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IIT JAM Biotechnology study materials
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IIT JAM Biotechnology practice questions

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Question 1Biotechnology (BT)

In the lac operon of Escherichia coli, which protein binds the operator region to block transcription when lactose is absent?

  1. LacZ beta-galactosidase
  2. LacY permease
  3. CAP (catabolite activator protein)
  4. LacI repressor Correct answer.

Explanation

The lacI gene product is a tetrameric repressor that binds the operator and physically blocks RNA polymerase from initiating transcription. When allolactose accumulates, it binds the repressor and reduces operator affinity, allowing induction. CAP instead acts at the promoter as an activator of the same operon.

Question 2Biotechnology (BT)

A double-stranded DNA molecule contains 30% adenine. What percentage of its bases is guanine?

  1. 70%
  2. 30%
  3. 20% Correct answer.
  4. 40%

Explanation

By Chargaff's rules, adenine equals thymine on a double-stranded molecule, so A = 30% and T = 30%, leaving 40% for G plus C combined. Since G equals C, guanine is half of 40%, namely 20%. The calculation depends on complementary base pairing, not on the sequence order.

Question 3Biotechnology (BT)

Which enzyme is used to join two DNA fragments with compatible sticky ends during construction of a recombinant plasmid?

  1. DNA ligase Correct answer.
  2. DNA polymerase I
  3. Restriction endonuclease
  4. Alkaline phosphatase

Explanation

DNA ligase seals the phosphodiester backbone between adjacent nucleotides at the annealed cohesive ends, covalently joining vector and insert. Restriction enzymes cut DNA but do not seal nicks, and DNA polymerase extends primers rather than joining preformed ends.

Question 4Biotechnology (BT)

During which phase of the eukaryotic cell cycle does chromosome condensation and the formation of the mitotic spindle first become clearly evident?

  1. G1 phase
  2. S phase
  3. G2 phase
  4. M phase Correct answer.

Explanation

Chromosome condensation, nuclear envelope breakdown, and spindle assembly are hallmarks of mitosis, which constitutes the M phase. S phase is restricted to DNA replication and G2 to preparation and checkpoint control before division.

Question 5Biotechnology (BT)

Which feature distinguishes archaeal membrane lipids from those of bacteria and eukaryotes?

  1. They contain isoprenoid chains attached by ether linkages. Correct answer.
  2. They contain glycerol linked to fatty acids by ester bonds.
  3. They lack any hydrophobic component.
  4. They contain cholesterol as the major sterol.

Explanation

Archaeal lipids typically use sn-glycerol-1-phosphate with branched isoprenoid chains joined by ether bonds, which resist hydrolysis and heat. Bacterial and eukaryotic lipids use ester-linked fatty acids on glycerol-3-phosphate. These differences underpin the separate phylogenetic placement of Archaea.

Question 6Biotechnology (BT)

In enzyme kinetics, what does the Michaelis constant (Km) represent?

  1. The total enzyme concentration needed for saturation
  2. The maximum velocity achieved at saturating substrate
  3. The substrate concentration at which the reaction velocity is half of Vmax Correct answer.
  4. The rate constant for product formation from the enzyme-substrate complex

Explanation

Km is defined as the substrate concentration giving half-maximal velocity under steady-state assumptions. It is an inverse index of the apparent affinity of enzyme for substrate, so a low Km indicates tight binding. Vmax and kcat describe capacity and turnover rather than affinity.

Question 7Biotechnology (BT)

Which reagent is commonly used to detect reducing sugars by forming a coloured precipitate upon heating?

  1. Fehling's solution Correct answer.
  2. Sudan III
  3. Biuret reagent
  4. Ninhydrin

Explanation

Fehling's solution contains copper(II) ions that are reduced to insoluble copper(I) oxide, giving a brick-red precipitate with reducing sugars on heating. Biuret and ninhydrin detect peptide bonds and amino groups respectively, while Sudan dyes stain lipids.

Question 8Biotechnology (BT)

Which class of immunoglobulin is the first to be produced during a primary humoral immune response?

  1. IgG
  2. IgE
  3. IgM Correct answer.
  4. IgA

Explanation

Naive B cells express surface IgM, and plasma cells secrete pentameric IgM early in a primary response before class switching occurs. IgG dominates the secondary response, while IgA and IgE are specialised mucosal and allergic isotypes.

