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GATE BT 2022 question paper with solutions

All 65 questions of the GATE 2022 Biotechnology paper, in order, with the official answer key from IIT Kharagpur. 34 have a full solution.

The full paper (PDF)

questions · 100 marks: 65questions · 100 marks
with a full solution: 34with a full solution
question types: 37 MCQ · 8 MSQ · 20 NATquestion types

GATE BT 2022 answer key

Official answers to every question of GATE BT 2022
Question	Type	Marks	Answer
Q.1	MCQ	1	D
Q.2	MCQ	1	A
Q.3	MCQ	1	D
Q.4	MCQ	1	B
Q.5	MCQ	1	B
Q.6	MCQ	2	B
Q.7	MCQ	2	B
Q.8	MCQ	2	B
Q.9	MCQ	2	B
Q.10	MCQ	2	Marks to all
Q.11	MCQ	1	B
Q.12	MCQ	1	C
Q.13	MCQ	1	C
Q.14	MCQ	1	D
Q.15	MCQ	1	D
Q.16	MCQ	1	D
Q.17	MCQ	1	A
Q.18	MCQ	1	A
Q.19	MCQ	1	D
Q.20	MCQ	1	A
Q.21	MCQ	1	A
Q.22	MCQ	1	C
Q.23	MCQ	1	D
Q.24	MCQ	1	D
Q.25	MCQ	1	A
Q.26	MCQ	1	C
Q.27	NAT	1	64
Q.28	NAT	1	102
Q.29	NAT	1	4
Q.30	NAT	1	1.2 to 1.4
Q.31	NAT	1	0.11 to 0.13
Q.32	NAT	1	1
Q.33	NAT	1	1
Q.34	NAT	1	20000
Q.35	NAT	1	6097 to 6223
Q.36	MCQ	2	D
Q.37	MCQ	2	A
Q.38	MCQ	2	B
Q.39	MCQ	2	B
Q.40	MCQ	2	B
Q.41	MCQ	2	C
Q.42	MCQ	2	B
Q.43	MCQ	2	D
Q.44	MCQ	2	A
Q.45	MCQ	2	B
Q.46	MCQ	2	A
Q.47	MSQ	2	A, B
Q.48	MSQ	2	A, C, D
Q.49	MSQ	2	A, B
Q.50	MSQ	2	A, D
Q.51	MSQ	2	A, C
Q.52	MSQ	2	A, B, C
Q.53	MSQ	2	A, C, D
Q.54	MSQ	2	A, C
Q.55	NAT	2	0.8
Q.56	NAT	2	1
Q.57	NAT	2	2
Q.58	NAT	2	150
Q.59	NAT	2	100
Q.60	NAT	2	1
Q.61	NAT	2	291 to 293
Q.62	NAT	2	2420 to 2430
Q.63	NAT	2	17 to 19
Q.64	NAT	2	2
Q.65	NAT	2	1

General Aptitude Q.1 to Q.10

Q.1MCQ · 1 markOfficial answer: D
Q.2MCQ · 1 markOfficial answer: A
Q.3MCQ · 1 markOfficial answer: D
Q.4MCQ · 1 markOfficial answer: B
Q.5MCQ · 1 markOfficial answer: B
Q.6MCQ · 2 marksOfficial answer: B
Q.7MCQ · 2 marksOfficial answer: B
Q.8MCQ · 2 marksOfficial answer: B
Q.9MCQ · 2 marksOfficial answer: B
Q.10MCQ · 2 marksOfficial answer: Marks to all

Biotechnology Q.11 to Q.65

Q.11MCQ · 1 markOfficial answer: B
Q.12MCQ · 1 markOfficial answer: C
Q.13MCQ · 1 markOfficial answer: C
Q.14MCQ · 1 markOfficial answer: D
Fundamentals of Biological Engineering › Upstream
A fermentation medium has a job for each component. Read the chemical formula to find what each one can supply: a sugar gives carbon and energy, an ammonium salt gives nitrogen, a metal salt added in micromolar amounts gives a trace element (an enzyme cofactor), and a water-insoluble, low-surface-tension oil breaks foam.
