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GATE Life Science Mock Test- 2 (Botany & Zoology) - GATE Life Sciences MCQ


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30 Questions MCQ Test GATE Life Sciences 2026 Mock Test Series - GATE Life Science Mock Test- 2 (Botany & Zoology)

GATE Life Science Mock Test- 2 (Botany & Zoology) for GATE Life Sciences 2025 is part of GATE Life Sciences 2026 Mock Test Series preparation. The GATE Life Science Mock Test- 2 (Botany & Zoology) questions and answers have been prepared according to the GATE Life Sciences exam syllabus.The GATE Life Science Mock Test- 2 (Botany & Zoology) MCQs are made for GATE Life Sciences 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Life Science Mock Test- 2 (Botany & Zoology) below.
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GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 1

The transparent square sheet depicted above is folded along the dashed line. What will be the appearance of the folded sheet?  

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 2

A person begins at a certain location and walks 1 km to the east, then 2 km to the north, followed by another 1 km to the east, 1 km to the north, 1 km to the east, and finally 1 km to the north before reaching his destination. What is the minimum distance (in km) from the starting point to this destination?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 2

Let the man start from point O and reach his destination at point A.
The shortest distance between O and his destination is

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 3

In a series of 10 consecutive integers arranged in increasing order, the total of the first 5 integers is 660. What is the total of the last 5 integers in this series?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 3

Let the first 5 consecutive integers be represented by x, x + 1, x + 2, x + 3 and x + 4. Then, since the sum of the integers is 660, x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 660.
Thus, 5x + 10 = 660
5x = 650
x = 130
The first integer in the sequence is 130, so the next integers are 131, 132 , 133 and 134. From this, the last 5 integers in the sequence, and thus their sum, can be determined. The sum of the 6th, 7th, 8th, 9th and 10th integers is, 135 + 136 + 137 + 138 + 139 = 685.
This problem can also be solved without algebra: The sum of the last 5 integers exceeds the sum of the first 5 integers by 1 + 3 + 5 + 7 + 9 = 25.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 4

Which of the following pairs of 'common name : chemical formula' is incorrect?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 4

Green vitriol: FeSO4.7H2O
Blue Vitriol : CuSO4.5H2O
Hence the answer is option (4)

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 5

The reaction rate between A and B is reduced to one-fourth when the concentration of B is increased twofold. What is the order of the reaction with respect to B?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 5

According to rate law equation:
(Rate)1 = K[A]m[B]n ...... (i)
According to the given conditions:
If [B] = 2[B], (Rate)2 = (rate)1/4 .....(ii)
Divide (ii) by (i).
n = -2
Therefore, the order of the reaction w.r.t. to B is -2.

*Answer can only contain numeric values
GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 6

A solution containing a compound exhibits an absorbance of 0.42 at a wavelength of 275 nm in a cuvette with a light path of 0.1 dm. The molar absorptivity of this compound is ε275 = 8.4 × 103 M–1 cm–1. What is the concentration of the compound in ______× 10–5 M (rounded to the nearest integer)?


Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 6

Beer-Lambert's Law:
A = ε × c × l

Where:

  • A is the absorbance (0.42)
  • ε is the molar absorptivity (8.4 × 10^3 M–1 cm–1)
  • c is the concentration (unknown)
  • l is the path length (0.1 dm = 1 cm, since 1 dm = 10 cm)

Now, we rearrange the equation to solve for the concentration (c):

c = A / (ε × l)

Substitute the known values:

c = 0.42 / (8.4 × 10^3 × 1)

Let's calculate this.

The concentration of the compound is 5 × 10^–5 M.

thus answer is 5

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 7

According to Wade's rule, the structures of [B10C2H12] and [B9C2H11]2 are respectively represented as

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 7

According to Wade's rule, carboranes are obtained by replacing BH unit by CH+ unit.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 8

If 4g of NaOH is dissolved in 100 ml, 500 ml, 1000 cm3, and 1900 cm3 of water respectively, four stock solutions labeled A, B, C, and D are created. Which of these solutions will exhibit a pH of 12?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 8

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 9

What is the bond order of the O–O bond in the O22– ion?

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 10

Four grams of NaOH are dissolved in volumes of 100 ml, 500 ml, 1000 cm3, and 1900 cm3 respectively to create four stock solutions designated as A, B, C, and D. If 99 ml of a 1 M HCl solution is added, which stock solution will have a pH closest to 7?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 10

4 g of NaOH in 100 ml = solution in 1 M NaOH
99 ml of 1 M HCℓ ≈100 ml of 1 M HCℓ
∴ Solution A will be closer to pH 7.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 11

Pair the items in list - A of reversible chemical reactions with those in list - B of corresponding conditions, and choose the correct answer from the options provided below:

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 11

Reaction (a) proceeds suitably at high temperature and low pressure, reaction (b) proceeds preferably at low temperature and high pressure. Reaction (c) proceeds suitably at low temperature and low pressure.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 12

The effectiveness of market-based breeding systems in agriculture depends on the genetic map containing a sufficient number of evenly spaced polymorphic markers to accurately identify the desired QTLs or major gene(s). Which of the following indicates such effectiveness?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 12

In usual, market-based breeding system's success in crops relies on the genetic map having sufficient number of equally-spaced polymorphic markers for properly locating desired QTLs or the major gene(s)- A capability for analyzing huger number of plants in the time-and-cost-effective manner, close relation between QTL or major gene of concern & adjacent markers & sufficient recombination between markers and rest of genome.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 13

Phytoalexins are crucial for plant defense mechanisms against pathogens. Identify the INCORRECT statement regarding phytoalexins.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 14

The diagram illustrates the germination rates of seeds that have absorbed water and were subjected to a specific sequence of red (R) and far-red (F) light, with each exposure lasting 5 minutes. The germination percentage was recorded after 72 hours of darkness at a temperature of 25°C. Based on this information, which ONE of the following statements is CORRECT?

