GATE BT 2019 question paper with solutions

GATE BT 2019 answer key

General Aptitude GA Q.1 to Q.10

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

Biotechnology Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: A
2. Q.2MCQ · 1 markOfficial answer: B
3. Q.3MCQ · 1 markOfficial answer: D

   Genetics, Cell and Molecular Biology › Gene expression

   Tetracycline is a bacteriostatic protein-synthesis inhibitor that binds the 30S ribosomal subunit and sterically blocks the A site, preventing the aminoacyl-tRNA-EF-Tu-GTP ternary complex from docking — it stops aminoacyl-tRNA from binding to the ribosome.

   WATCH OUT FOR

   Confusing tetracycline (blocks aminoacyl-tRNA binding at the A site) with chloramphenicol (blocks peptidyl transferase) or fusidic acid (blocks translocation/EF-G recycling).

   Full solution →
4. Q.4MCQ · 1 markOfficial answer: A

   Recombinant DNA Technology and Tools › Bioinformatics

   PROSITE is a curated database of protein domains, families and functional sites, represented as sequence patterns (regular-expression-like signatures) and profiles derived from conserved motifs — as opposed to UniProt/TrEMBL/Swiss-Prot (full sequence records) or PDB (3D structures).

   WATCH OUT FOR

   Confusing any UniProt-family database (TrEMBL, SWISS-PROT) with a motif database just because they are also "protein sequence" resources — they store full annotated sequences, not curated pattern/profile signatures of motifs.

   Full solution →
5. Q.5MCQ · 1 markOfficial answer: A
6. Q.6MCQ · 1 markOfficial answer: C

   Genetics, Cell and Molecular Biology › Cell biology

   The eukaryotic mitotic cell cycle is Interphase (G1 → S → G2) followed by M phase; DNA replication (synthesis of new DNA, doubling content from 2C to 4C) is confined to S phase, giving it its name ("Synthesis").

   WATCH OUT FOR

   Confusing G1 and G2 (both non-synthetic growth phases) with S phase, or assuming G0 (quiescence) is simply "before M phase" and therefore synthetic.

   Full solution →
7. Q.7MCQ · 1 markOfficial answer: C
8. Q.8MCQ · 1 markOfficial answer: B

   Recombinant DNA Technology and Tools › Assays & electrophoresis

   A DNA microarray carries thousands of gene-specific probe spots on a single chip; fluorescently labelled cDNA from two samples is hybridized to it simultaneously, so the relative signal at each spot gives a genome-scale, single-experiment comparison of gene expression between the two samples — something PCR, Northern and Southern hybridization each do only one (or a few) targets at a time.

   WATCH OUT FOR

   Confusing DNA microarray (measures RNA/gene expression via labelled cDNA hybridized to probe spots) with Southern hybridization (detects specific DNA sequences, not expression) — the word "DNA" in "DNA microarray" misleads some students into picking a genomic-DNA technique like (D) instead.

   Full solution →
9. Q.9MCQ · 1 markOfficial answer: C or B

   Genetics, Cell and Molecular Biology › Microbial genetics

   Bacteria acquire new genetic material through three natural horizontal gene transfer mechanisms — conjugation (direct transfer via pilus), transformation (uptake of free environmental DNA), and transduction (phage-mediated DNA transfer). Transfection is a laboratory technique for artificially introducing nucleic acid into cells and is not a process that occurs on its own in nature.

   WATCH OUT FOR

   Confusing transfection (lab-mediated artificial introduction of naked DNA/RNA into cells) with transformation (natural uptake of free DNA by competent bacteria) — both involve "naked" nucleic acid entering a cell, but only transformation is a process bacteria undergo spontaneously in nature.

   Full solution →
10. Q.10MCQ · 1 markOfficial answer: D
11. Q.11MCQ · 1 markOfficial answer: B

    Genetics, Cell and Molecular Biology › Mutation & repair

    A polypeptide that is simply shortened (truncated) at a fixed point, with nothing else about it changed, points to a stop codon appearing where a sense codon used to be — a nonsense mutation.

