.

(2) Protein X is a transcriptional repressor that

controls expression of Y and Z.

(3) Expression of protein X and loss of Y and Z are

unrelated.

(4) Information is not sufficient to distinguish

between the three possibilities stated above.

(2015)

Answer: (4) Information is not sufficient to distinguish

between the three possibilities stated above.

Explanation:

A pulse-chase experiment using 35S-labeled

methionine allows tracking of newly synthesized proteins over time.

The key observations here are:

Protein X accumulates shortly after drug stimulation.

Proteins Y and Z decrease afterward.

Possible interpretations:

(1) Protein X is a protease that degrades Y and Z. This is a

possibility if X functions as a proteolytic enzyme, leading to the

degradation of Y and Z. However, the experiment does not confirm

whether Y and Z are being degraded or just downregulated.

(2) Protein X is a transcriptional repressor that controls Y and Z

expression. This would mean that X suppresses transcription of Y

and Z, leading to a decrease in their new synthesis rather than direct

degradation.

(3) Expression of protein X and loss of Y and Z are unrelated. This

remains a possibility if their changes are coincidental rather than

causally linked.

(4) Information is not sufficient to distinguish between these

possibilities. The experiment only measures protein levels but does

not confirm whether X directly degrades or represses Y and Z.

Further studies, such as mRNA expression analysis, protease

inhibitors, or protein stability assays, are needed to determine the

exact relationship.

Why Not the Other Options?

❌

(1) Protein X is a protease which degrades Y and Z – Incorrect;

no direct evidence of proteolytic activity from the experiment.

❌

(2) Protein X is a transcriptional repressor that controls

expression of Y and Z – Incorrect; the study does not measure mRNA

levels to confirm repression.

❌

(3) Expression of protein X and loss of Y and Z are unrelated –

Incorrect; while possible, there is no evidence ruling out a

relationship.

168. Which of the following statement regarding

membrane transport is FASLE?

(1) Polar and charged solutes will not cross cell

membranes effectively without specific protein

carriers.

(2) Each protein carrier will only bind and transport

one (or a few very similar) type of solute.

(3) Sugars such as glucose are always transported by

active transport rather than facilitated diffusion

carrier.

(4) Ions are typically transported by special proteins

that form membrane channels.

(2014)

Answer:

(3) Sugars such as glucose are always transported by active tra

nsport rather than facilitated diffusion.

Explanation:

This statement is FALSE because glucose transport occurs through

both facilitated diffusion and active transport, depending on the cell

type and physiological conditions.

During Facilitated Diffusion (Passive Transport): Glucose

Transporters (GLUTs) allow glucose to cross the membrane

passively (without ATP) down its concentration gradient. Example:

GLUT1, GLUT2, GLUT4 transporters transport glucose into cells

when extracellular glucose is higher than intracellular glucose.

During Active Transport (Against Gradient): Sodium-Glucose Linked

Transporters (SGLTs) perform secondary active transport, which

requires energy (indirectly using Na⁺ gradients).Example: SGLT1,

SGLT2 in intestinal and kidney epithelial cells co-transport glucose

with sodium ions. Thus, glucose is not "always" transported by active

transport; facilitated diffusion is also a common mechanism.

Why Not the Other Options?

✅

(1) Polar and charged solutes require protein carriers to cross

the membrane.

True. The hydrophobic lipid bilayer prevents direct diffusion of polar

molecules (e.g., glucose, amino acids, ions), so they require specific

transport proteins.

✅

(2) Each protein carrier binds and transports only specific solutes.

True. Transport proteins exhibit specificity (e.g., GLUT transporters

for glucose, amino acid transporters for amino acids).

✅

(4) Ions are transported via special channel proteins.

True. Ions (e.g., Na⁺, K⁺, Cl⁻, Ca²⁺) require ion channels (voltage-

gated, ligand-gated) or pumps (Na⁺/K⁺ ATPase, Ca²⁺ pumps) for

transport across membranes.

169. What will happen if histones are depleted from a

metaphase chromosome and viewed under a

transmission electron microscope?

(1) 30 nm chromatin fibres will be observed.

(2) 10 nm chromatin fibres will be observed.

(3) A scaffold and huge number of loops of DNA

fibres will be observed

(4) A huge number of loops of DNA fibres without

scaffold will be observed.

(2014)

Answer: (3)A scaffold and a huge number of loops of DNA

fibres will be observed.

Explanation:

Histones are essential structural proteins that package DNA into

chromatin. During metaphase, chromatin is highly condensed into

chromosomes. If histones are depleted, the DNA will lose its

nucleosome-based structure but still maintain a higher-order

organization due to non-histone scaffold proteins.

Step-by-Step Breakdown:

Normal Metaphase Chromosome Structure:

DNA wraps around histone octamers to form nucleosomes (10 nm

fibers). Nucleosomes further fold into 30 nm chromatin fibers.These

fibers form looped domains attached to a scaffold of non-histone

proteins, creating a compact metaphase chromosome.

What Happens If Histones Are Depleted?

Nucleosomes (which require histones) will be removed.

10 nm and 30 nm chromatin fibers will no longer be observed.

However, the scaffold proteins will remain intact, organizing the

DNA into large loops.

This scaffold-loop structure is what will be seen under a

transmission electron microscope.

Thus, under a transmission electron microscope (TEM), one would

observe:

✔

A central scaffold structure

✔

Many loops of DNA extending outward

Why Not the Other Options?

❌

(1) 30 nm chromatin fibers will be observed → Incorrect because

30 nm fibers require histones (H1) to form, and removing histones

will disrupt this structure.

❌

(2) 10 nm chromatin fibers will be observed → Incorrect because

the 10 nm fiber (beads-on-a-string structure) consists of nucleosomes

(which contain histones). Removing histones means this structure

will not be visible.

❌

(4) A huge number of loops of DNA fibers without scaffold will be

observed → Incorrect because scaffold proteins remain even after

histone depletion.

170. Which of the following statements about meiosis is

NOT true?

(1) Kinetochores of sister chromatids attach to

opposite poles in Meiosis I.

(2) Kinetochores of sister chromatids attach to

opposite poles in Meiosis II.

(3) Chiasma is formed in Prophase I.

(4) Homologous chromosomes are segregated in

Meiosis I.

(2014)

Answer: (1) Kinetochores of sister chromatids attach to

opposite poles in Meiosis I.

Explanation:

Meiosis consists of two sequential divisions: Meiosis I and Meiosis II,

leading to the formation of haploid gametes. The key difference

between the two divisions lies in how chromosomes behave during

segregation.

✔

(2) Kinetochores of sister chromatids attach to opposite poles in

Meiosis II – TRUE, In Meiosis II, sister chromatids separate just like

in mitosis.

✔

(3) Chiasma is formed in Prophase I – TRUE, During Prophase I

(Pachytene stage), homologous chromosomes undergo crossing over,

leading to the formation of a chiasma, which increases genetic

diversity.

✔

(4) Homologous chromosomes are segregated in Meiosis I –

TRUE, In Anaphase I of Meiosis I, homologous chromosomes (not

sister chromatids) are segregated to opposite poles.

❌

(1) Kinetochores of sister chromatids attach to opposite poles in

Meiosis I – FALSE

In Meiosis I, the kinetochores of sister chromatids remain attached to

the same pole (monopolar attachment). This ensures that

homologous chromosomes, not sister chromatids, are separated

during Anaphase I. The correct attachment of sister chromatid

kinetochores to opposite poles happens in Meiosis II, not Meiosis I.