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GATE Biotechnology Mock Test - 8 - GATE Biotechnology MCQ


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30 Questions MCQ Test - GATE Biotechnology Mock Test - 8

GATE Biotechnology Mock Test - 8 for GATE Biotechnology 2025 is part of GATE Biotechnology preparation. The GATE Biotechnology Mock Test - 8 questions and answers have been prepared according to the GATE Biotechnology exam syllabus.The GATE Biotechnology Mock Test - 8 MCQs are made for GATE Biotechnology 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Biotechnology Mock Test - 8 below.
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GATE Biotechnology Mock Test - 8 - Question 1

Boni, Soni, Moni, Goni, and Joni are five siblings. Each of them is 2 years older than his next younger brother. The oldest brother, Joni, is twice the age of the youngest brother, Boni. What is Moni's age?

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

Let us assume the age of Boni (B) be x year, then age of Soni (S), Moni (M), Goni (G) and Joni (J) be x + 2, x + 4, x + 6, x + 8, respectively.
From question, age of Joni (J) = 2 × age of Boni (B)
It means that, x + 8 = 2x
So, x = 8
Age of Moni = x + 4 = 8 + 4 = 12 years.

GATE Biotechnology Mock Test - 8 - Question 2

A rectangular sheet of paper measuring 54 cm × 4 cm is taken. The two longer sides of the sheet are brought together to form a cylindrical tube. Additionally, a cube is taken whose surface area matches that of the sheet.
Then, what is the ratio of the volume of the cylindrical tube to the volume of the cube?

GATE Biotechnology Mock Test - 8 - Question 3

Ishaant prepared for a cross-country competition.
On Monday, he covered a specific distance.
On Tuesday, he ran twice the distance he covered on Monday.
On Wednesday, he ran half the distance he covered on Tuesday.
On Thursday, he ran half of what he ran on Wednesday.
On Friday, he ran double the distance he ran on Thursday.
If the shortest distance he ran on any of the five days is 5 km, what was the total distance he ran?

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 3

Suppose that Ishaant ran a distance of x km on Monday.
Then on Tuesday, he ran 2x km. On Wednesday, he ran x km. On Thursday, he ran   and on Friday he ran x km.
The shortest of any of his runs was on Thursday, so  or x = 10.
Therefore, his runs were 10 km, 20 km, 10 km, 5 km, and 10 km.
Total run = 10 + 20 + 10 + 5 + 10 = 55 km

GATE Biotechnology Mock Test - 8 - Question 4

A football team participated in three matches. Each match concluded with a win, loss, or draw. (A match is considered a draw if both teams score the same amount of goals.) Overall, the team netted more goals than those scored against them. Which of the following outcome combinations is impossible for this team?

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 4

When Team A played Team B, if Team B won, then Team B scored more goals than Team A, and if the game ended in a tie, then Team A and Team B scored the same number of goals.
Therefore, if a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its opponent twice (the 2 ties).
Combining this information, we see that the team must have scored fewer goals than were scored against them.
In other words, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.
We can also examine choices (A), (B), (D) to see that, in each case, it is possible that the team scored more goals than it allowed.
This will eliminate each of these choices, and allow us to conclude that option (C) must be correct.
(A): If the team won 2-0 and 3-0 and tied 1-1, then it scored 6 goals and allowed 1 goal.
(B): If the team won 4-0 and lost 1-2 and 2-3, then it scored 7 goals and allowed 5 goals.
(D): If the team won 4-0, lost 1-2, and tied 1-1, then it scored 6 goals and allowed 3 goals.
Therefore, it is only the case of 0 wins, 1 loss, and 2 ties where it is not possible for the team to score more goals than it allows.

GATE Biotechnology Mock Test - 8 - Question 5

Nimisha is located in room 1401. She contacts Mansi and asks, “Which room are you currently in?” Mansi responds: “The difference between the room number I’m in and the room number you’re in is 100 times the nth prime number, where n is the smallest integer that has six positive factors.” In which room is Mansi situated? (For clarity, if n = a * b, then a and b are considered factors of n. The positive factors of 6 are 1, 2, 3, and 6).

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 5

The prime numbers between 2 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89 and 97. The smallest number which has six positive factors is 12.
12th prime number = 37.
Mansi’s room number = 1401 + 100 × 37 = 1401 + 3700 = 5101

GATE Biotechnology Mock Test - 8 - Question 6

An instance of forced convection is represented by ________.

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 6

An instance of forced convection is represented by the flow of water in condenser tubes. Here's why:

  • Forced Convection: This occurs when a fluid is forced to flow over a surface or in a tube by external means such as a pump or fan.
  • Condenser Tubes: In these, water is pumped through tubes to remove heat, thus exemplifying forced convection.
  • Mechanism: The movement of water is not due to natural causes like temperature differences but by a mechanical pump.
GATE Biotechnology Mock Test - 8 - Question 7

Which of the following statements accurately describes membrane potential?

GATE Biotechnology Mock Test - 8 - Question 8

An immunoglobulin's heavy chain results from the recombination of one DNA segment from 200 distinct V segments, 12 D segments, and 4 J segments from the respective DNA sequence. In addition, a light chain is produced by the recombination of one segment from 200 various V segments and 5 different J segments. When a heavy chain combines with a light chain to create an immunoglobulin, what is the maximum number of unique immunoglobulins that can be generated?

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

Maximum number of different immunoglobulins = 200 × 12 × 4 × 200 × 5 = 9600 × 1000 = 9600000 = ~107.

GATE Biotechnology Mock Test - 8 - Question 9

What is the degree of similarity between two sequences that are separated by 30 PAM (i.e., using the PAM30 matrix for estimation)?

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 9

PAM (Percentage of Accepted Mutations) is a measure of sequence divergence. Two sequences 1 PAM apart have 99% identical residues, in other words, PAM1 matrix estimates the rate of substitution likely to be observed if 1% of the amino acids had changed. Similarly, two sequences 30 PAM apart would have approximately 75% identity.

GATE Biotechnology Mock Test - 8 - Question 10

What is the highest value of ?

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 10




So the function will have its maximum value at x = e
Required maximum value is

GATE Biotechnology Mock Test - 8 - Question 11

Determine the mass transfer coefficient in the gas phase for the removal of unbound moisture if the flux is 5 kg/sq.m sec and the difference in humidity between the liquid and the main stream is 0.5 units.

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 11

Flux = Mass transfer co-efficient × (Ys – Y)
Mass transfer co-efficient = 5/0.5 = 10.

GATE Biotechnology Mock Test - 8 - Question 12

A fair coin is flipped 10 times. What is the probability that heads will appear only in the first two flips?

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 12

p(only first two tosses are heads) = p(H, H, T, T, T, … T)
Now, each toss is independent. So, required probability = p(H) × p(H) × [p(T)]8 ……= (1/2)2(1/2)= (1/2)10  

GATE Biotechnology Mock Test - 8 - Question 13

Sequences of amino acids from cytochrome c and ribulose 5-phosphate epimerase were selected from 40 different organisms, and phylogenetic trees were constructed for each of these two protein families.
Assess the validity of the following Assertion [a] and Reason [r].
Assertion [a]: The two resulting trees will not be the same.
Reason [r]: The mutation types and their frequencies differ between the two families.

GATE Biotechnology Mock Test - 8 - Question 14

What is the degree of the differential equation illustrated below? 

Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 14

Degree refers to the highest power of the derivative however order is the highest derivative's order in a differential equation.

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 8 - Question 15

What is the value of the largest eigenvalue of the matrix ?


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 8 - Question 16

A variable Y depends on t. It is given that Y (t = 0) = 1 and Y (t = 1) = 2. The approximation of Y within the interval t = [0, 1] can be expressed as ______________.


GATE Biotechnology Mock Test - 8 - Question 17

Which of the following biophysical methods is not utilized for the analysis of protein structures?

GATE Biotechnology Mock Test - 8 - Question 18

Which equation represents the connection between μ and the substrate that limits residual growth?

GATE Biotechnology Mock Test - 8 - Question 19

Every single nerve cell exhibits a resting membrane potential averaging at –70 mv. A minimal threshold is essential to trigger an action potential. A synapse is observed where several neurons stimulate a single neuron, resulting in a post-synaptic potential of –10 mv. In this scenario, which of the following statements is accurate?

GATE Biotechnology Mock Test - 8 - Question 20

T-cells exhibit self MHC restriction. Both CD4 and CD8 T-cells are capable of recognizing an antigen exclusively when it is displayed within the groove of an MHC molecule. In a specific experiment, guinea pig macrophages from strain 2 were first treated with an antigen. After these antigen-pulsed macrophages processed the antigen and exhibited it on their surface, they were combined with T-cells derived from the same strain (strain 2), a different strain (strain 13), or their F1 hybrid offspring (2 × 13), and the extent of T-cell proliferation in response to the antigen-pulsed macrophages was assessed.
P. Macrophages from strain 2 that were pulsed with antigen stimulated T-cells from both strain 2 and F1 mice.
Q. Antigen-pulsed macrophages from strain 13 activated T-cells from strain 2 and F1 mice.
R. Macrophages from both strain 2 and strain 13 that were pulsed with antigen activated T-cells from the F1 mice.
S. T-cells from F1 mice were activated solely by antigen-pulsed macrophages from strain 2.

GATE Biotechnology Mock Test - 8 - Question 21

The Fourier law of thermal conduction is most accurately conveyed by

GATE Biotechnology Mock Test - 8 - Question 22

In modeling processes, what do experimental observations contribute?

GATE Biotechnology Mock Test - 8 - Question 23

Examine the following methods along with their respective uses:

Select the options that are accurately matched.

GATE Biotechnology Mock Test - 8 - Question 24

Assess the validity of the following Assertion [a] and Reason [r].
Assertion [a]: Embryonic stem cells are appropriate for creating knockout mice.
Reason [r]: The frequency of homologous recombination is greater in embryonic stem cells compared to somatic cells.

GATE Biotechnology Mock Test - 8 - Question 25

Pair the items from Column I with their appropriate uses listed in Column II.

GATE Biotechnology Mock Test - 8 - Question 26

Connect the elements in Column I with the relevant methods in Column II.

*Multiple options can be correct
GATE Biotechnology Mock Test - 8 - Question 27

Which of the following statement(s) is(are) ACCURATE concerning the lac operon in E. coli when it is cultured in an environment containing both glucose and lactose?

*Answer can only contain numeric values
GATE Biotechnology Mock Test - 8 - Question 28

A bacterial strain is grown in nutrient medium at 37 ˚C under aerobic conditions. The medium is inoculated with 102 cells from a seed culture. If the number of cells in the culture is 105 after 10 hours of growth, the doubling time of the strain (rounded off to nearest integer) is ____________h. 


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 8 - Question 29

In a cell, the concentrations of ATP, ADP, and inorganic phosphate are measured at 2.59, 0.73, and 2.72 mM, respectively. Given these values, the free energy change associated with the synthesis of ATP at 37 °C is ____ kJ/mol (rounded to two decimal places).
Given: the free energy change for the hydrolysis of ATP under standard conditions is -30.5 kJ/mol and R = 8.315 kJ/mol.K


*Answer can only contain numeric values
GATE Biotechnology Mock Test - 8 - Question 30

A fermentation medium is cooled from 121 °C to 30 °C using a double pipe heat exchanger. If cold water flows in the opposite direction and is warmed from 10 °C to 70 °C, what is the Log-Mean Temperature Difference (LMTD) in °C (rounded to the nearest integer)?


Detailed Solution for GATE Biotechnology Mock Test - 8 - Question 30

To calculate the Log-Mean Temperature Difference (LMTD) for a counterflow heat exchanger, we use the following formula:

LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)

Where:

ΔT1 is the temperature difference at one end of the heat exchanger.
ΔT2 is the temperature difference at the other end of the heat exchanger.

Given:
Hot fluid (fermentation medium) cools from 121 °C to 30 °C.
Cold fluid (water) warms from 10 °C to 70 °C.
Step 1: Calculate ΔT1 and ΔT2
For a counterflow heat exchanger:

ΔT1 is the difference between the hot fluid inlet temperature and the cold fluid outlet temperature: ΔT1 = 121 - 70 = 51 °C

ΔT2 is the difference between the hot fluid outlet temperature and the cold fluid inlet temperature: ΔT2 = 30 - 10 = 20 °C

Step 2: Calculate LMTD
Substitute ΔT1 and ΔT2 into the LMTD formula:

LMTD = (51 - 20) / ln(51 / 20)

LMTD = 31 / ln(2.55)

ln(2.55) ≈ 0.936

LMTD = 31 / 0.936 ≈ 33.12 °C

Step 3: Round to the nearest integer
LMTD ≈ 33 °C

Thus, the Log-Mean Temperature Difference (LMTD) is 33 °C.

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