MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 2

Download the MH SET April 2024 Life Science Answer Key with complete explanations. Analyze each answer and boost your exam preparation effectively. This is April 2024 MH SET Life Science Answer Key Part 2.

Q. 1 The maternal characteristics are inherited through cytoplasmic inheritance also. One of the following organell is involved in :

(A) Mitochondria
(B) Golgi body
(C) Lysosomes
(D) Endoplasmic reticulum

Answer – (A) Mitochondria
Explanation: Mitochondria contain their own DNA and are passed from mother to offspring, contributing to cytoplasmic (non-Mendelian) inheritance.

Q. 2 Trisomy of chromosome no. 21 is :

(A) Edwards syndrome
(B) Klinefelter syndrome
(C) Down syndrome
(D) Patau syndrome

Answer – (C) Down syndrome
Explanation: Down syndrome is caused by trisomy 21—an extra copy of chromosome 21.

Q. 3 A bacterial strain that is lys+ his+ val+ is used as a donor, and another strain that is lys– his– val– as the recipient for setting up an experiment on transformation. Initial transformants were isolated on a minimal medium with supplemented with lysine and valine. Which of the following maximum genotypes will grow on this medium ?

(A) lys–his+ val+ ; lys+ his+ val–; lys+ his+ val+ ; lys–his+val–
(B) lys+his+ val+ ; lys– his– val+
(C) lys– his– val– only
(D) lys+ his– val+ only

Answer – (A) lys⁻his⁺val⁺ ; lys⁺ his⁺ val⁻ ; lys⁺ his⁺ val⁺ ; lys⁻ his⁺ val⁻
Explanation: Medium provides lys and val, so only his⁺ is essential for growth. All genotypes with his⁺ can grow.

Q. 4 In addition to serving as the building blocks for nucleic acid, nucleotides have many functions. Which of the following is not a function of a nucleotide ?

(A) They work as carriers of energy
(B) They are serving as secondary messengers in cells
(C) They are the components of coenzymes
(D) They provide electrons to the ETC

Answer – (D) They provide electrons to the ETC
Explanation: Electrons in the ETC come from reduced coenzymes (like NADH, FADH₂), not nucleotides.

Q. 5 A DNA sample contains 30% adenine on molar basis. What will be the percentage of cytosine ?

(A) 30%
(B) 20%
(C) 40%
(D) 60%

Answer – (B) 20%
Explanation: A=T → 30% A = 30% T → total = 60%. Remaining 40% is C+G → C = 20%.

Q. 6 Which of the following would contribute to intrinsic fluorescence in a protein ?

(A) Aromatic amino acids
(B) Charged amino acids
(C) Branched chain amino acids
(D) Disulfide bonds

Answer – (A) Aromatic amino acids
Explanation: Aromatic residues like tryptophan, tyrosine, and phenylalanine fluoresce under UV light.

Q. 7 The screw sense of DNA can be right handed or left handed. Which of the following statements is correct with respect to the screw sense of A, B and Z type of DNA ?

(A) A and Z both right handed
(B) A and B both right handed
(C) A left handed and B right handed
(D) A right handed and B left handed

Answer – (B) A and B both right handed
Explanation: A-DNA and B-DNA are right-handed, Z-DNA is left-handed.

Q. 8 Which chiral angle in the peptide backbone does not undergo rotation ?

(A) Φ (Phi)
(B) ψ (Psi)
(C) χ (Chi)
(D) ω (Omega)

Answer – (D) ω (Omega)
Explanation: The omega bond (peptide bond) is planar due to partial double-bond character and does not rotate.

Q. 9 Efficiency of enzyme can be better measured in terms of :

(A) Km
(B) Km/Vmax
(C) Kcat
(D) Kcat/Km

Answer – (D) Kcat/Km
Explanation: Kcat/Km reflects catalytic efficiency—rate of product formation per substrate concentration.

Q. 10 What is the resultant pH of a phosphate buffer made by mixing 0.2 M NaH2PO4 and 0.2 M Na2HPO4 (pKa = 6.86)

(A) 5.86
(B) 6.86
(C) 7.86
(D) 7.00

Answer – (B) 6.86
Explanation: Henderson-Hasselbalch equation → pH = pKa + log([base]/[acid]) → pH = 6.86 + log(1) = 6.86.