    WATCH OUT FOR

    Confusing "nonsense" (produces a premature STOP codon, truncating the protein) with "missense" (changes to a DIFFERENT amino acid, full length) or "silent"/"synonymous" (no change to the protein at all).

    Full solution →
12. Q.12MCQ · 1 markOfficial answer: B
13. Q.13MCQ · 1 markOfficial answer: B
14. Q.14MCQ · 1 markOfficial answer: D
15. Q.15MCQ · 1 markOfficial answer: D

    Fundamentals of Biological Engineering › Growth kinetics

    Monod kinetics is an unstructured, black-box model. Its limiting substrate S is always a concentration in the medium, meaning a nutrient taken up from outside the cell. That can be a carbon source, a nitrogen source or dissolved oxygen, but never a pool inside the cell.

    WATCH OUT FOR

    Picking (C) because oxygen is thought of as an electron acceptor rather than a 'substrate'. Dissolved oxygen is routinely modelled as a Monod limiting substrate with its own saturation constant $K_{O_2}$. The option that fails is the intracellular one, because the model only sees what is in the medium.

    Full solution →
16. Q.16MCQ · 1 markOfficial answer: C
17. Q.17MCQ · 1 markOfficial answer: A

    Fundamentals of Biological Engineering › Growth kinetics

    Endogenous means the energy source is inside the cell. Exogenous means the source comes from the medium. Endogenous metabolism burns the cell's own internal reserves (storage polymers, then cell constituents) to meet maintenance demand. It becomes dominant in the stationary and death phases of a batch culture, once the external substrate has run out.

    WATCH OUT FOR

    Choosing (C) because cells in the stationary phase sometimes re-consume products they secreted earlier, such as acetate. That product is taken up from the medium, so its use is exogenous metabolism. Another mistake is to connect 'endogenous' with stress and pick (D). Whether a metabolism is endogenous depends only on where the material being catabolised comes from.

    Full solution →
18. Q.18MCQ · 1 markOfficial answer: B
19. Q.19NAT · 1 markOfficial answer: 3

    Recombinant DNA Technology and Tools › Bioinformatics

    For n labeled taxa, the number of distinct rooted bifurcating (binary) phylogenetic trees is $(2n-3)!!=1\times 3\times 5\times \cdots \times (2n-3)$. For $n=3$, this gives $3!!=3$: choose which 2 of the 3 species form a sister clade (root joins their common ancestor to the third species), and there are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ such choices.

    WATCH OUT FOR

    Confusing the ROOTED tree count with the UNROOTED tree count for 3 taxa: an unrooted tree topology for 3 species is unique (all three star-like arrangements are the same tree once the root is removed), so students sometimes answer 1. The question asks specifically for rooted trees, where placing the root creates 3 distinguishable topologies.

    Full solution →
20. Q.20NAT · 1 markOfficial answer: -6
21. Q.21NAT · 1 markOfficial answer: 16
22. Q.22NAT · 1 markOfficial answer: 20
23. Q.23NAT · 1 markOfficial answer: 4

    Fundamentals of Biological Engineering › Growth kinetics

    Degree of reduction is counted per carbon atom (per C-mole): $\gamma =(4w+x-2y-3z)/w$ for ${\text{C}}_w{\text{H}}_x{\text{O}}_y{\text{N}}_z$, using C = +4, H = +1, O = −2, N = −3. Acetic acid has the same C-mole formula as glucose (${\text{CH}}_2\text{O}$), so it has the same $\gamma =4$.

    WATCH OUT FOR

    Answering 8, the available electrons for the whole molecule, and forgetting to divide by the 2 carbon atoms. Another mistake is taking O as +2 or leaving out the oxygen term.

    Full solution →
24. Q.24NAT · 1 markOfficial answer: 31800 to 32000
25. Q.25NAT · 1 markOfficial answer: 50

    Recombinant DNA Technology and Tools › Assays & electrophoresis

    One enzyme unit (U) is defined as the amount of enzyme that converts 1 μmol of substrate (or produces 1 μmol of product) per minute under the stated assay conditions; specific activity is that activity expressed per mg of total protein in the same sample.

