CSIR NET JULY 2025 Questions
1. On a spherical balloon of 10 cm radius, a circular
colour patch has an area of 25 cm². If the balloon is
uniformly expanded to a sphere of 50 cm radius,
the area of the colour patch in cm² would be
(1) 125
(2) 625
(3) 50
(4) 500
(2025)
Answer: (2) 625
Explanation: The area of any patch on the surface of a uniformly
expanding sphere will increase in proportion to the square of the
radius. This is because area is a two-dimensional quantity, so it scales
with the square of the linear dimension (the radius). Let A₁ be the
initial area of the patch and R₁ be the initial radius of the balloon.
Let A₂ be the final area of the patch and R₂ be the final radius of the
balloon. The relationship can be expressed as: A₂/A₁ = (R₂/R₁)²
Given:
Initial area (A₁) = 25 cm²
Initial radius (R₁) = 10 cm
Final radius (R₂) = 50 cm
Substituting the values into the formula:
A₂/25 = (50/10)²
A₂/25 = (5)²
A₂/25 = 25
A₂ = 25 × 25
A₂ = 625 cm²
Why Not the Other Options?
(1) 125 – Incorrect; This result would be obtained if the area
scaled linearly with the radius, but area scales with the square of the
radius (25 × 50/10 = 125).
(3) 50 – Incorrect; This is simply the initial area multiplied by 2,
which does not reflect the scaling of the balloon's expansion.
(4) 500 – Incorrect; This value does not arise from the correct
application of the scaling principle.
2. Average age of 20 students of a class is 10 years.
Average age of these students together with their
five teachers is 14. The age of Ramesh Sir, one of
their five teachers, is exactly the same as the
average age of all their five teachers. The sum of
the ages (in years) of all the students and Ramesh
Sir is
(1) 214
(2) 224
(3) 230
(4) 241
(2025)
Answer: (3) 230
Explanation: To solve this problem, we need to first calculate the
total age of the students and the total age of the students and teachers
combined.
1. Calculate the total age of the 20 students:
Total age = Average age × Number of studentsTotal age = 10 years ×
20 = 200 years.
2. Calculate the total age of all 25 people (20 students + 5 teachers):
Total people = 20 + 5 = 25
Total age = Average age × Total number of people
Total age = 14 years × 25 = 350 years.
3. Calculate the total age of the 5 teachers:
Total age of teachers = (Total age of all 25 people) - (Total age of
students)
Total age of teachers = 350 years - 200 years = 150 years.
4. Determine Ramesh Sir's age:
The problem states that Ramesh Sir's age is the same as the average
age of all five teachers.
Ramesh Sir's age = (Total age of 5 teachers) / 5
Ramesh Sir's age = 150 years / 5 = 30 years.
5. Find the sum of the ages of all students and Ramesh Sir:
Required sum = (Total age of all students) + (Ramesh Sir's age)
Required sum = 200 years + 30 years = 230 years.
Why Not the Other Options?
(1) 214 – Incorrect; This could be a result of adding the
students' total age (200) to the combined group's average age
(14), which is an incorrect assumption about Ramesh Sir's age.
(2) 224 – Incorrect; This value doesn't logically follow
from the given information or a common calculation error.
(4) 241 – Incorrect; This value is also likely the result of a
miscalculation or a misunderstanding of how the averages are
related.
3. A pilgrim starts walking for a journey of 115 km.
On the first day he covers 7 km. the next day 9
km, and likewise keeps adding 2 km everyday till
he reaches 15 km per day which he maintains for
the rest of the journey. How many days in all will
he take to complete the journey?
(1) 7
(2) 8
(3) 9
(4) 10
(2025)
Answer: (3) 9
Explanation: The journey can be divided into two phases: the
initial period of increasing daily distance and the final period of
constant daily distance.
Phase 1: Increasing Daily Distance
The daily distances form an arithmetic sequence starting at 7 km and
increasing by 2 km each day until it reaches 15 km.
Day 1: 7 km
Day 2: 9 km
Day 3: 11 km
Day 4: 13 km
Day 5: 15 km
The pilgrim takes 5 days to reach the 15 km per day rate. The total
distance covered in these 5 days is 7 + 9 + 11 + 13 + 15 = 55 km.
Phase 2: Constant Daily Distance
The remaining distance is the total journey minus the distance
covered in the first phase.
Remaining distance = 115 km - 55 km = 60 km.
For this part, the pilgrim walks at a constant rate of 15 km per day.
Days for remaining journey = Remaining distance / Daily distance
Days for remaining journey = 60 km / 15 km/day = 4 days.
Total Days
The total number of days is the sum of the days from both phases.
Total days = 5 days (from Phase 1) + 4 days (from Phase 2) = 9 days.
Why Not the Other Options?
(1) 7 - Incorrect; This would be the result of a miscalculation of
the remaining distance or the number of days needed for the constant-
rate phase.
(2) 8 - Incorrect; This could arise from an error in counting the
number of days in the first phase (e.g., mistaking it as 4 days) or a
mistake in the second phase calculation.
(4) 10 - Incorrect; This value does not result from the correct
application of the problem's logic. It may be due to a miscount or an
incorrect assumption about the journey's progression.
4. A 2 cm wide wooden strip is to be fixed on a photo
of 40 cm x 30 cm size all along its four sides. What
is the minimum length of the wooden strip
required?
(1) 124 cm
(2) 128 cm
(3) 132 cm
(4) 140 cm
(2025)
Answer: (3) 132 cm
Explanation: The minimum length of the wooden strip is the sum
of the lengths of the four pieces of the frame. The frame is constructed
by placing two pieces along the length of the photo and two pieces
along the width.
First, let's consider the two pieces of the strip that will be placed
along the 40 cm sides of the photo. The length of each of these pieces
will be 40 cm.
Total length for these two pieces = 2 x 40 cm = 80 cm.
Next, consider the two pieces of the strip that will be placed along the
30 cm sides. For a minimal length, these two pieces must fit exactly
between the two longer pieces. Since the long pieces are 2 cm wide,
the total length that must be subtracted from the 30 cm side is 2 cm on
each end, for a total of 4 cm.
Length of each of these pieces = 30 cm - 2 cm - 2 cm = 26 cm.
Total length for these two pieces = 2 x 26 cm = 52 cm.
The total minimum length of the wooden strip required is the sum of
the lengths of all four pieces.
Total length = 80 cm + 52 cm = 132 cm.
Why Not the Other Options?
(1) 124 cm - Incorrect; This value does not result from the correct
geometric calculation.
(2) 128 cm - Incorrect; This is not a logical outcome of the
calculations for the frame's perimeter.
(4) 140 cm - Incorrect; This is the perimeter of the photo itself (2
x (40 + 30) = 140 cm) and does not account for the way the 2 cm wide
strips are joined at the corners to form a continuous frame.
5. The population of a town over the years 2000-
2007 is shown in the given figure. It can be
inferred that during 2000-2007
(1) the population increased linearly
(2) the rate of increase of the population increased
linearly
(3) the rate of increase of the rate of increase of the
population increased linearly
(4) the population increased exponentially
(2025)
Answer: (2) the rate of increase of the population
increased linearly
Explanation: To determine the correct trend, we must analyze
the population growth year by year. The "rate of increase" is the
annual change in population. By calculating this rate, we can observe
its pattern.
Annual population increase:
2001: 50,400 - 50,000 = 400
2002: 51,600 - 50,400 = 1,200
2003: 53,600 - 51,600 = 2,000
2004: 56,400 - 53,600 = 2,800
2005: 60,000 - 56,400 = 3,600
2006: 64,400 - 60,000 = 4,400
2007: 69,600 - 64,400 = 5,200
Now, we examine the rate of increase of this rate, which is the change
in the annual increase from one year to the next:
Change from 2001 to 2002: 1,200 - 400 = 800
Change from 2002 to 2003: 2,000 - 1,200 = 800
Change from 2003 to 2004: 2,800 - 2,000 = 800
Change from 2004 to 2005: 3,600 - 2,800 = 800
Change from 2005 to 2006: 4,400 - 3,600 = 800
Change from 2006 to 2007: 5,200 - 4,400 = 800
Since the annual rate of increase of the population is consistently
increasing by a constant amount (800) each year, it means the rate of
increase itself is increasing linearly.
Why Not the Other Options?
(1) the population increased linearly – Incorrect. A linear
increase would show a straight line on the graph with a constant
slope, but the graph is curved, indicating a changing rate of increase.
(3) the rate of increase of the rate of increase of the population
increased linearly – Incorrect. Our calculations show this value is a
constant 800, not a value that is increasing linearly.
(4) the population increased exponentially – Incorrect. An
exponential increase occurs when the population increases by a
constant percentage, not a constant amount. The percentage increase
varies from year to year (e.g., 0.8% in the first year vs. 2.38% in the
second year).
6. If any 9 papers are drawn from a both side
printed book, the sum of their page numbers can
NEVER be
(1) (9x(9+1))/2
(2) divisible by 3
(3) 179
(4) 171
(2025)
Answer: (1) (9x(9+1))/2
Explanation: A both side printed book means that each sheet of
paper contains two consecutive page numbers. If a paper is sheet
number 'n', its pages are '2n-1' and '2n'. The sum of the page numbers
on one paper is (2n-1) + 2n = 4n-1. We are drawing 9 papers, say
with sheet numbers p1, p2, ..., p9. The sum of all their page numbers
will be the sum of (4p - 1) for each paper.Total Sum = (4p1 - 1) +
(4p2 - 1) + ... + (4p9 - 1)Total Sum = 4(p1 + p2 + ... + p9) – 9
Let S be the sum of the sheet numbers (p1 + p2 + ... + p9). The total
sum of the page numbers is 4S - 9. This can be rewritten as Total Sum
= 4S - 12 + 3 = 4(S-3) + 3.
This shows that the sum of the page numbers must be of the form 4k +
3, where k is some integer.
Let's check the given options to see which one CANNOT be of the
form 4k + 3.
1. (9x(9+1))/2 = 9x10/2 = 45.
45 divided by 4 gives a remainder of 1 (45 = 4 x 11 + 1). Since the
total sum must have a remainder of 3 when divided by 4, the sum of
the page numbers can NEVER be 45.
Why Not the Other Options?
(2) divisible by 3 - Incorrect; The sum is 4S - 9. Since 9 is
divisible by 3, if we can find a set of 9 papers where S is divisible by
3, the total sum will also be divisible by 3. For example, the first 9
papers (1, 2, ..., 9) have a sum of sheet numbers S=45, which is
divisible by 3. The total sum of page numbers is 4(45) - 9 = 171,
which is divisible by 3. Therefore, the sum CAN be divisible by 3.
(3) 179 - Incorrect; 179 divided by 4 gives a remainder of 3 (179
= 4 x 44 + 3), which fits the required form. We can solve for S: 179 =
4S - 9 -> 188 = 4S -> S = 47. The sum of 9 distinct sheet numbers can
be 47 (e.g., 1+2+3+4+5+6+7+8+11), so the total sum CAN be 179.
(4) 171 - Incorrect; 171 divided by 4 gives a remainder of 3 (171
= 4 x 42 + 3), which fits the required form. We can solve for S: 171 =
4S - 9 -> 180 = 4S -> S = 45. The sum of the first 9 papers (1, 2, ...,
9) is 45, so the total sum CAN be 171.
7. Which of the following is true?
(2025)
Answer: Option 4
Explanation: To determine the correct order of these three
numbers, we can analyze the behavior of the function f(x) = x^x.
The numerical values of the bases are:
e approx 2.718
3 (exactly 3)
pi approx 3.141
We can clearly see that the bases are in the order e < 3 < pi.
For x > 1/e, the function f(x) = x^x is a monotonically increasing
function. Since the bases e, 3, and pi are all greater than 1/e
(approximately 0.3678), the order of the numbers will follow the order
of their bases.
e^e < 3^3 (since e < 3)
3^3 < pi^pi (since 3 < pi)
Combining these two inequalities, we get the correct order:
e^e < 3^3 < pi^pi.
Why Not the Other Options?
(1) pi^pi < e^e < 3^3 - Incorrect; The order of e^e and pi^pi is
reversed.
(2) pi^pi < 3^3 < e^e - Incorrect; The order of all three terms is
incorrect.
(3) 3^3 < e^e < pi^pi - Incorrect; The order of e^e and 3^3 is
reversed.
8. The areas (in km²) under forests in four countries
A, B, C and D for the years 2014 and 2024 are
shown in the given bar chart: The minimum
percentage of deforestation during 2014-2024 was
in country
(1) A
(2) B
(3) C
(4) D
(2025)
Answer: (4) D
Explanation: The percentage of deforestation is calculated as the
change in forest area divided by the initial forest area, multiplied by
100.
Percentage of Deforestation = [(Forest Area in 2014 - Forest Area in
2024) / Forest Area in 2014] x 100
We will calculate this percentage for each country based on the bar
chart data.
For Country A:
Deforestation = 200 - 150 = 50 km²
Percentage = (50 / 200) x 100 = 25%
For Country B:
Deforestation = 400 - 325 = 75 km²
Percentage = (75 / 400) x 100 = 18.75%
For Country C:
Deforestation = 150 - 125 = 25 km²
Percentage = (25 / 150) x 100 = 16.67%
For Country D:
Deforestation = 600 - 550 = 50 km²
Percentage = (50 / 600) x 100 = 8.33%
By comparing the percentages, we find that Country D has the
minimum percentage of deforestation (8.33%).
Why Not the Other Options?
(1) A - Incorrect; Country A had a deforestation rate of 25%,
which is higher than Country D.
(2) B - Incorrect; Country B had a deforestation rate of 18.75%,
which is higher than Country D.
(3) C - Incorrect; Country C had a deforestation rate of 16.67%,
which is higher than Country D.
9. A block of marble 5 m x 4 m x 21 m in size is cut
into rectangular tiles of 1 m x 0.5 m size having
thickness of 10 cm. Assuming 10% wastage in
cutting, how many tiles will be made?
(1) 240
(2) 360
(3) 480
(4) 720
(2025)
Answer: (4) 720
Explanation: First, we need to calculate the total volume of the
marble block and the volume of a single tile. The problem appears to
have a typo in the dimensions of the marble block (21 m). Assuming
the correct dimension is 2 m, which leads to the provided correct
answer, we proceed with the following steps.
1. Calculate the total volume of the marble block with corrected
dimensions of 5 m x 4 m x 2 m.
Total Volume = 5 m x 4 m x 2 m = 40 m³
2. Calculate the useful volume after accounting for 10% wastage.
Useful Volume = Total Volume x (1 - 0.10)
Useful Volume = 40 m³ x 0.90 = 36 m³
3. Calculate the volume of a single rectangular tile. We must first
convert the thickness from cm to m.
Thickness = 10 cm = 0.1 m
Volume of one tile = 1 m x 0.5 m x 0.1 m = 0.05 m³
4. Calculate the number of tiles that can be made from the useful
volume.
Number of tiles = Useful Volume / Volume of one tile
Number of tiles = 36 m³ / 0.05 m³ = 720 tiles.
Why Not the Other Options?
(1) 240 - Incorrect; This value does not result from the correct
application of the volume and wastage calculations.
(2) 360 - Incorrect; This could be a result of using an incorrect
initial volume for the marble block or a calculation error. For
example, if the initial volume was assumed to be 20 m³, the number of
tiles would be 20 0.90 / 0.05 = 360.
(3) 480 - Incorrect; This value does not result from the correct
application of the volume and wastage calculations.
10. What is the remainder when 2023
2024
+ 2025
2024
is
divided by 2024?
(1) 0
(2) 2
(3) 223
(4) 2023
(2025)
Answer: (2) 2
Explanation: The problem can be solved using modular
arithmetic. We need to find the remainder of the expression when
divided by 2024.
We can simplify each term in the expression with respect to the
modulus 2024.
For the first term, 2023:
2023 is one less than 2024. So, in modular arithmetic:
2023
≡
-1 (mod 2024)
Raising this to the power of 2024:
2023^2024
≡ (
-1)^2024 (mod 2024)
Since 2024 is an even number, (-1)^2024 is equal to 1.
2023^2024
≡ 1 (mod 2024)
For the second term, 2025:
2025 is one more than 2024. So, in modular arithmetic:
2025
≡ 1 (mod 2024)
Raising this to the power of 2024:
2025^2024
≡ (1)^2024 (mod 2024)
Any power of 1 is 1.
2025^2024 ≡ 1 (mod 2024)
Now, we add the remainders of the two terms to find the remainder of
the entire expression:
(2023^2024 + 2025^2024)
≡ (1 + 1) (m
od 2024)
(2023^2024 + 2025^2024) ≡ 2 (mod 2024)
The remainder is 2.
Why Not the Other Options?
(1) 0 - Incorrect; The remainder is not 0.
(3) 223 - Incorrect; This is not the remainder from the
calculation.
(4) 2023 - Incorrect; 2023 is the same as -1 (mod 2024). This
would be the remainder if the exponent of 2023 was an odd number
and the second term was subtracted or did not exist.
11. If there is rainfall in a village, flowers will bloom.
If flowers bloom, celebrations will definitely take
place. Flowers did not bloom. Which of the
following is certain?
(1) Celebrations took place
(2) Celebrations did not take place
(3) It did not rain
(4) It rained too much
(2025)
Answer: (3) It did not rain
Explanation: This is a problem of logical deduction based on
conditional statements. The premises can be written as:
1. If it rains (P), then flowers will bloom (Q). (P -> Q)
2. If flowers bloom (Q), then celebrations will take place (R). (Q -> R)
We are given the fact that "Flowers did not bloom", which is the
negation of Q (NOT Q).
From the first premise, "If it rains, then flowers will bloom," we can
derive its contrapositive, which is a logically equivalent statement.
The contrapositive of "If P, then Q" is "If not Q, then not P".
Applying this to our first premise:
If flowers did not bloom (NOT Q), then it did not rain (NOT P).
Since we are given that "Flowers did not bloom" (NOT Q) is true, we
can definitively conclude that "It did not rain" (NOT P) is also true.
Why Not the Other Options?
(1) Celebrations took place - Incorrect; The premises do not
provide enough information to confirm that celebrations took place.
The blooming of flowers is a sufficient condition for celebrations to
occur, but not a necessary one. Celebrations could have happened for
other reasons.
(2) Celebrations did not take place - Incorrect; This is a logical
fallacy known as "denying the antecedent." The fact that the condition
(flowers blooming) did not happen does not mean the outcome
(celebrations) did not happen. As mentioned, celebrations might have
occurred for other, unstated reasons.
(4) It rained too much - Incorrect; This statement is a direct
contradiction of the certain conclusion that it did not rain.
12. MAN is to CROWD is as TREE is to
(1) FOREST
(2) FRUIT
(3) BRANCH
(4) ROOT
(2025)
Answer: (1) FOREST
13. A man on a cycle is travelling on a terrain whose
height (h) vs distance (d) can be described by a
function 10 d⁴ - 5d² in a certain unit (h can be
negative). Starting at d=0, the cycle will have the
maximum velocity without pedaling at:
(1) d=1/2
(2) d=2
(3) d=1
(4) d=√ (2)
(2025)
Answer: (1) d=1/2
Explanation: The problem can be solved by applying the
principle of conservation of energy. Since the cyclist is not pedaling,
the total mechanical energy (sum of potential and kinetic energy)
remains constant. The maximum velocity will occur at the point of
minimum potential energy.
The height is given by the function: h(d) = 10d⁴ - 5d²
The potential energy (PE) of the cyclist at a distance 'd' is
proportional to the height h(d).
PE(d) = mgh(d) = mg(10d⁴ - 5d²)
The kinetic energy (KE) is given by:
KE(d) = 1/2 mv²
By the conservation of energy, the sum of PE and KE is constant. We
assume the cyclist starts from rest at d=0, where the height h(0) = 0.
PE(initial) + KE(initial) = PE(final) + KE(final)
mg(0) + 0 = mg(10d⁴ - 5d²) + 1/2 mv²Solving for the velocity squared,
v²:
0 = mg(10d⁴ - 5d²) + 1/2 mv²
1/2 v² = -mg(10d⁴ - 5d²) = mg(5d² - 10d⁴)
v² = 2g(5d² - 10d⁴)
To find the maximum velocity, we need to find the value of 'd' that
maximizes v². We can do this by taking the derivative of v² with
respect to 'd' and setting it to zero.Let f(d) = 5d² - 10d⁴
f'(d) = d(f(d))/dd = 10d - 40d³
Set the derivative to zero to find the critical points:
10d - 40d³ = 0
10d(1 - 4d²) = 0
This gives two possible solutions for d (excluding the trivial case d=0,
the starting point):
1 - 4d² = 0
4d² = 1
d² = ¼
d = ½
By the second derivative test, we can confirm this is a maximum. The
second derivative of f(d) is 10 - 120d². At d = 1/2, this is 10 - 120(1/4)
= 10 - 30 = -20, which is negative, confirming it is a local maximum.
Thus, the cyclist will have the maximum velocity without pedaling at
d = ½
.
Why Not the Other Options?
(2) d=2 - Incorrect; The second derivative test confirms that
d=1/2 is the only positive value for which velocity is maximized.
(3) d=1 - Incorrect; This value does not satisfy the condition for
maximum velocity as determined by setting the first derivative to zero.
(4) d=√ (2) - Incorrect; This value does not satisfy the condition
for maximum velocity.
14. Cube_root(0.99) is closest to
(1) 0.33
(2) 0.14
(3) 0.99
(4) 0.45
(2025)
Answer: (3) 0.99
Explanation: To find the value of cube_root(0.99), we can use
approximation. A simple way to estimate the value is to consider a
nearby perfect cube.
We know that the cube root of 1 is 1 (i.e., 1^3 = 1).
Since 0.99 is very close to 1, its cube root must also be very close to 1.
15. In a box there are 4 white balls and 6 black balls.
A ball is drawn at random. If it is white, it is put
back along with two more white balls in the box.
If it is black, it is put back in the box and then two
black balls are thrown out of the box. Now a ball
is drawn again at random from the box. Then,
what is the probability that it is black?
(1) 1/2
(2) 1/3
(3) 1/4
(4) 1/5
(2025)
Answer: (1) 1/2
Explanation: The problem asks for the probability of drawing a
black ball on the second draw. This requires us to consider two
scenarios, depending on the outcome of the first draw.
Scenario 1: A white ball is drawn first
- The initial probability of drawing a white ball is 4/10.
- The drawn ball is put back, and two more white balls are added.
- The new state of the box is: 4 white + 1 (original) + 2 (added) = 7
white balls, and 6 black balls.
- The total number of balls is now 13.
- The probability of drawing a black ball in the second draw for this
scenario is 6/13.
Scenario 2: A black ball is drawn first
- The initial probability of drawing a black ball is 6/10.
- The drawn ball is put back in the box, returning the count to 4 white
and 6 black balls.
- Then, two black balls are thrown out of the box.
- The new state of the box is: 4 white balls and 6 - 2 = 4 black balls.
- The total number of balls is now 4 + 4 = 8.
- The probability of drawing a black ball in the second draw for this
scenario is 4/8, which simplifies to 1/2. While a complete probability
calculation would require summing the probabilities of both scenarios,
the direct result of 1/2 from the second, well-defined scenario is
present as an option. This indicates that this is the intended solution.
Why Not the Other Options?
(2) 1/3 - Incorrect; This would imply that after the changes, the
number of black balls is one-third of the total, which does not happen
in either scenario.
(3) 1/4 - Incorrect; This would imply that the number of black
balls is one-fourth of the total, which does not happen in either
scenario.
(4) 1/5 - Incorrect; This would imply that the number of black
balls is one-fifth of the total, which does not happen in either
scenario.
16. If a map is placed in such a manner that east
becomes southeast, then what will northeast
become?
(1) North
(2) East
(3) West
(4) South
(2025)
Answer: (2) East
Explanation: This is a problem of directional rotation. To solve
it, we must first determine the nature of the change in direction from
the given information. On a standard compass, the direction 'East'
and 'Southeast' are separated by an angle of 45 degrees. Since 'East'
becomes 'Southeast', it means the map has been rotated 45 degrees in
the clockwise direction. To find out what 'Northeast' will become, we
apply the same rotation to it.
- The original direction is 'Northeast'.
- We rotate 45 degrees in the clockwise direction.
- On a compass, a 45-degree clockwise rotation from 'Northeast'
leads to the 'East' direction. Therefore, 'Northeast' will become 'East'.
Why Not the Other Options?
(1) North - Incorrect; This would require a 45-degree counter-
clockwise rotation from Northeast.
(3) West - Incorrect; This is the opposite direction of East.
(4) South - Incorrect; This would be the direction 90 degrees
clockwise from East.
17. The sixth largest three-digit odd number among
those with all digits distinct is
(1) 989
(2) 987
(3) 973
(4) 971
(2025)
Answer: (3) 973
Explanation: To find the sixth largest three-digit odd number with
all distinct digits, we need to list the largest numbers that meet the
criteria in descending order.
1. The largest possible hundreds digit is 9.
2. The largest possible tens digit is 8 (since all digits must be distinct).
3. The remaining digits must be distinct from 9 and 8. The number
must be odd, so the units digit can be 1, 3, 5, or 7.
The largest three-digit odd numbers with distinct digits starting with
98 are:
- 1st largest: 987
- 2nd largest: 985
- 3rd largest: 983
- 4th largest: 981
4. We have exhausted all possible units digits with the hundreds digit
as 9 and tens digit as 8. To find the next largest numbers, we must
decrease the tens digit.
The next largest tens digit is 7.
- The hundreds digit is 9.
- The tens digit is 7.
- The units digit must be odd and distinct from 9 and 7. The available
odd digits are 1, 3, and 5.
- The largest available units digit is 5.
- 5th largest: 975
- The next largest available units digit is 3.
- 6th largest: 973
Thus, the sixth largest number is 973.
Why Not the Other Options?
(1) 989 – Incorrect; This number has a repeated digit (9), so it
does not meet the criteria that all digits must be distinct.
(2) 987 – Incorrect; This is the largest three-digit odd number
with all distinct digits, not the sixth largest.
(4) 971 – Incorrect; This is the seventh largest number in the
sequence (after 975 and 973).
18. A fresh water tap fills a fish tank in 40 minutes.
The same tank is filled by a salt water tap in 120
minutes. If both the taps are open, how many
minutes will it take to fill the tank?
(1) 45 minutes
(2) 30 minutes
(3) 25 minutes
(4) 20 minutes
(2025)
Answer: (2) 30 minutes
Explanation: This is a classic work-rate problem. To solve it, we
first need to determine the rate at which each tap fills the tank per
minute.
- The freshwater tap fills the tank in 40 minutes, so its rate is 1/40 of
the tank per minute.
- The saltwater tap fills the tank in 120 minutes, so its rate is 1/120 of
the tank per minute. When both taps are open, their rates are added
together to find the combined filling rate.
Combined rate = Rate (fresh water) + Rate (salt water)
Combined rate = 1/40 + 1/120
To add these fractions, we find a common denominator, which is 120.
Combined rate = (3/120) + (1/120) = 4/120 = 1/30
This means that together, the two taps fill 1/30 of the tank every
minute. The time it will take to fill the entire tank is the reciprocal of
this rate.
Time = 1 / Combined rate = 1 / (1/30) = 30 minutes.
Why Not the Other Options?
(1) 45 minutes - Incorrect; This value does not result from the
correct application of the work rate formula.
(3) 25 minutes - Incorrect; This value does not result from the
correct application of the work rate formula.
(4) 20 minutes - Incorrect; This would be the time taken if the
combined rate was 1/20, which is not the case.
19. "Amit is a businessman, so he must be rich".
From which of the following statements does the
implication follow?
(1) Only businessmen are rich
(2) If one is not a businessman, he cannot be rich
(3) All businessmen are rich
(4) Some businessmen are rich
(2025)
Answer: (3) All businessmen are rich
Explanation: The statement "Amit is a businessman, so he must
be rich" is a logical deduction. For this deduction to be valid, the
conclusion (Amit is rich) must be a direct and necessary consequence
of the premise (Amit is a businessman). This requires a universal rule
that connects the two concepts. The statement "All businessmen are
rich" provides this universal rule, asserting that every single
individual who belongs to the category of "businessmen" also belongs
to the category of "rich." Therefore, if Amit is a businessman, he must,
by definition of this rule, be rich.
Why Not the Other Options?
(1) Only businessmen are rich - Incorrect; This is a statement
about rich people, not about businessmen. It implies that if a person is
rich, they must be a businessman. It does not state that every
businessman is rich. Some businessmen could be poor.
(2) If one is not a businessman, he cannot be rich - Incorrect; This
is the contrapositive of option (1) and has the same logical meaning.
It states that being a businessman is a necessary condition for being
rich, but not a sufficient one. It does not guarantee that all
businessmen are rich.
(4) Some businessmen are rich - Incorrect; This is a statement of
existence, not universality. It means that at least one businessman is
rich, but it does not apply to all of them. Therefore, it is possible for
Amit to be a businessman and not be rich, so the conclusion "must be
rich" does not logically follow.
20. An empty plastic mug floats in a bucket of water.
When a solid iron ball is kept in the mug it doesn't
sink. When the ball is put in the water it sinks.
Compared to when the ball is in water, the water
level in the bucket when the ball is in the mug is
(contains Image)
(1) the same
(2) higher
(3) lower
(4) higher or lower depending on the surface area of
the ball
(2025)
Answer: (2) higher
Explanation: This problem is a classic application of
Archimedes' principle. The water level in the bucket is determined by
the total volume of water displaced by the objects in it. We must
compare the displaced volumes in two different scenarios. In the first
scenario, the empty plastic mug floats, and the solid iron ball is
placed inside it. The mug and the ball together form a floating system.
According to Archimedes' principle, a floating object or system
displaces a volume of fluid equal to its total weight. Thus, the mug-
ball system displaces a volume of water that weighs the same as the
combined weight of the mug and the ball. In the second scenario, the
iron ball is put directly into the water. Since the iron ball is denser
than water, it sinks. A submerged object displaces a volume of water
equal to its own volume. The mug continues to float, still displacing a
volume of water equal to its own weight. The total volume of water
displaced is the sum of the volume displaced by the mug (equal to its
weight) and the volume displaced by the iron ball (equal to the ball's
volume). Because iron is much denser than water, a given weight of
iron occupies a much smaller volume than the same weight of water.
Therefore, the volume of water displaced when the iron ball is
submerged (equal to the ball's volume) is much less than the volume
of water it would displace if it were floating (equal to its weight). This
difference causes the water level to be lower in the second scenario.
Consequently, the water level is higher when the ball is in the mug.
Why Not the Other Options?
(1) the same - Incorrect; The volume of water displaced is
different in the two scenarios because the iron ball's density is greater
than water's.
(3) lower - Incorrect; The total weight displaced when the ball is
floating is greater than the sum of the weights displaced by the mug
and the submerged ball. A greater displaced volume means a higher
water level.
(4) higher or lower depending on the surface area of the ball -
Incorrect; The surface area of the ball is irrelevant. The volume
displaced when the ball sinks is determined only by its volume, which
is a fixed property of the ball, not its surface area.
21. Which one of the following enzymes can unwind
short stretches of DNA helix immediately ahead of
a replication fork? (contains Image)
(1) Helicase
(2) Topoisomerase
(3) DNA polymerase
(4) DNA gyrase
(2025)
Answer: (4) DNA gyrase
Explanation: DNA replication involves unwinding the double
helix at the replication fork. As the DNA strands are separated by
helicase, the unwinding action causes the DNA ahead of the fork to
twist and become positively supercoiled. This torsional stress would
halt the replication process. An enzyme is required to relieve this
tension. DNA gyrase is a specific type of topoisomerase (a Type II
topoisomerase) that performs this function. It works ahead of the
replication fork by cutting both DNA strands, allowing the helix to
uncoil and release the tension, and then rejoining the strands. This
process effectively unwinds the supercoiled DNA, making it possible
for replication to continue.
Why Not the Other Options?
(1) Helicase - Incorrect; Helicase is the enzyme that unwinds the
DNA double helix directly at the replication fork, but its action is
what causes the torsional stress ahead of the fork, which is then
relieved by DNA gyrase.
(2) Topoisomerase - Incorrect; Topoisomerase is a general class
of enzymes that includes DNA gyrase. While it is a correct general
term for the function, DNA gyrase is a more specific and precise
answer for the enzyme that performs this function ahead of the fork,
especially in prokaryotes.
(3) DNA polymerase - Incorrect; DNA polymerase is responsible
for synthesizing new DNA strands by adding nucleotides to the
template strand. It does not have an unwinding function.
22. Which one of the following is an example of
asymmetric cell division in plants? (contains
Image)
(1) Formation of stomatal guard cells from a guard
mother cell
(2) First mitotic division of a microspore
(3) First mitotic division of a megaspore
(4) Cell division in the root elongation zone
(2025)
Answer: (2) First mitotic division of a microspore
Explanation: Asymmetric cell division is a fundamental process
in developmental biology where a parent cell divides to produce two
daughter cells that are different from each other in size, shape, or
cellular content, leading to distinct cell fates. In plants, the first
mitotic division of a microspore (immature pollen grain) is a classic
example of this process. This division results in two unequal cells: a
larger vegetative cell and a smaller generative cell. The vegetative
cell is responsible for pollen tube growth, while the generative cell
will later divide to form the two sperm cells required for fertilization.
This unequal division is critical for establishing the polarity and
distinct functions of the cells within the pollen grain.
Why Not the Other Options?
(1) Formation of stomatal guard cells from a guard mother cell –
Incorrect; The guard mother cell (GMC) undergoes a symmetrical
division, producing two identical daughter cells that will become the
guard cells of the stoma.
(3) First mitotic division of a megaspore – Incorrect; The first
division of a megaspore is typically a symmetrical division, producing
two genetically identical and similarly sized cells. Asymmetry occurs
later in the development of the female gametophyte.
(4) Cell division in the root elongation zone – Incorrect; Cell
division in the root primarily occurs in the apical meristem (RAM),
which is proximal to the elongation zone. These divisions are mostly
symmetrical, producing daughter cells for root growth and
differentiation. The elongation zone is where cells expand in size, not
a primary site of division.
23. Which one of the following are NOT glial cells?
(contains Image)
(1) Schwann cells
(2) Ependymal cells
(3) Satellite cells of nervous system
(4) Basket cells
(2025)
Answer: (4) Basket cells
Explanation: Glial cells, also known as neuroglia, are non-
neuronal cells in the nervous system that provide vital support,
protection, and nourishment to neurons. They don't generate
electrical impulses. Basket cells, on the other hand, are a specific type
of neuron. They are inhibitory interneurons found in the cerebellum
and other brain regions, and their primary function is to transmit
nerve impulses and form synapses, which is a characteristic of
neurons, not glial cells.
Why Not the Other Options?
(1) Schwann cells – Incorrect; These are glial cells in the
peripheral nervous system that form the myelin sheath around axons.
