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30 Questions MCQ Test GATE Life Sciences 2025 Mock Test Series - GATE Life Sciences Mock Test - 1

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GATE Life Sciences Mock Test - 1 - Question 1

Directions: Choose the correct option to fill in the blank.

______________, the more they remain the same.

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 1

This is a part of a proverb. It means that as time and technology move forward, human behaviour remains the same.

GATE Life Sciences Mock Test - 1 - Question 2

If  then the value of x is:

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GATE Life Sciences Mock Test - 1 - Question 3

Pen : Write :: Knife : _________
Which one of the following options maintains a similar logical relation in the above?

GATE Life Sciences Mock Test - 1 - Question 4

273 - 272 - 271 is same as

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 4

Let X = 273 - 272 - 271
X = (22 × 271) - (21 × 271) - 271
= (4 × 271) - (2 × 271) - 271
= 271 (4 - 2 - 1)
= 271

GATE Life Sciences Mock Test - 1 - Question 5


The number of units of a product sold in three different years and the respective net profits are presented in the figure above. The cost/unit in Year 3 was 1, which was half the cost/unit in Year 2. The cost/unit in Year 3 was one-third of the cost/unit in Year 1. Taxes were paid on the selling price at 10%, 13% and 15% respectively for the three years. Net profit is calculated as the difference between the selling price and the sum of cost and taxes paid
in that year.

The ratio of the selling price in Year 2 to the selling price in Year 3 is ________. 

GATE Life Sciences Mock Test - 1 - Question 6

The radial wave function of hydrogen atom for 3s orbital is represented by the relationship __________.

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 6

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

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 7

The ground state energy of an electron in a hydrogen atom is –13.60 eV. The energy of the electron in the third excited state is _________eV (rounded off to two decimal places).


GATE Life Sciences Mock Test - 1 - Question 8

A particle is confined to a one – dimensional box of length 1 mm. If the length is changed by 10–9 m, the % change in the ground state energy is

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 8


Percent change in energy 

GATE Life Sciences Mock Test - 1 - Question 9

The CORRECT order of acidity of the following compounds is 

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 10

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


GATE Life Sciences Mock Test - 1 - Question 11

Which of the following combinations is true for sporophytic self–incompatibility?

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 11

To prevent self-fertilization, many angiosperms have developed a chemical system of self-incompatibility. The most common type is sporophytic self-incompatibility, in which the secretions of the stigmatic tissue or the transmitting tissue prevent the germination or growth of incompatible pollen. Other features of the sporophytic self–incompatibility are short pollen viability and dry stigma. A second type, gametophytic self-incompatibility, involves the inability of the gametes from the same parent plant to fuse and form a zygote or, if the zygote forms, then it fails to develop. These systems force outcrossing and maintain a wide genetic diversity.

GATE Life Sciences Mock Test - 1 - Question 12

The endopolyploidy occurs due to

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 12

The alkaloid colchicine inhibits the formation of mitotic spindle (inhibits polymerization of microtubules) and holds the cell in metaphase. The chromosomes and DNA undergo replication but remain with in the same cell. The nucleus does not divide. This increases the number of chromosomes sets per cell. This process leads to endopolyploidy or endomymitosis.

GATE Life Sciences Mock Test - 1 - Question 13

Which of the following statements is incorrect with respect to alternative oxidase activity in cyanide resistant respiration in plants?

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 13

The enzyme alternative oxidase (AOX) constitutes part of the electron transport chain in mitochondria. It provides an alternative route for electrons passing through the electron transport chain to reduce oxygen. 'Electrons are accepted directly from cytochrome C by alternative oxidase' is incorrect.

GATE Life Sciences Mock Test - 1 - Question 14

In binomial nomenclature, in case of two or more names are given, the oldest, i.e. the name given first is recognised as valid name and all other names are called

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 14

When there is one name for two different plants, it is homonym. When species and sub species names are same, it is antonyms. When generic and specific names are same, it is called tautonyms. In case of two or more names are given, the oldest i.e. the name given first is recognized as the valid name and all other called synonyms.

*Multiple options can be correct
GATE Life Sciences Mock Test - 1 - Question 15

Which of the following cellular component(s) is/are NOT part(s) of cytoskeleton in Angiosperms?

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 16

In spiral phyllotaxis, leaves are initiated sequentially on the meristem with two successive primordia being separated by golden angle. If a plant follows right-handed spiral phyllotaxis when looked down the meristem, then the angle between two successive leaves would be _________degrees (with correct sign, round off to one decimal place).