    WATCH OUT FOR

    Using the protein concentration (10 mg mL⁻¹) directly as "mg protein" instead of first converting it to the mass of protein actually present in the 10 μL aliquot that was assayed (0.1 mg), which would wrongly give 5/10 = 0.5 units mg⁻¹.

    Full solution →
26. Q.26MCQ · 2 marksOfficial answer: C
27. Q.27MCQ · 2 marksOfficial answer: B

    Recombinant DNA Technology and Tools › Editing & interactions

    In PCR-based site-directed mutagenesis (e.g. QuikChange), the parental plasmid template (isolated from a dam+ E. coli strain) is Dam-methylated at GATC sites, while the newly synthesized mutant strand made in vitro is unmethylated. DpnI is methylation-dependent and cleaves only methylated/hemimethylated GATC, so treating the PCR product with DpnI selectively destroys the parental template and enriches for the mutant plasmid.

    WATCH OUT FOR

    Confusing DpnI (cuts only methylated DNA) with the many restriction enzymes that are blocked by methylation (cut only unmethylated DNA), or picking a sequence-specific enzyme (KpnI, XhoI, MluI) that has no relationship to methylation status at all.

    Full solution →
28. Q.28MCQ · 2 marksOfficial answer: A

    Genetics, Cell and Molecular Biology › Cell biology

    Each organelle has a defining, distinguishing biochemical role: lysosomes digest material via hydrolytic enzymes, smooth ER synthesizes lipids (rough ER, not smooth, is associated with protein synthesis via bound ribosomes), the Golgi apparatus modifies, sorts, and targets proteins to their destinations (it does not synthesize proteins itself), and the nucleolus is the site of rRNA synthesis and ribosome subunit assembly.

    WATCH OUT FOR

    Confusing Golgi apparatus's role (protein targeting/sorting/packaging) with protein synthesis (which occurs on ribosomes, often on rough ER) — this trap is embedded in option (D), which correctly matches P, Q, and S but wrongly assigns R-4 (protein synthesis) to the Golgi instead of R-2 (protein targeting).

    Full solution →
29. Q.29MCQ · 2 marksOfficial answer: B

    Recombinant DNA Technology and Tools › Bioinformatics

    In BLAST, the E-value is the expected number of alignments of that score or better occurring purely by chance in a database search — so a SMALLER E-value means the match is less likely to be chance, i.e. more significant/homologous. The BLAST score, by contrast, is a direct measure of alignment quality (built from substitution-matrix rewards/penalties for matches, mismatches and gaps): a HIGHER score means a better, more similar alignment. The two measures move in opposite directions relative to "better," which is exactly why this question mixes them.

    WATCH OUT FOR

    Treating "larger/higher" as uniformly "better" for both quantities — assuming a larger E-value or a higher E-value threshold also signals stronger homology, when only a very small E-value (like P < 10^{-10}) indicates true homology, while a very large E-value (like > 10^{10}) indicates the match is essentially indistinguishable from random chance, i.e. no evidence of homology.

    Full solution →
30. Q.30MCQ · 2 marksOfficial answer: B

    Genetics, Cell and Molecular Biology › Mendelian genetics

    The ABO locus encodes a glycosyltransferase enzyme that attaches a terminal sugar residue onto the H antigen precursor on the red blood cell surface, generating the A or B antigen; this class of enzyme is grouped under "galactosyltransferase."

    WATCH OUT FOR

    Confusing this transferase (which ADDS a sugar to build the antigen) with β-galactosidase, a hydrolase that REMOVES/cleaves a sugar (as in the lacZ system) — the two enzyme classes catalyze opposite reactions and are easy to swap under time pressure.

    Full solution →
31. Q.31MCQ · 2 marksOfficial answer: A

    Recombinant DNA Technology and Tools › PCR & sequencing

    Fidelity is the polymerase's per-base copying accuracy, and specificity is where the primers bind. Reaction chemistry at the active site changes fidelity: free ${\text{Mg}}^{2+}$ in excess of dNTPs, pH, and an unbalanced dNTP pool. Annealing temperature changes primer–template stringency, which affects specificity and yield, not fidelity.

