Level 3 — ProductionBiomolecules

Biomolecules

60 marksprintable — key stays hidden on paper

Chapter: 4.5 Biomolecules Level: 3 — Production (from-scratch derivations, explain-out-loud, mechanism from memory) Time: 45 minutes Total Marks: 60

Instructions: Show all reasoning. Derive results from first principles where asked. Structures and equations must be drawn/written explicitly.

Q1. Michaelis–Menten derivation (from scratch). [12 marks]

(a) Starting from the reaction scheme $E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \xrightarrow{k_2} E + P,$ derive the Michaelis–Menten rate equation using the steady-state approximation. State every assumption. Define $K_M$ in terms of rate constants. [7]

(b) Show that when $[S] = K_M$ , the velocity equals $V_{max}/2$ . [2]

(c) A reaction has $V_{max} = 0.10\ \text{mM s}^{-1}$ and $K_M = 2.0\ \text{mM}$ . Compute $v$ at $[S] = 6.0\ \text{mM}$ . [3]

Q2. Amino acid zwitterion & pI (derive). [10 marks]

(a) For a neutral amino acid (glycine) with $pK_{a1}(\text{COOH}) = 2.3$ and $pK_{a2}(\text{NH}_3^+) = 9.6$ , draw the three protonation forms and derive the expression for the isoelectric point $pI$ . Compute $pI$ . [5]

(b) Explain, using the Henderson–Hasselbalch equation, why the molecule carries zero net charge exactly at $pI$ . [3]

(c) Lysine has side-chain $pK_a = 10.5$ (with $pK_{a1}=2.2$ , $pK_{a2}=9.0$ ). Explain which two $pK_a$ values are averaged for its $pI$ and compute it. [2]

Q3. Carbohydrate structure & mutarotation (explain out loud). [10 marks]

(a) Convert the open-chain Fischer projection of D-glucose to the α- and β-D-glucopyranose Haworth projections. Explain which carbon becomes the anomeric centre and why two anomers form. [5]

(b) Explain the mechanism of mutarotation and why the equilibrium specific rotation ( $+52.7°$ ) lies between that of pure α ( $+112°$ ) and pure β ( $+18.7°$ ). Given the equilibrium rotation, estimate the % of β-anomer present (ignore trace open form). [5]

Q4. Central dogma from memory. [10 marks]

(a) Write the base-pairing rules for DNA and for DNA→RNA. For the template strand 3'-TAC GGA TTC ACT-5', write the mRNA sequence (5'→3'). [4]

(b) Explain replication (semiconservative), transcription, and translation each in one precise sentence, naming the enzyme responsible for each. [3]

Q5. Lipids — saponification (derive + calculate). [10 marks]

(a) Write the general saponification reaction of a triglyceride with NaOH. Explain why the products are soap + glycerol. [4]

(b) Saponification value is the mg of KOH required to saponify 1 g of fat. Derive why a fat with lower average fatty-acid chain length has a higher saponification value. [3]

(c) Compute the saponification value of tristearin (glyceryl tristearate, $M = 890\ \text{g mol}^{-1}$ ), given $M_{KOH} = 56\ \text{g mol}^{-1}$ and 3 mol KOH per mol fat. [3]

Q6. Concept discrimination (explain out loud). [8 marks]

(a) Contrast lock-and-key vs induced-fit models of enzyme action. [3]

(b) Contrast peptide vs steroid hormones with respect to solubility, receptor location, and speed of action. [3]

Answer keyMark scheme & solutions

Q1 — Michaelis–Menten [12]

(a) [7]

Rate of product formation: $v = k_2[ES]$ . [1]
Steady-state assumption: $\dfrac{d[ES]}{dt}=0$ , so formation = breakdown: $k_1[E][S] = (k_{-1}+k_2)[ES].$ [1] (assumption stated: [ES] constant)
Solve: $[ES] = \dfrac{k_1[E][S]}{k_{-1}+k_2} = \dfrac{[E][S]}{K_M}$ , with $K_M \equiv \dfrac{k_{-1}+k_2}{k_1}$ . [1] (definition)
Mass balance: $[E]_0 = [E] + [ES]$ , so $[E] = [E]_0 - [ES]$ . [1]
Substitute: $[ES]K_M = ([E]_0-[ES])[S] \Rightarrow [ES] = \dfrac{[E]_0[S]}{K_M+[S]}$ . [1]
Then $v = k_2[ES] = \dfrac{k_2[E]_0[S]}{K_M+[S]}$ . With $V_{max}=k_2[E]_0$ : [1] $\boxed{v = \dfrac{V_{max}[S]}{K_M+[S]}}$ [1]

(b) [2] Set $[S]=K_M$ : $v = \dfrac{V_{max}K_M}{K_M+K_M} = \dfrac{V_{max}}{2}$ . Hence $K_M$ = [S] giving half-maximal velocity. [2]

(c) [3] $v = \dfrac{0.10 \times 6.0}{2.0+6.0} = \dfrac{0.60}{8.0} = 0.075\ \text{mM s}^{-1}$ . [3]

Q2 — Zwitterion & pI [10]

