GATE BT 2020 question paper with solutions

GATE BT 2020 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: B
3. GA Q.3MCQ · 1 markOfficial answer: D
4. GA Q.4MCQ · 1 markOfficial answer: D
5. GA Q.5MCQ · 1 markOfficial answer: B
6. GA Q.6MCQ · 2 marksOfficial answer: D
7. GA Q.7MCQ · 2 marksOfficial answer: C
8. GA Q.8MCQ · 2 marksOfficial answer: B
9. GA Q.9MCQ · 2 marksOfficial answer: B
10. GA Q.10MCQ · 2 marksOfficial answer: B

Biotechnology Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: B

   Genetics, Cell and Molecular Biology › Cell signalling

   A phosphomimetic substitution replaces the phosphorylatable residue (serine, threonine, or tyrosine) with a residue that is constitutively negatively charged, mimicking the negative charge added by the phosphate group without needing kinase activity.

   WATCH OUT FOR

   Confusing the phosphomimetic (aspartate or glutamate, mimics the always-on phosphorylated state) with the phospho-null/phospho-dead substitution (alanine, blocks phosphorylation and mimics the always-off unphosphorylated state); these are opposite tools and easy to swap under exam pressure.

   Full solution →
2. Q.2MCQ · 1 markOfficial answer: B or D

   Genetics, Cell and Molecular Biology › Cell signalling

   Ras is a small, monomeric GTPase (not a heterotrimeric G protein) that acts as a GDP/GTP-regulated molecular switch downstream of receptor tyrosine kinases, transmitting mitogenic signals from the cell surface to the nucleus through the Raf-MEK-ERK cascade.

   WATCH OUT FOR

   Confusing Ras (a monomeric/small GTPase) with heterotrimeric G proteins that couple to GPCRs, or conflating Ras's signalling role with the cytoskeleton-regulating role that belongs specifically to the Rho-family small GTPases (Rho, Rac, Cdc42) within the broader Ras superfamily.

   Full solution →
3. Q.3MCQ · 1 markOfficial answer: D

   Genetics, Cell and Molecular Biology › Population & evolution

   Cancer can arise from genetic changes (mutations, viral oncogene insertion) and from epigenetic/regulatory changes (promoter hypermethylation silencing tumor suppressors, dysregulated miRNA networks) — DNA sequence alteration is not the only route to malignant transformation.

   WATCH OUT FOR

   Assuming a tumor suppressor gene can only be inactivated by a mutation in its DNA sequence, and overlooking that epigenetic silencing (e.g., CpG-island hypermethylation of its promoter) achieves the same functional loss without altering the sequence — this leads students to wrongly mark Q as false and pick option (C) instead of (D).

   Full solution →
4. Q.4MCQ · 1 markOfficial answer: D
5. Q.5MCQ · 1 markOfficial answer: D

   Genetics, Cell and Molecular Biology › Cell signalling

   Ca2+ is the messenger ion because the cell keeps its cytosolic level extremely low (about 100 nM, compared with about 1 to 2 mM outside). A small influx then produces a large relative rise that sensors such as calmodulin and synaptotagmin act on. Cytosolic Na+ (about 10 mM) is roughly 10^5 times higher than cytosolic Ca2+, so the Reason reverses the true relation.

   WATCH OUT FOR

   Reading 'not Na+' as a claim that Na+ has no role in nerve conduction, and so marking the Assertion false and choosing (C). The Assertion is about the messenger ion that relays signals, and that is Ca2+, which triggers transmitter and hormone release and acts as a second messenger. Na+ only depolarises the membrane. A second mistake is accepting the Reason: cytosolic Ca2+ is the LOWER of the two, by about five orders of magnitude.

   The question and its official answer →
6. Q.6MCQ · 1 markOfficial answer: A
7. Q.7MCQ · 1 markOfficial answer: D

   Recombinant DNA Technology and Tools › Libraries & cloning

   Reverse transcriptase synthesizes DNA using an RNA template; this DNA product is called complementary DNA (cDNA).

   WATCH OUT FOR

   Confusing the direction of information flow — thinking "transcript" (RNA made from DNA) is the same process reversed, or confusing cDNA with recombinant DNA (which is made by joining DNA fragments from different sources, not by copying RNA).

