MH SET Previous Year Question Paper 2024 Life Science – Solved Questions with Explanations
If you're planning to appear for the Maharashtra State Eligibility Test (MH SET) 2025 in Life Science, one of the smartest ways to boost your preparation is by solving previous year question papers. This blog post brings you a carefully curated set of MH SET Previous Year Question Paper 2024 Life Science – Solved Questions with Explanations —each one explained in detail to help you grasp the concept, not just memorize the answer.
Whether you're a first-time aspirant or someone trying again for better results, these solved questions will help you understand the exam pattern, identify high-weightage topics, and improve your accuracy under time pressure. But more importantly, practicing past questions teaches you how to think like the examiners—what kind of traps they set, how they phrase questions, and what topics they emphasize year after year.
2024 MH SET Life Science Question Paper (April Session)
Hydrogen bonds are present in water structure. Presence of hydrogen bonds gives a characteristic property to water, which is
(A) High melting and high boiling points
(B) High melting and low boiling points
(C) Low melting and high boiling points
(D) Low melting and low boiling points
Answer: A) High melting and high boiling points
Explanation: Hydrogen bonds increase water's intermolecular forces, requiring more energy (higher temperature) to break bonds during melting/boiling.Which of the following membrane-associated process involves the step of Clathrin coated vesicle formation during its mechanism ?
(A) Phagocytosis
(B) Amino acid uptake
(C) Receptor mediated endocytosis
(D) Calcium uptake
Answer: C) Receptor mediated endocytosis
Explanation: Clathrin-coated vesicles internalize specific ligands (e.g., LDL) via receptor binding.Which one of the following is not a part of cell theory ?
(A) Organism(s) are made up of cell(s)
(B) Cells are functional unit(s) of multicellular organisms
(C) Cells in a multicellular organism arise from preexisting cells
(D) Cells in a multicellular organism are derived from non-cells
Answer: D) Cells in a multicellular organism are derived from non-cells
Explanation: Cell theory states cells arise only from preexisting cells (omnis cellula e cellula).Assembly of cytoplasmic microtubule from the tubulin dimers is inhibited if the cell is treated with …………..
(A) Streptomycin
(B) Calmodulin
(C) Colchicine
(D) Carbenicillin
Answer: C) Colchicine
Explanation: Colchicine binds tubulin, preventing polymerization into microtubules.Prokaryotes may be characterised by :
(A) Presence of nucleus or nucleoid
(B) Absence of DNA
(C) Absence of membrane surrounding organelles
(D) Presence of membrane around RNA
Answer: C) Absence of membrane surrounding organelles
Explanation: Prokaryotes lack membrane-bound organelles (e.g., mitochondria, nucleus).The following are the characteristics of which type of transposable element? Simplest transposable element found in bacteria that encode transposase, with less than 5 kb size and with terminal inverted repeats that can transfer DNA directly is :
(A) Composite transposons
(B) Retrotransposons
(C) IS element
(D) Tn3-like elements
Answer: C) IS element
Explanation: Insertion sequences (IS) are small (<2.5 kb), encode transposase, and have inverted repeats.The first chromosomal linkage map for five genes on the X chromosome of Drosophila was constructed by :
(A) S. Benzer
(B) T.H. Morgan
(C) A.H. Sturtevant
(D) G.J. Mendel
Answer: C) A.H. Sturtevant
Explanation: Sturtevant created the first genetic map in 1913 using recombination frequencies.A chromosomal site at which a group of genes affecting quantitative traits are located is called as :
(A) Qualitative trait locus
(B) Quantitative trait locus
(C) Multiple allele locus
(D) Dominant locus
Answer: B) Quantitative trait locus
Explanation: QTLs are genomic regions associated with traits like height or yield.In prion disease, classically, the :
(A) Liver becomes excessively enlarged
(B) Liver becomes sponge like
(C) Brain ventricles are squeezed
(D) Brain becomes sponge like
Answer: D) Brain becomes sponge like
Explanation: Prions cause spongiform encephalopathies (e.g., Creutzfeldt-Jakob disease).Which one of the following is not a diversity index of living organism ?
