GATE BT 2026 — Solved Paper

The second sentence ("False claims are not going to help") tells us the first blank describes a dishonest habit with numbers — so the first blank must mean overstates, not understates: rules out (C) and (D). Between (A) and (B), "excels the numbers" isn't idiomatic (excel is intransitive here), while "exaggerates the numbers" is exactly the dishonest habit the sentence is warning about. For the second blank, "honesty engenders trust" (engenders = brings about) is the standard sense — "honesty encourages trust" is grammatically fine but weaker/less precise than the intended pairing, and (A) reads cleanly as a matched pair.

Answer: (A) exaggerates; engenders

The dot counts are the first five prime numbers: 2, 3, ?, 7, 11. The missing (3rd) tile is the 3rd prime, 5.

Answer: (B) 5 dots

• $n$ — total number of items
• $k$ — number of groups the items are split into

This is a pigeonhole argument: 100 students spread over 10 standards average exactly 10 per standard, so at least one standard must have 10 or more — if every standard had 9 or fewer, the total would be at most 90 < 100, a contradiction. (B), (C), and (D) each describe one possible distribution, not one forced by the numbers (e.g. all 100 students could be in a single standard, satisfying (A) but violating (B), (C), and (D)).

Answer: (A)

• $n$ — total number of items available
• $r$ — number chosen and arranged (order matters)

The single-digit primes are 2, 3, 5, 7 — four digits. A 3-digit number using three distinct ones is an ordered selection of 3 from 4 (order matters, since 235 ≠ 253), and none of these digits is 0 so every arrangement is a valid 3-digit number. Count = $P(4,3) = 4 \times 3 \times 2 = 24$.

Answer: (B) 24

• $|A|$, $|B|$ — count of items with property $A$, with property $B$
• $|A \cap B|$ — count of items with both properties

By inclusion-exclusion, students who like at least one subject = $10 + 12 - 4 = 18$. Adding the 6 who like neither: total $= 18 + 6 = 24$.

Answer: (C) 24

The analogy pairs an act with the trait of a person who performs it: charity is done by someone who is magnanimous (generous), and retaliation is done by someone who is vindictive (bent on revenge). (A) pairs charity with its opposite trait (parsimonious = stingy), breaking the pattern; (B) and (C) don't consistently match either act to its performer's trait.

Answer: (D) P = Magnanimous; Q = Vindictive

The diamond-square sits two folds away from the top-left square in both directions, so it lands on the same face regardless of fold order, and both rendered options correctly put it on the front face — that part doesn't discriminate between (i) and (ii). The solid-black square and the two corner triangles are what differ between them, and tracing exactly which cube edge each lands on (by physically folding the net, or walking through it face by face) shows only option (i) reproduces the shading in Panel I correctly — (ii) has one of the corner triangles on the wrong face/orientation.

Answer: (A) Only (i)

The cube's vertex-edge graph is bipartite: split the 8 corners into two sets by parity — e.g. label each corner by how many of its 3 binary coordinates are 1; every edge of a cube connects a vertex of even parity to one of odd parity, never same-to-same. Coloring one parity class one color and the other a second color satisfies the constraint, and one color alone can't work since adjacent corners would clash — so the minimum is 2.

Answer: (D) 2

From (i) and (ii), H2 and H3 occupy the two middle west–east positions, leaving H1 and H4 to occupy the westernmost and easternmost ends (in some order). From (iv), exactly two hills lie west of H2, which fixes H2 at the 3rd position from the west; H3 then takes the 2nd position. From (v), the southernmost hill must have ≥2 hills to its east, which is only possible from the 1st or 2nd position (west–east). From (iii), the southernmost hill can't be at the westernmost (1st) or easternmost (4th) position. The only position satisfying both is the 2nd — which is H3.

Answer: (C) H3

• $O_1, O_2$ — centers of the two circles
• $r_1, r_2$ — their radii (consistent length units)

1. Set up coordinates. Place the square so $W$ is at the origin. Reading off the figure, going around the square the corners are $W=(0,0)$, $X=(4,0)$, $Y=(4,4)$, $V=(0,4)$ — a $4\times4$ square, since $\overline{VW}=\overline{VY}=4\text{ cm}$ are two of its sides.

2. Place $O_1$. $C_1$ sits in the $W$ corner, tangent to the two sides that meet there ($WV$, the left side $x=0$, and $WX$, the bottom side $y=0$). A circle tangent to a line is at a distance equal to its radius from that line — so being tangent to both $x=0$ and $y=0$ pins its center at $(r_1, r_1)$. With $r_1=1$: $O_1=(1,1)$.

