I85PREP
⌘ K
GATE BT study materials
GATE BT study resources

GATE BT practice questions

One question at a time. Build your GATE BT understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Biotechnology (BT)

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

  1. LacI repressor Correct answer.
  2. LacZ β-galactosidase
  3. CAP–cAMP complex
  4. LacY permease

Explanation

The lacI gene product is a tetrameric repressor that binds the operator and physically blocks RNA polymerase from initiating transcription. Its affinity for DNA drops when it binds allolactose, the inducer derived from lactose. CAP–cAMP acts as an activator at low glucose, not as the operator-binding repressor.

Question 2Biotechnology (BT)

Which enzyme converts hydrogen peroxide into water and oxygen and is widely used to remove peroxide from enzyme-linked assays?

  1. Catalase Correct answer.
  2. Peroxidase
  3. Glucose oxidase
  4. Superoxide dismutase

Explanation

Catalase catalyses the dismutation of two hydrogen peroxide molecules into water and molecular oxygen. It is routinely added to reaction mixtures to quench residual peroxide that would otherwise inhibit enzymes or interfere with detection. Superoxide dismutase acts on superoxide radicals rather than peroxide.

Question 3Biotechnology (BT)

A plasmid vector of 5.0 kb is digested with a restriction enzyme that cuts once. The linear product is analysed by agarose gel electrophoresis alongside a 1 kb ladder. Where should the band appear?

  1. At the 2.5 kb position
  2. At the 5.0 kb position Correct answer.
  3. At the 1.0 kb position
  4. At the well, because linear DNA does not migrate

Explanation

A single cut converts a circular plasmid into one linear molecule whose length equals the total plasmid size, so it migrates at 5.0 kb. Circular supercoiled DNA of the same mass migrates faster than linear DNA, but after digestion the linear form is the only species. The ladder simply provides size reference points.

Question 4Biotechnology (BT)

In a PCR cycle, the step performed at approximately 72 °C is responsible for which event?

  1. Annealing of primers to the template
  2. Inactivation of uracil-DNA glycosylase
  3. Extension of primers by thermostable polymerase Correct answer.
  4. Denaturation of the template duplex

Explanation

At about 72 °C, a thermostable DNA polymerase such as Taq extends primers along the template, synthesising the new strand. Denaturation occurs near 95 °C and primer annealing typically between 50 and 65 °C. The extension temperature is chosen to balance polymerase activity against primer–template stability.

Question 5Biotechnology (BT)

Which statement correctly describes the relationship between Km and substrate affinity for a Michaelis–Menten enzyme?

  1. Km changes proportionally with enzyme concentration
  2. Km equals Vmax divided by the substrate concentration
  3. A higher Km indicates higher affinity of the enzyme for substrate
  4. A lower Km indicates higher affinity of the enzyme for substrate Correct answer.

Explanation

Km is the substrate concentration at which the reaction proceeds at half of Vmax, so a small Km means only a little substrate is needed to half-saturate the enzyme, indicating tight binding. Km is an intrinsic constant of the enzyme–substrate pair and does not change with enzyme concentration. It is not defined as a ratio of Vmax to substrate concentration.

Question 6Biotechnology (BT)

Which pair of restriction enzymes produces compatible cohesive ends that can be ligated without prior modification?

  1. EcoRI and BamHI
  2. HindIII and SmaI
  3. BamHI and BglII Correct answer.
  4. EcoRI and SmaI

Explanation

BamHI (G^GATCC) and BglII (A^GATCT) both leave a 5'-GATC overhang, so their products anneal and can be sealed by DNA ligase. Compatibility depends on identical single-stranded overhangs, not on recognition-site identity. EcoRI leaves AATT ends and SmaI leaves blunt ends, so neither pairs with the others listed.

Question 7Biotechnology (BT)

During anaerobic growth, Saccharomyces cerevisiae converts pyruvate to ethanol primarily to achieve which outcome?