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

   Fundamentals of Biological Engineering › Downstream

   Match the separation technique to the one property that differs. Here size and mass are equal and both proteins carry the same His-tag, so size-based methods (SDS-PAGE, size exclusion) and tag-based affinity (Ni-IMAC) cannot tell them apart. Only charge differs: at a buffer pH between the two pIs, one protein is net negative and the other net positive, which is exactly what ion exchange separates.

   WATCH OUT FOR

   Picking nickel affinity chromatography because the proteins are His-tagged. IMAC captures every His-tagged protein, so both bind and elute together. It separates tagged from untagged proteins, not two tagged proteins from each other. A second trap is thinking SDS-PAGE separates by charge. SDS masks the native charge, so separation is by size only.

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

   Recombinant DNA Technology and Tools › Libraries & cloning

   Positive-negative selection in gene targeting places a negative marker (HSV/viral thymidine kinase) flanking the homology arms; homologous recombinants lose it (survive ganciclovir), while random integrants retain and express it, and viral (but not mammalian) thymidine kinase efficiently phosphorylates ganciclovir into a DNA-damaging, cytotoxic form, killing the randomly integrated cells.

   WATCH OUT FOR

   Confusing this with simple positive selection (e.g., neo/G418 resistance) or assuming any kinase can activate ganciclovir; only viral (HSV) thymidine kinase has the substrate specificity to phosphorylate this nucleoside analogue — mammalian thymidine kinase and unrelated protein kinases (serine/threonine, tyrosine kinases) cannot.

   Full solution →
10. Q.10MCQ · 1 markOfficial answer: C

    Recombinant DNA Technology and Tools › Enzymes & vectors

    Stable, high-frequency integration of a transgene into the host chromosome requires a viral vector encoding an integrase enzyme; among common animal-cell gene delivery vectors, only lentiviral (retroviral) vectors carry this and can integrate even into non-dividing cells.

    WATCH OUT FOR

    Confusing AAV's ability to persist long-term in cells (as episomal concatemers, or via rare/random integration) with true high-frequency, integrase-driven integration — AAV is favoured precisely because it is largely non-integrating and safer, the opposite property being tested here.

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

    Recombinant DNA Technology and Tools › Enzymes & vectors

    The Ti plasmid is functionally split: T-DNA (transferred into and expressed in the plant cell) carries phytohormone (auxin/cytokinin) biosynthesis genes and opine biosynthesis genes so the plant tumour makes hormones and opines for the bacterium; the vir genes (transfer machinery, act in trans) and opine catabolism genes (let the bacterium consume the opines) remain outside T-DNA on the bacterial backbone.

    WATCH OUT FOR

    Confusing the T-DNA (the segment actually transferred to and expressed in the plant genome — carries phytohormone and opine BIOSYNTHESIS genes) with the vir region and opine-catabolism region, which remain on the bacterial backbone of the Ti plasmid and are never transferred. Students often misplace vir genes or opine catabolism genes inside T-DNA, or think phytohormone genes stay outside it.

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

    Genetics, Cell and Molecular Biology › Population & evolution

    Different proteins in the same set of organisms accumulate substitutions at different rates and patterns because they are under different functional/structural constraints — this is why phylogenetic trees built from different protein families for the same taxa need not be identical, and often are not.

    WATCH OUT FOR

    Assuming that because the same 40 organisms are used, a "true" underlying species tree must force identical gene/protein trees regardless of which molecule is sequenced — ignoring that each protein's own mutation rate and pattern (its own effective molecular clock) independently shapes the inferred tree, so different proteins can legitimately give different topologies/branch lengths for the same species.

    Full solution →
16. Q.16MCQ · 1 markOfficial answer: D
17. Q.17NAT · 1 markOfficial answer: 15
18. Q.18NAT · 1 markOfficial answer: 2
19. Q.19NAT · 1 markOfficial answer: 51.01 to 52.99

    Fundamentals of Biological Engineering › Growth kinetics

    A biomass formula normalised to one carbon atom gives the C-mole weight when each subscript is multiplied by its atomic weight and the results are added. The carbon mass fraction is then 12 divided by that C-mole weight.

