Biomolecules

Question 1 (12 marks)

(a) [3]

• pH 1.0 (below both $pK_a$): –COOH and –$NH_3^+$; net charge +1. [1]
• At pI: –COO⁻ and –$NH_3^+$ (zwitterion); net charge 0. [1]
• pH 11.0 (above both): –COO⁻ and –$NH_2$; net charge –1. [1]

(b) [2] For an amino acid with no ionizable side chain, the zwitterion lies between the two relevant $pK_a$ values: $pI = \frac{pK_{a1}+pK_{a2}}{2} = \frac{2.2+9.6}{2} = 5.9$ Correct formula [1], answer 5.9 [1].

(c) [3] Henderson–Hasselbalch for the carboxyl group: $pH = pK_{a1} + \log\frac{[\text{COO}^-]}{[\text{COOH}]}$ $3.2 = 2.2 + \log\frac{[\text{COO}^-]}{[\text{COOH}]}\Rightarrow \log(\text{ratio}) = 1.0$ $\frac{[\text{COO}^-]}{[\text{COOH}]} = 10^{1.0} = 10$ Setup [1], substitution [1], answer 10:1 [1].

(d) [2] At pH 6.0, which is below Z's pI (7.6), Z carries a net positive charge [1]. Positively charged species migrate toward the cathode (negative electrode) [1].

(e) [2] At its pI the molecule is a zwitterion — it bears both a full positive and full negative charge simultaneously [1]. These ionic charges give strong ion–dipole interactions with water (solubility) and strong electrostatic lattice forces (high melting point) [1].

Question 2 (12 marks)

(a) [5] Michaelis–Menten: $v_0 = \dfrac{V_{max}[S]}{K_m+[S]}$.

Equation 1: $20 = \dfrac{2V_{max}}{K_m+2}$ → $20(K_m+2)=2V_{max}$ [1] Equation 2: $40 = \dfrac{8V_{max}}{K_m+8}$ → $40(K_m+8)=8V_{max}$ [1]

From (1): $V_{max}=10(K_m+2)$. Substitute into (2): $40(K_m+8)=8\cdot10(K_m+2)=80(K_m+2)$ $40K_m+320 = 80K_m+160 \Rightarrow 160 = 40K_m \Rightarrow K_m = 4.0\ \text{mM}$ [2] $V_{max}=10(4+2)=60\ \mu\text{mol min}^{-1}$ [1]

(b) [3] $0.75V_{max} = \dfrac{V_{max}[S]}{K_m+[S]}$ → $0.75(K_m+[S])=[S]$ $0.75K_m = 0.25[S] \Rightarrow [S] = 3K_m = 3(4.0)=12\ \text{mM}$ Setup [1], algebra [1], answer 12 mM [1].

(c) [2] Competitive inhibitor: apparent $K_m$ increases, $V_{max}$ unchanged [1]. Induced fit describes conformational change on substrate binding but not where the inhibitor binds (active site vs elsewhere), so it alone cannot classify inhibition type [1].

(d) [2] As $[S]\gg K_m$, $K_m+[S]\approx[S]$, so $v_0\approx \dfrac{V_{max}[S]}{[S]}=V_{max}$ [1]. Physically, all enzyme active sites are saturated with substrate; rate is limited by enzyme turnover, not substrate availability [1].

Question 3 (10 marks)

(a) [3] Given 5′-TAC GGT CAC ATT-3′. Complement (antiparallel): 3′-ATG CCA GTG TAA-5′, written 5′→3′: $\textbf{5'-AAT GTG ACC GTA-3'}$ Correct base pairing (A–T, G–C) [1], antiparallel orientation [1], correct labelling [1].

(b) [3] mRNA is complementary and antiparallel to the template, with U replacing T: Template 5′-TAC GGT CAC ATT-3′ → mRNA reads antiparallel: $\textbf{5'-AAU GUG ACC GUA-3'}$ Correct complement [1], U for T [1], labelling/antiparallel [1].

(c) [2] Chargaff's rule: A = T, G = C [1]. A = 32% → T = 32%; remaining 36% split equally: G = C = 18% each [1].

(d) [2] Each new duplex retains one parental (old) strand and one newly synthesised strand — hence "semiconservative" [1]. Enzyme: DNA polymerase [1].

Question 4 (10 marks)

(a) [3] Ester formation loses 3 H₂O (one per bond): $M = 92 + 3(284) - 3(18) = 92 + 852 - 54 = 890\ \text{g mol}^{-1}$ Formula/logic [1], arithmetic of fatty acid & water [1], answer 890 [1].

(b) [2] $\text{Triglyceride} + 3\,\text{NaOH} \rightarrow \text{glycerol} + 3\,\text{sodium stearate (soap)}$ Correct products [1], 3:1 stoichiometry [1].

(c) [3] Moles triglyceride $= 89/890 = 0.10\ \text{mol}$ [1]. NaOH needed $= 3\times0.10 = 0.30\ \text{mol}$ [1]. Mass $=0.30\times40 = \mathbf{12\ g}$ [1].

(d) [2] Stearic acid chains are saturated, straight, and pack tightly → strong van der Waals forces → solid [1]. Oleic acid has a cis C=C causing a kink, preventing close packing → weaker forces → liquid [1].

Question 5 (6 marks)

(a) [3] Phenomenon = mutarotation [1]. In solution the cyclic hemiacetal opens to the open-chain aldehyde form and recloses [1], generating both α (OH down) and β (OH up) anomers at the anomeric C-1 until equilibrium; the changing ratio changes optical rotation [1].

(b) [3] Starch has α-1,4 glycosidic linkages which human α-amylase can hydrolyse [1]. Cellulose has β-1,4 linkages [1]; humans lack the enzyme (cellulase/β-glucosidase) to cleave β-linkages, so cellulose passes undigested [1].

[
  {"claim":"pI of amino acid X = 5.9","code":"pI=(2.2+9.6)/2; result = (pI==5.9)"},
  {"claim":"COO-/COOH ratio at pH 3.2 is 10","code":"ratio=10**(3.2-2.2); result = (abs(ratio-10)<1e-9)"},
  {"claim":"Km=4 and Vmax=60 solve both MM equations","code":"Km,Vmax=4,60; e1=20*(Km+2)-2*Vmax; e2=40*(Km+8)-8*Vmax; result = (e1==0 and e2==0)"},
  {"claim":"[S] for 75% Vmax equals 3Km = 12 mM","code":"Km=4; S=3*Km; result = (S==12)"},
  {"claim":"NaOH mass to saponify 89 g glyceryl tristearate = 12 g","code":"M=92+3*284-3*18; n=89/M; mass=3*n*40; result = (M==890 and abs(mass-12)<1e-9)"},
  {"claim":"Chargaff: A=32% gives G=C=18%","code":"A=32; T=A; GC=100-A-T; G=GC/2; result = (T==32 and G==18)"}
]