  1. Net synthesis of ATP by oxidative phosphorylation
  2. Regeneration of NAD+ so glycolysis can continue Correct answer.
  3. Formation of acetyl-CoA for the TCA cycle
  4. Fixation of carbon dioxide into biomass

Explanation

Under anaerobiosis the electron transport chain cannot reoxidise NADH, so yeast reduces acetaldehyde to ethanol, regenerating NAD+ needed by glyceraldehyde-3-phosphate dehydrogenase. The energy yield comes from substrate-level phosphorylation in glycolysis, not from oxidative phosphorylation. Ethanol formation is therefore a redox-balancing strategy.

Question 8Biotechnology (BT)

A DNA sample has an absorbance ratio A260/A280 of 1.9. What does this value most likely indicate?

  1. Relatively pure double-stranded DNA Correct answer.
  2. Substantial RNA contamination
  3. Extensive DNA degradation to nucleotides
  4. Substantial protein contamination

Explanation

Pure double-stranded DNA typically gives an A260/A280 ratio of about 1.8 to 1.9, so this reading suggests little protein contamination. Protein absorbs strongly near 280 nm and lowers the ratio when present. RNA contamination is better detected by the A260/A230 ratio and by gel analysis.

Question 9Biotechnology (BT)

In a batch culture of E. coli, the specific growth rate is 0.35 h⁻¹. What is the doubling time?

  1. 0.35 h
  2. 2.86 h
  3. 1.98 h Correct answer.
  4. 0.69 h

Explanation

Doubling time equals ln2 divided by the specific growth rate, so 0.693/0.35 h⁻¹ gives about 1.98 h. The relationship assumes exponential growth with a constant specific rate. Using the rate itself or its reciprocal gives dimensionally or numerically incorrect answers.

Question 10Biotechnology (BT)

A protein of 60 kDa is to be separated from a 25 kDa contaminant using size-exclusion chromatography. Which outcome is expected?

  1. The 60 kDa protein binds the column matrix and elutes only after a salt gradient
  2. The 25 kDa protein elutes first because it enters more pores
  3. The 60 kDa protein elutes first because it is excluded from more pores Correct answer.
  4. Both proteins co-elute because size exclusion does not resolve by mass

Explanation

In size-exclusion chromatography, large molecules cannot enter the porous beads and travel through the interstitial volume, so they elute earlier. Smaller molecules penetrate the pores and follow a longer path, eluting later. The method therefore separates by hydrodynamic size rather than by binding affinity.

Question 11Biotechnology (BT)

Which primer design feature most directly reduces the risk of amplifying a non-target genomic region?

  1. A high GC content of about 80% across the primer
  2. A 3' end complementary to a repeated sequence
  3. A long self-complementary hairpin at the 5' end
  4. A primer length of 18–22 nucleotides with a unique match to the target Correct answer.

Explanation

Primers of roughly 18–22 nucleotides that match a unique sequence in the template give sufficient specificity and acceptable melting temperature. Uniqueness of the match is what prevents priming at related or repeated loci. Extremes of GC content or self-complementarity instead promote mispriming and primer dimers.

Question 12Biotechnology (BT)

In a chemostat operating at steady state, increasing the dilution rate while keeping the feed substrate concentration constant will, up to washout, cause which change?

  1. The biomass concentration increases indefinitely
  2. The specific growth rate falls below the dilution rate
  3. The residual substrate concentration decreases
  4. The residual substrate concentration increases Correct answer.

Explanation

At steady state the specific growth rate equals the dilution rate, so a higher dilution rate demands a higher growth rate. Because growth rate rises with substrate concentration according to Monod kinetics, the residual substrate must increase. Biomass declines as more substrate is diverted to maintenance and growth at the higher rate.

Question 13Biotechnology (BT)

A recombinant protein expressed in E. coli forms insoluble inclusion bodies. Which strategy is most appropriate to recover active protein?