3. Place $O_2$. Same logic, opposite corner: $C_2$ sits in the $Y$ corner, tangent to the top side ($y=4$) and the right side ($x=4$). Being at distance $r_2$ from each pins its center at $(4-r_2,\ 4-r_2)$ — $r_2$ in from each of those two sides.

4. Use the tangency condition between the circles. The two circles touch each other externally at $T$ (they're outside one another, just grazing at one point) — for two externally tangent circles, the distance between centers equals the sum of the radii:

   $$|O_1O_2| = r_1+r_2$$

   This is the one geometric fact that turns the picture into an equation in $r_2$ (everything else is just substitution).

5. Write out the distance. Using the distance formula between $O_1=(1,1)$ and $O_2=(4-r_2,\ 4-r_2)$:

   $$|O_1O_2| = \sqrt{(4-r_2-1)^2 + (4-r_2-1)^2} = \sqrt{2(3-r_2)^2} = \sqrt{2}\,(3-r_2)$$

   (both coordinate differences are the same, $3-r_2$, because $O_1$ and $O_2$ sit symmetrically on the square's diagonal — so the two squared terms are identical, and $\sqrt{2\cdot(\cdot)^2}$ simplifies to $\sqrt2$ times that quantity.)

6. Set the two expressions for $|O_1O_2|$ equal (step 4's requirement, using step 5's formula, with $r_1=1$):

   $$\sqrt{2}\,(3 - r_2) = 1 + r_2$$

7. Solve for $r_2$. Expand the left side, then collect every $r_2$ term on one side:

   $$3\sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2$$ $$3\sqrt{2} - 1 = r_2 + \sqrt{2}\,r_2 = r_2\left(1+\sqrt{2}\right)$$ $$r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}}$$

8. Rationalize the denominator, so the answer matches the form of the options (no surd on the bottom). Multiply top and bottom by $(\sqrt2-1)$ — chosen because $(1+\sqrt2)(\sqrt2-1)$ collapses to a plain number, clearing the denominator:

   $$(1+\sqrt{2})(\sqrt{2}-1) = \sqrt{2}-1+2-\sqrt{2} = 1$$ $$(3\sqrt{2}-1)(\sqrt{2}-1) = 3\sqrt2\cdot\sqrt2 - 3\sqrt2 - \sqrt2 + 1 = 6 - 4\sqrt{2} + 1 = 7-4\sqrt{2}$$

   So $r_2 = \dfrac{7-4\sqrt2}{1} = 7-4\sqrt{2}$.

9. Sanity-check the number, not just the algebra: $7-4\sqrt2 \approx 7 - 5.657 = 1.34\text{ cm}$. That's positive (a radius has to be), and less than 3 cm (needed for step 5's $3-r_2$ to be positive, i.e. for $O_2$ to still be past $O_1$ along the diagonal) — both checks pass, so the answer is geometrically sensible, not just an artifact of the algebra.

Answer: (C) $7 - 4\sqrt{2}$

$f_1$ has a removable discontinuity at $x=2$: factoring, $f_1(x) = \dfrac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$, so $\lim_{x\to 2} f_1(x) = 4$. $f_2$ is a polynomial (continuous everywhere), so $\lim_{x\to 2} f_2(x) = f_2(2) = 4-4+2 = 2$. Sum of the limits: $4+2=6$.

Answer: (B) 6

• $A$ — the square matrix
• $\lambda$ — an eigenvalue
• $I$ — the identity matrix of the same size as $A$

Solve $\det(A - \lambda I) = 0$: $(3-\lambda)(-3-\lambda) - 16 = 0 \Rightarrow \lambda^2 - 9 - 16 = 0 \Rightarrow \lambda^2 = 25 \Rightarrow \lambda = \pm 5$.

Answer: (C) 5

Each area is $\int_0^{10} y\,dx$: (A) gives $[x^2/2+8x]_0^{10}=50+80=130$; (B) gives $[3x^2/2]_0^{10}=150$; (C) gives $[x^2+x]_0^{10}=100+10=110$; (D) gives $[x^2/2+2x]_0^{10}=50+20=70$. (B) is largest.

Approximate syllabus mapping: the listed syllabus has no standalone "definite integration" line item, so this is filed under 1.2.1 (calculus) as the closest fit — flagging in case you'd rather leave it untagged.

