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

GATE Life Science Mock Test- 3 (Botany & Zoology) for GATE Life Sciences 2025 is part of GATE Life Sciences 2026 Mock Test Series preparation. The GATE Life Science Mock Test- 3 (Botany & Zoology) questions and answers have been prepared according to the GATE Life Sciences exam syllabus.The GATE Life Science Mock Test- 3 (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- 3 (Botany & Zoology) below.
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GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 1

Disagree : Protest : : Agree : _______
(By meaning of words)

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

Ankita needs to ascend 5 stairs starting from the ground level, adhering to these rules:

  1. At any moment, Ankita can advance either one or two stairs at a time.
  2. At no point can Ankita step down to a lower stair.

Let F(N) represent the total number of distinct ways Ankita can reach the Nth stair. For instance, F(1) = 1, F(2) = 2, and F(3) = 3.

The value of F(5) is _______.

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

Which of the figures labeled P, Q, R, and S depicts the graph of the following function?
f(x) = ||x 2| - |x - 1|

*Multiple options can be correct
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 4

Which reagent(s) can effectively facilitate the transformation of P to Q with a good yield?

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 5

In a reaction involving a single substrate, a competitive inhibitor was introduced to investigate its influence on Km and Vmax. Which of the following equations would be applicable?

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 5

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 6

Determine the product formed from the following Diels-Alder reaction.

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 6

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 7

Consider two distinct processes in which the volume of an ideal gas increases to double its original value isothermally:
(i) Reversible expansion (work done = wrev)
(ii) Irreversible expansion, where the external pressure matches the final pressure of the gas (work done = wirrev)
Here,  ____.

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 7

For a reversible isothermal expansion of an ideal gas, the work done is:

where:

  • V= 2Vi (final volume is double the initial volume).

Substituting V= 2Vi​:

For an irreversible isothermal expansion, where the external pressure (Pext) is constant and matches the final pressure of the gas, the work done is:

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 8

Pair the coordination complexes listed in Column I with their corresponding properties from Column II.
(Given: Atomic numbers of Mn: 25; Co: 27; Ni: 28)

*Multiple options can be correct
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 9

Compounds P and Q undergo E2 elimination with rate constants k1 and k2 respectively, as illustrated below. Which of the following option(s) is/are CORRECT?

*Multiple options can be correct
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 10

According to the Hard-Soft Acid-Base (HSAB) principle, which option(s) accurately reflect the trend of solubility in water?

*Answer can only contain numeric values
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 11

The reduction potentials of are 1.33 V and -0.74 V, respectively. What is the reduction potential of ?


Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 11


(ii) Cr+3 + 3e–1 → Cr (ε° = – 0.74 V) ………..(2)
Calculate ΔG° = -nFε° 
 = – 6 × 1.33 F = – 7.98 F
 = – 6 × (– 0.74) F - + 4.44 F

To calculate the reduction potential of  add equations (1) and (2).
 Cr2O72– + 14 H+ + 12e–1 → 2Cr + 7H2O

*Answer can only contain numeric values
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 12

The pH of a buffer solution containing 0.20 mole of NH4OH and 0.25 mole of NH4Cℓ per litre where the dissociation of NH4OH at room temperature is 1.81 × 10–5 will be


Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 12


As we know,
Pkb = – log kb = – log (1.81 × 10–5) = 0.7423

pH = POH = 14
pH = 14 – POH
pH = 9.16

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 13

Which of the following does not qualify as an edaphic factor?

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 14

Rotenone is a chemical frequently utilized to eliminate insect pests on agricultural crops and fish in bodies of water. It functions by blocking electron transport from the NADH dehydrogenase enzyme in Complex I to ubiquinone within the mitochondrial electron transport chain. Which of the following reasons accounts for the ability of plants to withstand the application of rotenone?

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 15

Which of the following possesses oil and sometimes starch as stored nutrients?

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 16

Associate the disease name with its corresponding causal organism.

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 17

Determine the floral formula along with the corresponding family and the specific plant species.

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 17


GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 18

The periderm is a protective tissue located in the stems and roots of gymnosperms and woody dicotyledons. It plays a role in increasing thickness through secondary growth. Match the components of periderm listed in Group I with the corresponding cell/tissue types provided in Group II.

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 19

Pair the rice diseases listed in Group I with their corresponding causal agents found in Group II.

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 20

Which of the following groups is/are spread through subterranean roots?

  • 1. Bryophyllum and Kalanchoe
  • 2. Ginger, Potato, Onion and Zamikand
  • 3. Sweet Potato and Dahlia
  • 4. Asparagus and Tapioca

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 21

In eukaryotic cells, there are three types of RNA polymerases (I, II, and III), each responsible for synthesizing a specific class of RNA molecule. Which of the following pairs is correctly matched?

*Answer can only contain numeric values
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 22

A female plant exhibiting cytoplasmic male sterility and possessing the nuclear genotype rr is mated with a male-fertile male plant having the genotype RR. Both genotypes RR and Rr are capable of restoring fertility, while the genotype rr is not. When an F1 female plant with the genotype Rr is test-crossed with a male-fertile male having the rr genotype, what percentage of the resulting population will be male fertile? (Provide your answer as an integer)


GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 23

Which hormone listed below is classified as a modified amino acid?

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 23

Epinephrine is synthesised from amino acid tyrosine while estrogen and progesterone are modified steroids and prostaglandins are basically fat. Progesterone and estrogen are secreted from corpus luteum of ovary.
Prostaglandins are derived from many tissues including the prostate gland (or can be made synthetically). Prostaglandins have been used in the induction of labour and abortion.

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 24

What do the fossil remains of Archaeopteryx reveal?

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 24

Archaeopteryx lived in late Jurassic period around 150-145 million years ago.

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 25

Which of the following options best characterizes an animal's “innate behavior”?

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 26

The afferent nerve fiber transmits impulses from

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 26

The efferent fiber is a long process projecting far from the neuron's body that carries nerve impulses away from the central nervous system toward the peripheral effector organs (mainly muscles and glands). A bundle of these fibers is called a motor nerve or an efferent nerve.

*Answer can only contain numeric values
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 27

In a population consisting of 1000 wild dogs inhabiting a grassland, there were 360 dogs with the black body color having the genotype BB, and 480 dogs with the genotype Bb. The remaining dogs in this population exhibited a white color with the genotype bb. Using this information, calculate the frequency of allele “b” in the population and round your answer to one decimal place.


GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 28

Match the items listed below:

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 28
  1. Down's syndrome - d. Presence of an extra chromosome (Down's syndrome is caused by trisomy 21, where there is an extra chromosome 21.)
  2. Cri-du-chat syndrome - b. Loss of part of chromosome 5 (This syndrome is caused by a deletion in part of chromosome 5.)
  3. Klinefelter's syndrome - a. An additional sex chromosome (Klinefelter's syndrome occurs when males have an extra X chromosome, resulting in XXY.)
  4. Turner's syndrome - c. Absence of a sex chromosome (Turner's syndrome occurs when females have only one X chromosome, so they are XO.)

The correct answer is:
a) 1 - d, 2 - b, 3 - a, 4 - c

GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 29

The injection of dinitrophenol (DNP) into a rat leads to a rapid increase in its body temperature due to which of the following reasons?

Detailed Solution for GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 29

DNP encouples electron transport from oxidative phosphorylation

*Multiple options can be correct
GATE Life Science Mock Test- 3 (Botany & Zoology) - Question 30

It has been established that an abundance of glucose can inhibit the activation of the lac operon, along with other operons that regulate enzymes related to carbohydrate metabolism in E. coli. Which of the following processes describes this phenomenon?

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