Question 9Biotechnology (BT)

In a polymerase chain reaction, what is the purpose of the initial denaturation step at about 95 degrees Celsius?

  1. To extend the primers using Taq polymerase
  2. To separate the two strands of the template DNA Correct answer.
  3. To allow primers to anneal to complementary sequences
  4. To inactivate the polymerase before cycling

Explanation

High temperature disrupts the hydrogen bonds between complementary bases, producing single-stranded templates for primer annealing. Annealing occurs at a lower, primer-specific temperature, and extension happens near 72 degrees Celsius. Without denaturation, primers cannot access the template.

Question 10Biotechnology (BT)

A culture of bacteria grows exponentially with a doubling time of 30 minutes. Starting from 1,000 cells, how many cells are present after 2 hours?

  1. 32,000
  2. 8,000
  3. 4,000
  4. 16,000 Correct answer.

Explanation

Two hours contains four 30-minute doubling intervals, so the population increases by a factor of 2^4 = 16. Multiplying 1,000 by 16 gives 16,000 cells. Exponential growth assumes constant doubling time and no lag or death phase.

Question 11Biotechnology (BT)

During glycolysis, which reaction is the principal site of regulation by phosphofructokinase-1?

  1. Conversion of glucose to glucose-6-phosphate
  2. Phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate Correct answer.
  3. Oxidation of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate
  4. Conversion of 3-phosphoglycerate to 2-phosphoglycerate

Explanation

Phosphofructokinase-1 catalyses the committed, rate-limiting step of glycolysis and is allosterically inhibited by ATP and citrate and activated by AMP and fructose-2,6-bisphosphate. The other listed steps are catalysed by hexokinase, phosphoglycerate mutase, and glyceraldehyde-3-phosphate dehydrogenase.

Question 12Biotechnology (BT)

A plasmid vector carries an ampicillin resistance gene containing a unique restriction site. After digestion and ligation with a foreign insert, which plating strategy directly selects for cells carrying a recombinant plasmid rather than the empty vector?

  1. Plate on ampicillin and screen for blue colonies with X-gal
  2. Plate on ampicillin and screen for white colonies with X-gal Correct answer.
  3. Plate on ampicillin only and count all colonies
  4. Plate on kanamycin and screen for blue colonies

Explanation

Insertional inactivation of the lacZ alpha fragment prevents beta-galactosidase activity, so recombinants remain white while empty vectors form blue colonies on X-gal. Ampicillin selects for any plasmid-bearing cell, so colour screening is needed to distinguish recombinants. Both blue and white colonies survive ampicillin.

Question 13Biotechnology (BT)

In a mating experiment with an Hfr strain, which process explains the transfer of chromosomal genes to a recipient cell?

  1. Transfer of a single-stranded DNA segment starting from the origin of transfer, often interrupted before completion Correct answer.
  2. Uptake of naked DNA from the surrounding medium
  3. Transduction mediated by a bacteriophage particle
  4. Transfer of the entire chromosome through a conjugation pilus within minutes

Explanation

In Hfr strains the F factor is integrated into the chromosome, so rolling-circle transfer begins at the origin of transfer and proceeds into chromosomal DNA. Because the full chromosome is large, the mating bridge usually breaks before transfer is complete, giving a gradient of gene transfer. This is why interrupted-mating mapping works.

Question 14Biotechnology (BT)

Which type of mutation is most likely to cause a complete loss of protein function by introducing a premature stop codon?

  1. Silent mutation
  2. Nonsense mutation Correct answer.
  3. Missense mutation
  4. Frameshift mutation that restores the reading frame

Explanation

A nonsense mutation changes a sense codon into a termination codon, producing a truncated polypeptide that usually lacks function. Silent mutations leave the amino acid unchanged, and missense mutations substitute one residue. A compensatory frameshift restoring the reading frame may partially recover function.

Question 15Biotechnology (BT)

Which characteristic makes 16S ribosomal RNA suitable for inferring phylogenetic relationships among bacteria?

