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GATE BT 2026 question paper with solutions

All 65 questions of the GATE 2026 Biotechnology paper, in order, with the official answer key from IIT Guwahati. 27 have a full solution; 2 more show the official answer only.

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

questions · 100 marks: 65questions · 100 marks
with a full solution: 27with a full solution
question types: 31 MCQ · 14 MSQ · 20 NATquestion types

GATE BT 2026 answer key

Official answers to every question of GATE BT 2026
Question	Type	Marks	Answer
Q.1	MCQ	1	A
Q.2	MCQ	1	B
Q.3	MCQ	1	A
Q.4	MCQ	1	B
Q.5	MCQ	1	C
Q.6	MCQ	2	D
Q.7	MCQ	2	A
Q.8	MCQ	2	D
Q.9	MCQ	2	C
Q.10	MCQ	2	C
Q.11	MCQ	1	B
Q.12	MCQ	1	C
Q.13	MCQ	1	B
Q.14	MCQ	1	B
Q.15	MCQ	1	C
Q.16	MCQ	1	B
Q.17	MCQ	1	A
Q.18	MCQ	1	B
Q.19	MCQ	1	B
Q.20	MCQ	1	C
Q.21	MCQ	1	C
Q.22	MCQ	1	B
Q.23	MCQ	1	C
Q.24	MCQ	1	A
Q.25	MCQ	1	C
Q.26	MCQ	1	D
Q.27	MSQ	1	B, C
Q.28	MSQ	1	B, D
Q.29	MSQ	1	B, D
Q.30	MSQ	1	A
Q.31	NAT	1	0.505
Q.32	NAT	1	20
Q.33	NAT	1	40000
Q.34	NAT	1	-106.12
Q.35	NAT	1	0.016 to 0.017
Q.36	MCQ	2	B
Q.37	MCQ	2	C
Q.38	MCQ	2	B
Q.39	MCQ	2	D
Q.40	MCQ	2	A
Q.41	MSQ	2	C, D
Q.42	MSQ	2	A, C
Q.43	MSQ	2	B, C
Q.44	MSQ	2	C
Q.45	MSQ	2	D
Q.46	MSQ	2	B, C
Q.47	MSQ	2	B
Q.48	MSQ	2	A, B
Q.49	MSQ	2	A, D
Q.50	MSQ	2	A, D
Q.51	NAT	2	0.69 to 0.71
Q.52	NAT	2	8
Q.53	NAT	2	1.414
Q.54	NAT	2	59 to 61
Q.55	NAT	2	2.2 to 2.4
Q.56	NAT	2	100
Q.57	NAT	2	1.3 to 1.4
Q.58	NAT	2	8.3 to 8.7
Q.59	NAT	2	30
Q.60	NAT	2	6
Q.61	NAT	2	2.4 to 2.5
Q.62	NAT	2	3.6 to 3.8
Q.63	NAT	2	96
Q.64	NAT	2	0.5
Q.65	NAT	2	0.31 to 0.32