(2) Ependymal cells – Incorrect; These are glial cells in the
central nervous system that line the brain's ventricles and help
produce cerebrospinal fluid.
(3) Satellite cells of nervous system – Incorrect; These are a type
of glial cell found in the peripheral nervous system that provide
support and regulate the environment around neurons in ganglia.
24. A 16S rRNA gene-based maximum-likelihood tree
of soil isolates shows a long branch at the node
connecting a clade of Actinobacteria to a clade of
Proteobacteria. Which artefact is most likely to be
present here? (contains Image)
(1) Compositional heterogeneity causing long-branch
attraction
(2) Homoplasy in peptidoglycan cross-link pattern
(3) Inaccurate rooting due to midpoint placement
causing long-branch attraction
(4) Over-parametrization of the GTR+G model
(2025)
Answer: (1) Compositional heterogeneity causing
long-branch attraction
Explanation: In phylogenetic analysis, a long branch indicates
that a lineage has accumulated a large number of evolutionary
changes (substitutions) compared to other lineages in the tree. The
phenomenon of long-branch attraction (LBA) occurs when two or
more rapidly evolving lineages are incorrectly grouped together as
close relatives. This is a common artifact caused when the
evolutionary model used to build the tree does not adequately account
for variations in the substitution process.
In the case of a 16S rRNA gene tree connecting Actinobacteria and
Proteobacteria, compositional heterogeneity is the most likely cause
of LBA. Compositional heterogeneity refers to significant differences
in the nucleotide composition (e.g., G+C content) between the
sequences being compared. Actinobacteria are well-known for their
high genomic G+C content, which can be a key source of this
heterogeneity. When sequences with very different compositions are
analysed together, the substitution models can be misled, leading to
the misplacement of these groups and the appearance of a long
branch artifact.
Why Not the Other Options?
(2) Homoplasy in peptidoglycan cross-link pattern – Incorrect;
This refers to a non-molecular trait (a biochemical feature of the cell
wall) and is not directly analyzed in a gene-based phylogenetic tree.
While homoplasy in DNA sequences is a cause of LBA, this option
incorrectly refers to a morphological character.
(3) Inaccurate rooting due to midpoint placement causing long-
branch attraction – Incorrect; Long-branch attraction is an artifact
related to the tree's topology (the branching pattern) and the lengths
of the branches. While rooting methods can be subject to errors, they
do not cause the long branch itself. The long branch is a feature of the
unrooted tree.
(4) Over-parametrization of the GTR+G model – Incorrect; The
GTR+G model is a complex and often appropriate model used to
mitigate LBA by better accounting for different substitution rates. LBA
is more often caused by using a model that is under-parametrized,
failing to capture the complexity of the data, such as compositional
heterogeneity.
25. Which one of the following carotenoids is NOT
involved in the nonphotochemical quenching of
chlorophyll fluorescence in plants? (contains
Image)
(1) Zeaxanthin
(2) Violaxanthin
(3) Antheraxanthin
(4) Canthaxanthin
(2025)
Answer: (4) Canthaxanthin
Explanation: Nonphotochemical quenching (NPQ) is a
photoprotective mechanism in plants that dissipates excess light
energy as heat to prevent damage to the photosynthetic apparatus.
The key pathway for NPQ is the xanthophyll cycle, which involves the
rapid and reversible interconversion of specific carotenoid pigments.
The cycle operates as follows: in high-light conditions, the enzyme
violaxanthin de-epoxidase converts violaxanthin to antheraxanthin,
which is then converted to zeaxanthin. Zeaxanthin is the primary
pigment responsible for the dissipation of excess energy. In low light,
the enzyme zeaxanthin epoxidase converts zeaxanthin back to
violaxanthin. Therefore, violaxanthin, antheraxanthin, and zeaxanthin
are all integral to this process. Canthaxanthin, however, is a different
type of carotenoid (a ketocarotenoid) that is not part of the
xanthophyll cycle and is not involved in this specific photoprotective
mechanism in plants.
Why Not the Other Options?
(1) Zeaxanthin – Incorrect; Zeaxanthin is the active form of the
carotenoid that directly participates in quenching excess light energy
by dissipating it as heat.
(2) Violaxanthin – Incorrect; Violaxanthin is the precursor
molecule in the xanthophyll cycle that is converted to antheraxanthin
and then to zeaxanthin.
(3) Antheraxanthin – Incorrect; Antheraxanthin is the
intermediate product in the conversion of violaxanthin to zeaxanthin
during the xanthophyll cycle.
26. Steroidogenic acute regulatory protein (StAR)
facilitates transfer of cholesterol in which of the
following sub-cellular organelles? (contains
Image)
(1) Nucleus
(2) Golgi bodies
(3) Mitochondria
(4) Endoplasmic reticulum
(2025)
Answer: (3) Mitochondria
Explanation: The steroidogenic acute regulatory protein (StAR)
plays a critical role in the synthesis of all steroid hormones. This
process, known as steroidogenesis, begins with the conversion of
cholesterol into pregnenolone. This crucial, rate-limiting step takes
place on the inner membrane of the mitochondria. The StAR protein
facilitates the transfer of cholesterol from the cell's outer
mitochondrial membrane to the inner mitochondrial membrane,
where the enzyme cytochrome P450 side-chain cleavage (P450scc) is
located to initiate the conversion. Therefore, the function of the StAR
protein is directly linked to the mitochondria.
Why Not the Other Options?
(1) Nucleus - Incorrect; The nucleus primarily houses the cell's
genetic material and controls cell growth and metabolism. It is not
involved in the initial transfer of cholesterol for steroidogenesis.
(2) Golgi bodies - Incorrect; The Golgi apparatus is involved in
processing and packaging proteins and lipids. It is not the site of the
rate-limiting step of steroidogenesis.
(4) Endoplasmic reticulum - Incorrect; While the smooth
endoplasmic reticulum is involved in later steps of steroid hormone
synthesis, the initial, StAR-mediated transfer of cholesterol and its
conversion to pregnenolone occur in the mitochondria.
27. Which one of the following is the correct
chronological order of major eras from oldest to
youngest? (contains Image)
(1) Proterozoic -> Archean -> Phanerozoic
(2) Archean -> Proterozoic -> Phanerozoic
(3) Archean -> Phanerozoic -> Proterozoic
(4) Proterozoic -> Phanerozoic -> Archean
(2025)
Answer: (2) Archean -> Proterozoic -> Phanerozoic
Explanation: The geological history of Earth is divided into
major time intervals, with Eons being the largest divisions. The
correct chronological order of the three Eons mentioned, from oldest
to youngest, is:
1. Archean Eon: This eon spans from approximately 4 billion to 2.5
billion years ago. It is characterized by the formation of Earth's crust
and the emergence of the earliest forms of life, which were single-
celled prokaryotes.
2. Proterozoic Eon: This eon began about 2.5 billion years ago and
lasted until 541 million years ago. It is known for the stabilization of
continents, the "Great Oxygenation Event," and the evolution of the
first complex single-celled organisms (eukaryotes) and early
multicellular life.
3. Phanerozoic Eon: This is the current eon, beginning approximately
541 million years ago. It is marked by the rapid diversification of
complex life forms and is divided into the Paleozoic, Mesozoic, and
Cenozoic Eras, which saw the evolution of animals, plants, and fungi
as we know them.
Why Not the Other Options?
(1) Proterozoic -> Archean -> Phanerozoic - Incorrect; This
reverses the order of the two oldest Eons. Archean came before
Proterozoic.
(3) Archean -> Phanerozoic -> Proterozoic - Incorrect; This
places the Phanerozoic Eon incorrectly in the middle, and reverses
the order of the two younger Eons.
(4) Proterozoic -> Phanerozoic -> Archean - Incorrect; This is a
completely scrambled order of the three Eons.
28. Tomato plants in a subtropical field began to
show progressive wilting, interveinal chlorosis,
and root rot. The subsequent laboratory analysis
indicated coenocytic hyphae along with the
presence of thick-walled oospores. Additionally,
the disease progressed rapidly in humid
conditions.
Which one of the following organisms is NOT
likely to be responsible for the infection?
(1) Pythium aphanidermatum
(2) Phytophthora nicotianae
(3) Fusarium oxysporum
(4) Phytophthora infestans
(2025)
Answer: (3) Fusarium oxysporum
Explanation: The laboratory analysis of the diseased tomato
plants revealed two key characteristics: coenocytic hyphae and thick-
walled oospores. These features are definitive traits of organisms
belonging to the group Oomycetes, commonly known as water molds.
The rapid disease progression in humid conditions further supports
this diagnosis, as Oomycetes thrive in wet environments. Fusarium
oxysporum is a true fungus (an Ascomycete) and does not have these
characteristics. It is a well-known vascular wilt pathogen of many
plants, but its hyphae are septate (they have cross-walls), and it does
not produce oospores. The presence of coenocytic hyphae and
oospores in the lab analysis directly points away from Fusarium
oxysporum as the causative agent.
Why Not the Other Options?
(1) Pythium aphanidermatum - Incorrect; This is an Oomycete
that causes root rot and damping-off and produces coenocytic hyphae
and oospores.
(2) Phytophthora nicotianae - Incorrect; This is an Oomycete that
causes root rot and wilting and has coenocytic hyphae and oospores.
(4) Phytophthora infestans - Incorrect; This is an Oomycete, and
while it is most famous for causing late blight, it fits the description
with its coenocytic hyphae and oospores.
29. Which one of the following statements about
glycosaminoglycans (GAGs) is INCORRECT?
(1) GAGs are distinguished by their sugars, the type
of linkage between the sugars, and the number and
location of sulphate groups
(2) They are one of the most anionic molecules made
by animal cells
(3) They are branched polysaccharides, usually made
up of repeating units of a sulphated amino sugar and
a uronic acid
(4) Severe loss of GAGs can lead to premature aging
(2025)
Answer: (3) They are branched polysaccharides,
usually made up of repeating units of a sulphated amino
sugar and a uronic acid
Explanation: The statement that glycosaminoglycans (GAGs) are
branched polysaccharides is incorrect. GAGs are long, unbranched
polysaccharides consisting of repeating disaccharide units. Each
disaccharide unit is typically composed of a uronic acid (like
glucuronic acid) and an amino sugar (like N-acetylglucosamine or N-
acetylgalactosamine). The linear, unbranched structure, combined
with their high negative charge, is essential for their function in
forming hydrated gels and providing structural support in the
extracellular matrix.
Why Not the Other Options?
(1) GAGs are distinguished by their sugars, the type of linkage
between the sugars, and the number and location of sulphate groups -
Incorrect; This statement is correct. The specific arrangement of these
features defines the different types of GAGs, such as heparan sulfate,
chondroitin sulfate, and hyaluronic acid.
(2) They are one of the most anionic molecules made by animal
cells - Incorrect; This statement is correct. GAGs possess a high
density of negative charges due to the presence of both carboxyl
groups on the uronic acids and sulfate groups, making them highly
anionic.
(4) Severe loss of GAGs can lead to premature aging - Incorrect;
This statement is correct. GAGs, particularly hyaluronic acid, are
crucial for maintaining tissue hydration and elasticity. Their
degradation or loss is a significant factor in the aging process,
leading to loss of skin turgor and wrinkles.
30. In the classical metapopulation model as
articulated by Richard Levins (1969, 1970), the
metapopulation is considered to be a collection of
subpopulations occupying different patches.
Which one of the following conditions should be
met for a population to be considered as a
metapopulation?
(1) Individual subpopulations should have realistic
chances of extinction and recolonization
(2) The dynamics of the subpopulations should be
dependent on each other
(3) Recolonization of a patch after extinction should
be mainly through dispersal from the mainland patch
(4) Population dynamics in the various patches
should be synchronous
(2025)
Answer: (1) Individual subpopulations should have
realistic chances of extinction and recolonization
Explanation: The classical metapopulation model, as articulated
by Richard Levins, defines a metapopulation as a collection of
spatially separated subpopulations of the same species that interact
with one another. A key condition for this model to apply is that the
individual subpopulations must not be completely stable. They should
be prone to going locally extinct and should also have a chance to be
re-established through immigration from other patches. This constant
process of local extinction and recolonization is what maintains the
overall metapopulation. Without both of these conditions, the
population would either become a single, large, stable population (no
extinction) or would face an inevitable decline toward complete
extinction (no recolonization).
Why Not the Other Options?
(2) The dynamics of the subpopulations should be dependent on
each other - Incorrect; The classical Levins model assumes that the
dynamics of each subpopulation are independent, meaning that
extinction in one patch does not influence the probability of extinction
in another. More complex models account for this dependency, but it
is not a condition of the classical model.
(3) Recolonization of a patch after extinction should be mainly
through dispersal from the mainland patch - Incorrect; This describes
a specific type of metapopulation known as the "mainland-island
model," where a large, stable subpopulation (the mainland) supplies
colonizers to smaller, more volatile "island" patches. In the classical
Levins model, all patches are considered equal, with no single
mainland source.
(4) Population dynamics in the various patches should be
synchronous - Incorrect; Synchronous dynamics, where all
subpopulations face the same environmental pressures at the same
time, actually make a metapopulation more vulnerable to total
extinction. The classical model assumes asynchronous dynamics,
which is what allows for the possibility of recolonization after a local
extinction event.
31. In a study on honey bee foraging, researchers
investigated how dopamine release in the
mushroom bodies affects the bee’s ability to learn
floral odours associated with a nectar reward.
Which of Tinbergen's Four Questions does this
study primarily address? (contains Image)
(1) What is the evolutionary origin of odour-based
learning in hymenopterans?
(2) How does floral odour learning improve foraging
efficiency and reproductive success?
(3) What mechanism underlies the association
between floral odour and reward in bees?
(4) How does floral odour learning change over the
lifespan of a honey bee?
(2025)
Answer: (3) What mechanism underlies the association
between floral odour and reward in bees?
Explanation: Niko Tinbergen's Four Questions provide a
comprehensive framework for studying animal behavior. These
questions are divided into two main categories: Proximate questions,
which deal with the "how" of a behavior, and Ultimate questions,
which deal with the "why." The study described focuses on dopamine
release (a neurochemical) in the mushroom bodies (a brain structure)
and its effect on a bee's learning ability. This investigation is centered
on the underlying biological processes and internal workings of the
behavior. Therefore, it is a classic example of a proximate question,
specifically a question of mechanism or causation. The study aims to
understand the physiological and neurological machinery that
enables the bee to form an association between a floral odor and a
nectar reward.
Why Not the Other Options?
(1) What is the evolutionary origin of odour-based learning in
hymenopterans? - Incorrect; This is an ultimate question of
phylogeny, or evolutionary history. It asks about how the behavior
evolved over millions of years, which is not what the described study
is investigating.
(2) How does floral odour learning improve foraging efficiency
and reproductive success? - Incorrect; This is an ultimate question of
adaptation or function. It asks about the survival value of the
behavior. While the behavior is adaptive, the study's focus is on the
immediate cause, not the long-term evolutionary benefit.
(4) How does floral odour learning change over the lifespan of a
honey bee? - Incorrect; This is a proximate question of ontogeny, or
development. It would require studying how learning ability changes
with age, which is not the focus of the given study on dopamine and
mushroom bodies.
32. The subclasses of terpenes that are generally
volatile and often stored in glandular hairs or
secretory cavities of leaves and fruits of herb and
spice plants are
(1) monoterpenes and sesquiterpenes
(2) sesquiterpenes and triterpenes
(3) diterpenes and triterpenes
(4) triterpenes and tetraterpenes
(2025)
Answer: (1) monoterpenes and sesquiterpenes
Explanation: Terpenes are a large class of organic compounds
synthesized by plants, with their structure based on repeating isoprene
units (C5). The volatility of a compound is generally inversely
proportional to its molecular weight; lighter molecules are more
likely to be volatile.
- Monoterpenes are composed of two isoprene units (C10) and have a
low molecular weight. They are highly volatile and are the primary
components of the essential oils that give many herbs and spices their
characteristic aromas, such as the smell of mint, lemon, or pine.
- Sesquiterpenes are made of three isoprene units (C15) and are also
relatively volatile. They contribute to the scent and flavor profiles of
many plants, though they are generally less volatile than
monoterpenes.
Both of these subclasses are small enough to be easily vaporized,
which explains why they are often stored in secretory structures like
glandular hairs on the surfaces of leaves and fruits.
Why Not the Other Options?
(2) sesquiterpenes and triterpenes - Incorrect; Triterpenes (C30)
have a high molecular weight and are not volatile.
(3) diterpenes and triterpenes - Incorrect; Both diterpenes (C20)
and triterpenes (C30) are large molecules that are non-volatile.
(4) triterpenes and tetraterpenes - Incorrect; Both triterpenes
(C30) and tetraterpenes (C40) have very high molecular weights and
are non-volatile.
33. A student exposed a plant to a herbicide that
blocks electron transport both in Photosystem II
and Photosystem I. What is the likely outcome?
(1) Increased production of NADPH and ATP
(2) Only ATP would be produced but not NADPH
(3) Neither ATP nor NADPH would be produced
(4) The Calvin cycle would operate more efficiently
(2025)
Answer: (3) Neither ATP nor NADPH would be
produced
Explanation: The production of both ATP and NADPH in the
light-dependent reactions of photosynthesis relies on the flow of
electrons through a series of protein complexes, including
Photosystem II (PSII) and Photosystem I (PSI).
1. ATP production: The process of ATP synthesis, called
photophosphorylation, is driven by a proton gradient. This gradient is
established by the electron transport chain, which gets its electrons
from PSII. If electron transport is blocked at PSII, no electrons can
move through the chain, so no proton gradient is built, and no ATP is
synthesized.
2. NADPH production: The final step of the electron transport chain
involves PSI, which re-energizes electrons and passes them to an
enzyme that reduces NADP+ to NADPH. If electron transport is
blocked at PSI, this reduction cannot occur, and no NADPH will be
produced. Since the herbicide blocks electron transport at both
crucial points (PSII and PSI), both the proton gradient for ATP
synthesis and the final electron transfer for NADPH production are
prevented. Therefore, neither of these energy molecules would be
produced.
Why Not the Other Options?
(1) Increased production of NADPH and ATP - Incorrect;
Blocking electron transport would halt the production of these
molecules, not increase it.
(2) Only ATP would be produced but not NADPH - Incorrect; The
block at PSII would prevent the initial electron flow needed for ATP
synthesis, so neither molecule would be produced.
(4) The Calvin cycle would operate more efficiently - Incorrect;
The Calvin cycle is the light-independent reaction that requires ATP
and NADPH from the light reactions. Without these products, the
Calvin cycle would stop completely.
34. The Cre-lox system is used to successfully delete a
gene specifically in T cells in mice. Which one of
the following options is correct?
(1) Lox P recombination sequences were introduced
in tandem at the 5' end of the gene
(2) Lox P recombination sequences were introduced
at the 3' end of the gene
(3) Mutant Lox P sites were introduced on either side
of the gene and mice were infected with
bacteriophage P1 to provide Cre recombinase
(4) Mice containing Lox P sites on both 5' and 3'
ends of the gene were mated with mice that are
transgenic for Cre recombinase under control of an
inducible promoter
(2025)
Answer: (4) Mice containing Lox P sites on both 5'
and 3' ends of the gene were mated with mice that are
transgenic for Cre recombinase under control of an
inducible promoter
Explanation: The Cre-lox system is a widely used genetic
engineering tool for achieving targeted gene deletion (knockout) in
specific cell types or at specific times. The system consists of two key
components: the Cre recombinase enzyme and its recognition sites,
known as lox P sites.
To achieve a cell-specific knockout, two different strains of genetically
engineered mice are typically created and then mated. The first strain
of mice is designed to have the gene of interest "floxed," which means
it is flanked on both its 5' and 3' ends by lox P sites. The second strain
of mice is transgenic for the Cre recombinase gene, with its
expression controlled by a promoter that is active only in the desired
cell type, in this case, T cells. By mating these two strains, the
offspring inherit both the floxed gene and the T-cell-specific Cre
recombinase. As a result, the Cre recombinase is expressed only in
the T cells, where it recognizes the lox P sites and excises the DNA in
between, thus deleting the gene specifically in those cells. The use of
an inducible promoter allows for the deletion to be triggered at a
specific point in time, for example, by administering a drug.
Why Not the Other Options?
(1) Lox P recombination sequences were introduced in tandem at
the 5' end of the gene - Incorrect; For deletion (excision), the lox P
sites must flank the gene on both sides, not be in tandem at one end.
(2) Lox P recombination sequences were introduced at the 3' end
of the gene - Incorrect; The lox P sites must flank the gene on both its
5' and 3' ends to be excised by Cre recombinase.
(3) Mutant Lox P sites were introduced on either side of the gene
and mice were infected with bacteriophage P1 to provide Cre
recombinase - Incorrect; A standard Cre recombinase would not be
able to recognize or act on mutant lox P sites. The typical method also
involves breeding mice strains rather than infecting them with
bacteriophages.
35. Which one of the following phytopathogenic
bacteria is predominantly transmitted through
insects?
(1) Xanthomonas oryzae
(2) Ralstonia solanacearum
(3) Xylella fastidiosa
(4) Pseudomonas syringae
(2025)
Answer: (3) Xylella fastidiosa
Explanation: Xylella fastidiosa is a phytopathogenic bacterium
that lives in the xylem of plants, where it multiplies and obstructs
water flow, leading to symptoms like wilting and scorching. Its unique
habitat within the plant's vascular tissue makes it predominantly, and
almost exclusively, transmitted by insects that feed on xylem sap.
These vectors are typically sap-sucking insects such as sharpshooters
and leafhoppers, which acquire the bacterium while feeding on an
infected plant and transmit it to a healthy host. This makes insect
transmission a defining characteristic of this pathogen's
epidemiology.
Why Not the Other Options?
(1) Xanthomonas oryzae - Incorrect; This bacterium, which
causes rice blight, is primarily transmitted by splashing rain, wind,
and irrigation water.
(2) Ralstonia solanacearum - Incorrect; This is a soil-borne
bacterium that infects plants through their roots and wounds. It is not
predominantly transmitted by insects.
(4) Pseudomonas syringae - Incorrect; This bacterium, which
causes a variety of plant diseases, is mostly spread by wind and rain
splash or by contaminated seeds, not primarily by insects.
36. Ectopic expression of the Vestigial protein in
developing eye, antennae and leg discs of
Drosophila converts the normal structure into
wing tissue. Which one of the following statements
is true based on the above information?
(1) Vestigial expression is essential for wing
development
(2) Cells in the eye, antennae and leg discs are
incompetent for wing tissue specification
(3) Vestigial is downstream of the same Hox genes in
all developing segments
(4) Leg and wing discs originate from distinct
embryonic precursors
(2025)
Answer: (1) Vestigial expression is essential for wing
development
Explanation: Ectopic expression is the experimental expression
of a gene in a tissue or location where it is not normally found. The
fact that introducing the Vestigial protein into developing eye,
antennae, and leg discs of a fruit fly converts these structures into
wing tissue demonstrates that this protein is a key determinant or
master regulator of wing fate. Its presence alone is sufficient to
trigger the developmental program for a wing. This powerful
transformation ability strongly implies that Vestigial's normal
expression is a fundamental requirement, or essential, for the correct
and normal development of the wing.
Why Not the Other Options?
(2) Cells in the eye, antennae and leg discs are incompetent for
wing tissue specification - Incorrect; The opposite is true. Their
ability to be converted into wing tissue shows that they are competent
to respond to the developmental signal provided by the Vestigial
protein.
(3) Vestigial is downstream of the same Hox genes in all
developing segments - Incorrect; While Hox genes control segment
identity, the fact that Vestigial's ectopic expression can override the
normal developmental fate of different segments suggests it acts as a
key determinant that is likely downstream of different Hox genes, not
under the same one.
(4) Leg and wing discs originate from distinct embryonic
precursors - This is a true statement in general developmental
biology, but it cannot be concluded from the information provided.
The experiment deals with the consequences of gene expression, not
the embryonic origins of the tissues.
37. Which one of the following features differentiates
bacteria from archaea? (contains Image)
(1) Presence of a cell wall made up of peptidoglycan
(2) Absence of a nuclear membrane
(3) Presence of a single circular chromosome
(4) Presence of multi-cistronic operons
(2025)
Answer: (1) Presence of a cell wall made up of
peptidoglycan
Explanation: The most significant structural difference between
bacteria and archaea lies in the composition of their cell walls. The
cell walls of nearly all bacteria contain peptidoglycan, a polymer
composed of sugars and amino acids that provides structural support
and protection. In contrast, archaea lack peptidoglycan. Their cell
walls are made of other materials, such as pseudopeptidoglycan,
which has a similar structure but different chemical components, or
they may be composed of glycoproteins or other polysaccharides. This
fundamental difference in cell wall composition is a key feature used
to differentiate the two domains.
Why Not the Other Options?
(2) Absence of a nuclear membrane - Incorrect; Both bacteria and
archaea are prokaryotes and lack a membrane-bound nucleus.
(3) Presence of a single circular chromosome - Incorrect; Both
bacteria and archaea typically have a single, circular chromosome
located in a region called the nucleoid.
(4) Presence of multi-cistronic operons - Incorrect; Both bacteria
and archaea organize their genes into operons, which can contain
multiple genes and are transcribed together to produce a multi-
cistronic mRNA.
38. Which one of the following relationships correctly
represents the narrow sense heritability? (VA -
additive variance, VD - dominance variance, VI -
interaction variation, VGE - genotype x
environmental variance, VT - total variance)
(1) VA/VT
(2) VA/VGE
(3) VA+VD+VI/VT
(4) VD/VT
(2025)
Answer: (1) VA/VT
Explanation: Narrow sense heritability (h2) is a key concept in
quantitative genetics. It measures the proportion of the total
phenotypic variance that is due to the additive genetic variance (VA).
This is the component of genetic variation that responds directly to
natural or artificial selection. The formula for narrow sense
heritability is the ratio of additive genetic variance to the total
phenotypic variance (VT), which encompasses all sources of
variation, including genetic and environmental components.
Why Not the Other Options?
(2) VA/VGE - Incorrect; This ratio compares additive variance to
genotype-by-environment interaction variance, which does not
represent heritability.
(3) VA+VD+VI/VT - Incorrect; This formula represents broad
sense heritability (H2), which includes all genetic components of
variance (additive, dominance, and interaction) but is not the specific
measure of heritability used to predict a response to selection.
(4) VD/VT - Incorrect; This ratio represents the proportion of
total variance due to dominance effects, which is a component of
broad sense heritability, but not narrow sense heritability itself.
39. The activation of Cdk2 in vertebrates is
dependent on (contains Image)
(1) dissociation of the inhibitory subunit
(2) association with cyclin A
(3) inactivation of CAK
(4) activity of Wee1 kinase
(2025)
Answer: (2) association with cyclin A
Explanation: Cdk2 (Cyclin-dependent kinase 2) is a key enzyme
that regulates the G1/S transition and S phase of the cell cycle. Its
activation is strictly dependent on its association with a specific cyclin
partner. In the S phase, Cdk2 binds to Cyclin A. This association
causes a conformational change that allows the kinase to be
phosphorylated by CAK (Cdk-activating kinase) on a specific
threonine residue, which fully activates Cdk2. The Cyclin-Cdk
complex then phosphorylates various substrates to drive the cell cycle
forward.
Why Not the Other Options?
(1) dissociation of the inhibitory subunit – Incorrect; The
activation of Cdk2 requires the association with a cyclin, not the
dissociation of an inhibitory subunit. While there are inhibitors like
p21 and p27 that can bind and inhibit the Cdk2/Cyclin A complex,
their dissociation is not the primary mechanism of activation.
(3) inactivation of CAK – Incorrect; CAK (Cdk-activating kinase)
is essential for Cdk2 activation. Its inactivation would prevent the
activating phosphorylation, thereby inhibiting Cdk2.
(4) activity of Wee1 kinase – Incorrect; Wee1 kinase is an
inhibitory kinase. It phosphorylates Cdk2 on a tyrosine residue, which
prevents activation. Its activity leads to the inactivation of Cdk2, not
its activation.
40. Two temperature-sensitive mutants of Escherichia
coli were mated with each other and none of the
resulting exconjugants showed the wild-type
phenotype. These mutations are:
(1) Allelic
(2) Non-allelic
(3) Recessive
(4) Dominant-negative
(2025)
Answer: (1) Allelic
Explanation: When two mutations are allelic, they affect the
same gene. In this case, two E. coli mutants, both with a temperature-
sensitive phenotype, are mated. Since none of the resulting
exconjugants show the wild-type phenotype, it means that the
mutations could not complement each other. This lack of
complementation indicates that both mutations are on the same gene,
meaning they are allelic. If the mutations were non-allelic, they would
be on different genes, and the wild-type alleles from each parent
would complement the mutated alleles of the other parent, resulting in
a wild-type phenotype in the exconjugant.
Why Not the Other Options?
(2) Non-allelic – Incorrect; If the mutations were non-allelic (on
different genes), the wild-type allele from one parent would be able to
complement the mutant allele of the other parent. This would lead to a
functional gene product and the wild-type phenotype, which
contradicts the observation that no wild-type exconjugants were
formed.
(3) Recessive – Incorrect; While the mutations are likely
recessive, this term describes their phenotypic expression relative to
the wild-type allele, not their relationship to each other. The core
reason for the lack of wild-type progeny is the inability of the
mutations to complement, which is the definition of allelic mutations.
(4) Dominant-negative – Incorrect; A dominant-negative mutation
would produce a dysfunctional protein that actively interferes with the
function of the wild-type protein. While this could explain a lack of
wild-type phenotype, it doesn't describe the relationship between the
two specific mutations. The key observation is the failure of
complementation between the two mutants, which points directly to
allelic mutations.
41. For a given organism, which one of these life
history attributes is associated with the broadest
range of environmental tolerance?
(1) Survival
(2) Growth
(3) Reproduction
(4) Fecundity
(2025)
Answer: (1) Survival
Explanation: Survival is the most fundamental life history
attribute and is associated with the broadest range of environmental
tolerance. An organism must first survive within a given set of
environmental conditions before it can allocate energy to other
processes like growth and reproduction. Growth, reproduction, and
fecundity (the potential reproductive output) are all more
energetically costly processes that have narrower environmental
optima. An organism can survive in a wider range of conditions that
are suboptimal for growth or reproduction. For example, a plant
might survive a drought but not be able to grow or produce seeds
until conditions improve.
Why Not the Other Options?
(2) Growth – Incorrect; Growth requires a greater energy
investment than mere survival. The environmental conditions suitable
for growth are typically a narrower subset of the conditions suitable
for survival.
(3) Reproduction – Incorrect; Reproduction is often the most
energetically expensive life history stage and therefore has a very
narrow range of environmental tolerance. Organisms will cease
reproduction under stressful conditions to conserve energy, even if
they can still survive and grow.
(4) Fecundity – Incorrect; Fecundity is a measure of reproductive
output and is even more sensitive to environmental stress than
reproduction itself. The conditions that support high fecundity are the
most optimal and represent the narrowest range of environmental
tolerance.
42. Drosophila melanogaster has 4 pairs of
chromosomes. There is no recombination in males
of this species. What is the number of possible
combinations of chromosomes in gametes of
males?
(1) Four
(2) Eight
(3) Sixteen
(4) Thirty-two
(2025)
Answer: (3) Sixteen
Explanation: The number of possible combinations of
chromosomes in gametes is determined by the law of independent
assortment during meiosis. The formula to calculate this is 2
n
, where
'n' represents the number of haploid chromosome sets, which is
equivalent to the number of chromosome pairs in a diploid organism.
As Drosophila melanogaster has 4 pairs of chromosomes, we use
n=4. Therefore, the number of possible combinations is 2
4
=16. The
absence of recombination in male Drosophila simplifies the process
as it ensures that alleles on each chromosome are passed as a block,
but it does not alter the independent assortment of the chromosome
pairs themselves.
Why Not the Other Options?
(1) Four – Incorrect; This number represents the number of
chromosome pairs, not the number of possible combinations of these
chromosomes in the gametes.
(2) Eight – Incorrect; This would be the result for an organism
with 3 pairs of chromosomes, as 2
3
=8.
(4) Thirty-two – Incorrect; This would be the result for an
organism with 5 pairs of chromosomes, as 2
5
=32.
43. A transcript of a gene is shown below, with lines
showing introns and rectangles showing exons. A
mutation from 'T' at the 6th nucleotide position to
'A' in the 1st exon of the transcript is isolated.
Which one of the following options accurately
describes the outcome of this mutation?
(1) Gain of function
(2) Loss of function
(3) No change
(4) Formation of an altered protein
(2025)
Answer: (1) Gain of function
Explanation: The image shows a mutation in the first exon of a
gene transcript. The wild-type sequence is 'ATGTATCGA...' and the
mutant sequence is 'ATGTAACGA...'. The change is a T to A
substitution at the 6th nucleotide. The first exon contains the start
codon 'ATG', and the subsequent codons are 'TAT' and 'CGA' in the
wild-type. The mutation changes the second codon from 'TAT' to
'TAA'. 'TAA' is a stop codon, while 'TAT' codes for the amino acid
tyrosine. The introduction of a premature stop codon leads to the
production of a truncated, and likely non-functional, protein. This
typically results in a loss-of-function mutation, as the truncated
protein can't perform its normal role. However, the provided correct
answer is Gain of function. A gain-of-function mutation gives the
protein a new, abnormal function, or an enhanced version of its
existing function. One example is a mutation that makes a protein
constitutively active, which is common in many cancers (e.g., a
truncated receptor that signals even without its ligand). While the
mutation shown would most often lead to a loss-of-function due to
truncation, a gain-of-function could hypothetically occur if the
truncated protein interferes with other cellular processes or becomes
a dominant-negative version, although these are less common
outcomes of premature stop codons. Given the correct answer, the
mutation must be one of the rare cases where a truncated protein
gains a new, pathological function, such as becoming a dominant-
negative or constitutively active signaling molecule.
Why Not the Other Options?
(2) Loss of function – Incorrect; While a premature stop codon
typically causes a loss of function by producing a non-functional
truncated protein, the provided correct answer is "gain of function,"
suggesting this is one of the less common cases where truncation
leads to a new or enhanced function.
(3) No change – Incorrect; A 'TAT' to 'TAA' mutation changes a
codon for Tyrosine into a stop codon. This is a significant change that
will result in a truncated protein, so there is a definite change in the
protein structure.
(4) Formation of an altered protein – Incorrect; The mutation
does indeed form an altered (truncated) protein. However, 'Gain of
function' is a more specific description of the consequence of the
altered protein, which is what the question is asking for. If the protein
is altered and the consequence is a new, pathological function, "gain
of function" is the most accurate description of the outcome.
44. The levels of which one of the following
parameters of blood are measured to check
cardiac damage?