  1. It is highly variable in every region, allowing species-level discrimination alone
  2. It evolves faster than any protein-coding gene
  3. It contains conserved regions flanking variable regions, allowing universal amplification and comparative sequencing Correct answer.
  4. It is encoded by a single-copy gene in all bacteria

Explanation

The 16S rRNA gene combines highly conserved stretches that permit binding of universal primers with variable regions that carry phylogenetic signal. This mosaic structure allows alignment and comparison across diverse taxa. Copy number and variable-region resolution vary, so it is not a perfect species-level marker.

Question 16Biotechnology (BT)

In SDS-PAGE, what is the principal role of sodium dodecyl sulfate?

  1. It selectively stains basic proteins
  2. It cross-links proteins covalently to the gel matrix
  3. It denatures proteins and imparts a uniform negative charge-to-mass ratio Correct answer.
  4. It reduces disulfide bonds between cysteine residues

Explanation

SDS is an anionic detergent that unfolds proteins and binds at a roughly constant ratio per unit mass, giving them a similar charge density. Migration in the electric field then depends mainly on molecular size through the gel mesh. Disulfide reduction is achieved separately by agents such as beta-mercaptoethanol or DTT.

Question 17Biotechnology (BT)

Which feature of an enzyme distinguishes a competitive inhibitor from a non-competitive inhibitor in kinetic analysis?

  1. Both inhibitors lower Vmax and raise Km equally
  2. Competitive inhibition raises apparent Km with unchanged Vmax; non-competitive inhibition lowers Vmax with unchanged Km Correct answer.
  3. Competitive inhibition lowers Vmax with unchanged Km; non-competitive inhibition raises Km with unchanged Vmax
  4. Competitive inhibition lowers both Km and Vmax; non-competitive inhibition raises both

Explanation

A competitive inhibitor binds the free enzyme active site and can be overcome by high substrate, so Vmax is unchanged but apparent Km increases. A pure non-competitive inhibitor binds equally to enzyme and ES complex, reducing Vmax without altering Km. These patterns are read from Lineweaver-Burk plots.

Question 18Biotechnology (BT)

In a dihybrid cross between two heterozygotes (AaBb x AaBb) with independent assortment and complete dominance, what fraction of the progeny shows both recessive phenotypes?

  1. 1/16 Correct answer.
  2. 1/4
  3. 3/16
  4. 9/16

Explanation

Each recessive phenotype appears with probability 1/4 in a monohybrid cross, and independent assortment allows multiplication: 1/4 x 1/4 = 1/16. The 9:3:3:1 ratio assigns 9/16 to both dominant traits and 3/16 to each single-recessive class.

Question 19Biotechnology (BT)

A restriction enzyme recognises the palindromic sequence GAATTC and cuts between G and A. How many fragments are produced when a circular plasmid of 5,000 bp containing two such sites is completely digested?

  1. Three fragments
  2. Four fragments
  3. Two fragments Correct answer.
  4. One fragment

Explanation

A circular molecule cut at two sites is opened at each position, yielding two linear fragments whose lengths sum to the plasmid size. In general, n cut sites in a circular DNA produce n fragments, whereas n sites in a linear molecule produce n+1 fragments. Complete digestion means every site is cleaved.

Question 20Biotechnology (BT)

Which statement best explains why Taq polymerase is used in standard PCR rather than a mesophilic DNA polymerase?

  1. It synthesises RNA primers needed for initiation
  2. It retains activity through the high-temperature denaturation steps of each cycle Correct answer.
  3. It can ligate Okazaki fragments during amplification
  4. It has proofreading exonuclease activity that ensures error-free synthesis

Explanation

Taq polymerase from Thermus aquaticus is thermostable and survives repeated exposure to about 95 degrees Celsius, so fresh enzyme is not needed each cycle. It lacks 3' to 5' proofreading, so it is not a high-fidelity enzyme. Its role is DNA-dependent DNA synthesis from primers.

Question 21Biotechnology (BT)

In the context of plant tissue culture, what is the role of the auxin-to-cytokinin ratio in the medium?

  1. It controls the pH of the medium during autoclaving
  2. It fixes the ploidy level of the regenerated plant
  3. It determines whether explants form roots or shoots during regeneration Correct answer.
  4. It selects against bacterial contamination

Explanation

A high auxin-to-cytokinin ratio favours root formation, whereas a low ratio favours shoot formation, and intermediate ratios promote callus proliferation. This balance is a central principle of organogenesis in plant tissue culture. pH and sterility are managed separately by buffers and sterilisation.