  1. Solubilise the pellet in urea and refold by controlled removal of denaturant Correct answer.
  2. Add a protease inhibitor cocktail and load the pellet on a gel-filtration column
  3. Increase the induction temperature to 42 °C to boost folding
  4. Lyse cells in isotonic buffer and purify the soluble fraction directly

Explanation

Inclusion bodies consist of misfolded, aggregated protein that must first be solubilised with a chaotrope such as urea or guanidinium chloride. The denaturant is then removed gradually to allow refolding into the native conformation. Simply purifying the pellet without denaturation yields inactive aggregates.

Question 14Biotechnology (BT)

Which property makes ethidium bromide useful for detecting DNA in agarose gels?

  1. It covalently binds DNA and terminates replication
  2. It absorbs visible light and precipitates DNA in the gel
  3. It fluoresces strongly when intercalated into double-stranded DNA under UV light Correct answer.
  4. It stains proteins selectively after transfer to membranes

Explanation

Ethidium bromide intercalates between stacked base pairs, and its fluorescence quantum yield increases markedly upon binding to double-stranded DNA. Illumination with ultraviolet light then reveals DNA bands as bright orange signals. The dye is not a covalent cross-linker and does not selectively stain protein.

Question 15Biotechnology (BT)

In a fed-batch fermentation, feeding glucose at a rate that keeps the residual concentration near zero is intended mainly to achieve which effect?

  1. Eliminate the need for oxygen supply
  2. Avoid catabolite repression and overflow metabolism such as acetate formation Correct answer.
  3. Shorten the lag phase to zero
  4. Increase the maximum specific growth rate beyond the organism's limit

Explanation

Keeping glucose low in fed-batch culture limits the flux through glycolysis, reducing overflow metabolism and acetate accumulation that arise from catabolite repression and limited respiratory capacity. This supports high biomass and product yields. Feeding does not raise the organism's intrinsic maximum growth rate or remove the oxygen requirement.

Question 16Biotechnology (BT)

Which technique would best determine the secondary-structure content of a purified protein in solution?

  1. Isoelectric focusing
  2. Sodium dodecyl sulfate polyacrylamide gel electrophoresis
  3. Circular dichroism spectroscopy in the far-UV range Correct answer.
  4. Enzyme-linked immunosorbent assay

Explanation

Far-UV circular dichroism between about 190 and 250 nm reports on the amide backbone environment and allows estimation of α-helix, β-sheet, and random-coil content. The measurement is made in solution without labelling. Electrophoresis and immunoassays give mass or binding information but not secondary-structure composition.

Question 17Biotechnology (BT)

A gene is cloned downstream of a T7 promoter and expressed in E. coli BL21(DE3). What role does isopropyl β-D-1-thiogalactopyranoside (IPTG) play?

  1. It binds the Lac repressor and induces expression of T7 RNA polymerase Correct answer.
  2. It acts as a substrate for T7 RNA polymerase
  3. It removes the T7 lysozyme inhibitor by proteolysis
  4. It directly activates the T7 promoter by binding its DNA sequence

Explanation

In DE3 lysogens the T7 RNA polymerase gene is controlled by a lac promoter, so IPTG binding to the Lac repressor relieves repression and allows polymerase expression. The T7 polymerase then transcribes the target gene from the T7 promoter. IPTG is a non-metabolisable inducer and does not bind DNA or serve as a substrate.

Question 18Biotechnology (BT)

A fed-batch process shows a sudden rise in dissolved oxygen and a simultaneous fall in oxygen uptake rate while biomass is still high. Which explanation is most plausible?

  1. The feed rate was increased and growth accelerated
  2. The temperature controller failed low, slowing metabolism
  3. The culture has been poisoned or lysed, so respiration has stopped Correct answer.
  4. A new carbon source was added that supports faster growth

Explanation

A sharp rise in dissolved oxygen together with a falling oxygen uptake rate indicates that the culture has stopped consuming oxygen, as occurs with lysis, contamination by a non-respiring organism, or toxic shock. Faster growth or a new carbon source would increase oxygen demand and lower dissolved oxygen. These two signals moving in opposite directions are diagnostic of a metabolic shutdown.

