MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 4
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 4.
Q.1 (I) The 3′ to 5′ exonuclease activity of eukaryotic DNA polymerase operates in reverse direction of DNA synthesis and participate in fidelity of replication.
(II) The ability of DNA polymerase to extend a primer only in 5′ to 3′ direction appears to make replication a complicated process, it is necessary for proof reading of duplicated DNA.
Choose the correct answer :
(A) (I) is correct, (II) is wrong
(B) (II) is correct, (I) is wrong
(C) Both (I) and (II) are correct
(D) Both (I) and (II) are wrong
Answer – (C) Both (I) and (II) are correct
Explanation:
(I): DNA polymerases have 3’ to 5’ exonuclease activity for proofreading, which removes incorrectly paired nucleotides, ensuring replication fidelity.
(II): DNA synthesis is only 5’ to 3’, and this directionality is necessary for proofreading to work correctly—if DNA synthesis went the other way, proofreading would be chemically unfavorable.
Q.2 Which of the following is a wrong statement in SOS repair system ?
(A) Lex A shows autorepression of lex A
(B) Lex A shows repression of rec A
(C) Negative regulation of Lex A blocks the SOS repair
(D) Rec A shows negative regulation of lex A
Answer – (C) Negative regulation of Lex A blocks the SOS repair
Explanation:
LexA represses SOS genes, including recA. Under DNA damage, RecA activates and induces self-cleavage of LexA, derepressing the genes needed for SOS repair. Thus, negative regulation of LexA does not block but initiates SOS repair when lifted.
Q.3 Which of the following statements about apoptosis is false ?
(A) Cell shrinks
(B) Chromatin condenses
(C) Cell contents are leaked
(D) Apoptotic bodies are formed, which are phagocytosed by macrophages
Answer – (C) Cell contents are leaked
Explanation:
Apoptosis is controlled cell death. It does not leak contents; instead, apoptotic bodies form and are phagocytosed. Leakage of contents is characteristic of necrosis, not apoptosis.
Q.4 Which of the following signalling molecules plays a key role in the patterning of neural tube along its anterior-posterior axis ?
(A) Wnt4
(B) BMP4
(C) FGF4
(D) Retinoic acid
Answer – (D) Retinoic acid
Explanation:
Retinoic acid (a derivative of Vitamin A) plays a central role in patterning the anterior-posterior (A-P) axis of the neural tube. It acts as a morphogen to regulate Hox genes, which are essential in specifying regional identity along the A-P axis during development.
Q.5 If you compare the B cell epitope against T cell epitope :
(A) B cell epitopes are based more on primary sequence of protein antigen
(B) B cell epitopes are based more on nature conformation of protein antigen
(C) B cell epitopes are based more on presentation of antigen by MHC molecule
(D) B cell epitopes are of continuous epitopes than the B cell epitope
Answer – (B) B cell epitopes are based more on native conformation of protein antigen
Explanation:
B-cell epitopes are recognized by antibodies in their native (3D) form, often conformational or discontinuous. T-cell epitopes are linear peptides presented by MHC molecules. Thus, B-cells depend more on the natural structure of the protein.
MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 4
Q.6 Mark the correct statement for antigenicity of molecule :
(A) Antigenicity is the ability of the molecule to provocate immune response in the body
(B) Antigenicity refers to the binding ability of Antibody to its antigen
(C) Antigenicity and immunogenicity are synonyms
(D) All antigens are immunogens
Answer – (B) Antigenicity refers to the binding ability of Antibody to its antigen
Explanation:
Antigenicity is the ability of an antigen to bind specifically to antibodies or T-cell receptors. It is distinct from immunogenicity, which is the ability to induce an immune response. Not all antigens are immunogens.
Q.7 Which of the following molecules acts as a general adhesive connecting one cell to another and cells to other substrates ?
(A) Fibronectin
(B) Heparin sulphate
(C) Collagen
(D) Laminin
Answer – (A) Fibronectin
Explanation:
Fibronectin is a glycoprotein in the extracellular matrix that binds to integrins and other matrix components like collagen and laminin, playing a critical role in cell adhesion, migration, and wound healing.
Q.8 Formation and migration of epiblast as a sheet of cells during gastrulation in zebra fish requires the expression of ………………..
(A) P-cadherins
(B) R-cadherins
(C) N-cadherins
(D) E-cadherins
Answer – (D) E-cadherins
Explanation:
During zebrafish gastrulation, epiboly requires the expression of E-cadherins, which mediate cell–cell adhesion in epithelial layers, allowing the blastoderm to spread over the yolk.