Question 22Biotechnology (BT)

A protein is to be expressed in E. coli, but the gene was cloned from a human cDNA containing introns. What is the most appropriate explanation for the failure to obtain the correct protein?

  1. E. coli lacks spliceosomes and cannot remove introns from the transcript Correct answer.
  2. E. coli ribosomes cannot translate eukaryotic codons
  3. E. coli lacks the ability to add a poly(A) tail
  4. E. coli RNA polymerase cannot transcribe human DNA

Explanation

Prokaryotes lack the spliceosomal machinery that removes introns, so a genomic eukaryotic gene with introns yields an incorrectly processed mRNA and a truncated or aberrant protein. Using cDNA, which is already intron-free, avoids this problem. Prokaryotic RNA polymerase and ribosomes can function with eukaryotic coding sequences.

Question 23Biotechnology (BT)

In a Lineweaver-Burk plot, a series of lines from experiments with increasing concentrations of a reversible inhibitor intersect at a single point on the x-axis. What does this pattern indicate?

  1. Irreversible inactivation, because the enzyme is destroyed
  2. Uncompetitive inhibition, because both Km and Vmax change
  3. Non-competitive inhibition, because Vmax falls while Km is unchanged
  4. Competitive inhibition, because apparent Km changes while Vmax is unchanged Correct answer.

Explanation

On a double-reciprocal plot the y-intercept equals 1/Vmax and the x-intercept equals -1/Km. Lines intersecting exactly on the x-axis share the same y-intercept, so Vmax is unchanged while the apparent Km increases, the signature of competitive inhibition. Non-competitive inhibition would instead give a common x-intercept with varying y-intercepts.

Question 24Biotechnology (BT)

A bacterial strain is grown in continuous culture at a fixed dilution rate, and a mutant with a higher maximum specific growth rate but a lower substrate affinity is introduced. Under which condition would the mutant most likely displace the original strain?

  1. When the dilution rate is zero and the culture is in batch mode
  2. At very low residual substrate concentration, where affinity dominates
  3. When the substrate is supplied in excess and growth is substrate-saturated
  4. At high dilution rates approaching washout, where the maximum growth rate dominates Correct answer.

Explanation

In continuous culture, the strain that maintains growth at the imposed dilution rate wins. At high dilution rates the specific growth rate must be large, so the strain with the higher maximum growth rate outcompetes the high-affinity strain. At low residual substrate, by contrast, the low-Ks strain is favoured.

Question 25Biotechnology (BT)

A diploid organism is heterozygous for a reciprocal translocation. During meiosis, which meiotic configuration and segregation pattern most often yields unbalanced gametes?

  1. A bivalent with normal disjunction, producing only balanced gametes
  2. A quadrivalent with adjacent segregation, producing duplication-deficiency gametes Correct answer.
  3. A quadrivalent with alternate segregation, producing only balanced gametes
  4. A univalent pair with random segregation, producing balanced gametes

Explanation

Homologous pairing in a reciprocal translocation heterozygote forms a cross-shaped quadrivalent. Alternate segregation sends homologous centromeres to opposite poles and yields balanced gametes, whereas adjacent segregation produces gametes with duplications and deletions. These unbalanced gametes are associated with reduced fertility.

Question 26Biotechnology (BT)

Which feature of the genetic code explains why a single-base substitution in a codon sometimes leaves the encoded amino acid unchanged?

  1. Universality of the code
  2. Presence of a single start codon
  3. Non-overlapping reading of codons
  4. Degeneracy of the code Correct answer.

Explanation

Degeneracy means that several different codons can specify the same amino acid, so a change at the third position often preserves the encoded residue. Universality and non-overlapping reading describe other properties of the code and do not by themselves allow silent substitutions.

Question 27Biotechnology (BT)

A bacterial culture is diluted serially and plated to obtain countable colonies. Which quantity does the viable plate count directly estimate?

  1. Total cell number including dead cells
  2. Mean generation time of the culture
  3. Number of colony-forming units per millilitre Correct answer.
  4. Intracellular ATP concentration

Explanation

A viable plate count enumerates only cells capable of dividing and forming a colony, reported as colony-forming units per millilitre. Direct microscopic counts or flow cytometry would also include non-viable cells, which the plate count does not.