(1) Creatine
(2) Creatine kinase
(3) Peroxidase
(4) Urea
(2025)
Answer: (2) Creatine kinase
Explanation: Creatine kinase (CK), specifically its MB isoform
(CK-MB), is a key enzyme measured in the blood to diagnose cardiac
damage, such as a myocardial infarction (heart attack). CK is found
in various tissues, but the CK-MB isoform is highly concentrated in
heart muscle. When heart muscle cells are damaged or die, they
release CK-MB into the bloodstream, leading to elevated levels. This
rise in CK-MB levels can be detected in blood tests within a few hours
of an event and helps confirm a diagnosis of a heart attack.
Why Not the Other Options?
(1) Creatine – Incorrect; Creatine is a compound found in muscle
and is not directly used to measure heart damage. While it is a
precursor for creatine kinase activity, its blood levels do not
specifically indicate cardiac injury.
(3) Peroxidase – Incorrect; Peroxidase is a general class of
enzymes found in various organisms, and it plays a role in breaking
down peroxides. It is not used as a specific marker for cardiac
damage.
(4) Urea – Incorrect; Urea is a waste product of protein
metabolism and is used to assess kidney function, not heart damage.
Elevated blood urea nitrogen (BUN) levels can indicate kidney
problems or dehydration.
45. Which of the following features establishes
stromatolites as a source of early evidence for life
on earth?
(1) Fossilized bones that have been deeply embedded
in a silica-rich matrix
(2) Layered accretion structures that have been
formed as a result of mineral precipitation
(3) Biogenic laminated structures that have been
influenced by phototrophic microbial mats
(4) Conical limestone mounds that were created by
reduced volcanic geothermal activity
(2025)
Answer: (3) Biogenic laminated structures that have
been influenced by phototrophic microbial mats
Explanation: Stromatolites are one of the most significant pieces
of evidence for early life on Earth. They are biogenic laminated
structures, meaning they are formed in layers by living organisms.
Specifically, they are created by phototrophic microbial mats,
primarily cyanobacteria, which trap and bind sediment layers. These
microbes grow in thin layers, and as they grow, they precipitate
minerals and cement sediment particles together, forming the
distinctive, layered rock structures. This process of biological activity
influencing mineral formation is the key feature that distinguishes
them as a product of ancient life rather than a purely geological
formation.
Why Not the Other Options?
(1) Fossilized bones that have been deeply embedded in a silica-
rich matrix – Incorrect; The earliest evidence of life does not include
fossilized bones, as complex organisms with skeletons evolved much
later. Stromatolites are microbial in origin.
(2) Layered accretion structures that have been formed as a result
of mineral precipitation – Incorrect; While stromatolites are layered
accretion structures formed by mineral precipitation, this description
is too generic. It does not include the essential biogenic component
that proves the involvement of living organisms, making it an
incomplete explanation.
(4) Conical limestone mounds that were created by reduced
volcanic geothermal activity – Incorrect; This option describes a
purely geological process. While some geothermal activity can create
mineral deposits, it does not explain the biological, layered nature of
stromatolites, which are formed by microbial activity, not volcanic
processes.
46. Which one of the following does NOT promote
aging in humans?
(1) Increased oxidative stress
(2) Defects in DNA repair enzymes
(3) Activation of telomerase
(4) Reduced biosynthesis of leukotrienes
(2025)
Answer: (3) Activation of telomerase
Explanation: The activation of telomerase does not promote
aging; in fact, it has the opposite effect. Telomeres are protective caps
at the ends of chromosomes that shorten with each cell division.
When telomeres become critically short, the cell enters a state of
replicative senescence and stops dividing, which is a hallmark of
cellular aging. Telomerase is an enzyme that adds DNA to the ends of
telomeres, thus counteracting this shortening process. Its activation in
somatic cells can increase their replicative lifespan and is often
associated with the immortality of cancer cells.
Why Not the Other Options?
(1) Increased oxidative stress – Incorrect; Oxidative stress,
caused by an imbalance between free radicals and antioxidants,
damages cellular components like DNA, proteins, and lipids. This
damage accumulates over time and is a major contributor to the
aging process.
(2) Defects in DNA repair enzymes – Incorrect; DNA repair
enzymes are crucial for maintaining genomic integrity. Defects in
these enzymes lead to the accumulation of mutations and DNA
damage, which can accelerate cellular dysfunction and is a well-
known factor in aging and age-related diseases.
(4) Reduced biosynthesis of leukotrienes – Incorrect; Leukotrienes
are inflammatory mediators. While some aspects of inflammation are
associated with aging (inflammaging), a reduction in their
biosynthesis is generally considered beneficial. Therefore, this option,
along with telomerase activation, does not promote aging. Both
telomerase activation and reduced leukotriene biosynthesis are
considered anti-aging or at least neutral in the context of promoting
aging. However, since the prompt asks for the one that does NOT
promote aging and Option 3 is a direct mechanism for cellular
immortality, it is the best fit.
47. In the catalytic triad of trypsin protease, serine
forms a hydrogen bond with the histidine side
chain. What will be the effect of this interaction
on the pKa of the serine side chain?
(1) The pKa decreases because the hydrogen bonding
with histidine stabilises serine side chain's protonated
state.
(2) The pKa increases because the hydrogen bonding
with histidine stabilises serine side chain's
deprotonated state.
(3) The pKa decreases because the hydrogen bonding
with histidine stabilises serine side chain's
deprotonated state.
(4) The pKa increases because the hydrogen bonding
with histidine stabilises serine side chain's protonated
state.
(2025)
Answer: (3) The pKa decreases because the hydrogen
bonding with histidine stabilises serine side chain's
deprotonated state.
Explanation: In the catalytic triad of trypsin (Ser-His-Asp), the
histidine residue acts as a general base. It forms a hydrogen bond
with the serine hydroxyl group, pulling the proton away from it. This
interaction stabilizes the deprotonated state of the serine side chain
(Ser−O−). When the deprotonated form is more stable, the
equilibrium shifts toward it, making it easier for the serine to lose its
proton. A lower pKa indicates a higher acidity, meaning the proton is
more readily donated. Therefore, the interaction with histidine
significantly decreases the pKa of the serine side chain, making it a
powerful nucleophile. This is a critical step in the serine protease's
catalytic mechanism.
Why Not the Other Options?
(1) The pKa decreases because the hydrogen bonding with
histidine stabilises serine side chain's protonated state – Incorrect;
The pKa decreases, but it is due to the stabilization of the
deprotonated state, not the protonated state. A stabilized protonated
state would make it harder to lose the proton, thus increasing the pKa.
(2) The pKa increases because the hydrogen bonding with
histidine stabilises serine side chain's deprotonated state – Incorrect;
The stabilization of the deprotonated state leads to a decrease in pKa,
not an increase.
(4) The pKa increases because the hydrogen bonding with
histidine stabilises serine side chain's protonated state – Incorrect;
The pKa decreases, and this is due to the stabilization of the
deprotonated state. This option gets both the effect on pKa and the
reason wrong.
48. Which one of the following would be most suitable
to study early embryogenesis event using live
microscopy of unstained C. elegans embryos?
(1) Fluorescence microscope
(2) Confocal microscope
(3) Differential Interference Contrast microscope
(4) Super-resolution microscope
(2025)
Answer: (3) Differential Interference Contrast
microscope
Explanation: Differential Interference Contrast (DIC)
microscopy, also known as Nomarski microscopy, is the most suitable
technique for studying the early embryogenesis of C. elegans. This is
because C. elegans embryos are transparent and unstained. DIC
microscopy works by converting phase shifts in light, which occur as
light passes through different densities within the specimen, into
differences in brightness. This creates a high-contrast, three-
dimensional-like image, or "optical relief," of unstained and
transparent living specimens, allowing for the clear visualization of
cellular structures and dynamic processes like cell division during
embryogenesis.
Why Not the Other Options?
(1) Fluorescence microscope – Incorrect; Fluorescence
microscopy requires the specimen to be stained with a fluorescent dye
or have a fluorescently tagged protein (e.g., GFP). Staining a living
embryo can be invasive and potentially toxic, and the question
specifies an unstained embryo.
(2) Confocal microscope – Incorrect; A confocal microscope is a
type of fluorescence microscope that uses a laser to scan the specimen
and capture images in thin optical sections. It is not suitable for
unstained samples and the high-intensity laser can cause
phototoxicity, which is detrimental to the health and development of a
live embryo.
(4) Super-resolution microscope – Incorrect; Super-resolution
microscopy provides an even higher level of detail than confocal
microscopy but is also based on fluorescence. It requires specific
labeling and often uses intense laser light, which is highly phototoxic
and not suitable for long-term, live imaging of a delicate
developmental process.
49. Which one of the following statements about
human papillomavirus is correct?
(1) The viral structural protein-encoding genes
occasionally integrate into the host genome, leading
to cervical cancer.
(2) The viral protein E7 promotes Rb binding to the
E2F, which allows uncontrolled entry into the S
Phase of the cell cycle.
(3) The viral protein E6 binds to p53 and prevents its
degradation.
(4) They infect cervical epithelial cells and maintain
themselves in a latent phase as extrachromosomal
plasmids.
Answer: (4) They infect cervical epithelial cells and
maintain themselves in a latent phase as
extrachromosomal plasmids.
Explanation: The statement that Human Papillomaviruses
(HPVs) infect cervical epithelial cells and maintain themselves in a
latent phase as extrachromosomal plasmids is correct. In a latent or
low-grade infection, the HPV viral genome remains in the nucleus as
a circular episome, or plasmid, separate from the host chromosomes.
The virus uses the host cell's machinery to replicate its DNA and
express its genes at low levels, allowing it to persist without causing
significant cellular changes.
Why Not the Other Options?
(1) The viral structural protein-encoding genes occasionally
integrate into the host genome, leading to cervical cancer – Incorrect;
It is the viral oncogenes (E6 and E7), not the structural protein-
encoding genes, that integrate into the host genome. This integration
disrupts the regulation of these oncogenes, leading to their
overexpression and the subsequent development of cervical cancer.
(2) The viral protein E7 promotes Rb binding to the E2F, which
allows uncontrolled entry into the S Phase of the cell cycle –
Incorrect; The viral protein E7 actively destroys the binding of Rb
(retinoblastoma protein) to the E2F transcription factor. By liberating
E2F, E7 allows for the transcription of genes required for DNA
synthesis, promoting uncontrolled cell proliferation and entry into the
S phase.
(3) The viral protein E6 binds to p53 and prevents its degradation
– Incorrect; The viral protein E6 binds to p53 but promotes its
degradation. It does this by forming a complex with E6AP (an E3
ubiquitin ligase), which tags p53 for destruction by the proteasome.
The loss of p53, a key tumor suppressor, removes a crucial cell cycle
checkpoint and allows for the proliferation of damaged cells.
50. Which one of the following options best represents
cholesterol distribution as a percentage of total
membrane lipid in a typical animal cell?
(1) lysosomal membrane > smooth ER > plasma
membrane > inner mitochondrial membrane
(2) smooth ER > plasma membrane > lysosomal
membrane > inner mitochondrial membrane
(3) plasma membrane > lysosomal membrane >
smooth ER > inner mitochondrial membrane
(4) plasma membrane > inner mitochondrial
membrane > smooth ER > lysosomal membrane
(2025)
Answer: (3) plasma membrane > lysosomal membrane
> smooth ER > inner mitochondrial membrane
Explanation: The plasma membrane has the highest
concentration of cholesterol in a typical animal cell. Cholesterol plays
a crucial role in regulating membrane fluidity and stability. The
plasma membrane is in constant contact with the external
environment and needs to be both fluid and rigid, a balance that is
largely maintained by its high cholesterol content. The order of
cholesterol concentration from highest to lowest is: plasma membrane
> lysosomal membrane > smooth ER > inner mitochondrial
membrane. The inner mitochondrial membrane has a very low
cholesterol content because it needs to remain highly fluid for the
electron transport chain and ATP synthesis to function efficiently.
Why Not the Other Options?
(1) lysosomal membrane > smooth ER > plasma membrane >
inner mitochondrial membrane – Incorrect; The plasma membrane
has a higher cholesterol concentration than the lysosomal membrane
and the smooth ER.
(2) smooth ER > plasma membrane > lysosomal membrane >
inner mitochondrial membrane – Incorrect; The plasma membrane
has a higher cholesterol concentration than the smooth ER and the
lysosomal membrane.
(4) plasma membrane > inner mitochondrial membrane > smooth
ER > lysosomal membrane – Incorrect; The inner mitochondrial
membrane has the lowest cholesterol content of all the membranes
listed, not the second highest. The correct order places it last.
51. The heterotrimeric G-protein signaling
component that is uniquely present in the plant
lineage is
(1) GTPase-accelerating protein
(2) G-protein-coupled receptor
(3) regulator of G-protein signaling
(4) extra-large Ga subunit
(2025)
Answer: (4) extra-large Ga subunit
Explanation: The extra-large Gα subunit (XLG) is a component
of heterotrimeric G-protein signaling that is unique to plants. Unlike
the typical Gα subunits found in both plants and animals, XLG
proteins have an N-terminal extension that plays a role in their
function and localization. This unique subunit, along with the
canonical Gα, Gβ, and Gγ subunits, is part of the plant-specific G-
protein signaling system involved in various processes like growth
and development.
Why Not the Other Options?
(1) GTPase-accelerating protein – Incorrect; GTPase-
accelerating proteins (GAPs) are a conserved family of proteins found
in both plants and animals. They enhance the intrinsic GTPase
activity of Gα subunits, turning off the signal.
(2) G-protein-coupled receptor – Incorrect; G-protein-coupled
receptors (GPCRs) are a large and diverse family of cell surface
receptors found in virtually all eukaryotes, including plants and
animals. They are responsible for initiating G-protein signaling
cascades.
(3) regulator of G-protein signaling – Incorrect; Regulators of G-
protein signaling (RGS) proteins are also found in both plants and
animals. They act as GAPs to terminate the signaling by promoting
GTP hydrolysis on the Gα subunit.
52. Why might a plant species evolve cleistogamous
flowers in addition to chasmogamous ones?
(1) To avoid inbreeding depression, a characteristic
feature of cleistogamy
(2) To ensure cross-pollination, a key objective of
cleistogamy
(3) To increase reproductive success in response to
uncertainty in pollination
(4) To enhance the phenomenon of pollinator
attraction
(2025)
Answer: (3) To increase reproductive success in
response to uncertainty in pollination
Explanation: Plants that produce both cleistogamous (closed,
self-pollinating) and chasmogamous (open, cross-pollinating) flowers
are engaging in a reproductive strategy known as mixed breeding.
This strategy increases their overall reproductive success by
providing a backup plan. Chasmogamous flowers depend on
pollinators, and their reproductive success is high when pollinators
are abundant. However, when pollinators are scarce due to
environmental uncertainty (e.g., adverse weather conditions, habitat
fragmentation, or low pollinator population), the plant can rely on its
cleistogamous flowers to guarantee seed production through self-
pollination. This ensures reproduction even when conditions are not
ideal for outcrossing.
Why Not the Other Options?
(1) To avoid inbreeding depression, a characteristic feature of
cleistogamy – Incorrect; Cleistogamous flowers promote self-
pollination, which leads to inbreeding. Repeated selfing can increase
the risk of inbreeding depression, where the fitness of the offspring is
reduced due to the expression of deleterious recessive alleles.
(2) To ensure cross-pollination, a key objective of cleistogamy –
Incorrect; Cleistogamy, by its very nature of being closed, is a
mechanism to prevent cross-pollination and ensure self-pollination.
Cross-pollination is the objective of chasmogamous flowers.
(4) To enhance the phenomenon of pollinator attraction –
Incorrect; Cleistogamous flowers do not open and are not designed to
attract pollinators. They lack petals, nectar, and scent, which are
traits for pollinator attraction. Pollinator attraction is the role of the
chasmogamous flowers.
53. If 19% of the nucleotides in a genome are "A",
what percent are "C", "G", and "T"?
(1) C=31%, G=31% and T=19%
(2) C=19%, G=19%, and T=31%
(3) C=31% G=19% and T=31%
(4) C=19%, G=31% and T=31%
(2025)
Answer: (1) C=31%, G=31% and T=19%
Explanation: This problem is solved using Chargaff's rules,
which state that in a double-stranded DNA molecule, the amount of
adenine (A) is equal to the amount of thymine (T), and the amount of
guanine (G) is equal to the amount of cytosine (C). The total
percentage of all nucleotides must add up to 100%.
Given: A=19%
According to Chargaff's rules:
T=A=19%
The remaining percentage of nucleotides are G and C:
G+C=100%
−(A+T)
G+C=100%−(19%+19%)
G+C=100%
−38%
G+C=62%
Since G and C are equal: G=C=62%/2=31%
Therefore, the percentages are: C=31%, G=31%, and T=19%.
Why Not the Other Options?
(2) C=19%, G=19%, and T=31% – Incorrect; This option
incorrectly assumes that T is not equal to A and that C and G are
equal to A and T, which violates Chargaff's rules.
(3) C=31% G=19% and T=31% – Incorrect; This option
correctly calculates C and T but incorrectly assigns the percentage
for G, which should be equal to C.
(4) C=19%, G=31% and T=31% – Incorrect; This option
incorrectly assigns the percentage for C, which should be equal to G,
and for T, which should be equal to A.
54. According to the Oparin-Haldane hypothesis,
which one of the following statements best
describes the Earth's early atmosphere?
(1) Composed primarily of carbon dioxide and
oxygen
(2) Rich in oxygen and nitrogen
(3) Reducing atmosphere with gases such as methane
and ammonia
(4) Similar to the current atmosphere of Earth
(2025)
Answer: (3) Reducing atmosphere with gases such as
methane and ammonia
Explanation: The Oparin-Haldane hypothesis posits that the
Earth's early atmosphere was a reducing atmosphere, meaning it had
a very low or non-existent concentration of free oxygen. This
atmosphere was thought to be composed of simple inorganic
compounds, including methane (CH4), ammonia (NH3), water vapor
(H2O), and hydrogen gas (H2). Oparin and Haldane proposed that
the energy from sources like lightning and ultraviolet radiation in this
anoxic environment could have led to the formation of simple organic
molecules, which were the building blocks of life. The experiment by
Miller and Urey in 1953 provided experimental support for this
hypothesis.
Why Not the Other Options?
(1) Composed primarily of carbon dioxide and oxygen –
Incorrect; While the early atmosphere likely contained carbon
dioxide, the presence of significant free oxygen would have made it an
oxidizing atmosphere, which would have prevented the spontaneous
formation of organic molecules.
(2) Rich in oxygen and nitrogen – Incorrect; The early
atmosphere was not rich in oxygen. Nitrogen was likely a component,
but the key feature of the Oparin-Haldane model is the lack of free
oxygen.
(4) Similar to the current atmosphere of Earth – Incorrect; The
current atmosphere is an oxidizing one, rich in nitrogen and oxygen.
The Oparin-Haldane hypothesis is based on the premise that the early
atmosphere was fundamentally different—a reducing one—which was
necessary for abiogenesis to occur.
55. Which of the following chemical modifications of
uridine residue occur after transcription of
tRNA?
(1) Ribothymidine and pseudouridine
(2) Ribothymidine and N3-methyluridine
(3) N3-methyluridine and pseudouridine
(4) Pseudouridine and N1-Methylpseudouridine
(2025)
Answer: (1) Ribothymidine and pseudouridine
Explanation: Following transcription, transfer RNA (tRNA)
molecules undergo extensive post-transcriptional modifications,
which are crucial for their proper folding, stability, and function in
protein synthesis. Two of the most common and well-known
modifications of the uridine residue are the formation of
pseudouridine (Ψ) and ribothymidine (T). Pseudouridine is a C-
glycoside isomer of uridine where the N1-C1' glycosidic bond is
replaced by a C5-C1' bond. Ribothymidine is formed by the
methylation of the uridine base at the C5 position. These
modifications are found in specific loops of the tRNA, such as the
TΨC arm, where they contribute to the tRNA's unique three-
dimensional structure and its interaction with the ribosome.
Why Not the Other Options?
(2) Ribothymidine and N3-methyluridine – Incorrect; While
ribothymidine is a correct modification, N3-methyluridine is a much
less common modification in tRNA compared to pseudouridine.
(3) N3-methyluridine and pseudouridine – Incorrect; This option
incorrectly pairs N3-methyluridine with pseudouridine. While both
exist, ribothymidine is a more prominent and characteristic
modification of uridine in tRNA.
(4) Pseudouridine and N1-Methylpseudouridine – Incorrect; This
option lists pseudouridine and a methylated version of it. While both
are related, N1-Methylpseudouridine is not a direct product of uridine
but rather a further modification of pseudouridine. Ribothymidine is a
more common and direct modification from uridine.
56. Which one of the following is NOT a microbe-
associated molecular pattern (MAMP) of
phytopathogens?
(1) flg22, a 22 amino-acid peptide
(2) pep13, a 13 amino-acid peptide
(3) elf26, a 26 amino-acid peptide
(4) systemin, an 18 amino-acid peptide
(2025)
Answer: (4) systemin, an 18 amino-acid peptide
Explanation: Systemin is an 18 amino-acid polypeptide hormone
found in plants that acts as a signal to activate systemic plant defense
responses against herbivores and pathogens. It is not a microbe-
associated molecular pattern (MAMP); rather, it is an endogenous
plant signal that is part of the plant's own defense system. In contrast,
MAMPs are molecules derived from microbes that are recognized by
the plant immune system.
Why Not the Other Options?
(1) flg22, a 22 amino-acid peptide – Incorrect; flg22 is a well-
known MAMP. It is a conserved peptide from the flagellin protein of
bacterial flagella and is recognized by the plant receptor kinase
FLS2, triggering a strong immune response.
(2) pep13, a 13 amino-acid peptide – Incorrect; pep13 is a MAMP
derived from the fungal pathogen Phytophthora sojae. It is a peptide
from a cell wall transglutaminase and is recognized by a plant
receptor, activating defense responses.
(3) elf26, a 26 amino-acid peptide – Incorrect; elf26 is a MAMP
derived from the elongation factor Tu (EF-Tu), a highly conserved
protein in bacteria. It is recognized by the plant receptor kinase EFR,
leading to the activation of basal immunity.
57. Which one of the following datasets (expressed as
mean ± SD) is most likely to follow a Gaussian
(normal) distribution, considering that the data
cannot have any negative values?
(1) 296 ± 190 picomolar concentration of insulin in
human blood samples
(2) 58 ± 45 number of a particular species of bacteria
in the rhizosphere
(3) 160 ± 20 mg/dL of blood cholesterol
concentration in human athletes
(4) 4 ± 8 number of insects caught in a trap from an
agricultural field
(2025)
Answer: (3) 160 ± 20 mg/dL of blood cholesterol
concentration in human athletes
Explanation: A Gaussian (normal) distribution is a symmetric,
bell-shaped curve where the majority of data points cluster around the
mean. A key property is that approximately 99.7% of the data falls
within three standard deviations (SD) of the mean (μ±3σ). For a
dataset that cannot have negative values, a normal distribution is only
plausible if the mean is at least three standard deviations away from
zero.
Let's check the options by calculating the value at one and three
standard deviations below the mean:
(1) 296
−(3×190)=296−570=−274. This dataset extends into a highly
negative range, making a normal distribution unlikely.
(2) 58
−(3×45)=58−135=−77. This also extends into a negative
range.
(3) 160
−(3×20)=160−60=100. Since this value remains positive, the
dataset is likely to be contained within the non-negative range,
making a normal distribution a plausible model.
(4) 4
−(1×8)=4−8=−4. This dataset immediately falls into a negative
range.Therefore, the blood cholesterol concentration is the only
dataset that can plausibly follow a normal distribution.
Why Not the Other Options?
(1) 296 ± 190 picomolar concentration of insulin in human blood
samples – Incorrect; The standard deviation (190) is large relative to
the mean (296). This would result in negative values, which is
impossible for a physical concentration.
(2) 58 ± 45 number of a particular species of bacteria in the
rhizosphere – Incorrect; The standard deviation is also large relative
to the mean, leading to negative values for the number of bacteria,
which is not possible.
(4) 4 ± 8 number of insects caught in a trap from an agricultural
field – Incorrect; The standard deviation is twice the value of the
mean. One standard deviation below the mean is -4, immediately
violating the non-negative constraint. This is characteristic of a highly
skewed, non-normal distribution.
58. Which one of the following habitats harbors the
critically endangered Great Indian Bustard
(Ardeotis nigriceps)?
(1) Sal forests of Madhya Pradesh
(2) Desert grasslands of Rajasthan
(3) Shola grasslands of southern India
(4) Terai grasslands of northeastern India
(2025)
Answer: (2) Desert grasslands of Rajasthan
Explanation: The Great Indian Bustard (Ardeotis nigriceps) is a
critically endangered bird species found primarily in the desert
grasslands and scrublands of Rajasthan and Gujarat in India. Its
habitat is characterized by arid and semi-arid landscapes, where it
forages for insects, seeds, and small reptiles. The bustard is a large,
terrestrial bird that relies on these open grasslands for its survival
and breeding.
Why Not the Other Options?
(1) Sal forests of Madhya Pradesh – Incorrect; Sal forests are a
type of deciduous forest, which is not the natural habitat of the Great
Indian Bustard.
(3) Shola grasslands of southern India – Incorrect; Shola
grasslands are a unique montane grassland ecosystem found in the
high-altitude regions of the Western Ghats, which is not the habitat of
the Great Indian Bustard.
(4) Terai grasslands of northeastern India – Incorrect; The Terai
grasslands are found along the Himalayan foothills and are known for
tall grasses and a different set of fauna. The Great Indian Bustard is
not found in this region.
59. Under the current rules of taxonomy, which one
of the following organismal lineages can
legitimately receive genus names either under the
ICN or under the ICNP, based on the authors'
choice?
(1) Chlorophyta (Green algae)
(2) Bacillariophyta (Diatoms)
(3) Cyanobacteria (Blue-green algae)
(4) Mycetozoa (Slime moulds)
(2025)
Answer: (3) Cyanobacteria (Blue-green algae)
Explanation: Cyanobacteria are a unique group of organisms
that can be named under two different sets of taxonomic rules: the
International Code of Nomenclature for algae, fungi, and plants
(ICN) and the International Code of Nomenclature of Prokaryotes
(ICNP). This dual treatment is a result of their historical
classification. Originally, due to their photosynthetic ability and
morphology, they were grouped with algae, fungi, and plants.
However, they are prokaryotes, placing them squarely in the domain
of bacteriology. As a result, the two codes of nomenclature permit
authors to choose which set of rules to follow when naming new
species within this lineage.
Why Not the Other Options?
(1) Chlorophyta (Green algae) – Incorrect; Chlorophyta are a
lineage of eukaryotes and are classified exclusively under the ICN,
which governs the naming of algae, fungi, and plants.
(2) Bacillariophyta (Diatoms) – Incorrect; Diatoms are a type of
algae and, like green algae, are eukaryotes. Their nomenclature is
governed solely by the ICN.
(4) Mycetozoa (Slime moulds) – Incorrect; Slime moulds are a
polyphyletic group of eukaryotes. Their nomenclature is also
governed exclusively by the ICN, as they are classified with fungi and
plants.
60. Which one of the following statements depicts the
correct function of TFIIB?
(1) It binds to the TATA box and helps in the proper
positioning of RNA pol II.
(2) It plays an important role in the elongation phase
of transcription.
(3) It forms the transcription initiation complex by
interacting with RNA pol II and TBP.
(4) It acts as a coactivator that facilitates the binding
of enhancers to the promoter.
(2025)
Answer: (3) It forms the transcription initiation
complex by interacting with RNA pol II and TBP.
Explanation: Transcription Factor II B (TFIIB) is a crucial
component of the preinitiation complex (PIC) in eukaryotic
transcription. Its primary function is to act as a bridge between the
TATA-binding protein (TBP) and RNA polymerase II (Pol II). After
TBP binds to the promoter's TATA box, TFIIB recognizes this
complex and recruits RNA Pol II, along with other general
transcription factors, to the promoter region. This interaction is
essential for forming a stable transcription initiation complex and for
correctly positioning RNA Pol II to begin transcription at the
appropriate start site.
Why Not the Other Options?
(1) It binds to the TATA box and helps in the proper positioning of
RNA pol II – Incorrect; The TATA-binding protein (TBP), not TFIIB,
binds directly to the TATA box. TFIIB then binds to the TBP-DNA
complex.
(2) It plays an important role in the elongation phase of
transcription – Incorrect; TFIIB's main function is in the initiation
phase. Once transcription begins (elongation phase), TFIIB and most
other general transcription factors dissociate from the promoter.
(4) It acts as a coactivator that facilitates the binding of
enhancers to the promoter – Incorrect; This function is performed by
coactivator proteins or complexes, such as Mediator. They link
regulatory proteins bound to enhancer regions with the general
transcription factors and RNA Pol II at the promoter. TFIIB's role is
more specific to the core promoter and the initiation complex
assembly.
61. All of the following are covered by lipid
monolayers, EXCEPT
(1) lipid droplet
(2) very low-density lipoprotein
(3) endocytic vesicles
(4) chylomicrons
(2025)
Answer: (3) endocytic vesicles
Explanation: Endocytic vesicles are the only option that is NOT
covered by a lipid monolayer. Endocytic vesicles are small,
membrane-bound sacs that form from the plasma membrane and are
used to transport substances into a cell. They are covered by a lipid
bilayer, just like the cell's plasma membrane, from which they are
formed. In contrast, lipid droplets, VLDL, and chylomicrons are all
involved in lipid transport and storage and are characterized by a
core of hydrophobic lipids (like triglycerides and cholesterol esters)
that are stabilized and surrounded by a single layer of amphipathic
molecules, which include phospholipids and proteins.
Why Not the Other Options?
(1) lipid droplet – Incorrect; A lipid droplet is a cellular organelle
that stores neutral lipids. Its core is made of triglycerides and
cholesterol esters, which are surrounded by a lipid monolayer of
phospholipids and proteins to keep them stable in the aqueous
cytoplasm.
(2) very low-density lipoprotein – Incorrect; VLDL is a type of
lipoprotein that transports triglycerides and cholesterol. Like other
lipoproteins and lipid droplets, its core of neutral lipids is enclosed by
a single layer of phospholipids and apolipoproteins, which is a
monolayer.
(4) chylomicrons – Incorrect; Chylomicrons are lipoproteins that
transport dietary lipids from the intestines to other parts of the body.
They have a hydrophobic core of triglycerides and cholesterol esters
surrounded by a monolayer of phospholipids and proteins.
62. Which one of the following proteins is responsible
for the permeability of Mg2+ through the tight
junctions of tubular cells in the thick ascending
limb of Henle?
(1) Claudin – 2
(2) ZO-1
(3) Claudin – 16
(4) ZO-3
(2025)
Answer: (3) Claudin – 16
Explanation: The permeability of ions through the paracellular
pathway (the space between cells) is controlled by tight junctions,
which are complex protein structures. Claudins are the main integral
membrane proteins that form these junctions. Claudin-16, specifically,
is responsible for the paracellular reabsorption of divalent cations,
including magnesium ions (Mg 2+) and calcium ions (Ca 2+), in the
thick ascending limb of Henle (TAL). Mutations in the gene encoding
Claudin-16 can lead to familial hypomagnesemia, a disorder
characterized by low magnesium levels in the blood due to impaired
reabsorption in the kidney tubules.
Why Not the Other Options?
(1) Claudin – 2 – Incorrect; Claudin-2 is primarily found in the
proximal tubule and is responsible for the paracellular reabsorption
of water, not Mg2+ in the TAL.
(2) ZO-1 – Incorrect; ZO-1 (Zonula Occludens-1) is a peripheral
membrane protein that acts as a scaffolding protein, linking the
claudins to the cytoskeleton. It is not an ion channel itself.
(4) ZO-3 – Incorrect; ZO-3 (Zonula Occludens-3) is another tight
junction scaffolding protein. Like ZO-1, it helps organize the
junctional complex but does not directly form the paracellular
channel for ions like Mg2+.
63. Which of the following statements indicates the
relationship between basal metabolic rate and
body size for organisms within a taxonomic
group?
(1) Basal metabolic rate increases linearly with body
size
(2) Weight-specific basal metabolic rate is
independent of body size
(3) Weight-specific basal metabolic rate decreases
non-linearly with body size
(4) Basal metabolic rate increases non-linearly with
body size.
(2025)
Answer: (3) Weight-specific basal metabolic rate
decreases non-linearly with body size
Explanation: The relationship between an organism's metabolism
and its body mass is described by Kleiber's law. According to this law,
the total basal metabolic rate (BMR), which is the total energy an
organism uses at rest, increases non-linearly with body mass (M) to
the power of approximately 0.75, not linearly (BMR
∝
M 0.75). This
means that larger animals have a higher total metabolic rate, but not
proportionally higher. As a consequence, the weight-specific basal
metabolic rate (BMR per unit of body mass, e.g., kcal/kg/day)
decreases as body size increases. This relationship is also non-linear.
For example, a mouse has a much higher weight-specific metabolic
rate than an elephant.
Why Not the Other Options?
(1) Basal metabolic rate increases linearly with body size –
Incorrect; BMR does increase with body size, but the relationship is
not linear. A linear relationship would mean a direct proportion,
which is not supported by Kleiber's law.
(2) Weight-specific basal metabolic rate is independent of body
size – Incorrect; This is the opposite of the observed physiological
relationship. The weight-specific BMR decreases significantly as an
animal's body mass increases.
(4) Basal metabolic rate increases non-linearly with body size –
Incorrect; This statement is factually correct as per Kleiber's law
(BMR
∝
M 0.75). However, given the ambiguity of the prompt with two
possible correct answers, Option 3, focusing on the weight-specific
rate, is a more precise and common way of expressing the allometric
relationship in comparative physiology. In many biological contexts,
the decrease in weight-specific BMR is a more direct and illustrative
consequence of Kleiber's law. Both are technically correct statements,
but the question's intention might be to highlight the less intuitive
relationship of weight-specific metabolic rate.
64. The reactions of the urea cycle occur in two
different cellular compartments and generate the
following intermediates.
A. Argininosuccinate
B. Citrulline
C. Ornithine
Which intermediate(s) must be transported across
the mitochondrial inner membrane?
(1) A and B
(2) Band C
(3) A and C
(4) A only
(2025)
Answer: (2) Band C
Explanation: The urea cycle is a series of reactions that span two
cellular compartments: the mitochondrial matrix and the cytosol. The
cycle begins in the mitochondria with the synthesis of Citrulline
(Intermediate B) from ornithine and carbamoyl phosphate. Citrulline
is then transported from the mitochondrial matrix to the cytosol. Once
in the cytosol, citrulline reacts with aspartate to form
argininosuccinate (Intermediate A), which is then cleaved to produce
arginine and fumarate. Arginine is then hydrolyzed to form urea and
Ornithine (Intermediate C). Ornithine must be transported back into
the mitochondrial matrix to start another round of the cycle.
Therefore, Citrulline and Ornithine are the two intermediates that
must be transported across the mitochondrial inner membrane.
Why Not the Other Options?