General Aptitude Q.1 to Q.10

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

Biotechnology Q.11 to Q.65

Q.11MCQ · 1 markOfficial answer: B
Q.12MCQ · 1 markOfficial answer: C
Q.13MCQ · 1 markOfficial answer: B
Q.14MCQ · 1 markOfficial answer: B
Recombinant DNA Technology and Tools › Assays & electrophoresis
Single-cell-resolved assays (flow cytometry) preserve heterogeneity information that bulk/pooled assays (immunoblotting, ELISA, qRT-PCR) inherently average away.
WATCH OUT FOR
Picking immunoblotting or ELISA because they are the standard protein-detection techniques for TP53, without noticing that both require a pooled/homogenized lysate and therefore report only an averaged, sample-level value — they cannot reveal cell-to-cell (intra-tumoral) heterogeneity.
Full solution
Q.15MCQ · 1 markOfficial answer: C
Fundamentals of Biological Engineering › Process control
A controller transfer function always acts on the error $E (s)$ , and $\frac{1}{τ _{I} s}$ means integrate in time. So the integral term adds up $e = y_{s p} - y_{m}$ , the set point minus the measured variable. It keeps changing the controller output until that error is zero, and that is how PI control removes steady-state offset.
WATCH OUT FOR
Picking (A) because 'manipulated variable' is confused with 'measured variable'. The manipulated variable is what the controller adjusts, not what it compares with the set point. Picking (D) because 'input minus output' sounds like an error. The error is specifically set point minus measured value, not the difference between any input and output.
Full solution
Q.16MCQ · 1 markOfficial answer: B
Q.17MCQ · 1 markOfficial answer: A
Q.18MCQ · 1 markOfficial answer: B
Q.19MCQ · 1 markOfficial answer: B
Fundamentals of Biological Engineering › Downstream
Gel filtration separates by size. Large molecules are kept out of the bead pores and come out first. Small molecules enter the pores and come out last. For globular proteins, $V_{e}$ falls roughly linearly with $lo g M$ , so the elution order is simply the order of decreasing molecular mass.
WATCH OUT FOR
Getting the elution order backwards. In ion-exchange or affinity chromatography, a molecule is held back by binding. In gel filtration nothing binds: the small proteins are the ones slowed down, because they wander through the pores. Students who think big molecules are 'held up' pick Thrombin (C) as the third to elute. That is the second protein counting from the small end.
Full solution
Q.20MCQ · 1 markOfficial answer: C
Genetics, Cell and Molecular Biology › Gene expression
Peptidyl transferase is the ribozyme activity of the bacterial 50S subunit that catalyzes peptide bond formation; chloramphenicol specifically inhibits this step, distinguishing it from other translation-targeting antibiotics that act at different points (A-site tRNA binding, translocation, etc.).
WATCH OUT FOR
Confusing tetracycline (blocks aminoacyl-tRNA binding at the A site) with chloramphenicol (directly inhibits the peptidyl transferase reaction itself) — both act on the ribosome during elongation, but only chloramphenicol targets peptidyl transferase specifically.
Full solution
Q.21MCQ · 1 markOfficial answer: C
Fundamentals of Biological Engineering › Enzyme kinetics
When $[S] ≪ K_{M}$ , the Michaelis-Menten equation reduces to $v = (k_{cat} / K_{M}) [E]_{0} [S]$ . The ratio $k_{cat} / K_{M}$ therefore measures both turnover ( $k_{cat}$ ) and binding ( $1/ K_{M}$ ) in one number, and diffusion caps it at about $1 0^{8}$ to $1 0^{9} M^{- 1} s^{- 1}$ .
WATCH OUT FOR
Choosing $k_{cat}$ alone (A). That is the turnover number: how fast a substrate-saturated enzyme works, with binding ignored. Options (B) and (D) reduce to $[E]_{0}$ and $1/ [E]_{0}$ because $V_{m a x} = k_{cat} [E]_{0}$ . Both depend on how much enzyme is present, so neither describes the enzyme's efficiency.
Full solution
Q.22MCQ · 1 markOfficial answer: B
Genetics, Cell and Molecular Biology › Gene expression
In bacteria, DNA Polymerase I is the only replicative polymerase with 5'→3' exonuclease activity, letting it simultaneously excise RNA primers and fill the gap with DNA (nick translation); DNA ligase then seals the remaining nick.
WATCH OUT FOR
Confusing DNA Polymerase III (the main chain-extending replicative enzyme) with DNA Polymerase I (the primer-removal/gap-filling enzyme) — students often forget Pol III lacks 5'→3' exonuclease activity and cannot excise RNA primers.
Full solution
Q.23MCQ · 1 markOfficial answer: C
Q.24MCQ · 1 markOfficial answer: A
Q.25MCQ · 1 markOfficial answer: C
Q.26MCQ · 1 markOfficial answer: D
Recombinant DNA Technology and Tools › Editing & interactions
Natural CRISPR/Cas9 in bacteria uses Cas9 protein, a CRISPR repeat-spacer array, and PAM sequences on target DNA — all naturally occurring. The single-guide RNA (sgRNA) is a synthetic, laboratory-engineered fusion of the natural crRNA and tracrRNA into one chimeric molecule, created for genome-editing applications; it does not occur in this fused form in nature.
WATCH OUT FOR
Assuming any RNA involved in Cas9 targeting must be "natural" because RNA-guided cleavage itself is a natural bacterial process, and missing that the single-molecule sgRNA construct specifically is an engineered fusion (crRNA + tracrRNA) not found in native bacterial CRISPR systems.
Full solution
Q.27MSQ · 1 markOfficial answer: B, C
Q.28MSQ · 1 markOfficial answer: B, D
Q.29MSQ · 1 markOfficial answer: B, D
Genetics, Cell and Molecular Biology › Cell biology
The extracellular matrix (ECM) is built mainly of structural fibrous proteins (like collagen) and adhesive glycoproteins (like fibronectin) that are secreted by cells into the space outside them; motor proteins (myosin) and nuclear lamina proteins (lamins) are intracellular, not ECM components.
WATCH OUT FOR
Confusing "structural protein" in general with "extracellular matrix protein" specifically — students sometimes mark Lamin because it is also a structural (intermediate filament) protein, forgetting that lamins line the INNER nuclear membrane and are strictly intracellular, not secreted into the ECM.
Full solution
Q.30MSQ · 1 markOfficial answer: A
Recombinant DNA Technology and Tools › Bioinformatics
BLAST aligns sequences with sequences. Its extensions differ only in whether the query and the database are nucleotide or protein, with six-frame translation (blastx, tblastn, tblastx) linking the two. A protein query against nucleotide data is exactly what tblastn does. Structures, ligands and functional categories are not sequence databases.
WATCH OUT FOR
Ticking (C) because BLAST can be run against the sequences of PDB entries. That searches a sequence database taken from structure entries, not the structures themselves. Comparing a protein with 3D structures needs structure-alignment or threading methods, not BLAST.
The question and its official answer
Q.31NAT · 1 markOfficial answer: 0.505
Q.32NAT · 1 markOfficial answer: 20
Fundamentals of Biological Engineering › Enzyme kinetics
Turnover number $k_{c a t} = V_{ma x} / [E]_{T}$ . It is the rate at saturation divided by the concentration of enzyme doing the work. Once the concentration units cancel, only a time unit is left, so convert it to what the question asks for.
WATCH OUT FOR
Stopping at $1200$ because the question asks for $s^{- 1}$ and $V_{ma x}$ was given per minute. Dividing by 60 is required. A second slip is multiplying by 60 instead of dividing, which gives $72000 s^{- 1}$ .
Full solution
Q.33NAT · 1 markOfficial answer: 40000
Q.34NAT · 1 markOfficial answer: -106.12
Q.35NAT · 1 markOfficial answer: 0.016 to 0.017
Recombinant DNA Technology and Tools › Assays & electrophoresis
One International Unit (IU) of enzyme activity is defined as the amount of enzyme that converts 1 µmol of substrate per minute under the specified assay conditions. Specific activity is total enzyme activity (IU) divided by the total mass of enzyme protein (µg) present in the sample, so it is independent of the assay volume chosen for the calculation — any consistent volume basis (1 mL, 1 L, etc.) gives the same ratio.
WATCH OUT FOR
A common error is to divide by 1000 twice (once converting mM to M and again converting mmol to µmol) or to forget the hour-to-minute conversion (dividing/multiplying by 60 incorrectly), which changes the final specific activity by one or two orders of magnitude. Another common slip is reporting activity in IU/mL (total activity concentration) instead of IU/µg (specific activity, i.e., normalized to enzyme mass).
Full solution
Q.36MCQ · 2 marksOfficial answer: B
Fundamentals of Biological Engineering › Process control
For a first-order process $G_{p} (s) = K_{p} / (τ_{p} s + 1)$ driven by a step of size $M$ ( $U (s) = M / s$ ), $Y (s) = M K_{p} / [s (τ_{p} s + 1)]$ and $y (t) = M K_{p} (1 - e^{- t / τ_{p}})$ . The output starts at 0, reaches 63.2% of its final value at $t = τ_{p}$ , and settles at gain times step size, $M K_{p}$ .
WATCH OUT FOR
Using the unit-step response $K_{p} (1 - e^{- t / τ_{p}})$ (option A) and forgetting that the input here has magnitude $M$ . Because the system is linear, $M$ scales the whole response, both the constant term and the exponential term. Options (C) and (D) put $M$ on only one of the two terms, so they fail $y (0) = 0$ .
Full solution
Q.37MCQ · 2 marksOfficial answer: C
Q.38MCQ · 2 marksOfficial answer: B
Genetics, Cell and Molecular Biology › Population & evolution
Each disorder maps to one specific molecular/chromosomal defect: Klinefelter (47,XXY, extra X in a male), Turner (45,X, X insufficiency), Bloom syndrome (BLM helicase mutation causing genomic instability), and Xeroderma pigmentosum (NER pathway defect impairing repair of UV damage).
WATCH OUT FOR
Confusing Bloom syndrome (a RecQ helicase defect causing chromosome instability/sister-chromatid exchange) with Xeroderma pigmentosum (an NER defect specific to UV-damage repair) — both are DNA-repair-related disorders, but only XP is a nucleotide excision repair defect, and only Bloom is a helicase mutation.
Full solution
Q.39MCQ · 2 marksOfficial answer: D
Q.40MCQ · 2 marksOfficial answer: A
Fundamentals of Biological Engineering › Growth kinetics
Ash is inorganic and is left out of the formula. Taking it out scales every element's percentage by the same factor (1/0.88), so the atom ratios stay the same. Convert each mass to moles (divide by atomic mass), then divide every mole count by the moles of N: C/N = (50/12)/(10/14) = 5.83, H/N = 8/(10/14) = 11.2, O/N = (20/16)/(10/14) = 1.75.
WATCH OUT FOR
Using mass ratios instead of mole ratios. The mass ratio C:N = 50/10 = 5.0, which is exactly the trap in option (B). You must divide by atomic masses first. Two other mistakes: thinking the 12% ash changes the atom ratios (it only rescales every element by the same factor), and ruling out (A) because 1.75 would round to 1.8. The option truncates to 1.7, and no other option comes close on C, H or O.
Full solution
Q.41MSQ · 2 marksOfficial answer: C, D
Q.42MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Gene expression
Mitochondrial translation reuses the universal genetic code except that AGA (and AGG) are reassigned from Arg codons to stop codons; so the same mRNA can yield different-length products in the two compartments depending on whether an in-frame AGA is read through (cytosol) or halts translation (mitochondria).
WATCH OUT FOR
Assuming there is only one reading frame (only the first AUG) to consider, and thus missing that a second, shorter-cytosolic-product ORF (from the internal AUG) becomes the LONGEST mitochondrial product once the first ORF is truncated by the AGA-as-stop rule; also assuming AGA truncation only removes the AGA and downstream residues without checking whether AGA itself even occurs in the alternate frame (it doesn't — that frame reads CGA instead).
Full solution
Q.43MSQ · 2 marksOfficial answer: B, C
Genetics, Cell and Molecular Biology › Mutation & repair
Classify a point mutation on two independent axes at once: (1) its effect on the encoded amino acid/reading (silent / missense / nonsense) by translating the codon before and after the change, and (2) the chemical nature of the base substitution itself (transition: purine↔purine or pyrimidine↔pyrimidine, vs transversion: purine↔pyrimidine) — an MSQ can require both classifications to be marked correct simultaneously.
WATCH OUT FOR
Forgetting that the printed 30-nt segment is the LAST 10 codons of a 330-nt (110-codon) message and mis-numbering which codon contains position 322 — or picking only the coding-consequence answer (nonsense, B) while overlooking that the question also independently asks about the transversion/transition nature of the base change (C), missing the second correct option in this MSQ.
Full solution
Q.44MSQ · 2 marksOfficial answer: C
Fundamentals of Biological Engineering › Bioreactor operation
A DO reading shows the balance between oxygen transfer (OTR) and oxygen uptake (OUR). With aeration running, a zero reading means the cells take up oxygen as fast as the largest rate at which it can dissolve, $k_{L} a C^{*}$ . The culture is alive and respiring hard, and its growth rate is capped by oxygen transfer rather than by $μ_{max}$ .
WATCH OUT FOR
Reading "zero oxygen" as "no oxygen reached the cells, so they died" (A). Dead cells stop consuming oxygen, so DO would rise towards saturation. A second mistake is confusing a zero reading with carbon exhaustion (B), which actually makes DO spike upwards because respiration stops.
The question and its official answer
Q.45MSQ · 2 marksOfficial answer: D
Q.46MSQ · 2 marksOfficial answer: B, C
Fundamentals of Biological Engineering › Bioreactor operation
$k_{L} a$ depends on hydrodynamics. Anything that raises the interfacial area $a = 6 ε_{G} / d_{b}$ or the interfacial turbulence (more gas flow, smaller bubbles, more power input) raises it. Changing the gas composition changes only the saturation concentration $C^{*}$ , so it changes the driving force in $OTR = k_{L} a (C^{*} - C_{L})$ and leaves $k_{L} a$ alone.
WATCH OUT FOR
Ticking (A) because pure oxygen does raise the oxygen transfer rate. It does that by raising $C^{*}$ (the driving force), not the volumetric mass transfer coefficient $k_{L} a$ . The question asks about $k_{L} a$ , not OTR.
Full solution
Q.47MSQ · 2 marksOfficial answer: B
Q.48MSQ · 2 marksOfficial answer: A, B
Genetics, Cell and Molecular Biology › Cell signalling
Posttranslational modifications (PTMs) act on proteins (here, histones) after translation; only histone marks — not chemical changes to DNA bases — qualify as PTMs, even though DNA methylation is also an epigenetic mechanism.
WATCH OUT FOR
Treating "epigenetic" as synonymous with "posttranslational modification" and picking cytosine methylation (C) because it is a well-known epigenetic mark — it modifies DNA, not a protein, so it fails the PTM criterion in the stem even though it is epigenetic.
Full solution
Q.49MSQ · 2 marksOfficial answer: A, D
Fundamentals of Biological Engineering › Ideal reactors
In ideal plug flow every fluid element has the same residence time. That means a flat velocity profile, complete mixing across each cross-section and zero mixing along the flow direction (dispersion number $D_{a x} / (uL) = 0$ ). The tanks-in-series RTD has variance $1/ N$ , which goes to zero as $N \to \infty$ and gives the same RTD.
WATCH OUT FOR
Mixing up the two directions of mixing. An ideal PFR is perfectly mixed RADIALLY, across each cross-section, but has NO mixing AXIALLY, along the flow. So 'significant axial mixing' and 'significant back mixing' are false even though mixing across the section is complete. Another mistake is to reject (A) because a single CSTR is the opposite of a PFR. It is the limit of MANY CSTRs in series that approaches plug flow.
Full solution
Q.50MSQ · 2 marksOfficial answer: A, D
Q.51NAT · 2 marksOfficial answer: 0.69 to 0.71
Q.52NAT · 2 marksOfficial answer: 8
Q.53NAT · 2 marksOfficial answer: 1.414
Q.54NAT · 2 marksOfficial answer: 59 to 61
Q.55NAT · 2 marksOfficial answer: 2.2 to 2.4
Fundamentals of Biological Engineering › Growth kinetics
Net specific growth rate = Monod growth rate minus the endogenous decay rate, $μ_{n e t} = \frac{μ _{ma x} S}{K _{s} + S} - k_{d}$ . First put every quantity in the same units: S and K_s both in mg L^-1, and both rates in day^-1. Here S is 23000 times K_s, so the Monod term is almost exactly $μ_{ma x}$ .
WATCH OUT FOR
Mixing units. If you subtract 0.1 day^-1 from 0.1 h^-1 you get zero. If you use S = 23 with K_s = 1 without converting g to mg, you get 0.0958 h^-1 and a net rate near 2.2 day^-1, which only lands in the range by luck. The right way is to convert S to 23000 mg L^-1 and mu_max to 2.4 day^-1, then subtract k_d.
Full solution
Q.56NAT · 2 marksOfficial answer: 100
Fundamentals of Biological Engineering › Bioreactor operation
Mean cell retention time is the biomass held in the reactor divided by the rate at which biomass leaves it: $θ_{c} = \frac{V X}{F X _{e}}$ when cells leave only in the effluent. With cell recycle, $θ_{c}$ equals the hydraulic retention time $V / F$ multiplied by $X / X_{e}$ .
WATCH OUT FOR
Answering with the hydraulic retention time $V / F = 10$ days, which is the mean cell retention time only when there is no recycle and $X_{e} = X$ . A second mistake is to divide by $F X$ , using the reactor concentration in place of the effluent concentration, which also gives 10 days.
Full solution
Q.57NAT · 2 marksOfficial answer: 1.3 to 1.4
Q.58NAT · 2 marksOfficial answer: 8.3 to 8.7
Q.59NAT · 2 marksOfficial answer: 30
Fundamentals of Biological Engineering › Upstream
First-order death gives $ln (N_{0} / N) = k_{d} t$ at constant temperature. The time depends only on the ratio $N_{0} / N$ (here $1 0^{9}$ , a 9-log reduction), so the medium volume does not matter. A quick check: the decimal reduction time is $D = ln 10/ k_{d} = 2.3026/0.69 = 3.34$ min, and 9 log cycles take $9 \times 3.34 \approx 30$ min.
WATCH OUT FOR