Question 19Biotechnology (BT)

Which feature distinguishes an expression vector from a simple cloning vector?

  1. A selectable marker for plasmid maintenance
  2. A promoter and ribosome-binding site positioned to drive insert expression Correct answer.
  3. An origin of replication for propagation in the host
  4. A multiple cloning site with several unique restriction sites

Explanation

Expression vectors carry regulatory elements such as a promoter, operator, and ribosome-binding site upstream of the insert so that the cloned gene is transcribed and translated. Cloning vectors generally lack these elements and serve only to carry and propagate DNA. Both vector types typically share markers, origins, and cloning sites.

Question 20Biotechnology (BT)

In a two-stage continuous culture with a recycle loop, which parameter is fixed by the operator rather than by the organism's kinetics?

  1. The saturation constant for the limiting substrate
  2. The specific growth rate of the cells Correct answer.
  3. The maintenance coefficient
  4. The yield coefficient for substrate

Explanation

In continuous culture at steady state the specific growth rate is set equal to the dilution rate, which the operator controls through the flow and working volume. Yield, maintenance, and saturation constants are biological parameters of the organism and medium. This is why continuous culture is used to study growth-rate-dependent physiology.

Question 21Biotechnology (BT)

A protein binds its ligand with a dissociation constant Kd of 10 nM. In an equilibrium binding assay at a ligand concentration of 10 nM, what fraction of binding sites is occupied, assuming simple one-site binding?

  1. 0.50 Correct answer.
  2. 0.25
  3. 0.10
  4. 0.90

Explanation

For one-site binding the fractional occupancy equals [L] divided by (Kd + [L]). Substituting 10 nM for both terms gives 10/20, or 0.50. At a ligand concentration equal to Kd, half the sites are saturated by definition. Higher or lower fractions correspond to ligand concentrations away from Kd.

Question 22Biotechnology (BT)

In a plasmid with a pMB1 origin, a ColE1-derived copy-number control mechanism is disrupted. Which consequence is most likely?

  1. Selection markers are lost because antibiotic resistance genes are silenced
  2. Plasmid copy number rises, increasing metabolic burden and instability Correct answer.
  3. The plasmid integrates into the chromosome at high frequency
  4. The plasmid becomes unable to replicate because the origin is inactivated

Explanation

ColE1 replication is controlled by an antisense RNA and a small protein that limit primer formation, so disrupting this control raises copy number. Higher copy number increases the metabolic burden and can promote plasmid loss or structural instability. The origin remains functional, and integration is not a direct consequence of copy-number dysregulation.

Question 23Biotechnology (BT)

A fed-batch culture is described by Monod kinetics with a maximum specific growth rate of 0.5 h⁻¹ and a saturation constant of 0.2 g/L. If the operator wishes to hold the specific growth rate at 0.25 h⁻¹, what residual substrate concentration must be maintained?

  1. 0.10 g/L
  2. 0.05 g/L
  3. 0.20 g/L Correct answer.
  4. 0.40 g/L

Explanation

Rearranging the Monod equation gives S equal to Ks multiplied by μ, divided by (μmax − μ). Substituting 0.2 g/L, 0.25 h⁻¹, and 0.5 h⁻¹ yields 0.2 × 0.25 / 0.25, which is 0.20 g/L. This illustrates that half of the maximum growth rate is achieved at a substrate concentration equal to Ks.

Question 24Biotechnology (BT)

In a protein A affinity purification, the eluate is collected at low pH and immediately neutralised. Which phenomenon is being prevented by the rapid neutralisation?

  1. Non-specific binding of host-cell proteins to the resin
  2. Aggregation and loss of activity from prolonged acid exposure Correct answer.
  3. Cleavage of the Fc region by resin-bound proteases
  4. Oxidation of methionine residues by dissolved oxygen

Explanation

Low pH is needed to dissociate the antibody from protein A, but prolonged exposure can unfold the Fc region and drive aggregation. Rapid neutralisation into a buffered solution limits this denaturation. Host-cell protein binding and oxidation are separate concerns not addressed by pH neutralisation timing.