Q.9 Which of the following groups of transcription factors (TFs) is responsible for maintaining the pluripotency of stem cells ?
(A) Oct 4, Nanog and Stat 3
(B) Oct 4, Nanog and SOX 2
(C) Eomesodermin, Cdx 2 and SOX 2
(D) Cdx 2, Nanog and SOX 2
Answer – (B) Oct 4, Nanog and SOX 2
Explanation:
These core transcription factors maintain the undifferentiated state and pluripotency of embryonic stem cells.
- Oct4 and Sox2 regulate Nanog.
- Together, they form a self-reinforcing regulatory network.
Q.10 Yersinia pestis is the causative agent for …………….
(A) Gas gangrene
(B) Tularemia
(C) Lyme disease
(D) Plague
Answer – (D) Plague
Explanation:
Yersinia pestis is a Gram-negative bacterium that causes bubonic, pneumonic, and septicemic plague. It is transmitted by fleas from rodents to humans.
Q.11 Pseudohermaphroditism is a condition in which the individual is insensitive to ………………..
(A) Androgen
(B) Estrogen
(C) Progesterone
(D) Corticosterone
Answer – (A) Androgen
Explanation:
In androgen insensitivity syndrome, genetically male (XY) individuals have mutations in the androgen receptor gene, making cells unresponsive to androgens, leading to female-like external features.
Q.12 In plants, when the apical cell of the 2-celled pro-embryo divides longitudinally and further the basal cell plays only a minor role or none in the development of the embryo proper then the embryo is ………… type.
(A) Asterad
(B) Onagrad
(C) Solanad
(D) Chenopodial
Answer – (B) Onagrad
Explanation:
In Onagrad type embryogenesis (common in Arabidopsis), the apical cell forms the embryo proper and the basal cell plays a minimal or no role, usually contributing to the suspensor only.
Q.13 Which of the following is necessary for the development of epididymis in mammals ?
(A) Dihydrotestosterone
(B) Methyl testosterone
(C) Testosterone
(D) Androsterone
Answer – (C) Testosterone
Explanation:
Testosterone, produced by Leydig cells in the testes, is essential for the development of Wolffian ducts into structures like epididymis, vas deferens, and seminal vesicles.
Q.14 The mid-blastula transition is the point in the development when ……….
(A) Translation of maternal mRNA is initiated
(B) Cell determination is fixed
(C) Cell-division in the zygote begins
(D) Transcription of zygotic genes begins
Answer – (D) Transcription of zygotic genes begins
Explanation:
MBT is a developmental stage (seen in organisms like Xenopus) where:
- Cells become more autonomous and start differentiation
- Cell cycles slow down
- Zygotic transcription begins (previously only maternal mRNAs were used)
Q.15 In the chick limb, the signal for condensation of mesenchymal cells to give rise to cartilage comes from the appearance of :
(A) N-cadherins
(B) E-cadherins
(C) C-cadherins
(D) B-cadherins
Answer – (A) N-cadherins
Explanation:
In chick limb development, N-cadherin mediates mesenchymal condensation, the precursor to cartilage formation. This is crucial in chondrogenesis.
MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 4
Q.16 Which of the following is not a property of stem cells ?
(A) Ability to self-renew
(B) Ability to differentiate into different types of cells
(C) They are unspecialised cells with cell markers
(D) They are actively dividing cells with cell marker changes
Answer – (D) They are actively dividing cells with cell marker changes — Incorrect
Explanation:
Not all stem cells are actively dividing (some are quiescent). Correct features include:
- Self-renewal
- Differentiation potential
- Unspecialized with specific markers (but not necessarily changing constantly)
Q.17 Which of the following proteins acts as a death receptor that triggers the extrinsic pathway of apoptosis ?
(A) Fas ligand
(B) FADD
(C) FaS
(D) Procaspase 9
Answer – (C) FaS
Explanation:
FaS (CD95) is a death receptor on the cell surface. Binding of FasL activates FADD, which then activates caspase 8, initiating extrinsic apoptosis.
Q.18 With reference to normal ECG, the T-wave is :
(A) Negative in all standard bipolar limb leads recording
(B) Caused by repolorization of apex and outer surfaces of the ventricles
(C) Abnormal when normal sequence of depolarization does not occur
(D) Caused by depolarization of the septum and the ventricles
Answer – (B) Caused by repolarization of apex and outer surfaces of ventricles
Explanation:
The T-wave represents ventricular repolarization, beginning at the apex and outer regions and progressing inward.