Using $lo g_{10}$ where the natural log belongs: $lo g_{10} (1 0^{9}) /0.69 = 13$ min is wrong, because $k_{d}$ is defined through $e^{- k_{d} t}$ . Another slip is taking the ratio as $1 0^{10} /1$ , or mixing up which concentration goes on top, which changes the number of log cycles.
Full solution
Q.60NAT · 2 marksOfficial answer: 6
Fundamentals of Biological Engineering › Enzyme kinetics
In a zero-order reaction the rate is constant, so concentration falls linearly with time: $S = S_{0} - k_{0} t$ . Get $k_{0}$ from the drop you are told about, then complete consumption takes $t = S_{0} / k_{0}$ .
WATCH OUT FOR
Answering 2 h, which is only the time left after the first 4 h, when the question asks for the total time from the start. The other mistake is using first-order kinetics (a logarithmic decay), which never reaches zero concentration and so cannot give a finite time for complete utilization.
Full solution
Q.61NAT · 2 marksOfficial answer: 2.4 to 2.5
Fundamentals of Biological Engineering › Bioreactor operation
With sterile feed, a chemostat washes out once the dilution rate D = F/V reaches D_crit = mu_max S0/(Ks + S0). This is the growth rate the cells would have if the reactor held undiluted feed. Here Ks is much smaller than S0, so D_crit is almost exactly mu_max, and F_crit = D_crit V.
WATCH OUT FOR
Mixing units: Ks is given in mg L^-1 and S0 in g L^-1. Adding 0.1 to 10 without converting gives D_crit = 0.495 h^-1 and F = 2.48 L h^-1. That happens to fall inside the key range here, but it is the wrong method. Other slips: reporting the dilution rate (0.5 h^-1) instead of the flow rate (multiply by V = 5 L), or using the approximate formula D_crit = mu_max without checking that Ks is much smaller than S0.
Full solution
Q.62NAT · 2 marksOfficial answer: 3.6 to 3.8
Fundamentals of Biological Engineering › Upstream
This is an elemental mass balance. The formula $C_{5} H_{7} O_{2} N$ gives the mass fraction of nitrogen in the cells, $14/113$ . That fraction times the cell mass is the least nitrogen the medium must supply. Phosphorus is then 20% of that nitrogen, by weight.
WATCH OUT FOR
Taking 20% of the cell mass (30 mg) instead of 20% of the nitrogen mass. Another slip is to work on a mole basis, where 0.2 mol P per mol N would give $0.2 \times 150/113 \times 31 \approx 8.2 mg$ . The question says weight/weight, so the 20% applies to the mass of nitrogen.
Full solution
Q.63NAT · 2 marksOfficial answer: 96
Genetics, Cell and Molecular Biology › Cell biology
Track cell number through exponential doubling, applying the one-time apoptotic loss exactly when it occurs (at t = 48 h), then resume doubling from the reduced population for the remaining time.
WATCH OUT FOR
Applying the apoptosis loss at the END of the 7-day period (168 − 1 = ... ) instead of at t = 48 h, or forgetting that "no more cell death occurs" means only ONE cell is ever lost, not a repeated loss at every doubling.
Full solution
Q.64NAT · 2 marksOfficial answer: 0.5
Genetics, Cell and Molecular Biology › Mendelian genetics
For an X-linked recessive disease, a son receives his only X chromosome from his mother (his Y comes from the father), so the father's genotype is irrelevant to a son's risk — only the mother's carrier (heterozygous) genotype matters.
WATCH OUT FOR
Trying to factor in the father's genotype for a son's risk (fathers cannot pass an X-linked allele to sons), or confusing this with a daughter's risk, which would require the father's X allele and give a different probability.
Full solution
Q.65NAT · 2 marksOfficial answer: 0.31 to 0.32
Fundamentals of Biological Engineering › Bioreactor operation
Superficial velocity is the flow rate divided by the area of the empty column, $u_{0} = Q / A$ . The liquid actually moves only through the voids, so the interstitial velocity is higher: $u_{i} = u_{0} / ε$ . The void fraction is therefore the ratio $u_{0} / u_{i}$ .
WATCH OUT FOR
Writing $ε = u_{i} / u_{0}$ (about 3.14, which cannot be right because a void fraction must be below 1), or dividing $Q$ by $A$ without first converting litres to cm $^{3}$ (1 L = 1000 cm $^{3}$ ). Some candidates also get distracted by the 1 m bed length and try to use a bed volume, but the length plays no part in the void fraction.
Full solution

WHERE THE MARKS CAME FROM

Marks by subject in GATE BT 2026
Subject	Qs	Marks
Fundamentals of Biological Engineering	16	28
Genetics, Cell and Molecular Biology	9	15
Recombinant DNA Technology and Tools	4	4
Not on a subject page yet	36	53

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

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