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GATE BT 2024 question paper with solutions

All 65 questions of the GATE 2024 Biotechnology paper, in order, with the official answer key from IISc Bengaluru. 32 have a full solution.

The full paper (PDF)Official answer key (IISc Bengaluru)

questions · 100 marks: 65questions · 100 marks
with a full solution: 32with a full solution
question types: 40 MCQ · 10 MSQ · 15 NATquestion types

GATE BT 2024 answer key

Official answers to every question of GATE BT 2024
Question	Type	Marks	Answer
Q.1	MCQ	1	A
Q.2	MCQ	1	D
Q.3	MCQ	1	A
Q.4	MCQ	1	C
Q.5	MCQ	1	A
Q.6	MCQ	2	C
Q.7	MCQ	2	A
Q.8	MCQ	2	C
Q.9	MCQ	2	A
Q.10	MCQ	2	A
Q.11	MCQ	1	D
Q.12	MCQ	1	A
Q.13	MCQ	1	D
Q.14	MCQ	1	D
Q.15	MCQ	1	D
Q.16	MCQ	1	B
Q.17	MCQ	1	A
Q.18	MCQ	1	D
Q.19	MCQ	1	A
Q.20	MCQ	1	A
Q.21	MCQ	1	A
Q.22	MCQ	1	B
Q.23	MCQ	1	A
Q.24	MCQ	1	D
Q.25	MCQ	1	C
Q.26	MCQ	1	A
Q.27	MCQ	1	A
Q.28	MCQ	1	C
Q.29	MCQ	1	D
Q.30	MCQ	1	C
Q.31	MCQ	1	C
Q.32	MCQ	1	B
Q.33	NAT	1	1
Q.34	NAT	1	2
Q.35	NAT	1	2.3
Q.36	MCQ	2	A
Q.37	MCQ	2	B
Q.38	MCQ	2	D
Q.39	MCQ	2	A
Q.40	MCQ	2	D
Q.41	MCQ	2	B
Q.42	MCQ	2	C
Q.43	MCQ	2	A
Q.44	MSQ	2	B, C
Q.45	MSQ	2	A, C, D
Q.46	MSQ	2	A, D
Q.47	MSQ	2	A, B, C
Q.48	MSQ	2	B, D
Q.49	MSQ	2	A, D
Q.50	MSQ	2	A, C
Q.51	MSQ	2	A, D
Q.52	MSQ	2	A, B, C
Q.53	MSQ	2	C
Q.54	NAT	2	33 to 35
Q.55	NAT	2	80
Q.56	NAT	2	625
Q.57	NAT	2	-24
Q.58	NAT	2	13.00 to 14.00
Q.59	NAT	2	110
Q.60	NAT	2	0.1
Q.61	NAT	2	50
Q.62	NAT	2	6
Q.63	NAT	2	1012
Q.64	NAT	2	3
Q.65	NAT	2	3.11 to 3.13

