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GATE BT 2025 question paper with solutions

All 65 questions of the GATE 2025 Biotechnology paper, in order, with the official answer key from IIT Roorkee. 27 have a full solution; 1 more shows the official answer only.

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

questions · 100 marks: 65questions · 100 marks
with a full solution: 27with a full solution
question types: 33 MCQ · 15 MSQ · 17 NATquestion types

GATE BT 2025 answer key

Official answers to every question of GATE BT 2025
Question	Type	Marks	Answer
Q.1	MCQ	1	D
Q.2	MCQ	1	D
Q.3	MCQ	1	C
Q.4	MCQ	1	A
Q.5	MCQ	1	B
Q.6	MCQ	2	B
Q.7	MCQ	2	C
Q.8	MCQ	2	A
Q.9	MCQ	2	B
Q.10	MCQ	2	B
Q.11	MCQ	1	B
Q.12	MCQ	1	A
Q.13	MCQ	1	A
Q.14	MCQ	1	D
Q.15	MCQ	1	A
Q.16	MCQ	1	A
Q.17	MCQ	1	A
Q.18	MCQ	1	A
Q.19	MCQ	1	D
Q.20	MCQ	1	D
Q.21	MSQ	1	B, D
Q.22	MSQ	1	A, D
Q.23	MSQ	1	A, C, D
Q.24	MSQ	1	B, D
Q.25	MSQ	1	A, D
Q.26	MSQ	1	A, B, C
Q.27	MSQ	1	B, D
Q.28	NAT	1	0.20 to 0.25
Q.29	NAT	1	15450 to 15750
Q.30	NAT	1	0.5
Q.31	NAT	1	300 to 310
Q.32	NAT	1	0.94 to 0.96
Q.33	NAT	1	0.72 to 0.75
Q.34	NAT	1	4
Q.35	NAT	1	0.16 to 0.18
Q.36	MCQ	2	A
Q.37	MCQ	2	A
Q.38	MCQ	2	A
Q.39	MCQ	2	C
Q.40	MCQ	2	A
Q.41	MCQ	2	A
Q.42	MCQ	2	C
Q.43	MCQ	2	A
Q.44	MCQ	2	C
Q.45	MCQ	2	B
Q.46	MCQ	2	C
Q.47	MCQ	2	B
Q.48	MCQ	2	A
Q.49	MSQ	2	A, B, D
Q.50	MSQ	2	A, C
Q.51	MSQ	2	A, D
Q.52	MSQ	2	B, C
Q.53	MSQ	2	A, C
Q.54	MSQ	2	A, B, C
Q.55	MSQ	2	A, B, C
Q.56	MSQ	2	B, C
Q.57	NAT	2	0.76 to 0.82
Q.58	NAT	2	3.30 to 3.60
Q.59	NAT	2	11
Q.60	NAT	2	1
Q.61	NAT	2	48000 to 49000
Q.62	NAT	2	3.4 to 3.8
Q.63	NAT	2	1
Q.64	NAT	2	23.2 to 23.8
Q.65	NAT	2	-32.00 to -30.00