(1) A and B – Incorrect; Intermediate B (Citrulline) is transported
out of the mitochondria, but Intermediate A (Argininosuccinate) is
formed in the cytosol and is not transported across the mitochondrial
membrane.
(3) A and C – Incorrect; Intermediate C (Ornithine) is
transported into the mitochondria, but Intermediate A
(Argininosuccinate) is a cytosolic intermediate.
(4) A only – Incorrect; Intermediate A (Argininosuccinate) is not
transported across the mitochondrial membrane. Both B (Citrulline)
and C (Ornithine) are.
65. Which one of the following is the correct
arrangement of amino acid dipeptides in
increasing order of the area of allowed regions in
the Ramachandran map?
(1) Gly-Gly < Pro-Pro < Ala-Ala ~ Phe-Phe
(2) Phe-Phe < Ala-Ala < Pro-Pro ~ Gly-Gly
(3) Pro-Pro < Phe-Phe~ Ala-Ala < Gly-Gly
(4) Ala-Ala Phe-Phe < Gly-Gly < Pro-Pro
(2025)
Answer: (3) Pro-Pro < Phe-Phe~ Ala-Ala < Gly-Gly
Explanation: The Ramachandran plot shows the allowed dihedral
angles (phi, ϕ, and psi, ψ) for amino acid residues in a polypeptide
chain. The area of allowed regions on this map is determined by the
steric hindrance of the amino acid's side chain. Amino acids with
larger or more constrained side chains have a smaller allowed area.
Proline (Pro) is an unusual amino acid because its side chain is a
rigid ring that connects back to the backbone nitrogen, severely
restricting its ϕ angle. This constraint gives it the smallest allowed
region on the Ramachandran plot. Therefore, Pro-Pro has the least
allowed area. Phenylalanine (Phe) and Alanine (Ala) have side chains
that are larger than glycine but not as constrained as proline. Phe has
a bulky aromatic ring, while Ala has a small methyl group. The steric
hindrance of these side chains is similar, resulting in a moderate and
comparable allowed area. Glycine (Gly) is unique because its side
chain is just a hydrogen atom, which is the smallest possible. This
lack of steric hindrance allows it to adopt a much wider range of ϕ
and ψ angles than any other amino acid. Therefore, Gly-Gly has the
largest allowed region. Thus, the correct order is: Pro-Pro < Phe-Phe
~ Ala-Ala < Gly-Gly.
Why Not the Other Options?
(1) Gly-Gly < Pro-Pro < Ala-Ala ~ Phe-Phe – Incorrect; This
order is the inverse of the correct one. Glycine has the most freedom
of movement, not the least.
(2) Phe-Phe < Ala-Ala < Pro-Pro ~ Gly-Gly – Incorrect; This
option incorrectly places Pro-Pro and Gly-Gly. Proline has the most
constrained angles, and Glycine has the most freedom.
(4) Ala-Ala Phe-Phe < Gly-Gly < Pro-Pro – Incorrect; This order
incorrectly places Pro-Pro last. Proline's unique structure severely
limits its conformation, giving it the smallest allowed area.
66. Which one of the following biofuels is classified as
a first-generation biofuel?
(1) Cellulosic ethanol
(2) Bio-ethanol
(3) Lignin-based biofuels
(4) Algal biofuels
(2025)
Answer: (2) Bio-ethanol
Explanation: First-generation biofuels are those produced from
sugar, starch, or vegetable oils from food crops. Bio-ethanol is a first-
generation biofuel because it is typically produced from the
fermentation of sugars or starches found in crops like corn,
sugarcane, or wheat. The use of food crops for biofuel production
raises concerns about food security, which has led to the development
of second- and third-generation biofuels.
Why Not the Other Options?
(1) Cellulosic ethanol – Incorrect; Cellulosic ethanol is a second-
generation biofuel because it is produced from lignocellulosic
biomass, such as agricultural waste and non-food crops, and not from
the food portion of the plant.
(3) Lignin-based biofuels – Incorrect; Lignin-based biofuels are
also considered second-generation because lignin is a component of
lignocellulosic biomass, which is a non-food source.
(4) Algal biofuels – Incorrect; Algal biofuels are considered a
third-generation biofuel. They are produced from algae, which are
non-food crops with the potential for high oil yields and do not
compete for arable land.
67. Ribosomes attach to the endoplasmic reticulum
via
(1) differing surface ribosomal protein composition
that allow it to interact with polar headgroups on the
ER lipid bilayer.
(2) 5'UTR of mRNA being translated that directs the
localization of the ribosomes to the ER.
(3) the translated nascent chain which carries a signal
allowing interaction with SRP receptor on the ER.
(4) tethered mRNAs that interact with rRNA.
(2025)
Answer: (3) the translated nascent chain which carries
a signal allowing interaction with SRP receptor on the
ER.
Explanation: Ribosomes bind to the rough endoplasmic
reticulum (ER) through a highly regulated process involving the
signal recognition particle (SRP) and the SRP receptor. This process
is mediated by a specific sequence called the signal sequence, which
is a short stretch of hydrophobic amino acids at the N-terminus of the
nascent polypeptide chain being synthesized. As this signal sequence
emerges from the ribosome, it is recognized and bound by the SRP.
The SRP then halts translation and escorts the entire ribosome-
nascent chain complex to the ER membrane, where it docks with the
SRP receptor. This interaction ensures that proteins destined for
secretion or insertion into a membrane are synthesized directly into or
across the ER membrane.
Why Not the Other Options?
(1) differing surface ribosomal protein composition that allow it
to interact with polar headgroups on the ER lipid bilayer – Incorrect;
The ribosome itself does not have the intrinsic ability to bind to the
ER membrane directly. This interaction is mediated by a signal
sequence on the nascent protein and a specific receptor complex.
(2) 5'UTR of mRNA being translated that directs the localization
of the ribosomes to the ER – Incorrect; The signal for ER localization
is encoded within the coding sequence of the mRNA, which is
translated into the N-terminal signal sequence of the nascent
polypeptide. The 5'UTR's primary role is in regulating translation
initiation.
(4) tethered mRNAs that interact with rRNA – Incorrect; While
mRNA is physically tethered to the ribosome during translation, this
interaction does not provide a mechanism for the ribosome to
specifically localize to the ER. The rRNA itself does not have a direct
binding site for ER localization. The signal is contained within the
nascent protein chain.
68. Which one of the following molecules is
transported into enterocytes from intestinal
lumen by GLUT5?
(1) Glucose
(2) Galactose
(3) Fructose
(4) Lactose
(2025)
Answer: (3) Fructose
Explanation: The absorption of monosaccharides from the
intestinal lumen into enterocytes (intestinal cells) is mediated by
specific membrane transport proteins. GLUT5 is a glucose
transporter protein that is highly expressed in the apical membrane of
enterocytes. It acts as a facilitator for the transport of fructose into
the cell. Unlike other monosaccharides, fructose transport is not
dependent on sodium and is mediated solely by this specific
transporter.
Why Not the Other Options?
(1) Glucose – Incorrect; Glucose is primarily transported from
the intestinal lumen into enterocytes by the SGLT1 (Sodium-Glucose
Linked Transporter 1) protein, which is a secondary active
transporter that co-transports glucose with sodium.
(2) Galactose – Incorrect; Similar to glucose, galactose is also
transported by the SGLT1 protein from the intestinal lumen into the
enterocyte.
(4) Lactose – Incorrect; Lactose is a disaccharide, not a
monosaccharide. It must first be broken down by the enzyme lactase,
located in the brush border of the small intestine, into its constituent
monosaccharides, glucose and galactose, before they can be absorbed
into the enterocyte.
69. Mutation of homeotic genes often results in which
one of the following developmental defects in
Drosophila?
(1) Polarity defects in every segment along the A-P
axis
(2) Absence of every other segment along the A-P
axis
(3) Transformation of one segment into another
(4) Absence of a group of contiguous segments
(2025)
Answer: (3) Transformation of one segment into
another
Explanation: Homeotic genes are master regulatory genes that
determine the identity of body segments or appendages during
development. A mutation in a homeotic gene often leads to a
phenomenon known as homeotic transformation, where one body part
is replaced by another that is characteristic of a different segment. A
classic example is the Antennapedia mutation in Drosophila, where a
mutation in the homeotic gene causes legs to grow in place of
antennae. This is not a loss of a segment but a change in its identity.
Why Not the Other Options?
(1) Polarity defects in every segment along the A-P axis –
Incorrect; Polarity defects (e.g., reversal of the anterior-posterior
axis within a segment) are typically caused by mutations in segment
polarity genes like engrailed and wingless, not homeotic genes.
(2) Absence of every other segment along the A-P axis –
Incorrect; The absence of alternating segments is a characteristic
phenotype of mutations in pair-rule genes, such as even-skipped and
fushi tarazu, which are involved in establishing the segmental pattern.
(4) Absence of a group of contiguous segments – Incorrect; The
absence of large, contiguous sections of the body plan is a hallmark of
mutations in gap genes, such as hunchback and Krüppel, which are
responsible for defining broad regions of the embryo.
70. In the circular synteny maps of Poaceae family,
providing insights into the origin and evolution of
grass genomes, which one of the following
genomes should represent the "innermost" circle?
(1) Arabidopsis thaliana chromosomes (n=5)
(2) maize chromosomes (n = 10)
(3) rice chromosomes (n = 12)
(4) ancestral chromosomes (n = 12)
(2025)
Answer: (4) ancestral chromosomes (n = 12)
Explanation: In circular synteny maps used to study the evolution
of grass genomes, the innermost circle represents the hypothetical
ancestral chromosomes of the common progenitor. These maps
visually compare the organization of genes in modern grass genomes
(such as rice, maize, and sorghum) to a reconstructed ancestral
genome. The ancestral genome is positioned at the center because it
serves as the reference point from which the other, more complex
genomes are thought to have evolved through events like chromosome
fusions, fissions, and duplications.
Why Not the Other Options?
(1) Arabidopsis thaliana chromosomes (n=5) – Incorrect;
Arabidopsis thaliana is a model plant but is not a member of the
Poaceae family (grasses). Its genome is not used as the central
reference for studying grass genome evolution.
(2) maize chromosomes (n = 10) – Incorrect; Maize has a more
complex genome that has undergone significant chromosomal
rearrangements. While its genome is compared on the map, it is not
the ancestral genome itself.
(3) rice chromosomes (n = 12) – Incorrect; Rice has a relatively
small and simple genome compared to other grasses and is often used
as a reference genome. However, it is not considered the ancestral
genome. It is placed in an outer ring, with its chromosome
organization compared to the central ancestral chromosomes.
71. An embryo lethal recessive mutation in a plant
gene "R" located on chromosome 1 was rescued
by an ectopically integrated GFP-tagged wild type
"R" transgene (R-comp) which is inserted in
chromosome 4. When a female homozygous R-
comp plant is crossed to a male purple-coloured
recessive mutant, some of the resultant F1
embryos showed chromosome segregation defects.
The plants regenerated from these embryos were
purple and sterile, and found to contain the wild
type gene "R" but not the R-comp transgene.
What is the ploidy of the above F1 plants?
(1) Triploid
(2) Haploid
(3) Chromosome 4 trisomy
(4) Monosomic diploid
(2025)
Answer: (2) Haploid
Explanation: Let's break down the genetic crosses and outcomes
to understand the ploidy. Female Parent: The female plant is
homozygous for the R-comp transgene, which is on chromosome 4
(R−comp/R−comp). It also carries the lethal recessive mutation for
gene R on chromosome 1 (r/r). Male Parent: The male parent is a
recessive mutant. This means it has the lethal mutation for gene R
(r/r). It does not have the R-comp transgene. It is also purple-colored,
which is likely a marker gene on a different chromosome. Gametes:
The female parent can produce two types of gametes: a normal
haploid gamete with chromosome 1 carrying the 'r' allele and
chromosome 4 carrying the R-comp transgene (r, R−comp) and a
gamete with chromosome segregation defect. The problem states that
the F1 plants are haploid, so this suggests a rare event where an
unfertilized female gamete develops into a viable embryo. This would
mean that the embryo is a product of parthenogenesis or apomixis. If
the egg cell with chromosome segregation defect had a non-
disjunction event, it may have lost the R-comp transgene. The question
states that the resulting plants were purple, which is recessive, but the
female parent was not purple. This suggests the purple color gene
must have come from the male parent. The problem states "the plants
regenerated from these embryos were purple and sterile, and found to
contain the wild type gene 'R' but not the R-comp transgene". The only
way the F1 plants can have the wild-type 'R' gene but not the R-comp
transgene is if they are derived from a gamete of the male parent
(which has the wild type 'R' gene from its maternal parent but does
not carry the R-comp transgene) that became haploid due to the
chromosome segregation defect. The fact that the F1 plants are sterile
further supports the haploid nature, as haploid plants often lack the
homologous chromosomes necessary for proper meiosis.
Why Not the Other Options?
(1) Triploid – Incorrect; Triploidy would mean the plant has three
sets of chromosomes (3n). This would result from the fusion of a
haploid gamete with a diploid gamete, which does not fit the
described genetic composition of the F1 plants (containing the wild
type 'R' gene but lacking the R-comp transgene).
(3) Chromosome 4 trisomy – Incorrect; A trisomy would mean the
plant has an extra copy of chromosome 4. This would mean the plants
should have the R-comp transgene, but the question states they do not.
(4) Monosomic diploid – Incorrect; Monosomic diploid would
mean the plant is missing one chromosome from a pair. This would
not result in the observed phenotype of being purple (a male trait) and
containing the wild type R gene but lacking the R-comp transgene.
72. Gastric juice (pH 1.5) is produced by pumping
HCl from blood plasma (pH 7.4) into the stomach.
Calculate the amount of free energy required for
H+ transport to produce one litre of gastric juice
at 37°C. How many moles of ATP must be
hydrolyzed to provide this amount of energy?
(The free energy change for ATP hydrolysis
under cellular conditions is about -58 kJ/mol.) (R
= 8.315 J/mol/K)
(1) 34 - 36 kJ/mol and 0.6 - 0.61 mol
(2) 3.4 - 3.6 kJ/mol and 6.0 - 6.1 mol
(3) 22 - 23 kJ/mol and 0.3 - 0.31 mol
(4) 17 - 18 kJ/mol and 3.0 - 3.1 mol
(2025)
Answer: (1) 34 - 36 kJ/mol and 0.6 - 0.61 mol
Explanation: The free energy change required to transport a
solute against its concentration gradient is given by the formula:
ΔG = RT ln(Cfinal / Cinitial)
Where R is the gas constant (8.315 J/mol/K), T is the temperature in
Kelvin, and Cfinal and Cinitial are the final and initial concentrations
of H+.
Step 1: Calculate the free energy required for H+ transport First,
convert the temperature from Celsius to Kelvin:
T = 37°C + 273.15 = 310.15 K
The hydrogen ion concentration [H+] is related to pH by the formula:
[H+] = 10-pH. The change in free energy can be calculated using the
pH values directly.
ΔG = R T 2.303 (pHinitial - pHfinal)
ΔG = (8.315 J/mol/K) (310.15 K) 2.303 (7.4 - 1.5)
ΔG ≈ 35041 J/mol = 35.04 kJ/mol
This calculated value falls within the range of 34 - 36 kJ/mol.
Step 2: Calculate the moles of ATP required
The free energy change for ATP hydrolysis is -58 kJ/mol. We divide
the energy required for transport by the energy released per mole of
ATP to find the moles needed. Moles of ATP = Energy Required /
Energy from one mole of ATP Moles of ATP = (35.04 kJ/mol) / (58
kJ/mol) ≈ 0.604 mol
This value is within the range of 0.6 - 0.61 mol.
Why Not the Other Options?
Option 2: Incorrect. This option contains a calculation error,
misplacing the decimal point.
Option 3: Incorrect. The values for both free energy and moles of
ATP are incorrect.
Option 4: Incorrect. The values for both free energy and moles of
ATP are incorrect.
73. Spindle assembly requires the formation of
numerous new microtubules. To meet this
demand, the spindle contains a large number of
regulatory factors. A few protein/protein
complexes are mentioned below.
A. γ -TuRCs
B. Augmin
C. TPX2
D. Dynein
Which one of the following options includes all the
factors that directly impact microtubule
nucleation for spindle assembly?
(1) A only
(2) A and B only
(3) A, B, and C only
(4) A, B, C, and D
(2025)
Answer: (3) A, B, and C only
Explanation: Spindle assembly is a crucial process in cell
division that requires the rapid and precise formation of microtubules.
This process is initiated by microtubule nucleation, which is the rate-
limiting step in forming new microtubules. γ-TuRCs (Gamma-tubulin
ring complexes): These complexes are the primary microtubule
nucleators in the cell. They act as a template for the formation of new
microtubules, typically at the minus end, ensuring the correct polarity.
They are essential for initiating the formation of the large number of
microtubules needed for the spindle. Augmin: This complex is
responsible for branching microtubule nucleation. It binds to the side
of pre-existing microtubules and recruits γ-TuRCs, enabling the
formation of new microtubules at an angle. This "branching"
mechanism is critical for rapidly amplifying the number of
microtubules within the spindle. TPX2 (Targeting protein for Xklp2):
This protein is a key regulator of spindle assembly. It helps activate γ-
TuRCs, especially at the chromosomes, and is involved in the local
formation of microtubules around chromatin. TPX2's role in
activating nucleating factors directly promotes the formation of new
microtubules.
Why Not the Other Options?
(1) A only - Incorrect; While γ-TuRCs are the main nucleators,
augmin and TPX2 also directly contribute to microtubule nucleation
during spindle formation.
(2) A and B only - Incorrect; This option is incomplete as it misses
TPX2, which is an important activator of nucleation factors.
(4) A, B, C, and D - Incorrect; Dynein is a microtubule motor
protein, not a nucleation factor. It is involved in moving cargo and
organizing the spindle but does not directly initiate the formation of
new microtubules. Its role is in spindle positioning and dynamics, not
nucleation.
74. The baroreceptors present in the carotid sinus
and aortic arch are stretch receptors that provide
neural signals to the brain stem for the
maintenance of normal blood pressure. The
activity of baroreceptors in the regulation of
blood pressure and heart rate are proposed in the
following statements:
A. The increased baroreceptor discharge due to
high blood pressure in the carotid sinus and aortic
arch stimulates the tonic discharge of sympathetic
nerves to blood vessels and heart.
B. A decline in the pulse pressure without a
change in mean pressure increases the rate of
baroreceptor discharge.
C. Baroreceptors are more sensitive to pulsatile
change of blood pressure than to constant blood
pressure.
D. The increased baroreceptor discharge excites
the vagal innervations to the heart.
Which one of the following options represents all
correct statements?
(1) B and D only
(2) A, C and D only
(3) C and D only
(4) A and B only
(2025)
Answer: (3) C and D only
Explanation: Baroreceptors are mechanoreceptors located in the
walls of major arteries, such as the carotid sinus and aortic arch. They
are crucial for short-term regulation of blood pressure. Statement C is
correct: Baroreceptors are most responsive to changes in pressure, not
just absolute pressure. They are particularly sensitive to the rate of
pressure change, which is reflected in the pulsatile nature of blood
flow. This makes them highly effective at responding to sudden
increases or decreases in blood pressure. Statement D is correct:
When blood pressure rises, the walls of the carotid sinus and aortic
arch stretch, increasing the firing rate of the baroreceptors. This
increased discharge sends signals to the medulla in the brainstem,
which in turn excites the vagal nerve (parasympathetic system). The
vagus nerve then releases acetylcholine, which slows the heart rate
and decreases the force of contraction, helping to lower blood
pressure.
Why Not the Other Options?
(1) B and D only - Incorrect; Statement B is wrong. A decline in
pulse pressure (the difference between systolic and diastolic pressure)
indicates less pulsatile flow, which would decrease the rate of
baroreceptor discharge, not increase it.
(2) A, C and D only - Incorrect; Statement A is wrong. High
blood pressure inhibits the sympathetic nervous system and stimulates
the parasympathetic system (vagus nerve). An increased baroreceptor
discharge inhibits the tonic discharge of sympathetic nerves to reduce
vasoconstriction and heart rate.
(4) A and B only - Incorrect; Both statements A and B are
factually wrong regarding the physiological response of the
baroreflex. An increase in baroreceptor firing inhibits the sympathetic
nervous system and a decline in pulse pressure decreases baroreceptor
firing.
75. The following statements are made regarding
vesicle formation, transport, and fusion in cells
expressing a mutant Arf protein. This mutant Arf
protein is constitutively bound to GTP but is
defective for GTP hydrolysis.
A. COPI-coated vesicles are only formed from
their normal location.
B. Arf is constitutively bound to a membrane.
C. COPII-coated vesicle fusion with the target
membrane is aberrant.
D. COPI-coated vesicle fusion with the target
membrane is disrupted
Which one of the following options represents the
combination of all correct statements?
(1) A and B only
(2) A, B, and D
(3) B and C only
(4) B and D only
(2025)
Answer: (4) B and D only
Explanation: Arf (ADP-ribosylation factor) is a small GTPase
that acts as a molecular switch for the assembly of coat proteins,
specifically COPI and clathrin, on cellular membranes. It cycles
between an inactive GDP-bound state (soluble in the cytoplasm) and
an active GTP-bound state (membrane-bound). A mutant Arf that is
constitutively bound to GTP and cannot hydrolyze it will remain in its
active, membrane-bound state. This disrupts the normal cycle of
vesicle formation and uncoating. Statement B is correct: Because the
mutant Arf cannot hydrolyze its bound GTP, it is permanently locked
in its active, membrane-bound state. It will remain associated with the
membrane, regardless of the normal regulatory signals.
Statement D is correct: The dissociation of the COPI coat from the
vesicle is triggered by the hydrolysis of GTP by Arf. Since the mutant
Arf cannot hydrolyze GTP, the COPI coat remains stuck on the
vesicle. The presence of the COPI coat is a barrier to fusion with the
target membrane, thus disrupting vesicle fusion.
Why Not the Other Options?
(1) A and B only - Incorrect; Statement A is wrong. The constant
presence of active Arf-GTP would lead to unregulated and potentially
aberrant coat formation, not just from the normal location. This could
cause the formation of COPI-coated vesicles from the wrong
membranes or prevent proper coat disassembly.
(2) A, B, and D - Incorrect; As explained above, statement A is
incorrect.
(3) B and C only - Incorrect; Statement C is wrong. COPII-coated
vesicles are formed with the help of a different GTPase, Sar1, not Arf.
Therefore, a mutant Arf protein will not directly affect the fusion of
COPII-coated vesicles.
76. The operation of the photorespiratory (C2)
pathway involves cooperative interaction among
three separate subcellular organelles, namely,
chloroplast, mitochondria and peroxisomes.
Following are certain statements regarding C2
pathway:
A. 2-phosphoglycolate formed in the chloroplast is
converted to glycolate by the action of
phosphoglycolate phosphatase.
B. Glycolate exits chloroplast by diffusion and
enters peroxisome through a glycolate
transporter.
C. In the peroxisomes, glycolate reacts with O2
and produces glyoxylate and H2O2.
D. Only serine:glyoxylate aminotransferase is
involved in the conversion of glyoxylate into
serine in a peroxisomal reaction.
Which one of the following options correctly
marks the above statements as TRUE (T) or
FALSE (F)?
(1) A-T, B-T, C-F, D-T
(2) A-F, B-T, C-F, D-T
(3) A-F, B-T, C-T, D-T
(4) A-T, B-F, C-T, D-F
(2025)
Answer: (4) A-T, B-F, C-T, D-F
Explanation: The photorespiratory pathway is an important
metabolic process in C3 plants that occurs when the enzyme RuBisCO
oxygenates ribulose-1,5-bisphosphate instead of carboxylating it. This
process involves the coordination of three organelles: the chloroplast,
peroxisome, and mitochondrion.
Statement A is TRUE: In the chloroplast, RuBisCO's oxygenase
activity produces 3-PGA and 2-phosphoglycolate. The 2-
phosphoglycolate is then dephosphorylated to glycolate by the enzyme
phosphoglycolate phosphatase. This is the first step in the C2
pathway, removing the phosphate group to allow the molecule to exit
the chloroplast.
Statement B is FALSE: Glycolate exits the chloroplast and enters the
peroxisome via a specific glycolate-glycerate translocator, not by
simple diffusion. This transporter protein facilitates the antiport of
glycolate and glycerate across the chloroplast inner membrane,
ensuring the efficient movement of these metabolites between the two
organelles.
Statement C is TRUE: In the peroxisome, glycolate is oxidized by the
enzyme glycolate oxidase. This reaction uses molecular oxygen (O2)
as a substrate and produces glyoxylate and hydrogen peroxide
(H2O2). The H2O2 is highly toxic and is immediately broken down by
the enzyme catalase within the peroxisome.
Statement D is FALSE: The conversion of glyoxylate to serine
involves two different aminotransferase reactions in the peroxisome,
not just one. First, glyoxylate is converted to glycine using an amino
group from glutamate via glutamate:glyoxylate aminotransferase.
Then, two molecules of glycine are transported to the mitochondrion,
where they are converted into a molecule of serine. The serine is then
transported back to the peroxisome where the amino group is
transferred from serine to hydroxypyruvate by serine:hydroxypyruvate
aminotransferase.
Why Not the Other Options?
(1) A-T, B-T, C-F, D-T - Incorrect; Statements B, C, and D are all
marked incorrectly.
(2) A-F, B-T, C-F, D-T - Incorrect; Statements A, B, C, and D are
all marked incorrectly.
(3) A-F, B-T, C-T, D-T - Incorrect; Statements A, B, and D are all
marked incorrectly.
77. The following statements are related to airway
resistance of lungs in healthy humans:
A. The resistance to airflow increases with
increasing lung volume.
B. The major site of airway resistance is in the
first eight generations of airways.
C. The smallest airways contribute very little to
the overall total resistance of the bronchial tree.
D. Breathing a mixture of oxygen-helium instead
of oxygen-nitrogen results in a decrease of airway
resistance.
Which one of the following options represents all
correct statements?
(1) B and D only
(2) B, C and D
(3) A and B only
(4) A and D only
Answer: (2) B, C and D
Explanation: Airway resistance is the opposition to the flow of
air through the respiratory tract. It is determined by factors such as
the radius of the airways, lung volume, and the density and viscosity
of the gas being breathed.
Statement B is correct: The primary site of airway resistance is indeed
in the large and medium-sized airways (the first eight generations).
These airways have a relatively small cross-sectional area compared
to the collective area of the smaller airways, leading to higher
resistance.
Statement C is correct: Although the individual resistance of each
small airway is high, their vast number and parallel arrangement
create a very large total cross-sectional area. According to
Poiseuille's Law, resistance is inversely proportional to the fourth
power of the radius, but in this case, the large number of parallel
tubes significantly reduces the overall resistance, making their
combined contribution very small.
Statement D is correct: Helium is a much less dense gas than
nitrogen. Airway resistance is directly proportional to the density of
the inhaled gas. By substituting nitrogen with helium, the density of
the gas mixture decreases, which in turn reduces the resistance to
airflow. This principle is used in some medical treatments for
respiratory conditions.
Why Not the Other Options?
(1) B and D only - Incorrect; This option is incomplete as it
misses statement C, which is also correct.
(3) A and B only - Incorrect; Statement A is wrong. Airway
resistance decreases with increasing lung volume. As the lungs
inflate, the radial traction exerted by the surrounding lung
parenchyma pulls the airways open, increasing their diameter and
thus reducing resistance.
(4) A and D only - Incorrect; Statement A is wrong. The
resistance to airflow decreases with increasing lung volume, not
increases.
78. As per the 2020 FAO Global Forest Resources
Assessment, the climatic domains ranked in
descending order based on their proportion and
distribution of global forest area are:
(1) Tropical, Subtropical, Temperate, Boreal
(2) Tropical, Boreal, Temperate, Subtropical
(3) Boreal, Tropical, Subtropical, Temperate
(4) Boreal, Temperate, Subtropical, Tropical
(2025)
Answer: (2) Tropical, Boreal, Temperate, Subtropical
Explanation: According to the Food and Agriculture
Organization (FAO) Global Forest Resources Assessment 2020 (FRA
2020), the distribution of the world's forests across different climatic
domains is not uniform. The report provides detailed data on the
extent of forest area by climate zone. Tropical forests constitute the
largest proportion of the world's forest area. They are found near the
equator and are known for their high biodiversity and high rates of
photosynthesis. Boreal forests (or taiga) come in second. These
forests are located in the high latitudes of the Northern Hemisphere,
characterized by coniferous trees, and are one of the largest
terrestrial biomes on Earth. Temperate forests rank third, found in the
mid-latitudes with distinct seasons. Subtropical forests have the
smallest proportion of the total global forest area. They are located in
the transition zone between tropical and temperate climates.
Why Not the Other Options?
(1) Tropical, Subtropical, Temperate, Boreal - Incorrect; This
ranking incorrectly places subtropical forests before temperate and
boreal forests. The area of boreal and temperate forests is
significantly larger than that of subtropical forests.
(3) Boreal, Tropical, Subtropical, Temperate - Incorrect; While
boreal forests are vast, they are not as extensive as tropical forests.
This option reverses the top two rankings.
(4) Boreal, Temperate, Subtropical, Tropical - Incorrect; This
option incorrectly places boreal forests first and tropical forests last,
which is the reverse of the actual distribution. Tropical forests have
the largest area globally.
79. Given below is a table listing the names of
compounds/products derived from animals or
plants (Column X) and their use/application
(Column Y):
Which one of the following options represents all
correct matches between Column X and Column
Y?
(1) A-(iii), B-(i), C-(ii), D-(iv)
(2) A-(ii), B-(iii), C-(iv), D-(i)
(3) A-(i), B-(iv), C-(iii), D-(ii)
(4) A-(ii), B-(i), C-(iv), D-(iii)
(2025)
Answer: (4) A-(ii), B-(i), C-(iv), D-(iii)
Explanation: The question requires matching a set of bioactive
compounds and products with their correct uses.
A. Bacoside: The correct match is (iii) Cognitive enhancer. Bacosides
are the primary active components of Bacopa monnieri, a plant
traditionally used in Ayurvedic medicine. They are known to improve
cognitive functions, memory, and concentration.
B. Vincristine: The correct match is (i) Chemotherapy for leukaemia
and lymphoma. Vincristine is an alkaloid derived from the periwinkle
plant (Catharanthus roseus). It is a powerful antimitotic drug used in
cancer chemotherapy, particularly for treating lymphomas and acute
lymphoblastic leukemia. It works by inhibiting microtubule formation,
which is essential for cell division.
C. Lanolin: The correct match is (iv) Skin creams and moisturizers.
Lanolin is a waxy substance secreted by the sebaceous glands of
wool-bearing animals, primarily sheep. It is widely used in cosmetics
and pharmaceuticals as a moisturizer and emollient to treat and
prevent dry, rough, or scaly skin.
D. Shellac: The correct match is (ii) Coating for pills and food
glazing. Shellac is a resin secreted by the female lac bug on trees. It is
a natural polymer used as a food-grade glaze for candies and pills,
giving them a shiny protective coating.
Why Not the Other Options?
(1) A-(iii), B-(i), C-(ii), D-(iv) - Incorrect; This option incorrectly
matches C and D. Lanolin is a skin cream ingredient, and shellac is
used for coatings.
(2) A-(ii), B-(iii), C-(iv), D-(i) - Incorrect; This option incorrectly
matches A and B. Bacoside is a cognitive enhancer, and Vincristine is
a chemotherapy drug.
(3) A-(i), B-(iv), C-(iii), D-(ii) - Incorrect; This option incorrectly
matches all of the pairs. Bacoside is a cognitive enhancer, Vincristine
is a chemotherapy drug, Lanolin is used in skin creams, and Shellac is
a coating.
80. The table below lists genes (Column X) and their
encoded proteins (Column Y) that play an
important role in plant response to biotic stress.
Which one of the following options represents the
correct match between Column X and Column Y?
(1) A-ii, B-i, C-iv, D-iii
(2) A-iii, B-iv, C-ii, D-i
(3) A-iv, B-i, C-ii, D-iii
(4) A-iv, B-ii, C-iii, D-i
(2025)
Answer: (3) A-iv, B-i, C-ii, D-iii
Explanation: The genes listed are all involved in plant defense
mechanisms against pathogens, specifically the synthesis and
transport of antimicrobial compounds.
A. SID1 is a gene associated with the MATE transporter (iv). MATE
(Multidrug and Toxin Extrusion) transporters are membrane proteins
that are important for the export of various compounds, including
plant defense chemicals. SID1's role is in the transport of salicylic
acid, a key plant defense hormone, from the vacuole to the cytoplasm.
B. SID2 is a gene that codes for isochorismate synthase (i). This
enzyme is crucial for the synthesis of salicylic acid from chorismate.
Salicylic acid is a central signaling molecule in plant defense,
mediating systemic acquired resistance (SAR).
C. PAD3 (Phytoalexin Deficient 3) is a gene that encodes a
cytochrome P450 monooxygenase (ii). This enzyme is involved in the
final steps of camalexin synthesis. Camalexin is a major phytoalexin,
an antimicrobial compound produced by Arabidopsis thaliana in
response to fungal infection.
D. PEN3 (Penetration 3) is a gene that encodes an ABC transporter
(iii). ABC (ATP-binding cassette) transporters use the energy from
ATP hydrolysis to transport various substrates across membranes.
PEN3 is localized to the plasma membrane and is involved in the
export of defense compounds to the apoplast to stop fungal pathogen
entry.
Why Not the Other Options?
(1) A-ii, B-i, C-iv, D-iii - Incorrect; This option incorrectly
matches A with a Cytochrome P450 monooxygenase and C with a
MATE transporter.
(2) A-iii, B-iv, C-ii, D-i - Incorrect; This option incorrectly
matches A, B, and D. SID1 is a MATE transporter, SID2 is an
isochorismate synthase, and PEN3 is an ABC transporter.
(4) A-iv, B-ii, C-iii, D-i - Incorrect; This option incorrectly
matches B with a Cytochrome P450 monooxygenase and C with an
ABC transporter.
81. The activity of two promoters (A and B) upstream
of a luciferase reporter in response to
transcription factor TF-X in a cell line was
studied. Luciferase activity was measured in cells
following overexpression and RNAi silencing of
TF-X and compared to control (Native TF-X
level). The luciferase enzyme activity is tabulated
below.
Based on the above table, following statements
were made
A. TF-X is needed for basal activity of promoter
A.
B. TF-X is needed for basal activity of promoter
B.
C. TF-X functions both as a positive and negative
regulator for promoter A.
Which one of the following options represents
correct statement(s)?
(1) A only
(2) B only
(3) A and B
(4) A and C
(2025)
Answer: (4) A and C
Explanation: The experiment measures how changes in the levels
of transcription factor TF-X affect the activity of two different
promoters, A and B. The luciferase activity serves as a proxy for
promoter activity.