(A) Shannon index
(B) Simpson index
(C) Berger-Parker index
(D) MPN index
Answer: D) MPN index
Explanation: MPN (Most Probable Number) estimates microbial abundance, not diversity.Articulation of both the jaws with skull is termed as Jaw suspension. One of the following types is observed in the teleost fishes :
(A) Craniostylic
(B) Autostylic
(C) Holostylic
(D) Hyostylic
Answer: D) Hyostylic
Explanation: Teleosts (e.g., bony fish) have hyostylic suspension where the jaw is braced by the hyomandibular bone.In the representatives of order Cestoda, the mature proglottids get separated from main body by a process known as :
(A) Apoptosis
(B) Apolysis
(C) Homolysis
(D) Genolysis
Answer: B) Apolysis
Explanation: Apolysis is the detachment of gravid proglottids in tapeworms for egg dispersal.Taxonomic hierarchy refers to :
(A) Classification of species based on fossil record
(B) A list of taxonomists worked on taxonomy of a species
(C) A group of taxonomists who decide nomenclature of animals and plants
(D) Stepwise arrangement of all categories for classification of animals and plants
Answer: D) Stepwise arrangement of all categories for classification of animals and plants
Explanation: Hierarchy includes ranks like domain, kingdom, phylum, etc.Mycoplasmas, rickettsiae, and chlamydiae are :
(A) Types of fungi
(B) Small bacteria
(C) Species of protozoa
(D) Forms of viruses
Answer: B) Small bacteria
Explanation: These are obligate intracellular or cell-wall-deficient bacteria.When biological oxygen demand becomes very high ?
(A) The water becomes suitable for growth of aerobic microbes
(B) The amount of dissolved oxygen increases
(C) The water becomes anaerobic
(D) Both concentration of microbes and dissolved oxygen decreases
Answer: C) The water becomes anaerobic
Explanation: High BOD depletes dissolved oxygen, creating anoxic conditions.Which one of the following extra embryonic membrane found as unique feature in terrestrial vertebrates only ?
(A) Yolk sac
(B) Chorion
(C) Allantois
(D) Amnion
Answer: D) Amnion
Explanation: Amnion forms a fluid-filled sac to prevent desiccation in amniotes (reptiles, birds, mammals).Since humans are having such a notable impact on the Earth and its life—a new, youngest epoch of the quaternary period is termed as :
(A) The Holocene
(B) The Paleocene
(C) The Anthropocene
(D) The Eocene
Answer: C) The Anthropocene
Explanation: Proposed epoch reflecting dominant human influence on geology/ecology.What largest bacterium of > 600 μm size is known to inhabit the gut of Acanthurus nigrofuscus ?
(A) Vibrio fischeri
(B) Spirillum volutans
(C) Epulopiscium fishelsoni
(D) Borrelia burgdorferi
Answer: C) Epulopiscium fishelsoni
Explanation: This symbiotic bacterium in surgeonfish gut is visible to the naked eye.Heliophytes are the plants :
(A) Growing under shade
(B) Growing under direct sunlight
(C) Growing under waterlogged conditions
(D) Growing under saline conditions
Answer: B) Growing under direct sunlight
Explanation: Heliophytes are sun-adapted (e.g., cacti, sunflower).The largest ‘carbon sink’ on the earth is :
(A) Forests
(B) Soils
(C) Oceans
(D) Animals
Answer: C) Oceans
Explanation: Oceans absorb ~50% of anthropogenic CO₂ via phytoplankton and dissolution.Nomenclature is governed by certain universal rules. Which one of the following is contrary to the rules of nomenclature ?
(A) The first word represents genus and second is a specific epithet
(B) The names are written in Latin and italicised
(C) Biological names can be written in any language
(D) When written by hand, names are to be underlined
Answer: C) Biological names can be written in any language
Explanation: Binomial names must be Latinized and italicized (e.g., Homo sapiens).Which among the following is functionally more diverse ?
(A) DNA
(B) RNA
(C) Protein
(D) Lipid
Answer: C) Protein
Explanation: Proteins perform catalysis, structure, transport, signaling, etc.River blindness disease occurs due to :
(A) Loa loa
(B) Mansonella perstans
(C) Mansonella streptocerca
(D) Onchocerca volvulus
Answer: D) Onchocerca volvulus
Explanation: This nematode causes onchocerciasis, transmitted by blackflies.Cri du chat syndrome is caused due to which of the following ?
(A) Deletion of short arm of chromosome no. 5
(B) Translocation of long arm of chromosome no. 5
(C) Duplication of short arm of chromosome no. 5
(D) Inversion of short arm of chromosome no. 5
Answer: A) Deletion of short arm of chromosome no. 5
Explanation: Partial monosomy of 5p (5p15.2) causes the disorder.If the genotypes of a husband and a wife are IAIB, how many different genotypes and phenotypes are possible among the blood types of their children?