Answer: (B) $y=3x$

Immunoblotting, ELISA, and qRT-PCR all report a single bulk average across the whole sample, which by construction cannot reveal cell-to-cell heterogeneity. Flow cytometry measures TP53 signal in each dissociated single cell and reports the full distribution across the population — no upfront sorting/separation into subpopulations is needed to see that heterogeneity, since it shows up directly as a spread in the per-cell histogram.

Answer: (B) Flow cytometry

• $K_C$ — controller gain
• $\tau_I$ — integral time constant ($\text{s}$ or $\text{min}$)
• $s$ — Laplace variable

A controller acts on the error signal, $e(t) =$ set point − measured (process) variable. The integral term in a PI controller accumulates (integrates) this error over time to drive out steady-state offset — that's specifically the set point vs. measured-variable difference, not the manipulated variable, disturbance, or a generic input/output pair.

Answer: (C)

Glucose-6-phosphate dehydrogenase (G6PD) is the committed, rate-limiting step of the pentose phosphate pathway, and its product is the NADPH used to regenerate reduced glutathione (GSH) — the cell's main buffer against reactive oxygen species. Hexokinase, enolase, and pyruvate kinase are glycolytic enzymes with no direct role in redox defense.

Answer: (B) Glucose 6-phosphate dehydrogenase

In the nitrogenase complex, ferredoxin donates electrons to the Fe-protein (dinitrogenase reductase), which uses ATP hydrolysis to pass those electrons one at a time to the MoFe-protein (dinitrogenase, i.e. "nitrogenase" proper), which is where N₂ is actually reduced to ammonia. So the order is ferredoxin → reductase → nitrogenase.

Answer: (A)

Hybridoma selection uses HAT medium (hypoxanthine, aminopterin, thymidine). Aminopterin blocks de novo purine and thymidylate synthesis, so surviving cells must run the salvage pathway instead. Hypoxanthine feeds that salvage pathway specifically for purines, via HGPRT (hypoxanthine-guanine phosphoribosyltransferase) — myeloma fusion partners are chosen to be HGPRT-deficient, so only fused hybridomas (which inherit HGPRT from the B cell) can use hypoxanthine to survive. (A) describes thymidine's role in the same medium, not hypoxanthine's.

Answer: (B)

In gel filtration (size-exclusion) chromatography, larger proteins are excluded from the pores of the resin and elute first, while smaller proteins get retained longer and elute last — the opposite order to molecular weight. Elution order here is Albumin (66 kDa, 1st) → Thrombin (37 kDa, 2nd) → Myoglobin (17 kDa, 3rd) → Insulin (6 kDa, 4th/last).

Answer: (B) Myoglobin (17 kDa)

Chloramphenicol binds the 50S ribosomal subunit and directly inhibits peptidyl transferase, blocking peptide bond formation during elongation. Bleomycin damages DNA (not a translation inhibitor); rifampicin inhibits bacterial RNA polymerase (transcription, not translation); and tetracycline blocks aminoacyl-tRNA from binding the ribosomal A site — a different elongation-blocking mechanism, not peptidyl transferase inhibition.

Answer: (C) Chloramphenicol

• $k_{cat}$ — turnover number ($\text{s}^{-1}$)
• $K_M$ — Michaelis constant ($\text{mol L}^{-1}$)

Catalytic efficiency is the standard Michaelis-Menten figure of merit for how effectively an enzyme converts substrate at low (non-saturating) concentration, defined as $k_{cat}/K_M$ — $k_{cat}$ alone just measures turnover rate at saturation, without accounting for how tightly the enzyme binds substrate ($K_M$).

Answer: (C) $k_{Cat}/K_M$

Primase (C) lays down the RNA primer, and DNA Polymerase III (D) extends it into new DNA — neither removes it. DNA gyrase (A) relieves supercoiling ahead of the replication fork. DNA Polymerase I carries a 5'→3' exonuclease activity that excises the RNA primer and then fills the resulting gap with DNA, before DNA ligase seals the nick.

Answer: (B) DNA Polymerase I

Amylose is a linear glucose polymer joined entirely by α(1→4) bonds → 2. Amylopectin is the branched form of starch: α(1→4) along each chain plus α(1→6) at branch points → 4. Cellulose is a linear glucose polymer joined by β(1→4) bonds → 1. Sucrose is glucose linked to fructose through their anomeric carbons, an α1→β2 glycosidic bond → 3. So P-2; Q-3; R-4; S-1.

Answer: (C)

Anaerobic digestion breaks complex organics down in stages: hydrolysis first splits polymers (carbohydrates, proteins, lipids) into soluble monomers; acidogenesis ferments those into volatile fatty acids, alcohols, CO₂, and H₂; acetogenesis converts those products into acetate, CO₂, and H₂; methanogenesis then converts the acetate/CO₂/H₂ pool into methane — necessarily last, since it consumes the products of the stage before it.

