PAPERS · GATE BT

GATE BT 2023 question paper with solutions

All 65 questions of the GATE 2023 Biotechnology paper, in order, with the official answer key from IIT Kanpur. 29 have a full solution; 1 more shows the official answer only.

The full paper (PDF)Official answer key (IIT Kanpur)

questions · 100 marks: 65questions · 100 marks
with a full solution: 29with a full solution
question types: 32 MCQ · 7 MSQ · 26 NATquestion types

GATE BT 2023 answer key

Official answers to every question of GATE BT 2023
Question	Type	Marks	Answer
Q.1	MCQ	1	B
Q.2	MCQ	1	C
Q.3	MCQ	1	B
Q.4	MCQ	1	D
Q.5	MCQ	1	C
Q.6	MCQ	2	D
Q.7	MCQ	2	A
Q.8	MCQ	2	C
Q.9	MCQ	2	A
Q.10	MCQ	2	D
Q.11	MCQ	1	A
Q.12	MCQ	1	A
Q.13	MCQ	1	D
Q.14	MCQ	1	B
Q.15	MCQ	1	A
Q.16	MCQ	1	D
Q.17	MCQ	1	A
Q.18	MCQ	1	A
Q.19	MCQ	1	A
Q.20	MCQ	1	A
Q.21	MCQ	1	A
Q.22	MCQ	1	A
Q.23	MCQ	1	A
Q.24	MCQ	1	A
Q.25	MCQ	1	A
Q.26	MCQ	1	C
Q.27	MCQ	1	C
Q.28	MCQ	1	C
Q.29	MCQ	1	A
Q.30	NAT	1	25
Q.31	NAT	1	6
Q.32	NAT	1	2500
Q.33	NAT	1	0.10 to 0.12
Q.34	NAT	1	0.166 to 0.168
Q.35	NAT	1	1.40 to 1.42
Q.36	MCQ	2	A
Q.37	MCQ	2	A
Q.38	MCQ	2	A
Q.39	MSQ	2	B, C
Q.40	MSQ	2	A, B, C
Q.41	MSQ	2	A, B, D
Q.42	MSQ	2	A, C
Q.43	MSQ	2	A, B, C
Q.44	MSQ	2	A, B, C
Q.45	MSQ	2	A, B
Q.46	NAT	2	0.41 to 0.44
Q.47	NAT	2	2.24 to 2.26
Q.48	NAT	2	10
Q.49	NAT	2	94 to 96
Q.50	NAT	2	-2.74 to -2.71
Q.51	NAT	2	15
Q.52	NAT	2	945
Q.53	NAT	2	4.6 to 4.9
Q.54	NAT	2	20
Q.55	NAT	2	74
Q.56	NAT	2	1.0 to 1.2
Q.57	NAT	2	0.165 to 0.168
Q.58	NAT	2	10551
Q.59	NAT	2	-2
Q.60	NAT	2	5
Q.61	NAT	2	0.45
Q.62	NAT	2	1
Q.63	NAT	2	1437 to 1439
Q.64	NAT	2	4
Q.65	NAT	2	1600

General Aptitude Q.1 to Q.10

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

Biotechnology Q.11 to Q.65

Q.11MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Gene expression
Transcription is the DNA-templated synthesis of RNA; in eukaryotes this is carried out by DNA-dependent RNA polymerases (RNA Pol I, II and III), each transcribing a distinct class of genes (rRNA, mRNA/most genes, tRNA/5S rRNA respectively).
WATCH OUT FOR
Confusing the four polymerase/enzyme classes by template-versus-product: DNA-dependent DNA polymerase (B) performs replication, not transcription; RNA-dependent DNA polymerase (C) is reverse transcriptase, used by retroviruses to convert their RNA genome into DNA; RNA-dependent RNA polymerase (D) is used by RNA viruses to replicate RNA genomes — none of these describe normal eukaryotic transcription.
Full solution
Q.12MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Cell signalling
In SA-node pacemaker cells, parasympathetic acetylcholine acts on M2 muscarinic receptors — GPCRs coupled to an inhibitory G protein (Gi/Go), not on any GABA-gated channel. Binding activates this Gi/Go protein, which then inhibits adenylyl cyclase (lowering cAMP) and opens GIRK K+ channels, hyperpolarizing the cell and slowing the pacemaker rate (negative chronotropy).
WATCH OUT FOR
Confusing "activates the inhibitory G protein" with "inhibits the G protein" — the inhibition in this pathway happens downstream (Gi inhibits adenylyl cyclase), not at the G protein itself, which is what makes distractor (C) attractive but wrong.
Full solution
Q.13MCQ · 1 markOfficial answer: D
Genetics, Cell and Molecular Biology › Gene expression
This is an Assertion-Reason question requiring two independent judgments: (1) is [a] true or false on its own biological merits, and (2) is [r] true or false on its own merits — only if both are true do you then ask whether [r] correctly explains [a]. Differential/lineage-specific gene expression (via differential transcription factor activity, chromatin state, epigenetic marks, etc.) is the well-established basis of cellular differentiation in multicellular organisms, making [a] true. Protein diversity, however, arises from multiple independent mechanisms — alternative splicing, alternative promoter usage/transcription start sites, alternative polyadenylation, RNA editing, and post-translational modifications (phosphorylation, glycosylation, proteolytic cleavage) — so the absolute claim that alternative splicing is "the only" mechanism is false, regardless of alternative splicing's real importance.
WATCH OUT FOR
Treating alternative splicing as if it were "a" or "the main" source of protein diversity and conflating that with "the only" mechanism — since alternative splicing genuinely does create isoform diversity, students may misread the exaggerated exclusivity claim in [r] as true and pick (B) or (C) instead of recognizing that "only" makes the reason factually false.
Full solution
Q.14MCQ · 1 markOfficial answer: B
Genetics, Cell and Molecular Biology › Population & evolution
Chromosome (structural) mutations — deletions, duplications, inversions, translocations — change chromosome STRUCTURE and arise from breakage-and-faulty-rejoining; genomic (numerical) mutations — aneuploidy, polyploidy — change chromosome NUMBER and arise from nondisjunction. The two mechanisms and the two categories are distinct, so Reason [r] conflates them.
WATCH OUT FOR
Treating "chromosome mutation" as a blanket term covering any chromosome-related abnormality, and then assuming nondisjunction (the most commonly taught mutational mechanism) must be the cause — missing that nondisjunction specifically produces numerical (genomic), not structural, changes.
Full solution
Q.15MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Mutation & repair
The C-value paradox: an organism's haploid genome size (its C-value) does not correlate with its morphological or organismal complexity — e.g., some amoebae, lungfish, and even some plants have vastly larger genomes than humans, largely due to variable amounts of non-coding/repetitive DNA rather than a greater number of functional genes.
WATCH OUT FOR
Confusing the C-value paradox (genome size vs. complexity mismatch) with related but distinct concepts also tested as distractors: transposable/selfish DNA propagation (a proposed cause of genome size variation, not the paradox itself), multiple alleles at a locus (a population-genetics idea unrelated to genome size), or gene functional redundancy (unrelated to genome size).
Full solution
Q.16MCQ · 1 markOfficial answer: D
Q.17MCQ · 1 markOfficial answer: A
Q.18MCQ · 1 markOfficial answer: A
Q.19MCQ · 1 markOfficial answer: A
Q.20MCQ · 1 markOfficial answer: A
Q.21MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Cell signalling
Covalent tagging of a target protein with (poly)ubiquitin chains is the signal recognized by the 26S proteasome for ATP-dependent proteolytic degradation.
WATCH OUT FOR
Confusing peptidase (an enzyme that carries out proteolysis, including proteasomal subunits) with the actual conjugated targeting tag; peptidase is never itself covalently attached to a substrate as a degradation signal.
Full solution
Q.22MCQ · 1 markOfficial answer: A
Recombinant DNA Technology and Tools › Assays & electrophoresis
ELISA detection relies on antibody-enzyme conjugates whose enzyme activity produces a measurable colored/chemiluminescent signal upon substrate addition; only alkaline phosphatase and horseradish peroxidase are standard ELISA reporter enzymes among the four listed.
WATCH OUT FOR
Confusing any hydrolytic/digestive enzyme (trypsin, amylase) with a chromogenic reporter enzyme; only enzymes with a colorimetric/chemiluminescent substrate reaction (ALP, HRP) are used as ELISA labels.
Full solution
Q.23MCQ · 1 markOfficial answer: A
Q.24MCQ · 1 markOfficial answer: A
Recombinant DNA Technology and Tools › PCR & sequencing
Southern blotting detects DNA (named after Edwin M. Southern); Northern blotting detects RNA (a naming pun on "Southern"); Western blotting detects proteins. The question asks in the order DNA, then RNA, so the pairing must be Southern, then Northern — option A.
WATCH OUT FOR
Confusing the mnemonic order — swapping Southern/Northern (option C) because the "Northern" name is a play on "Southern" and students misremember which technique targets DNA vs RNA, or wrongly pairing Western blotting (a protein-detection technique) with a nucleic acid.
Full solution
Q.25MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Gene expression
Each RNA class has a distinct, non-overlapping functional role: mRNA carries coding sequence, rRNA is the ribosome's structural and catalytic (ribozyme) core, miRNA post-transcriptionally silences gene expression, and tRNA is the adaptor linking codons to amino acids.
WATCH OUT FOR
Confusing tRNA's adaptor role (item 1) with miRNA's regulatory role (item 2), as in distractor (B), since both are "small" non-coding RNAs — but only tRNA physically carries amino acids and reads codons via its anticodon; miRNA acts on mRNA stability/translation, not amino acid delivery.
Full solution
Q.26MCQ · 1 markOfficial answer: C
Recombinant DNA Technology and Tools › Bioinformatics
PSI-BLAST (Position-Specific Iterated BLAST) builds and iteratively refines a position-specific scoring matrix (profile) from initial hits, which amplifies weak, distant sequence similarity — making it the standard BLAST variant for detecting remote protein homologs, unlike the single-pass BLASTN/BLASTX/TBLASTX searches.
WATCH OUT FOR
Picking BLASTX or TBLASTX because they sound more "thorough" (translated searches), while missing that PSI-BLAST's iterative profile refinement — not translation of nucleotide frames — is what specifically boosts sensitivity to distant/remote homologs.
Full solution
Q.27MCQ · 1 markOfficial answer: C
Recombinant DNA Technology and Tools › Bioinformatics
Needleman-Wunsch is the classic dynamic-programming algorithm for global sequence alignment (aligns full length, end to end); Smith-Waterman is its local-alignment counterpart (best-scoring subregion only). Chou-Fasman and GOR are unrelated secondary-structure prediction methods, not alignment algorithms.
WATCH OUT FOR
Confusing Needleman-Wunsch (global) with Smith-Waterman (local) since both are DP-based pairwise alignment algorithms — remembering "N for 'k(N)ow the whole thing' = global" or simply memorizing the pairing (Needleman-Wunsch = Global, Smith-Waterman = Local) avoids the mix-up. A secondary mistake is confusing alignment algorithms with secondary-structure prediction methods (Chou-Fasman, GOR), which solve a completely different problem.
Full solution
Q.28MCQ · 1 markOfficial answer: C
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
Secondary structure content (%α-helix, %β-sheet, %random coil) of a protein is determined by spectroscopic/structural methods that probe backbone conformation and H-bonding — CD (far-UV signature bands), FTIR (amide I band deconvolution), and X-ray crystallography (atomic-resolution structure). Mass spectrometry measures mass-to-charge ratio and reports mass, composition, sequence and PTMs, not backbone conformation, so it cannot determine secondary structure content.
WATCH OUT FOR
Students sometimes assume any advanced structural/analytical technique (including MS) can report "structure" in a general sense, without distinguishing between primary-structure/mass information (MS) and conformational/secondary-structure information (CD, FTIR, X-ray).
Full solution
Q.29MCQ · 1 markOfficial answer: A
Q.30NAT · 1 markOfficial answer: 25
Genetics, Cell and Molecular Biology › Population & evolution
Fabry disease is X-linked recessive. A normal father ( $X^{A} Y$ ) crossed with a carrier mother ( $X^{A} X^{a}$ ) produces four equally likely offspring classes; only one of the four ( $X^{a} Y$ , an affected son) has the disease, giving a probability of $\frac{1}{4} = 25%$ among all offspring.
WATCH OUT FOR
Confusing "probability among all offspring" (1/4) with "probability among sons only" (1/2); the question asks for the former.
Full solution
Q.31NAT · 1 markOfficial answer: 6
Q.32NAT · 1 markOfficial answer: 2500
Q.33NAT · 1 markOfficial answer: 0.10 to 0.12
Q.34NAT · 1 markOfficial answer: 0.166 to 0.168
Q.35NAT · 1 markOfficial answer: 1.40 to 1.42
Q.36MCQ · 2 marksOfficial answer: A
Q.37MCQ · 2 marksOfficial answer: A
Q.38MCQ · 2 marksOfficial answer: A
Q.39MSQ · 2 marksOfficial answer: B, C
Genetics, Cell and Molecular Biology › Population & evolution
Tumor suppressor genes are inactivated by loss-of-function events on both alleles (Knudson's two-hit hypothesis): genetic hits such as loss of heterozygosity (deletion, mitotic recombination, or chromosome loss of the remaining wild-type allele) and epigenetic hits such as promoter hypermethylation, which silences transcription. Oncogenes, by contrast, are activated (not tumor suppressors inactivated) by gene amplification, and gene silencing is generally associated with histone deacetylation, not acetylation.
WATCH OUT FOR
Confusing gene amplification (an oncogene-activation mechanism, e.g. HER2/MYC copy number gain) with tumor suppressor inactivation, or confusing histone acetylation (which opens chromatin and activates transcription) with histone deacetylation (which is the actual silencing mechanism associated with HDACs).
Full solution
Q.40MSQ · 2 marksOfficial answer: A, B, C
Genetics, Cell and Molecular Biology › Gene expression
Promoter CpG methylation silences a gene in three ways that act together. Directly, it stops activators, and through them RNA polymerase, from binding. Indirectly, it recruits methyl-CpG-binding repressors such as MeCP2 and MBD2. Those repressors bring in histone deacetylases, which turn the region into heterochromatin.
WATCH OUT FOR
Rejecting (A) on the grounds that RNA polymerase does not bind methylated DNA directly. The question says "can inhibit". Methylation stops the transcription factors that recruit RNA polymerase II from binding, so the polymerase is never loaded onto the promoter. That is still inhibition by preventing RNA polymerase binding. The other common error is mixing up heterochromatin (silent) with euchromatin (active).
The question and its official answer
Q.41MSQ · 2 marksOfficial answer: A, B, D
Genetics, Cell and Molecular Biology › Cell biology
Induced pluripotent stem cells (iPSCs) are self-renewing, pluripotent cells generated from somatic cells by forced expression of reprogramming transcription factors. Like other pluripotent stem cells, their undifferentiated, self-renewing state is not a passive default — it must be actively maintained by specific extrinsic signals/culture conditions, and withdrawing these signals triggers spontaneous differentiation. Because iPSCs arise from a genetic reprogramming procedure in the first place, they are inherently amenable to further genetic manipulation (e.g., CRISPR editing, reporter knock-ins, gene correction) and are widely used to derive three-dimensional organoids (e.g., brain, intestinal, kidney, cardiac organoids) in culture.
WATCH OUT FOR
Confusing "genetically manipulated" with "genetically abnormal" or assuming pluripotent cells are somehow protected from manipulation because they are so tightly regulated; in fact iPSCs are created BY genetic manipulation (forced expression of reprogramming factors such as Oct4, Sox2, Klf4, c-Myc) and are routinely further engineered (e.g. CRISPR gene editing, reporter knock-ins), so (C) is clearly false. A second common error is treating "self-renewal" as needing no external input, which would wrongly make (B) look false or contradictory to (A) — self-renewal of pluripotent stem cells is not spontaneous/unconditional; it is actively maintained by specific signalling inputs, so (A) and (B) are both true and not in conflict.
Full solution
Q.42MSQ · 2 marksOfficial answer: A, C
Q.43MSQ · 2 marksOfficial answer: A, B, C
Q.44MSQ · 2 marksOfficial answer: A, B, C
Recombinant DNA Technology and Tools › Enzymes & vectors
Direct (physical/chemical, vector-independent) DNA delivery methods for plants include microparticle bombardment (biolistics/gene gun), electroporation, and PEG-mediated uptake (typically into protoplasts); Agrobacterium-mediated transformation is an indirect, biological-vector method relying on the natural Ti-plasmid T-DNA transfer machinery.
WATCH OUT FOR
Students sometimes misclassify Agrobacterium-mediated transformation as a "direct" method because it is the most common and efficient transformation technique taught, forgetting that "direct" specifically means vector-independent (no biological carrier); Agrobacterium's Ti-plasmid T-DNA system makes it a vector-mediated (indirect) method regardless of its ubiquity.
Full solution
Q.45MSQ · 2 marksOfficial answer: A, B
Recombinant DNA Technology and Tools › Enzymes & vectors
Cloning vector choice depends on insert-capacity range: plasmids (e.g. pUC19) hold only a few kb, cosmids are capped by lambda phage packaging at ~30-45 kb, while BACs (F-plasmid based, 100-300 kb) and YACs (linear yeast chromosome with telomeres/centromere, 100 kb to several Mb) are the vectors built for large (>100 kb) inserts such as 220 kb.
WATCH OUT FOR
Confusing cosmids (packaging-limited by the lambda phage head to ~30-45 kb total) with BACs, or assuming any "artificial chromosome" vector name alone (without checking insert-capacity range) is enough — some students pick cosmids because they associate "large insert" loosely with cosmids rather than knowing the phage-packaging ceiling.
Full solution
Q.46NAT · 2 marksOfficial answer: 0.41 to 0.44
Fundamentals of Biological Engineering › Growth kinetics
Respiratory quotient is moles of CO2 evolved divided by moles of O2 consumed. Both numbers can be read straight off the growth stoichiometry, so no elemental balancing is needed. Very reduced substrates such as alkanes give RQ well below 1.
WATCH OUT FOR
Inverting the ratio (O2/CO2 = 12.5/5.37 = 2.33). Another is dividing by the moles of biomass or NH3 instead of O2. A third is assuming RQ is about 1, as it is for glucose. For a highly reduced substrate like hexadecane most of the oxygen goes to water and biomass, so RQ is far below 1.
Full solution
Q.47NAT · 2 marksOfficial answer: 2.24 to 2.26
Fundamentals of Biological Engineering › Enzyme kinetics
Divide the Arrhenius equation at one temperature by the same equation at the other. The pre-exponential factor cancels, leaving $ln (k_{2} / k_{1}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ . Only $E_{a}$ , $R$ and the two absolute temperatures are needed, provided $E_{a}$ and $R$ use the same energy unit.