    WATCH OUT FOR

    Dropping the small S and P terms, or leaving out carbon's own 12 g from the denominator. Dropping S and P gives 12/22.8 = 52.63 %, which still falls inside the key range but does not answer the question as it was asked. Another mistake is to use atom counts in place of masses, which gives 1/3.2194 = 31.06 %.

    Full solution →
20. Q.20NAT · 1 markOfficial answer: 60

    Fundamentals of Biological Engineering › Downstream

    In single-stage liquid-liquid extraction between two immiscible solvents, the solute divides so that the ratio of its concentrations follows the partition coefficient: $K=C_A/C_B$. Combine that with a mass balance on the solute. When the two solvent masses are equal, the solute masses themselves come out in the ratio $K:1$, here $1.5:1$, so A gets $\frac{1.5}{2.5}\times 100=60\text{g}$.

    WATCH OUT FOR

    Reading "1.5 times more soluble" as $K=2.5$ (that is, $1+1.5$). That gives $100\times 2.5/3.5\approx 71.4\text{g}$, which is wrong: GATE means $C_A/C_B=1.5$. A second mistake is to give the whole 100 g to the better solvent, or to split it 50:50. A third is to apply $K$ to the solute masses when the solvent masses are unequal. That shortcut works here only because both phases hold 2 kg of solvent.

    Full solution →
21. Q.21NAT · 1 markOfficial answer: 5.99 to 6.05
22. Q.22NAT · 1 markOfficial answer: 2

    Fundamentals of Biological Engineering › Bioreactor operation

    The gas volume stays the same when the bubble breaks up. Splitting one sphere into $N$ equal spheres makes each radius $r/N^{1/3}$. Since the area-to-volume ratio of a sphere is $3/\text{radius}$, it rises by a factor of $N^{1/3}$. Here $N=8$, so the factor is $8^{1/3}=2$. This is why smaller bubbles give a larger specific interfacial area, $a$, and so a larger $k_{L}a$ in a bioreactor.

    WATCH OUT FOR

    One mistake is to assume the radius stays at $r$ and multiply the area by 8, which gives $n=8$. Another is to divide the radius by 8 instead of by $8^{1/3}=2$, which also gives $n=8$. The gas volume is fixed, so the new radius is $r/2$ and the ratio only doubles.

    Full solution →
23. Q.23NAT · 1 markOfficial answer: 3
24. Q.24NAT · 1 markOfficial answer: 0.49 to 0.51
25. Q.25NAT · 1 markOfficial answer: 1
26. Q.26MCQ · 2 marksOfficial answer: A
27. Q.27MCQ · 2 marksOfficial answer: B

    Recombinant DNA Technology and Tools › PCR & sequencing

    In PCR, the forward primer has the same sequence (5'->3') as the top/sense strand at the left end of the target region (it primes synthesis off the bottom strand template), while the reverse primer has the same sequence as the bottom strand's 5' end — i.e., it is the reverse complement of the top strand's 3' end (it primes synthesis off the top strand template). A primer that merely copies the printed top-strand sequence at the right end, without reverse-complementing it, cannot function as a reverse primer.

    WATCH OUT FOR

    Picking a "primer" that is simply the top strand's own 3' end sequence copied as printed (option D's second primer, TGGGCAATAATGTAGCGC) instead of its reverse complement — this sequence is identical in polarity to the template strand it needs to hybridize to and therefore cannot anneal and prime synthesis in the required direction.

    Full solution →
28. Q.28MCQ · 2 marksOfficial answer: A
29. Q.29MCQ · 2 marksOfficial answer: D

    Genetics, Cell and Molecular Biology › Cell biology

    CDK activity is controlled both positively (cyclin binding) and negatively (inhibitory phosphorylation, e.g. Wee1-mediated Tyr15/Thr14 phosphorylation of CDK1), and progression past checkpoints (such as the metaphase-to-anaphase transition and mitotic exit) requires the APC/C-mediated ubiquitination and proteasomal degradation of cyclins.

    WATCH OUT FOR

    Assuming CDKs act only through activation (cyclin binding, activating dephosphorylation by Cdc25) and forgetting that phosphorylation can also switch a CDK off (Wee1 pathway), which leads to wrongly rejecting statement Q; also assuming chromosome condensation is driven only by condensin loading and not recognizing that CDK1-cyclin B phosphorylates condensin subunits and histone H3, which would wrongly validate statement S.