(A) 2 genotypes : 3 phenotypes
(B) 3 genotypes : 4 phenotypes
(C) 4 genotypes : 3 phenotypes
(D) 4 genotypes : 4 phenotypes
Answer: D) 4 genotypes: 4 phenotypes
Explanation: Cross: IAIB × IAIB. Genotypes: IAIA, IAIB, IBIA, IBIB. Phenotypes: A, AB, BA, B (4 distinct).The maternal characteristics are inherited through cytoplasmic inheritance also. One of the following organelle is involved in :
(A) Mitochondria
(B) Golgi body
(C) Lysosomes
(D) Endoplasmic reticulum
Answer: A) Mitochondria
Explanation: Mitochondrial DNA is maternally inherited (e.g., in humans).Trisomy of chromosome no. 21 is :
(A) Edwards syndrome
(B) Klinefelter syndrome
(C) Down syndrome
(D) Patau syndrome
Answer: C) Down syndrome
Explanation: Trisomy 21 causes Down syndrome (47,XX/XY,+21).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 transformations were isolated on a minimal medium 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 supplies lys/val, so his+ transformants grow regardless of lys/val status.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: Nucleotides (e.g., ATP, cAMP) do not donate electrons; NADH/FADH₂ do.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: Chargaff's rules: A=T (30%), so G+C=40%. G=C, thus C=20%.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: Tryptophan, tyrosine, phenylalanine fluoresce under UV light.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.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: ω-angle (peptide bond) is planar (180°) due to partial double-bond character.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 measures catalytic efficiency (specificity constant).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: For equimolar weak acid/base, pH = pKa (Henderson-Hasselbalch equation).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: D) C, A and B
Explanation: SDS-PAGE separates by mass: largest (C, 79 kDa) at top, smallest (B, 28 kDa) at bottom.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 have multiple subspecies across regions (e.g., Panthera tigris).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 (e.g., Cooksonia) appeared ~430 MYA in the Silurian.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: Inclusive fitness includes direct + indirect (kin) fitness (Hamilton's rule).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: B) Record the absolute value of a signal
Explanation: Standards calibrate instruments to quantify unknowns (e.g., protein standards in ELISA).Why sexual selection can lead to extreme phenotypes ?
(A) Because the extreme phenotype incurs lower costs than the mean trait 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: Runaway selection amplifies traits via mate preference coevolution.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: C) 10 mya
Explanation: Molecular clock data showed hominid divergence ~10 MYA (Miocene).Beaker ‘A’ has 100 ml of some fluid at 80°C. Beaker B contains the same fluid 200 ml at 20°C. If both the fluids are mixed, what would be the temperature of the resultant mixture ?
(A) 25°C
(B) 65°C
(C) 40°C
(D) 50°C
Answer: C) 40°C
Explanation: Heat balance: (100 × 80) + (200 × 20) = (100 + 200) × T → T = 40°C.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 series provides high-resolution imagery for urban planning.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: Laser light is used in confocal microscopy, not phase contrast.What will be the radioactivity of 125 I 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: After 4 half-lives (240 days), activity = 2 mCi × (½)<sup>4</sup> = 0.125 mCi.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: Plants immobilize contaminants via root absorption or precipitation.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 act as chaperones to prevent denaturation under stress.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 uses fluorophore-labeled antibodies; enzymes (e.g., HRP) are for colorimetric assays.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: 4x × 2x yields 3x seedless fruit due to odd chromosome number.51. Amplification of a DNA fragment by PCR requires primers. Which of the following statements are correct for primers?
(i) Primers with hairpin structure formation are not desirable
(ii) Primer sequence should not have long runs (>3) of a single nucleotide
(iii) GC content of the primer should be less than 20%
(iv) Both the forward and reverse primers must have high percentage of base complementarities
(A) (i), (ii) and (iii)
(B) (ii), (iii) and (iv)
(C) (i) and (ii)
(D) (iii) and (iv)
Answer: C) (i) and (ii)
Explanation: Hairpins and homopolymers cause mispriming; GC content should be 40–60%.52. When subcultured into fresh medium, a bacterium grows with a lag phase of 2 hours followed by log phase and a stationary phase. If the bacterium was treated with ethidium bromide for 3 hours in all the phases of growth to cure a plasmid, it is more likely to lose the plasmid in :
(A) Early stationary phase
(B) Late stationary phase
(C) Logarithmic phase
(D) Lag phase
Answer: C) Logarithmic phase
Explanation: Actively dividing cells in log phase are more susceptible to plasmid-curing agents.53. The free hydroxyl groups present on either end of chemically synthesized oligonucleotides are 5'-phosphorylated with ATP by the enzyme :
(A) Polynucleotide phosphatase
(B) Polynucleotide phosphorylase
(C) Polynucleotide ATPase
(D) Polynucleotide kinase
Answer: D) Polynucleotide kinase
*Explanation: PNK transfers γ-phosphate from ATP to 5'-OH of DNA/RNA.*54. Which one of the following is not a fluorophore?