General Aptitude Q.1 to Q.10

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

Biotechnology Q.11 to Q.65

Q.11MCQ · 1 markOfficial answer: D
Fundamentals of Biological Engineering › Downstream
In chromatography, the mechanism follows from the kind of interaction. With net charges on solute and matrix you have ion exchange (ionic or electrostatic binding). With a covalent bond you have covalent chromatography (chemisorption). Uncharged solutes on silica or alumina are held by weak, reversible physisorption, which means van der Waals forces, with hydrogen bonding to the surface silanols helping polar solutes.
WATCH OUT FOR
Picking (B) because silica silanols are polar and "adsorption is electrostatic". In this question, electrostatic and ionic binding mean attraction between net charges, which is ion-exchange chromatography, and the solutes are stated to be uncharged. Another mistake is picking (A) because adsorption sounds like "bonding". A covalent bond would not let the solute come off again when the eluent is changed.
Full solution
Q.12MCQ · 1 markOfficial answer: A
Fundamentals of Biological Engineering › Process control
A process's order is the highest power of $s$ in its transfer-function denominator. $\frac{K _{p}}{τ _{p} s + 1}$ is the standard first-order lag: one pole at $s = - 1/ τ_{p}$ , and its step response settles at $K_{p} M$ after reaching 63.2% of that value at $t = τ_{p}$ .
WATCH OUT FOR
Calling it "purely capacitive" because a single tank is a capacity. A purely capacitive (integrating) process is $K / s$ , with no $+ 1$ term and a pole at the origin, and its output ramps without bound after a step. $\frac{K _{p}}{τ _{p} s + 1}$ is a capacity with a resistance, and its output settles.
Full solution
Q.13MCQ · 1 markOfficial answer: D
Q.14MCQ · 1 markOfficial answer: D
Fundamentals of Biological Engineering › Bioreactor operation
$N_{R e} = ρN D_{i}^{2} / μ$ rises with impeller speed and falls with viscosity. When the dimensionless mixing time $N t_{m}$ is plotted against $N_{R e}$ , it falls through the laminar and transitional ranges and then levels off to a constant, set by the geometry, in the turbulent range.
WATCH OUT FOR
Picking C because it mixes up where the terms sit ( $μ$ is in the denominator, $N$ is in the numerator). Picking B because more vigorous flow feels like it should need longer to settle. Rejecting D because $t_{m}$ still falls when $N$ is raised: the independence is in the number of revolutions needed, $N t_{m}$ , and it means that a change in $N_{R e}$ through viscosity or density leaves $t_{m}$ unchanged.
Full solution
Q.15MCQ · 1 markOfficial answer: D
Q.16MCQ · 1 markOfficial answer: B
Genetics, Cell and Molecular Biology › Mendelian genetics
The law of segregation is a single-locus rule: the two alleles at a gene (on homologous chromosomes) separate from each other during meiosis I, so each gamete gets exactly one allele.
WATCH OUT FOR
Confusing the law of segregation (single-locus: the two alleles of one gene separate) with the law of independent assortment (multi-locus: alleles of different genes assort independently), leading students to pick "linked" or "unlinked genes" options that describe assortment/linkage, not segregation.
Full solution
Q.17MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Cell biology
Co-translational translocation means the nascent polypeptide is threaded into/across a membrane while translation is still ongoing; this is the defining feature of the SRP-Sec61 pathway into the rough ER, whereas mitochondrial and peroxisomal import (and Golgi delivery) occur post-translationally.
WATCH OUT FOR
Confusing co-translational translocation (ER, via SRP/Sec61, threading the chain in as it is synthesized) with post-translational import (mitochondria via TOM/TIM, peroxisomes via PTS1/PTS2/Pex5-Pex7), where the full-length or nearly full-length protein is imported only after release from the ribosome.
Full solution
Q.18MCQ · 1 markOfficial answer: D
Q.19MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Cell biology
Digit individualization during limb development occurs by programmed cell death (apoptosis) of the interdigital mesenchyme/webbing, a classic, well-characterized example of developmental apoptosis (regulated by BMP signaling in the interdigital regions).