WATCH OUT FOR
Mixing kcal and cal. Putting $E_{a} = 15$ over $R = 1.9872$ gives a ratio of about 1.00. Also watch the sign and order of $(\frac{1}{T _{1}} - \frac{1}{T _{2}})$ : reversing it gives the inverse ratio, about 0.44. Approximating the bracket as $Δ T / T^{2}$ with $T = 300 K$ inflates the answer to about 2.31. Finally, rounding $R$ to 2 gives 2.2400, which only just sits on the edge of the key, so keep $R = 1.9872$ as printed.
Full solution
Q.48NAT · 2 marksOfficial answer: 10
Fundamentals of Biological Engineering › Bioreactor operation
In a chemostat at steady state, $μ = D$ . The oxygen supplied must equal the oxygen the cells consume, so $OTR = OUR = \frac{D X}{Y _{X / O}}$ .
WATCH OUT FOR
Multiplying by the yield instead of dividing, which gives $D X Y_{X / O} = 0.4$ . Another is reporting the specific uptake rate $q_{O_{2}} = μ / Y_{X / O} = 1 h^{- 1}$ without multiplying by the biomass concentration $X$ . A yield is biomass made per unit of oxygen used, so the oxygen needed is biomass made divided by the yield.
Full solution
Q.49NAT · 2 marksOfficial answer: 94 to 96
Fundamentals of Biological Engineering › Downstream
In a two-phase partition, the fraction of solute in the upper phase depends only on the product $K R$ (partition coefficient times the upper-to-lower volume ratio): $Y_{U} = \frac{K R}{1 + K R}$ . A large upper-phase volume raises the yield even when $K$ is modest. Here $K R = 20$ , so only $1/21$ of the enzyme stays in the lower phase.
WATCH OUT FOR
Using $K$ alone and writing $\frac{K}{1 + K} = 80%$ , which ignores the volume ratio. Another slip is inverting $R$ ( $V_{L} / V_{U} = 0.2$ ), which gives $\frac{0.8}{1.8} = 44%$ . A third is reporting the lower-phase fraction, $\frac{1}{1 + K R} = 4.76%$ , as the yield.
Full solution
Q.50NAT · 2 marksOfficial answer: -2.74 to -2.71
Q.51NAT · 2 marksOfficial answer: 15
Recombinant DNA Technology and Tools › Bioinformatics
An unrooted bifurcating tree for $n$ taxa has $(2 n - 3)$ branches, and placing a root on any one of these branches produces a distinct rooted tree. So rooted trees = (number of unrooted trees) $\times (2 n - 3)$ .
WATCH OUT FOR
Confusing the two standard counting formulas — using $(2 n - 3)!!$ (rooted count) as if it directly gave the unrooted count, or forgetting that the multiplier linking unrooted to rooted trees is the number of branches ( $2 n - 3$ ), not the number of taxa ( $n$ ) or internal nodes.
Full solution
Q.52NAT · 2 marksOfficial answer: 945
Q.53NAT · 2 marksOfficial answer: 4.6 to 4.9
Fundamentals of Biological Engineering › Bioreactor operation
All you need is a cell balance around the cell separator. Nothing grows there, so at steady state the cells coming in with the reactor outflow, $(F_{0} + F_{r}) X_{1}$ , must equal the cells leaving in the recycle, $F_{r} X_{0}$ , plus those leaving in the product, $F_{0} X$ . No growth kinetics ( $μ$ , $D$ , $K_{s}$ ) are needed. Dividing by $F_{0}$ gives $X = X_{1} (1 + α - α C)$ , with $α = F_{r} / F_{0}$ and $C = X_{0} / X_{1}$ .
WATCH OUT FOR
Three mistakes come up often. (1) Giving the stream into the separator a flow of $F_{0}$ instead of $F_{0} + F_{r}$ . That gives $X = X_{1} (1 - α C) = - 0.37 X_{1}$ , a negative concentration, which is impossible. (2) Writing the balance over the reactor instead. That brings in growth and needs $μ$ , which the question does not give. (3) Getting the sign wrong on the recycle term, e.g. $X = X_{1} (1 + α + α C) = 2.75 X_{1}$ . With recycle concentrated at the separator, the product concentration must come out below $X_{1}$ , not above it.
Full solution
Q.54NAT · 2 marksOfficial answer: 20
Fundamentals of Biological Engineering › Enzyme kinetics
$V_{ma x} = k_{c a t} [E_{t}]$ , then solve Michaelis-Menten for $K_{m}$ : $K_{m} = [A] (\frac{V _{ma x}}{V _{0}} - 1)$ .
WATCH OUT FOR
Forgetting to convert $[E_{t}]$ from nM to $μ$ M. Using $30$ directly gives $V_{ma x} = 15000 μ M s^{- 1}$ and a $K_{m}$ that is wrong by orders of magnitude. The other slip is treating $V_{0}$ as $V_{ma x}$ .
Full solution
Q.55NAT · 2 marksOfficial answer: 74