(A) Acridine orange
(B) Acrylamide
(C) Quinacrine
(D) Ethidium bromide
Answer: B) Acrylamide
Explanation: Acrylamide is a gel matrix polymer; others fluoresce under UV.55. Which of the following best describes principle of western blotting technique?
(A) A means of separation of restriction fragments on the basis of size
(B) It is based on attraction to a specific chemical group
(C) The proteins have a net charge zero at a particular pH
(D) It is an immunoassay technique preceded by electrophoresis
Answer: D) It is an immunoassay technique preceded by electrophoresis
Explanation: Proteins are separated by SDS-PAGE, then detected with antibodies.56. Which of the following is resulted due to photorespiration reaction catalysed by oxygenase activity of rubisco?
(A) 3-phosphoglycerate only
(B) 2-phosphoglycolate only
(C) Fructose 6-phosphate
(D) 3-phosphoglycerate and 2-phosphoglycolate
Answer: D) 3-phosphoglycerate and 2-phosphoglycolate
Explanation: Rubisco oxygenase produces both compounds in C2 cycle.57. Mutation in ACC synthase enzyme will inhibit synthesis of ......
(A) Gibberellins
(B) Salicyclic acid
(C) Ethylene
(D) Auxin
Answer: C) Ethylene
Explanation: ACC synthase converts SAM to ACC (ethylene precursor).58. Which of the following phytohormone accumulates during Systemic Acquired Resistance (SAR) in plants ?
(A) Salicylic acid
(B) Jasmonic acid
(C) Abscisic acid
(D) Ethylene
Answer: A) Salicylic acid
Explanation: SA is a key signaling molecule for pathogen defense in SAR.59. The hormone auxin was first discovered in :
(A) Mustard
(B) Pea
(C) Oats
(D) Rice
Answer: C) Oats
Explanation: Went isolated auxin from oat (Avena) coleoptiles in 1926.60. How many times more energy gained in aerobic respiration over anaerobic respiration?
(A) 8
(B) 12
(C) 18
(D) 32
Answer: C) 18
Explanation: Aerobic yields 36 ATP; anaerobic yields 2 ATP (net), ratio ≈18:1.61. Which of the following is NOT involved in electron transport during Photosystem I and II ?
(A) Ferredoxin
(B) Plastocyanin
(C) Cytochrome b6f complex
(D) Cytochrome oxidase
Answer: D) Cytochrome oxidase
Explanation: Cytochrome oxidase is in mitochondrial ETC; others are in chloroplasts.62. The treatment of abscisic acid is associated with ......
(A) Internode elongation
(B) Expansion of leaf area
(C) Closure of stomata
(D) Axillary root elongation
Answer: C) Closure of stomata
Explanation: ABA triggers stomatal closure under water stress.63. Action potential is generated due to :
(A) Influx of Na+ and efflux of K+
(B) Influx of Ca2+ and efflux of K+
(C) Influx of Na+and efflux of Ca2+
(D) Influx of K+ and efflux of Na+
Answer: A) Influx of Na+ and efflux of K+
Explanation: Depolarization by Na+ influx; repolarization by K+ efflux.64. Hookworm infection can lead to deficiency of :
(A) Vitamin B12
(B) Iron
(C) Vitamin B6
(D) Folic acid
Answer: B) Iron
Explanation: Ancylostoma causes blood loss, leading to iron-deficiency anemia.65. What kind of microbes grow best at higher CO2 tension (~ 10%) than are normally present in the atmosphere?
(A) Microaerophilic microorganisms
(B) Capnophilic microorganisms
(C) Aerotolerant anaerobic microorganisms
(D) Obligate anaerobic microorganisms
Answer: B) Capnophilic microorganisms
Explanation: Capnophiles (e.g., Neisseria) require 5–10% CO2 for growth.66. Which free living microorganisms occupy a niche to feed primarily on organic detritus from dead organisms and unable to adapt to the body of live host?