GATE Life Sciences Mock Test - 1 - Question 17

Match the following:

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 17

Karsurin, a metabolites, obtained from Trichoxanthus roseusis an immuno suppressant, and induces abortion. Barberin, obtained from coptis japonicais an anti-bacterial and anti-inflammatory vincristine obtained from Catharanthus roseus and is antilukaemic. Codein obtained from papaver somniferum is an antifertility agent. Taxol obtained from Taxux sp. is used in breast and ovarian cancer treatment.

GATE Life Sciences Mock Test - 1 - Question 18

9 : 3 : 4 is the dihybrid phenotypic ratio for the genetic interactions of

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 18

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 Sciences Mock Test - 1 - Question 19

Match the modified organs in GROUP I with their corresponding prototypic forms in GROUP II and choose the CORRECT option.

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 20

According to Hardy Weinberg equation, if allelic frequency for PTC TASTER is 0.2, then what will be the number of non-taster individuals in a population of 1000?
(Give your answer up to 3-digit whole numbers.)


Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 20

PTC tasting is a dominant trait.
So, p = 0.2
As p + q = 1,
q = 1 - 0.2 = 0.8, q2 = (0.8)2 = 0.64
So, 1000 × 0.64 = 640 individuals

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 21

Plant genetics is an extremely immense term. It actually concerns with heredity particularly the mechanism of heredity transmission & differentiations of derived properties among identical or related organisms.One of Mendel's pure pea plant's strains had green peas. How many several types of eggs might such a plant form with respect to the colour of pea?


*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 22

In a population of a diploid plant species obeying Hardy-Weinberg equilibrium, a locus regulating flower color has two alleles R and r. Individuals with RR, Rr and rr genotypes produce red, pink and white flowers, respectively. If the ratio of red, pink and white flower-producing individuals in the population is 6:3:1, then the frequency of r allele in the population would be __________%. (Round-off to two decimal places.)


GATE Life Sciences Mock Test - 1 - Question 23

Archaeopteryx became extinct during

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 23

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 Sciences Mock Test - 1 - Question 24

Infectivity develops in microfilariae inside

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 24

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 Sciences Mock Test - 1 - Question 25

A population of snakes in an isolated island is in Hardy–Weinberg equilibrium for a gene with only two alleles (A and a). If the allelic frequency of A is 0.6, then the genetic frequency of Aa is _________ (round off to 2 decimal places).


GATE Life Sciences Mock Test - 1 - Question 26

Match the following:

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 26

Down syndrome (21 trisomy) is a condition in which a person is born with an extra copy of chromosome 21. People with Down syndrome can have physical problems, as well as intellectual disabilities. Every person born with Down syndrome is different.
Cri-du-chat syndrome (5p-) is a group of symptoms that result from missing a piece of chromosome number 5. The syndrome's name is based on the infant's cry, which is high-pitched and sounds like a cat.
Klinefelter syndrome (47, XXY) is a condition that occurs in men who have an extra X chromosome in most of their cells. The most common symptom is infertility. Teenagers with Klinefelter syndrome may have less facial and body hair and may be less muscular than other boys.
Turner syndrome (45, X) is a genetic disorder that affects a girl's development. The cause is a missing or incomplete X chromosome. Most women with Turner syndrome are infertile. They are at risk for health difficulties such as high blood pressure, kidney problems, diabetes, cataract, osteoporosis and thyroid problems.

GATE Life Sciences Mock Test - 1 - Question 27

Which of the following cyclins functions during M phase in cell cycle?

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 27

There are two main groups of cyclins:
G1/S cyclins – essential for the control of the cell cycle at the, G1/ S transition
Cyclin A/CDK2 – active in S phase.
Cyclin D/CDK4, Cyclin D/ CDK6 and Cyclin E/ CDK2 – regulates transition from G1 to S phase.
G2/M cyclins – essential for the control of the cell cycle at the G2/M transition. G2/M cyclins accumulate steadily during G2 and are abruptly destroyed as cells exit from mitosis (at the end of the M phase)
Cyclin B/CDK1 – regulates progression from G2 to M phase.
Thus, cyclin B functions during M phase in cell cycle.

GATE Life Sciences Mock Test - 1 - Question 28

CheA, a protein that participates in bacterial chemotaxis
P. is a histidine kinase
Q. shows autophosphorylation
R. transfers phosphoryl groups to conserved aspartate residues in the CheY
S. is a tyrosine kinase

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 28

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 Sciences Mock Test - 1 - Question 29

Iodoacetate inhibits glycolytic enzyme

Detailed Solution for GATE Life Sciences Mock Test - 1 - Question 29

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.

*Answer can only contain numeric values
GATE Life Sciences Mock Test - 1 - Question 30

An enzyme catalyzes the conversion of 30 µM of a substrate to product at reaction velocity of 9.0 µM s-1. When [Et] = 30 nM and Km = 10 µM, Kcat / Km of enzyme will be n × 107 M-1s-1. The value of n is _________ (in integer).


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