Question 25Biotechnology (BT)

A chemostat is run at a dilution rate of 0.4 h⁻¹ with an organism whose maximum specific growth rate is 0.35 h⁻¹. What is the predicted steady-state outcome?

  1. Higher biomass concentration than at a dilution rate of 0.2 h⁻¹
  2. Stable steady state with residual substrate above the feed value
  3. Washout of biomass because growth cannot match the dilution rate Correct answer.
  4. Oscillating biomass with a constant mean value

Explanation

When the dilution rate exceeds the maximum specific growth rate, the organism cannot grow fast enough to replace cells removed in the outflow, so biomass washes out. A true steady state is impossible because the required growth rate exceeds μmax. Residual substrate then approaches the feed concentration rather than exceeding it.

Question 26Biotechnology (BT)

In the Sanger dideoxy sequencing reaction, chain termination occurs because a dideoxynucleotide lacks which chemical group?

  1. The 2'-hydroxyl group of the sugar
  2. The 5'-phosphate group of the sugar
  3. The 3'-hydroxyl group of the sugar Correct answer.
  4. The nitrogenous base

Explanation

Incorporation of a dideoxynucleotide stops chain elongation because the absence of the 3'-hydroxyl prevents formation of the next phosphodiester bond. The 5'-triphosphate is still present and is required for incorporation, and the base itself is unchanged.

Question 27Biotechnology (BT)

Which type of RNA carries the amino acid to the ribosome during translation?

  1. Ribosomal RNA
  2. Small nuclear RNA
  3. Messenger RNA
  4. Transfer RNA Correct answer.

Explanation

Transfer RNA (tRNA) is charged with a specific amino acid by aminoacyl-tRNA synthetase and delivers it to the ribosome by pairing its anticodon with the mRNA codon. Ribosomal RNA forms the core of the ribosome, snRNA participates in splicing, and mRNA carries the coding sequence.

Question 28Biotechnology (BT)

In the polymerase chain reaction, what is the principal role of the initial denaturation step at about 95 °C?

  1. To allow primers to anneal to the template
  2. To separate the two strands of the template DNA Correct answer.
  3. To extend the primers with a thermostable polymerase
  4. To inactivate contaminating proteins

Explanation

Heating to roughly 95 °C breaks the hydrogen bonds between complementary bases, converting double-stranded template into single strands that can be accessed by primers. Primer annealing and polymerase extension occur at the lower temperatures of the subsequent steps.

Question 29Biotechnology (BT)

A bacterial culture is enumerated by plating serial dilutions on agar and counting colonies. If 0.1 mL of a 10⁻⁵ dilution yields 80 colonies, what is the estimated concentration of colony-forming units in the original culture?

  1. 8.0 × 10⁵ CFU/mL
  2. 8.0 × 10⁸ CFU/mL
  3. 8.0 × 10⁷ CFU/mL Correct answer.
  4. 8.0 × 10⁶ CFU/mL

Explanation

The concentration equals the colony count divided by the plated volume and multiplied by the dilution factor: 80 / 0.1 mL × 10⁵ = 8.0 × 10⁷ CFU/mL. Counting only plates with 30–300 colonies keeps the estimate statistically reliable.

Question 30Biotechnology (BT)

Which class of enzymes catalyses the cleavage of internal phosphodiester bonds in a DNA molecule?

  1. Ligases
  2. Topoisomerases
  3. Endonucleases Correct answer.
  4. Exonucleases

Explanation

Endonucleases hydrolyse phosphodiester bonds within the DNA chain, generating internal breaks. Exonucleases remove nucleotides from the ends, ligases join ends, and topoisomerases change supercoiling rather than acting as general internal nucleases.