(A) Obligate saprobic microorganisms
(B) Parasitic microorganisms
(C) Facultative parasitic microorganisms
(D) Auxotrophic microorganisms
Answer: A) Obligate saprobic microorganisms
Explanation: Saprobes decompose dead matter and cannot infect hosts.67. The DNA of each chromosome occupies a defined volume of the nucleus and only overlaps with its immediate neighbours is called as:
(A) Chromosome territories
(B) Nuclear scaffolds
(C) Nucleosome assembly
(D) Chromatin
Answer: A) Chromosome territories
Explanation: Chromosomes occupy discrete nuclear regions with minimal overlap.68. Amphibians contain 30X more DNA than human beings. However human beings are more complex than amphibians. This lack of correlation is called as:
(A) Species specificity
(B) C-value paradox
(C) D-value paradox
(D) P-value paradox
Answer: B) C-value paradox
Explanation: C-value paradox describes genome size vs. organismal complexity disconnect.69. Which one of the following is not true in relation to eukaryotic gene expression?
(A) RNA Pol I synthesizes pre-rRNA
(B) RNA Pol II synthesizes rRNA
(C) RNA Pol III synthesizes tRNA
(D) All three polymerase contains β', β, α and ω subunits
Answer: B) RNA Pol II synthesizes rRNA
Explanation: RNA Pol II transcribes mRNA; rRNA is by Pol I.70. The triple helix domain of collagen consists of repeats of the X-Y-Z amino acids. The amino acid present at ‘X’ position is required in every third position in order for the polypeptide chain is:
(A) Lysine
(B) Proline
(C) Hydroxy proline
(D) Glycine
Answer: D) Glycine
Explanation: Collagen repeats Gly-X-Y; glycine fits in the helix core.71. Which DNA binding domain is present in steroid hormone receptors family of ligand inducible transcription factors?
(A) Helix-loop-helix
(B) Zinc fingers
(C) Helix-turn-helix
(D) Leucine zipper
Answer: B) Zinc fingers
Explanation: Steroid receptors use zinc finger motifs to bind DNA response elements.72. Which of the following elongation factor is called as translocase?
(A) EF-G
(B) EF-2
(C) EF-G and EF-2
(D) EF-Tu and EF-Ts
Answer: C) EF-G and EF-2
*Explanation: EF-G (prokaryotes) and EF-2 (eukaryotes) catalyze ribosomal translocation.*73. Which of the following post-translational modification is not related with epigenetic regulation?
(A) Histone acetylation
(B) Sumoylation
(C) Ubiquitination
(D) DNA methylation
Answer: B) Sumoylation
Explanation: SUMO modification regulates protein function, not chromatin structure.74. The disease which arises because of aberrant splicing is:
(A) Huntington’s disease
(B) Cystic fibrosis
(C) Sickle cell anemia
(D) Thalassemia
Answer: D) Thalassemia
Explanation: Mutations in splice sites of globin genes cause β-thalassemia.75. Identify the enzymes involved in DNA replication for their function :
(Enzyme)
(P) DNA polymerase I
(Q) Primase
(R) Helicase
(S) Gyrase
(Function)
(1) Unzipping the DNA helix
(2) Supercoiling of DNA
(3) Remove primer and close gaps
(4) Synthesize RNA primer
(A) (P)-(2),(Q)-(3),(R)-(4),(S)-(1)
(B) (P)-(3),(Q)-(4),(R)-(1),(S)-(2)
(C) (P)-(1),(Q)-(4),(R)-(2),(S)-(3)
(D) (P)-(3),(Q)-(1),(R)-(2),(S)-(4)
Answer: B) (P)-(3),(Q)-(4),(R)-(1),(S)-(2)
*Explanation: Pol I: Exonuclease + gap filling (3); Primase: RNA primer synthesis (4); Helicase: Unwinds DNA (1); Gyrase: Relieves supercoiling (2).
76. (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: 3'→5' exonuclease proofreads errors; 5'→3' synthesis direction enables proofreading.
77. 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: D) Rec A shows negative regulation of lex A
Explanation: RecA activates LexA cleavage (inactivation), not negative regulation.
78. 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 prevents content leakage (unlike necrosis); phagocytosis avoids inflammation.
79. 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: RA gradients specify hindbrain/cervical regions via Hox gene expression.