WATCH OUT FOR
Picking (C) because zinc chloride and ammonium sulphate are both 'salts', and swapping their roles. $ZnCl_{2}$ contains no nitrogen, so it cannot be a nitrogen source. Ammonium sulphate is added in grams per litre as the main nitrogen supply, not in trace amounts.
Full solution
Q.15MCQ · 1 markOfficial answer: D
Q.16MCQ · 1 markOfficial answer: D
Fundamentals of Biological Engineering › Growth kinetics
Atoms are conserved, so balance each element (C, H, O, N) separately. Whatever the reactants supply and the $CO_{2}$ and $H_{2} O$ do not take away ends up in the biomass. Divide that remainder by the biomass coefficient, $1.5$ , to get each subscript.
WATCH OUT FOR
The common slips are counting $O_{2}$ as one oxygen atom instead of two (that gives $c = 5.33$ ), forgetting the hydrogen that $NH_{3}$ brings in (that gives $b = 17.33$ ), and forgetting to divide by the biomass coefficient $1.5$ (that gives $C_{15}$ ). None of these results is among the options, so each one sends you to a wrong guess.
Full solution
Q.17MCQ · 1 markOfficial answer: A
Q.18MCQ · 1 markOfficial answer: A
Q.19MCQ · 1 markOfficial answer: D
Fundamentals of Biological Engineering › Enzyme kinetics
Each class of reversible inhibitor is defined by what it binds. Competitive binds free $E$ at the active site, so $K_{m}$ rises and $V_{ma x}$ stays the same. Uncompetitive binds only $E S$ , so $K_{m}$ and $V_{ma x}$ fall by the same factor. Pure non-competitive binds $E$ and $E S$ equally at a separate site, so $K_{m}$ stays the same and $V_{ma x}$ falls.
WATCH OUT FOR
Mixing up uncompetitive and non-competitive because the names sound alike, and so rejecting P (thinking uncompetitive inhibitors also bind free E) or rejecting Q. Another mistake is to drop a statement because it contains the word "only". Here "only" matches the textbook definition in both P and Q.
Full solution
Q.20MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Cell biology
Each organelle listed is matched to its single best-known signature function: lysosome = intracellular digestion (hydrolytic enzymes), peroxisome = general detoxification (catalase breaking down H2O2), glyoxysome = the glyoxylate cycle converting stored fatty acids to sugars (plant seeds), cytoskeleton = structural/motility roles (microtubules, actin filaments).
WATCH OUT FOR
Confusing peroxisome's general detoxification role (H2O2 breakdown via catalase, found in nearly all cell types) with the glyoxysome's specialized fatty-acid-to-sugar conversion (glyoxylate cycle, restricted to plant seeds/germinating tissue) — swapping Q and R, or picking option (C) which assigns lysosome to the glyoxylate function instead of digestion.
Full solution
Q.21MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Gene expression
Endogenous miRNAs in animal cells silence gene expression post-transcriptionally: mature miRNA loaded into the RNA-induced silencing complex (RISC) base-pairs, typically with imperfect complementarity, to the 3'-untranslated region (3'-UTR) of target mRNAs, causing translational repression and/or mRNA degradation — not transcriptional blocking.
WATCH OUT FOR
Confusing miRNA-mediated post-transcriptional silencing with transcriptional repression mechanisms (e.g., 'blocking mRNA synthesis' or 'binding to the operator site', the latter being a prokaryotic operon-repressor concept like the lac repressor, not applicable to eukaryotic miRNA action). Some students also mistakenly think miRNAs act on rRNA rather than mRNA.
