Biomolecules — Proteins & Nucleic Acids

Question 1

(a) (5)

• Central α-carbon bonded to: amino group $\text{NH}_3^+$, carboxyl group $\text{COO}^-$, hydrogen atom, and R (variable side chain). (2, all four)
• Zwitterion: at pH 7.4 the $-\text{COOH}$ loses H⁺ → $-\text{COO}^-$ and $-\text{NH}_2$ gains H⁺ → $-\text{NH}_3^+$; net charge zero. (2)
• Elements in all proteins: C, H, O, N; some also contain S (sulfur, e.g. cysteine). (1)

(b) (4)

• $-\text{COOH}$ of one + $-\text{NH}_2$ of next react; water ($\text{H}_2\text{O}$) is removed (–OH from carboxyl, –H from amino). (2)
• Condensation = two monomers joined with elimination of a small molecule (water). (1)
• Bond formed = peptide bond (amide, C–N linkage). (1)

(c)(i) (3)

• Joining $n$ residues needs $n-1$ links → peptide bonds $= n-1$, water molecules released $= n-1$. (Each bond removes one water.) (3)

(c)(ii) (3)

• Per chain: $148 - 1 = 147$ peptide bonds and $147$ waters.
• 4 chains: $4 \times 147 = 588$ peptide bonds; $588$ water molecules.
• (Quaternary assembly is non-covalent, adds no peptide bonds.) (3)

(d) (4)

• α-helix: stabilised by H-bonds between C=O and N–H of residues 4 apart along the same chain; coiled, ~3.6 residues/turn. (1)
• β-sheet: H-bonds between adjacent segments/strands lying side by side; pleated/zig-zag repeat. (1)
• Denaturation cause 1: heat — increased kinetic energy breaks H-bonds/ionic bonds, tertiary & secondary structure unfold (primary sequence intact). (1)
• Cause 2: pH extremes / strong acid or alkali — alters charges on R-groups (or heavy metals/organic solvents), breaking ionic & H-bonds, shape lost. (1)

(e) (3) — collagen = structural; haemoglobin = transport; amylase = enzymatic; antibody = defence. (¾ each, round to 3)

Question 2

(a) (5)

• Nucleotide = pentose sugar + phosphate group + nitrogenous base, correctly linked. (3)
• DNA has deoxyribose; RNA has ribose. DNA uses thymine, RNA uses uracil (both share A, G, C). (1)
• Nucleic acids contain N and P (nitrogen & phosphorus) absent from pure carbohydrate. (1)

(b) (3)

• Purines (double ring, fused 6+5): adenine, guanine.
• Pyrimidines (single 6-membered ring): cytosine, thymine, uracil. (3)

(c)(i) (3)

• Rule: A=T and G=C (Chargaff). T = 22% → A = 22%.
• A+T = 44% → G+C = 56% → G = C = 28%. (3)

(c)(ii) (3)

• G = 34% → C = 34%; G+C = 68% → A+T = 32% → A = T = 16%.
• Purines (A+G) = 16+34 = 50%; Pyrimidines (T+C) = 16+34 = 50% → ratio 1:1. (2)
• Fixed because each purine always pairs with a pyrimidine across the double helix. (1)

(d)(i) (2)

• Template 5′-TACGGCATTAGC-3′. Complement (antiparallel): 3′-ATGCCGTAATCG-5′ → written 5′→3′: 5′-GCTAATGCCGTA-3′. (2)

(d)(ii) (4) — pseudocode:

function transcribe(dna):
    map = {A:U, T:A, C:G, G:C}
    rna = ""
    for base in dna:      # dna read as template
        rna += map[base]
    return rna

• Applied to template 5′-TACGGCATTAGC-3′ → mRNA (antiparallel) reads: A→U, T→A... giving 3′-AUGCCGUAAUCG-5′ = 5′-GCUAAUGCCGUA-3′. (Full marks for correct base substitution logic + product.) (4)

(e) (2) — G–C pairs held by 3 hydrogen bonds, A–T by 2; more bonds ⇒ more energy needed to separate. (2)

(f) (2) — DNA double-stranded, RNA usually single-stranded; DNA long/stable stores genetic info, RNA shorter and functions in protein synthesis. (any 2)

Question 3

(a) (4)

• ATP = adenine base + ribose sugar + 3 phosphate groups (adenosine triphosphate). (2)
• Energy currency: hydrolysis of the terminal phosphate bond (ATP → ADP + Pi) releases usable energy that drives cellular work; readily re-synthesised. (2)

(b)(i) (2) — $915 / 30.5 = 30$ mol ATP.

(b)(ii) (2) — $30 \times 6.02\times10^{23} = 1.806\times10^{25}$ molecules.

(c) (4)

• Benedict's → brick-red/orange (from blue) = reducing sugar.
• Iodine → blue-black (from orange) = starch.
• Biuret → purple/violet (from blue) = protein.
• Sudan III → red-stained layer / red droplets = lipid. (1 each)

(d) (2)

• Negative Benedict's then positive after acid hydrolysis = non-reducing sugar (e.g. sucrose). Acid + heat hydrolyses glycosidic bond into reducing monosaccharides (glucose/fructose) which then reduce Benedict's. (2)

[
  {"claim":"148-residue chain gives 147 peptide bonds and 4 chains give 588",
   "code":"per=148-1; total=4*per; result = (per==147 and total==588)"},
  {"claim":"DNA 22% T gives G=C=28%, A=22%",
   "code":"T=22; A=T; GC=100-(A+T); G=GC/2; C=GC/2; result = (A==22 and G==28 and C==28)"},
  {"claim":"DNA 34% G gives A=T=16 and purine:pyrimidine = 1:1",
   "code":"G=34; C=34; AT=100-(G+C); A=AT/2; T=AT/2; pur=A+G; pyr=T+C; result = (A==16 and pur==pyr)"},
  {"claim":"915 kJ / 30.5 = 30 mol ATP and 1.806e25 molecules",
   "code":"mol=915/30.5; N=mol*6.02e23; result = (mol==30 and abs(N-1.806e25) < 1e21)"}
]