GATE Biotechnology Mock Test - 1 - GATE Biotechnology MCQ

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.

The assertion that 'Some human beings are not cruel creatures' is false implies that all human beings must be cruel, as the negation of 'some are not' is 'all are.' This directly supports statements (i) and (ii). Since all humans are cruel, it follows that some cruel creatures are indeed human beings, making statement (iii) true.

Therefore, the correct answer includes:

• Statements (i)
• Statements (ii)
• Statements (iii)

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.

Leghemoglobin is a protein found in the root nodules of leguminous plants, where it plays a crucial role in nitrogen fixation. It binds to oxygen molecules (O₂) to regulate oxygen levels within the nodule. This regulation is essential because:

• The enzyme nitrogenase, which catalyses the conversion of atmospheric nitrogen (N₂) into ammonia (NH₃), is highly sensitive to oxygen.
• If exposed to too much oxygen, nitrogenase would be deactivated.

By binding oxygen, leghemoglobin ensures that nitrogenase remains functional, allowing the plant to fix nitrogen effectively. Thus, the correct statement is:

It binds oxygen to protect nitrogenase.

Eigen values of a skew-symmetric matrix are either zero or pure imaginary, occuring in conjugate pairs.

Iodoacetate is a potent inhibitor of glyceraldehyde-3-phosphate dehydrogenase because it forms a covalent derivative of the essential –SH group of the enzyme active site, rendering it inactive.

The cloning capacity of YAC is about 3000 kb per molecule of YAC. The other vectors have much less cloning capacities with the maximum being 45 kb for cosmid vectors.

The Needleman-Wunsch algorithm is designed for global alignment, aligning entire sequences end-to-end.

Chou-Fasman and GOR methods predict secondary structures, while Smith-Waterman performs local alignments.

Thus, the correct answer is C.

Assertion [a] states that mutations in chromosomes can alter their structure. This is accurate because chromosomal mutations, such as:

• Deletions
• Duplications
• Inversions
• Translocations

do change chromosome structure. However, Reason [r] incorrectly attributes all chromosomal mutations to nondisjunction during cell division. In reality, mutations also occur due to:

• Environmental factors like mutagens
• Replication errors

Therefore, [a] is true but [r] is false, making option B correct.

Hepatitis B is the only hepatitis virus to have a double stranded DNA (dsDNA) as its genome. All other hepatitis viruses have single stranded RNA as their genome.

Fabry disease is inherited in an X-linked recessive pattern. The inheritance pattern can be summarised as follows:

• The father is phenotypically normal and must have a normal X chromosome (since males have only one X chromosome).
• The mother is a carrier with one normal and one mutated X chromosome, represented as genotype X_cX.
• Sons inherit their X chromosome from their mother and the Y chromosome from their father.
• Therefore, there is a 50% chance that a son will inherit the X chromosome with the mutation (X_c) from the mother, resulting in him being affected by Fabry disease.

Percentage of defective items = 10% or p = 0.1
Probability of non-defective items q = 1 - 0.1 = 0.9
So, probability that exactly 2 of the chosen items are defective = ¹⁰C₂p²q⁸
= ¹⁰C₂(0.1)²(0.9)⁸
= 0.194

To find the distance between the points where the equations x² + y = 7 and x + y = 7 intersect, follow these steps:

• Express both equations in terms of y:
  • From the first equation: y = 7 - x².
  • From the second equation: y = 7 - x.
• Set the equations equal to each other:
  • 7 - x² = 7 - x
  • Simplify to find: x² = x
• Factor the equation:
  • x(x - 1) = 0
  • So, x = 0 or x = 1
• Find corresponding y-values:
  • If x = 0, then y = 7.
  • If x = 1, then y = 6.
• Calculate the distance between points (0, 7) and (1, 6):
  • Use the distance formula: √((x₂ - x₁)² + (y₂ - y₁)²)
  • Plug in values: √((1 - 0)² + (6 - 7)²)
  • Simplify: √(1 + 1) = √2
  • Result: Approximately 1.41

The distance between the two points is 1.41, rounded to two decimal places.

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° 

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

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.

The number of generations, n = Total time passed/doubling time
n = 2.5/0.5 = 5
N_o = N/2ⁿ = 32 × 10⁶/25 = 10 × 10⁵ cells

Induced pluripotent stem cells (iPSCs) possess several key characteristics:

• Self-renewal: They have the ability to self-renew, which is a fundamental trait of stem cells.
• Undifferentiated state: iPSCs require specific signals to maintain their undifferentiated state, such as certain growth factors or signalling pathways.
• Organoid formation: They are capable of forming organoids in vitro, which are three-dimensional structures that mimic organs.

Therefore, the following statements are true:

• Statement A
• Statement B
• Statement D

However, statement C is false because iPSCs can indeed be genetically manipulated for research purposes.

The methylation of CpG islands near a gene's promoter typically prevents transcription by facilitating the formation of heterochromatin. This process involves:

• The attraction of proteins that recognise methyl groups.
• The recruitment of histone deacetylases and other chromatin-modifying enzymes.
• The creation of a more condensed chromatin structure (heterochromatin), which makes the DNA less accessible to transcription machinery.

As a result, this process inhibits transcription. While options A and B might play secondary roles or occur in specific contexts, option C represents the primary mechanism by which methylation silences gene expression. Therefore, the most accurate answer is C.

Tumor suppressor genes can be inactivated through several mechanisms:

• Promoter methylation (Option B):
  • Leads to gene silencing by preventing transcription factor binding.
  • Reduces gene expression.
• Loss of heterozygosity (Option C):
  • Results in the loss of one allele.
  • The remaining defective allele is often unable to function properly.
  • This inactivates the tumor suppressor gene.
• Gene amplification (Option A):
  • More commonly associated with oncogene activation.
  • Not typically linked to tumor suppressor inactivation.
• Histone acetylation (Option D):
  • Generally activates genes rather than inactivates them.

Thus, the correct answers are B and C.

Bacterial Artificial Chromosomes (BACs) are suitable for cloning large DNA fragments up to 300 kb.

Yeast Artificial Chromosomes (YACs) can accommodate even larger fragments, often exceeding 220 kb.

Cosmids typically hold inserts up to 40-50 kb, which is insufficient for 220 kb.

The pUC19 plasmid has a much smaller capacity, around 10 kb.

Thus, both BACs and YACs can replicate a 220 kb DNA fragment.

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.

2.303 log (N/N0) = -k_dt
Or
k_d = - (2.303/t) log (N/N_o)
Putting the values, we get
kd = - (2.303/1) log (10⁴/10⁶) = 4.606 hr^-1

The time required to kill 90% of the original population is decimal reduction time (D).
D = 1/k_d
2.303 log (N/N₀) = -k_dt
Or
kd = - (2.303/t) log (N/N_o)
Putting the values, we get
kd = - (2.303/1) log (10⁴/10⁶) = 4.606 hr^-1
kd = 4.606 hr^-1 = 0.077 min^-1
So, D = 1/0.077
D = 13 min

When pH > pI, then amino acid is negatively charged. In this state, the carboxylate ion accepts a proton to become COOH.

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

The oxygen transfer rate (OTR) is calculated using the formula:

OTR = (μ · X) / Y_bioxygen.

Given:

• μ = 0.2 h^-1
• X = 10 g.L^-1
• Y_bioxygen = 0.2 g.g^-1

Substituting the values:

OTR = (0.2 · 10) / 0.2 = 10 g.L^-1.h^-1.

• The partition coefficient (K) is the ratio of enzyme concentration in the upper phase to the lower phase.
• Given K = 4.0, the enzyme prefers the upper phase.
• Volume ratio of upper to lower phase is 5.0, indicating the upper phase is larger.
• The recovery (R) of enzyme in the upper phase is calculated using R = (K * V_u) / (K * V_u + V_l).
• Substitute V_u/V_l = 5, K = 4, we get R = (4 * 5) / (4 * 5 + 1).
• R = 20 / 21, approximately 95%.