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GATE Life Science Mock Test- 1 (Botany & Zoology) - GATE Life Sciences MCQ


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

According to research, listening to music while exercising enhances performance and alleviates discomfort. Scientists investigated if listening to music during study sessions could improve students' learning outcomes, but the findings were inconclusive. It was observed that students who required external stimulation for studying performed worse, whereas those who did not require any external stimulation experienced benefits from music.

Which of the following statements is the CORRECT inference from the passage above?

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 2

The expression for the radial wave function of the hydrogen atom corresponding to the 3s orbital is given by __________.

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 2

The radial wave function of hydrogen atom for 3s orbital is given by the relationship

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 3

The proper sequence of Cr2O3, CrO3, MgO, and MnO, arranged by their increasing base strength, is

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 3

The oxide of alkali and alkaline metals are more basic than transition metal oxides because basic character of the oxides decreases across the period. Among transition of metal oxides, the acidic character increases with increase in oxidation state of the transition metal. The order of increasing basic character of the oxides is

CrO3 < Cr2O3 < MnO < MgO
+6 +3 +2 +2

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 4

The major product Q resulting from the following reaction has the structure shown below:

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 5

The energy of an electron in its ground state within a hydrogen atom is –13.60 eV. What is the energy of the electron in the third excited state, expressed in eV (rounded to two decimal places)?


GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 6

A particle is restricted to a one-dimensional box measuring 1 mm in length. If the length is altered by 10–9 m, what is the percentage change in the ground state energy?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 6


Percent change in energy 

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 7

The proper order of acidity for the compounds depicted below is 

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 8

Consider a reaction represented as X → Products.
Group I presents three graphs depicting the concentrations of the reactant over time, where x signifies the concentration of reactant X at time t, and x0 denotes the initial concentration of reactant X at time t = 0. Group II lists various reaction orders. Match each graph with the corresponding order of reaction.

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 9

Given the following standard heats of formation, ∆fH(P, g) = 314.6 kJ mol−1, ∆fH(PH3, g) = 5.4 kJ mol−1, and ∆fH(H, g) = 218.0 kJ mol−1, the average bond enthalpy of a P–H bond in PH3(g) is __________ kJ mol−1 (rounded off to one decimal place).


*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 10

Considering the standard reduction potentials, and , determine the potential of the following electrochemical cell: Ag (aq., 1 mM) Mg(s) ↔ Ag(s) Mg2 (aq., 0.2 M) at 25°C, which is _______ V (rounded to two decimal places).
(Given: Faraday constant = 96500 C mol−1, Gas constant R = 8.314 JK−1mol−1)


*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 11

The freezing point of 80 g of acetic acid (freezing point constant 3.9 K kg mol−1) was depressed by 7.8 K upon adding 20 g of a certain compound. The molar mass of this compound is _____________ g mol−1 (rounded to the nearest integer).


GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 12

In Mendel's experiments, the dwarf pea mutant exhibited a deficiency in which of the following enzymes that play a role in the synthesis of gibberellins?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 12

The dwarf pea mutant as used by Mendel in his experiment had a defect in BA3β-hydroxylase involved in the biosynthesis of gibberellins.

*Multiple options can be correct
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 13

Which of the following enzyme(s), when expressed in excess, would lead to an increase in the β-carotene content of rice grains?

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 14

The dihybrid phenotypic ratio of 9 : 3 : 4 is associated with which type of genetic interactions?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 14

Here the recessive allele in homogenous condition marks the effect of dominant allele. For e.g. In mice, the wild body colour is known as agouti (greyish) and is controlled by gene say A which is hypostatic to recessive allele C. The dominant allele C in the presence of 'a' gives coloured mice. In the presence of dominant allele C, A gives rise to agouti. So, ccaa will be coloured and ccAA will be albino. When coloured mice (CCaa) are crossed with albino (ccAA), agouti mice (CcAa) appear in F1. Small cc mask the effect of AA and is therefore epistatic. Consequently, ccAA is albino. The ratio 9 : 3 : 3 : 1 is modified to 9 : 3 : 4. The combination ccAA is also albino due to the absence of both the dominant alleles.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 15

Pair the altered organs in GROUP I with their respective prototypical forms in GROUP II and select the CORRECT option.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 16

Associate each genetically modified crop from GROUP I with its relevant genetic element in GROUP II.

*Multiple options can be correct
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 17

Which of the following option(s) is/are CORRECT regarding the generation of hybrid plants utilizing male sterile lines based on Barnase/Barstar technology?

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 18

In a diploid species of plants, the T allele results in tall individuals and is entirely dominant over the t allele, which leads to short individuals. Likewise, the W allele causes the formation of round seeds and is fully dominant over the w allele that produces wrinkled seeds (assuming the T and W loci are not linked). If a plant with a TTWW genotype is crossed with another plant having a ttww genotype, what fraction of the F2 generation generated from the combination of both recombinant gametes would be ___________? (Round your answer to two decimal places.)


GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 19

During which period did Archaeopteryx face extinction?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 19

Geological time scale shows the ages of various eras and periods together with the major groups of plants and animals that are believed to have existed during that period. It has been divided into 6 major era. An era may be further divided into periods and periods into epoch. Mesozoic era is divided into 3 period namely cretaceous, jurassic and triassic. First bird archaeopteryx appeared in jurassic period. Archaeopteryx is a link between reptile and bird, having reptilian long tail, teeth and avian feathers and beak. In cretaceous period, the first modern bird appeared and the previous archaeopteryx became extinct.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 20

Infectivity in microfilariae occurs within

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 20

Wuchereria bancroftii passes its life cycle in 2 hosts: man (definitive host) and mosquito (intermediate host). The embryonic stage of Wuchereria bancroftii is known as microfilariae. Sheathed microfilariae are ingested by the mosquito during its blood meal. They cast off the sheath quickly, penetrate the gut wall within an hour or two, and migrate to the thoracic muscle. Here they rest and begin to metamorphose. Within 2 weeks, the microfilariae develop into infective larvae inside the mosquito thoracic muscle through 3 larval stages. The infective stage of microfilariae then enters the proboscis sheath of mosquito.

