Biomolecules — Carbohydrates & Lipids

Question 1

(a) (6 marks)

• Monomer = small repeating molecular subunit; polymer = large molecule built from many covalently linked monomers. (2)
• Each glycosidic bond forms by dehydration synthesis, removing one $H_2O$ ($18.02$ g/mol). For $n$ residues, $n-1$ bonds form, so $n-1$ waters are lost. (2)
• Therefore mass is not $180.16n$ because water is removed at each linkage. $M(n) = 180.16\,n - 18.02\,(n-1)\ \text{g/mol}$ (2)

(b) (4 marks)

• Moles of water = $1.20/18.02 = 0.0666\ \text{mol}$. (2)
• Each hydrolysed glycosidic bond consumes one water, so bonds broken $= 0.0666 \times 6.022\times10^{23} \approx 4.01\times10^{22}$ bonds. (2)
• Assumption: negligible correction for chain ends / assume long chains where bonds ≈ residues, and complete hydrolysis. (within marks)

(c) (6 marks — 1.5 each)

• Bond type: starch & glycogen = $\alpha(1\to4)$ (+ $\alpha(1\to6)$ branches); cellulose = $\beta(1\to4)$.
• Branching: amylose unbranched, amylopectin moderately branched, glycogen highly branched (rapid mobilisation), cellulose unbranched straight chains.
• Role: starch = plant energy storage; glycogen = animal energy storage; cellulose = plant cell-wall structural support.
• Indigestibility: the $\beta(1\to4)$ linkage gives straight H-bonded microfibrils; most animals lack cellulase enzymes able to hydrolyse $\beta$ bonds. (2 of the 6 for this point)

(d) (4 marks)

function chain_mass(n):
    return 180.16*n - 18.02*(n-1)

• For $n=1000$: $180.16(1000) - 18.02(999) = 180160 - 18001.98 = 162158.02\ \text{g/mol}$. (2 for code, 2 for value)

Question 2

(a) (6 marks)

• Glycerol + 3 fatty acids → triglyceride + 3 water (one ester bond per fatty acid, each by dehydration). (3)
• Phospholipid: hydrophilic phosphate head + two hydrophobic fatty-acid tails = amphipathic; forms bilayers/micelles in water (heads out, tails in). Triglyceride has three tails, no polar head → wholly hydrophobic, forms fat droplets, not bilayers. (3)

(b) (6 marks)

• Saturated stearic acid chains are straight, pack tightly → maximal surface contact → many van der Waals interactions → high melting point. (3)
• A cis double bond introduces a rigid kink/bend, preventing tight packing, reducing contact area and van der Waals forces → lower energy needed to disrupt → much lower melting point (liquid at room temp). (3)

(c) (6 marks)

• Model $T_m = T_0 - kd$.
• $d=0$: $T_0 = 69\,^\circ\text{C}$. (2)
• $d=1$: $13 = 69 - k \Rightarrow k = 56\,^\circ\text{C}$ per double bond. (2)
• Prediction $d=2$: $T_m = 69 - 112 = -43\,^\circ\text{C}$. (1)
• Comment: linear model gives plausible ordering (more unsaturation → lower $T_m$) but the effect saturates/is non-linear in reality; extrapolation to $d=3$ ($-99\,^\circ\text{C}$) is not physically reliable. (1)

(d) (4 marks)

def predict_Tm(d):
    return 69 - 56*d

• Outputs: $d=0 \to 69$; $d=1 \to 13$; $d=2 \to -43$; $d=3 \to -99$ (°C). (2 code, 2 values)

Question 3

(a) (6 marks)

• Steroid = four fused carbon rings (three six-membered + one five-membered), with functional groups; cholesterol has a hydroxyl (—OH) head. (2)
• At high T: cholesterol's rigid ring restrains phospholipid tail motion → reduces excessive fluidity/order it. (2)
• At low T: it wedges between phospholipids, disrupting tight packing/crystallisation → maintains fluidity. Thus "fluidity buffer." (2)

(b) (8 marks)

• Outer leaflet surface at $r_{out}=50$ nm: $A_{out} = 4\pi r^2 = 4\pi(50)^2 = 31416\ \text{nm}^2$. (2)
• Inner leaflet surface at $r_{in}=50-5=45$ nm: $A_{in}=4\pi(45)^2 = 25447\ \text{nm}^2$. (2)
• Molecules per leaflet = area / 0.70:
  • Outer: $31416/0.70 \approx 44880$.
  • Inner: $25447/0.70 \approx 36353$. (2)
• Total $\approx 44880 + 36353 \approx 8.12\times10^{4}$ phospholipids. (2) (Acceptable range $\sim 8\times10^4$.)

(c) (4 marks)

• Chitin monomer = N-acetylglucosamine (glucose derivative with an acetylamino group); cellulose monomer = plain glucose — difference is the —NH-COCH₃ group replacing an —OH. (2)
• Waxes: long saturated hydrocarbon chains are highly hydrophobic and tightly packed, forming an impermeable water-repellent layer → reduces water loss/entry on cuticles/exoskeletons. (2)

[
  {"claim":"Amylose molar mass for n=1000 residues", "code":"n=1000; M=180.16*n-18.02*(n-1); result=abs(M-162158.02)<0.1"},
  {"claim":"Bonds broken from 1.20 g water hydrolysis ~4.01e22", "code":"N=(1.20/18.02)*6.022e23; result=abs(N-4.01e22)/4.01e22<0.01"},
  {"claim":"Fatty acid melting model k=56 and Tm(d=2)=-43", "code":"T0=69; k=69-13; Tm2=T0-k*2; result=(k==56) and (Tm2==-43)"},
  {"claim":"Liposome total phospholipids ~8.1e4", "code":"import math; Aout=4*math.pi*50**2; Ain=4*math.pi*45**2; N=(Aout+Ain)/0.70; result=abs(N-8.12e4)/8.12e4<0.02"}
]