Q. 11 A protein mixture contains three polypeptides A, B and C, whose masses are 63, 28 and 79 kDa with pI values of 6.5, 7.0 and 8.0, respectively, were subjected to standard reducing SDS-PAGE. The order of their separation from top to bottom would be :

(A) A, B and C
(B) C, B and A
(C) A, C and B
(D) C, A and B

Answer – (B) C, B, A
Explanation: SDS-PAGE separates by size—largest moves least. So order: 79 (C), 28 (B), 63 (A).

Q. 12 A geographically variable species often divided into many subspecies is referred to as ………………….

(A) Monotypic species
(B) Sister species
(C) Sibling species
(D) Polytypic species

Answer – (D) Polytypic species
Explanation: Polytypic species are classified into multiple subspecies due to geographical or genetic variation.

Q. 13 First fossil record of vascular plants appeared in the ……………… period.

(A) Silurian
(B) Jurassic
(C) Triassic
(D) Cambrian

Answer – (A) Silurian
Explanation: Early vascular plants like Cooksonia appeared in the Silurian period (~425 mya).

Q. 14 Inclusive fitness refers to ………….

(A) Reproductive success of an individual
(B) Reproductive success of an individual along with its neighbour’s reproductive success
(C) Reproductive success of an individual along with its relatives’ reproductive success
(D) Reproductive success of an individual and its partners

Answer – (C) Reproductive success of an individual along with its relatives’ reproductive success
Explanation: It includes personal reproduction and aid to relatives’ reproduction (kin selection).

Q. 15 In a classical experiment, a standard is used to :

(A) Add background noise
(B) Record the absolute value of a signal
(C) Compare and eliminate non-specific background noise
(D) Record the non-specific error in the experiment

Answer – (C) Compare and eliminate non-specific background noise

Explanation:
In experiments like ELISA, Western blot, or spectrophotometry, a standard is a sample with a known concentration or known properties. It is used to create a reference point. Comparing your test results with the standard helps in identifying real signals versus background noise or non-specific signals. This ensures the measurements are accurate and reliable.

Q. 16 Why sexual selection can lead to extreme phenotypes ?

(A) Because the extreme phenotype incurs lower costs than the mean train in the population
(B) Because the partner quickly learns and favours new images
(C) Because a positive feedback exists between the gene favouring the trait and the gene coding for it
(D) Because a negative feedback exists between the trait and the risk of predation

Answer – (C) Because a positive feedback exists between the gene favouring the trait and the gene coding for it
Explanation:
In sexual selection, individuals with certain traits (like brighter feathers or larger antlers) are chosen more often as mates. Over time, genes for both the trait and preference for it spread through the population. This creates a positive feedback loop, where the trait becomes more exaggerated in each generation. This is known as runaway sexual selection.

Q. 17 Sarich and Wilson (1967) demonstrated that human, Gorilla and Chimpanzee were genetically equidistant and distinct for Orangutan and the divergence time is approximately :

(A) 30 mya
(B) 05 mya
(C) 10 mya
(D) 100 mya

Answer – (B) 5 million years ago (mya)
Explanation:
Sarich and Wilson used molecular clocks, comparing immunological differences in proteins between species. They found that humans, gorillas, and chimpanzees are genetically very similar and estimated that they shared a common ancestor around 5 million years ago. Orangutans were found to be more distant.

Q.18 Beaker ‘A’ has 100 ml of some fluid at 80°C. Beaker B contains the some fluid 200 ml at 20ºC. If both the fluids are mixed, what would be the temperature of the resultant mixture ?

(A) 25ºC
(B) 65ºC
(C) 40ºC
(D) 50ºC

Answer – (C)

Explanation –

This is based on the principle of heat balance:

Final Temperature =

(100 × 80) + (200 × 20)
-----------------------
      100 + 200
  

Now calculate:

(8000 + 4000) / 300 = 12000 / 300 = 40°C
  

The larger volume of cooler liquid pulls the temperature down more, resulting in a final temperature of 40°C.

Q.19 With the advent of Remote Sensing and Geographical Information System, the city planners have begun to use the Indian satellite data. Which one of the following satellite data are useful for such tasks ?

(A) CARTOSAT
(B) LANDSAT
(C) OCEANSAT
(D) RESOURCESAT

Answer – (A) CARTOSAT
Explanation:
CARTOSAT satellites provide high-resolution images of Earth’s surface, useful for mapping, urban planning, and infrastructure development. These satellites are specifically designed for cartographic applications, making them ideal for city planners and GIS professionals.