General Aptitude Q.1 to Q.10

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

Biotechnology Q.11 to Q.65

Q.11MCQ · 1 markOfficial answer: B
Q.12MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Microbial genetics
Diphtheria toxin production by Corynebacterium diphtheriae is not encoded by the bacterial chromosome — it requires lysogenic conversion by the temperate corynephage β (bearing the tox gene), a classic textbook example of phage/lysogenic conversion in bacterial pathogenesis.
WATCH OUT FOR
Confusing this with other toxin-conversion systems (e.g., epsilon toxin of Clostridium perfringens) or with well-known model phages (T4, a lytic E. coli phage; lambda, a lysogenic E. coli phage) that have no role in C. diphtheriae toxigenicity.
Full solution
Q.13MCQ · 1 markOfficial answer: A
Q.14MCQ · 1 markOfficial answer: D
Q.15MCQ · 1 markOfficial answer: A
Q.16MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Gene expression
piRNAs (Piwi-interacting RNAs) are small non-coding RNAs (~24-31 nt) found chiefly in germline cells, where they associate with Piwi-clade Argonaute proteins to silence transposable elements and protect genome integrity during gametogenesis.
WATCH OUT FOR
Confusing the "pi" in piRNA (Piwi-interacting) with "P-element interacting," or confusing which molecule methylates which — piRNAs themselves are 2'-O-methylated at their 3' end by the methyltransferase Hen1, they are not agents that methylate other molecules.
Full solution
Q.17MCQ · 1 markOfficial answer: A
Q.18MCQ · 1 markOfficial answer: A
Q.19MCQ · 1 markOfficial answer: D
Q.20MCQ · 1 markOfficial answer: D
Recombinant DNA Technology and Tools › Libraries & cloning
CHO (Chinese hamster ovary) cells are the industry-standard mammalian host cell line for producing glycosylated recombinant therapeutic proteins (e.g., monoclonal antibodies, EPO) because they perform human-compatible N- and O-linked glycosylation and other post-translational modifications that non-mammalian systems cannot replicate.
WATCH OUT FOR
Confusing "can glycosylate proteins" with "is a mammalian host": Pichia pastoris (yeast) and Sf9 cells (insect, baculovirus system) both glycosylate proteins but produce non-human-type glycan patterns and are not mammalian cell lines, so they fail the question's explicit "mammalian cell culture" requirement even though they are valid non-mammalian expression hosts in other contexts.
Full solution
Q.21MSQ · 1 markOfficial answer: B, D
Q.22MSQ · 1 markOfficial answer: A, D
Genetics, Cell and Molecular Biology › Gene expression
miRNA biogenesis requires two sequential RNase III-family endonucleases: nuclear Drosha (with DGCR8) converts pri-miRNA to pre-miRNA, and cytoplasmic Dicer converts pre-miRNA to the mature miRNA duplex, which is then loaded into RISC (Argonaute).
WATCH OUT FOR
Confusing Cas9 (a CRISPR genome-editing nuclease with no role in endogenous miRNA processing) or XRCC4 (a DNA double-strand break repair protein) with the RNase III enzymes that actually process miRNA precursors.
Full solution
Q.23MSQ · 1 markOfficial answer: A, C, D
Fundamentals of Biological Engineering › Downstream
Name the physical property each process uses to tell molecules apart. Gel filtration uses pore exclusion, ultrafiltration uses a membrane MWCO, and ultracentrifugation uses sedimentation rate ( $v \propto d_{p}^{2}$ ). All three are size-based. Ion exchange uses net charge.
WATCH OUT FOR
Leaving out ultracentrifugation on the grounds that centrifugation is 'density-based'. Only isopycnic (equilibrium density-gradient) runs separate by buoyant density. In differential and rate-zonal runs the sedimentation velocity scales with $d_{p}^{2}$ , so the separation is by size and mass. The other common slip is to think ion exchange depends on size because larger proteins carry more charges, but the separation is set by net charge (pI relative to pH), not size.
Full solution
Q.24MSQ · 1 markOfficial answer: B, D
Q.25MSQ · 1 markOfficial answer: A, D
Q.26MSQ · 1 markOfficial answer: A, B, C
Fundamentals of Biological Engineering › Growth kinetics
The degree of reduction per C-mole depends only on the C-mole formula, $γ = 4 + a - 2 b - 3 c$ for $CH_{a} O_{b} N_{c}$ . So any compounds that divide down to the same C-mole formula (here $CH_{2} O$ ) must have the same $γ$ .
