Genetic Engineering & CRISPR

Question 1

(a) (5 marks)

• Two different enzymes leave two different, non-complementary overhangs (1) → the insert can ligate into the vector in only ONE orientation (directional cloning) (1).
• Prevents insert self-ligation and vector re-circularisation (1).
• DNA ligase catalyses formation of phosphodiester bonds (1) between the 3′-OH and 5′-phosphate of the annealed sticky ends, covalently sealing the nicks in both strands (1).

(b) (5 marks)

• Ideal doubling: $N_n = N_0 \cdot 2^n$ (2).
• Require $5\times10^3 \cdot 2^n > 10^9 \Rightarrow 2^n > 2\times10^5$ (1).
• $n > \log_2(2\times10^5) = \dfrac{\ln(2\times10^5)}{\ln 2} = 17.61$ (1).
• Smallest integer $n = 18$ cycles. (1) Check: $2^{18}=262144$, $N_{18}=1.31\times10^9 > 10^9$. ✓

(c) (5 marks)

• Model: $N_{C_t} = N_0 (1+E)^{C_t}$ (1).
• $(1+E)^{24} = \dfrac{10^9}{5\times10^3} = 2\times10^5$ (1).
• $1+E = (2\times10^5)^{1/24}$ (1).
• $\ln(2\times10^5)=12.206$; $/24 = 0.50859$; $e^{0.50859}=1.663$ (1).
• $E = 0.663$ → $E \approx 0.663$ (66.3%). (1)

(d) (5 marks)

• fold-change $= 2^{\Delta C_t} = 2^{3.32}$ (1).
• $2^{3.32} = e^{3.32\ln 2}=e^{2.301}=9.99 \approx 10$ (2).
• Hence $\Delta C_t \approx 3.32$ ⇒ 10-fold difference in initial template (this is the origin of the "3.3 cycles per log" rule) (1).
• RT-PCR: mRNA cannot serve as PCR template directly; reverse transcriptase first converts mRNA → cDNA, which is then amplified/quantified — needed because qPCR measures DNA, and expression level is an mRNA quantity (1).

(e) (4 marks)

function melting_temp(seq):
    A = count(seq, 'A');  T = count(seq, 'T')
    G = count(seq, 'G');  C = count(seq, 'C')
    Tm = 2*(A + T) + 4*(G + C)
    if 55 <= Tm <= 65:
        return (Tm, "acceptable")
    else:
        return (Tm, "reject")

• Correct counting of each base (1); correct Wallace formula (1); correct threshold test 55–65 (1); returns both value and verdict (1).

Question 2

(a) (4 marks)

• DNA is uniformly negatively charged (phosphate backbone) so charge-to-mass ratio is ~constant → charge alone doesn't separate them (1).
• Gel acts as a molecular sieve / matrix of pores (1); smaller fragments navigate the pores more easily / with less friction (1); so they travel further per unit time toward the positive electrode (anode) (1).

(b) (6 marks) Two equations from $d = a - b\log_{10}L$:

• $20 = a - b\log_{10}(1000) = a - 3b$ (1)
• $50 = a - b\log_{10}(100) = a - 2b$ (1)
• Subtract: $(50-20) = (a-2b)-(a-3b) = b \Rightarrow b = 30$ ... wait: $50-20 = -2b+3b = b$, so $b=30$? Recheck sign: $50 - 20 = (a-2b)-(a-3b)= b$ ⇒ $b = 30$. (2) But $d$ decreases with $L$, so $b>0$ ✓.
• From $50 = a - 2(30) = a - 60 \Rightarrow a = 110$ (2).
• $a = 110$ mm, $b = 30$ mm per decade.

(c) (4 marks)

• $32 = 110 - 30\log_{10}L$ (1)
• $30\log_{10}L = 78 \Rightarrow \log_{10}L = 2.6$ (1)
• $L = 10^{2.6} = 398$ bp (1)
• ≈ 400 bp (nearest 10 bp) (1).

(d) (2 marks)

• Very large fragments (> ~20 kb) all migrate together / co-migrate — resolution is lost because they can no longer be sieved by pore size (1).
• Pulsed-field gel electrophoresis (PFGE) is used instead (1).

Question 3

(a) (5 marks)

• sgRNA: contains a 20-nt spacer complementary to the target DNA, guiding Cas9 by base-pairing (1); the scaffold portion binds Cas9 (1).
• PAM (NGG): short motif adjacent to target required for Cas9 to bind/cut; absent in the sgRNA so the CRISPR array itself is not cut (1).
• Cas9 nuclease domains: RuvC and HNH each cut one strand (1).
• Result: a blunt double-strand break ~3 bp upstream of the PAM (1).

(b) (6 marks)

• Probability of a specific 20-nt sequence at one position $= (1/4)^{20} = 4^{-20}$ (2).
• $4^{20} = 2^{40} \approx 1.10\times10^{12}$, so $P \approx 9.1\times10^{-13}$ (1).
• Expected occurrences in genome $\approx 3\times10^9 \times 9.1\times10^{-13} = 2.7\times10^{-3}$ (2).
• Since expected count ≪ 1, a 20-nt target is essentially unique → but sequences with only partial complementarity/mismatches can still be cut, causing off-target edits; the effective search space is smaller, and PAM abundance (NGG) is high, so off-target risk is non-negligible (1).

(c) (4 marks)

• Knockout: DSB repaired by NHEJ (error-prone) (1) → insertions/deletions (indels) cause frameshift → non-functional gene (1).
• Knock-in: DSB repaired by HDR using a supplied donor template (1) → inserts/corrects a defined sequence, adding new/desired function (1).

(d) (3 marks)

• Cytosine base editor = catalytically impaired Cas9 (nickase) fused to a cytidine deaminase (1); deaminase converts C→U on the exposed non-target strand within the R-loop; U is read as T, and after replication/repair gives C·G → T·A (1).
• Advantage: no double-strand break ⇒ fewer indels, lower toxicity, no need for HDR/template (1).

(e) (2 marks)

• Somatic edits affect only the treated individual and are not heritable (1); germline edits alter gametes/embryos and are passed to all descendants — raising consent (future generations cannot consent) and irreversibility/equity concerns (1). (Any one clear distinction = full marks.)

[
  {"claim":"Cycles to exceed 1e9 from 5e3 is 18","code":"import math; n=math.ceil(math.log(2e5,2)); result = (n==18)"},
  {"claim":"PCR efficiency E ~ 0.663","code":"E=(2e5)**(1/24)-1; result = abs(E-0.663)<0.005"},
  {"claim":"2^3.32 approx 10","code":"result = abs(2**3.32 - 10) < 0.05"},
  {"claim":"Gel calibration a=110,b=30 and unknown ~400bp","code":"b=30; a=110; L=10**((a-32)/b); result = (a==110 and b==30 and abs(L-398)<5)"},
  {"claim":"Expected 20-nt occurrences in 3e9 bp genome ~2.7e-3","code":"P=(Rational(1,4))**20; exp=3e9*float(P); result = abs(exp-2.7e-3)<3e-4"}
]