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

    Genetics, Cell and Molecular Biology › Mendelian genetics

    A biosynthetic-pathway mutant blocked at a given step can grow only if supplied with the product of that step or any later intermediate (it can still make everything upstream itself); it cannot use anything upstream of its own block. Map each mutant's Yes/No growth pattern to a block position accordingly, then read off the pathway order.

    WATCH OUT FOR

    Reversing the growth-rescue logic — assuming a mutant is rescued by intermediates upstream of its block (since it can still "process" earlier steps) rather than by the blocked step's product and everything downstream. This flips the mapping and can make Z1 look like the earliest block instead of Z4.

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

    Genetics, Cell and Molecular Biology › Mendelian genetics

    When two independently assorting traits are involved, decompose the progeny into single-trait totals first; the ratio each trait shows on its own (3:1, 1:1, etc.) tells you the cross type at that one locus, and the product of the two single-trait ratios must equal the observed four-class ratio.

    WATCH OUT FOR

    Jumping straight to a 9:3:3:1 dihybrid expectation (as in Option B) without first checking that BOTH traits individually segregate 3:1 — here the stem-color trait alone is 1:1 (a test cross, Pp × pp), not 3:1, which rules out a heterozygote × heterozygote cross at that locus.

    Full solution →
32. Q.32MCQ · 2 marksOfficial answer: D
33. Q.33MCQ · 2 marksOfficial answer: B

    Genetics, Cell and Molecular Biology › Population & evolution

    Carl Woese pioneered the use of small-subunit ribosomal RNA (16S rRNA in prokaryotes, 18S rRNA in eukaryotes) sequence comparisons to build phylogenetic trees, which led to the discovery of Archaea and the three-domain system of life (Bacteria, Archaea, Eukarya).

    WATCH OUT FOR

    Confusing rRNA (nucleic acid) with ribosomal proteins as the molecule compared, or choosing the large subunit rRNA (23S/28S) instead of the small subunit (16S/18S), which is the one Woese specifically used because of its universal presence, moderate size, and mix of conserved and variable regions.

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

    Recombinant DNA Technology and Tools › Enzymes & vectors

    Dam methylase is the modification arm of a restriction–modification system: it methylates GATC on the host's own chromosome so the host's OWN restriction endonuclease (which cuts unmethylated GATC) does not degrade self DNA — the enzyme it protects against is native to E. coli, not phage-encoded.

    WATCH OUT FOR

    Choosing (A) or (B) by assuming that because [r] is textbook-correct, [a] must also be true (or true-but-unrelated) simply because Dam methylation and restriction protection are both real, related phenomena — without noticing that the assertion misidentifies WHOSE endonuclease is being evaded (the bacterium's own restriction enzyme, not a phage-encoded one).

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

    Recombinant DNA Technology and Tools › Libraries & cloning

    Knockout mice are made by targeting a gene in embryonic stem (ES) cells via homologous recombination, then using the modified ES cells to generate germline-transmitting chimeras; ES cells are chosen for this because they support homologous recombination at a usefully higher relative frequency than ordinary somatic cells, and because only they can be returned to an embryo to contribute to the germline.

    WATCH OUT FOR

    Assuming [r] is a "true but irrelevant" statement (picking C) because homologous recombination sounds like a generic molecular fact unconnected to the assertion, without recognizing that this comparatively higher HR frequency in ES cells is exactly the mechanistic reason gene targeting for knockouts is feasible in this cell type.

    Full solution →
37. Q.37MCQ · 2 marksOfficial answer: B

    Recombinant DNA Technology and Tools › Libraries & cloning

    Closing a nicked/gapped circular plasmid needs, in order: (1) a polymerase to extend the available 3'-OH primer across the gap using the intact strand as template, (2) if the resulting nick's downstream terminus is 5'-OH rather than 5'-phosphate, a kinase to phosphorylate it (ligase cannot join a 5'-OH), and only then (3) ligase to seal the phosphodiester backbone into a covalently closed circle.

