Biomolecules

Question 1

(a) Derivation (7 marks)

• Rate of ES formation = $k_1[E][S]$; rate of breakdown = $(k_{-1}+k_2)[ES]$. (1)
• Steady state: $k_1[E][S] = (k_{-1}+k_2)[ES]$. (1)
• Substitute $[E]=[E]_0-[ES]$: $k_1([E]_0-[ES])[S]=(k_{-1}+k_2)[ES]$. (1)
• Solve: $[ES]=\dfrac{[E]_0[S]}{\frac{k_{-1}+k_2}{k_1}+[S]}=\dfrac{[E]_0[S]}{K_M+[S]}$. (1)
• Rate of product formation $v=k_2[ES]=\dfrac{k_2[E]_0[S]}{K_M+[S]}$. (1)
• Identify $V_{max}=k_2[E]_0$ ⇒ $v=\dfrac{V_{max}[S]}{K_M+[S]}$. (1)
• Physical meaning: when $k_2\ll k_{-1}$, $K_M\approx k_{-1}/k_1 = K_d$, the dissociation constant of ES; low $K_M$ ⇒ high substrate affinity. (1)

(b) Fitting (8 marks)

Take two points to solve the linear form $1/v = (K_M/V_{max})(1/S) + 1/V_{max}$.

Using $[S]=1.0, v=30$: $1/v=0.03333$; $1/S=1.0$. Using $[S]=4.0, v=48$: $1/v=0.020833$; $1/S=0.25$.

Slope $= \dfrac{0.03333-0.020833}{1.0-0.25}=\dfrac{0.0125}{0.75}=0.016667$ (= $K_M/V_{max}$). (2)

Intercept $=1/v - \text{slope}\cdot(1/S)=0.020833-0.016667(0.25)=0.016667$ (= $1/V_{max}$). (2)

So $V_{max}=1/0.016667=60.0$ µM s⁻¹. (1) $K_M=\text{slope}\times V_{max}=0.016667\times60=1.0$ mM. (1)

Predict at $[S]=8.0$: $v=\dfrac{60\times8}{1+8}=\dfrac{480}{9}=53.3$ µM s⁻¹. (2)

(Check consistency with $[S]=0.5$: $v=60(0.5)/1.5=20$ ✓, and $[S]=2$: $60(2)/3=40$ ✓.)

(c) Coding (4 marks)

def predict_v(S, Vmax, Km):
    return Vmax * S / (Km + S)

• 2 marks correct formula. Least squares: fit $y=1/v$ vs $x=1/S$ to a straight line by minimising $\sum(y_i-(mx_i+c))^2$; slope $m=K_M/V_{max}$, intercept $c=1/V_{max}$. Using all points reduces the impact of measurement error in any single point (2). (2)

(d) Models (3 marks)

• Lock-and-key: active site is rigid, pre-shaped complementary to substrate; no conformational change (1.5).
• Induced fit: active site is flexible; substrate binding induces a conformational change that moulds the site around it, optimising catalysis (1.5).

Question 2

(a) Definitions (5 marks)

• Zwitterion: a species with both a positive ($-NH_3^+$) and negative ($-COO^-$) charge but net charge zero (2).
• pI: pH at which the amino acid has zero net charge / exists predominantly as zwitterion (1).
• For acidic amino acid, moving up in pH the species pass through charges $+1 \to 0 \to -1 \to -2$. The neutral (zwitterion) species lies between the two lowest $pK_a$ dissociations, so $pI = \frac{1}{2}(pK_{a1}+pK_{a2})$ — the two protonations that flank the neutral form (2).