Q.19 Increased muscle activity will increase generation of lactic acid. In this condition, blood coming from the lungs to the muscles, would show :
(A) Increased O2 release from haemoglobin
(B) Decreased O2 release from haemoglobin
(C) No change in O2 release property
(D) Increased degradation of haemoglobin
Answer – (A) Increased O2 release from haemoglobin
Explanation:
High lactic acid → lowers blood pH → Bohr effect → decreased hemoglobin affinity → increased O₂ release to tissues like muscle.
Q.20 The capability of vertebrate kidney and produce hypertonic urine is a function of :
(A) Reduced filtration at the glomerulus
(B) Increased reabsorption at Bowman’s capsule
(C) The loop of Henle
(D) Addition of waste molecules in the Bowman’s capsule
Answer – (C) The loop of Henle
Explanation:
The loop of Henle establishes a counter-current multiplier system, enabling water reabsorption and concentration of urine, especially in juxtamedullary nephrons.
MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 4
Q.21 For long distance runner, the muscle needs increased O2 supply. At the muscle, this is formed by :
(A) Increased and frequent muscle contraction
(B) Increased demand made by the muscles
(C) Increased squeezing of the muscles
(D) Increased accumulation of lactic acid causing demand blood pH
Answer – (D) Increased accumulation of lactic acid causing demand blood pH
Explanation:
Exercise → lactic acid increases → pH drops → hemoglobin offloads more O₂ due to Bohr effect, meeting muscle demand.
Q.22 Which of the following heat shock proteins has both protein folding and degradation activities ?
(A) HSP 60
(B) HSP 70
(C) HSP 90
(D) HSP 100
Answer – (D) HSP 100
Explanation:
HSP100 family members (e.g., Clp proteins) are involved in unfolding, refolding, and protein degradation, especially under stress conditions.
Q.23 The sequential order of various stages of a fermentation process are :
(A) Fermentation, removal of waste, inoculation, downstream processing
(B) Inoculation, downstream processing, fermentation, removal of waste
(C) Inoculation, fermentation, downstream processing, removal of waste
(D) Removal of waste, inoculation, fermentation, downstream processing
Answer – (C) Inoculation, fermentation, downstream processing, removal of waste
Explanation:
Correct sequence:
- Inoculation (adding microbes)
- Fermentation (product formation)
- Downstream processing (product recovery)
- Waste removal
Q.24 Match the following product/process to the microorganism involved :
Product/Process
(L) Biopol
(M) Biopesticide
(N) Bioleaching
(O) Bioremediation of oil
Microorganism
(1) Thiobacillus ferroxidans
(2) Pseudomonas putida
(3) Bacillus sphaericus
(4) Alcaligenes eutrophus
(A) (L)-(1),(M)-(4),(N)-(2),(O)-(3)
(B) (L)-(2),(M)-(1),(N)-(3),(O)-(4)
(C) (L)-(3),(M)-(1),(N)-(2),(O)-(4)
(D) (L)(-4),(M)-(3),(N)-(1),(O)-(2)
Answer – (D) (L)(-4),(M)-(3),(N)-(1),(O)-(2)
Explanation-
Biopol – Alcaligenes eutrophus (makes bioplastic)
Biopesticide – Bacillus sphaericus (kills insect larvae)
Bioleaching – Thiobacillus ferroxidans (extracts metals)
Bioremediation of oil – Pseudomonas putida (breaks hydrocarbons)
Q.25 In golden rice, the gene phytoene desaturase (crtl) was obtained from :
(A) Narcissus species
(B) Erwinia uredovera
(C) Zea mays
(D) Oryza sativa
Answer – (B) Erwinia uredovora
Explanation:
The phytoene desaturase (crtI) gene was introduced from Erwinia uredovora to create Golden Rice, allowing β-carotene biosynthesis in rice endosperm to combat vitamin A deficiency.
Next: MH SET April 2024 Life Science Answer Key with Detailed Explanation
- MH SET April 2024 Life Science Answer Key: Part 3 (Will be uploaded on 10th June 2025)
- MH SET April 2024 Life Science Answer Key: Part 4 (Will be uploaded 11th June 2025)
- MH SET April 2024 Life Science Answer Key with Detailed Explanation: Part 1
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