Question 31Biotechnology (BT)

In a stirred-tank bioreactor, the dissolved oxygen concentration is maintained by which operating variable?

  1. Temperature set point
  2. pH set point
  3. Agitation rate and aeration rate Correct answer.
  4. Back-pressure only

Explanation

Oxygen transfer from gas to liquid depends on the volumetric mass transfer coefficient kLa and the concentration driving force, and kLa rises with agitation and aeration. pH and temperature affect metabolism but do not directly supply oxygen, and back-pressure changes solubility only modestly.

Question 32Biotechnology (BT)

Which biological agent is the causative organism of tuberculosis?

  1. Neisseria meningitidis
  2. Mycobacterium tuberculosis Correct answer.
  3. Corynebacterium diphtheriae
  4. Streptococcus pneumoniae

Explanation

Tuberculosis is caused by Mycobacterium tuberculosis, an acid-fast bacillus with a lipid-rich cell wall that resists decolourisation. The other organisms cause pneumonia, diphtheria, and meningococcal disease respectively.

Question 33Biotechnology (BT)

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

  1. The substrate concentration at which the enzyme is fully saturated
  2. The maximum velocity achieved at saturating substrate
  3. The rate constant for product formation
  4. The substrate concentration at which the reaction velocity is half of Vmax Correct answer.

Explanation

Km is the substrate concentration giving half the maximal velocity under the Michaelis-Menten model and is often used as an inverse index of substrate affinity. Vmax is the plateau velocity, and full saturation occurs at substrate concentrations far above Km.

Question 34Biotechnology (BT)

A protein mixture is loaded onto an anion-exchange column at pH 8.0. Which protein is expected to bind most strongly?

  1. A protein with pI 7.0
  2. A protein with pI 10.0
  3. A protein with pI 5.0 Correct answer.
  4. A protein with pI 8.0

Explanation

At pH 8.0 a protein whose pI is 5.0 carries a net negative charge and binds strongly to a positively charged anion-exchange matrix. As pI approaches or exceeds the running pH the net negative charge falls, and at pI 10.0 the protein is net positive and is repelled.

Question 35Biotechnology (BT)

During fed-batch cultivation of a recombinant yeast, a sudden drop in dissolved oxygen accompanied by a rise in carbon dioxide evolution rate most likely indicates which event?

  1. A decrease in viable biomass
  2. A loss of plasmid selection pressure
  3. A failure of the pH controller
  4. An increase in metabolic activity and oxygen demand Correct answer.

Explanation

A falling dissolved oxygen signal together with rising carbon dioxide evolution indicates that oxygen uptake is outpacing supply, usually because growth or metabolism has accelerated. Plasmid loss, pH faults, or declining viability would not produce a simultaneous increase in respiratory activity.

Question 36Biotechnology (BT)

A bioprocess engineer needs to concentrate a dilute protein solution and exchange it into a low-salt buffer without denaturation. Which technique is most suitable?

  1. Affinity chromatography
  2. Precipitation with ammonium sulphate
  3. Lyophilisation
  4. Tangential-flow ultrafiltration with diafiltration Correct answer.

Explanation

Ultrafiltration across a membrane with an appropriate molecular weight cut-off concentrates the protein while diafiltration replaces the buffer by continuous dilution and filtration. It operates under mild conditions and avoids the phase changes or elution steps of the alternatives.

Question 37Biotechnology (BT)

Which statement best explains why the lac promoter of E. coli is described as inducible?

  1. It is activated only when glucose is abundant
  2. It is repressed by cyclic AMP and the catabolite activator protein
  3. It is repressed by LacI and derepressed when allolactose binds the repressor Correct answer.
  4. It is constitutively active regardless of carbon source

Explanation

In the absence of lactose the LacI repressor binds the operator and blocks transcription; allolactose binding releases the repressor and allows expression. Catabolite activator protein with cyclic AMP acts as a positive regulator when glucose is scarce.

