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GATE Biotechnology Mock Test - 1 - GATE Biotechnology MCQ


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30 Questions MCQ Test GATE Biotechnology 2025 Mock Test Series - GATE Biotechnology Mock Test - 1

GATE Biotechnology Mock Test - 1 for GATE Biotechnology 2024 is part of GATE Biotechnology 2025 Mock Test Series preparation. The GATE Biotechnology Mock Test - 1 questions and answers have been prepared according to the GATE Biotechnology exam syllabus.The GATE Biotechnology Mock Test - 1 MCQs are made for GATE Biotechnology 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Biotechnology Mock Test - 1 below.
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GATE Biotechnology Mock Test - 1 - Question 1

Which of the following is an antonym of the word PROFESSIONAL?

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 1

Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.

GATE Biotechnology Mock Test - 1 - Question 2

References : ______ : : Guidelines : Implement
(By word meaning)

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GATE Biotechnology Mock Test - 1 - Question 3

In the given figure, PQRS is a parallelogram with PS = 7 cm, PT = 4 cm and PV = 5 cm. What is the length of RS in cm? (The diagram is representative.)

GATE Biotechnology Mock Test - 1 - Question 4

The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 4

A = 2( l + b)h
A = 2(3 + 2) × 1 = 10
New area = 2(6 + 1) × 1/2 = 7
Decrease = (3/10) × 100
= 30%

GATE Biotechnology Mock Test - 1 - Question 5

Deepika sells her goods 10% cheaper than those of Nidhi and 10% costlier than those of Gurpreet. A customer purchases goods from Nidhi worth Rs. 100. How much would he have saved if he bought the same goods from Gurpreet?

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 5

Selling price of goods sold by Nidhi = Rs. 100
Selling price of goods sold by Deepika = Rs. 90
Now, according to the question:
110% of selling price of goods sold by Gurpreet = Rs. 90
Selling price of goods sold by Gurpreet = (100/110) x 90 = Rs. 900/11
So, Gurpreet's goods are cheaper than Nidhi's goods by

Therefore, the customer would have saved 18.18%, if he bought the goods from Gurpreet instead of buying them from Nidhi.

GATE Biotechnology Mock Test - 1 - Question 6

Which one of the following is required for the development of B-cells in the bone marrow?

GATE Biotechnology Mock Test - 1 - Question 7

In hybridoma technology, which one of the following enzymes is absent in the myeloma cells that are used for monoclonal antibody production?

GATE Biotechnology Mock Test - 1 - Question 8

The process of addition of selected bacteria to an area for breakdown of contaminants is known as

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 8

Bioaugmentation is an in-situ bioremediation technique in which selected standardized bacteria are added to an area that has been contaminated with an unwanted substance. These bacteria breakdown the contaminants. Biostimulation involves modification of the environment to stimulate the existing bacteria for bioremediation. In biosparging, air (oxygen) and nutrients are injected in a saturated zone to increase microbial biological activity. Phytoremediation is a technique that employs plants for bioremediation.

GATE Biotechnology Mock Test - 1 - Question 9

Which one of the following programs is used for finding distantly related (or remote) protein homologs?

GATE Biotechnology Mock Test - 1 - Question 10

The phospholipid which is present in the plasma membrane of human cells but not in those of E. coli is

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 10

PC is present in the plasma membranes of human cells such as the RBCs and neurons, but it is totally absent from that of the E. coli. The major phospholipids present in the membrane of E. coli are PE and PS that collectively account for 85% of its weight.

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 11

For the function f(x) = x2e-x, the maximum occurs when x is equal to ____.(Answer up to the nearest integer)


Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 11

f(x) = x2e-x
Or f'(x) = 2xe-x - x2e-x
= xe-x (2 - x)
f''(x) = (x2 - 4x+ 2) e-x
Now for maxima and minima, f'(x) = 0
xe-x (2 - x) = 0
or x = 0, 2
at x = 0   f''(0) = 2 (+ ve)
at x = 2   f''(2) = - 2e-2(- ve)
Now f''(0) = 2 and f''(2) = - 2e-2 < 0. Thus, x = 2 is the point of maxima. 

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 12

The distance between the two points of intersection of x 2 +y= 7 and x +y= 7 (rounded off to two decimal places) is ____________.


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 13

If 73x = 216 , the value of 7− x (rounded off to three decimal places) is  ____________.


GATE Biotechnology Mock Test - 1 - Question 14

The point of maxima of the function sin x + 2 cos x over the interval [0, x] is

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 14

We have 
f(x) = sin x + 2cos x
f'(x) = cos x - 2 sin x ....(i)
f"(x) = - sin x - 2 cos x ....(ii)
For maxima and minima, put f'(x) = 0
cos x = 2 sin x
tan x= 1/2 
x = 26.56° 
At x = 26.56° 
f''(x) = - 2.236(-ve)
Thus f(x) is maximum at x = 26.56° 

GATE Biotechnology Mock Test - 1 - Question 15

If the region of convergence of x1 [n] + x2 [n] is  then the region of convergence of x1 [n] - x2 [n] includes

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 15

The ROC of addition or subtraction of two functions x1(n) and x2(n) is R1 ∩ R2.
We have been given ROC of addition of two function and has been asked ROC of subtraction of two function. It will be same.

GATE Biotechnology Mock Test - 1 - Question 16

A fair coin is tossed n times and let X denote the number of heads obtained. If P(X = 4), P(X = 5) and P(X = 6) are in A.P., then n is equal to

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 16

Given: 2P (X = 5) = P (X = 4) + P (X = 6)

⇒ 2 nC5 = nC4 + nC6
⇒ n = 7 or 14.