    WATCH OUT FOR

    Treating 'fidelity' as meaning 'anything that gives a wrong PCR product', and so choosing (D). A low annealing temperature causes mispriming and non-specific bands. That is a failure of specificity, not a change in how accurately the polymerase inserts nucleotides.

    Full solution →
32. Q.32MCQ · 2 marksOfficial answer: C

    Recombinant DNA Technology and Tools › Microscopy & spectroscopy

    Each spectroscopic method has one signature application: IR probes bond vibrations (functional groups), CD probes chiral secondary-structure content, and multidimensional NMR provides distance/angle restraints for full 3-D structure determination; molecular weight is estimated by other techniques (e.g., mass spectrometry), not by these three.

    WATCH OUT FOR

    Confusing CD with 3-D structure determination (since it is a "structural" technique) instead of secondary structure, or assuming NMR is used for molecular weight estimation because NMR-active nuclei are also used in mass-related contexts (e.g., MS), leading to a mismatch with option 3.

    Full solution →
33. Q.33MCQ · 2 marksOfficial answer: A

    Recombinant DNA Technology and Tools › Assays & electrophoresis

    A protein/peptide run in a gel above its pI carries a net negative charge and migrates towards the anode; the size of that net negative charge (and hence migration speed) depends on how far the running pH is above the pI. Removing an acidic (negatively charged) residue raises the pI, which — at a fixed running pH above both pI values — reduces the net negative charge and therefore the migration speed towards the anode.

    WATCH OUT FOR

    Confusing which peptide is "more negative": students often assume that because Q has one fewer negative side chain it must be the more mobile one, or they send the negatively charged species towards the cathode instead of the anode (cathode attracts cations, anode attracts anions).

    Full solution →
34. Q.34MCQ · 2 marksOfficial answer: B
35. Q.35MCQ · 2 marksOfficial answer: C
36. Q.36MCQ · 2 marksOfficial answer: C

    Genetics, Cell and Molecular Biology › Gene expression

    mRNA 5' capping installs m^7G via a unique 5'–5' triphosphate linkage (not the standard 3'–5' phosphodiester bond of the rest of the chain), formed when guanylyltransferase adds GMP to the diphosphate 5' end left after RNA triphosphatase removes the γ-phosphate; N7-methylation of the added guanosine follows and does not alter the linkage.

    WATCH OUT FOR

    Assuming the cap is attached by the usual 3'–5' phosphodiester bond used for all other internucleotide linkages in RNA, rather than recognizing the atypical 5'-to-5' triphosphate bridge unique to capping.

    Full solution →
37. Q.37MCQ · 2 marksOfficial answer: A
38. Q.38MCQ · 2 marksOfficial answer: B
39. Q.39MCQ · 2 marksOfficial answer: D
40. Q.40MCQ · 2 marksOfficial answer: D

    Fundamentals of Biological Engineering › Growth kinetics

    The lag phase is the time cells spend adapting before they divide. Anything that changes how much adapting they must do changes its length. A small inoculum dilutes the metabolites cells leak out, an old inoculum needs repair first, and an unfamiliar medium forces the cells to make new enzymes. So all three factors lengthen or shorten the lag.

    WATCH OUT FOR

    Treating the lag phase as a property of the organism alone and dropping inoculum size (P), on the idea that more cells only shift the curve up. A small inoculum really does lengthen the lag, because leaked metabolites and cofactors are diluted. Some students also drop medium composition (R), forgetting that a change of medium between seed and fermentor is the main reason cells have to adapt at all.

    Full solution →
41. Q.41MCQ · 2 marksOfficial answer: B

    Fundamentals of Biological Engineering › Process control

    A P controller is a pure gain, $c-\bar{c}=K_{c}e$. It acts at once and in proportion to the error, so a bigger error gets a bigger correction. It still acts on a constant error, and it cannot remove offset, because holding a new steady state needs a nonzero error.