Answer: (A)

The light reactions split water (releasing O₂), and use that electron flow plus the proton gradient to generate both ATP (via photophosphorylation) and NADPH (via reduction of NADP⁺ at the end of the electron transport chain) — both feed the Calvin cycle. CO₂ is a Calvin cycle input, not a light-reaction product, and NADH (not NADPH) is the respiratory carrier, ruling out (A), (B), and (D).

Answer: (C) O₂/NADPH/ATP

Cas9 protein, CRISPR repeats, and the PAM sequence are all native to the bacterial/archaeal CRISPR immune system. The single guide RNA (sgRNA) is a laboratory construct: it's an engineered fusion of the natural crRNA and tracrRNA into one chimeric molecule, built specifically to simplify Cas9 targeting for genome editing — it doesn't exist as such in nature.

Answer: (D) sgRNA

• $p, q$ — constant coefficients of the ODE
• $A$ — a root of the characteristic equation (an exponent of a valid solution)

Substituting $y=e^{Ax}$ gives $A^2 e^{Ax} - e^{Ax} = 0 \Rightarrow A^2 = 1 \Rightarrow A = \pm 1$. Both roots are valid — this is a second-order linear ODE with constant coefficients, and its characteristic equation has two real roots, each giving a valid exponential solution.

Answer: (B) 1 and (C) -1

ATP-ADP translocase (the adenine nucleotide translocator, ANT) shuttles ADP into and ATP out of the mitochondrial matrix. Atractyloside blocks it by locking it in the cytosol-facing conformation, and bongkrekic acid blocks it by locking it in the matrix-facing conformation instead — both are classic, textbook ANT inhibitors. Oligomycin acts elsewhere, on the $F_o$ subunit of ATP synthase; amytal blocks Complex I of the electron transport chain. Neither touches the translocase itself.

Answer: (B) Atractyloside and (D) Bongkrekic acid

Fibronectin and collagen are both secreted structural components of the extracellular matrix. Myosin is an intracellular motor protein (cytoskeletal, not secreted), and lamin is a structural protein of the nuclear lamina, inside the cell — neither is part of the ECM.

Answer: (B) Fibronectin and (D) Collagen

A protein query can be searched with tblastn against a database of translated nucleotide sequences (A) — this is a standard BLAST extension, and it's the only option that names a real BLAST-searchable database type. (B), (C), and (D) don't correspond to standard BLAST targets — a protein structure database (C) is searched by structural-alignment tools (e.g. DALI), not BLAST's sequence-comparison model, even though the PDB's sequences are themselves BLAST-searchable via pdbaa.

Answer: (A)

• $a_1$ — first term of the series
• $a_n$ — last term of the series

This is an arithmetic series (step 0.01, from 0.01 to 1.00 — 100 terms). For an arithmetic series, the mean equals the average of the first and last terms: $(0.01+1.00)/2 = 0.505$.

Answer: 0.505

• $V_{max}$ — maximum reaction velocity
• $[E]_{total}$ — total enzyme concentration

Turnover number $k_{cat} = V_{max}/[E]_{total} = 1800/1.5 = 1200\ \text{min}^{-1}$. Converting to per-second: $1200/60 = 20\ \text{s}^{-1}$.

Answer: 20

• $\text{dilution factor}$ — reciprocal of the dilution fraction (e.g. 100 for a 100-fold dilution)

Colony count in the plated volume gives the diluted sample's concentration first: $20\ \text{colonies} / 0.05\ \text{mL} = 400\ \text{CFU mL}^{-1}$ in the diluted sample. Since the sample was diluted 100-fold before plating, the original culture's concentration is 100× that: $400 \times 100 = 40{,}000\ \text{CFU mL}^{-1}$.

Answer: 40000

• $\Delta G_{donor}$ — the electron-donor half-reaction, written as an oxidation
• $\Delta G_{acceptor}$ — the electron-acceptor half-reaction, written as a reduction

This is asking for a full aerobic oxidation reaction, so it needs both half-reactions combined, not just the acetate one on its own — that's why the O₂ half-reaction is given at all.