Question 28Biotechnology (BT)

In a bacterial cloning workflow, which property of a plasmid is essential for it to be maintained in a host population under antibiotic selection?

  1. A multiple cloning site
  2. A selectable marker gene
  3. A polyhistidine tag
  4. An origin of replication recognised by the host Correct answer.

Explanation

A plasmid must carry an origin of replication that the host's machinery recognises so that it is replicated and inherited by daughter cells. The selectable marker provides the pressure for retention, but without replication the plasmid would be diluted out.

Question 29Biotechnology (BT)

Which structural feature distinguishes a Gram-positive bacterial cell envelope from a Gram-negative one?

  1. Presence of a capsule as the sole outermost layer
  2. A thick peptidoglycan layer without an outer membrane Correct answer.
  3. Absence of peptidoglycan and presence of pseudopeptidoglycan
  4. A thin peptidoglycan layer with an outer membrane

Explanation

Gram-positive cells retain crystal violet because they have a thick peptidoglycan layer and no outer membrane. Gram-negative cells have a thin peptidoglycan layer plus an outer membrane containing lipopolysaccharide, which is lost during decolourisation.

Question 30Biotechnology (BT)

Which nucleic acid hybridisation technique is used to transfer and detect specific DNA fragments separated by gel electrophoresis?

  1. Northern blotting
  2. Southern blotting Correct answer.
  3. Immunoprecipitation
  4. Western blotting

Explanation

Southern blotting transfers size-fractionated DNA from a gel onto a membrane and detects specific sequences with a labelled probe. Northern blotting is the analogous method for RNA, while western blotting detects proteins.

Question 31Biotechnology (BT)

In a tryptophan operon of Escherichia coli, what is the effect of abundant tryptophan on transcription of the trp genes?

  1. It has no effect because the operon is constitutive
  2. It increases transcription by acting as an inducer
  3. It causes premature termination of the mRNA at the promoter
  4. It decreases transcription by enabling the repressor to bind the operator Correct answer.

Explanation

Tryptophan acts as a co-repressor: when abundant, it binds the trp repressor and enables it to bind the operator, blocking transcription. This is negative regulation, in contrast to induction systems such as the lac operon.

Question 32Biotechnology (BT)

Which enzyme is responsible for synthesising the RNA primer needed by DNA polymerase during chromosomal replication?

  1. Primase Correct answer.
  2. DNA ligase
  3. Helicase
  4. Topoisomerase

Explanation

Primase is an RNA polymerase that lays down a short RNA primer, providing the free 3'-OH that DNA polymerase requires to begin synthesis. DNA ligase seals nicks, helicase unwinds duplex DNA, and topoisomerase relieves supercoiling.

Question 33Biotechnology (BT)

Which property of an antigen determines its ability to elicit an immune response without needing a carrier protein?

  1. Its ability to bind MHC molecules and be recognised by T-helper cells Correct answer.
  2. Its molecular mass being below one kilodalton
  3. Its solubility in organic solvents
  4. Its lack of repetitive epitopes

Explanation

Thymus-dependent antigens must be processed and presented on MHC molecules to activate T-helper cells, which then help B cells. Small molecules (haptens) are not immunogenic alone and require a carrier, and repetitive epitopes often favour T-independent responses.

Question 34Biotechnology (BT)

A DNA sample has an absorbance ratio A260/A280 of about 1.8. What does this indicate about the sample?

  1. It is relatively pure DNA Correct answer.
  2. It is heavily contaminated with protein
  3. It contains mostly RNA
  4. It is degraded into short fragments

Explanation

Pure double-stranded DNA typically shows an A260/A280 ratio near 1.8, whereas protein contamination lowers the ratio and pure RNA is closer to 2.0. A value near 1.8 therefore indicates relatively protein-free DNA.

Question 35Biotechnology (BT)

In an affinity chromatography step using immobilised nickel ions, which protein feature is exploited for purification?

  1. A polyhistidine tag Correct answer.
  2. A disulfide bond
  3. A signal peptide
  4. An N-linked glycan

Explanation

Immobilised metal affinity chromatography uses nickel or cobalt ions to bind the imidazole rings of a polyhistidine tag, allowing selective capture and imidazole-based elution. Glycans, disulfide bonds, and signal peptides do not provide this specific metal-binding interaction.