    WATCH OUT FOR

    Assuming DNA ligase alone (with polymerase) is sufficient to close any gap/nick, without checking the chemical nature of the 5' terminus; a 5'-OH end (as printed here) cannot be ligated until polynucleotide kinase converts it to 5'-phosphate. A related trap is confusing alkaline phosphatase (removes 5'-phosphates, used to prevent vector self-ligation) with polynucleotide kinase (adds 5'-phosphates) — they have opposite effects, so including alkaline phosphatase here is wrong.

    Full solution →
38. Q.38MCQ · 2 marksOfficial answer: C

    Genetics, Cell and Molecular Biology › Cell biology

    This is a one-to-one matching question (each Group II feature used exactly once), and two of the four rows have only one biologically valid feature each — solve those first, then use the printed options (not biology alone) to break the remaining tie between mitochondrion and chloroplast.

    WATCH OUT FOR

    Assuming there is one "textbook-clean" one-to-one mapping between organelle and feature on biological grounds alone. In reality, mitochondria and chloroplasts BOTH have double membranes, BOTH arose by endosymbiosis, and BOTH are predominantly maternally inherited in most organisms — so P and Q cannot be split between "endosymbiotic origin" and "maternal inheritance" by biology alone. The split is only fixed by first solving the two unambiguous rows (R and S) and then using the four printed options to eliminate every internally inconsistent match.

    Full solution →
39. Q.39MCQ · 2 marksOfficial answer: D
40. Q.40MCQ · 2 marksOfficial answer: D

    Genetics, Cell and Molecular Biology › Cell biology

    Totipotency is the capacity of a cell to differentiate into every cell type of the organism — including extra-embryonic tissues like the placenta — not a limited subset; that broader capacity is what distinguishes totipotent cells from pluripotent, multipotent, or unipotent cells.

    WATCH OUT FOR

    Confusing totipotent with pluripotent — assuming "differentiate into a few cell types" describes totipotency, when in fact totipotent cells can form ALL cell types (embryonic and extra-embryonic), while only a limited subset (pluripotent, multipotent, unipotent) is restricted.

    Full solution →
41. Q.41MCQ · 2 marksOfficial answer: C
42. Q.42MCQ · 2 marksOfficial answer: C
43. Q.43NAT · 2 marksOfficial answer: 4096

    Recombinant DNA Technology and Tools › Enzymes & vectors

    For a random DNA sequence with equiprobable bases, the probability that a specific n-nucleotide sequence occurs starting at any given position is $(1/4{)}^n$; the average spacing between successive occurrences of that sequence is the reciprocal of this probability, $4^n$ nucleotides — independent of total genome length (which only needs to be large enough for the site to occur a reasonable number of times).

    WATCH OUT FOR

    Confusing the expected spacing formula: some students multiply (1/4)^6 by the genome length instead of taking its reciprocal, or forget that a 6-cutter's probability is (1/4)^6 regardless of genome size, mistakenly trying to bring the 1 Mb figure into the final calculation.

    Full solution →
44. Q.44NAT · 2 marksOfficial answer: 0.11 to 0.14

    Genetics, Cell and Molecular Biology › Gene expression

    In semiconservative replication starting from one fully-¹⁵N duplex, only the original two heavy parental strands ever end up paired with a new ¹⁴N strand — so the number of hybrid (intermediate-density) duplexes stays fixed at 2 in every generation after the first, while the total number of duplexes doubles each generation (2^n after n generations). Hence the hybrid fraction after n generations (n ≥ 1) is 2/2^n, not 1/2 (which is only true at n = 1) and not something that keeps halving indefinitely toward the "half the DNA is hybrid forever" myth.

    WATCH OUT FOR

    Assuming the fraction of hybrid DNA stays at 1/2 for all generations (true only after generation 1), or conversely assuming hybrid duplexes double along with total duplexes (they don't — the two original heavy strands are a fixed, non-renewable pool, since a strand is never destroyed or re-paired with another old strand). Both errors are avoided by explicitly tracking that hybrid count = 2 (constant) and total = 2^n (growing) separately.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 526.01 to 529.99

    Fundamentals of Biological Engineering › Growth kinetics

    In a batch culture, the time to reach a target biomass is the lag time plus the exponential-growth time. The exponential part is $t_{exp}=\frac{1}{\mu }\ln \frac{X}{X_0}$. Because $8=2^3$, the exponential part is exactly three doubling times, $3\ln 2/\mu$.