Question 38Biotechnology (BT)

A plasmid carries an ampicillin-resistance gene and a tetracycline-resistance gene, with a cloning site inside the tetracycline gene. After transformation, which phenotype identifies cells carrying a recombinant plasmid?

  1. Ampicillin sensitive and tetracycline resistant
  2. Ampicillin resistant and tetracycline sensitive Correct answer.
  3. Ampicillin resistant and tetracycline resistant
  4. Ampicillin sensitive and tetracycline sensitive

Explanation

Insertional inactivation of the tetracycline gene by the cloned fragment leaves transformants resistant to ampicillin but unable to grow on tetracycline. Non-recombinant plasmids retain both resistances, and cells without plasmid are sensitive to both.

Question 39Biotechnology (BT)

In a chemostat, a culture is grown at steady state with a dilution rate of 0.2 h⁻¹ and a feed substrate concentration of 10 g/L. If the residual substrate concentration is 0.5 g/L and the biomass yield is 0.5 g biomass per g substrate, what is the steady-state biomass concentration?

  1. 5.00 g/L
  2. 9.50 g/L
  3. 2.50 g/L
  4. 4.75 g/L Correct answer.

Explanation

Biomass equals the yield multiplied by the substrate consumed: 0.5 × (10 − 0.5) = 4.75 g/L. The residual substrate must be subtracted because it is not converted into biomass.

Question 40Biotechnology (BT)

Which method directly measures the number of viable cells by detecting their ability to reduce a tetrazolium dye?

  1. Dry cell weight determination
  2. Turbidimetric measurement at 600 nm
  3. MTT assay Correct answer.
  4. Colony counting on a spread plate

Explanation

Viable cells reduce tetrazolium salts such as MTT to a coloured formazan, giving a signal proportional to metabolic activity. OD600 and dry weight measure total biomass including dead cells, and plate counting enumerates only cells able to form colonies.

Question 41Biotechnology (BT)

In Southern blotting, what is the purpose of transferring separated DNA fragments from a gel to a nitrocellulose or nylon membrane?

  1. To convert double-stranded DNA into single-stranded RNA
  2. To separate fragments further by size
  3. To amplify the fragments before detection
  4. To immobilise the fragments so that a labelled probe can hybridise to them Correct answer.

Explanation

Blotting fixes the size-fractionated DNA to a solid support so that a specific labelled probe can hybridise to complementary sequences and be detected. The separation has already occurred in the gel, and no amplification is involved.

Question 42Biotechnology (BT)

A fermentation is scaled up from a 10 L to a 1000 L stirred tank while keeping the power per unit volume constant. Which quantity is held constant by this scale-up criterion?

  1. Dissolved oxygen concentration
  2. Power input per unit liquid volume Correct answer.
  3. Reynolds number
  4. Impeller tip speed

Explanation

Constant P/V scale-up fixes the power dissipated per unit volume, which influences oxygen transfer and mixing intensity. Tip speed and Reynolds number change with impeller diameter and speed and are not held constant by this criterion.

Question 43Biotechnology (BT)

Which feature of a plasmid vector allows it to be maintained in a bacterial population even when it imposes a metabolic burden?

  1. A selectable marker such as an antibiotic-resistance gene Correct answer.
  2. A multiple cloning site
  3. A promoter for the cloned gene
  4. A ribosome-binding site

Explanation

A selectable marker lets only plasmid-bearing cells survive under selective pressure, so the plasmid is retained despite any growth penalty. A multiple cloning site, promoter, and ribosome-binding site aid cloning and expression but do not enforce retention.

Question 44Biotechnology (BT)

A protein is to be purified by immobilised metal affinity chromatography using a polyhistidine tag. Which interaction mediates binding to the resin?

  1. Disulphide bond formation with the resin
  2. Coordination between histidine imidazole groups and immobilised nickel ions Correct answer.
  3. Ionic attraction between the tag and a positively charged matrix
  4. Hydrophobic interaction between the tag and the resin matrix

Explanation

The imidazole nitrogens of consecutive histidine residues coordinate nickel or cobalt ions chelated on the resin, giving reversible binding. Elution is typically achieved with imidazole or a pH drop, which competes for or releases the metal coordination.