1. Acetate is the electron donor, and it's being oxidized — the reverse of the given "acetate synthesis" (reduction) half-reaction. Reversing a half-reaction flips the sign of its ΔG:

   $$\Delta G_{\text{acetate oxidation}} = -27.4\ \text{kJ e}^-\text{eq}^{-1}$$

2. O₂ is the electron acceptor, and it's being reduced — exactly as given, no sign flip needed:

   $$\Delta G_{\text{O}_2\text{ reduction}} = -78.72\ \text{kJ e}^-\text{eq}^{-1}$$

3. Combine the two per-electron-equivalent half-reactions into the overall reaction. Because both are already normalized to one electron equivalent, the overall reaction's free energy is simply their sum (the electron donor supplies the electron, the acceptor consumes it — add the two ΔG's to get the net driving force of the coupled redox reaction):

   $$\Delta G_{\text{overall}} = \Delta G_{\text{donor (oxidation)}} + \Delta G_{\text{acceptor (reduction)}} = -27.4 + (-78.72) = -106.12\ \text{kJ e}^-\text{eq}^{-1}$$

Answer: -106.12

• $\text{IU}$ — international unit, 1 µmol of substrate converted per minute

Working per mL: the reaction rate is $10\ \text{mM/hr} = 10\ \mu\text{mol mL}^{-1}\text{hr}^{-1}$ (since 1 mM = 1 µmol/mL), which is $10/60 = 0.1\overline{6}\ \mu\text{mol mL}^{-1}\text{min}^{-1} = 0.1\overline{6}\ \text{IU mL}^{-1}$ (1 IU = 1 µmol/min). Specific activity is activity per unit mass of enzyme: $0.1\overline{6}\ \text{IU mL}^{-1} / 10\ \mu g\ mL^{-1} = 0.01\overline{6}\ \text{IU}\ \mu g^{-1}$.

Answer: 0.017

• $M$ — step input magnitude
• $K_p$ — process gain
• $\tau_p$ — process time constant

$Y(s) = G_p(s)\,U(s) = \dfrac{MK_p}{s(\tau_p s + 1)}$. This is exactly $M K_p$ times the standard first-order step response $\dfrac{1}{s(\tau_p s+1)}$, whose inverse Laplace transform is $1-e^{-t/\tau_p}$. So $y(t) = MK_p\left(1-e^{-t/\tau_p}\right)$ — the step magnitude $M$ scales the whole response, it doesn't sit inside the exponential.

Answer: (B)

The J (joining) gene segment is one of the V(D)J recombination building blocks that generates antibody diversity → 4. Cyclosporin is an immunosuppressant that works by blocking calcineurin, which normally activates T cells → 1. An epitope is the specific site on a (large) antigen that an antibody binds → 2. A hapten is a small molecule that's only immunogenic once attached to a larger carrier → 3. So P-4; Q-1; R-2; S-3.

Answer: (C)

Klinefelter syndrome is caused by an extra X chromosome in a male (typically XXY) → 4. Turner syndrome is monosomy X (a single X, i.e. X-chromosome insufficiency) → 1. Bloom syndrome is caused by mutations in BLM, a RecQ-family DNA helicase → 2. Xeroderma pigmentosum is a defect in nucleotide excision repair → 3. So P-4; Q-1; R-2; S-3.

Answer: (B)

Lectins are carbohydrate-binding proteins → 3. P-glycoprotein is an ATP-dependent efflux pump (the multidrug-resistance transporter) →

1. Digitoxigenin is a cardiac glycoside that inhibits the Na⁺/K⁺-ATPase → 4. Mannose-6-phosphate is the tag that routes lysosomal enzymes to the lysosome → 2. So P-3; Q-1; R-4; S-2.

Answer: (D)

• $n_i$ — moles of element $i$ per basis mass
• $n_{reference}$ — moles of the element chosen to normalize against (usually N)

Per 100 g of biomass, the ash-free portion is C:50g, H:8g, O:20g, N:10g (these already sum to 88g, consistent with 12% ash separately). Converting to moles ($C{=}12$, $H{=}1$, $O{=}16$, $N{=}14$): C: $50/12=4.167$, H: $8/1=8.0$, O: $20/16=1.25$, N: $10/14=0.714$. Dividing every value by the N moles (0.714) to normalize to N=1: C: $4.167/0.714 = 5.83$, H: $8/0.714=11.2$, O: $1.25/0.714=1.75$.

Answer: (A) $C_{5.8}H_{11.2}O_{1.7}N$

• $a,b,c,d,e,f,g,h,i$ — matrix entries, read row by row

Compute the determinant by cofactor expansion along the first row: $3[(-1)(9)-(0)(2)] - 1[(2)(9)-(0)(5)] + 5[(2)(2)-(-1)(5)] = 3(-9) - 1(18) + 5(9) = -27-18+45 = 0$. The determinant is 0, so the matrix is singular — not full rank and not invertible, ruling out (A) and (B). But (C) and (D) are trivially true of any square matrix, regardless of whether it happens to be singular: every matrix has a determinant (even when that value is zero), and every matrix has a transpose.