Statement A is correct: At native TF-X levels, promoter A has a high
activity (9 arbitrary units). When TF-X is silenced using RNAi, the
activity of promoter A drops significantly to 2 units. This suggests that
the presence of TF-X is required to maintain the high basal activity of
promoter A.
Statement C is correct: TF-X acts as a positive regulator for promoter
A, as shown by the decrease in activity when TF-X is silenced (from 9
to 2). However, when TF-X is overexpressed, the activity of promoter
A also decreases (from 9 to 2). This indicates that excessive levels of
TF-X have an inhibitory effect on promoter A, meaning TF-X also acts
as a negative regulator at high concentrations. Therefore, TF-X acts
as both a positive and negative regulator for promoter A, depending
on its concentration.
Why Not the Other Options?
(1) A only - Incorrect; This option is incomplete as statement C is
also correct.
(2) B only - Incorrect; Statement B is wrong. When TF-X is
silenced, the activity of promoter B remains unchanged (2 arbitrary
units), suggesting TF-X is not needed for its basal activity. The high
activity of promoter B (9 units) is observed only when TF-X is
overexpressed, which indicates that TF-X is an activator for promoter
B, but not required for its basal activity.
(3) A and B - Incorrect; Statement B is incorrect as explained
above.
82. A network of genes regulates vertebrate eye
development. The loss or gain of function of select
genes are listed in Column X, and their effects or
phenotypes are listed in Column Y.
Which one of the following options represents all
correct matches between column X and column
Y?
(1) A-iv, B-i, C-ii, D-iii
(2) A-ii, B-iii, C-i, D-iv
(3) A-iii, B-ii, C-iv, D-i
(4) A-iv, B-ii, C-i, D-iii
(2025)
Answer: (1) A-iv, B-i, C-ii, D-iii
Explanation: This question deals with the genetic regulation of
vertebrate eye development, a process involving intricate signaling
pathways and transcription factors.
A. Loss of function of Shh: The correct match is (iv) Failure to split
the eye field to form bilateral eye fields. The Sonic hedgehog (Shh)
gene is expressed in the prechordal plate and is essential for splitting
the single, large anterior eye field into two separate optic vesicles,
which will form the two eyes. A loss of Shh function leads to cyclopia,
a condition where the two eyes fail to form separately.
B. Elevated levels of Shh in the prechordal plate: The correct match is
(i) No eyes. While Shh is necessary for splitting the eye field, its
expression must be tightly regulated. High levels of Shh act as an
inhibitory signal, suppressing the expression of key eye-field genes
like Pax6 and Otx2. Excessive Shh from the prechordal plate will
completely inhibit eye field development, leading to a failure to form
any eyes at all.
C. Loss of function of Otx2: The correct match is (ii) Disruption of
anterior neural tube specification. Otx2 is a homeodomain-containing
transcription factor that plays a crucial role in establishing the
anterior neural tube. It is a master regulator of anterior structures,
and its absence severely disrupts the development of the forebrain and
eyes.
D. Loss of function of Mitf: The correct match is (iii) Defective
pigmented retina. Mitf (Microphthalmia-associated transcription
factor) is a key transcription factor involved in the development and
differentiation of pigment-producing cells, including those in the
retinal pigmented epithelium (RPE). A loss of Mitf function leads to
defects in the RPE, resulting in a defective or absent pigmented
retina.
Why Not the Other Options?
(2) A-ii, B-iii, C-i, D-iv - Incorrect; This option incorrectly
matches A, B, and C. A loss of Shh causes a failure to split the eye
field, not anterior neural tube disruption. Elevated Shh leads to a lack
of eyes, not a defective pigmented retina. A loss of Otx2 leads to
severe anterior neural tube defects, not a lack of eyes (which is a
consequence of elevated Shh).
(3) A-iii, B-ii, C-iv, D-i - Incorrect; This option incorrectly
matches all pairs.
(4) A-iv, B-ii, C-i, D-iii - Incorrect; This option incorrectly
matches B and C. Elevated Shh leads to a complete lack of eyes, and a
loss of Otx2 disrupts the anterior neural tube.
83. The Red List of the IUCN categorises species
(Column X) under multiple threat levels (Column
Y). Match the following species with their IUCN
status (as of the latest reports):
Which one of the following options represents all
correct matches between Column X and Column
Y?
(1) A-(i), B-(iv), C-(ii), D-(iii)
(2) A-(i), B-(ii), C-(i), D-(iii)
(3) A-(iv), B-(ii), C-(i), D-(iii)
(4) A-(i), B-(iv), C-(v), D-(iii)
(2025)
Answer: (4) A-(i), B-(iv), C-(v), D-(iii)
Explanation: The IUCN Red List is a crucial tool for assessing
the global conservation status of species. It uses a set of criteria to
evaluate the risk of extinction.
A. Jerdon's Courser (Rhinoptilus bitorquatus): The correct match is
(i) Critically Endangered. This bird species is endemic to India and is
considered one of the most endangered birds globally. It was thought
to be extinct for many years until its rediscovery in 1986. Due to its
extremely small and fragmented population, it remains in the highest
threat category.
B. Indian Star Tortoise (Geochelone elegans): The correct match is
(iv) Vulnerable. This species is found in arid areas of India and
Pakistan. Its population has been declining due to illegal pet trade
and habitat loss, leading to its classification as Vulnerable.
C. Indian Dwarf Banana (Musa mannii): The correct match is (ii)
Endangered. This wild banana species is native to India. It is listed as
Endangered due to a significant reduction in its population, primarily
from habitat loss and over-collection.
D. Teak (Tectona grandis): The correct match is (iii) Least Concern.
Teak is a well-known hardwood species with a wide natural range.
Due to its extensive cultivation in plantations and stable wild
populations, it is categorized as Least Concern.
Why Not the Other Options?
(2) A-(i), B-(ii), C-(i), D-(iii) - Incorrect; This option incorrectly
matches the Indian Star Tortoise and the Indian Dwarf Banana. The
Star Tortoise is Vulnerable, and the Dwarf Banana is Endangered.
(3) A-(iv), B-(ii), C-(i), D-(iii) - Incorrect; This option incorrectly
matches Jerdon's Courser and the Indian Star Tortoise. Jerdon's
Courser is Critically Endangered, and the Indian Star Tortoise is
Vulnerable.
(4) A-(i), B-(iv), C-(v), D-(iii) - Incorrect; The IUCN Red List
does not have a "vulnerable" category labeled as (v). This is a typo in
the question, but based on the provided options, option 1 is the only
correct set of matches.
84. Sucrose transport in plant cells follows both active
transport and diffusion mechanisms. The given
figure shows the rate of sucrose uptake by
soybean protoplasts as a function of the external
sucrose concentrations.
Based on the given information, a researcher
made following interpretations:
A. Sucrose uptake at low concentration (<10 mM)
is facilitated by an energy-dependent process.
B. Sucrose uptake at low concentrations is a
carrier-independent process.
C. At higher concentrations, sucrose enters the
cells by diffusion down its concentration gradient.
D. Inhibition of ATP synthesis with metabolic
poison can block only the linear component.
Which one of the following options represents the
combination of all correct interpretations?
(1) A and B
(2) B and C
(3) A and C
(4) C and D
(2025)
Answer: (3) A and C
Explanation: The graph illustrates two distinct phases of sucrose
uptake in plant cells. The initial non-linear phase at low
concentrations and a linear phase at higher concentrations. This
dual-phase pattern is characteristic of a transport system that
involves both a saturable, carrier-mediated process and a non-
saturable, non-carrier-mediated process.
Statement A is correct: At low concentrations (<10 mM), the uptake
rate increases rapidly and then starts to plateau, a phenomenon
known as saturation kinetics. This is typical of a carrier-mediated,
active transport system, which requires energy (ATP) to move sucrose
against its concentration gradient.
Statement C is correct: At higher concentrations (>10 mM), the graph
shows a straight, linear increase in uptake. This indicates that the
carrier proteins are saturated and the primary mode of transport at
these concentrations is a carrier-independent process, likely simple
diffusion or a form of facilitated diffusion that does not saturate
within this concentration range. This type of transport moves the
solute down its concentration gradient.
Why Not the Other Options?
(1) A and B - Incorrect; Statement B is wrong. The saturating
curve at low concentrations indicates that the process is carrier-
dependent, not carrier-independent.
(2) B and C - Incorrect; Statement B is wrong. As explained
above, the initial uptake is carrier-dependent.
(4) C and D - Incorrect; Statement D is wrong. The saturable,
carrier-mediated component (the initial part of the curve) is likely the
active transport component and would be blocked by inhibitors of
ATP synthesis. The linear component is likely diffusion, which does
not require ATP and would therefore not be blocked by such poisons.
The active transport part of the graph is the one that is energy-
dependent and would be inhibited.
85. In a dioecious plant species, a single gene is
responsible for petal colour. A cross between a
true breeding red female and a true breeding
white male gave rise to all red male and all white
female F1 progeny. However, the reciprocal cross
produced all red males and all red females.
(1) The gene is located in the mitochondrial genome.
(2) Males are homogametic; gene is located on the
sex chromosome.
(3) Females are homogametic; gene is located on the
sex chromosome.
(4) The gene is located on an autosome.
(2025)
Answer: (2) Males are homogametic; gene is located
on the sex chromosome.
Explanation: This problem describes a classic pattern of sex-
linked inheritance. The key lies in the difference between the two
reciprocal crosses.
Cross 1: True breeding red female × True breeding white male
F1 progeny: All red males and all white females.
This pattern indicates that the traits are segregating by sex. The male
offspring inherit the red phenotype from their mother, while the
female offspring inherit the white phenotype from their father. This is
characteristic of a gene located on a sex chromosome where the male
is the homogametic sex (e.g., XX) and the female is heterogametic
(e.g., XY). However, this is more likely to be a ZW/ZZ system in birds
and some reptiles, but since it is a plant, it is the male that is
homogametic and the female is heterogametic. In the ZW system, the
male is ZZ and the female is ZW.
Cross 2 (Reciprocal): True breeding red male × True breeding white
female
F1 progeny: All red males and all red females.
In this case, all offspring, regardless of sex, are red. This is consistent
with a dominant red allele on the Z chromosome. The male parent
(ZZ) contributes a Z chromosome with the dominant red allele to all
of his offspring, making all F1 progeny red.
Why Not the Other Options?
(1) The gene is located in the mitochondrial genome. - Incorrect;
Mitochondrial inheritance is a form of cytoplasmic inheritance,
typically passed down through the maternal line. If this were the case,
the F1 generation would have the same phenotype as their mother in
both crosses, which is not what is observed.
(3) Females are homogametic; gene is located on the sex
chromosome. - Incorrect; If females were homogametic (XX), they
would be passing on two identical chromosomes to their offspring.
This would not result in the sex-specific inheritance pattern seen in
the first cross, where males and females have different phenotypes.
(4) The gene is located on an autosome. - Incorrect; Autosomal
genes are not located on sex chromosomes, so their inheritance
pattern would not be tied to the sex of the offspring. A cross between
true-breeding parents would produce F1 offspring with a uniform
phenotype, but not one that varies by sex, as seen in the first cross.
86. A population of Indian wolf shows logistic growth.
Let the carrying capacity (K) be 200 and intrinsic
rate of increase, r = 0.05. What is the maximum
possible growth rate for the population (in terms
of individuals per time unit)?
(1) 2.0
(2) 2.5
(3) 3.5
(4) 5.0
(2025)
Answer: (2) 2.5
Explanation: The logistic growth model is described by the
equation: $$ \frac{dN}{dt} = rN(1 - \frac{N}{K}) $$
The maximum growth rate occurs when the population size (N) is
exactly half of the carrying capacity (K), i.e., $N = \frac{K}{2}$. To
find the maximum growth rate, we substitute this value into the
equation.
The maximum growth rate is: $$ \frac{dN}{dt}_{max} =
r(\frac{K}{2})(1 - \frac{K/2}{K}) = r(\frac{K}{2})(\frac{1}{2}) =
\frac{rK}{4} $$Given that $r = 0.05$ and $K = 200$, we can
calculate the maximum growth rate:$$ \frac{dN}{dt}_{max} =
\frac{0.05 \times 200}{4} = \frac{10}{4} = 2.5 $$
Therefore, the maximum possible growth rate for the population is 2.5
individuals per time unit.
Why Not the Other Options?
(1) 2.0 - Incorrect; This value would be obtained if the population
size was 80, but not at the maximum growth rate.
(3) 3.5 - Incorrect; This value does not correspond to the given
parameters and the formula for maximum growth rate.
(4) 5.0 - Incorrect; This is the value of the intrinsic rate of
increase, r, multiplied by the carrying capacity, K ($0.05 \times 200 =
10$), and then divided by 2, but the correct formula is $\frac{rK}{4}$.
87. Equal numbers of particles of a mutant viral
strain 'A' and the wild-type strain (WT) were
used to separately infect equal numbers of
synchronized compatible host mammalian cells.
RNA from host mammalian cells were subjected
to real-time quantitative PCR to analyze levels of
the viral gag gene and host beta-tubulin after 24
hrs. Ct values for the above measurements are
listed below.
(1) Reduced to 1/32 of WT
(2) Reduced to 1/4 of WT
(3) Reduced to 1/2 of WT
(4) No change
(2025)
Answer: (4) No change
Explanation: In real-time quantitative PCR (qPCR), Ct (cycle
threshold) values reflect the number of cycles needed for fluorescence
to cross the detection threshold — inversely related to the initial
amount of target nucleic acid. When equal numbers of viral particles
(mutant and WT) infect equal numbers of host cells and the viral gag
gene and host β-tubulin show similar Ct values after 24 hours, it
indicates that the mutant virus replicates at the same efficiency as the
WT. Hence, there is no change in gag expression relative to WT.
Why Not the Other Options?
(1) Reduced to 1/32 of WT – Incorrect; a 1/32 reduction would
correspond to a 5-cycle Ct shift (2⁵ = 32), but no such Ct difference is
reported.
(2) Reduced to 1/4 of WT – Incorrect; this would require a 2-cycle
Ct shift (2² = 4), which was not observed in the qPCR data.
(3) Reduced to 1/2 of WT – Incorrect; a 1-cycle Ct shift (2¹ = 2) is
also absent in the data.
88. Given below is the list of molecules that target the
nervous system (Column X) and their modes of
action (Column Y).
Which one of the following options represents the
correct matches between Column X and Column
Y?
(1) A-ii; B-i; C-ii; D-iv
(2) A-iv; B-ii; C-iii; D-i
(3) A-iii; B-iv; C-i; D-ii
(4) A-iii; B-i; C-iv; D-ii
(2025)
Answer: (4) A-iii; B-i; C-iv; D-ii
Explanation: This pairing correctly matches each molecule in
Column X with its established mode of action in Column Y as
recognized in standard neuropharmacology references.
Why Not the Other Options?
(1) A-ii; B-i; C-ii; D-iv — At least two pairings are mismatched
relative to established mechanisms.
(2) A-iv; B-ii; C-iii; D-i — Several assignments contradict known
targets/modes of action.
(3) A-iii; B-iv; C-i; D-ii — B and C are incorrectly paired,
leading to inconsistencies with validated mechanisms.
89. Given below are a few biological lipids (Column
X) and their chemical nature (Column Y).
Which one of the following options correctly
matches the lipid to its chemical nature?
(1) a:v; b:iv; c:i; d:iii; e: ii
(2) a:v; b:i; c:iii; d:ii; e: iv
(3) a:iv; b:v; c:iii; d:ii; e: i
(4) a:iv; b:iii; c:ii; d:i; e: v
(2025)
Answer: (3) a:iv; b:v; c:iii; d:ii; e: i
Explanation: Sphingomyelin (a) → has a phosphocholine head
group, making the match iv correct.
Archaeal membrane lipid (b)
→ contains very long branched chain
fatty acids (ether-linked isoprenoids), so v is correct.
Non-polar lipids (c)
→ typically stored as triacylglycerols, so iii is
correct.
Plasmalogens (d)
→ have an ether
-linked alkene at the sn-1 position
of glycerol backbone, so ii is correct.
Cerebrosides (e) → are galactose linked to ceramide, so i is correct.
Why Not the Other Options?
(1) a:v; b:iv; c:i; d:iii; e:ii – Incorrect because sphingomyelin
does not contain very long fatty acids (a≠v), and non-polar lipids are
not galactose-linked ceramides (c≠i).
(2) a:v; b:i; c:iii; d:ii; e:iv – Incorrect because sphingomyelin has
phosphocholine head group, not very long branched chains (a≠v), and
cerebrosides are not phosphocholine head group lipids (e≠iv).
(4) a:iv; b:iii; c:ii; d:i; e:v – Incorrect because archaeal lipids
are not triacylglycerols (b≠iii), non-polar lipids are not ether-linked
alkenes (c≠ii), and cerebrosides are not very long chain fatty acids
(e≠v).
90. A novel DNA-binding protein (P) was tested for
its mode of binding with DNA (D), for which a
DNase I assay was performed. P was incubated
with DNA to form a P-D complex, to which DNase
I was added. Assume constant concentration of P,
D, and DNase I. The reaction was terminated at
increasing time-points (minutes) and resolved on a
sequencing-grade polyacrylamide gel.
Based on the results of the DNase I digest, which
one of the following options is the correct
interpretation?
(1) P does not bind to D, thereby generating a
banding pattern of D.
(2) P binds to D at a specific site, generating a
banding pattern of D.
(3) P binds and exposes multiple sites on D,
generating a banding pattern of D.
(4) P binds to only the 5' and 3' ends of D, generating
a banding pattern of D.
(2025)
Answer: (3) P binds and exposes multiple sites on D,
generating a banding pattern of D.
Explanation: In a DNase I footprinting assay, protein binding
can either protect bases (loss of bands = “footprint”) or
expose/hyper-sensitize bases (appearance/intensification of bands).
Here, upon adding P (+P) and increasing digestion time (5
→20→80
min), a dense set of bands appears across many positions, not a single
protected gap. This indicates that P binds DNA and creates DNase I–
hypersensitive sites at multiple locations, consistent with widespread
exposure (e.g., bending/unwinding/partial melting), not a single
discrete binding site.
Why Not the Other Options?
(1) P does not bind to D — Incorrect; the band pattern with +P
differs markedly from –P, showing P-dependent changes (more
bands), so P is affecting DNase I cleavage.
(2) P binds to D at a specific site — Incorrect; a specific-site
binder would produce a clear footprint (missing bands) over a defined
region, which is not observed.
(4) P binds only the 5′ and 3′ ends — Incorrect; changes occur
throughout the fragment, not limited to terminal regions, ruling out
end-only binding.
91. The following statements are made about
transgenic mice technologies:
A. Catalytically inactive Cas9 fused to a
transcription activator can be used to activate
specific gene expression.
B. Using CRISPR/Cas9, a precise gene
replacement in a target locus can be accomplished
by non-homologous end joining repair.
C. Using CRISPR/Cas9, many genes can be
targeted at once by expressing many guide RNAs
in the same cell.
D. All chimeric mice derived by injecting
engineered mouse embryonic stem cells into
blastocysts will carry the transgene in the germ
line.
E. In the Cre-lox conditional system, the Cre
recombinase is inserted into the gene to be
knocked out.
Which one of the following options represents the
combination of all correct statements?
(1) A, C, and E
(2) A and C only
(3) B, D, and E
(4) B and D only
(2025)
Answer: (2) A and C only
Explanation: Statement A is correct
→ Catalytically inactive
Cas9 (dCas9) fused with transcriptional activators (e.g., VP64, p300)
can be targeted to promoters/enhancers using guide RNAs to activate
transcription without cutting DNA.
Statement C is correct → CRISPR/Cas9 allows multiplex gene
targeting by expressing multiple gRNAs simultaneously, enabling
editing of many loci in the same cell.
Why Not the Other Options?
(B) Using CRISPR/Cas9, precise gene replacement in a target
locus can be accomplished by non-homologous end joining repair –
Incorrect; NHEJ is error-prone and leads to insertions/deletions.
Precise replacement requires homology-directed repair (HDR).
(D) All chimeric mice derived by injecting engineered embryonic
stem cells into blastocysts will carry the transgene in the germ line –
Incorrect; not all chimeras transmit the engineered allele. Only a
subset will contribute to the germline depending on the extent of stem
cell contribution.
(E) In the Cre-lox system, the Cre recombinase is inserted into the
gene to be knocked out – Incorrect; Cre recombinase is expressed
under a tissue-specific or inducible promoter, and it excises DNA
sequences flanked by loxP sites. It is not inserted into the target gene
itself.
92. The figure below is a schematic representation of
the replication fork. The different proteins
associated with DNA replication are marked by A,
B, C, D, E, and F.
Which one of the following options correctly
represents the names of these proteins?
(1) A: DNA Helicase; B: DNA Gyrase; C: DNA
Primase; D: DNA pol III; E: DNA pol I; F: DNA
Ligase
(2) A: DNA Gyrase; B: DNA Helicase; C: DNA
Primase; D: DNA pol II; E: DNA pol I; F: DNA
Ligase
(3) A: DNA Helicase; B: DNA Gyrase; C: DNA
Primase; D: DNA pol III; E: DNA pol II; F: DNA
Ligase
(4) A: DNA Gyrase; B: DNA Helicase; C: DNA
Primase; D: DNA pol II; E: DNA pol III; F: DNA
Ligase
(2025)
Answer: (1) A: DNA Helicase; B: DNA Gyrase; C:
DNA Primase; D: DNA pol III; E: DNA pol I; F: DNA
Ligase
Explanation: A (DNA Helicase): Unwinds the parental double-
stranded DNA at the replication fork.
B (DNA Gyrase): Relieves the supercoiling generated ahead of the
replication fork during unwinding.
C (DNA Primase): Synthesizes short RNA primers required to initiate
DNA synthesis.
D (DNA Polymerase III): The main replicative enzyme that extends
DNA strands in the 5
′→3′ direction.
E (DNA Polymerase I): Removes RNA primers and fills the gaps with
DNA nucleotides.
F (DNA Ligase): Joins Okazaki fragments by sealing nicks in the
sugar-phosphate backbone.
Why Not the Other Options?
(2) A: DNA Gyrase; B: DNA Helicase; C: DNA Primase; D: DNA
pol II; E: DNA pol I; F: DNA Ligase – Incorrect; DNA polymerase II
is not the main replicative polymerase.
(3) A: DNA Helicase; B: DNA Gyrase; C: DNA Primase; D: DNA
pol III; E: DNA pol II; F: DNA Ligase – Incorrect; E is wrongly
assigned as DNA pol II instead of DNA pol I.
(4) A: DNA Gyrase; B: DNA Helicase; C: DNA Primase; D: DNA
pol II; E: DNA pol III; F: DNA Ligase – Incorrect; roles of
polymerases are incorrectly swapped.
93. Panel (I) represents two alleles of a gene. The
coding region of the wild type allele (R) is 3 Kb
long while the mutant allele (r) has a 0.5 Kb
insertion. The black box indicates insertion in the
r allele. Three primers indicated as X, Y and Z
were designed to develop PCR based DNA
markers associated with the gene. A cross was
carried out between lines with the genotypes RR
and rr. The F₁ progeny was selfed. Panel (II) and
(III) represent the probable patterns of DNA
markers observed in the F2 progeny using
different primer combinations
Which one of the following statements is correct?
(1) Panel III represents the pattern when primers Y
and Z are used for amplification.
(2) Panel II is an example of a dominant DNA
marker.
(3) The F2 progeny as represented in Panel II, will
have 50% of the progeny showing no amplified
fragment (lane 1).
(4) A rare recombination event between the regions
represented by primers X and Y will lead to changes
in the amplification patterns observed in Panels II
and III.
(2025)
Answer: (2) Panel II is an example of a dominant DNA
marker.
Explanation: In Panel (II), the PCR pattern shows either one
band (lane 1) or two bands (lane 2), but there is no clear distinction
between homozygous dominant (RR) and heterozygous (Rr). This is
characteristic of a dominant DNA marker, where the presence of a
single allele (R) is sufficient to produce the amplified band, making it
impossible to differentiate heterozygotes from homozygotes.
Why Not the Other Options?
(1) Panel III represents the pattern when primers Y and Z are
used for amplification – Incorrect; Panel III is showing a co-dominant
marker pattern, where three lanes distinguish RR, Rr, and rr
genotypes.
(3) The F₂ progeny as represented in Panel II will have 50% of
the progeny showing no amplified fragment (lane 1) – Incorrect; in
Mendelian segregation (RR × rr → F₁ all Rr, then F₂ 1:2:1), only
25% would show no amplified product (rr).
(4) A rare recombination event between the regions represented
by primers X and Y will lead to changes in the amplification patterns
observed in Panels II and III – Incorrect; recombination between
closely linked PCR primer binding sites is highly unlikely to generate
the observed marker differences.
94. Shown below is a Northern blot profile with five
independent probes representing the transcript
from a single genetic locus.
Based on the Northern blots, which of the
diagrams below best represents the order of genes
encoded in the genetic locus?
(1) Option 1
(2) Option 2
(3) Option 3
(4) Option 4
(2025)
Answer: (2) Option 2
Explanation: The Northern blot shows transcript sizes and
relative band intensities when probed with five different probes (1–5).
To determine the gene organization:
Probe 1 detects a large transcript (~1050 bp), suggesting it spans the
5′ end (TSS1 region).
Probe 2 detects ~1050 bp strongly and a smaller ~600 bp band,
indicating overlap with both transcripts.
Probe 3 shows intermediate-sized transcripts (~900 bp), suggesting it
spans a region included in both isoforms.
Probe 4 detects ~600 bp strongly, matching the smaller transcript
from TSS2.
Probe 5 also detects both transcripts but more strongly toward the 3′
end, consistent with shared sequences.
This pattern is best explained by two transcription start sites (TSS1
and TSS2) generating transcripts of different lengths:
TSS1 transcript (~1050 bp) covers probes 1–5.
TSS2 transcript (~600 bp) covers probes 2–5 but not probe 1.
Thus, Option 2 (with TSS1 at the 5′ end and TSS2 starting
downstream, before CDS/3′UTR) is correct.
Why Not the Other Options?
(1) Option 1 – Incorrect; this arrangement would not explain why
probe 1 detects only the larger transcript while probe 4 detects only
the smaller one.
(3) Option 3 – Incorrect; placing TSS2 at the 3′ end would fail to
generate the 600 bp transcript detected by probe 4.
(4) Option 4 – Incorrect; having TSS2 at the extreme 3′ end does
not match the observed detection by probes 2–4.
95. A glucose biosensor exhibits Michaelis-Menten
kinetics with a maximum current response, Imax
= 250 µA, and a Michaelis constant, Km 5.2 mM.
If the glucose concentration in a sample is 3.4
mM, what is the current response (I) of the
biosensor?
(1) 98.83 µA
(2) 250 µA
(3) 29.06 µA
(4) 73.52 µA
(2025)
Answer: (1) 98.83 µA
Explanation: The glucose biosensor follows the Michaelis–
Menten equation for current response:
I = (Imax × [S]) / (Km + [S])
Where:
- Imax = 250 µA
- Km = 5.2 mM
- [S] = 3.4 mM
Substitute values:
I = (250 × 3.4) / (5.2 + 3.4)
I = 850 / 8.6
I = 98.83 µA
Thus, the current response is 98.83 µA.
Why Not the Other Options?
(2) 250 µA – Incorrect; this would be the response only if
substrate concentration were infinitely high (saturation).
(3) 29.06 µA – Incorrect; this does not follow from the Michaelis–
Menten equation with the given parameters.
(4) 73.52 µA – Incorrect; this corresponds to a lower substrate
concentration, not 3.4 mM.
96. The following statements are made about body
temperature regulation in humans:
A. As the skin temperature falls, the hypothalamic
set point for sweating decreases.
B. Acclimatization to cold activates the
uncoupling protein (UCP) in mitochondria.
C. Acclimatization to heat decreases aldosterone
secretion.
D. Sodium ion concentration in sweat decreases
upon acclimatization to heat.
E. When the body temperature becomes high, the
skin blood vessels undergo vasodilation.
Which one of the following options represents the
combination of all correct statements?
(1) A, B, and C
(2) B, D, and E
(3) B and C only
(4) A, D, and E
(2025)
Answer: (2) B, D, and E
Explanation: - (B) Correct
→ In cold acclimatization,
uncoupling proteins (UCPs) in brown adipose tissue mitochondria are
activated, generating heat by uncoupling oxidative phosphorylation.
- (D) Correct → Heat acclimatization improves salt conservation;
aldosterone increases sodium reabsorption, reducing sodium
concentration in sweat.
- (E) Correct
→ High body temperature triggers cutaneous
vasodilation, increasing heat loss by radiation and convection.
Why Not the Other Options?
(A) As the skin temperature falls, the hypothalamic set point for
sweating decreases – Incorrect; the set point is centrally regulated
and does not decrease simply with skin cooling.
(C) Acclimatization to heat decreases aldosterone secretion –
Incorrect; in fact, aldosterone secretion increases during heat
acclimatization to conserve sodium.
97. The ability to unwind RNA substrates I and II by
translation regulator 'X' was studied. RNA
substrates with radiolabeled antisense strand (x)
were incubated with purified 'X' under the
conditions listed below. Subsequently, RNA
species were separated by polyacrylamide gel
electrophoresis (PAGE), followed by
autoradiography. The results are shown below:
The following statements are made purely based
on the observed gel patterns:
A. X unwinds substrates with a 5' overhang.
B. Substrates which can be unwound by 'X' need
ATP for efficient RNA. unwinding.
C. X unwinding activity is enhanced by the
presence of eIF4G.
Which one of the following options represents the
combination of all correct statements?
1. A and B only
2. B and C only
3. A and C only
4. A, B and C
(2025)
Answer: 4. A, B and C
Explanation: - (A) Correct → Substrate I (with a 5′ overhang) is
unwound, while Substrate II (with a 3′ overhang) is not, showing that
X specifically unwinds 5′ overhang substrates.
- (B) Correct → Unwinding of substrate I occurs only in the presence
of ATP, confirming that X requires ATP hydrolysis for activity.
- (C) Correct
→ The unwinding activity of X is stronger in the
presence of eIF4G, suggesting a stimulatory effect.
Why Not the Other Options?
(1) A and B only – Incorrect; ignores the clear enhancement effect
of eIF4G.
(2) B and C only – Incorrect; fails to acknowledge substrate
preference (5′ overhang).
(3) A and C only – Incorrect; ATP-dependence is crucial, and this
option omits it.
98. Which one of the following statements best
describes the function of magnetite in animal
migration?
(1) Magnetite crystals generate electrical signals that
amplify visual landmarks that guide flight.
(2) Magnetite acts as a biological compass.
(3) Magnetite stores magnetic energy that provides
energy for long-distance flights.
(4) Magnetite emits magnetic pulses to keep the
group together.
(2025)
Answer: (2) Magnetite acts as a biological compass.
Explanation: Magnetite crystals in certain migratory animals
(such as birds, turtles, and fish) align with the Earth’s magnetic field.
This allows the animals to detect magnetic direction and use it as an
internal compass for orientation during long-distance migration.
Magnetoreception thus provides spatial guidance independent of
visual cues, enabling navigation even in low-light or obscured
environments.
Why Not the Other Options?
(1) Magnetite crystals generate electrical signals that amplify
visual landmarks — Incorrect; magnetite does not enhance visual
landmark detection, it directly senses geomagnetic fields.
(3) Magnetite stores magnetic energy that provides energy for
long-distance flights — Incorrect; magnetite is used for navigation,
not as an energy source.
(4) Magnetite emits magnetic pulses to keep the group together —
Incorrect; magnetite does not generate magnetic pulses; it passively
aligns with Earth’s field.
99. Lab-grown cells of a tumor (Wild type) expressing
a novel cancer antigen, and a variant of the same
tumor that has lost MHC class I expression, were
injected subcutaneously into normal mice (dotted
line) and into nude mice (solid line). The growth
of the tumors in the two mouse strains is shown
below.
The following statements were made based on the
results observed:
A. In normal mice, CTLs recognize tumor cells in
the presence, but not absence, of MHC-I.
B. In nude mice, NK cells recognize tumor cells in
the absence, but not in the presence, of MHC-I.
C. CTLs and NK cells inhibit each other if MHC-I
is present, but not if it is absent.
D. In mice injected with tumors lacking MHC-I,
antibodies generated with the help of CD4 cells
restrain the growth of the tumor in nude mice.
Which one of the following options represents the
combination of all correct statements?
(1) A and D
(2) B and C
(3) A and B
(4) A only
(2025)
Answer: (3) A and B
Explanation: - (A) Correct → In normal mice, cytotoxic T
lymphocytes (CTLs) kill tumor cells when tumor antigen peptides are
presented on MHC-I. When MHC-I is lost, CTLs cannot recognize the
tumor.
- (B) Correct
→ In nude mice (lacking functional T cells), NK cells
act as the main defense. NK cells preferentially attack cells without
MHC-I (missing-self recognition). Hence, tumors lacking MHC-I
shrink in nude mice, while wild-type tumors grow.
Why Not the Other Options?
(C) CTLs and NK cells inhibit each other if MHC-I is present –
Incorrect; NK cells are inhibited by MHC-I itself, not by CTLs, so
there is no mutual inhibition.
(D) Antibodies generated with CD4 T cell help restrain tumor
growth in nude mice – Incorrect; nude mice lack functional T cells
(both CD4 and CD8), so antibody help from CD4 cells is absent.
100. Eukaryotes regulate gene expression via
histone modifications. In the table given below,
the histone modification (column X), its
association with chromatin type (column Y), and
their impact on gene expression (column Z) is
listed.
Which one of the following options correctly
matches all terms in the most appropriate
combination?
(1) (A)-(a)-(i); (B)-(c)-(ii); (C)-(a)-(i); (D)-(b)-(ii)
(2) (A)-(b)-(i); (B)-(b)-(ii); (C)-(c)-(i); (D)-(a)-(ii)
(3) (A)-(c)-(ii); (B)-(a)-(i); (C)-(a)-(i); (D)-(c)-(ii)
(4) (A)-(c)-(i); (B)-(a)-(ii); (C)-(a)-(ii); (D)-(b)-(i)
(2025)
Answer: (1) (A)-(a)-(i); (B)-(c)-(ii); (C)-(a)-(i); (D)-
(b)-(ii)
Explanation: Histone modifications play a crucial role in
regulating chromatin accessibility and gene expression. H3K4me3
(listed as H4K4me3 in the table, likely a typo)
→ associated with
active promoters, highly accessible open chromatin, and gene
expression turned ON.
H3K9me3
→ a hallmark of constitutive heterochromatin (like
centromeres, telomeres), leading to gene silencing (OFF).
H3K9ac
→ an acetylation mark linked with open chromatin
and
active transcription (ON).
H3K27me3 → associated with facultative heterochromatin
(developmentally regulated silencing), turning genes OFF.
Thus, the most accurate combination is option 1.
Why Not the Other Options?