Question 45Biotechnology (BT)

A culture of E. coli is grown in a medium containing lactose and glucose. Which pattern of lac operon expression is expected during the early exponential phase?

  1. Low expression, because glucose lowers cyclic AMP and the operon is catabolite repressed Correct answer.
  2. High expression, because glucose directly activates the promoter
  3. No expression, because the repressor cannot bind the operator
  4. High expression, because lactose is present and induces the repressor

Explanation

Glucose uptake reduces cyclic AMP, so catabolite activator protein does not activate the promoter even though lactose may relieve repression. This diauxic pattern means the operon is expressed strongly only after glucose is exhausted.

Question 46Biotechnology (BT)

An enzyme follows Michaelis-Menten kinetics with Vmax of 100 µmol/min and Km of 5 mM. At a substrate concentration of 20 mM, what is the initial velocity?

  1. 100 µmol/min
  2. 50 µmol/min
  3. 80 µmol/min Correct answer.
  4. 20 µmol/min

Explanation

Applying v = Vmax·S/(Km + S) gives 100 × 20 / (5 + 20) = 80 µmol/min. The velocity approaches but never reaches Vmax at finite substrate concentration.

Question 47Biotechnology (BT)

In an anaerobic digester, a stable population converts organic matter to methane. If the acetoclastic methanogens are selectively inhibited while acidogens remain active, which outcome is most likely?

  1. Volatile fatty acids accumulate and the pH falls Correct answer.
  2. The pH rises because carbon dioxide is consumed
  3. Methane production increases because acetate is spared
  4. The digester becomes aerobic

Explanation

Acetoclastic methanogens consume acetate and keep volatile fatty acids low; if they are inhibited, acidogenesis outpaces consumption and acids accumulate, lowering pH. This imbalance can cause digester souring and failure.

Question 48Biotechnology (BT)

A two-substrate enzyme follows a ping-pong bi-bi mechanism. Which kinetic pattern is expected in a Lineweaver-Burk plot when the first substrate is varied at several fixed concentrations of the second?

  1. A family of lines intersecting on the y-axis
  2. A single straight line independent of the second substrate
  3. A family of parallel lines Correct answer.
  4. A family of lines intersecting on the x-axis

Explanation

In ping-pong kinetics the apparent Km and Vmax change in a coordinated way so that reciprocal plots at different fixed concentrations of the second substrate give parallel lines. Intersecting patterns are characteristic of sequential mechanisms.

Question 49Biotechnology (BT)

A mammalian cell line is engineered to secrete a glycoprotein, but the product shows reduced terminal sialylation. Which process change would most directly increase sialylation?

  1. Lowering the dissolved oxygen set point
  2. Supplementing the medium with a precursor such as galactose and maintaining culture viability Correct answer.
  3. Adding a protease inhibitor
  4. Increasing the agitation rate

Explanation

Sialylation depends on nucleotide-sugar availability and on residence time in the Golgi, and it declines when cells lose viability and release glycosidases. Supplying precursors and limiting lysis favours complete glycan processing, whereas agitation, oxygen, and protease changes act indirectly or not at all.

Question 50Biotechnology (BT)

A bioprocess uses a recombinant strain whose product gene is under a temperature-sensitive promoter. Shifting from 30 °C to 37 °C induces product formation but reduces the specific growth rate. Which explanation best accounts for this trade-off?

  1. The higher temperature inactivates the antibiotic-resistance protein
  2. Plasmid copy number falls to zero at 37 °C
  3. Cellular resources are redirected from growth to product synthesis Correct answer.
  4. The promoter becomes repressed at the higher temperature

Explanation

Inducing high-level expression imposes a metabolic burden, so carbon and energy are diverted from biomass formation to product synthesis. This burden response lowers the observed specific growth rate without requiring plasmid loss or repressor action.