Question 36Biotechnology (BT)

A reaction mixture contains a competitive inhibitor of an enzyme. Which kinetic parameter changes in the presence of the inhibitor compared with the uninhibited reaction?

  1. Km increases while Vmax is unchanged Correct answer.
  2. Vmax increases while Km decreases
  3. Vmax decreases while Km is unchanged
  4. Both Vmax and Km decrease

Explanation

A competitive inhibitor binds the free enzyme at the active site and can be outcompeted by high substrate concentrations. This raises the apparent Km without changing Vmax, because at saturating substrate the inhibitor is displaced.

Question 37Biotechnology (BT)

During the polymerase chain reaction, why is the annealing temperature set below the melting temperature of the primers?

  1. To activate the polymerase by removing magnesium
  2. To prevent extension of the primers
  3. To allow primers to hybridise specifically to the template Correct answer.
  4. To denature the template DNA completely

Explanation

The annealing temperature is chosen a few degrees below the primer melting temperature so that primers can base-pair with complementary template sequences while non-specific binding is minimised. Denaturation occurs at the earlier high-temperature step, and extension follows annealing.

Question 38Biotechnology (BT)

A mutation changes a codon from GAG to GUG in the coding strand. What type of amino acid substitution results, given that GAG codes for glutamate and GUG for valine?

  1. A frameshift mutation
  2. A conservative substitution
  3. A nonsense mutation
  4. A missense mutation Correct answer.

Explanation

A single-base change that replaces one amino acid with a different one is a missense mutation. Because glutamate and valine differ in charge and polarity, the change is non-conservative, and no reading frame is shifted.

Question 39Biotechnology (BT)

In a batch culture, which phase immediately follows the exponential phase and is characterised by a constant cell number?

  1. Stationary phase Correct answer.
  2. Log phase
  3. Lag phase
  4. Death phase

Explanation

In the stationary phase, growth rate equals death rate, so the viable count remains roughly constant as nutrients become limiting and waste products accumulate. The lag phase precedes exponential growth, and the death phase follows the stationary phase.

Question 40Biotechnology (BT)

Which feature of a plasmid-based expression system allows tight control of recombinant protein production in Escherichia coli?

  1. Use of a low-copy-number origin
  2. Use of an inducible promoter such as the lac or T7 system Correct answer.
  3. Presence of a multiple cloning site
  4. Use of a constitutive promoter

Explanation

Inducible promoters allow expression to be switched on by adding an inducer such as IPTG, preventing premature expression and toxicity during biomass accumulation. A constitutive promoter drives continuous expression and offers no such control.

Question 41Biotechnology (BT)

A plant explant is cultured on a medium with a high auxin-to-cytokinin ratio. Which developmental outcome is favoured?

  1. Root formation Correct answer.
  2. Somatic embryogenesis only
  3. Shoot formation
  4. Callus proliferation without organogenesis

Explanation

In plant tissue culture, a high auxin-to-cytokinin ratio promotes root initiation, whereas a high cytokinin-to-auxin ratio favours shoot formation. An intermediate ratio favours callus proliferation.

Question 42Biotechnology (BT)

In a transformation experiment, competent Escherichia coli cells are heat-shocked after incubation with a plasmid. What is the purpose of the heat shock?

  1. To select for transformants
  2. To increase membrane permeability and allow plasmid entry Correct answer.
  3. To degrade the plasmid DNA
  4. To induce the lac operon

Explanation

The brief heat shock, following incubation with divalent cations, transiently increases the permeability of the cell envelope and promotes uptake of plasmid DNA. Selection of transformants is achieved separately with antibiotics.

Question 43Biotechnology (BT)

Which type of restriction endonuclease recognises a specific sequence and cleaves DNA at a defined position within or near that sequence?

  1. Type II Correct answer.
  2. Type III
  3. Type I
  4. Type IV

Explanation

Type II restriction endonucleases cleave at defined positions within or near their recognition sequences, which makes them useful for cloning. Type I and III enzymes cut at variable distances from the recognition site, and Type IV enzymes target modified DNA.