(a) [5] Forms: cationic $\text{H}_3\text{N}^+\!-\!CHR\!-\!COOH$ → zwitterion $\text{H}_3\text{N}^+\!-\!CHR\!-\!COO^-$ → anionic $\text{H}_2\text{N}\!-\!CHR\!-\!COO^-$ . [2]

pI is where zwitterion dominates and net charge = 0, midway between the two $pK_a$ flanking the zwitterion: $pI = \tfrac12(pK_{a1}+pK_{a2}) = \tfrac12(2.3+9.6) = 5.95.$ [3]

(b) [3] By H–H: $pH = pK_a + \log\frac{[\text{base}]}{[\text{acid}]}$ . At pI the concentration of the cation equals that of the anion (symmetric about the zwitterion), so the positive and negative species cancel and average charge = 0. [3]

(c) [2] Lysine is basic (extra amino group). The zwitterion (net 0) is flanked by $pK_{a2}=9.0$ (α-NH₃⁺) and the side-chain $pK=10.5$ . Average the two highest pK's: $pI = \tfrac12(9.0+10.5)=9.75.$ [2]

Q3 — Carbohydrate [10]

(a) [5] C1 (aldehyde carbon) attacks the C5–OH → forms hemiacetal ring; C1 becomes the anomeric carbon. [2] Ring closure creates a new stereocentre at C1, giving α (OH down, trans to C6 CH₂OH) and β (OH up) anomers. [2] Correct Haworth with CH₂OH up, α-OH down/β-OH up. [1]

(b) [5] Mutarotation: ring opens to open-chain aldehyde then recloses, interconverting α and β until equilibrium. [2] Let fraction β = $x$ : $18.7x + 112(1-x) = 52.7$ . [1] $112 - 93.3x = 52.7 \Rightarrow 93.3x = 59.3 \Rightarrow x = 0.635$ . [1] So ≈ 64% β, 36% α — β favoured (equatorial OH more stable). [1]

Q4 — Central dogma [10]

(a) [4] DNA pairing: A–T, G–C. DNA→RNA: A–U, T–A, G–C, C–G. [1] Template 3'-TAC GGA TTC ACT-5' → mRNA read 5'→3'complement: 5'-AUG CCU AAG UGA-3'. [3]

(b) [3] Replication: DNA is copied semiconservatively, each daughter keeping one parental strand — DNA polymerase. [1] Transcription: DNA template copied into mRNA — RNA polymerase. [1] Translation: ribosome reads mRNA codons to build a polypeptide (tRNA delivers amino acids) — ribosome/peptidyl transferase. [1]

(c) [3] mRNA AUG CCU AAG UGA = 4 codons. [1] Start codon = AUG (Met). [1] (UGA = stop) [1]

Q5 — Lipids/Saponification [10]

(a) [4] Triglyceride + 3 NaOH → glycerol + 3 RCOO⁻Na⁺ (soap). [2] Base hydrolyses the three ester bonds; carboxylate salts (soap) form and glycerol is released as the alcohol backbone. [2]

(b) [3] Sap. value ∝ mol of ester (KOH) per gram of fat. Shorter chains → lower molar mass per triglyceride → more moles of fat per gram → more mol ester bonds per gram → more KOH per gram → higher value. [3]

(c) [3] Per mol tristearin: 3 mol KOH = $3\times56 = 168$ g KOH $= 168000$ mg per 890 g fat. $\text{Sap value}=\frac{168000}{890}=188.8\ \text{mg KOH/g}.$ [3]

Q6 — Concepts [8]

(a) [3] Lock-and-key: rigid complementary active site fits substrate exactly. Induced-fit: active site is flexible and changes shape upon substrate binding to optimise catalysis. (1 each + 1 for correct contrast) [3]

(b) [3] Peptide: water-soluble, bind surface (membrane) receptors, fast (second-messenger) action. Steroid: lipid-soluble, cross membrane to bind intracellular/nuclear receptors, slower (gene-level) action. [3]

(c) [2] Vitamin C is water-soluble → excreted in urine, not stored. Vitamin A is fat-soluble → stored in liver/adipose tissue → accumulates to toxicity. [2]

[
  {"claim":"MM velocity at S=6, Vmax=0.10, KM=2.0 is 0.075", "code":"Vmax=Rational(1,10); KM=2; S=6; v=Vmax*S/(KM+S); result = (v==Rational(3,40)) and (float(v)==0.075)"},
  {"claim":"Glycine pI = 5.95", "code":"pI=(Rational(23,10)+Rational(96,10))/2; result = (float(pI)==5.95)"},
  {"claim":"Lysine pI = 9.75", "code":"pI=(Rational(90,10)+Rational(105,10))/2; result = (float(pI)==9.75)"},
  {"claim":"Beta anomer fraction ~0.635 from rotation balance", "code":"x=symbols('x'); sol=solve(Eq(Rational(187,10)*x+112*(1-x), Rational(527,10)), x)[0]; result = (round(float(sol),3)==0.636 or abs(float(sol)-0.635)<0.002)"},
  {"claim":"Tristearin saponification value = 168000/890 ~188.8", "code":"sv=168000/Integer(890); result = (round(float(sv),1)==188.8)"}
]