(b) pI (3 marks) $pI=\frac{1}{2}(1.99+3.90)=\frac{5.89}{2}=2.945\approx 2.95$. (3)

(c) Henderson–Hasselbalch (6 marks) $pH=pK_a+\log\frac{[A^-]}{[HA]}$ $7.4=1.99+\log\frac{[A^-]}{[HA]}$ ⇒ $\log\text{ratio}=5.41$ ⇒ ratio $=10^{5.41}=2.57\times10^{5}$. (4) Net charge sign: pH 7.4 ≫ pI (2.95), so the molecule is deprotonated at both COOH groups and NH₃⁺ largely intact ⇒ net negative charge. (2)

(d) Peptide bond (6 marks)

• Reaction: $H_2N–CHR_1–COOH + H_2N–CHR_2–COOH \rightarrow H_2N–CHR_1–CO–NH–CHR_2–COOH + H_2O$ (condensation, loss of water) (2).
• Partial double-bond character: lone pair on amide N delocalises into the C=O, giving resonance ⇒ C–N bond is partially double (2).
• Consequence: the peptide bond is planar and rotation about it is restricted; this planarity plus H-bonding between backbone C=O and N–H dictates regular secondary structures (α-helix, β-sheet) (2).

Question 3

(a) Chargaff (6 marks) Total bases $=2\times3000=6000$. (1) A = 22% of 6000 $=1320$; by Chargaff A=T ⇒ T $=1320$. (2) A+T = 44% ⇒ G+C = 56% ⇒ 3360 bases; G=C ⇒ each $=1680$. (3) Answer: A = 1320, T = 1320, G = 1680, C = 1680.

(b) Melting temperature (4 marks) G–C pairs share 3 hydrogen bonds; A–T pairs share 2 (2). More H-bonds per pair means greater thermal energy needed to separate strands, so G–C-rich regions have higher $T_m$ (2).

(c) Saponification (8 marks)

(i) Balanced equation: $C_3H_5(OOC\,C_{17}H_{35})_3 + 3\,NaOH \rightarrow C_3H_5(OH)_3 + 3\,C_{17}H_{35}COONa$ (tristearin + 3 NaOH → glycerol + 3 sodium stearate). (2)

(ii) Moles tristearin $=89.05/890.5=0.1000$ mol. (1) NaOH needed $=3\times0.1000=0.300$ mol ⇒ mass $=0.300\times40.0=12.0$ g. (2) Soap: $3\times0.1000=0.300$ mol sodium stearate ⇒ mass $=0.300\times306.5=91.95$ g. (2) (Glycerol $=0.1\times92=9.2$ g — mass balance check: $89.05+12.0=101.05 = 91.95+9.2$ ✓.) (1)

[
 {"claim":"MM fit: Vmax=60, Km=1 from two points; predict v at S=8 is 53.33",
  "code":"slope=(Rational(1,30)-Rational(1,48))/(1-Rational(1,4)); intercept=Rational(1,48)-slope*Rational(1,4); Vmax=1/intercept; Km=slope*Vmax; v8=Vmax*8/(Km+8); result=(Vmax==60 and Km==1 and simplify(v8-Rational(160,3))==0)"},
 {"claim":"pI of aspartic acid = 2.945",
  "code":"pI=(Float('1.99')+Float('3.90'))/2; result=abs(pI-2.945)<1e-9"},
 {"claim":"HH ratio A-/HA at pH7.4 for pKa 1.99 approx 2.57e5",
  "code":"ratio=10**(Float('7.4')-Float('1.99')); result=abs(ratio-2.57e5)/2.57e5<0.02"},
 {"claim":"Chargaff base counts A=T=1320, G=C=1680 for 3000 bp, 22% A",
  "code":"total=6000; A=Rational(22,100)*total; T=A; GC=total-2*A; G=GC/2; C=G; result=(A==1320 and T==1320 and G==1680 and C==1680)"},
 {"claim":"Saponification: 0.1 mol tristearin needs 12.0 g NaOH and gives 91.95 g soap",
  "code":"n=Float('89.05')/Float('890.5'); mNaOH=3*n*40; mSoap=3*n*Float('306.5'); result=(abs(mNaOH-12.0)<1e-6 and abs(mSoap-91.95)<1e-6)"}
]