Question 44Biotechnology (BT)

A gene is cloned into a plasmid downstream of a promoter, but the protein produced is truncated and non-functional. Which factor most likely explains this outcome?

  1. Absence of a polyhistidine tag
  2. Use of a defined minimal medium
  3. High copy number of the plasmid
  4. Presence of a premature stop codon within the cloned insert Correct answer.

Explanation

A premature stop codon within the cloned coding sequence causes premature termination of translation, yielding a truncated protein. Copy number, tag absence, and medium composition affect yield or purification but not the length of the translated product.

Question 45Biotechnology (BT)

Two genes are located 20 map units apart on the same chromosome. A dihybrid in coupling phase is testcrossed. What proportion of progeny is expected to be recombinant?

  1. 20 percent Correct answer.
  2. 40 percent
  3. 80 percent
  4. 10 percent

Explanation

Recombination frequency equals the map distance in centimorgans, so genes 20 map units apart recombine in 20 percent of meiotic products. In a testcross, this corresponds to 20 percent recombinant progeny, split evenly between the two recombinant classes.

Question 46Biotechnology (BT)

Which type of enzyme is used to join two DNA fragments that have compatible cohesive ends?

  1. Alkaline phosphatase
  2. DNA ligase Correct answer.
  3. Exonuclease III
  4. DNA polymerase I

Explanation

DNA ligase catalyses the formation of a phosphodiester bond between adjacent nucleotides, sealing nicks between compatible cohesive or blunt ends. DNA polymerase extends primers, alkaline phosphatase removes terminal phosphates, and exonucleases degrade DNA.

Question 47Biotechnology (BT)

A mutant bacterial strain shows constitutive expression of the lac operon even in the absence of lactose. Which genetic lesion most plausibly explains this phenotype?

  1. A mutation in the lacY permease gene
  2. Loss of function of the lac repressor gene Correct answer.
  3. Loss of function of the cAMP receptor protein
  4. A defect in the lacZ gene

Explanation

If the lac repressor cannot bind the operator, transcription proceeds even without an inducer, giving a constitutive phenotype. A defect in the catabolite activator protein would instead reduce expression when glucose is present, and lacZ or lacY mutations affect enzyme activity or uptake rather than regulation.

Question 48Biotechnology (BT)

In an enzyme-catalysed reaction, a mutation lowers the turnover number kcat but leaves the Michaelis constant Km essentially unchanged. What does this indicate about the mutation?

  1. It impairs substrate binding but not catalysis
  2. It increases the affinity for the substrate
  3. It converts the enzyme into a competitive inhibitor
  4. It impairs the catalytic step but not substrate binding Correct answer.

Explanation

Km reflects the balance of substrate binding and catalysis, while kcat reflects the rate of product formation per active site. An unchanged Km with a reduced kcat suggests the mutation affects the chemical step rather than the affinity for substrate.

Question 49Biotechnology (BT)

A bacterial strain is grown in continuous culture at a fixed dilution rate. A mutant with a higher maximum specific growth rate but a higher substrate affinity constant (Ks) is introduced. Under which condition would the mutant outcompete the wild type?

  1. When both strains have identical Ks values
  2. When the dilution rate exceeds the mutant's maximum growth rate
  3. At high residual substrate concentrations where growth rate is near maximal Correct answer.
  4. At low residual substrate concentrations where the wild type has the higher affinity

Explanation

At high residual substrate concentrations, growth rate approaches the maximum specific growth rate, so the strain with the higher maximum growth rate wins. At low substrate concentrations, the strain with the lower Ks has the advantage, which favours the wild type here.

Question 50Biotechnology (BT)

A heterozygous individual carries a balanced reciprocal translocation. During meiosis, which configuration is formed to allow homologous pairing, and what segregation pattern yields viable balanced gametes?

  1. A univalent, with random segregation producing balanced gametes
  2. A quadrivalent, with adjacent-1 segregation producing balanced gametes
  3. A quadrivalent, with alternate segregation producing balanced gametes Correct answer.
  4. A bivalent, with adjacent segregation producing balanced gametes

Explanation

A reciprocal translocation heterozygote forms a quadrivalent during pairing. Alternate segregation sends one normal and one translocated chromosome to each pole, producing balanced gametes, whereas adjacent segregation yields unbalanced products.