Answer: (C) and (D)

There are two in-frame AUG codons here, in different reading frames, so each needs to be checked separately as a possible start:

1. Start at position 7 (AUG-ACC-CAU-GUG-GCG-AGA-CGG-UAG): in the cytosol, AGA just codes Arg (not a stop there), so translation runs all the way to the UAG stop — Met-Thr-His-Val-Ala-Arg-Arg, 7 amino acids. In the mitochondria, AGA is a stop, so this same start only reaches Met-Thr-His-Val-Ala before terminating — 5 amino acids. That's a valid mitochondrial product, but not the longest one available (see next).

2. Start at position 14 (AUG-UGG-CGA-GAC-GGU-AGU-UAA): this reading frame never hits an AGA codon at all (note CGA and AGU are not AGA), so it runs to the universal UAA stop in both compartments — Met-Trp-Arg-Asp-Gly-Ser, 6 amino acids, identically in cytosol and mitochondria.

The question asks for the longest possible polypeptide in each compartment. In the cytosol, that's the 7-aa product from start 1 (A) — longer than the 6-aa alternative from start 2. In the mitochondria, the AGA-truncated 5-aa product from start 1 is not the longest available, because start 2 avoids the AGA stop entirely and reaches 6 amino acids instead (C) — so (B) describes a real but non-maximal product, and isn't selected.

Answer: (A) and (C)

The shown 30-nucleotide fragment is the last 30 bases of a 330-base message, so it spans positions 301–330. Position 322 is the 22nd base of this fragment, which falls on the first base of the 8th codon: AGA (coding Arg). The mutation changes this to UGA — a stop codon. Turning a sense codon into a stop codon mid-message is, by definition, a nonsense mutation (B). Separately, classifying the substitution itself: A is a purine and U is a pyrimidine, so a purine→pyrimidine change is a transversion (C), not a transition (which would be purine→purine or pyrimidine→pyrimidine).

Answer: (B) and (C)

A DO reading of zero under continuous aeration means oxygen is being consumed as fast as (or faster than) it's being supplied — the oxygen transfer rate can't keep up with the culture's oxygen uptake rate, so growth becomes oxygen-transfer-limited (C). It says nothing about the culture being dead (A — cells can grow perfectly well right up to the point oxygen becomes limiting, that's the definition of the limitation), nothing about carbon source (B — an unrelated, unsupported inference), and directly contradicts (D).

Answer: (C)

• $\tau$ — shear stress
• $\mu$ — viscosity
• $\dot\gamma$ — shear rate

A Newtonian fluid is defined by $\tau = \mu\dot\gamma$ — shear stress is directly proportional to shear rate, with viscosity $\mu$ (always positive) as the constant of proportionality. That's a straight line through the origin with slope $\mu > 0$.

Answer: (D)

Approximate syllabus mapping: the listed syllabus has no dedicated fluid mechanics/rheology line item, so this and Q47 are filed under 4.1.10 (reactor design/scale-up, which rheology directly feeds into) as the closest fit.

$k_La$ is a physical/hydrodynamic property of the gas-liquid system — how much interfacial area is available and how fast oxygen crosses it — not a statement about the driving force. Switching to pure oxygen (A) increases the concentration gradient (raises $C^*$), which increases the actual oxygen transfer rate, but doesn't change $k_La$ itself, so it's excluded here. Increasing air flow rate (B) raises gas holdup and turbulence, increasing interfacial area $a$. Reducing bubble diameter (C) directly increases interfacial area per unit volume ($a \propto 1/d$). Decreasing agitation (D) reduces turbulence and bubble breakup, decreasing $k_La$ — the opposite of what's asked.

Answer: (B) and (C)

• $\rho$ — fluid density
• $v$ — velocity
• $D$ — pipe diameter
• $\mu$ — viscosity

The laminar–turbulent transition is governed entirely by the Reynolds number, $Re = \rho v D/\mu$ — density, velocity, diameter, and viscosity. Pipe length doesn't appear anywhere in that expression: a longer or shorter pipe carrying the same fluid at the same velocity and diameter transitions at the same $Re$, so length is the one factor that doesn't matter.