WATCH OUT FOR
Working out the degree of reduction per mole instead of per C-mole (glucose 24, lactic acid 12, acetic acid 8) and deciding the three are different. A second slip is to treat formic acid like the others and miss that it carries two oxygens on its single carbon, which gives $γ = 2$ , not 4.
Full solution
Q.27MSQ · 1 markOfficial answer: B, D
Fundamentals of Biological Engineering › Downstream
Where the product is decides the downstream train. An intracellular product needs cell disruption, and if it forms inclusion bodies it also needs solubilisation and refolding. A soluble extracellular product skips both: remove the cells, concentrate the dilute broth (ultrafiltration/diafiltration), then purify it by chromatography.
WATCH OUT FOR
Picking (C) because recombinant proteins made in E. coli often form inclusion bodies, or (A) because downstream processing is taught as starting with cell disruption. Both steps apply only to intracellular products. This question says outright that the protein is secreted and soluble.
Full solution
Q.28NAT · 1 markOfficial answer: 0.20 to 0.25
Fundamentals of Biological Engineering › Growth kinetics
In exponential growth $X = X_{0} e^{μ t}$ , so the specific growth rate and the doubling time are linked by $μ t_{d} = ln 2$ . That gives $μ = 0.693/ t_{d}$ .
WATCH OUT FOR
Taking $μ = 1/ t_{d} = 0.33 h^{- 1}$ , which mixes up the specific growth rate with the number of doublings per hour. Using $lo g_{10} 2$ instead of $ln 2$ , which gives $0.10 h^{- 1}$ . Both answers fall outside the key. The natural log is needed because $μ$ is defined through $e^{μ t}$ .
Full solution
Q.29NAT · 1 markOfficial answer: 15450 to 15750
Q.30NAT · 1 markOfficial answer: 0.5
Genetics, Cell and Molecular Biology › Mendelian genetics
A test cross between a homozygous recessive individual (cc) and a heterozygous carrier (Cc) produces offspring in a 1:1 ratio of heterozygous (Cc, normal phenotype) to homozygous recessive (cc, albino) genotypes, since the recessive parent contributes only the c allele and the heterozygous parent contributes C or c with equal probability.
WATCH OUT FOR
Confusing this test cross with a standard Cc × Cc monohybrid cross (which gives a 3:1 phenotypic ratio) and answering 0.75 instead of recognizing that one parent is homozygous recessive, giving a 1:1 ratio and an answer of 0.5.
Full solution
Q.31NAT · 1 markOfficial answer: 300 to 310
Genetics, Cell and Molecular Biology › Mutation & repair
Convert protein mass to number of amino acid residues (mass ÷ average residue mass), use the triplet genetic code to get the number of coding base pairs (residues × 3), then multiply by the B-DNA rise per base pair (0.34 nm) to get contour length.
WATCH OUT FOR
Forgetting that each amino acid is encoded by a codon of 3 nucleotides (3 base pairs), and instead multiplying the number of amino acids directly by the rise per base pair (300 × 0.34 = 102 nm), which omits the triplet code entirely and gives a value far outside the key range.
Full solution
Q.32NAT · 1 markOfficial answer: 0.94 to 0.96
Fundamentals of Biological Engineering › Enzyme kinetics
Divide the Michaelis-Menten equation by $V_{m a x}$ to get the fractional saturation $v_{0} / V_{m a x} = [S] / (K_{m} + [S])$ . When $[S] = n K_{m}$ , $K_{m}$ cancels and the ratio is simply $n / (n + 1)$ . Here $n = 20$ , so the ratio is $20/21$ .
WATCH OUT FOR
Assuming the enzyme is fully saturated at $[S] ≫ K_{m}$ and answering 1.00. At $20 K_{m}$ the rate is still only $20/21$ of $V_{m a x}$ . The other common slip is dropping the $K_{m}$ term from the denominator. Writing $20/20$ or $1/20$ instead of $20/21$ also gives an answer outside the key.
Full solution
Q.33NAT · 1 markOfficial answer: 0.72 to 0.75
Q.34NAT · 1 markOfficial answer: 4
Q.35NAT · 1 markOfficial answer: 0.16 to 0.18
Q.36MCQ · 2 marksOfficial answer: A
Q.37MCQ · 2 marksOfficial answer: A
Q.38MCQ · 2 marksOfficial answer: A
Q.39MCQ · 2 marksOfficial answer: C
Q.40MCQ · 2 marksOfficial answer: A
Genetics, Cell and Molecular Biology › Mutation & repair