Genetics, Cell and Molecular Biology › Mutation & repair
In double-stranded DNA, Chargaff's rule gives %A = %T and %G = %C, and all four bases sum to 100%, so %G + %C = 100% − 2(%A).
WATCH OUT FOR
Assuming A = G = C = T = 25% by default instead of applying Chargaff's pairing rule to the given 13% adenine value, or forgetting to double the adenine percentage before subtracting from 100%.
Full solution
Q.56NAT · 2 marksOfficial answer: 1.0 to 1.2
Genetics, Cell and Molecular Biology › Mutation & repair
Convert the Earth–Moon distance to nanometres, divide by the 0.34 nm rise per base pair to get the total number of base pairs needed, then multiply by the mass per base pair (derived from the given 1×10⁻¹⁸ g per 1000 bp) to get the total mass, converting grams to milligrams at the end.
WATCH OUT FOR
Forgetting to convert km to nm before dividing by the 0.34 nm rise (mixing units gives a wildly wrong bp count), or dividing by 1000 bp instead of multiplying back up when converting mass-per-1000-bp to mass-per-bp, or missing the final g-to-mg conversion (leaving the answer as 0.0011 instead of 1.1).
Full solution
Q.57NAT · 2 marksOfficial answer: 0.165 to 0.168
Recombinant DNA Technology and Tools › Editing & interactions
In a FRET (Förster resonance energy transfer) system, the donor's total decay rate is $k_{0} = k_{r} + k_{n r}$ without any acceptor. Each acceptor nearby opens one additional non-radiative decay pathway (energy transfer) with rate $k_{T}$ . Since all three sites sit at the vertices of an equilateral triangle, the donor is equidistant from both other vertices, so the transfer rate to two independent acceptors adds: total transfer rate becomes $2 k_{T}$ , not some geometry-dependent combination.
WATCH OUT FOR
Assuming energy transfer to two acceptors combines non-additively (e.g., only the nearer acceptor matters, or the rates combine as some average) instead of simply summing the two independent FRET pathways, since Förster transfer rates from a single donor to multiple non-interacting acceptors are additive.
Full solution
Q.58NAT · 2 marksOfficial answer: 10551
Q.59NAT · 2 marksOfficial answer: -2
Q.60NAT · 2 marksOfficial answer: 5
Q.61NAT · 2 marksOfficial answer: 0.45
Q.62NAT · 2 marksOfficial answer: 1
Q.63NAT · 2 marksOfficial answer: 1437 to 1439
Q.64NAT · 2 marksOfficial answer: 4
Fundamentals of Biological Engineering › Process control
Proportional band and gain are reciprocals when both are expressed in percent of span: $K_{c} = 100/ PB (%)$ . A 25% band means a quarter of the input span swings the output through its full range, so the gain is 4. The span of the measured variable cancels out.
WATCH OUT FOR
Using the temperature range directly. For example, $100%/17.5^{\circ} C \approx 5.71 % /^{\circ} C$ is a gain with units, not the dimensionless gain the question asks for. Other slips are $70/25 = 2.8$ , or reading PB = 25% as $K_{c} = 0.25$ (confusing the band with the gain, which is its reciprocal).
Full solution
Q.65NAT · 2 marksOfficial answer: 1600
Recombinant DNA Technology and Tools › PCR & sequencing
Track dilution in two stages using $C_{1} V_{1} = C_{2} V_{2}$ separately: (1) each 20 mM dNTP stock is diluted from 40 μL into the pooled 160 μL master-mix, giving each species a concentration of 5 mM, so the four species sum to a total dNTP concentration of 20 mM in the master-mix; (2) this master-mix is itself diluted 4 μL into a 50 μL PCR reaction, scaling the total 20 mM down by the factor 4/50.
WATCH OUT FOR
Confusing "total dNTPs" with a single dNTP's concentration — some solvers compute only 400 μM (one species) and forget to multiply by 4 for the sum of dATP+dCTP+dGTP+dTTP, or forget the mM→μM conversion factor of 1000.
Full solution

WHERE THE MARKS CAME FROM

Marks by subject in GATE BT 2023
Subject	Qs	Marks
Genetics, Cell and Molecular Biology	13	18
Recombinant DNA Technology and Tools	10	15
Fundamentals of Biological Engineering	7	14
Not on a subject page yet	35	53

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

MORE GATE BT

Every GATE BT paper GATE qualifying marks by paper How to read the GATE answer key The most repeated GATE topics GATE BT coaching in Kolkata