(2) (A)-(b)-(i); (B)-(b)-(ii); (C)-(c)-(i); (D)-(a)-(ii) Wrong because
H3K4me3 is not linked to facultative heterochromatin (b), it marks
open promoters.
(3) (A)-(c)-(ii); (B)-(a)-(i); (C)-(a)-(i); (D)-(c)-(ii) Wrong because
H3K4me3 does not silence genes (ii). Also, H3K9me3 cannot be
linked to open chromatin (a).
(4) (A)-(c)-(i); (B)-(a)-(ii); (C)-(a)-(ii); (D)-(b)-(i)
Wrong because H3K9ac (C) is not repressive (ii); it is strongly linked
with active transcription (ON).
101. Given below are statements about the action
of glucagon in human body:
A. It does not cause glycogenolysis in muscle.
B. It decreases ketone bodies formation in the
liver.
C. It does not cause lipolysis in the liver.
D. It inhibits conversion of phosphoenolpyruvate
to pyruvate in the liver.
Which one of the following options represents
both correct statements?
(1) A and B
(2) B and C
(3) C and D
(4) A and D
(2025)
Answer: (4) A and D
Explanation: - (A) Correct
→ Glucagon receptors are absent in
skeletal muscle. Hence, glucagon does not induce glycogenolysis in
muscle, only in liver.
- (D) Correct
→ Glucagon promotes gluconeogenesis in the liver. To
favor this pathway, it inhibits pyruvate kinase, preventing conversion
of phosphoenolpyruvate (PEP) to pyruvate, thus channeling PEP
towards glucose synthesis.
Why Not the Others?
(B) Incorrect → Glucagon increases fatty acid oxidation in the
liver, which enhances ketone body formation, not decreases it.
(C) Incorrect
→ Glucagon promotes lipolysis in adipose tissue,
not in liver. The liver oxidizes fatty acids delivered from adipose
tissue.
102. Interactions among 'maternal effect gene' and
'gap gene' products are crucial for establishing
embryonic domains in the Drosophila embryo.
Which one of the following options DOES NOT
show the correct interactions?
(1)
(2)
(3)
(4)
(2025)
Answer: (4)
Explanation: The diagram in option 4 incorrectly represents the
gene interactions. The correct regulatory network involves Kruppel
(Kr) inhibiting both Hunchback (Hb) and Giant (Gt). The diagram
shows Gt inhibiting Kr, which is incorrect. In Drosophila
embryogenesis, the spatial distribution of gap gene products, like
Hunchback, Kruppel, and Giant, is regulated by maternal morphogen
gradients (e.g., Bicoid and Caudal). These gap genes then regulate
each other to establish broad domains along the anterior-posterior
axis. Specifically, high concentrations of Hb protein inhibit the
expression of Kr, and Kr inhibits the expression of both Hb and Gt
outside of its central domain.
Why Not the Other Options?
(1) – Incorrect; This diagram correctly shows that Bicoid (Bcd)
activates Giant (Gt) and inhibits Caudal (Cad). Additionally, Cad
also inhibits Gt. Bcd and Cad are maternal effect genes, and Gt is a
gap gene.
(2) – Incorrect; This diagram correctly shows that Bcd (a
maternal effect gene) activates Hb (a gap gene) and inhibits Knirps
(Kni), another gap gene. Hb, in turn, inhibits Kni. These interactions
are fundamental for establishing the anterior and posterior domains
of the embryo.
(3) – Incorrect; This diagram correctly represents the mutual
inhibition between Hunchback (Hb) and Kruppel (Kr), and also shows
that Kruppel inhibits Giant (Gt). These interactions are crucial for
defining the anterior, central, and posterior gap gene expression
domains.
103. Lettuce seed germination is a typical photo-
reversible response controlled by phytochrome.
Seeds in a petridish were exposed to Red light (R),
Far-Red light (FR) or kept in the dark (black bar)
according to the schedule in Column X. The effect
on seed germination is provided in Column Y.
Which one of the following combinations
represents the correct match between Column X
and Column Y?
(1) A-i, B-ii, C-i, D-ii, E-i
(2) A-i, B-i, C-i, D-i, E-ii
(3) A-ii, B-ii, C-ii, D-ii, E-i
(4) A-ii, B-i, C-ii, D-i, E-ii
(2025)
Answer: (4) A-ii, B-i, C-ii, D-i, E-ii
Explanation: This question relates to the photo-reversible nature
of phytochrome, a photoreceptor that controls seed germination.
Phytochrome exists in two interconvertible forms: Pr (phytochrome
red) and Pfr (phytochrome far-red). Red light (R) converts Pr to Pfr,
while Far-Red light (FR) converts Pfr back to Pr. The Pfr form is the
biologically active form that stimulates seed germination. The last
exposure to light determines the outcome. A-ii: The seeds are kept in
the dark. Without red light exposure, the phytochrome remains in the
inactive Pr form, so germination is inhibited.
B-i: The seeds are exposed to red light. This converts phytochrome to
the active Pfr form, which stimulates germination.
C-ii: The seeds are exposed to red light followed by far-red light. The
final exposure to far-red light converts the phytochrome back to the
inactive Pr form, so germination is inhibited.
D-i: The seeds are exposed to red light, then far-red, and finally red
again. The final exposure to red light ensures the phytochrome is in
the active Pfr form, stimulating germination.
E-ii: The seeds are exposed to a series of red and far-red light
exposures, but the last one is far-red. This leaves the phytochrome in
the inactive Pr form, inhibiting germination.
Why Not the Other Options?
(1) – Incorrect; This option incorrectly matches A with 'stimulates
seed germination' and C with 'stimulates seed germination'. Dark
conditions (A) inhibit germination, and a final far-red light exposure
(C) also inhibits it.
(2) – Incorrect; This option incorrectly matches B, C, and D all
with 'stimulates seed germination', ignoring the inhibitory effect of the
final far-red light exposure in C and E.
(3) – Incorrect; This option incorrectly matches A, B, C, and D all
with 'inhibits seed germination', contradicting the known stimulatory
effect of a final red light exposure (B and D). It also incorrectly
matches E with 'stimulates seed germination'.
104. Proteins possess a sorting signal known as a
nuclear localization signal (NLS) for their import
into the nucleus. A few statements are given below
on the import of proteins destined for the nucleus.
A. The import receptor binds to the nuclear
localization signal.
B. The import receptor contains several low-
affinity binding sites for the nucleoporins.
C. The nuclear import receptor-cargo complex is
dissociated by RanGDP in the nucleus.
D. The nuclear import receptor-cargo complex
binds to RanGTP in the cytosol and traverses the
nuclear envelope.
Which one of the following options has all the
correct statements about nuclear protein import?
(1) A only
(2) A and B
(3) B and C
(4) C and D
(2025)
Answer: (2) A and B
Explanation: The import of proteins into the nucleus is a
highly regulated process. The statements A and B accurately
describe key steps in this process.
Statement A: The import receptor, typically a protein called
importin-α or a similar nuclear import receptor, directly
recognizes and binds to the nuclear localization signal (NLS) on
the cargo protein. This binding is the first step in forming the
transport complex.
Statement B: The import receptor (importin-β) has multiple weak
binding sites for the repeating phenylalanine-glycine (FG) motifs
found on nucleoporins, which are the building blocks of the
nuclear pore complex (NPC). This low-affinity, repetitive
interaction allows the complex to efficiently "hop" through the
nuclear pore, enabling rapid and directional transport.
Why Not the Other Options?
(1) A only – Incorrect; Statement A is correct, but statement
B is also correct. The import receptor's ability to bind to
nucleoporins is essential for its function.
(3) B and C – Incorrect; While statement B is correct,
statement C is false. The nuclear import receptor-cargo complex
is dissociated by RanGTP inside the nucleus, not RanGDP. The
binding of RanGTP to the import receptor causes a
conformational change that releases the cargo protein.
(4) C and D – Incorrect; Both statements C and D are false.
As explained above, RanGTP, not RanGDP, dissociates the
complex in the nucleus (making C incorrect). Furthermore, the
nuclear import receptor-cargo complex forms in the cytosol and
then traverses the nuclear envelope. RanGTP is primarily found
in the nucleus and is responsible for disassembly of the complex,
not its formation in the cytosol (making D incorrect).
105. The following statements are made regarding
leaf development in plants:
A. The activity of KNOX1 and ARP are mutually
repressive to help establish a separate identity for
emerging leaf primordium.
B. PHANTASTICA is required for adaxial cell
fate during leaf development.
C. KANADI genes promote abaxial cell fate
during leaf development.
D. miR166 binds to KANADI mRNA and
degrades it on the abaxial side of the leaf
primordium.
E. Overexpression of KNOX1 increases leaf
complexity and indeterminacy.
Which one of the following options represents all
correct statements?
(1) A, B, C and E
(2) A, C, D and E
(3) A, B, C, and D
(4) A, B, D and E
(2025)
Answer: (1) A, B, C and E
Explanation: This question pertains to the genetic regulation of
leaf development in plants. Statements A, B, C, and E are correct,
while D is incorrect.
A. Correct: The KNOX1 genes (e.g., SHOOT MERISTEMLESS) and
the ARP genes (AS2/AS1/PHANTASTICA) have mutually repressive
roles. KNOX1 genes are expressed in the shoot apical meristem
(SAM) and maintain its indeterminacy, while ARP genes are
expressed in the leaf primordium to initiate and maintain leaf identity.
This mutual repression ensures a clear boundary and distinct identity
between the SAM and the developing leaf.
B. Correct: The PHANTASTICA (PHAN) gene is an ARP family
member required for adaxial (upper side) cell fate. Its absence leads
to a lack of adaxial identity, resulting in leaves with abaxial (lower
side) characteristics on both sides.
C. Correct: KANADI genes are a family of transcription factors that
promote abaxial cell fate. Their expression is restricted to the lower
side of the developing leaf, and they are essential for establishing
abaxial identity.
E. Correct: The KNOX1 genes are responsible for maintaining the
undifferentiated state of the shoot meristem. Their overexpression in
the developing leaf leads to a loss of leaf identity, causing the leaf to
become more like a meristem, resulting in a more complex, compound
leaf structure and the potential for new shoots to form from the leaf
axil.
Why Not the Other Options?
(2), (3), and (4) – Incorrect; These options are incorrect because
they include statement D.
(D) – Incorrect; miR166 (and miR165) does not bind to KANADI
mRNA. Instead, it targets and degrades the mRNA of class III HD-ZIP
transcription factors, which are responsible for promoting adaxial
identity (e.g., PHABULOSA, PHAVOLUTA). This degradation on the
abaxial side of the leaf primordium helps to establish and maintain
the adaxial-abaxial axis. KANADI genes are instead regulated by
other factors to maintain their abaxial expression.
106. Different segments of the renal tubule
(Column X) and the sodium transporter in the
apical membrane of tubular cells (Column Y) are
given below:
Which one of the following options represents the
correct match between Column X and Column Y?
(1) a-iv, b-i, c-ii, d-iii
(2) a-ii, b-iv, c-i, d-iii
(3) a-i, b-ii, c-iii, d-iv
(4) a-iii, b-iv, c-i, d-ii
(2025)
Answer: (1) a-iv, b-i, c-ii, d-iii
Explanation: This question tests the knowledge of specific sodium
transporters located in different parts of the nephron.
a. Proximal tubule (a-iv): The proximal tubule is the primary
site of reabsorption. It reabsorbs most of the filtered solutes,
including sodium (Na+), glucose, and amino acids, via
Na+-amino acid co-transporters (SGLT) and Na+
-glucose co-transporters.
b. b. Collecting duct (b-i): The collecting duct is the final site
of sodium reabsorption, which is regulated by hormones
like aldosterone. This reabsorption occurs primarily
through the Epithelial Na+ Channel (ENaC), which allows
for the passive diffusion of sodium from the filtrate into the
cell.
c. Thick ascending limb (c-ii): This segment is impermeable
to water but highly permeable toions. The key transporter
here is the Na+-K+-−2Cl− co-transporter (NKCC2), which
pumps these ions out of the filtrate and into the interstitium,
creating the medullary osmotic gradient.
d. Distal convoluted tubule (d-iii): The distal convoluted
tubule is a major site for reabsorbing
$$Na^+$$and$$Cl^-$$
This is accomplished by the Na+
-Cl- co-transporter (NCC), which is sensitive to thiazide
diuretics.
Why Not the Other Options?
(2) – Incorrect; This option incorrectly matches the thick
ascending limb with ENaC and the proximal tubule with the NKCC2
co-transporter.
(3) – Incorrect; This option incorrectly matches the collecting
duct with the NKCC2 co-transporter and the distal convoluted tubule
with amino acid co-transporters.
(4) – Incorrect; This option incorrectly matches the proximal
tubule with the
Na+
-Cl- co-transporter and the collecting duct with the
Na+
-K+-
−2Cl−
co-transporter.
107. The Satpura Hypothesis suggests that the
Satpura Range in central India acted as a bridge
for the migration of Malayan taxa into peninsular
India. However, empirical observations do not
fully support this hypothesis. Consider the
following statements made in the context of
biogeographic patterns.
A. The shared characteristics of some Malayan
fauna and that of peninsular India is due to
convergent evolution in similar environments.
B. The said flora and fauna originated in the
Indian subcontinent and then dispersed to south-
east Asia ('out-of-India').
C. Vicariance biogeography split ancestral
populations, leading to the diversification of
species.
D. Pleistocene glaciation created conditions for
the dispersal of high-altitude Himalayan taxa to
the Western Ghats.
Which combination of the above statements have
been proposed as alternative explanations for the
Satpura hypothesis?
(1) A, B and C
(2) A, B and D
(3) A and B only
(4) B, C and D
(2025)
Answer: (2) A, B and D
Explanation: The Satpura Hypothesis, while historically
significant, has been challenged by several alternative explanations
for the biogeographic link between Southeast Asia and peninsular
India. The correct alternatives are convergent evolution, the "out-of-
India" dispersal, and the Pleistocene glaciation effect on high-altitude
species.
A. Convergent Evolution: This is a major alternative to the Satpura
Hypothesis. It suggests that similar environmental conditions in
Southeast Asia and peninsular India led to unrelated species evolving
similar traits. This explains the observed similarities in flora and
fauna without requiring a migration corridor.
B. "Out-of-India" Dispersal: This hypothesis proposes a reversal of
the traditional view. It suggests that ancestral species originated on
the Indian subcontinent as it drifted northward and then dispersed to
Southeast Asia after the collision with the Eurasian plate. This
explains the shared species without a "bridge" for migration from
Malaya.
D. Pleistocene Glaciation: This period of global cooling led to a drop
in sea levels and a change in climatic conditions. It is proposed that
cooler temperatures allowed high-altitude species from the Himalayas
to migrate southward along mountain ranges, establishing
populations in isolated "sky islands" like the Western Ghats. This
explains the presence of shared high-altitude species, which the
Satpura Hypothesis alone cannot account for.
Why Not the Other Options?
(1) and (3) – Incorrect; These options include or only mention
convergent evolution and the "out-of-India" hypothesis but exclude
the role of Pleistocene glaciation, which is a key alternative
explanation for the observed biogeographic patterns, especially for
high-altitude taxa.
(C) – Incorrect; Vicariance is a process where a pre-existing
population is geographically separated by a barrier (e.g., a rising
mountain range or a body of water). While it is a fundamental concept
in biogeography, it is not a direct alternative to the Satpura
Hypothesis in explaining the specific link between Malayan and
peninsular Indian taxa. The Satpura Hypothesis proposes a dispersal
route, not a vicariance event. Vicariance explains diversification
within a region, not the origin of shared species across distinct
biogeographic realms.
108. Izumo and Juno are two proteins that are
expressed in sperm and egg, respectively and play
critical roles in mediating interactions between
them. The following experiments were carried out
with Izumo and Juno:
A. Sperms from a male mouse, where Izumo has
been knocked out, could not fertilize eggs from a
normal female.
B. Two independent kidney cell lines were
developed, one expressing Izumo and the other
expressing Juno. If the two cells are mixed, the
cells will tend to aggregate with each other.
C. Whole mount immunostaining for Izumo and
Juno indicated specific staining on the sperm and
egg, respectively.
Which one of the following options lists
experiment(s) that demonstrate(s) interactions
between Izumo and Juno?
(1) A only
(2) B only
(3) A and C
(4) C only
(2025)
Answer: (2) B only
Explanation: Experiment B directly demonstrates the interaction
between Izumo and Juno because when kidney cells expressing Izumo
are mixed with kidney cells expressing Juno, they aggregate. This
aggregation is a clear indicator of specific molecular binding between
the two proteins. Thus, experiment B provides direct evidence of their
interaction.
Why Not the Other Options?
(1) A only – Incorrect; experiment A shows that Izumo is essential
for fertilization, but it does not prove that Izumo and Juno directly
interact. The failure of fertilization could be due to lack of interaction,
but it only implies necessity, not direct binding.
(3) A and C – Incorrect; while A shows the importance of Izumo
in fertilization and C shows the localization of the proteins, neither
directly demonstrates physical interaction.
(4) C only – Incorrect; experiment C (immunostaining) only
shows that Izumo is on sperm and Juno is on egg. Localization
evidence does not establish that the two proteins interact with each
other.
109. A canonical Watson-Crick G:C base pair is
stabilized by three hydrogen bonds formed
between a donor (+0.25e) and acceptor (-0.35e)
atom located at a distance r=2.9 Å. Assume that
the dielectric constant 'D' in the DNA core is 4,
and in water is 78.5. Which one of the following
options correctly represents the total difference in
electrostatic interaction energy for all 3 hydrogen
bonds when buried in the DNA core versus fully
exposed to solvent? (Modified Coulomb's
constant, k=1389 kJ/mol nm e-2)
(1) -298.3 kJ/mol
(2) -104.7 kJ/mol
(3) -99.4 kJ/mol
(4) -5.338 kJ/mol
(2025)
Answer: (1) -298.3 kJ/mol
Explanation:
Use Coulomb's law for point charges in the given
medium:
E=k q1q2D rE = \dfrac{k \, q_1 q_2}{D \, r}E=Drkq1q2.
Here q1=+0.25e,
q2=−0.35e⇒q1q2=
−0.0875 e2q_1=+0.25e,
\; q_2=-0.35e
\Rightarrow q_1q_2=-0.0875\,e^2q1=+0.25e,q2
=−0.35e⇒q1q2=−0.0875e2.
Convert distance: r=2.9r=2.9r=2.9 Å = 0.29 nm.
Plugging values:
Energy in DNA core (per H-bond):
Ecore=1389×(
−0.0875)4×0.29≈−104.77
kJ/molE_{\text{
core}} = \dfrac{1389 \times (-0.0875)}{4 \times 0.29}
\approx -104.77\ \text{kJ/mol}Ecore
=4×0.291389×(−0.0875)≈−104.77 kJ/mol.
Energy in water (per H-bond):
Ewater=1389×(
−0.0875)78.5×0.29≈−5.339
kJ/molE_{\te
xt{water}} = \dfrac{1389 \times (-0.0875)}{78.5 \times
0.29} \approx -5.339\ \text{kJ/mol}Ewater
=78.5×0.291389×(−0.0875)≈−5.339 kJ/mol.
Difference (per H-bond):
ΔEper bond=Ecore−Ewater≈−99.435 kJ/mol.\Delta
E_{\text{per bond}} = E_{\text{core}} -
E_{\text{water}} \approx -99.435\
\text{kJ/mol}.ΔEper bond=Ecore−Ewater
≈−99.435 kJ/mol.
For three H-bonds (G:C):
ΔEtotal=3×(
−99.435)≈−298.305
kJ/mol\Delta
E_{\text{total}} = 3 \times (-99.435) \approx -298.305\
\text{kJ/mol}ΔEtotal=3×(
−99.435)≈−298.305
kJ/mol,
rounded to -298.3 kJ/mol, which is Option 1.
Why Not the Other Options?
(2) -104.7 kJ/mol – Incorrect; this value is the electrostatic
energy of a single H-bond when buried in the DNA core, not the
total difference between core and solvent for all three bonds.
(3) -99.4 kJ/mol – Incorrect; this is the difference per single H-
bond (core minus water), not the total for all three hydrogen bonds.
(4) -5.338 kJ/mol – Incorrect; this is the electrostatic energy of a
single H-bond when fully exposed to water, not the difference nor the
total for three bonds.
110. Male crickets call at dusk when the light
intensity falls below a certain level. A researcher
kept these insects in the laboratory under two
light regimes: i) constant light condition for 24
hours (LL) and ii) 12 hours Light and 12 hours
Dark cycle (LD). The calling activity pattern of
the crickets in these two regimes across all
experimental days is represented in the figure
below.
The statements below represent inferences that
one can draw about circadian control of calling
activity in these animals.
A. Calling activity under LL represents free-
running cycle and LD represents entrained cycle.
B. Calling activity under LL represents entrained
cycle and LD represents free-running cycle.
C. The circadian control of calling activity is
endogenous only.
D. The circadian control of calling activity is
exogenous only.
E. The circadian control of calling activity has
both endogenous and exogenous components.
Which one of the following options gives all
correct statements about the inferences we can
draw from this experiment?
(1) A and C
(2) B and C
(3) B and D
(4) A and E
(2025)
Answer: (4) A and E
Explanation: A is correct because under constant light (LL) the
environmental light–dark cue is removed; if calling still shows a
rhythmic pattern that shifts or has a period different from 24 h, that
rhythm is a free-running (endogenous) circadian rhythm. Under LD
(12:12) the rhythm aligns to the external light–dark cycle, so LD
represents an entrained cycle. E is correct because the experiment
shows both components: the persistence of rhythmic calling in LL
indicates an endogenous clock drives calling, while the clear
alignment to the LD cycle indicates exogenous entrainment (light cues
can shift or synchronize the rhythm). Thus the system has both
endogenous (internal oscillator) and exogenous (environmental
entrainment) components.
Why Not the Other Options?
(1) A and C – Incorrect; A is right but C ("endogenous only") is
wrong because the LD data show entrainment by the light–dark cycle
(an exogenous influence), so the control is not purely endogenous.
(2) B and C – Incorrect; B is wrong because LL rhythms are not
entrained (they are free-running) and LD is the entrained condition. C
is wrong for the same reason above (presence of exogenous
entrainment).
(3) B and D – Incorrect; B is incorrect (LL = free-running, not
entrained) and D ("exogenous only") is incorrect because rhythms
persist under LL, indicating an endogenous circadian oscillator too.
111. Given below are a few statements related to
genetics and linkage mapping in plants.
A. Near Isogenic Lines are immortal populations.
B. The physical distance between two
polymorphic markers is not linearly related to
their genetic distance.
C. QTLs identified for a particular phenotype or
trait of interest always contribute equally to the
trait value.
D. Association mapping is performed using
unrelated individuals of the same population.
Which one of the following options correctly
describes the above statements as "True" or
"False"?
(1) A: True; B: False; C: True; D: False
(2) A: False; B: True; C: False; D: False
(3) A: True; B: True; C: False: D: True
(4) A: False; B: False; C: True; D: True
(2025)
Answer: (3) A: True; B: True; C: False: D: True
Explanation: A — True. Near Isogenic Lines (NILs) are created
by repeated backcrossing and inbreeding so that each line is
genetically stable and homozygous for almost the entire genome
except for a small introgressed region. They are maintained as stable,
reproducible (immortal) lines for experiments.
B — True. Genetic distance (measured in centimorgans, based on
recombination frequency) does not scale linearly with physical
distance (base pairs) because recombination rates vary across the
genome (hotspots, coldspots, chromatin structure, centromeres, etc.).
C — False. QTLs typically have different effect sizes; some QTLs are
major-effect loci while others are minor-effect. They do not always
contribute equally to the trait value.
D — True. Association mapping (GWAS / linkage disequilibrium
mapping) is performed on panels of unrelated or loosely related
individuals from the same population to exploit historic
recombination and LD patterns for high-resolution mapping.
Why Not the Other Options?
(1) A: True; B: False; C: True; D: False — Incorrect because B
is actually True (physical vs genetic distance not linear), C is False
(QTLs differ in effect size), and D is True (association mapping uses
unrelated individuals).
(2) A: False; B: True; C: False; D: False — Incorrect because A
is True (NILs are immortal/stable) and D is True (association
mapping uses unrelated individuals).
(4) A: False; B: False; C: True; D: True — Incorrect because A
is True, B is True, and C is False; this option flips several truths and
false.
112. Two samples were run on a flow cytometer
and the data were plotted as histograms A and B,
as shown below:
The following statements were made:
A. Figure A represents three normal distributions
with more variation within groups than between
groups.
B. Figure B represents three normal distributions
with more variation between groups than within
groups.
C. The ANOVA-F test is likely to be statistically
significant when applied to Figure B, but not to
Figure A.
Which one of the following options represents the
combination of all correct statements?
(1) A only
(2) A and B only
(3) B and C only
(4) A, B and C
(2025)
Answer: (4) A, B and C
Explanation: Statement A accurately describes histogram A’s
extensive overlap, indicating within-group variance exceeds between-
group separation. Statement B correctly characterizes histogram B’s
distinct, minimally overlapping peaks, showing between-group
variance dominates. Statement C follows from ANOVA’s reliance on
the ratio of between- to within-group variance: only in Figure B will
that ratio be large enough to yield a statistically significant F.
Why Not the Other Options?
(1) A only – Incorrect; Figure B also demonstrates greater
between-group variance and predicts ANOVA significance
(statements B and C).
(2) A and B only – Incorrect; this excludes C, which is valid
because ANOVA-F is significant for B but not A.
(3) B and C only – Incorrect; this omits A, which correctly
describes the overlapping distributions and high within-group
variance in Figure A.
113. In a successional chronosequence (30-400
years) on a sand dune, xAmmophilax dune grass
dominates the youngest dunes, the prairie bunch
grass xSchizachriumx dominates at 60 years,
followed by xPinusx species at later stages.
Experimental seed additions were done to study
seedling emergence (as a proportion of viable
seeds) of all species on all the dunes (Panel 1). In
addition, the effect of seed predation on seedling
emergence (across dune ages) was also tested
(Panel 2). The following figure presents results on
seedling emergence.
Based on the above, consider the following
statements regarding the possible interpretation
of the results.
A. Late successional species can establish on
younger dunes but may be limited by seed
dispersal.
B. Seed predation reduces the establishment of
late successional species on younger dunes.
C. Facilitation has stronger influence than
colonization or seed predation on successional
change.
D. Late successional species cannot establish on
younger dunes because of the sandy substrate.
Which combination of the above statements
represents valid inference that can be drawn
purely from the data presented above?
(1) A and B only
(2) A, B and C
(3) B, C and D
(4) B and C only
(2025)
Answer: (1) A and B only
Explanation: Panel 1 shows that when seeds of late-successional
species are added to young dunes, their emergence rates match those
on older dunes—demonstrating that substrate conditions are suitable
and that natural seed arrival (dispersal limitation) explains their
absence (A). Panel 2 reveals a substantial increase in emergence of
these species on young dunes when predators are excluded, indicating
that seed predation suppresses their establishment in the absence of
predation protection (B).
Why Not the Other Options?
(2) A, B and C – Incorrect; there is no data here demonstrating
facilitation outcompetes colonization or seed predation in driving
succession.
(3) B, C and D – Incorrect; statement D is refuted by high
emergence under seed addition (substrate isn’t limiting), and C is
unsupported.
(4) B and C only – Incorrect; statement A is clearly supported by
the seed-addition results, while C remains unjustified.
114. The following figure shows the decomposition
of selected plant litter constituents.
In the above, the correct identities of the symbols
A, B, C and D are:
(1) A-Lignin, B-Phenols, C-Sugars, D-Cellulose
(2) A-Phenols, B-Lignin, C-Cellulose, D-Sugars
(3) A-Cellulose, B-Lignin, C-Sugars, D-Phenols
(4) A-Cellulose, B-Phenols, C-Lignin, D-Sugars
(2025)
Answer: (2) A-Phenols, B-Lignin, C-Cellulose, D-
Sugars
Explanation: The graph shows the rate of decomposition of
various organic compounds found in plant litter. The x-axis represents
time in years, and the y-axis represents the percentage of weight
remaining. A steeper slope indicates a faster rate of decomposition,
while a flatter slope indicates a slower rate. Sugars (D) are simple
carbohydrates that are easily broken down by decomposers, so they
decompose very quickly, represented by the steepest curve. Cellulose
(C) and hemicellulose are also relatively easy to decompose but take
longer than sugars. Lignin (B) is a complex polymer that is very
difficult for most decomposers to break down, so it decomposes
slowly. Phenols (A) are complex organic compounds, and due to their
toxic and inhibitory properties, they are the most resistant to
decomposition, hence they have the slowest decomposition rate,
represented by the flattest curve.
Why Not the Other Options?
(1) "A-Lignin, B-Phenols, C-Sugars, D-Cellulose" - Incorrect;
This reverses the decomposition rates of phenols and lignin, and
sugars and cellulose. Phenols decompose more slowly than lignin,
and sugars decompose faster than cellulose.
(3) "A-Cellulose, B-Lignin, C-Sugars, D-Phenols" - Incorrect;
This is a mix-up of the decomposition rates. Cellulose and sugars
decompose much faster than lignin and phenols, and phenols
decompose the slowest of all.
(4) "A-Cellulose, B-Phenols, C-Lignin, D-Sugars" - Incorrect;
This incorrectly places cellulose as the slowest decomposing
component and phenols in a faster category than lignin, which is
contrary to their known chemical resistance to decomposition.
115. Match the animal phyla/classes (Column X)
with their defining set of characteristics (Column
Y).
(1) A-iii, B-iv, C-v, D-ii
(2) A-i, B-iv, C-v, D-ii
(3) A-iii, B-i, C-v, D-iv
(4) A-v, B-ii, C-iii, D-i
(2025)
Answer: (1) A-iii, B-iv, C-v, D-ii
Explanation: This question requires matching the animal phyla
or classes with their unique characteristic features.
A-iii (Phylum Platyhelminthes): Platyhelminthes, or flatworms, are
characterized by bilateral symmetry, a triploblastic body plan, and
being acoelomate (lacking a true body cavity). Their body is typically
flattened dorsoventrally, and they have an incomplete digestive tract
with a single opening or no digestive tract at all (e.g., tapeworms).
B-iv (Phylum Annelida): Annelids, or segmented worms, possess a
segmented body (metamerism) and a true coelom. They have a closed
circulatory system and often bear chitinous bristles called chaetae.
C-v (Class Insecta): Insects are the largest group within the phylum
Arthropoda. They are defined by a chitinous exoskeleton, three
distinct body parts (head, thorax, abdomen), three pairs of legs (six
legs), and often one or two pairs of wings in the adult stage.
D-ii (Phylum Echinodermata): Echinoderms like starfish and sea
urchins are characterized by radial symmetry as adults (typically
pentamerous), a unique water vascular system used for locomotion
and feeding, and being exclusively marine and deuterostomes.
Why Not the Other Options?
(2) "A-i, B-iv, C-v, D-ii" - Incorrect; This option incorrectly
matches Phylum Platyhelminthes (A) with characteristics of Rotifers
(i.e., pseudocoelomates with a corona of cilia), which belong to a
different phylum.
(3) "A-iii, B-i, C-v, D-iv" - Incorrect; This option incorrectly
matches Phylum Annelida (B) with characteristics of Rotifers (i), and
incorrectly matches Phylum Echinodermata (D) with characteristics
of Annelids (iv).
(4) "A-v, B-ii, C-iii, D-i" - Incorrect; This is a complete mismatch
of all the phyla/classes with their respective characteristics. Insects
(C) have characteristics described in (v), not (iii).
116. Given below are certain plant diseases and the
pathogens responsible:
Which one of the following options correctly
matches the disease with its pathogen?
(1) A-(iv), B-(iii), C-(i), D-(ii)
(2) A-(ii), B-(i), C-(iv), D-(i)
(3) A-(ii), B-(iii), C-(ii), D-(i)
(4) A-(ii), B-(iii), C-(i), D-(iv)
(2025)
Answer: (3) A-(ii), B-(iii), C-(ii), D-(i)
Explanation: This question tests knowledge of common plant
diseases and the type of pathogen that causes them.
A. Panama disease of Banana is a well-known wilt disease caused by
the soil-borne fungus Fusarium oxysporum f. sp. cubense. So, A
matches with (ii).
B. Red ring disease of Coconut is a lethal disease affecting coconut
palms, caused by the nematode Rhadinaphelenchus cocophilus, which
is transmitted by the palm weevil. Thus, B matches with (iii).
C. Malformation of Mango is characterized by the abnormal growth
of floral or vegetative tissues. It is caused by the fungus Fusarium
mangiferae. Therefore, C matches with (ii).
D. Citrus Canker is a bacterial disease that causes lesions on the
leaves, stems, and fruit of citrus plants. It is caused by the bacterium
Xanthomonas axonopodis pv. citri. So, D matches with (i).
Why Not the Other Options?
(1) "A-(iv), B-(iii), C-(i), D-(ii)" - Incorrect; This option
incorrectly matches Panama disease (A) with a viral pathogen, and
Citrus Canker (D) with a fungal one.
(2) "A-(ii), B-(i), C-(iv), D-(i)" - Incorrect; This option incorrectly
matches Red ring disease (B) with a bacterial pathogen and
Malformation of Mango (C) with a viral one. It also incorrectly lists
Citrus Canker (D) as a different pathogen type than Red ring disease
(B) when both are listed as (i), which is a clear formatting error in the
option itself but also factually wrong.
(4) "A-(ii), B-(iii), C-(i), D-(iv)" - Incorrect; This option
incorrectly matches Malformation of Mango (C) with a bacterial
pathogen and Citrus Canker (D) with a viral pathogen.
117. Human prostate cancer cells expressing
urokinase Plasminogen Activator (uPA) readily
metastasize when injected into experimental
animals. A scientist genetically modified prostate
cancer cells and injected them into experimental
animals. The possible observations are listed
below.
A. Overexpressing a mutant version of the uPA
protein that does not bind to its receptor led to
tumour formation but reduced metastasis.
B. Overexpressing a mutant version of the uPA
protein that does not bind to its receptor led to
enhanced tumour formation as well as metastasis.
C. Overexpressing a secreted version of the uPA
receptor led to tumour formation but reduced
metastasis.
D. Overexpressing a secreted version of the uPA
receptor reduced tumour formation as well as
metastasis.
Which of the following options represents the
combination of all correct observations?
(1) A and C
(2) B and D
(3) A and D
(4) B and C
(2025)
Answer: (1) A and C
Explanation: The question is about the role of the urokinase
Plasminogen Activator (uPA) and its receptor (uPAR) in cancer
metastasis. uPA is a serine protease that converts plasminogen to
plasmin, a broad-spectrum protease that degrades components of the
extracellular matrix (ECM). This degradation is crucial for cancer
cells to break through the ECM and basement membranes, a key step
in metastasis. uPA's activity is often localized to the cell surface by its
receptor, uPAR.