Full solution
Q.22MCQ · 1 markOfficial answer: C
Q.23MCQ · 1 markOfficial answer: D
Q.24MCQ · 1 markOfficial answer: D
Recombinant DNA Technology and Tools › Bioinformatics
Optimal pairwise alignment algorithms (Needleman-Wunsch for global alignment, Smith-Waterman for local alignment) fill a dynamic-programming matrix using a fixed scoring scheme consisting only of match scores, mismatch scores, and gap penalties (gap-open/gap-extension). The overall nucleotide composition (%A/%T/%G/%C) of a sequence is a separate, global compositional statistic that plays no role in the cell-by-cell recursion or traceback of the alignment algorithm.
WATCH OUT FOR
Confusing "used in sequence analysis generally" with "used in the alignment scoring recursion specifically" — nucleotide/base composition is relevant to other analyses (e.g., GC-content, codon usage bias, compositional adjustment for statistical significance) but is not one of the three parameters (match, mismatch, gap penalty) that actually build the alignment scoring matrix itself.
Full solution
Q.25MCQ · 1 markOfficial answer: A
Recombinant DNA Technology and Tools › Enzymes & vectors
For a palindromic (double-stranded-symmetric) restriction site, an enzyme produces sticky (cohesive) ends when its cleavage point on each strand is offset from the site's centre of symmetry (leaving a single-stranded overhang), and blunt ends when the cleavage point sits exactly at that centre (both strands cut in register).
WATCH OUT FOR
Assuming any recognition site with a "↓" symbol off-centre in the way it is typeset must be sticky, without actually locating the site's own centre of symmetry for each different site length (6 bp vs 4 bp) — e.g. mistakenly treating CTG↓CAG (cut after base 3 of 6, i.e. exactly centred) as asymmetric because the arrow "looks" placed differently from G↓GATCC.
Full solution
Q.26MCQ · 1 markOfficial answer: C
Recombinant DNA Technology and Tools › PCR & sequencing
A DNA sequence variant (SNP or other minor change) at a restriction enzyme recognition sequence creates or destroys that cut site; digesting genomic DNA with the enzyme and comparing the resulting fragment sizes (by Southern blot or electrophoresis) between individuals reveals the variation — this is RFLP analysis.
WATCH OUT FOR
Confusing RFLP (which detects DNA sequence variation via changes in restriction-digest fragment sizes) with PCR-based quantification or amplification methods (real-time PCR, RT-PCR), which detect gene expression/RNA levels or amplify DNA but do not by themselves reveal restriction-site polymorphisms.
Full solution
Q.27NAT · 1 markOfficial answer: 64
Genetics, Cell and Molecular Biology › Mendelian genetics
With independent assortment, each homologous chromosome pair segregates independently, so a gamete can receive the maternal or paternal member of each pair; the number of distinct combinations is $2^{n}$ , where $n$ is the haploid chromosome number.
WATCH OUT FOR
Using the diploid number (12) directly as the exponent instead of first halving it to get the haploid number of pairs (6), which gives a wrong answer of $2^{12}$ .
Full solution
Q.28NAT · 1 markOfficial answer: 102
Fundamentals of Biological Engineering › Growth kinetics
The yield $Y_{X / S}$ links the biomass formed to the substrate consumed. With no endogenous (maintenance) metabolism it stays constant, so the most biomass you can get is the inoculum plus the yield times all of the substrate: $X_{ma x} = X_{0} + Y_{X / S} S_{0}$ .
WATCH OUT FOR
Answering $100 g L^{- 1}$ ( $0.5 \times 200$ ) and leaving out the initial biomass $X_{0} = 2 g L^{- 1}$ . The yield gives only the biomass newly formed, not the total. A second slip is using $Y_{X / S}$ as substrate per biomass and getting $400$ .
Full solution
Q.29NAT · 1 markOfficial answer: 4
Fundamentals of Biological Engineering › Growth kinetics
Degree of reduction means available electrons per carbon: $γ = (4 c + h - 2 o - 3 n) / c$ , with $CO_{2}$ , $H_{2} O$ and $NH_{3}$ as the zero reference. Any compound whose C-mole formula is $CH_{2} O$ (glucose, lactic acid, acetic acid) has $γ = 4$ .
WATCH OUT FOR
Giving the electron count per molecule (12) and forgetting to divide by the 3 carbons. Another slip is using O = $+ 2$ or leaving oxygen out, which gives 6 instead of 4.
Full solution
Q.30NAT · 1 markOfficial answer: 1.2 to 1.4
Q.31NAT · 1 markOfficial answer: 0.11 to 0.13
Q.32NAT · 1 markOfficial answer: 1
Q.33NAT · 1 markOfficial answer: 1
Fundamentals of Biological Engineering › Growth kinetics
In exponential growth, $X = X_{0} e^{μ t}$ , so the doubling time and the specific growth rate are linked by $t_{d} = \frac{ln 2}{μ}$ . The doubling time given here, 0.693 h, is exactly $ln 2$ hours, which makes $μ = 1 h^{- 1}$ .
WATCH OUT FOR
Taking $μ = 1/ t_{d}$ (which gives about $1.44 h^{- 1}$ ) and forgetting the $ln 2$ . Another error is using $lo g_{10} 2 = 0.301$ in place of $ln 2$ , which gives about $0.43 h^{- 1}$ . Growth is $e^{μ t}$ , so the natural log is the right one.
Full solution
Q.34NAT · 1 markOfficial answer: 20000
Fundamentals of Biological Engineering › Bioreactor operation
For a stirred tank, the Reynolds number is built from the impeller: $R e_{i} = ρN D_{i}^{2} / μ$ . $N$ is in revolutions per second and $D_{i}$ is the impeller diameter. The tank volume only tests whether you know which length to use.
WATCH OUT FOR
Three slips are common. The first is turning $N = 2 s^{- 1}$ into angular speed ( $2 π N$ ), which gives about $1.26 \times 1 0^{5}$ ; the stirred-tank Reynolds number uses revolutions per second. The second is writing $D_{i}$ instead of $D_{i}^{2}$ , which gives $200$ . The third is working a tank diameter out of the $100 L$ volume and using it as the length scale.
Full solution
Q.35NAT · 1 markOfficial answer: 6097 to 6223
Q.36MCQ · 2 marksOfficial answer: D
Genetics, Cell and Molecular Biology › Population & evolution
A gene pool (its allele frequencies) stays stable across generations only under Hardy-Weinberg conditions: an infinitely large (in practice, sufficiently large) population, no mutation, random mating, and no selection (also no migration/gene flow).
WATCH OUT FOR
Assuming random mating and non-random mating are interchangeable, or forgetting that Hardy-Weinberg equilibrium requires RANDOM mating specifically — so "non-random mating" (R) is a disturbing force, not a stabilizing condition, and must be excluded.
Full solution
Q.37MCQ · 2 marksOfficial answer: A
Q.38MCQ · 2 marksOfficial answer: B
Q.39MCQ · 2 marksOfficial answer: B
Q.40MCQ · 2 marksOfficial answer: B
Fundamentals of Biological Engineering › Downstream
A stationary phase is named after what it carries. A specific biological ligand (Protein A binding the Fc region of IgG) means affinity. A porous cross-linked gel with no binding groups (Sephadex) means size exclusion. Hydrophobic groups on agarose (phenyl on Sepharose) mean hydrophobic interaction. Charged groups (the protonated DEAE amine, an anion exchanger) mean ion exchange.
WATCH OUT FOR
Seeing "Sepharose" in Phenylsepharose and matching it to size exclusion, because plain Sepharose (agarose) is also used as a gel-filtration medium. The phenyl groups are what count here: they make it a hydrophobic interaction resin. Another slip is calling DEAE a cation exchanger. Its amine is positively charged, so it binds anions.
Full solution
Q.41MCQ · 2 marksOfficial answer: C
Fundamentals of Biological Engineering › Process control
Offset is set by the loop's gain at $s = 0$ . With P control, $G_{c} G_{p} (0) = K_{c} K_{p}$ is finite, so $y (\infty) = K_{c} K_{p} / (1 + K_{c} K_{p}) < 1$ . Integral action puts a $1/ s$ in the loop, so $G_{c} G_{p} \to \infty$ as $s \to 0$ , $y (\infty) = 1$ , and the offset disappears.
WATCH OUT FOR
Mixing up integral and derivative action. Q describes derivative control ( $\propto d ε / d t$ ), not integral control. The other common slip is thinking a large enough $K_{c}$ removes the P-controller offset. It only shrinks it, as $1/ (1 + K_{c} K_{p})$ , and never reaches zero at any finite gain.
Full solution
Q.42MCQ · 2 marksOfficial answer: B
Q.43MCQ · 2 marksOfficial answer: D
Genetics, Cell and Molecular Biology › Cell signalling
Ubiquitin-mediated proteolysis of cell-cycle regulators (cyclins, securin, CDK inhibitors) by the proteasome always proceeds through the same enzymatic cascade: E1 (activation) → E2 (conjugation) → E3 (ligation, substrate specificity), so all three are required.
WATCH OUT FOR
Students often stop at "E1 activates, E3 ligates" and forget that E2 is an obligatory carrier intermediate between them (option B), or mistake E4 (a chain-elongation factor present only in some substrates) for a core, universal component (option C).
Full solution
Q.44MCQ · 2 marksOfficial answer: A
Recombinant DNA Technology and Tools › Bioinformatics
Maximum parsimony selects, among all possible tree topologies, the one(s) requiring the fewest total character-state changes (evolutionary steps) to explain the observed sequence data — it minimizes tree length, not node count, and uses no probability model.
WATCH OUT FOR
Confusing the parsimony criterion (minimum number of character-state changes/steps) with the number of branch points or nodes in the tree (option C), or mistaking parsimony for a probability-based method like maximum likelihood (option D).
Full solution
Q.45MCQ · 2 marksOfficial answer: B
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
A technique can identify every functional group only if every group gives it a signal. IR does this through bond vibrations, and NMR does it through the chemical shifts of $^{1} H$ and $^{13} C$ nuclei. CD needs a chiral chromophore and UV-Vis needs a chromophore, so both miss most groups.
WATCH OUT FOR
Choosing (A) because IR is thought of as the only functional-group technique. NMR chemical shifts are just as diagnostic of functional groups ( $^{13} C$ NMR detects even carbonyl and quaternary carbons), and IR with NMR is the standard pair for structure elucidation. Choosing (D) is the opposite mistake: CD and UV-Vis give signals only from chiral chromophores and chromophores respectively.
Full solution
Q.46MCQ · 2 marksOfficial answer: A
Genetics, Cell and Molecular Biology › Mutation & repair
Spontaneous deamination of adenine to hypoxanthine (which base-pairs like guanine, i.e. with cytosine) creates a base-pair mismatch that is corrected by base excision repair (BER): a DNA glycosylase excises the aberrant base, an AP endonuclease (a nuclease) nicks the backbone at the resulting abasic site, DNA polymerase fills the single-nucleotide gap, and DNA ligase seals the nick.
WATCH OUT FOR
Confusing this single-base lesion repair with replication machinery: pulling in helicase (unwinds the duplex ahead of a replication fork) or primase (lays down RNA primers for new strand synthesis) as in options B/D, neither of which is needed to excise and replace one damaged base.
Full solution
Q.47MSQ · 2 marksOfficial answer: A, B
Q.48MSQ · 2 marksOfficial answer: A, C, D
Fundamentals of Biological Engineering › Immobilisation
The Thiele modulus $ϕ = \frac{R}{3} V_{ma x} / (K_{m} D_{e})$ measures reaction speed against internal diffusion speed. The effectiveness factor $η$ (rate with diffusion limitation divided by rate without it) equals 1 when $ϕ$ is small and falls roughly as $1/ ϕ$ when $ϕ$ is large. Since $ϕ \propto R$ , making the particle smaller raises $η$ .
WATCH OUT FOR
Getting the direction of the Thiele modulus backwards and marking B: a LOW $ϕ$ means diffusion is fast compared with reaction ( $η \approx 1$ , reaction controls). A second slip is reading the effectiveness factor the other way up (rate without limitation over rate with it), which would give a value above 1.
Full solution
Q.49MSQ · 2 marksOfficial answer: A, B
Q.50MSQ · 2 marksOfficial answer: A, D
Fundamentals of Biological Engineering › Enzyme kinetics
At low substrate ( $[S] ≪ K_{m}$ ) the Michaelis-Menten equation reduces to first-order kinetics, $v_{0} \approx (V_{ma x} / K_{m}) [S]$ . A first-order rate law gives two shapes: rate linear in $[S]$ through the origin, and substrate decaying exponentially in time, $[S] = [S]_{0} e^{- (V_{ma x} / K_{m}) t}$ .
WATCH OUT FOR
Picking B because Michaelis-Menten kinetics 'saturates'. The hyperbolic plateau only appears at $[S] ≫ K_{m}$ , and a sigmoid (with an inflection point) belongs to cooperative enzymes, not Michaelis-Menten ones. Another trap is rejecting D because $v_{0}$ is defined as an initial velocity. The rate law holds at every instant, and $[S]$ only decreases, so the low-substrate condition stays true and first-order decay follows.
Full solution
Q.51MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Gene expression
The lac operon has two controls: the repressor on the operator (negative control) and cAMP-CAP (positive control). Glucose lowers cAMP and blocks lactose uptake (inducer exclusion). The molecule that actually binds the repressor is allolactose, an isomer that beta-galactosidase makes from lactose.
WATCH OUT FOR
Many students mark (D) because textbooks loosely say "lactose induces the operon". But the repressor binds allolactose, not lactose. Others reject (C) because lactose is present. While glucose is high, inducer exclusion keeps lactose out of the cell, so the repressor stays on the operator.
Full solution
Q.52MSQ · 2 marksOfficial answer: A, B, C
Genetics, Cell and Molecular Biology › Population & evolution
Viral emergence (the appearance/rise of "new" epidemic viruses like SARS-CoV-2) is a multi-step process: genetic change generates variant viruses, spillover moves them into a new host species, and sustained transmission within that new host population turns a spillover event into an epidemic. All three are recognized drivers; a virus replicating outside a living host cell is not a real biological process, since viruses are obligate intracellular parasites.
WATCH OUT FOR
Treating "jumping to a new host" (B) as sufficient by itself and forgetting that sustained spread within the new host population (C) is a separate, necessary step for an epidemic to result — a single spillover infection that doesn't transmit further is a dead end, not an epidemic. A second common error is assuming any exotic-sounding process (D) must be a distractor testing obscure virology rather than recognizing it contradicts the basic obligate intracellular nature of viruses.
Full solution
Q.53MSQ · 2 marksOfficial answer: A, C, D
Recombinant DNA Technology and Tools › Enzymes & vectors
Foreign gene delivery into plant cells uses methods suited to a rigid cell wall — Agrobacterium's natural T-DNA transfer, electroporation of wall-reduced protoplasts, or physical bombardment via gene gun — never the CaCl2/heat-shock method, which is specific to making bacterial cells chemically competent.
WATCH OUT FOR
Students sometimes select CaCl2-mediated uptake by pattern-matching it to "plasmid" and "DNA transfer" language, forgetting that this technique specifically relies on transiently destabilizing a bacterial cell membrane during heat shock and has no established role in transforming plant cells, which have a cellulosic wall and different membrane biology.
Full solution
Q.54MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Cell biology
Dynamin is a large GTPase that pinches off vesicles (endocytosis/budding); dynein is an ATPase motor that moves cargo/organelles along microtubules (typically toward the minus end). Kinesin, by contrast, generally moves cargo toward the microtubule plus end, and neither motor's nucleotide is interchangeable with the other's.
WATCH OUT FOR
Confusing the nucleotide used by motor/mechanoenzyme proteins — assuming all motor and trafficking proteins use ATP (missing that dynamin is a GTPase) or, conversely, assigning GTP to dynein (option D) by analogy with other GTP-binding trafficking regulators (e.g., Rab or Arf GTPases). Also mistakenly crediting dynamin with a general cytoskeletal remodelling role (option B) rather than its specific, well-defined function in vesicle scission.
Full solution
Q.55NAT · 2 marksOfficial answer: 0.8
Q.56NAT · 2 marksOfficial answer: 1
Q.57NAT · 2 marksOfficial answer: 2
Fundamentals of Biological Engineering › Growth kinetics
In metabolic flux analysis at pseudo-steady state, no intracellular metabolite accumulates, so at each branch point the fluxes in equal the fluxes out. Work node by node, from the uptake flux towards the product: $A_{i}$ gives $r_{2} = r_{1} - r_{3}$ , then $B_{i}$ gives $r_{4} = r_{2} - r_{6}$ .
WATCH OUT FOR
Forgetting the second branch at $B_{i}$ and taking $r_{4} = r_{2} = 3$ . Part of $B_{i}$ is diverted to $D_{i}$ by $r_{6}$ , so it never leaves the cell as B. Another mistake is to treat $D_{i}$ as a balanced node. No reaction out of it is shown, so it only acts as a drain.
Full solution
Q.58NAT · 2 marksOfficial answer: 150
Fundamentals of Biological Engineering › Bioreactor operation
At quasi-steady state in fed-batch operation, the residual substrate stays near zero, so all substrate fed becomes biomass at the yield. Total biomass then rises linearly: $X_{t} = X_{0} + Y_{X S}^{M} F S_{0} t$ .
WATCH OUT FOR
Answering 100 g by leaving out the 50 g already in the reactor when feeding starts. Another mistake is working with biomass concentration and dividing by a volume the question never gives, when it asks for the total amount of biomass.
Full solution
Q.59NAT · 2 marksOfficial answer: 100
Fundamentals of Biological Engineering › Ideal reactors
In an ideal mixed flow reactor the contents are uniform and identical to the outlet stream, so the whole volume reacts at the rate evaluated at the EXIT concentration: $V = v_{0} (C_{A 0} - C_{A}) / (- r_{A})_{exit}$ . No integration is needed, even for a Michaelis-Menten rate law.
WATCH OUT FOR
Evaluating the rate at the feed concentration ( $- r_{A} = 10/15 = 0.667$ , giving 75 L) instead of at the exit concentration, or using the integrated plug-flow/batch expression $τ = [K_{m} ln (C_{A 0} / C_{A}) + (C_{A 0} - C_{A})] / V_{ma x} = 5 ln 2 + 5 \approx 8.47 min$ (about 84.7 L), which applies to a PFR, not an MFR.
Full solution
Q.60NAT · 2 marksOfficial answer: 1
Fundamentals of Biological Engineering › Bioreactor operation
At steady state, oxygen transfer equals oxygen uptake: $k_{L} a (C^{*} - C_{L}) = OUR$ . Divide the OUR by the driving force to get $k_{L} a$ . Then divide by the specific interfacial area $a$ to get $k_{L}$ .
WATCH OUT FOR
There are three common slips. (1) Using $C^{*}$ alone as the driving force instead of $C^{*} - C_{L}$ , which gives $k_{L} a = 0.9 s^{- 1}$ . (2) Leaving $a$ as $100 m^{- 1}$ , which gives $k_{L} = 0.01 m s^{- 1}$ , then entering 0.01 even though the blank asks for $cm s^{- 1}$ . (3) Getting the m-to-cm conversion of $a$ upside down, which gives $100 cm^{- 1}$ .