Answer: (B)

Arginine methylation and lysine acetylation are both classic histone post-translational modifications — protein modifications made after translation, which is specifically what the question asks for. Cytosine methylation (C) is a genuine epigenetic mechanism, but it's a direct modification of the DNA base itself, not a protein — not a "posttranslational" modification at all. Cytosine deamination (D) is a DNA-damage/mutagenic event, not a regulated epigenetic mark.

Answer: (A) and (B)

An ideal PFR has, by definition, no back-mixing and no axial dispersion — ruling out (B) and (C) directly. It does have a flat ("plug") velocity profile across every cross-section, by definition (D). And a classic, textbook result is that a PFR is the limiting case of an infinite number of ideal CSTRs connected in series — so a large number of CSTRs-in-series is a standard way to approximate one (A).

Answer: (A) and (D)

• $\Delta G$ — standard Gibbs free energy change
• $R$ — gas constant
• $T$ — absolute temperature

$\Delta G$ is a thermodynamic quantity — it tells you the direction and extent of a reaction at equilibrium, not how fast it gets there. Rate is governed by kinetics (activation energy, $k_{cat}$), which $\Delta G$ says nothing about directly — a highly favorable reaction can still be kinetically slow (that's exactly why cells need enzymes at all). So (A) is correct, and (B)/(C) are both unjustified guesses in either direction. Separately, $K_{eq} = e^{-\Delta G/RT}$: for $\Delta G=-100\text{ kJ/mol}$ at room temperature, $K_{eq}$ is astronomically large, meaning the reverse reaction is negligible — thermodynamically, that's effectively irreversible (D), even though nothing here says anything about its rate.

Answer: (A) and (D)

• $h$ — step size between points
• $x_0,\dots,x_n$ — equally spaced points from $a$ to $b$

Since $dy/dx = 1/x$, the integral $\int_1^2 dy$ is really $\int_1^2 \frac{1}{x}\,dx$, approximated numerically with 4 trapezoidal strips ($h=0.25$, points $x=1, 1.25, 1.5, 1.75, 2$):

(Sanity check: the exact value is $\ln 2 = 0.6931$ — trapezoidal over-estimates slightly here because $1/x$ is convex, which matches.)

Answer: 0.70

• $A_1,B_1,C_1$ — coefficients of the first line
• $A_2,B_2,C_2$ — coefficients of the second line

Two lines $A_1x+B_1y=C_1$ and $A_2x+B_2y=C_2$ are parallel exactly when their coefficients are proportional: $A_1/A_2 = B_1/B_2$. Here:

Answer: 8

• $(a,b)$ — circle center
• $r$ — circle radius

The circle is $(x-1)^2+(y-1)^2=1$. Substituting $y=x-1$ and letting $u=x-1$:

giving $x=1$ (so $y=0$) and $x=2$ (so $y=1$) — the two intersection points are $(1,0)$ and $(2,1)$. The chord length is the distance between them:

Answer: 1.414

Approximate syllabus mapping: this and Q52 are pure coordinate-geometry questions with no dedicated syllabus line item — filed under 1.1.2 (systems of linear equations) as the closest fit, since solving them comes down to solving a system.

• $T_{env}$ — ambient/final temperature
• $T_0$ — initial temperature
• $k$ — cooling rate constant

This is Newton's law of cooling: $\frac{dT}{dt}=-k(T-30)$, giving $T(t)-30 = (T_0-30)e^{-kt} = 60\,e^{-kt}$ with $T_0=90$.

1. Find $k$ from the given data point ($T=70$ at $t=30$):

   $$70-30=60\,e^{-30k} \;\Rightarrow\; e^{-30k}=\tfrac{2}{3} \;\Rightarrow\; k = \frac{\ln(3/2)}{30} \approx 0.01352\ \text{min}^{-1}$$

2. Evaluate at $t=51.5$ min:

   $$T-30 = 60\,e^{-0.01352 \times 51.5} = 60\,e^{-0.6960} \approx 60(0.4985) \approx 29.9$$ $$T \approx 30+29.9 = 59.9 \approx 60\text{°C}$$

Answer: 60

• $\mu_{max}$ — maximum specific growth rate
• $K_S$ — Monod (half-saturation) constant
• $S$ — limiting substrate concentration
• $k_d$ — endogenous decay rate

Net specific growth rate accounts for endogenous decay: $\mu_{net} = \mu_{max}\dfrac{S}{K_S+S} - k_d$. Everything needs to be in the same time base (days) before combining terms:

1. Convert $\mu_{max}$ to day⁻¹: $0.1\text{ h}^{-1} \times 24 = 2.4\text{ day}^{-1}$.
2. Evaluate the Monod term, with $S=23{,}000\text{ mg L}^{-1}$ and $K_S=1\text{ mg L}^{-1}$ (converting $S$ to the same units as $K_S$): $$\mu = 2.4 \times \frac{23{,}000}{1+23{,}000} \approx 2.4 \times 0.99996 \approx 2.4\text{ day}^{-1}$$ ($S \gg K_S$ here, so the culture is essentially at $\mu_{max}$.)
3. Subtract decay: $\mu_{net} = 2.4 - 0.1 = 2.3\text{ day}^{-1}$.