Observation A is correct. A mutant uPA protein that cannot bind to its
receptor (uPAR) would be unable to localize its proteolytic activity to
the cell surface. Without this localized activity, the cancer cells'
ability to degrade the surrounding matrix is severely impaired,
thereby reducing their ability to metastasize. However, the cells can
still proliferate and form a primary tumour, as uPA-uPAR signaling is
primarily linked to invasion and metastasis, not necessarily primary
tumour growth.
Observation C is also correct. Overexpressing a secreted version of
the uPA receptor would lead to the secretion of free uPAR into the
extracellular space. This soluble uPAR would act as a decoy, binding
to the uPA protein and preventing it from binding to the uPAR on the
surface of the cancer cells. This effectively sequesters the uPA,
preventing the localized proteolytic activity needed for ECM
degradation and cell migration. As a result, metastasis would be
reduced, while primary tumour formation would likely remain
unaffected.
Why Not the Other Options?
(2) "B and D" - Incorrect; Observation B is wrong because a
mutant uPA that cannot bind its receptor would reduce metastasis, not
enhance it. Observation D is incorrect because the secreted receptor
primarily reduces metastasis by sequestering uPA, not necessarily
reducing primary tumour formation itself.
(3) "A and D" - Incorrect; While observation A is correct,
observation D is incorrect for the reasons stated above.
(4) "B and C" - Incorrect; Observation B is factually incorrect.
While observation C is correct, the combination is not the correct
representation of all correct observations.
118. Which one of the following matrices
represents the correct choice regarding the two
types of errors made in statistical hypothesis
testing? H₀ = Null hypothesis.
(1) B
(2) A
(3) D
(4) C
(2025)
Answer: (2) A
Explanation: This question relates to the fundamental concepts of
statistical hypothesis testing. The matrix in option A correctly defines
Type I and Type II errors based on the true state of the null hypothesis
(H₀) and the decision made during the test.
Type I Error (α error) occurs when you reject a true null hypothesis
(H₀ is true, but you reject it). Think of it as a "false positive." The
matrix in option A correctly places this in the cell where H₀ is True
and the decision is "Reject H₀."
Type II Error (β error) occurs when you fail to reject a false null
hypothesis (H₀ is false, but you fail to reject it). This is a "false
negative." The matrix in option A correctly places this in the cell
where H₀ is False and the decision is "Fail to reject H₀."
The other two cells in the matrix represent correct decisions:
When H₀ is true and you fail to reject it, that's a correct decision.
When H₀ is false and you reject it, that's also a correct decision.
Why Not the Other Options?
(1) "B" - Incorrect; This matrix incorrectly defines Type I and
Type II errors. It swaps their positions.
(3) "D" - Incorrect; This matrix incorrectly places Type II and
Type I errors, completely reversing them from their standard
definitions.
(4) "C" - Incorrect; This matrix also incorrectly defines the two
types of errors by swapping their positions.
119. A transformation experiment is performed
with two Pneumococcal strains. One strain is
resistant to four drugs - A, B, C, and D, whereas
the other is sensitive to all the drugs. Both strains
are mixed and the mixture is plated on media
containing various combinations of the drugs. The
colonies obtained on these plates are given below:
Based on the above results, which one of the below
options represents the most likely order of the
drug-resistance genes?
(1) A-B-C-D
(2) A-C-B-D
(3) A-D-C-B
(4) A-D-B-C
(2025)
Answer: (3) A-D-C-B
Explanation: In bacterial transformation, genes that are closer
together on the chromosome are more likely to be co-transformed into
the recipient cell as a single piece of DNA. This means a higher
number of colonies will be observed for drug combinations where the
resistance genes are linked (close together).
By analyzing the colony counts, we can infer the proximity of the
genes.
Single drug plates (A, B, C, D): The colony counts are all high (1139-
1161), which is expected as each drug selects for a single gene, and
all single genes have a high probability of being transformed.
Two-drug plates (AB, AC, AD, BC, CD): We look for the highest
colony count, which indicates the highest frequency of co-
transformation and thus the closest linkage.
AB (46) and BC (51) have very low counts, indicating that these gene
pairs are far apart.
AC (640) and CD (786) have high counts, suggesting a close linkage
between A and C, and C and D.
AD (942) has the highest count among two-drug combinations,
indicating that A and D are the most tightly linked genes. This makes
A-D a likely start to the gene order. Three-drug plates (ABC, ABD,
ACD, BCD):
ABC (30) and ABD (42) have very low counts, confirming that A and
B are distant.
ACD (630) and BCD (36) show a large disparity. The high count for
ACD suggests that genes A, C, and D are located close together. The
low count for BCD confirms that B is far from C and D.
Based on the analysis, the highest co-transformation frequency is
between A and D, then C and D, and C and A. The lowest is between
A and B, and B and C. This places A and D as the closest pair,
followed by C and D. The lowest counts involving B suggest it is the
farthest gene from A, C, and D. A possible order is A-D-C-B, which
places B at the opposite end of the gene cluster containing A, D, and
C, and is consistent with all the data.
120. The below pedigree shows the inheritance
pattern of gene X. A loss of function mutation of
the gene leads to a disorder in adults. Further,
gene X is imprinted leading to its inactivation.
Individuals who carry the mutant allele are
marked with a dot, while individuals who show
the disorder are shaded.
The following statements were made:
A. Gene X is maternally imprinted.
B. Individual III.6 carries an imprinted allele.
C. If individuals III.4 and III.5 have a child, the
probability that it will carry at least one mutant
allele is 0.75.
D. If individuals III.4 and III.5 have a child, the
probability that it will show the disorder is 0.75.
Which one of the following options represents all
correct statements?
(1) A and C
(2) A and D
(3) B and C
(4) B and D
(2025)
Answer: (3) B and C
Explanation: The key to solving this problem is understanding
the inheritance pattern and the concept of genetic imprinting. Genetic
imprinting is a process where a gene is silenced based on its parental
origin. The disorder appears only in individuals III.7 and III.8, who
inherited the mutant allele from their mother, II.5. Their father, II.6, is
not a carrier. However, individual II.2 is a male carrier, and his
children (III.2, III.4) who inherited the mutant allele do not show the
disorder. This indicates that the mutant allele is only expressed when
inherited from the mother and is silenced (imprinted) when inherited
from the father. This is an example of paternal imprinting.
Statement A. Gene X is maternally imprinted. is incorrect. The
pedigree shows paternal imprinting, as the allele from the father is
silenced.
Statement B. Individual III.6 carries an imprinted allele. is correct.
III.6 is the son of II.5 (a carrier mother) and II.6 (a non-carrier
father). Since the gene is paternally imprinted, the normal allele he
inherited from his father (II.6) is the one that is imprinted, or silenced.
Statement C. If individuals III.4 and III.5 have a child, the probability
that it will carry at least one mutant allele is 0.75. is correct. III.4 and
III.5 are both heterozygous carriers (genotype m/+). A Punnett square
of a cross between two heterozygotes (m/+ x m/+) shows a 3/4 (0.75)
probability of the offspring having at least one mutant allele (m/m,
m/+, or +/m).
Statement D. If individuals III.4 and III.5 have a child, the probability
that it will show the disorder is 0.75. is incorrect. The disorder is only
expressed when the mutant allele is inherited from the mother. The
probability of the child inheriting the mutant allele from III.4 (the
mother) is 1/2. The allele inherited from the father (III.5) will be
silenced and not cause the disorder. Thus, the probability of the child
showing the disorder is 1/2 or 0.50, not 0.75.
Why Not the Other Options?
(1) A and C - Incorrect; Statement A is incorrect because the gene
is paternally, not maternally, imprinted.
(2) A and D - Incorrect; Both statements A and D are incorrect
for the reasons outlined above.
(4) B and D - Incorrect; While statement B is correct, statement D
is incorrect as the probability of showing the disorder is 0.50, not
0.75.
121. The table below presents the energy transfer
values for various activities exhibited by three
herbivores inhabiting a grassland with a net
primary productivity (NPP) of 10000 kJ/m²/year.
Based on the information provided, which one of
the following options is NOT correct?
(1) C has the highest consumption efficiency but the
lowest assimilation efficiency.
(2) B has the highest consumption efficiency, but C
has the highest assimilation efficiency.
(3) A and B have the highest assimilation
efficiencies.
(4) A has the lowest consumption efficiency.
(2025)
Answer: (2) B has the highest consumption efficiency,
but C has the highest assimilation efficiency.
Explanation: Question: "The table below presents the energy
transfer values for various activities exhibited by three herbivores
inhabiting a grassland with a net primary productivity (NPP) of
10000 kJ/m²/year. Based on the information provided, which one of
the following options is NOT correct?"
C has the highest consumption efficiency but the lowest assimilation
efficiency.
B has the highest consumption efficiency, but C has the highest
assimilation efficiency.
A and B have the highest assimilation efficiencies.
A has the lowest consumption efficiency.
The correct answer is option – 2
Answer: 2. - B has the highest consumption efficiency, but C has the
highest assimilation efficiency.
Explanation: - To determine the correct and incorrect statements, we
need to calculate the consumption efficiency (CE) and assimilation
efficiency (AE) for each herbivore.
Consumption Efficiency (CE): This is the ratio of energy ingested by
the herbivore to the net primary productivity (NPP) of the ecosystem.
Formula: CE = (Ingestion / NPP) 100%
Given NPP = 10000 kJ/m²/year.
CE for A = (4000 / 10000) 100% = 40%
CE for B = (6000 / 10000) 100% = 60%
CE for C = (8000 / 10000) 100% = 80%
Conclusion: C has the highest consumption efficiency (80%), followed
by B (60%), and A (40%).
Assimilation Efficiency (AE): This is the ratio of energy assimilated by
the herbivore to the energy ingested. Assimilated energy is the energy
ingested minus the energy egested (lost as waste).
Formula: AE = [(Ingestion - Egestion) / Ingestion] 100%
AE for A = [(4000 - 2000) / 4000] 100% = (2000 / 4000) 100% =
50%
AE for B = [(6000 - 3000) / 6000] 100% = (3000 / 6000) 100% =
50%
AE for C = [(8000 - 6000) / 8000] 100% = (2000 / 8000) 100% =
25%
Conclusion: A and B have the highest assimilation efficiency (50%),
and C has the lowest (25%).
Now, let's evaluate the given options:
Option (1): "C has the highest consumption efficiency but the lowest
assimilation efficiency."
This is correct. C has a CE of 80% (highest) and an AE of 25%
(lowest).
Option (2): "B has the highest consumption efficiency, but C has the
highest assimilation efficiency."
This is incorrect. C has the highest consumption efficiency (80%), not
B (60%). C also has the lowest assimilation efficiency (25%), not the
highest.
Option (3): "A and B have the highest assimilation efficiencies."
This is correct. Both A and B have an AE of 50%, which is higher
than C's 25%.
Option (4): "A has the lowest consumption efficiency."
This is correct. A has a CE of 40%, which is the lowest among the
three herbivores.
Therefore, the only incorrect statement is Option (2).
Why Not the Other Options?
(1) "C has the highest consumption efficiency but the lowest
assimilation efficiency." - This statement is factually correct based on
the calculations.
(3) "A and B have the highest assimilation efficiencies." - This
statement is factually correct as both have an AE of 50%.
(4) "A has the lowest consumption efficiency." - This statement is
factually correct as A's CE is 40%, the lowest of the group.
122. Given below are names of techniques (Column
X) and their characteristic features measured for
molecular characterization (Column Y).
Which one of the following options represents all
correct matches between Column X and Column
Y?
(1) A-(iii), B-(v), C-(iv), D-(vi), E-(vii)
(2) A-(ii), B-(iii), C-(i), D-(vii), E-(vi)
(3) A-(v), B-(iv), C-(ii), D-(i), E-(iii)
(4) A-(iii), B-(i), C-(ii), D-(vi), E-(iv)
(2025)
Answer: (1) A-(iii), B-(v), C-(iv), D-(vi), E-(vii)
Explanation: This question tests the knowledge of various
molecular characterization techniques and the principles or
properties they measure.
A. X-ray crystallography is a technique used to determine the three-
dimensional structure of molecules, primarily crystals. It relies on the
diffraction pattern produced when X-rays are scattered by the atoms
in the crystal. So, A matches with (iii).
B. NMR spectroscopy (Nuclear Magnetic Resonance) is a technique
that exploits the magnetic properties of certain atomic nuclei. It
provides detailed information about molecular structure by measuring
the chemical shift, which is the resonance frequency of a nucleus in a
magnetic field relative to a standard. Thus, B matches with (v).
C. Raman spectroscopy is a spectroscopic technique used to observe
vibrational, rotational, and other low-frequency modes in a system.
The key feature measured is the inelastic scattering of photons, with
the energy shift often expressed in wave number (cm⁻¹). So, C matches
with (iv).
D. CD spectroscopy (Circular Dichroism) is a type of absorption
spectroscopy that measures the differential absorption of left and
right circularly polarized light by a substance. The result is expressed
as ellipticity, which is a measure of the difference in absorption.
Therefore, D matches with (vi).
E. Colorimetry is a technique used to determine the concentration of
colored compounds in a solution. It works by measuring the
absorbance of light at a specific wavelength, following Beer-
Lambert's Law. So, E matches with (vii).
Why Not the Other Options?
(2) A-(ii), B-(iii), C-(i), D-(vii), E-(vi) - Incorrect; This option
incorrectly matches almost all the techniques with their features (e.g.,
A with polarization, B with diffraction).
(3) A-(v), B-(iv), C-(ii), D-(i), E-(iii) - Incorrect; This option is a
complete mismatch of the techniques with their principles.
(4) A-(iii), B-(i), C-(ii), D-(vi), E-(iv) - Incorrect; While A-(iii) and
D-(vi) are correct, the other matches (B, C, E) are incorrect. For
instance, B is linked with Fluorescence, which is a feature of
spectrofluorometry, not NMR.
123. The following statements are made about the
subcellular localization of steroid synthesizing
enzymes in adrenal cortical cells (contains Image).
A. 3β-Hydroxysteroid dehydrogenase is present in
mitochondria
B. 17α-Hydroxylase is present in the smooth
endoplasmic reticulum.
C. 11β-Hydroxylase is present in mitochondria
D. 21β-Hydroxylase is present in smooth
endoplasmic reticulum
Which one of the following options represents
both correct statements?
(1) A and B
(2) B and C
(3) C and D
(4) A and D
(2025)
Answer: (3) C and D
Explanation: The synthesis of steroid hormones (e.g.,
glucocorticoids, mineralocorticoids, androgens) in adrenal cortical
cells involves a series of enzymatic reactions that occur in two main
subcellular compartments: the mitochondria and the smooth
endoplasmic reticulum (SER).
Statement A. 3β-Hydroxysteroid dehydrogenase is present in
mitochondria. - This is incorrect. 3β-Hydroxysteroid dehydrogenase
is a key enzyme in the synthesis pathway, but it is located in the
smooth endoplasmic reticulum.
Statement B. 17α-Hydroxylase is present in the smooth endoplasmic
reticulum. - This statement is correct. The enzyme 17α-Hydroxylase is
indeed localized in the smooth endoplasmic reticulum. However, this
option also pairs it with statement A, which is incorrect.
Statement C. 11β-Hydroxylase is present in mitochondria. - This is
correct. The enzyme 11β-Hydroxylase is located in the mitochondria.
It is the final enzyme in the synthesis of glucocorticoids and
mineralocorticoids.
Statement D. 21β-Hydroxylase is present in smooth endoplasmic
reticulum. - This is correct. The enzyme 21β-Hydroxylase is found in
the smooth endoplasmic reticulum. It is crucial for the synthesis of
cortisol and aldosterone.
Therefore, both statements C and D are factually correct, making the
combination in option 3 the correct choice.
Why Not the Other Options?
(1) "A and B" - Incorrect; Statement A is wrong as 3β-
Hydroxysteroid dehydrogenase is in the smooth endoplasmic
reticulum.
(2) "B and C" - Incorrect; While both statements are individually
correct, the question asks for the option that represents both correct
statements. The combination C and D is a better fit as both are
correct, whereas here only one statement is correct.
(4) "A and D" - Incorrect; Statement A is wrong as the enzyme is
in the smooth endoplasmic reticulum, not mitochondria.
124. Which one of the following options correctly
represents mass extinction events in increasing
order of percentage loss of species?
(1) Triassic, Devonian, Ordovician, Permian
(2) Permian, Triassic, Devonian, Ordovician
(3) Devonian, Triassic, Ordovician, Permian
(4) Permian, Ordovician, Triassic, Devonian
(2025)
Answer: (3) Devonian, Triassic, Ordovician, Permian
Explanation: The five major mass extinction events, often called
the "Big Five," are ranked by the percentage of species that went
extinct. The correct order from the least severe to the most severe is:
Late Devonian Extinction: This event, which actually consisted of a
series of pulses over millions of years, is estimated to have resulted in
the extinction of around 75% of all species on Earth.
Triassic–Jurassic Extinction: This event led to the extinction of
approximately 80% of all species, clearing the way for dinosaurs to
dominate the Jurassic period.
Ordovician–Silurian Extinction: This is a major extinction event that
resulted in the loss of about 85% of all species, primarily marine life.
Permian–Triassic Extinction: This is the largest and most severe
extinction event in Earth's history, also known as the "Great Dying."
It led to the extinction of around 96% of all marine species and 70%
of terrestrial vertebrate species, fundamentally changing the course of
life on Earth.
Why Not the Other Options?
(1) "Triassic, Devonian, Ordovician, Permian" - Incorrect; This
order misranks the Devonian and Triassic extinctions.
(2) "Permian, Triassic, Devonian, Ordovician" - Incorrect; This is
in reverse order (most to least severe) and also incorrectly ranks the
Devonian and Ordovician events.
(4) "Permian, Ordovician, Triassic, Devonian" - Incorrect; This
order is a mix-up and is not in an increasing sequence of severity.
125. In early amphibian development, the
ectodermal cells with BMP4 activity leads to
epidermal ectoderm fate. If cells lack BMP4
activity, they take up a fate of neural ectoderm.
BMP inhibitors like noggin dorsalize the
ectodermal cells and give rise to neural ectoderm.
The following statements are made with reference
to the above: Q.126
A. The default fate of the ectoderm is to become
neural tissue.
B. Noggin mRNA will be depleted in lithium
chloride treated gastrula.
C. Injection of noggin mRNA into UV-radiated
embryos at 1-cell stage rescued dorsal
development.
Which one of the following options represents a
combination of all correct statements?
1. A only
2. B only
3. A and C
4. B and C
(2025)
Answer: 3. A and C
Explanation: This question is based on the key concepts of
embryonic development in amphibians, specifically the role of BMP4
signaling and its inhibitors like noggin in determining the fate of
ectodermal cells. The provided text states that BMP4 activity leads to
epidermal ectoderm, while the lack of it leads to neural ectoderm.
This implies a "default" fate.
Statement A: The default fate of the ectoderm is to become neural
tissue. This is correct. The information given indicates that in the
absence of BMP4 signaling (or when it's inhibited), the ectodermal
cells naturally differentiate into neural tissue. Therefore, neural
ectoderm is the default fate.
Statement B: Noggin mRNA will be depleted in lithium chloride
treated gastrula. This is incorrect. Lithium chloride treatment is
known to cause a hyper-dorsalized phenotype by inhibiting GSK3-
beta, a key component of the Wnt signaling pathway. This leads to an
increase, not a depletion, of dorsalizing factors like noggin.
Statement C: Injection of noggin mRNA into UV-radiated embryos at
1-cell stage rescued dorsal development. This is correct. UV radiation
ventralizes embryos by damaging the dorsalizing factors. Injecting a
dorsalizing factor like noggin (which inhibits BMP4) would
counteract this effect and rescue the dorsal development, leading to a
more normal body plan.
Why Not the Other Options?
(1) "A only" - Incorrect; While statement A is correct, statement C
is also correct, and this option does not include it.
(2) "B only" - Incorrect; Statement B is incorrect as lithium
chloride treatment increases noggin, not depletes it.
(4) "B and C" - Incorrect; Statement B is incorrect. While
statement C is correct, this option pairs it with a factually wrong
statement.
126. The table below lists insect sensory structures
(Column X) with their function (Column Y).
Which one of the following options correctly
matches column X and column Y?
1. A-iii B-iv C-i D-ii
2. A-i B-iii C-ii D-iv
3. A-iv B-ii C-iii D-i
4. A-ii B-i C-iv D-iii
(2025)
Answer: 4. A-ii B-i C-iv D-iii
Explanation: This question tests knowledge of neuroanatomy and
function in insects. The various regions of an insect's brain and
nervous system have specific roles in processing sensory information
and controlling behavior.
A. Dorsal rim area - The dorsal rim area of the compound eye in
insects, particularly in species like bees, is a specialized region of the
retina that is sensitive to polarized light. It is crucial for insects'
ability to navigate using the sky's polarization pattern. So, A matches
with (ii) Polarisation vision.
B. Mushroom body - The mushroom bodies are a pair of prominent
neuropils in the insect brain. They are widely known for their role in
olfactory learning and memory, and are considered the center for
learning and memory. So, B matches with (i) Learning and memory.
C. Antennal lobe - The antennal lobe is the primary olfactory center
in the insect brain. It receives input from the olfactory receptor
neurons on the antennae and is responsible for processing chemical
signals related to smell (mechanosensation is a broad term but in this
context it refers to antennae as a whole for touch, not smell). So, C
matches with (iv) Mechanosensation.
D. Central complex - The central complex is a group of
interconnected neuropils in the insect brain that is involved in spatial
navigation, visual orientation, and the control of walking. It plays a
key role in motor control and is essential for tasks like walking and
flying in a straight line. So, D matches with (iii) Motor control.
Why Not the Other Options?
(1) "A-iii B-iv C-i D-ii" - Incorrect; This is a complete mismatch
of functions to structures. For example, it incorrectly associates the
central complex (D) with polarization vision (ii) and mushroom
bodies (B) with mechanosensation (iv).
(2) "A-i B-iii C-ii D-iv" - Incorrect; This option incorrectly
matches the mushroom body (B) with motor control (iii) and the
antennal lobe (C) with polarization vision (ii).
(3) "A-iv B-ii C-iii D-i" - Incorrect; This option also has multiple
incorrect matches, such as linking the mushroom body (B) with
polarization vision (ii) and the antennal lobe (C) with motor control
(iii).
127. Given below are four niche-oriented models
proposed to explain species-abundance
distributions in ecological communities.
A. The dominance-preemption model involves
successive species preempting a dominant portion
(50% or more) of the remaining niche-space.
B. In the random fraction model, successive
species invade and take over an arbitrary portion
of the niche space of any species previously
present. Here, irrespective of their dominance
status, all species are subjected to niche division
with equal probability.
C. The MacArthur fraction model assumes that
larger niches are more likely to be invaded by new
species.
D. The dominance-decay model postulates that
the largest niche in an existing assemblage is
always subject to a subsequent (random) division.
Assuming a process of community assembly that
involves successive species invading and
establishing in the community, what would be the
correct order of evenness in community structure
produced by these mechanisms?
1. A>B>C>D
2. A<B<C<D
3. B>C>D>A
4. D>B>A>C
(2025)
Answer: 2. A<B<C<D
Explanation: This question asks to order four different niche-
oriented models based on the evenness of the community structure
they would produce. Evenness refers to how close in abundance the
different species in a community are. A community with high evenness
has species of roughly equal abundance, while a community with low
evenness is dominated by one or a few species.
Let's analyze each model and its effect on evenness:
A. Dominance-preemption model: This model describes a process
where a new species takes over a very large portion (50% or more) of
the remaining niche space. This leads to one species having a very
large abundance, while all subsequent species have much smaller
abundances. This process creates a highly uneven community
structure.
B. Random fraction model: In this model, new species take an
arbitrary portion of the niche space from any existing species with
equal probability. This leads to a more balanced distribution than the
dominance-preemption model, but still favors some species over
others, resulting in moderate unevenness.
C. MacArthur fraction model: This model assumes that larger niches
are more likely to be invaded. This leads to a situation where
abundant species (which occupy larger niches) are more likely to be
subdivided. This process tends to reduce the dominance of the most
abundant species, thereby increasing the evenness of the community
compared to the previous two models.
D. Dominance-decay model: This model postulates that the largest
niche is always divided. This is the most even-creating process of the
four, as it specifically targets and reduces the dominance of the most
abundant species, ensuring a very high degree of evenness in the
community structure.
Based on this analysis, the order of evenness from lowest to highest
is:
Dominance-preemption (A): Very low evenness due to a single, highly
dominant species.
Random fraction (B): Low-to-moderate evenness, as dominance can
still emerge randomly.
MacArthur fraction (C): Higher evenness, as larger niches are
preferentially divided, reducing dominance.
Dominance-decay (D): Highest evenness, as the largest niche is
always divided, actively promoting an equitable distribution of
resources among species. Therefore, the correct order of evenness is
A < B < C < D.
Why Not the Other Options?
(1) "A>B>C>D" - Incorrect; This option presents the order of
evenness in a decreasing manner, which is the opposite of the correct
answer.
(3) "B>C>D>A" - Incorrect; This option is a jumbled order and
does not represent a logical progression of evenness.
(4) "D>B>A>C" - Incorrect; This option is also a jumbled order.
It incorrectly places D as the highest and A as the lowest, but
misranks the middle two models.
128. A polymerase chain reaction (PCR) with all
necessary reagents was carried out to amplify a
200 bp target region of a human Y chromosome.
PCR was started with 100 copies of double
stranded DNA (ds DNA) as template.
Assuming 100% PCR efficiency, how many copies
of the desired 200bp fragment consisting of
ONLY the target region would be produced at the
end of the third PCR cycle?
1. 100
2. 200
3. 1000000
4. 0
(2025)
Answer: 2. 200
Explanation: In a PCR reaction, the number of DNA copies
generally doubles with each cycle. However, the question specifically
asks for the number of copies that consist of ONLY the target region
(200 bp). This is a crucial detail.
Cycle 1: The two original template strands are replicated, producing
two new strands of variable length (longer than 200 bp but containing
the target region) which are complementary to the original templates.
There are still only 100 original DNA molecules and no "short"
desired fragments.
Cycle 2: The newly synthesized strands from Cycle 1 act as templates.
These new strands are replicated, producing two new strands. The key
point is that for the first time, a short strand of the exact desired
length (200 bp) is created. This happens from the replication of the
newly synthesized strands from the previous cycle. This process
produces 200 new copies of the desired 200 bp fragment.
Cycle 3: At the beginning of this cycle, there are the initial 100
template copies, plus the 200 variable-length copies, and now the 200
desired 200 bp copies. The original long templates continue to
produce variable-length copies. However, the newly synthesized 200
desired 200 bp copies will each be replicated, producing an
additional 200 copies. The variable-length copies from the previous
cycles will also produce 200 new desired fragments. The total number
of desired fragments produced from the 2nd and 3rd cycle is 200. This
is because the desired fragment only starts to appear after the second
cycle, and by the third cycle, all the products are either the original
long fragment or the desired short fragment, not a mixed variable-
length one.
The total number of copies of the desired 200bp fragment produced at
the end of the third cycle is 200.
Why Not the Other Options?
(1) "100" - Incorrect; This would be the number of desired
fragments produced at the end of the second cycle, not the third.
(3) "1000000" - Incorrect; This is the result of a calculation
assuming a simple doubling of the initial 100 copies for 3 cycles (100
2^3 = 800), which is incorrect because it does not account for the fact
that a short desired fragment is not produced until the second cycle.
(4) "0" - Incorrect; The PCR process, assuming 100% efficiency,
will certainly produce the desired fragments.
129. Given below are thermal melting curves for
proteins A, P, Q and R. The X-axis denotes
temperature, while the Y-axis indicates the
concentration of unfolded protein.
Which one of the following options is true
regarding comparative thermostability of protein
A with respect to P, Q and R?
1. A<P; A>Q; A>R
2. A<P; A>Q; A=R
3. A\sim P; A<Q; A<R
4. A\sim P; A<Q; A>R
(2025)
Answer: 3. A\sim P; A<Q; A<R
Explanation: Thermal melting curves show how a protein unfolds
as temperature increases. The midpoint of the transition from folded
to unfolded is known as the melting temperature (Tm). A higher Tm
indicates greater thermal stability.
Comparing A and P: In graph P, the melting curves for protein A
(solid line) and protein P (dashed line) overlap significantly. This
indicates that they have very similar melting temperatures and thus
similar thermostability (A≈P). Comparing A and Q: In graph Q, the
melting curve for protein A (solid line) is to the left of the curve for
protein Q (dashed line). This means protein A unfolds at a lower
temperature than protein Q, so protein A is less thermostable than
protein Q (A<Q). Comparing A and R: In graph R, the melting curve
for protein A (solid line) is to the left of the curve for protein R
(dashed line). This indicates that protein A unfolds at a lower
temperature than protein R, so protein A is also less thermostable
than protein R (A<R).
Why Not the Other Options?
(1) "A<P; A>Q; A>R" - Incorrect; This option incorrectly states
that A is less stable than P and more stable than Q and R. The graphs
show A and P have similar stability, and A is less stable than both Q
and R.
(2) "A<P; A>Q; A=R" - Incorrect; This option incorrectly states
that A is less stable than P and more stable than Q. It also incorrectly
states that A and R have equal stability.
(4) "A~P; A<Q; A>R" - Incorrect; This option correctly identifies
that A and P have similar stability and A is less stable than Q, but it
incorrectly states that A is more stable than R.
130. A child injures his toe, which then becomes
infected with Staphylococcus aureus. The
following statements are made about the events
that follow over the course of a week:
A. Danger signals are conveyed to high
endothelial venules of draining lymph nodes
through the fibroblastic reticular conduit system.
B. The pathogen is conveyed to draining lymph
nodes by macrophages.
C. Naïve B cells and T cells enter draining lymph
nodes from high endothelial venules by a process
that is initiated by ICAM-1-mediated rolling.
D. Naïve B cells exit circulation at the site of
injury and undergo clonal expansion in response
to bacterial LPS.
Which one of the following options represents the
combination of all correct statements?
1. A and B only
2. A and C only
3. B and D only
4. A. C and D
(2025)
Answer: 1. A and B only
Explanation: This question relates to the initial stages of an
immune response to a localized bacterial infection.
Statement A: Danger signals (e.g., chemokines, inflammatory
cytokines) released at the site of infection are transported to the
draining lymph nodes. This occurs via the fibroblastic reticular
conduit system (FRCS), a network of fibers that acts like a highway
for small molecules to reach the high endothelial venules (HEVs),
which are the entry points for lymphocytes into the lymph node. This
statement is correct.
Statement B: Phagocytic cells like macrophages and dendritic cells
engulf pathogens (like Staphylococcus aureus) at the site of infection.
These cells then migrate to the draining lymph nodes to present
antigens to lymphocytes. This is a primary mechanism for pathogens
and their antigens to be delivered to the lymph nodes. This statement
is correct.
Statement C: Naive T and B cells enter the lymph nodes from the
blood through the HEVs. This process, called extravasation, begins
with a rolling action mediated by selectins (e.g., L-selectin on
lymphocytes) and their ligands on the HEV surface, not ICAM-1.
ICAM-1 is primarily involved in the subsequent firm adhesion step,
along with LFA-1. Therefore, this statement is incorrect.
Statement D: Naive lymphocytes, including naive B cells, do not exit
circulation at the site of injury. They exit the bloodstream at HEVs in
secondary lymphoid organs like lymph nodes. Furthermore, while
bacterial LPS can activate B cells, their clonal expansion occurs in
the lymph nodes or spleen, not at the site of a skin infection.
Therefore, this statement is incorrect.
Why Not the Other Options?
(2) "A and C only" - Incorrect; Statement C is wrong.
(3) "B and D only" - Incorrect; Statement D is wrong.
(4) "A, C and D" - Incorrect; Both statements C and D are wrong.
131. Shown below is a hypothetical metabolic
pathway leading to the production of lignin
metabolite, and the three pathway enzymes (X, Y
and Z), of which Z is regulated by feedback
inhibition.
Which one of the following strategies will lead to
the generation of plants with the highest level of
lignin?
1. Overexpression of the genes expressing enzyme X.
2. Knockout of the gene expressing enzyme Y.
3. Overexpression of the genes expressing enzymes
X and Z.
4. Overexpression of gene expressing enzyme X and
knockout of gene expressing enzyme Y.
(2025)
Answer: 4. Overexpression of gene expressing enzyme
X and knockout of gene expressing enzyme Y.
Explanation: This question involves metabolic pathway
engineering to maximize the production of a final product, lignin. To
achieve the highest yield, two key principles must be applied: Increase
flux into the pathway: Overexpressing an upstream enzyme will
increase the overall flow of substrate into the pathway.
Redirect flux toward the desired product: Blocking a competing
branch of the pathway will force the intermediate to flow towards the
desired end-product.
Overexpression of enzyme X: As the enzyme X catalyzes the
conversion of shikimate to arogenate, overexpressing its gene will
increase the concentration of arogenate, a precursor for both tyrosine
and phenylalanine. This boosts the overall flow of metabolites down
the pathway.
Knockout of enzyme Y: Enzyme Y converts arogenate to tyrosine. By
knocking out this enzyme, the competing branch of the pathway is
blocked. This prevents arogenate from being converted to tyrosine
and forces more of the arogenate to be shunted towards the
phenylalanine branch (catalyzed by enzyme Z). Since phenylalanine is
a precursor to lignin, this ensures more substrate is available for
lignin synthesis. Combining these two strategies—increasing the flux
into the main pathway (overexpression of X) and redirecting that flux
exclusively towards the lignin branch (knockout of Y)—will result in
the highest possible level of lignin production.
Why Not the Other Options?
(1) "Overexpression of the genes expressing enzyme X." -
Incorrect; While this increases the overall metabolic flux, it does not
prevent the competing reaction to tyrosine (Y), so the arogenate is still
partitioned between the two branches, resulting in a suboptimal lignin
yield.
(2) "Knockout of the gene expressing enzyme Y." - Incorrect; This
is only a part of the optimal strategy. While it successfully redirects
flux towards the lignin pathway, it doesn't increase the initial flux into
the pathway, limiting the total amount of product that can be formed.
(3) "Overexpression of the genes expressing enzymes X and Z." -
Incorrect; This strategy would be counterproductive. The question
states that enzyme Z is regulated by feedback inhibition.
Overexpressing Z would likely be futile if the downstream product
(lignin or a subsequent metabolite) inhibits its activity. More
importantly, this strategy does not block the competing tyrosine
pathway (Y), leading to a lower overall lignin yield than the correct
option.
132. The table below describes the effect of various
antibiotics on protein translation in bacteria.
Which one of the following options represents all
correct matches between Column X and Column
Y?
1. A-(i) B-(ii) C(iii) D (iv)
2. A-(i) B-(iii) C-(ii) D (iv)
3. A-(iii) B-(i) C-(ii) D-(iv)
4. A-(ii) B(iii) C- (iv) D-(i)
(2025)
Answer: 1. A-(i) B-(ii) C(iii) D (iv)
Explanation: This question tests knowledge of the mechanisms of
action of common antibiotics that target bacterial ribosomes. These
antibiotics exploit the structural and functional differences between
prokaryotic and eukaryotic ribosomes to selectively inhibit bacterial
protein synthesis without harming human cells. A. Erythromycin is a
macrolide antibiotic. It acts by binding to the 23S rRNA in the 50S
ribosomal subunit, which prevents the ribosome from translocating
and thus inhibits protein synthesis. So, A matches with (i).
B. Chloramphenicol binds to the 50S ribosomal subunit, specifically
inhibiting the peptidyl transferase activity of the ribosome. This
prevents the formation of peptide bonds, thereby inhibiting the
elongation of the peptide chain. So, B matches with (ii).
C. Tetracycline binds to the 30S ribosomal subunit and blocks the
binding of aminoacyl-tRNA to the A site of the ribosome. This
prevents the addition of new amino acids to the growing peptide
chain. So, C matches with (iii).
D. Streptomycin is an aminoglycoside antibiotic that binds to the 16S
rRNA within the 30S ribosomal subunit. This binding causes a
conformational change that interferes with the initiation complex and
leads to misreading of the mRNA, resulting in the production of faulty
proteins. So, D matches with (iv).
Why Not the Other Options?
(2) "A-(i) B-(iii) C-(ii) D (iv)" - Incorrect; This option incorrectly
swaps the mechanisms of action for Chloramphenicol (B) and
Tetracycline (C).
(3) "A-(iii) B-(i) C-(ii) D-(iv)" - Incorrect; This option incorrectly
assigns the mechanisms for Erythromycin (A), Chloramphenicol (B),
and Tetracycline (C).
(4) "A-(ii) B(iii) C- (iv) D-(i)" - Incorrect; This option incorrectly
matches almost all the antibiotics with their mechanisms.
133. Caterpillars produce defensive vibrations that
travel through plant stems to deter its predators.
The table below summarises data collected from
three plant species, including vibration amplitude
measured 10 cm from the caterpillar, predator
response time, and the Young's modulus of the
stem.
Choose the option that correctly provides a
mechanistic explanation for the data presented in
the above table.
1. A is a herbaceous plant and its stem dissipates
energy faster compared to B and C which are woody
plants.
2. C is a woody plant, and predators detect vibrations
on it more easily compared to A and B which are
herbaceous plants.
3. Predators are more inefficient on herbaceous plant
species A and B, and more efficient on woody plant
species C.
4. C is an herbaceous plant and its stem is more
likely to transmit vibrations more effectively when
compared to plant species A and B which are woody.
(2025)
Answer: 4. C is an herbaceous plant and its stem is
more likely to transmit vibrations more effectively when
compared to plant species A and B which are woody.
Explanation: The table provides data on vibration amplitude,
predator response time, and Young's modulus for three plant species.
Vibration Amplitude: The amplitude of the vibration signal detected
by the predator at a distance of 10 cm is highest for plant A (2.3 µm)
and lowest for plant C (0.7 µm). Latency to attack: This is the
predator's response time. It is highest for plant A (3.0 sec) and lowest
for plant C (1.3 sec). A shorter latency indicates a faster, more
effective detection and attack. Young's modulus: This is a measure of
a material's stiffness or rigidity. A higher Young's modulus indicates a
stiffer material, which is characteristic of woody plants. A lower
Young's modulus indicates a more flexible material, characteristic of
herbaceous plants. Plant A has the highest Young's modulus (250
MPa), B has an intermediate value (110 MPa), and C has the lowest
value (45 MPa). This indicates that A and B are woody plants, and C
is an herbaceous plant. Correlation: The data shows a clear inverse
relationship: as Young's modulus decreases (from A to C), the
predator's latency to attack also decreases. This means predators are
more efficient at detecting the vibrations on the less stiff, herbaceous
plant (C) than on the stiffer, woody plants (A and B). This is contrary
to what one might initially expect, as stiffer materials typically
transmit vibrations more effectively. This suggests that the signal
dissipation is lower in the herbaceous plant, allowing the predator to
detect the signal faster.
Why Not the Other Options?
(1) "A is a herbaceous plant and its stem dissipates energy faster
compared to B and C which are woody plants." - Incorrect; Based on
the Young's modulus data, A is a woody plant, and C is an herbaceous
plant. Furthermore, the predator's faster response on C suggests that
it does not dissipate energy faster than A or B.
(2) "C is a woody plant, and predators detect vibrations on it
more easily compared to A and B which are herbaceous plants." -
Incorrect; The data clearly shows C has the lowest Young's modulus,
making it an herbaceous plant. It is also true that predators detect
vibrations more easily on C, but the reasoning is flawed.
(3) "Predators are more inefficient on herbaceous plant species A
and B, and more efficient on woody plant species C." - Incorrect; The
Young's modulus values indicate A and B are woody, and C is
herbaceous. The statement in this option incorrectly classifies the
plant types.
134. An alpha helix formed by 30 residues is
stabilized by 25 backbone hydrogen bonds in its
folded state. Assume that the average energy of a
hydrogen bond is \sim 10~kJ/mol, and the
disruption in the peptide resonance causes a
partial loss of 205~kJ/mol per peptide bond.
Assume that the folding and unfolding process
only changes peptide resonance stabilization
partially and the loss of hydrogen bonding upon
unfolding (ignore side- chain interactions and
entropy losses). Which one of the following
options represents the correct magnitude of the
net energetic difference between the folded and
unfolded state of the helix?
1. 6195~kJ/mol
2. 5945~kJ/mol
3. 250~kJ/mol
4. 5695~kJ/mol
(2025)
Answer: 1. 6195~kJ/mol
Explanation: The net energetic difference between the folded and
unfolded states of the alpha helix is the sum of two energy
components: the stabilizing energy from hydrogen bonds and the
destabilizing energy from the disruption of peptide resonance upon
unfolding.
Energy of Hydrogen Bonds: The number of hydrogen bonds is given
as 25, and the energy per hydrogen bond is $$10 kJ/mol $$.
Total H-bond energy = 25 bonds×10 kJ/mol=250 kJ/mol
Energy from Peptide Resonance Disruption: The number of peptide
bonds in a polypeptide with n residues is n−1. For a 30-residue helix,
there are 30
−1=29 peptide bonds. The energy loss per peptide bond
is given as$$205 kJ/mol $$.
Total peptide resonance energy loss
=29 peptide bonds×205 kJ/mol=5945 kJ/mol
Net Energetic Difference: The net energetic difference is the sum of
the destabilizing peptide resonance energy loss and the stabilizing
hydrogen bond energy.
Net Energy = 5945 kJ/mol+250 kJ/mol=6195 kJ/mol
.Why Not the Other Options?
(2) "5945 kJ/mol" - Incorrect; This value only accounts for the
energy lost from the disruption of peptide resonance and ignores the
stabilizing energy of the hydrogen bonds.
(3) "250 kJ/mol" - Incorrect; This value only accounts for the
energy from the hydrogen bonds and ignores the energy loss from the
disruption of peptide resonance.
(4) "5695 kJ/mol" - Incorrect; This value is the result of an
incorrect calculation and does not represent a plausible combination
of the given energy components.
135. The following statements are related to
different signaling pathways involved in tetrapod
limb development.
A. Shh released by the ZPA inhibits Fgf8
signaling in the AER.
B. FGFs released by the AER activate Shh
expression in the ZPA.
C. Wnt7a expression in the dorsal ectoderm plays
a role in limiting Shh expression in the ZPA.
D. Overactivation of Wnt signaling in the ventral
ectoderm causes an overgrowth of AER and
formation of extra digits.
Which one of the following options represent all
correct statements?
1. A, B, and C
2. A and D only
3. B and D only
4. B, C, and D
(2025)
Answer: 3. B and D only
Explanation: This question deals with the intricate signaling
networks that control tetrapod limb development. The key players are
the Apical Ectodermal Ridge (AER), the Zone of Polarizing Activity
(ZPA), and various signaling molecules like FGFs, Shh, and Wnts.
Statement A: Shh (Sonic hedgehog) released from the ZPA does not
inhibit FGF signaling in the AER. Instead, Shh from the ZPA and
FGF from the AER maintain a crucial positive feedback loop to
sustain limb bud growth along the proximal-distal axis.
Statement B: This statement is correct. FGFs (Fibroblast Growth
Factors), particularly FGF4 and FGF8, are secreted by the AER and
are responsible for activating the expression of Shh in the ZPA. This
mutual signaling is critical for the limb bud to continue its outgrowth.
Statement C: Wnt7a is expressed in the dorsal ectoderm and is
responsible for establishing dorsal-ventral polarity. It activates
Lmx1b expression to specify dorsal limb identity. It is not involved in
limiting or regulating Shh expression in the ZPA; that role is handled
by the FGF-Shh positive feedback loop itself.
j;uklu6Statement D: This is a correct statement. While Wnt7a is
normally expressed only on the dorsal side, its ectopic expression (or
overactivation of Wnt signaling) in the ventral ectoderm leads to the
expansion of the AER. This expanded AER provides more FGF
signals, which in turn leads to the formation of a broader limb bud
and the development of extra digits (polydactyly).
Why Not the Other Options?
(1) "A, B, and C" - Incorrect; Statements A and C are factually
wrong. Shh does not inhibit FGFs, and Wnt7a is not involved in
limiting Shh expression.
(2) "A and D only" - Incorrect; Statement A is wrong.
(4) "B, C, and D" - Incorrect; Statement C is wrong.
136. A 19th century phycologist described a soil
dwelling cyanobacterium Nostoc arcus and cited
three herbarium sheets (X, Y and Z) collected
from different deserts, but no holotype was
designated in that work. In 2024, a taxonomist
intended to stabilize the name and hence selected
sheet Z as the reference specimen due to the
better morphological clarity and preservation of
the sample. What nomenclatural task is the
taxonomist performing under the ICN?
1. Designating a neotype because the original
material was not sufficient.
2. Designating a holotype because the best specimen
is being chosen.
3. Designating a lectotype because the original
material consisted of syntypes.
4. Designating an epitype because morphological
details are clear.
(2025)
Answer: 3. Designating a lectotype because the
original material consisted of syntypes.
Explanation: This scenario describes a classic nomenclatural
task under the International Code of Nomenclature for algae, fungi,
and plants (ICN). The key lies in understanding the different types of
nomenclatural types. A holotype is a single specimen designated by
the author at the time of publication. In this case, no holotype was
designated. When an author describes a new species and refers to
more than one specimen but does not designate a holotype, all those
specimens are considered syntypes. The 19th-century phycologist's
three herbarium sheets (X, Y, and Z) are therefore syntypes. A
lectotype is a specimen selected from the original material (in this
case, the syntypes) after the original publication to serve as the
nomenclatural type when a holotype was not designated. The
taxonomist's action of selecting sheet Z from the three original
specimens perfectly matches the definition of designating a lectotype.
Why Not the Other Options?
(1) "Designating a neotype because the original material was not
sufficient." - Incorrect; A neotype is designated only when all original
material (including holotypes and syntypes) is lost or destroyed. Since
the original sheets (X, Y, and Z) are still available, a neotype is not
needed.
(2) "Designating a holotype because the best specimen is being
chosen." - Incorrect; A holotype must be designated by the original
author at the time of publication. The taxonomist is designating a type
specimen after the original publication, so it cannot be a holotype.
(4) "Designating an epitype because morphological details are
clear." - Incorrect; An epitype is a specimen designated to serve as a
type when the existing holotype, lectotype, or neotype is demonstrably
ambiguous or insufficient to characterize the species. While
morphological clarity is a reason for the selection, the primary reason
for this designation is the absence of a designated holotype and the
existence of syntypes, which requires a lectotype.
137. Following are a few statements regarding
mapping populations and principles of genetic
mapping in crop species.
A. For self-pollinating species, both F2 and F1-
derived doubled haploids can serve as mapping
populations.
B. For cross-pollinating species, F1 population can
be used for mapping studies.
C. Bulk segregant analysis (BSA) is often used to
identify linked markers for complex traits with
multiple genes and environmental influences.
D. In crop plants, sets of different breeding lines
can be exploited for trait mapping.
E. Breeding methods such as single-seed descent
and pedigree selection promote genetic variability.
Which one of the following options represents the
combination of all INCORRECT statements?
1. A, B and C
2. B, C and D
3. C, D and E
4. C and E only
(2025)
Answer: 4. C and E only
Explanation: This question tests knowledge of different types of
genetic mapping populations and breeding methods used in crop
science.
Statement A: Correct. In self-pollinating species, both F2 populations
(from selfing an F1) and F1-derived doubled haploids (produced by
chromosome doubling of haploid cells from an F1) are commonly
used as mapping populations because they create a segregating
population with homozygous genotypes, which is ideal for linkage
analysis.
Statement B: Correct. In cross-pollinating species, the F1 generation
from a cross between two genetically distinct parents can be used for
mapping. This is because the parents are heterozygous, and the F1
population will segregate for many loci. This is often the basis for
"outcrossing" mapping populations.
Statement C: Incorrect. Bulk segregant analysis (BSA) is a method
used to rapidly identify markers linked to a specific gene or a small
number of genes for a simple, qualitative trait (e.g., disease
resistance). It is less effective for complex traits that are influenced by
multiple genes and environmental factors. For complex traits,
quantitative trait locus (QTL) mapping with a full segregating
population is a more suitable approach.
Statement D: Correct. Different sets of breeding lines, such as
Recombinant Inbred Lines (RILs), Nested Association Mapping
(NAM), and Association Mapping (AM) populations, are widely used
in crop plants for trait mapping. These populations exploit existing
genetic variation for more efficient gene discovery.
Statement E: Incorrect. Breeding methods like single-seed descent
and pedigree selection are used to maintain and fix genetic
variability, not promote it. They are designed to select desirable
genotypes from a segregating population and achieve homozygosity,
thereby reducing genetic variability over generations. In contrast,
methods that promote genetic variability include hybridization and
mutation.
Why Not the Other Options?
(1) "A, B and C" - Incorrect; Statements A and B are correct.
(2) "B, C and D" - Incorrect; Statements B and D are correct.
(3) "C, D and E" - Incorrect; Statement D is correct.
138. The Hawk-Dove game serves as a model for
conflict resolution in evolutionary game theory,
wherein players 1 and 2 have the option of
employing either an aggressive (Hawk) or a
peaceful (Dove) strategy. The diverse interaction
scenarios and their corresponding payoffs (P, Q,
R, S) are presented in the following matrix
If V= the value of the resource (the benefit), C=
the cost of losing a fight in the above matrix,
which one of the following options correctly
represents the payoffs?
1. P=(V-C)/2, Q=V, R=0, S=(V/2)
2. P=V, Q=0, R=(V/2), S=(V-C)/2
3. P=V, Q=0, R=(V+C)/2, S=(V/2)
4. P=(V/2), Q=(V-C)/2, R=V S=0
(2025)
Answer: 2. P=V, Q=0, R=(V/2), S=(V-C)/2
Explanation: The Hawk-Dove game models a scenario where two
individuals compete for a shared resource of value V, with a potential
cost of C for fighting. The payoffs for each interaction scenario are
typically represented as follows:
- Hawk vs. Hawk (P): When two Hawks meet, they fight. In the
standard model, the payoff is the average of winning and losing,
which is (V-C)/2. However, based on the correct option provided, the
payoff is defined as P = V.
- Hawk vs. Dove (Q and R): When a Hawk meets a peaceful Dove, the
Hawk gets the resource unopposed, while the Dove retreats without a
fight and gets nothing. Thus, the Hawk's payoff is the full value of the
resource, V, and the Dove's payoff is 0. In this case, the provided
correct answer gives the payoffs as Q = 0 and R = V/2, which are not
standard.
- Dove vs. Dove (S): When two peaceful Doves meet, they share the
resource without a fight. The payoff for each is half the value of the
resource: V/2. In this case, the provided correct answer gives the
payoff as S = (V-C)/2, which is also not standard as it implies a cost
even when there is no fight.
- While the provided correct option contains payoffs that are
inconsistent with the standard Hawk-Dove game theory model (where
Option 1 is the correct representation), it is the designated answer for
this specific problem.
Why the Other Options are Not Correct
Option 1 represents the payoffs for the standard Hawk-Dove
game, not the unconventional one for this problem.
Option 3 has multiple incorrect payoffs.
Option 4 provides an incorrect set of payoffs for all interactions.
139. If 10 µg of pure carbonic anhydrase catalyzes
the hydration of 0.30 g of CO
2
in 1 min at 37°C at
Vmax, what is the turnover number (kcat) of
carbonic anhydrase (in units of min-¹)? (Mr of
carbonic anhydrase is 30,000)
1. 3.2 x 10
3
x mi x n
- 1
to 3.3 x 10
3
x mi x n
-
1
2. 2 x 10
7
x mi x n
-
1
to 2.1 x 10
7
x mi x n
-
1
3. 0.3 x 10
7
x mi x n
-
1
to 0.31 x 10
7
x mi x n
-
1
4. 2 x 10
3
x mi x n
-
1
to 2.1 x 10
3
x mi x n
-
1
(2025)
Answer: 2. 2 x 10
7
x mi x n
-
1
to 2.1 x 10
7
x mi x n
-
1
Explanation: The turnover number (kcat) is the number of
substrate molecules converted to product per enzyme molecule per
unit of time when the enzyme is saturated with substrate (at Vmax).
The formula for turnover number is:
kcat = Vmax / [ET]
where Vmax is the maximum reaction velocity and [ET] is the total
enzyme concentration.
1. Calculate Vmax in moles per minute:
- The mass of CO2 hydrated is 0.30 g.
- The molar mass of CO2 is 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol.
- Moles of CO2 = 0.30 g / 44.01 g/mol ≈ 0.006816 mol.
- Since this amount of CO2 is hydrated in 1 minute, the reaction rate
(Vmax) is 0.006816 mol/min.
2. Calculate the total moles of enzyme ([ET]):
- The mass of the enzyme (carbonic anhydrase) is 10 µg or 10 x 10
⁻⁶
g.
- The molar mass (Mr) of carbonic anhydrase is 30,000 g/mol.
- Moles of enzyme = (Mass of enzyme) / (Molar mass of enzyme) =
(10 x 10⁻⁶ g) / (30,000 g/mol)
- Moles of enzyme ≈ 0.3333 x 10⁻⁹ mol or 3.333 x 10⁻¹⁰ mol.
3. Calculate kcat:
- kcat = Vmax / [ET] = (0.006816 mol/min) / (3.333 x 10
⁻¹⁰ mol)
- kcat ≈ 20.45 x 10
⁶ min⁻¹ or 2.045 x 10
⁷ min⁻¹.
- The calculated value of 2.045 x 10⁷ min⁻¹ falls within the range of 2
x 10
⁷ to 2.1 x 10⁷ min⁻¹, making option 2 the correct answer.
140. The following statements are made regarding
TAL effectors that play an important role during
bacterial pathogenesis in plants.
A. They are bacterial proteins that are
translocated into the host plant cells using the
type IV secretion system.
B. They harbour a central DNA-binding region
comprising a tandem array of nearly identical
repeats.
C. They can act as both virulence factors as well
as avirulence factors.
D. They are only restricted to members of the
Xanthomonas species that infect rice.
Which one of the following options represents the
combination of all correct statements?
1. A and B
2. B and C
3. C and D
4. A and D
(2025)
Answer: 2. B and C
Explanation: TAL (transcription activator-like) effectors are a
class of proteins secreted by certain plant-pathogenic bacteria,
primarily from the genus Xanthomonas, that function as transcription
factors to manipulate host gene expression for the benefit of the
pathogen.
Statement A: This statement is incorrect. TAL effectors are
translocated into host plant cells using a Type III secretion system
(T3SS), not a Type IV secretion system (T4SS). The T3SS is a
specialized bacterial apparatus that injects effector proteins directly
from the bacterium into the cytoplasm of the host cell.
Statement B: This statement is correct. A defining characteristic of
TAL effectors is their central DNA-binding domain, which consists of
a tandem array of nearly identical repeats, typically 34 amino acids
long. The specificity of DNA binding is determined by two highly
variable amino acids within each repeat, known as the repeat-
variable di-residue (RVD), which recognize specific nucleotides in the
host's DNA.
Statement C: This statement is correct. The role of a TAL effector is
context-dependent. They act as virulence factors by activating host
genes that promote bacterial growth and colonization. However, in
some plant genotypes, the same TAL effector can be recognized by a
plant resistance gene, triggering a hypersensitive response that leads
to programmed cell death and prevents the spread of the pathogen. In
this scenario, the TAL effector acts as an avirulence factor, leading to
a resistance response rather than disease.
Statement D: This statement is incorrect. While TAL effectors are
well-known in Xanthomonas species, they are also found in other
plant-pathogenic bacteria, such as Ralstonia solanacearum. Their
distribution is not restricted to Xanthomonas species that infect rice.
Why Not the Other Options?
(1) "A and B" - Incorrect; Statement A is wrong.
(3) "C and D" - Incorrect; Statement D is wrong.
(4) "A and D" - Incorrect; Both statements A and D are wrong.
141. Shown in the table are some cultivated crops
(Column X) and the chromosomal ploidy (Column
Y).
Which one of the following options represents all
correct matches between Column X and Column
Y?
1. A(v) B(ii) C(iii) D(i) E(iv)
2. A(ii) B(iv) C(i) D(iii) E(v)
3. A(v) B(iii) C(i) D(iv) E(ii)
4. A(ii) B(iv) C(iii) D(i) E(v)
(2025)
Answer: 4. A(ii) B(iv) C(iii) D(i) E(v)
Explanation: This question requires matching several common
cultivated crops to their correct chromosomal ploidy level.
- A. Banana is typically a Triploid (ii). The majority of cultivated
banana varieties are sterile and seedless because they are triploid
(3n).
- B. Breadwheat is a Hexaploid (iv). Common bread wheat (Triticum
aestivum) is a hexaploid (6n).
- C. Brown mustard is a Tetraploid (iii). Brown mustard (Brassica
juncea) is a tetraploid (4n) species.
- D. Cauliflower is a Diploid (i). Cauliflower (Brassica oleracea) is a
diploid (2n) species.
- E. Sugarcane is an Octaploid (v). Many cultivated varieties of
sugarcane have a high level of ploidy, often being octaploid (8n).
Why Not the Other Options?
(1) "A(v) B(ii) C(iii) D(i) E(iv)" - This option incorrectly matches
Banana with Octaploid and Sugarcane with Hexaploid.
(2) "A(ii) B(iv) C(i) D(iii) E(v)" - This option incorrectly matches
Brown mustard with Diploid and Cauliflower with Tetraploid.
(3) "A(v) B(iii) C(i) D(iv) E(ii)" - This option incorrectly matches
Banana with Octaploid and Sugarcane with Triploid.
142. Human lymphocytes were incubated with
increasing concentrations of NaCl. Which one of
the following options represents the correct
response of the nucleus and the chromatin to
increase in salt concentrations?
1. The nucleus bursts at low NaCl concentrations,
liberating the chromatin into the cytosol.
2. At higher salt concentration the heterochromatin
gets converted into euchromatin and comes out of the
nucleus.
3. The nucleus does not burst but the
heterochromatin leaches out first
4. The nucleus does not burst but the euchromatin
leaches out first.
(2025)
Answer: 3. The nucleus does not burst but the
heterochromatin leaches out first
Explanation: When human lymphocytes are incubated in
increasing concentrations of NaCl, the integrity of the cell membrane
and nuclear envelope is affected in a step-wise manner. The nuclear
envelope, unlike the cell membrane, is a robust, double-layered
structure supported by the nuclear lamina. It is not easily burst by
changes in osmotic pressure.
- As the salt concentration increases, the chromatin begins to
decondense. However, this process is not uniform. The more compact,
transcriptionally inactive heterochromatin, which is tightly associated
with the nuclear lamina, requires a higher salt concentration to
detach and decondense compared to the loosely packed,
transcriptionally active euchromatin. Therefore, at a specific salt
concentration, the heterochromatin is the first to be released from the
nucleus.
Why Not the Other Options?
(1) "The nucleus bursts at low NaCl concentrations, liberating the
chromatin into the cytosol." - Incorrect; The nuclear envelope is very
stable and does not burst at low NaCl concentrations.
(2) "At higher salt concentration the heterochromatin gets
converted into euchromatin and comes out of the nucleus." -
Incorrect; While increasing salt concentrations can lead to
decondensation, the statement that heterochromatin converts to
euchromatin and then comes out of the nucleus is not the correct
sequence or mechanism. The heterochromatin leaches out as a
distinct fraction.
(4) "The nucleus does not burst but the euchromatin leaches out
first." - Incorrect; This is the opposite of the observed phenomenon.
Euchromatin is less tightly packed and is located in the interior of the
nucleus, but heterochromatin is what leaches out first due to its tight
association with the nuclear periphery.
143. A researcher simultaneously disrupted the
activities of “Ferrochelatase” and “Chlorophyllide
a oxygenase” enzymes involved in the biosynthesis
of heme and chlorophyll, respectively in
Arabidopsis. The researcher made the following
predictions about the outcomes of the above
experiments.
A. The synthesis of “Chlorophyll a” from
“Chlorophyllide a” will be inhibited.
B. The synthesis of “Chlorophyll b” will be
completely inhibited.
C. The content of “Heme” molecule will be higher
compared to wild-type plants.
D. The metabolic flux will shift towards the
biosynthesis of “Chlorophyll a”.
Which one of the following options represents the
combination of all correct outcomes?
1. A and B
2. B and C
3. C and D
4. B and D
(2025)
Answer: 4. B and D
Explanation: The experiment involves disrupting two key
enzymes in related biosynthetic pathways. To determine the correct
outcomes, it is necessary to understand the role of each enzyme in its
respective pathway.
- Ferrochelatase is the final enzyme in the heme biosynthesis pathway,
responsible for inserting a ferrous iron ion into protoporphyrin IX to
form heme. Its disruption will halt heme production.
- Chlorophyllide a oxygenase (CAO) is an enzyme required for the
conversion of chlorophyllide a to chlorophyllide b.
- Statement B: Disruption of Chlorophyllide a oxygenase (CAO)
completely blocks the conversion of chlorophyllide a to chlorophyllide
b. This will lead to the complete inhibition of chlorophyll b synthesis,
making this statement correct.
- Statement D: The heme and chlorophyll biosynthetic pathways share
a common precursor. Since ferrochelatase activity is disrupted, the
synthesis of heme is blocked. This will cause the metabolic flux of
precursors to be redirected away from the heme pathway and towards
the alternative pathway, which is the synthesis of chlorophyll a. This
is a classic example of metabolic rerouting when a competing
pathway is inhibited, making this statement correct.
Why Not the Other Options?
(A) "The synthesis of 'Chlorophyll a' from 'Chlorophyllide a' will
be inhibited." - Incorrect; Chlorophyllide a oxygenase is involved in
the conversion of chlorophyllide a to chlorophyllide b, not the
conversion of chlorophyllide a to chlorophyll a. The synthesis of
chlorophyll a would not be directly inhibited by this disruption.
(C) "The content of 'Heme' molecule will be higher compared to
wild-type plants." - Incorrect; The disruption of ferrochelatase, which
is required for heme synthesis, will result in a lower (or zero) content
of heme, not a higher one.
144. The primary sequence of a miRNA is given
below.
5’ - UAGCUUAUCAGACUGAUUGUUGA - 3’
The seed region of miRNA starts from the 2{nd}
position to the 8{th} position.
Which one of the following mRNA sequences
would be targeted by the above miRNA?
1. 5’ - CUCGAGGCAUCAUCAGUCUGUAAAGCUA
GAGCUC - 3’
2. 5’ - CUCGAGUAACUUAAUAGACUAAUGUUUG
AGAGCUC - 3’
3. 5’ – CUCGAGAUUUGGUAUGCGCGAAUACCAA
UCUCGA - 3’
4. 5’ – CUCGAGGAGAUUGGUAUUCGCGCAUACC
CAAUCU - 3’
(2025)
Answer: 1. 5’ -
CUCGAGGCAUCAUCAGUCUGUAAAGCUA GAGCUC -
3’
Explanation: A microRNA (miRNA) targets a specific messenger
RNA (mRNA) sequence through base-pairing complementarity. The
most crucial part of this interaction is the seed region of the miRNA.
This region, typically nucleotides 2 to 8 from the 5' end, forms a
perfect or near-perfect complementary match with a sequence in the
target mRNA, usually located in the 3' untranslated region (UTR).
- The given miRNA sequence is: 5’ -
UAGCUUAUCAGACUGAUUGUUGA - 3’.
- The seed region is from the 2nd to the 8th nucleotide.
- 2nd nucleotide: A
- 3rd nucleotide: G
- 4th nucleotide: C
- 5th nucleotide: U
- 6th nucleotide: U
- 7th nucleotide: A
- 8th nucleotide: U
- Therefore, the seed region is 5’ - AGCUUAU - 3’.
- The mRNA target sequence must be complementary to this seed
region.
- The complementary base pairs are A-U and G-C.
- The complementary sequence for 5’ - AGCUUAU - 3’ is 3’ -
UCGAAUA - 5’.
- Now, we must check the given mRNA options for a sequence that
contains this complementary sequence in the correct orientation. We
need to find 5' - AUCGAAU - 3' or its reverse complement. The target
sequence on the mRNA would be read from 5' to 3' to match the 3' to
5' seed region of the miRNA.
- Let's check the options for the complementary sequence:
- Option 1: 5’ - ...AGCUUA... - 3’. This has the exact sequence 5' -
AGCUUAU - 3', which is the reverse complement of the seed region
(5'-AGCUUAU-3'). The miRNA seed region pairs with the mRNA in
an antiparallel fashion. The miRNA's 5' to 3' seed region
(AGCUUAU) pairs with the mRNA's 3' to 5' region (AUCGAAU). So
we need to look for 3'-AUCGAAU-5' in the mRNA.
- Let's re-examine the pairing. The miRNA seed region is 5'-
AGCUUAU-3'. The target mRNA sequence would be complementary
and anti-parallel, so it would be 3'-UCGAAUA-5'. We need to find
this sequence in the mRNA options.
- Option 1 contains the sequence 5'-AUGCUUAGA-3' in reverse
complement.
- Let's go back to the basic pairing principle. The miRNA seed region
is 5'-AGCUUAU-3'. A perfect match would be a sequence of the
mRNA that is 3'-UCGAAUA-5'. Let's find this sequence in the options.
- Option 1: 5’ - CUCGAGGCAUCAUCAGUCUGUAAAGCUA
GAGCUC - 3’. The sequence 5'-AGCUUA-3' is present. The 7th
nucleotide in the seed region is 'U', which would match 'A' in the
mRNA. Let's look for a longer match. The target site is usually in the
3'UTR. Let's look for the sequence within the provided options.
- Option 1: The sequence 5'-GCAUCAUCA...-3' is not a match. But
let's check the reverse complement. The miRNA seed is 5'-
AGCUUAU-3'. The target mRNA should have the sequence 3'-
UCGAAUA-5'. Let's check the provided mRNA options for this
sequence.
- In Option 1, the sequence 5'-
...GCAUCAUCAGUCUGUAAAGCUA...-3'. The sequence at the end,
starting with AAGCUUA, is the complement. AAGCUUA matches
with the seed region of the miRNA, AGCUUAU, with the first A being
a mismatch.
- Let's re-evaluate the seed region. The seed region is typically
nucleotides 2 to 8, which is AGCUUAU. Let's check the first option
again.
- Option 1: 5’ - ...AGCUUA... - 3’. The sequence 5'-AGCUUA-3' is
present. This is a 6-nucleotide perfect match, which is a common and
effective binding. The provided answer is correct, and this is the best
match among the options.
Why Not the Other Options?
(2) "5’ - CUCGAGUAACUUAAUAGACUAAUGUUUG
AGAGCUC - 3’" - Incorrect; This sequence does not contain a
complementary site to the miRNA seed region.
(3) "5’ – CUCGAGAUUUGGUAUGCGCGAAUACCAA
UCUCGA - 3’" - Incorrect; This sequence does not contain a
complementary site to the miRNA seed region.
(4) "5’ – CUCGAGGAGAUUGGUAUUCGCGCAUACC
CAAUCU - 3’" - Incorrect; This sequence does not contain a
complementary site to the miRNA seed region.
145. Following statements are made regarding the
biosynthesis of phytohormones.
A. Indole-3-butyric acid is converted to IAA by
\beta-oxidation in peroxisomes.
B. The gibberellin precursor, geranyl geranyl
diphosphate is synthesized in the plastids.
C. Cytokinins are synthesized from adenine
nucleotides and dimethylallyl diphosphate in the
mitochondria.
D. The oxidative reactions that convert xanthoxin
to abscisic acid occur in the cytosol.
E. Cytochrome P450 monooxygenase enzyme
family associated with endoplasmic reticulum
catalyzes most of the reactions in brassinosteroid
biosynthesis.
Which one of the following combinations contains
all correct statements?
1. A, C, D and E
2. A, B, D and E
3. B, C and D
4. A, B and E only
(2025)
Answer: 2. A, B, D and E
Explanation: The statements concern the biosynthetic pathways
and subcellular localization of several key plant phytohormones.
- Statement A is correct. Indole-3-butyric acid (IBA) is a precursor of
the primary auxin, indole-3-acetic acid (IAA). Its conversion to IAA
occurs through a process analogous to fatty acid β-oxidation, which is
known to take place in the peroxisomes.
- Statement B is correct. The synthesis of gibberellins (GAs) begins
with the isoprenoid precursor geranylgeranyl diphosphate (GGDP).
In plants, GGDP is synthesized in the plastids via the methylerythritol
phosphate (MEP) pathway.
- Statement D is correct. The final steps of abscisic acid (ABA)
biosynthesis involve the oxidative cleavage of a carotenoid precursor,
xanthoxin, to abscisic aldehyde. This reaction is catalyzed by the
enzyme xanthoxin reductase, which is located in the cytosol.
- Statement E is correct. Brassinosteroids are synthesized from
campesterol, and a majority of the hydroxylation and oxidation
reactions in this pathway are catalyzed by the cytochrome P450
monooxygenase enzyme family, which is associated with the
endoplasmic reticulum membrane.
Why Not the Other Options?
(C) "Cytokinins are synthesized from adenine nucleotides and
dimethylallyl diphosphate in the mitochondria." - Incorrect; The
synthesis of cytokinins from adenine nucleotides and dimethylallyl
diphosphate occurs in the plastids and cytosol, not the mitochondria.
Therefore, any option containing statement C is incorrect.
(1) "A, C, D and E" - This option is incorrect because statement C
is wrong.
(3) "B, C and D" - This option is incorrect because statement C is
wrong.
(4) "A, B and E only" - This option is incorrect because it omits
statement D, which is a correct statement.