Match each DNA/RNA-processing enzyme to its precise molecular function: gyrase relieves torsional strain by acting as a swivel, deadenylase is an exoribonuclease that removes the mRNA poly(A) tail, glycosylase excises a damaged base in base-excision repair, and DNA ligase seals nicks by joining a 3'-OH to a 5'-phosphate.
WATCH OUT FOR
Confusing glycosylase (removes only the damaged base, leaving an abasic site) with an endonuclease/exonuclease that removes nucleotides, or confusing gyrase's swivel/supercoiling-relief role with helicase's strand-unwinding role.
Full solution
Q.41MCQ · 2 marksOfficial answer: A
Q.42MCQ · 2 marksOfficial answer: C
Fundamentals of Biological Engineering › Process control
After a step change, a first-order instrument has covered the fraction $1 - e^{- t / τ}$ of its final change by time $t$ . Solving $1 - e^{- t / τ} = 0.95$ gives $t = τ ln 20 \approx 3 τ$ . So an instrument's 95% response time is about three time constants.
WATCH OUT FOR
Taking the time constant itself (40 s) as the settling time. One time constant only gets the instrument to 63.2% of its final value. Other mistakes are picking $2 τ = 80 s$ (86.5%) or $4 τ = 160 s$ (98.2%), or solving $e^{- t / τ} = 0.95$ instead of $0.05$ , which gives about 2 s.
Full solution
Q.43MCQ · 2 marksOfficial answer: A
Fundamentals of Biological Engineering › Process control
With zero initial conditions, take the Laplace transform term by term ( $d y / d t \to s Y (s)$ ) and form $G (s) = Y (s) / F (s)$ . A first-order process always has the standard form $K_{p} / (τ_{p} s + 1)$ : the gain is $G (0) = K_{p}$ and the single pole is at $s = - 1/ τ_{p}$ .
WATCH OUT FOR
Picking D, $K_{p} / [τ_{p} (s + 1)]$ , by dividing the whole equation by $τ_{p}$ and then factoring wrongly. That form changes the steady-state gain to $K_{p} / τ_{p}$ and moves the pole to $s = - 1$ . It is also dimensionally inconsistent, because $s$ has units of $time^{- 1}$ and cannot be added to a pure number $1$ unless it is multiplied by $τ_{p}$ .
Full solution
Q.44MCQ · 2 marksOfficial answer: C
Fundamentals of Biological Engineering › Bioreactor operation
The power number $N_{P} = P / (ρ N_{i}^{3} D_{i}^{5})$ is constant in the turbulent regime (about 5 to 6 for a baffled Rushton turbine). So for a fixed impeller diameter and fluid density, ungassed power goes as $N_{i}^{3}$ (and as $D_{i}^{5}$ ). In the laminar regime $N_{P} \propto 1/ R e_{i}$ , so power goes as $N_{i}^{2}$ instead.
WATCH OUT FOR
Choosing (B) by carrying over the laminar result $P = k_{1} μ N_{i}^{2} D_{i}^{3} \propto N_{i}^{2}$ into the turbulent regime. Another mistake is treating $N_{P}$ as if it still changes with speed in turbulent flow. Once $R e_{i}$ is above about $1 0^{4}$ , $N_{P}$ is flat, and all of the speed dependence sits in the $N_{i}^{3}$ of its definition.
Full solution
Q.45MCQ · 2 marksOfficial answer: B
Q.46MCQ · 2 marksOfficial answer: C
Q.47MCQ · 2 marksOfficial answer: B
Recombinant DNA Technology and Tools › Bioinformatics
Match each bioinformatics tool/database to its core function: BLAST = sequence similarity search; Bowtie = short-read genome alignment; AlphaFold = AI-based structure prediction; PDB = repository of solved 3D structures.
WATCH OUT FOR
Confusing BLAST (sequence similarity search against databanks) with a structure database, or confusing PDB (a repository of experimentally solved structures) with AlphaFold (a predictive AI tool) — both concern protein structure but serve opposite roles (storage vs. prediction).
Full solution
Q.48MCQ · 2 marksOfficial answer: A
Q.49MSQ · 2 marksOfficial answer: A, B, D
Q.50MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Gene expression
Eukaryotic origin licensing (ORC → Cdc6/Cdt1 → Mcm2-7 double hexamer loading) occurs in late M/G1 phase; origin firing (activation of the loaded Mcm helicase into the active CMG complex) is triggered at the G1/S transition by DDK and S-phase Cdks. Licensing and firing are temporally separated so each origin fires only once per cycle.
WATCH OUT FOR
Confusing the order of events — thinking Mcm loading happens in S phase (it happens in G1, forming the pre-replication complex) or thinking Mcm loads/recruits ORC (it is the reverse: ORC binds first and recruits Cdc6/Cdt1 to load Mcm).
Full solution