80. 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 nature conformation of protein antigen
Explanation: B cells recognize conformational epitopes; T cells recognize linear peptides via MHC.
81. 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: A) Antigenicity is the ability of the molecule to provocate immune response in the body
Explanation: Antigenicity = ability to bind immune receptors; immunogenicity = ability to provoke response.
82. 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 binds integrins, collagen, and fibrin for cell-matrix adhesion.
83. 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: E-cadherin maintains epithelial integrity during gastrulation movements.
84. 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 TFs form the core pluripotency network in embryonic stem cells.
85. Yersinia pestis is the causative agent for ......
(A) Gas gangrene
(B) Tularemia
(C) Lyme disease
(D) Plague
Answer: D) Plague
Explanation: Y. pestis causes bubonic/pneumonic plague.
86. Pseudohermaphroditism is a condition in which the individual is insensitive to ......
(A) Androgen
(B) Estrogen
(C) Progesterone
(D) Corticosterone
Answer: A) Androgen
Explanation: Androgen insensitivity syndrome (AIS) causes XY females.
87. 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: D) Chenopodial
Explanation: Chenopodial type (e.g., beet) has minimal basal cell contribution.
88. 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 (not DHT) promotes epididymis development.
89. 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: Zygotic genome activation occurs at MBT.
90. 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: N-cadherin mediates condensation of mesenchyme into cartilage.
91. 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
Explanation: Stem cells may be quiescent; markers define potency, not division status.
92. 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) binds FasL to initiate caspase cascade.
93. With reference to normal ECG, the T-wave is:
(A) Negative in all standard bipolar limb leads recording
(B) Caused by repolarization 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 the ventricles
Explanation: T-wave reflects ventricular repolarization.
94. 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 O<2 release property
(D) Increased degradation of haemoglobin
Answer: A) Increased O2 release from haemoglobin
*Explanation: Lactic acid lowers pH (Bohr effect), shifting O2-Hb curve right, enhancing O2 unloading.*
95. The capability of vertebrate kidney to 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: Countercurrent multiplier in the loop creates medullary osmotic gradient.
96. 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: B) Increased demand made by the muscles
Explanation: Metabolic demand triggers vasodilation and higher O2 extraction.
97. 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 (e.g., ClpB) disassembles aggregates for refolding/degradation.*
98. 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: Standard sequence: inoculate → ferment → recover product → dispose waste.
99. Match the following product/process to the microorganism involved :
Product/Process
(L) Biopol
(M) Biopesticide
(N) Bioleaching
(O) Bioremediation of oil
Microorganism
(1) Thiobacillus ferrooxidans
(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 (PHB plastic); Biopesticide: Bacillus (larvicide); Bioleaching: Thiobacillus (metal extraction); Oil bioremediation: Pseudomonas (hydrocarbon degradation).
100. In golden rice, the gene phytoene desaturase (crtI) was obtained from:
(A) Narcissus species
(B) Erwinia uredovera
(C) Zea mays
(D) Oryza sativa
Answer: B) Erwinia uredovera
Explanation: crtI from bacterium Erwinia enhances β-carotene synthesis.
Why Solving MH SET Previous Year Papers is a Game-Changer
Let’s be real—just reading textbooks or coaching materials isn't enough anymore. The competition is high, and every mark counts. That’s why solving actual previous year papers gives you a competitive edge. Here's why it matters:
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Familiarity with Exam Pattern: The MH SET follows a specific structure. Practicing old papers helps you get used to the types of questions asked, the marking scheme, and the time required per section.
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Spotting Repeated Concepts: Many core topics appear repeatedly across years. Going through previous papers lets you identify which units and chapters demand extra focus.
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Real-Time Practice: Treat each paper like a mock test. It trains your brain to handle pressure and manage time better.
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Boosts Confidence: The more familiar you are with the question style, the more confident you'll feel on exam day.
How to Use This Post to Your Advantage
This post isn’t just a dump of old questions—it’s a strategic learning tool. Each of the 100 questions comes with a clear explanation, breaking down the logic behind the correct answer. Here's how to get the most from it:
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Attempt the Question First: Don’t just read the answer—try solving it yourself before checking the explanation.
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Understand the Reasoning: Read the explanations thoroughly. They’re designed to teach you the concept, not just give you a shortcut.
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Track Topics: Make a list of recurring themes and subjects that show up often. This will help you prioritize your revision.
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Revise Smart: Revisit the tricky questions later. The ones you got wrong the first time are goldmines for learning.