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 15

Wheat plants that are subjected to extended cold temperatures during the seedling phase bloom sooner than those in the untreated control group. However, the seeds harvested from these cold-treated plants produce offspring that do not exhibit early flowering. This occurrence is referred to as

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 16

While all pigment molecules present in a photosystem have the ability to absorb photons, only a select few chlorophyll molecules that are linked to the photochemical reaction center are specialized for converting light into chemical energy. The photochemical process involved in photosynthesis, which necessitates both PS-I and PS-II, also encompasses

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 16

Photochemical process in the photosynthesis, which requires both PS-I & PS-II also includes assimilatory power synthesis, water photolysis and Z-scheme of the electron transfer. Energy from Sun which arrive earth surface is preserved as energy-rich chemicals by green vegetations like plants, algae and few bacteria. Maximum life on earth inherits their nutrients from photochemical process photosynthesis where 3 X 1011 metric tons of carbon is converted into carbohydrate a year.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 17

Identify the CORRECT correspondence among the plant species listed in GROUP I, the main phytochemical in GROUP II, and the economic or medical application in GROUP III. 

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 18

While exploring a forest, you come across an unusual fungus. You excise a reproductive structure and bring it home for further examination. Upon observing it under a microscope, you identify it as a zygosporangium. Based solely on this finding, this fungus is classified as a/an ________________.

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 18

This fungus is a zygomycete, which is a filamentous fungus of phylum Zygomycota that sexually reproduce by means of zygospores having bread mould. Their names derive from zygosporangia where antagonistic spherical spores are produced at the time of sexual reproduction.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 19

The term used to describe the compatibility phenomenon occurring in a flower with heterostyly is known as

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 19

Incompatibility involves morphological or physiological. Morphological incompabitily occurs in flowers having heterostyly. Flowers are dimorphic or trimorphic with regard to the length of style. Thus, facilitate cross pollination. e,g. Jasminum lythrum.

*Multiple options can be correct
GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 20

In order to investigate the mechanism behind systemic acquired resistance (SAR), a research team has isolated a mutant exhibiting a diminished SAR response. The sequencing of this mutant indicated homozygous mutations in two specific genes, X and Y. Which of the following experiment(s) would be appropriate to determine if the mutant phenotype results from mutations in either one or both of these genes?

*Multiple options can be correct
GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 21

The results from an experiment examining seed germination across different genotypes and light conditions are presented, where √ represents germination and X signifies its absence, respectively.

Based on these results, which of the following option(s) is/are CORRECT?

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 22

Assume that the anticodon for an unknown amino acid is 3ʹ AUG 5ʹ. The corresponding code on DNA sequence would be

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 23

The genes that are homologous and found in different species are referred to as

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 24

Which of the following statements is not correct?

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 24

Linking number of light handed dsDNA is regarded as negative. Linking number is the number of times every curve winds around each other. It is always an integer but can be positive or negative on the basis of the orientation of 2 curves.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 25

The phrase innate behavior refers to a type of behavior exhibited by animals that is

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 26

Associate the cell organelles listed in Column I with their corresponding functions provided in Column II.

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 27

The interaction of an antigen with the receptor on B-cells

Detailed Solution for GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 27

Each B-cell is specific for a particular antigen. The specificity of binding resides in the BCR (B-cell receptor) for antigen. They are integral membrane proteins. They are present in thousands of identical copies exposed at the cell surface. They are made before the cell ever encounters an antigen. B-cell receptor complex usually consists of an antigen-binding subunit (the membrane immunoglobulin or MIg), which is composed of two IgHs (Immunoglobulin Heavy Chains) and two IgLs (Immunoglobulin Light Chains), and a signalling subunit, which is a disulfide-linked heterodimer of Ig-Alpha (CD79A) and Ig-Beta (CD79B) proteins. Thus, antigen binding to the B-cell receptor transduces a signal through the signalling subunit (Igá and Igâ).

GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 28

Pair the autoimmune disorders in Column I with their corresponding self-antigens listed in Column II. 

*Multiple options can be correct
GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 29

Which of the following amino acids possess more than one chiral center?

*Answer can only contain numeric values
GATE Life Science Mock Test- 2 (Botany & Zoology) - Question 30

A cross is performed between two organisms with genotypes AaBb x AaBb, where the A and B loci assort independently. The offspring from this dihybrid cross were subsequently allowed to self-fertilize. The percentage of the offspring that exhibited segregation at locus A (i.e., producing A- and aa offspring) will be _________ (in integer).


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