Q.51MSQ · 2 marksOfficial answer: A, D
Genetics, Cell and Molecular Biology › Cell biology
Intraflagellar transport (IFT) is the bidirectional movement of protein cargo (axonemal precursors, signaling proteins) along the axoneme of a cilium/flagellum, between its base and tip. The axoneme itself (the transport track) is built of microtubule doublets. Anterograde (base→tip) movement of IFT trains is powered by kinesin-2 motors walking along these microtubules; retrograde (tip→base) movement uses cytoplasmic dynein-2. Actin and myosin belong to a separate cytoskeletal transport system and are not part of the ciliary/flagellar IFT machinery.
WATCH OUT FOR
Confusing the microtubule/kinesin-based IFT system (cilia/flagella) with the actin/myosin-based cytoskeletal transport system used elsewhere in the cell (e.g., cortical transport, melanosome movement) — actin and myosin play no role inside the flagellum/IFT machinery.
Full solution
Q.52MSQ · 2 marksOfficial answer: B, C
Genetics, Cell and Molecular Biology › Gene expression
Telomerase is a ribonucleoprotein reverse transcriptase: its RNA subunit (TERC) is the template it copies, and its protein subunit (TERT) extends chromosome ends 5'\to3' using that RNA template — i.e. RNA-dependent DNA polymerase activity.
WATCH OUT FOR
Confusing telomerase with a standard DNA-dependent DNA polymerase (choosing A) because both "make DNA", while missing that telomerase's template is its own RNA, not a DNA strand — and forgetting that all polymerases (including telomerase) synthesise only 5'\to3', which rules out D on direction alone even before considering the template type.
Full solution
Q.53MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Mendelian genetics
A phenotype that can come from more than one genotype has to be averaged over those genotypes. A group A mother can be $I^{A} I^{A}$ or $I^{A} i$ , and with no other information each gets a probability of $\frac{1}{2}$ . The father's AB phenotype fixes him as $I^{A} I^{B}$ . He can never give an $i$ allele, so an O child is impossible. A B child needs the mother to be $I^{A} i$ and to pass $i$ : $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .
WATCH OUT FOR
Assuming a group A mother must be heterozygous ( $I^{A} i$ ). That gives $P (B) = P (AB) = 0.25$ , so option C gets marked wrong. The other slip is treating Rh-positive parents as possibly Rr and multiplying by $\frac{3}{4}$ . The question gives no Rh genotype information, and it expects every child to be Rh-positive.
The question and its official answer
Q.54MSQ · 2 marksOfficial answer: A, B, C
Fundamentals of Biological Engineering › Immobilisation
The effectiveness factor depends only on the Thiele modulus $ϕ = \frac{R}{3} \frac{k _{c a t} E _{0}}{K _{m} D _{e}}$ , and it falls as $ϕ$ rises (in the strongly diffusion-limited regime, $η \approx 1/ ϕ$ ). Anything that lowers $R$ , $k_{c a t}$ or $E_{0}$ , or raises $D_{e}$ , increases $η$ .
WATCH OUT FOR
Treating the effectiveness factor as if it were the reaction rate. More enzyme (D) raises the overall rate, but it lowers $η$ because it raises $ϕ$ . A lower turnover number (C) lowers the rate but raises $η$ , so students often reject C and pick D.
Full solution
Q.55MSQ · 2 marksOfficial answer: A, B, C
Recombinant DNA Technology and Tools › Assays & electrophoresis
Histone post-translational modifications (acetylation, methylation, phosphorylation, etc.) can be identified by antibody-based methods (ChIP-seq for genome-wide location, immunofluorescence for cellular localization) and by direct physical/chemical detection (mass spectrometry, which measures the mass shift the modification causes). Patch-clamp electrophysiology measures ion-channel currents across membranes and has no readout for a covalent chromatin modification.
WATCH OUT FOR
Confusing "identifying/detecting a modification" with "measuring gene expression" — some students wrongly include RT-PCR/qPCR-type reasoning or wrongly exclude ChIP-seq thinking it only maps genomic location, not the modification itself, when the antibody step is precisely what identifies the modification.
Full solution
Q.56MSQ · 2 marksOfficial answer: B, C
Recombinant DNA Technology and Tools › Assays & electrophoresis