    WATCH OUT FOR

    Mixing up the three control modes. Zero action for a constant error describes derivative control (D), and eliminating offset describes integral control (C). A slow initial response belongs to lag-type or integral action (A). None of these describes proportional control.

    Full solution →
42. Q.42MCQ · 2 marksOfficial answer: C

    Fundamentals of Biological Engineering › Process control

    Match each instrument to the physical quantity it senses. A manometer reads a liquid-column height, so it gives a pressure difference. A rotameter reads the height of a float in a tapered tube, so it gives a flow rate (on a fermenter, the air flow to the sparger). A tachometer reads the speed of a rotating shaft (the agitator). A haemocytometer is a counting chamber of known volume, so it gives the cell number.

    WATCH OUT FOR

    Mixing up the rotameter and the tachometer because both names suggest rotation, which leads to option (D). The rotameter's float spins, but what it measures is flow rate. The tachometer is the instrument that measures rotational speed (rpm).

    Full solution →
43. Q.43MCQ · 2 marksOfficial answer: D

    Fundamentals of Biological Engineering › Bioreactor operation

    In a steady-state chemostat $\mu =D$. So $S=K_{s}D/({\mu }_{max}-D)$ rises with $D$, and $X=Y_{X/S}(S_0-S)$ falls with $D$. Productivity $DX$ is zero at $D=0$ and at washout, so it peaks strictly below $D_{crit}$. Cell recycle keeps $\mu$ below $D$, which is what lets the reactor run above the simple chemostat's washout rate.

    WATCH OUT FOR

    Picking R because a higher dilution rate brings in more substrate. In fact, raising $D$ forces cells to grow faster, which needs a higher residual $S$, and that leaves less substrate to turn into biomass. The other common mistake is to confuse the optimal dilution rate (maximum productivity $DX$) with the critical one (washout) and so reject Q.

    Full solution →
44. Q.44MCQ · 2 marksOfficial answer: B
45. Q.45NAT · 2 marksOfficial answer: 2

    Recombinant DNA Technology and Tools › Microscopy & spectroscopy

    The Beer-Lambert law (A = εcl) directly relates absorbance to concentration through the molar extinction coefficient and path length; a spectrophotometer's detection limit in absorbance units translates into a minimum measurable concentration via this same relation.

    WATCH OUT FOR

    Forgetting to convert the resulting mol/L answer into μM (a factor of 10⁶), or mixing up which quantity is being solved for (rearranging incorrectly to isolate ε or l instead of c).

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 280 to 295
47. Q.47NAT · 2 marksOfficial answer: 0.75

    Genetics, Cell and Molecular Biology › Mendelian genetics

    Each offspring's genotype is determined independently at fertilization (each is an independent Mendelian trial), so the phenotype of one sibling gives no information about another; the probability for the third offspring is simply the parental cross's phenotypic ratio.

    WATCH OUT FOR

    Treating the white first offspring as changing the probability for later offspring (e.g., trying to condition on "since one was already white, the remaining must balance to a 3:1 ratio" or halving/adjusting 3/4), when in fact fertilizations are independent Bernoulli trials and each offspring's probability of being purple is always 3/4 regardless of siblings' phenotypes.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 0.6

    Genetics, Cell and Molecular Biology › Gene expression

    The average molecular mass of an amino acid residue in a protein is taken as ~110 Da; each amino acid is encoded by one codon (3 nucleotides) in the coding (cDNA) sequence.

    WATCH OUT FOR

    Forgetting to convert the answer from base pairs (600 bp) to kilobases (0.6 kb), or using an incorrect average amino acid residue mass instead of the standard ~110 Da approximation, or forgetting that each amino acid needs 3 nucleotides (one codon).

    Full solution →
49. Q.49NAT · 2 marksOfficial answer: 128 to 130
50. Q.50NAT · 2 marksOfficial answer: 2.20 to 2.50
51. Q.51NAT · 2 marksOfficial answer: 12500 to 14500

    Fundamentals of Biological Engineering › Growth kinetics

    The coefficients in a balanced growth equation are in moles. Multiply each one by its molar mass to get a mass yield: $Y_{X/S}=(0.48\times 144)/180=0.384{\text{g g}}^{-1}$. The glucose needed is then the total biomass divided by $Y_{X/S}$. The equation balances for C, H, O and N, so its coefficients can be used as printed. Only the biomass and glucose coefficients matter here; NH3, O2, CO2 and H2O play no part in the answer.

    WATCH OUT FOR

    Treating 0.48 as a mass ratio, which gives $5000/0.48\approx 10417$ kg (outside the key), or forgetting to convert the molar ratio to a mass ratio with the two molar masses (144 and 180). Another slip is misreading the volume: 1,00,000 L in Indian digit grouping is $10^5$ L. Reading it as $10^6$ L makes every figure ten times too large.

    Full solution →
52. Q.52NAT · 2 marksOfficial answer: 970 to 1010

    Fundamentals of Biological Engineering › Bioreactor operation

    At steady state in a chemostat with cell death, the biomass balance gives $D=\mu -k_d$, where $\mu$ is the Monod rate at the effluent (reactor) concentration. Once you have $D$, the volume is $V=F/D$. The inlet concentration $S_0$ only sets how much biomass is formed. It does not change the dilution rate.

    WATCH OUT FOR

    Ignoring death and taking $D=\mu$ gives $V=880\text{L}$, which is outside the key. Adding death with the wrong sign ($D=\mu +k_d$) gives about $793\text{L}$. Putting the influent $10,000\text{ppm}$ into the Monod equation instead of the effluent $10\text{ppm}$ is also wrong, because in a well-mixed chemostat the cells only ever see the outlet concentration.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 1

    Fundamentals of Biological Engineering › Downstream

    In cross-flow filtration, $\Delta P=P_i-P_o$ is the pressure drop along the channel and $\Delta P_m=\frac{P_i+P_o}{2}-P_f$ is the average pressure difference across the membrane. Put together, $\Delta P_m=P_i-\frac{\Delta P}{2}-P_f$.

    WATCH OUT FOR

    Taking atmospheric pressure as $P_f=0$ (a gauge value) while $P_i=3\text{atm}$ is an absolute value. That gives $2\text{atm}$. The other common slip is subtracting $P_f$ from the inlet pressure alone ($3-1=2\text{atm}$) instead of from the average feed-side pressure, which ignores the pressure drop along the membrane.

    Full solution →
54. Q.54NAT · 2 marksOfficial answer: 9 to 10

    Fundamentals of Biological Engineering › Upstream

    In batch sterilization, $N_0$ is the TOTAL spore count in the whole vessel (concentration × volume), and the target $N$ is the contamination probability, read as the expected number of survivors per batch. With first-order death at constant temperature, $\ln (N_0/N)=k_{d}t$. Since heating and cooling kill nothing here, the holding stage has to deliver the whole Del factor.

    WATCH OUT FOR

    Using the spore concentration ($10^6$ per mL) as $N_0$ instead of the total count in 10,000 L ($10^{13}$), which gives $\ln 10^9/4=5.2$ min. Two other slips: forgetting that 1 L = 1000 mL, and using ${\log }_{10}$ instead of $\ln$ (this gives $16/4=4$ min).

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 30

    Fundamentals of Biological Engineering › Bioreactor operation

    Tip speed is $\pi N{D}_i$. Keeping it constant makes $N$ inversely proportional to $D_i$. In a geometrically similar vessel, $D_i$ grows as $(V_2/V_1{)}^{1/3}$, so $N_2=N_1(V_1/V_2{)}^{1/3}$.

    WATCH OUT FOR

    The usual slips are scaling the speed by the volume ratio (600/8000) instead of the length ratio (600/20), or using the constant power-per-volume rule ($N\propto D_i^{-2/3}$, which gives about 81 rpm) when the question asks for constant tip speed. The 20 cm vessel diameter and the 0.3 impeller ratio are not needed, because both vessels keep the same impeller ratio.

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