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 21

A man, whose parents have blood groups A and O respectively, marries a woman with blood group AB. If the man possesses blood group A, then the total number of distinct blood groups that their children could inherit will be _________ (in integer).


*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 22

In a polypeptide's structure, a single α-helix (3.613 helix) consists of 32 intrachain hydrogen bonds. The total number of turns present in the helix will be _________ (in integer).


*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 23

In a population of snakes residing on a remote island, the gene under consideration is in Hardy–Weinberg equilibrium and consists of two alleles (A and a). Given that the frequency of allele A is 0.6, what is the frequency of genotype Aa? Please round your answer to two decimal places.


GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 24

The high frequency of the allele linked to sickle-cell anemia in certain human populations is likely attributed to

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 24

The sickle-cell allele causes a change in the structure of blood cells that does not allow the malarial parasite to reproduce. Thus, there is some advantage to heterozygotes where malaria is a danger. Homozygotes for non-sickled blood cells would be more likely to suffer from malaria and thus less likely to reproduce. Homozygotes for sickled blood cells do not have enough oxygen carrying capacity and thus are less likely to survive to reproduce. Heterozygotes are likely to survive better than either alternative. Thus, they are more likely to survive and transfer either allele to their offspring. This will allow the sickle allele to remain at high frequency in the population. In populations where malaria is not a danger, there is no advantage of carrying the sickle allele. Thus, it will gradually decrease in frequency in the population.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 25

The process of ubiquitination of mitotic cyclins is carried out by which of the following?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 25

Mitotic cyclin ubiquitination is performed by APC (Anaphase Promoting Complex), which is activated in late mitosis by the association of Cdc20. The destruction of cyclin B (Mitotic cyclin) is required for exit from mitosis, and is mediated by the ubiquitin pathway. Recently, a 20S complex, termed the anaphase-promoting complex (APC) or the cyclosome, has been genetically and biochemically identified as the cyclin-specific ubiquitin ligase (E3). In addition, a ubiquitin-conjugating enzyme (E2), UBC4, was shown to be involved in cyclin ubiquitination in xenopus egg extracts. Another E2 activity, designated UBCx, can independently support cyclin ubiquitination in xenopus. A similar activity (E2 - C) has also been observed in clams.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 26

CheA, a protein involved in bacterial chemotaxis
P. acts as a histidine kinase
Q. exhibits autophosphorylation
R. transfers phosphoryl groups to conserved aspartate residues in CheY
S. functions as a tyrosine kinase

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 26

Bacterial chemotaxis is the most widely studied two component sensory system. Chemotaxis is the cell's response to a stress environment. It is controlled by more than 40 genes in over 14 operons that produce proteins for the structural components of the flagella, the flagellar motor, transmembrane receptors, and signal transduction. In a typical two component system, a receptor protein transduces environmental signals into metabolic changes though phosyphorylation of a second protein. The signal cascade requires the transfer of a phosphate group from an autophosphorylating protein kinase to a regulator protein. In bacterial chemotaxis, transmembrane chemoreceptors, the CheA histidine kinase, and the CheW coupling protein assembles into signalling complexes that allow bacteria to modulate their swimming behaviour in response to environmental stimuli. The rate of autophosphorylation of the protein CheA is controlled by the transmembrane receptor proteins. CheA then in turn transfers the phosphate group at aspartate residues to another protein, CheY, which interacts directly with the flagellar switch to affect a change in swimming pattern.

GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 27

Which glycolytic enzyme is inhibited by iodoacetate?

Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 27

Oxidation of glyceraldehyde - 3 - phosphate to 1, 3 - bisphosphoglycerate is catalysed by glyceraldehyde - 3 - phosphate dehydrogenase. The aldehyde group of glyceraldehyde - 3 - phosphate first reacts with the – SH group of an essential Cys residue in the active site of the enzyme. Iodoacetate reacts with the – SH group of Cys residue in glyceraldehyde - 3 - phosphate dehydrogenase. Hence, iodoacetate is a potent inhibitor of glyceraldehyde - 3 - phosphate dehydrogenase.

*Multiple options can be correct
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 28

Which of the following options identifies the animals that are endemic to India?

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 29

In a testcross experiment, the distances between locations A and B are 10, and between B and C are 20. If a double crossover occurs at a rate of 1.6%, determine the coefficient of coincidence.
(Calculate your answer to one decimal place.)


Detailed Solution for GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 29

From the given data, we can calculate that the expected double crossover if there is no interference.
(0.1 x 0.2) = 0.02 or 2%
Coefficient of coincidence is the ratio of the observed to the expected double crossovers.
Therefore, coefficient of coincidence = 1.6/2.0 = 0.8

*Answer can only contain numeric values
GATE Life Science Mock Test- 1 (Botany & Zoology) - Question 30

An enzyme facilitates the transformation of 30 µM of a substrate into product at a reaction velocity of 9.0 µM s-1. Given that [Et] = 30 nM and Km = 10 µM, the ratio Kcat / Km for the enzyme will be n × 107 M-1s-1. What is the value of n in integer form?


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