WATCH OUT FOR
Confusing apoptosis (programmed, regulated cell death used in developmental sculpting) with necrosis (unregulated, injury-induced cell death causing inflammation) — necrosis is never the mechanism of normal developmental tissue remodeling.
Full solution
Q.20MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Mendelian genetics
In somatic cell nuclear transfer (SCNT), the clone's nuclear genome comes entirely from the somatic-nucleus donor; the enucleated-ovum donor contributes only mitochondria (excluded here by the question), and the surrogate mother contributes no genetic material at all, only the uterine environment for gestation.
WATCH OUT FOR
Assuming the surrogate mother R, "of the same strain as Q," must therefore also be genetically related to Q (or to S) and choosing (D) Both P and R — surrogacy provides gestation only, not genetic contribution.
Full solution
Q.21MCQ · 1 markOfficial answer: A
Q.22MCQ · 1 markOfficial answer: B
Recombinant DNA Technology and Tools › Enzymes & vectors
Among common laboratory cell lines, Sf9 (from Spodoptera frugiperda, fall armyworm) is the standard insect cell line, used with the baculovirus expression vector system (BEVS) for recombinant protein production — distinct from mammalian lines (HEK 293, CHO) and bacterial strains (DH5α).
WATCH OUT FOR
Confusing DH5α (a bacterial, i.e. prokaryotic, E. coli strain) with a "cell line," or not knowing that HEK 293 and CHO are both mammalian lines used for recombinant protein expression, leading to a wrong pick among the two mammalian options.
Full solution
Q.23MCQ · 1 markOfficial answer: A
Recombinant DNA Technology and Tools › PCR & sequencing
Sanger (dideoxy chain-termination) sequencing works by randomly incorporating 2',3'-dideoxynucleotides (ddNTPs), which lack the 3'-OH needed to form the next phosphodiester bond, so DNA synthesis terminates at that position — producing a nested set of fragments whose lengths, when separated, reveal the sequence.
WATCH OUT FOR
Confusing "chain elongation" (option B) with the correct mechanism; students sometimes forget that a ddNTP's missing 3'-OH group is what blocks further elongation, mistakenly thinking any nucleotide incorporation event described in an option could be the "principle," rather than recognizing termination specifically.
Full solution
Q.24MCQ · 1 markOfficial answer: D
Q.25MCQ · 1 markOfficial answer: C
Q.26MCQ · 1 markOfficial answer: A
Q.27MCQ · 1 markOfficial answer: A
Genetics, Cell and Molecular Biology › Mutation & repair
A pseudogene is, by definition, a gene copy (arising from duplication or retrotransposition) that has lost the ability to produce the ancestral gene's original, functional protein through mutations, premature stop codons, frameshifts, or loss of essential regulatory elements — this loss of original protein-coding capacity is the one property that is true of every pseudogene without exception, unlike the other listed outcomes (altered-function proteins from rare intact ORFs, retained coding-like RNA sequence, or regulatory/antisense ncRNA roles), each of which is documented in only a subset of pseudogenes.
WATCH OUT FOR
Assuming "pseudogene" means "produces no RNA or protein product at all," and therefore picking an option about RNA (C or D) as the "impossible" one — when in fact many pseudogenes are actively transcribed (some even translated into truncated/altered products or functioning as regulatory ncRNAs); the only property universal to ALL pseudogenes is the definitional loss of the capacity to yield the original, functional protein.
Full solution
Q.28MCQ · 1 markOfficial answer: C
Q.29MCQ · 1 markOfficial answer: D
Q.30MCQ · 1 markOfficial answer: C
Q.31MCQ · 1 markOfficial answer: C
Q.32MCQ · 1 markOfficial answer: B
Q.33NAT · 1 markOfficial answer: 1
Fundamentals of Biological Engineering › Growth kinetics
Degree of reduction is counted per carbon atom (per C-mol), not per molecule. Add up the valences C $= 4$ , H $= 1$ , O $= - 2$ , N $= - 3$ and divide by the number of carbons: $γ = (4 w + x - 2 y - 3 z) / w$ .
WATCH OUT FOR
Stopping at the per-molecule count of 2 available electrons and forgetting to divide by the two carbon atoms. Using $+ 2$ for oxygen instead of $- 2$ is another slip, and it gives $(8 + 2 + 8) /2 = 9$ .
Full solution
Q.34NAT · 1 markOfficial answer: 2
Fundamentals of Biological Engineering › Growth kinetics
A division rate $k$ counts generations per unit time, and one generation doubles the population, so the doubling time is simply its reciprocal: $t_{d} = 1/ k$ . The formula $t_{d} = ln 2/ μ$ applies only when the rate given is the specific growth rate $μ$ in $d X / d t = μ X$ . The two are related by $μ = k ln 2$ .
WATCH OUT FOR
Treating $0.5 h^{- 1}$ as the specific growth rate $μ$ and writing $t_{d} = ln 2/ μ = 0.693/0.5 = 1.39 h$ . The question gives the rate at which the cells divide, which is a division rate, so $t_{d} = 1/ k = 2 h$ .
Full solution
Q.35NAT · 1 markOfficial answer: 2.3
Fundamentals of Biological Engineering › Upstream
In first-order death kinetics, the D-value (time for a 90% kill, a one-log reduction) depends only on the death rate constant: $D = \frac{ln 10}{k _{d}} \approx \frac{2.303}{k _{d}}$ . The temperature (120 °C) only tells you which $k_{d}$ applies. It does not enter the calculation.
WATCH OUT FOR
Using $lo g_{10}$ where $ln$ belongs gives $D = 1/ k_{d} = 1.0$ min. Using $ln 2$ , which gives the half-life of 0.7 min, is the other common slip. $k_{d}$ is a natural-log rate constant, so a tenfold reduction needs $ln 10 = 2.303$ . Some students also try to use 120 °C in an Arrhenius correction, but $k_{d}$ is already given at that temperature.
Full solution
Q.36MCQ · 2 marksOfficial answer: A
Q.37MCQ · 2 marksOfficial answer: B
Q.38MCQ · 2 marksOfficial answer: D
Genetics, Cell and Molecular Biology › Gene expression
Classify each enzyme by its template (DNA or RNA) and its product (DNA or RNA): Primase = DNA-dependent RNA synthesis (RNA primer on a DNA template); Reverse transcriptase = RNA-dependent DNA synthesis (retroviral cDNA from an RNA genome); RNA Replicase = RNA-dependent RNA synthesis (RNA virus genome replication, e.g. Qβ phage); DNA Polymerase III = DNA-dependent DNA synthesis (the main bacterial replicative polymerase).
WATCH OUT FOR
Confusing reverse transcriptase with RNA replicase because both act on RNA — the distinguishing feature is the PRODUCT: reverse transcriptase's product is DNA (hence "RNA-dependent DNA synthesis"), while RNA replicase's product is RNA (hence "RNA-dependent RNA synthesis").
Full solution
Q.39MCQ · 2 marksOfficial answer: A
Q.40MCQ · 2 marksOfficial answer: D
Recombinant DNA Technology and Tools › PCR & sequencing
Match each reagent to its standard molecular-biology use: DEPC inactivates RNases (protects RNA in solution), CsCl forms a density gradient for DNA separation by ultracentrifugation, ethidium bromide intercalates into nucleic acids and is used for staining in gels, and EDTA chelates Mg²⁺ to inhibit nuclease activity during DNA purification.
WATCH OUT FOR
Confusing ethidium bromide's role (a DNA/nucleic-acid intercalating stain used generically, including with agarose gels for RNA when specified) with EDTA's chelation function, or forgetting that CsCl requires ultracentrifugation to form the density gradient that separates DNA by buoyant density — leading to swapped R/S or Q/S assignments.
Full solution
Q.41MCQ · 2 marksOfficial answer: B
Recombinant DNA Technology and Tools › Microscopy & spectroscopy
Match each physical component/label to the biophysical technique that relies on it: a blue (488 nm) laser excites fluorophores in FACS; a tungsten filament is the thermionic electron source in electron microscopy; $^{15}$ N isotope labelling enables multidimensional NMR of proteins; polyacrylamide gels are the standard matrix for electrophoretic separation.
WATCH OUT FOR
Confusing which physical component belongs to which technique — e.g., assuming a laser is used in electron microscopy (it isn't; EM uses an electron beam from a filament) or that polyacrylamide is used in NMR rather than electrophoresis.
Full solution
Q.42MCQ · 2 marksOfficial answer: C
Genetics, Cell and Molecular Biology › Population & evolution
Each genetic disorder in Column I is matched to its correct molecular/genetic basis in Column II by recalling the specific gene or chromosomal defect underlying each disease.
WATCH OUT FOR
Confusing xeroderma pigmentosum (a DNA repair defect, NER) with a numerical chromosomal abnormality, or mixing up Tay-Sachs disease (an enzyme-deficiency lysosomal storage disorder) with Down Syndrome (a trisomy), since both involve well-known eponymous disease names that students sometimes swap.
Full solution
Q.43MCQ · 2 marksOfficial answer: A
Genetics, Cell and Molecular Biology › Population & evolution
Convergent evolution = unrelated/distantly related lineages independently evolving analogous (not homologous) structures under similar selective pressure — the hallmark being different developmental/anatomical origin but similar function. Bat wings (modified tetrapod forelimb) and insect wings (exoskeletal/cuticular outgrowth) are the classic textbook pair illustrating exactly this.
WATCH OUT FOR
Confusing convergent evolution with parallel evolution: parallel evolution requires related lineages that started from a similar ancestral structure/genetic background and then evolved similar traits independently under similar pressures, whereas convergent evolution involves distantly related lineages with fundamentally different starting structures (homology absent) converging on a similar function. Since bats (chordates) and insects (arthropods) share no winged common ancestor and their wings arise from entirely different tissues, this is convergence, not parallelism — a distinction students often blur.
Full solution
Q.44MSQ · 2 marksOfficial answer: B, C
Fundamentals of Biological Engineering › Enzyme kinetics
An uncompetitive inhibitor binds only to the ES complex. Both apparent constants are divided by $α^{'} = 1 + [I] / K_{I}^{'}$ , so $V_{ma x}$ and $K_{m}$ fall together, their ratio stays the same, and Lineweaver-Burk lines come out parallel.
WATCH OUT FOR
Mixing up uncompetitive and noncompetitive inhibition. Choosing (D) describes noncompetitive or mixed inhibition, where the inhibitor binds both E and ES. Choosing (A) describes competitive inhibition. Another common slip is to think that K_m goes up under uncompetitive inhibition. It actually goes down, by the same factor as V_max.
Full solution
Q.45MSQ · 2 marksOfficial answer: A, C, D
Q.46MSQ · 2 marksOfficial answer: A, D
Recombinant DNA Technology and Tools › PCR & sequencing
RFLP distinguishes alleles based on differences in the length of DNA fragments produced by restriction enzyme digestion; anything that changes the position or number of restriction sites (point mutations altering recognition sites, or variable-number tandem repeats lying between two fixed sites) changes fragment length and is detectable by RFLP.
WATCH OUT FOR
Confusing RFLP (a DNA-fragment-size-based marker technique) with genetics processes like recombination or segregation, which describe allele behaviour during meiosis but do not themselves alter restriction fragment sizes.
Full solution
Q.47MSQ · 2 marksOfficial answer: A, B, C
Q.48MSQ · 2 marksOfficial answer: B, D
Genetics, Cell and Molecular Biology › Cell biology
The eukaryotic cell cycle has surveillance checkpoints (G1/S, intra-S, G2/M, and the spindle-assembly checkpoint) that halt progression into or through mitosis whenever the genome is not yet safely ready to be segregated — most relevantly here, when DNA replication is not finished (G2/M / replication checkpoint, ATR-Chk1 mediated) or when DNA damage cannot be repaired (DNA-damage checkpoint, ATM/ATR-p53 mediated, often followed by apoptosis or senescence). Conditions that are simply not part of mammalian mitotic biology (plant-specific cell-plate cytokinesis; meiosis-specific chiasmata) are not "conditions that cause mitotic failure" — they are non-sequiturs for a somatic mitotic cell.
WATCH OUT FOR
Confusing cell-cycle checkpoint failures (which block mitosis) with processes that are simply absent or irrelevant in mammalian somatic mitosis — picking (A) by forgetting cell plate formation is a plant-cytokinesis mechanism (not part of animal mitosis at all), or picking (C) by forgetting chiasmata form during meiotic prophase I, not mitosis, so it has no bearing on whether a somatic cell enters or completes mitosis.
Full solution
Q.49MSQ · 2 marksOfficial answer: A, D
Q.50MSQ · 2 marksOfficial answer: A, C
Genetics, Cell and Molecular Biology › Gene expression
A reading frame (open reading frame) is defined by the start codon AUG, not by the first nucleotide of the sequence. Find the first AUG from the 5' end, read triplets from there up to the first in-frame stop codon, and count the stop codon as a codon but not as an amino acid.
WATCH OUT FOR
The usual error is chopping the sequence into triplets from nucleotide 1 (UGA UGA GCC …), which gives 10 codons and suggests option (B). That frame starts with a stop codon and has no AUG, so it is not a reading frame. The second trap is counting the stop codon UAA as an amino acid, which gives 9 amino acids instead of 8.
Full solution
Q.51MSQ · 2 marksOfficial answer: A, D
Genetics, Cell and Molecular Biology › Gene expression
The lac operon is under dual control: negative control by the LacI repressor (relieved by allolactose when lactose is present) and positive control by the CAP–cAMP complex (active when glucose is absent/low). Full induction of β-galactosidase expression needs both conditions simultaneously.
WATCH OUT FOR
Picking only (D) and forgetting that full/appreciable induction of the lac operon also needs low glucose (so that cAMP–CAP can positively activate the promoter) — students often treat lac operon control as repressor-only and ignore the CAP/cAMP catabolite-repression arm, missing option (A).
Full solution
Q.52MSQ · 2 marksOfficial answer: A, B, C
Fundamentals of Biological Engineering › Growth kinetics
A batch reactor is closed to liquid flow. Only the conditions inside the vessel (pH, osmolarity, temperature, dissolved oxygen, substrate concentration through Monod kinetics) can change growth. Flow variables such as feed rate or dilution rate belong to fed-batch and continuous operation.
WATCH OUT FOR
Ticking D because substrate supply clearly matters for growth. In a batch, substrate is supplied only once, as the starting concentration, which is option C. A feed rate exists only in fed-batch or continuous culture. The opposite slip is leaving out B because osmolarity seems less important than nutrients, but water activity and osmotic stress directly change the growth rate and the lag time.
Full solution
Q.53MSQ · 2 marksOfficial answer: C
Fundamentals of Biological Engineering › Bioreactor operation
In a chemostat, washout means $D$ is higher than the largest growth rate the culture can reach, $μ (S_{0})$ . The cells are flushed out ( $X = 0$ ). With no cells there is no substrate uptake, so the reactor acts as a plain flow-through vessel and the exit substrate concentration equals the feed concentration ( $S = S_{0}$ ).
WATCH OUT FOR
Mixing up washout with maximum biomass productivity. Productivity $D X$ is highest at an optimum dilution rate below $D_{cr i t}$ , but at washout the biomass is zero. Another mistake is to think the exit substrate must still be lower than the feed, or even zero. With no cells, nothing consumes substrate, so $S = S_{0}$ .
Full solution
Q.54NAT · 2 marksOfficial answer: 33 to 35
Fundamentals of Biological Engineering › Upstream
The LMTD is the log mean of the temperature differences at the two ends of the exchanger, $Δ T_{l m} = (Δ T_{1} - Δ T_{2}) / ln (Δ T_{1} /Δ T_{2})$ . The flow direction decides which terminal temperatures are paired. In counter-current flow, each stream's inlet sits at the same end as the other stream's outlet.
WATCH OUT FOR
(1) Pairing the ends as if the flow were co-current: $121 - 10 = 111$ and $30 - 70 = - 40$ . The negative difference shows that co-current flow cannot deliver this duty at all, because the water leaves hotter than the medium. (2) Using $lo g_{10}$ in place of $ln$ , which gives $31/0.4065 \approx 76^{\circ} C$ . (3) Using the arithmetic mean $(51 + 20) /2 = 35.5$ , which rounds to 36 and falls outside the key.
Full solution
Q.55NAT · 2 marksOfficial answer: 80
Fundamentals of Biological Engineering › Bioreactor operation
When oxygen supply is the limit, the dissolved oxygen falls to zero and the maximum transfer rate $k_{L} a C^{*}$ just matches the total uptake $q_{O_{2}} X$ . So $X_{ma x} = k_{L} a C^{*} / q_{O_{2}}$ , with $k_{L} a$ and $q_{O_{2}}$ in the same time unit.
WATCH OUT FOR
Leaving $k_{L} a$ in $min^{- 1}$ while $q_{O_{2}}$ is in $s^{- 1}$ gives $60 \times 8 \times 1 0^{- 3} /1 0^{- 4} = 4800$ . Converting the wrong way ( $k_{L} a = 1/60 s^{- 1}$ ) gives about 1.33. A third slip is multiplying by the 10,000 L volume, which has no place here because both rates are already per unit volume.
Full solution
Q.56NAT · 2 marksOfficial answer: 625
Fundamentals of Biological Engineering › Bioreactor operation
First use the impeller Reynolds number $R e_{i} = ρN D_{i}^{2} / μ$ to decide which flow regime you are in, and so which Power number to read from the table. Then rearrange $N_{P} = P / (ρ N^{3} D_{i}^{5})$ to get the power. In fully turbulent flow $N_{P}$ is constant, so $P$ goes as $N^{3} D_{i}^{5}$ .
WATCH OUT FOR
Using the speed in rpm (300) rather than rev/s (5): this gives a power $6 0^{3} = 216, 000$ times too large. A second mistake is leaving the viscosity as 1 instead of converting 1 cp to $1 0^{- 3} Pa s$ . That gives $R e_{i} = 5000$ , which falls in the gap between 500 and $1 0^{5}$ where the table gives no value. The 10,000 L volume is a distractor: power is fixed by $N$ , $D_{i}$ , $ρ$ and $N_{P}$ alone.
Full solution
Q.57NAT · 2 marksOfficial answer: -24
Q.58NAT · 2 marksOfficial answer: 13.00 to 14.00
Fundamentals of Biological Engineering › Growth kinetics
Binary fission doubles the population every generation, so $N = N_{0} \cdot 2^{n}$ and $n = lo g_{10} (N / N_{0}) / lo g_{10} 2 \approx 3.32 lo g_{10} (N / N_{0})$ . Only the fold-increase $N / N_{0}$ matters, not the absolute counts.
WATCH OUT FOR
Using the final count alone ( $n = lo g_{2} 1 0^{6} \approx 19.9$ ) and forgetting the initial $1 0^{2}$ cells. Other slips are mixing logarithm bases (for example $ln 1 0^{4} / lo g_{10} 2$ , or $4/ ln 2 \approx 5.8$ ), and dividing the fold-increase by 2 instead of taking $lo g_{2}$ of it.
Full solution
Q.59NAT · 2 marksOfficial answer: 110
Fundamentals of Biological Engineering › Bioreactor operation
This is an unsteady-state volume balance, dV/dt = F_in − F_out, with an outflow that changes over time. Measure t from the moment the leak starts. The feed of 1 L/min keeps running, so V(t) = 200 + t − t², and V(10) = 110 L.
WATCH OUT FOR
The usual mistake is to assume filling stops when the leak starts. That gives 200 − ∫₀¹⁰ 2t dt = 100 L instead of 110 L, but the question only says a leak develops. Another mistake is multiplying the rate by the time (2 × 10 × 10 = 200 L lost) when the leak rate grows with time and has to be integrated. A third is measuring t from the start of filling: the leak would then already be about 400 L/min, which empties the tank almost at once and makes no physical sense.
Full solution
Q.60NAT · 2 marksOfficial answer: 0.1
Fundamentals of Biological Engineering › Bioreactor operation
At quasi-steady state in a fed-batch reactor, $μ = D = F / V$ , using the volume at that moment. Putting this into Monod kinetics gives $S = K_{S} D / (μ_{m} - D)$ , the same result as a chemostat but with a dilution rate that falls as the volume grows.
WATCH OUT FOR
Using the starting volume or a time-averaged volume instead of the volume at the stated moment (500 L at 2 h). Another mistake is thinking the feed glucose concentration is needed: it fixes the biomass concentration, not the residual substrate. Some students also swap numerator and denominator and write $S = μ_{m} D / (K_{S} - D)$ .
Full solution
Q.61NAT · 2 marksOfficial answer: 50
Fundamentals of Biological Engineering › Bioreactor operation
With geometric similarity, $D \propto V^{1/3}$ . Constant shear means constant impeller tip speed $π N D$ , so $N \propto 1/ D \propto V^{- 1/3}$ . A 1000-fold volume increase gives a 10-fold larger diameter and one tenth of the speed.
WATCH OUT FOR
Scaling the speed by the volume ratio instead of the diameter ratio (giving 0.5 rpm), or using a different criterion by mistake. Constant power per volume gives $N \propto D^{- 2/3}$ , about 108 rpm. Keeping $N$ itself constant gives 500 rpm. Another slip is taking $D_{2} / D_{1} = 1000$ without the cube root.
Full solution
Q.62NAT · 2 marksOfficial answer: 6
Q.63NAT · 2 marksOfficial answer: 1012
Q.64NAT · 2 marksOfficial answer: 3
Q.65NAT · 2 marksOfficial answer: 3.11 to 3.13

WHERE THE MARKS CAME FROM

Marks by subject in GATE BT 2024
Subject	Qs	Marks
Fundamentals of Biological Engineering	16	26
Genetics, Cell and Molecular Biology	11	17
Recombinant DNA Technology and Tools	5	8
Not on a subject page yet	33	49

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

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