GATE Biotechnology Mock Test - 1 - Question 17

Determine the number of bonds that can be hydrolysed by the use of trypsin and chymotrypsin in the following oligopeptide sequence:

Tyr-Gly-Gly-Phe-Met-Lys-Lys-Tyr

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 17

Trypsin cleaves carboxyl side of basic residues, Lys and Arg, while chymotrypsin cleaves the carboxyl sides of aromatic amino acids Phe, Trp and Tyr.

GATE Biotechnology Mock Test - 1 - Question 18

Match the cofactors in Group I to the corresponding enzymes in Group II.

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 18

Many enzymes need a cofactor for their activity. A specific enzyme is functional in association to a specific metal ion cofactor.
A carboxypeptidase is a protease enzyme that hydrolyzes (cleaves) a peptide bond at the carboxy-terminal (C-terminal) end of a protein or peptide. Carboxypeptidase needs Zn ions for its activity.
Superoxide dismutase is an enzyme that helps break down potentially harmful oxygen molecules in cells. Superoxide dismutase requires Cu ions as a co factor.
Enolase, also known as phosphopyruvate hydratase, is a metalloenzyme responsible for the catalysis of the conversion of 2-phosphoglycerate (2-PG) to phosphoenolpyruvate (PEP). Enolase requires Mg ions as a cofactor.
Urease is an enzyme that catalyzes the hydrolysis of urea, forming ammonia and carbon dioxide. Urease needs Ni ions for its activity and functionality.

GATE Biotechnology Mock Test - 1 - Question 19

A bacterial culture contains 32 × 106 cells after 2.5 hours of exponential growth. What was the initial bacterial count if the doubling time of the bacteria is 30 minutes?

Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 19

The number of generations, n = Total time passed/doubling time
n = 2.5/0.5 = 5
No = N/2n = 32 × 106/25 = 10 × 105 cells

GATE Biotechnology Mock Test - 1 - Question 20

Match the immune tolerance mechanisms in Group I with their respective outcomes in Group II.

*Multiple options can be correct
GATE Biotechnology Mock Test - 1 - Question 21

The event(s) that lead(s) to inactivation of tumor suppressor genes in cancer cells is(are)

*Multiple options can be correct
GATE Biotechnology Mock Test - 1 - Question 22

Which of the following vector(s) is(are) used to clone a DNA fragment of size 220 kb?

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 23

What would be the recombination frequency of the genes (A and B)?
(Give your answer upto 1 decimal place.)


Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 23

The map distance between the genes is equal to the recombination frequency of the genes.
Recombination frequency = Number of recombinant offsprings/Total number of offsprings.
Recombination frequency = (36 + 39)/(36 + 39 + 160 + 165) = 0.1875 or 18.75%
Hence, the genes A and B are 18.8 map units apart.

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 24

What is the value of the rate constant kd for the process?
(Give your answer upto 3 decimal places.)


Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 24

2.303 log (N/N0) = -kdt
Or
kd = - (2.303/t) log (N/No)
Putting the values, we get
kd = - (2.303/1) log (104/106) = 4.606 hr-1

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 25

A diploid species has 48 chromosomes during the M phase of the cell cycle. How many chromosomes would it have during the G1 phase?


Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 25

During the M phase, the number of chromosomes(48) in a cell is double that of the actual number so as to ensure equal distribution of chromosomes in the daughter cells. The cell enters the G1 phase after cell division in the M phase is over. So, the number of chromosomes in the G1 phase would be half of what was in the M phase i.e 24.

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 26

In a population study, 1000 individuals are chosen at random and their genotypes for a trait are determined. The genotype distribution was found to be as follows:
AA = 800
Aa = 185
aa = 15

Determine the frequency of occurrence of the allele "a" in the population.
(Give your answer upto 3 decimal places.)


Detailed Solution for GATE Biotechnology Mock Test - 1 - Question 26

Frequency of allele a = q = [f(Aa) + (2 × f(aa))]/(2 × total number of individuals)
q = [185 + (2 × 15)]/2000
= 0.1075

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 27

Aqueous two-phase extraction is used to recover α-amylase from a solution. A polypropylene glycol-dextran mixture is added and the solution separates into upper and lower phases. The partition coefficient is 4.0 and the ratio of upper to lower phase volume is 5.0. The enzyme recovery or yield (in percentage, rounded off to the nearest integer) is ___________.


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 28

If 1000 bp of a double-helical DNA weighs 1x10-18 gm and distance between two bp is 0.34 nm , the total amount of DNA (in mg, rounded off to one decimal place) required to stretch from Earth to Moon (assuming the distance between Earth and Moon to be 3,74,000 km) is ___________.


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 1 - Question 29

The values of the consistency index ‘ K ’ and the flow behavior index ‘ n ’ of a dilatant fluid are 0.415 (in CGS units) and 1.23, respectively. The value of the apparent viscosity (in g. cm −1. s−1 ) of this fluid at a shear rate of 60 s −1 (rounded off to the nearest integer) is ___________.


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

A protein has three identical sites arranged at the vertices of an equilateral triangle. If one site is filled with a dye (donor), the measured quantum yield (ØD) is 0.5. Filling one site with a donor dye and a second site with an acceptor dye results in ØD of 0.25. The measured ØD of one site filled with donor and the other two sites filled with acceptor dye (rounded off to three decimal places) is ___________.


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