Answer: 2.3

• $V$ — reactor volume
• $X$ — in-reactor biomass concentration
• $Q$ — flow rate of the stream leaving the system
• $X_e$ — biomass concentration in that leaving stream

Mean cell retention time (MCRT, $\theta_c$) is cells held in the system divided by the rate cells leave it: $\theta_c = \dfrac{V\,X}{Q\,X_e}$, where $X$ is the in-reactor biomass concentration and $X_e$ is the concentration in the stream actually leaving the system.

Answer: 100

• $n_C, n_H, n_O$ — number of carbon, hydrogen, oxygen atoms per molecule
• $\gamma$ — electron equivalents per mole

1. Molar mass of acetate ion (C₂H₃O₂⁻): $2(12)+3(1)+2(16) = 59\text{ g/mol}$.
2. Molar concentration: $10 / 59 = 0.1695\text{ mol L}^{-1}$.
3. Electron equivalents per mole. Full combustion of one mole of acetate/acetic acid consumes 2 mol O₂ (C₂H₄O₂ + 2O₂ → 2CO₂ + 2H₂O), and each mole of O₂ accepts 4 electrons ($\text{O}_2+4\text{H}^++4e^-\to2\text{H}_2\text{O}$) — so acetate carries $2\times4=8$ electron equivalents per mole.
4. Combine: $0.1695 \times 8 = 1.356\text{ e}^-\text{eq L}^{-1}$.

Answer: 1.4

Approximate syllabus mapping: no dedicated "electron equivalents / degree of reduction" line item exists in the syllabus — filed under 4.1.6 as the closest fit, since this is the stoichiometric machinery behind growth/substrate balances.

The plotted line runs from $(0,1)$ to $(3,4)$, i.e. slope 1 and intercept 1, so $\dfrac{dy}{dx} = 1+x$. Integrating:

Use $y(2)=5$ to find $C$: $2 + 2 + C = 5 \Rightarrow C=1$. Then:

Answer: 8.5

• $N_0$ — initial viable count
• $k_d$ — death rate constant
• $t$ — time

First-order thermal death: $N(t) = N_0\,e^{-k_dt}$.

Answer: 30

• $C_0$ — initial concentration
• $k$ — zero-order rate constant

Zero-order kinetics: concentration falls linearly, $C = C_0 - kt$. From the given data, the rate constant is $k = \dfrac{42-14}{4} = 7\text{ g L}^{-1}\text{h}^{-1}$. "Complete utilization" means $C\to0$ starting from $C_0=42$:

Answer: 6

• $D_{crit}$ — critical (washout) dilution rate
• $S_0$ — feed substrate concentration

At steady state $D=\mu=\mu_{max}\frac{S}{K_S+S}$, and $S$ can rise at most to the feed concentration $S_0$ (as $X\to0$). So washout happens once the dilution rate reaches the maximum growth rate achievable at $S_0$:

(with $S_0 \gg K_S$, the Monod term is essentially 1). The minimum feed flow rate for washout is then:

Answer: 2.5

1. Molar mass of C₅H₇O₂N: $5(12)+7(1)+2(16)+14 = 113\text{ g/mol}$.
2. Nitrogen's mass fraction in the cells: $14/113 = 0.1239$ (12.39%).
3. Nitrogen needed for 150 mg cells: $150 \times 0.1239 = 18.58\text{ mg N}$.
4. Phosphorus needed, at 20% of the N requirement: $$0.2 \times 18.58 = 3.72\text{ mg P}$$

Answer: 3.7

• $N_0$ — starting cell number
• $t_d$ — doubling time

Track the population by doubling time (24 h):

• $t=0$: 1 cell
• $t=24$h: 2 cells
• $t=48$h: 4 cells, then one dies → 3 cells (the "after the first 48 hours" apoptosis)

From here, no more death — the population of 3 just keeps doubling every 24 h. From $t=48$h to $t=168$h (7 days) is $(168-48)/24 = 5$ more doublings: