Biomolecules

Q1. (3 marks)

• Monosaccharides — cannot be hydrolysed to simpler sugars; e.g. glucose, fructose. (1)
• Disaccharides — yield 2 monosaccharides on hydrolysis; e.g. sucrose, maltose. (1)
• Polysaccharides — yield many monosaccharide units on hydrolysis; e.g. starch, cellulose. (1)

Q2. (4 marks)

• Mutarotation: the gradual change in specific rotation of a freshly prepared solution of a sugar until it reaches a constant equilibrium value, due to interconversion of $\alpha$- and $\beta$-anomers via the open-chain form. (2)
• $\alpha$-D-glucose: $+112^\circ$; $\beta$-D-glucose: $+19^\circ$. (1)
• Equilibrium value: $+52.7^\circ$ (≈ $+52.5^\circ$). (1)

Q3. (5 marks) (a) Zwitterion: $\text{H}_3\overset{+}{\text{N}}-\text{CHR}-\text{COO}^-$ (net charge zero, –NH$_3^+$ and –COO$^-$). (2) (b) pI = the pH at which the amino acid exists mainly as the zwitterion with zero net charge (equal +ve and –ve, no net migration in an electric field). (1) $pI=\frac{pK_{a1}+pK_{a2}}{2}=\frac{2.3+9.7}{2}=6.0$ (2)

Q4. (5 marks)

• Peptide bond: the amide $-\text{CO-NH-}$ linkage formed between the –COOH of one amino acid and the –NH$_2$ of another with loss of water (condensation). (2)
• Primary — sequence of amino acids. (¾)
• Secondary — local folding: $\alpha$-helix / $\beta$-pleated sheet (H-bonds). (¾)
• Tertiary — overall 3-D folding of a polypeptide. (¾)
• Quaternary — assembly of two or more polypeptide subunits. (¾)

Q5. (5 marks) (a) Lock-and-key: active site has a rigid, pre-formed shape exactly complementary to the substrate (Fischer). Induced-fit: active site is flexible and changes shape on substrate binding to fit it (Koshland). (3: 1½ each) (b) Substituting $[S]=K_m$: $v=\frac{V_{max}K_m}{K_m+K_m}=\frac{V_{max}K_m}{2K_m}=\frac{V_{max}}{2}$ (2)

Q6. (3 marks) $v=\frac{V_{max}[S]}{K_m+[S]}=\frac{6.0\times6.0}{2.0+6.0}=\frac{36}{8}=4.5\ \mu\text{mol min}^{-1}$ Set-up (1), substitution (1), answer $4.5\ \mu\text{mol min}^{-1}$ (1).

Q7. (5 marks)

• Base pairing: A–T (2 H-bonds), G–C (3 H-bonds). (2)
• (In RNA, A pairs with U.)
• Differences (any two, 1½ each): DNA has deoxyribose, RNA has ribose; DNA uses thymine, RNA uses uracil; DNA is double-stranded (helix), RNA is usually single-stranded. (3)

Q8. (4 marks)

• Saponification: alkaline hydrolysis of a fat/oil (ester) to give soap (salt of fatty acid) + glycerol. (2)
• Glyceryl tristearate + 3 NaOH → 3 sodium stearate ($\text{C}_{17}\text{H}_{35}\text{COONa}$) + glycerol ($\text{C}_3\text{H}_5(\text{OH})_3$). (2)

Q9. (3 marks — ½ each)

• Fat-soluble: A, D, E, K.
• Water-soluble: B$_{12}$, C.

Q10. (5 marks) Differences (any two, 2 marks each) + examples (1 mark):

• Peptide hormones: made of amino acids; water-soluble; bind cell-surface receptors; act via second messengers; fast/short action. e.g. insulin.
• Steroid hormones: derived from cholesterol; lipid-soluble; cross membrane and bind intracellular receptors; act on gene transcription; slower/longer action. e.g. testosterone / estrogen / cortisol. (4 + 1)

[
  {"claim":"pI of amino acid with pKa1=2.3, pKa2=9.7 is 6.0","code":"pI=(2.3+9.7)/2\nresult = (pI==6.0)"},
  {"claim":"Michaelis-Menten velocity at S=6mM with Vmax=6, Km=2 is 4.5","code":"Vmax=6.0; Km=2.0; S=6.0\nv=Vmax*S/(Km+S)\nresult = (v==4.5)"},
  {"claim":"At S=Km, v equals Vmax/2","code":"Vmax,Km=symbols('Vmax Km',positive=True)\nv=Vmax*Km/(Km+Km)\nresult = simplify(v-Vmax/2)==0"},
  {"claim":"Saponification of tristearate needs 3 mol NaOH giving 3 soap","code":"NaOH_per_triglyceride=3\nsoap=NaOH_per_triglyceride\nresult = (NaOH_per_triglyceride==3 and soap==3)"}
]