    WATCH OUT FOR

    Leaving out the 30 min lag phase gives 498.67 min, which is outside the key range. Two other slips are using ${\log }_{10}8$ instead of $\ln 8$ (which gives $0.90309/0.00417=216.57$ min for the growth part), and treating the growth as linear.

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 1
47. Q.47NAT · 2 marksOfficial answer: 864

    Genetics, Cell and Molecular Biology › Gene expression

    Because the genetic code is degenerate, each amino acid position in a peptide can independently be encoded by any one of its synonymous codons. By the multiplication principle, the total number of distinct coding (DNA) sequences for the whole peptide is the product of the number of codon choices at every position in the chain, counting a repeated amino acid once per occurrence in the sequence.

    WATCH OUT FOR

    Treating repeated amino acids (Leu appears twice) as contributing only once, or adding the codon counts instead of multiplying them.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 1196

    Genetics, Cell and Molecular Biology › Cell biology

    A kinetochore forms at the centromere of every chromosome present. A resting (G1-state) cell contains 46 unreplicated chromosomes (single chromatid each), so it has exactly 46 kinetochores — the same number whether the cell has just been born, is about to divide, or never divides at all. So the answer reduces to (total number of cells now present) × 46.

    WATCH OUT FOR

    Confusing a chromosome's kinetochore count with its DNA/chromatid content: assuming a cell mid-S-phase or in G2 (duplicated chromosomes, two sister chromatids) has 92 kinetochores per cell instead of 46. In this problem all cells being counted are freshly-divided or never-divided cells sitting in G1, so each one — dividing lineage or not — carries only 46 kinetochores; forgetting this and, e.g., doubling the daughter-cell count for "two chromatids" gives 2392 instead of 1196.

    Full solution →
49. Q.49NAT · 2 marksOfficial answer: 0

    Fundamentals of Biological Engineering › Bioreactor operation

    A chemostat can hold cells at steady state only if $D=\mu$ for some reachable substrate level, so $D$ must be below the washout limit $D_{crit}={\mu }_{max}{S}_0/(K_s+S_0)$. Always compare $D=F/V$ with $D_{crit}$ before using the steady-state formulas. If $D\ge D_{crit}$, the only steady state is $X=0$, $S=S_0$.

    WATCH OUT FOR

    Plugging straight into $S=K_{s}D/({\mu }_{max}-D)$ and $X=Y_{X/S}(S_0-S)$ without checking for washout. That gives $S\approx -0.84{\text{g L}}^{-1}$ and a meaningless $X\approx 0.4(20+0.84)\approx 8.3{\text{g L}}^{-1}$. A negative $S$ means no non-trivial steady state exists. Taking $D=V/F=0.9{\text{h}}^{-1}$ instead of $F/V$ is also wrong, even though here it still leads to washout.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: 2
51. Q.51NAT · 2 marksOfficial answer: 2.20 to 2.30
52. Q.52NAT · 2 marksOfficial answer: 25

    Recombinant DNA Technology and Tools › Assays & electrophoresis

    SDS denatures non-covalent (secondary/tertiary/quaternary) structure in every lane regardless of reducing agent, but only beta-mercaptoethanol breaks a covalent (disulfide) bond. So the "without reducing agent" band mass is the sum of the covalently linked subunits, and the "with reducing agent" band mass is that of the separated monomer(s).

    WATCH OUT FOR

    Assuming SDS-PAGE always separates a complex into its true individual subunit masses regardless of reducing agent, and so reading 50 kDa directly as one subunit's mass — this ignores that an interchain disulfide bond is a covalent link that SDS alone cannot break, so without beta-mercaptoethanol A and B still run together as a single 50 kDa species.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 48.50 to 49.50, or 18521.20 to 18521.30
54. Q.54NAT · 2 marksOfficial answer: 83.30 to 83.40

    Recombinant DNA Technology and Tools › Bioinformatics

    Sensitivity (recall / true positive rate) in a bioinformatics search is the fraction of all real positives in the database that the algorithm successfully retrieves.

    WATCH OUT FOR

    Using the number of retrieved sequences (72) as the denominator instead of the total actual globins present in the database (78) — this computes precision (65/72 = 90.28%), not sensitivity.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: -219.30 to -219.20