MH SET April 2024 Life Science Answer Key with Detailed Explanation

Q.20 Which of the following statements is not true about phase contrast microscopy ?

(A) It is a modified version of the bright field microscope that is fitted with a condenser (annular ring) and objective (phase plate)
(B) It uses a special condenser and objective that accentuate small differences in the refractive index of various structures within the organism
(C) It can be used to observe living cells in their natural state
(D) It uses laser light to obtain focal level sections through a specimen

Answer – (D) It uses laser light to obtain focal level sections through a specimen
Explanation:
Phase contrast microscopy is designed to view live, unstained cells by enhancing contrast using optical phase differences. It does not use laser light — that is used in confocal laser scanning microscopy. So, statement D is false.

Q.21 What will be the radioactivity of ¹²⁵I labelled 2 mCi thyroxine after 4 generations (t½ = 60 days) :

(A) 0.25 mCi
(B) 0.125 mCi
(C) 0.125 µCi
(D) 1 mCi

Answer – (B) 0.125 mCi
Explanation:

In each half-life, radioactivity is reduced by half:

• 1st half-life: 2 → 1 mCi
• 2nd: 1 → 0.5 mCi
• 3rd: 0.5 → 0.25 mCi
• 4th: 0.25 → 0.125 mCi
  So after 4 half-lives (i.e., 240 days), the remaining activity is 0.125 mCi.

Q.22 Among the different types of phytoremediation ………….. is the process in which plants limit contaminated soil movement and migration.

(A) Phytostabilization
(B) Phytotransformation
(C) Phytodegradation
(D) Phytoextraction

Answer – (A) Phytostabilization
Explanation:
Phytostabilization involves planting specific plants that can immobilize contaminants in the soil by absorbing or binding them in their roots. This prevents the spread of harmful substances through water or wind. It does not remove the contaminants but stabilizes them in place.

Q.23 Which of the following proteins helps to maintain and stabilizes protein tertiary structure during temperature stress in plants ?

(A) Defensins
(B) Thionins
(C) Late embryogenesis abundant (LEA)
(D) Osmotin

Answer – (C) Late embryogenesis abundant (LEA)
Explanation:
LEA proteins are produced in large amounts during late embryogenesis and under stress conditions like drought or heat. They act as molecular chaperones, stabilizing other proteins and preventing their aggregation or denaturation, especially during dehydration or heat shock.

Q. 24 Enzyme labelled antibodies are not generally used for which of the following techniques ?

(A) FACS
(B) Immunohistochemistry
(C) ELISA
(D) Western blot

Answer – (A) FACS
Explanation:
FACS (Fluorescence-Activated Cell Sorting) uses fluorescent-labeled antibodies, not enzyme-labeled ones. Techniques like ELISA, Western blot, and Immunohistochemistry commonly use enzyme-labeled antibodies to produce color or light signals when a substrate is added.

Q.25 In watermelons, to induce seedlessness in triploid population, which of the following strategies is useful ?

(A) Self-pollinate resistant triploids
(B) Cross unrelated triploids
(C) Select for unreduced gametes in triploids
(D) Cross tetraploids with diploids to produce triploids

Answer – (D) Cross tetraploids with diploids to produce triploids
Explanation:
When tetraploid (4n) and diploid (2n) plants are crossed, they produce triploid (3n) offspring. These triploids are sterile and cannot complete normal meiosis, leading to seedless fruits. This is a common agricultural strategy for growing seedless watermelons.

More: MH SET April 2024 Life Science Answer Key with Detailed Explanation

• MH SET April 2024 Life Science Answer Key Explanation: Part 3 (Will be uploaded on 10th June 2025)
• MH SET April 2024 Life Science Answer Key Explanation: Part 4 (Will be uploaded on 11th June 2025)
• MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 1

MH SET Life Science Question Paper PDF Download

Read more:

MH-SET Question Papers last 7 years and Answer Keys:

We hope you liked MH SET April 2024 Life Science Answer Key with Detailed Explanation

Join SACHIN’S BIOLOGY on Instagram or Facebook to receive timely updates and important notes about exams directly on your mobile device. Connect with Mr. Sachin Chavan, the founder of Sachin’s Biology and author of biologywala.com, who holds an M.Sc., NET JRF (AIR 21), and GATE qualifications. With SACHIN’S BIOLOGY, you can have a direct conversation with a knowledgeable and experienced professional in the field of biology. Don’t miss out on this opportunity to enhance your exam preparation!