A carbon is chiral only if all four groups attached to it are different. In every amino acid the α-carbon (–NH2, –COOH, –H, side chain R) is chiral whenever R ≠ H; a second chiral centre exists only if some side-chain carbon also carries four distinct substituents.
WATCH OUT FOR
Forgetting to check the side-chain β-carbon for chirality and stopping after confirming only the α-carbon — this misses both Threonine and Isoleucine, or wrongly credits Leucine/Asparagine based on chain length alone rather than checking whether all four substituents on a given carbon are actually distinct.
Full solution
Q.57NAT · 2 marksOfficial answer: 0.76 to 0.82
Q.58NAT · 2 marksOfficial answer: 3.30 to 3.60
Fundamentals of Biological Engineering › Growth kinetics
Convert the carbon-basis yield to biomass C-mol (0.4 × 6 = 2.4). Then use the available-electron (degree of reduction) balance: the electrons in the glucose go either into the biomass or to oxygen. With NH3 as the nitrogen source, $γ_{S} = 24$ per mol glucose and $γ_{X} = 4.2$ per C-mol, so $4 n_{O_{2}} = 24 - 4.2 (2.4)$ .
WATCH OUT FOR
Treating 0.4 as mol biomass per mol glucose instead of per C-mol, which gives 0.4 C-mol of biomass instead of 2.4. Using 4 as the degree of reduction for a whole mole of glucose (4 is per C-mol; per mole it is 24). Forgetting that N in NH3 counts as −3, which makes the biomass value 5.0 instead of 4.2.
Full solution
Q.59NAT · 2 marksOfficial answer: 11
Q.60NAT · 2 marksOfficial answer: 1
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
Absorbance and percentage transmittance are related by $A = - lo g_{10} (T)$ , where $T$ is the fraction (0 to 1) of incident light transmitted; percentage transmittance is $T \times 100$ .
WATCH OUT FOR
Confusing absorbance with percentage absorbed directly (e.g. assuming $A = 2$ means 2% or 98% transmittance), instead of using the logarithmic relation $T = 1 0^{- A}$ .
Full solution
Q.61NAT · 2 marksOfficial answer: 48000 to 49000
Q.62NAT · 2 marksOfficial answer: 3.4 to 3.8
Fundamentals of Biological Engineering › Bioreactor operation
At steady state in a chemostat with sterile feed, the specific growth rate equals the dilution rate ( $μ = D = F / V$ ). Putting that into the Monod equation gives the substrate concentration in the reactor, $S = K_{s} D / (μ_{ma x} - D)$ . The yield then gives the biomass, $X = Y_{X / S} (S_{0} - S)$ . Check first that $D$ is below the washout dilution rate.
WATCH OUT FOR
Taking $D$ as the flow rate itself (0.8 h $^{- 1}$ ), which is above $μ_{ma x}$ and so wrongly predicts washout. Another is using the feed concentration $S_{0}$ in place of $S_{0} - S$ and getting $X = 0.4 \times 10 = 4.0$ g/L. It is also possible to swap $K_{s}$ and $μ_{ma x}$ when rearranging the Monod equation for $S$ .
Full solution
Q.63NAT · 2 marksOfficial answer: 1
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
For any square matrix, the trace equals the sum of its eigenvalues and the determinant equals the product of its eigenvalues — two independent checks on the same unknown.
WATCH OUT FOR
Forgetting that the trace–sum-of-eigenvalues relation holds for any square matrix (not just diagonalizable ones), or making a sign slip while expanding the determinant along the row/column containing k.
Full solution
Q.64NAT · 2 marksOfficial answer: 23.2 to 23.8
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
The NMR resonance condition (Larmor equation) relates the applied magnetic field to the resonance (Larmor) frequency through the gyromagnetic ratio of the nucleus: the angular frequency ω = γB₀, and ω = 2πν, so ν = γB₀/(2π), giving B₀ = 2πν/γ.
WATCH OUT FOR
Using ν directly in the Larmor equation ω = γB0 without converting ν to angular frequency (forgetting the factor of 2π), which gives a wrong field strength off by a factor of ~6.28.
Full solution
Q.65NAT · 2 marksOfficial answer: -32.00 to -30.00

WHERE THE MARKS CAME FROM

Marks by subject in GATE BT 2025
Subject	Qs	Marks
Fundamentals of Biological Engineering	11	17
Genetics, Cell and Molecular Biology	10	15
Recombinant DNA Technology and Tools	7	13
Not on a subject page yet	37	55

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

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