Full solution
Q.61NAT · 2 marksOfficial answer: 291 to 293
Q.62NAT · 2 marksOfficial answer: 2420 to 2430
Fundamentals of Biological Engineering › Growth kinetics
When heats of combustion share one reference state ( $CO_{2}$ , $H_{2} O$ , $O_{2}$ , $N_{2}$ all zero), the heat a reaction releases is (sum over reactants of $n Δ H_{c}$ ) minus (sum over products of $n Δ H_{c}$ ). The equation forms 1 mol of biomass, so this difference is the heat yield $Y_{Q / X}$ per mol of biomass. Multiplying it by the specific growth rate $μ$ gives the heat released per mol of biomass per hour: $q_{Q} = μ Y_{Q / X}$ .
WATCH OUT FOR
Three slips give answers outside the key. (1) Leaving out the ammonia term, which gives $0.5 (5604 - 833.2) \approx 2385$ . (2) Leaving out the ethanol by-product, which gives $0.5 (5680.6 - 560) \approx 2560$ . (3) Using glucose's heat of combustion once instead of twice. Two other errors are stopping at the heat yield (4847) without multiplying by $μ$ , and getting the sign backwards by subtracting reactants from products. In that last case the magnitude is right but the reaction looks heat-absorbing, when it actually releases heat.
Full solution
Q.63NAT · 2 marksOfficial answer: 17 to 19
Fundamentals of Biological Engineering › Upstream
Spore death is first order, so the Del factor you need is $\nabla = ln (N_{0} / N)$ , where $N_{0}$ and $N$ are the spore counts in the WHOLE vessel. If heating and cooling kill nothing, all of it has to come from holding, so $t_{h o l d} = \nabla/ k_{d}$ .
WATCH OUT FOR
Using concentrations instead of total counts: taking $ln (1 0^{8} /1)$ gives $9.2$ min. The limit of 1 spore applies to the whole $100 m^{3}$ vessel, so the volume must go into $N_{0}$ ( $1 0^{8} mL$ , not $1 0^{5}$ L mixed with a per-mL concentration). Another slip is using $lo g_{10}$ instead of $ln$ , which gives $16/2 = 8$ min.
Full solution
Q.64NAT · 2 marksOfficial answer: 2
Recombinant DNA Technology and Tools › Enzymes & vectors
On a circular DNA molecule, the number of linear fragments produced equals the number of cut sites used (not sites + 1, as it is for linear DNA); co-migrating fragments of identical size appear as a single band on the gel.
WATCH OUT FOR
Applying the linear-DNA rule (fragments = sites + 1) to a circular plasmid, which would wrongly give 4 fragments instead of 3; or forgetting that two fragments of identical size run as one band, over-counting the bands as 3 instead of 2.
Full solution
Q.65NAT · 2 marksOfficial answer: 1
Fundamentals of Biological Engineering › Growth kinetics
In exponential growth the cell count doubles once per doubling time, so the number of generations is $n = lo g_{2} (N / N_{0})$ and $t_{d} = t / n$ . The same result comes from $t_{d} = ln 2/ μ$ , with $μ = ln (N / N_{0}) / t$ . A 1000-fold rise is almost exactly 10 doublings, because $2^{10} = 1024$ .
WATCH OUT FOR
Dividing by the wrong quantity. The common slips are using $lo g_{10}$ in place of $lo g_{2}$ (which gives 3 "generations" and $t_{d} \approx 3.3$ h), using the cell increase $1 0^{5} - 1 0^{2}$ in place of the ratio, or reporting $μ$ ( $0.69 h^{- 1}$ ) as the doubling time. The temperature and aerobic conditions are context only and do not enter the calculation.
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WHERE THE MARKS CAME FROM

Marks by subject in GATE BT 2022
Subject	Qs	Marks
Fundamentals of Biological Engineering	18	29
Genetics, Cell and Molecular Biology	9	15
Recombinant DNA Technology and Tools	7	11
Not on a subject page yet	31	45

We sorted the questions into subjects ourselves; the exam paper does not label them.

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