Genetic Engineering & CRISPR

Question 1 (14 marks)

(a) (4) Use EcoRI and HindIII (1). These flank the MCS and are absent from GENX so they cut only at the intended ends (1). BamHI must be excluded because GENX contains an internal BamHI site (1) — cutting there would fragment the gene / destroy it (1).

(b) (3) Two different enzymes produce two non-complementary sticky ends (1). This prevents the vector self-ligating/re-circularising without insert (1) and forces the insert in a single fixed (directional) orientation (1).

(c) (5)

• (i) No plasmid → no ampR → no growth on ampicillin (1).
• (ii) Empty re-ligated plasmid → intact lacZ → functional β-galactosidase cleaves X-gal → blue colonies (1)(1 for reasoning).
• (iii) Plasmid + insert → insert disrupts lacZ → no β-galactosidase → white colonies (1)(1 for reasoning).

(d) (2) Blue/white screening only reports lacZ disruption; a wrong-sized fragment or non-target DNA could also disrupt lacZ (1), or partial digestion/frameshift may give white colonies without the correct insert — colony PCR/sequencing needed to confirm (1).

Question 2 (12 marks)

(a) (3) Copies $= N_0 \cdot 2^{n}$ (1). $= 500 \times 2^{25}$ (1) $= 500 \times 33{,}554{,}432 = 1.678 \times 10^{10}$ copies (1).

(b) (4) $N = N_0 (1+E)^{n}$, so $(1+E)^{25} = \frac{1.2\times10^{7}}{500} = 24{,}000$ (1). $1+E = 24000^{1/25}$ (1) $= e^{\ln(24000)/25} = e^{10.086/25} = e^{0.4034} = 1.497$ (1). So amplification factor ≈ 1.497 per cycle; efficiency $E = 0.497 \approx$ 49.7% of the maximum (doubling = 100%) (1).

(c) (2) No primers → no site for DNA polymerase to initiate extension (1) → no amplification / no product (1).

(d) (3) Annealing temp too high → primers cannot anneal/hybridise to the template (1) → few or no primer–template duplexes form (1) → greatly reduced or zero yield (1).

Question 3 (10 marks)

(a) (2) mRNA cannot be amplified by DNA polymerase directly (1); reverse transcriptase first converts mRNA into cDNA, which is then amplified (1).

(b) (5)

• $\Delta Ct_{control} = 24.0 - 20.0 = 4.0$ (1)
• $\Delta Ct_{treated} = 22.0 - 20.0 = 2.0$ (1)
• $\Delta\Delta Ct = \Delta Ct_{treated} - \Delta Ct_{control} = 2.0 - 4.0 = -2.0$ (1)
• Fold change $= 2^{-\Delta\Delta Ct} = 2^{-(-2.0)} = 2^{2}$ (1) $= \textbf{4-fold}$ (1).

(c) (3) Target gene is up-regulated (1). Lower Ct = threshold crossed earlier = more initial template (1); the treated sample crosses 2 cycles earlier, i.e. ~4× more starting target transcript (1).

Question 4 (14 marks)

(a) (4) A cytosine base editor performs C→T (with G→A on opposite strand) (1). Reversing GTG→GAG requires changing the T·A pair to A·T = an A→G on the sense strand (1), which a C→T editor cannot do (1). This is a transition (purine↔purine / pyrimidine↔pyrimidine), specifically requiring an adenine base editor (A→G) (1).

(b) (4) The PAM (e.g. 5′-NGG-3′ for SpCas9) is recognised by Cas9 and is required for DNA unwinding and cleavage (1); without PAM recognition Cas9 will not cut even with a matching guide (1). If no PAM lies near the target, the guide cannot be positioned to place the edit within the editing/cut window (1) → use a different Cas variant with altered PAM, or a different target strand/window (1).

(c) (4) Knock-out: disrupts/inactivates a gene (1), typically via NHEJ producing indels/frameshifts after a double-strand break (1). Knock-in: inserts or precisely replaces a defined sequence (1), typically via HDR using a supplied repair template (1).

(d) (2) Prime editing installs the precise edit without requiring a double-strand break (uses nickase + reverse transcriptase and pegRNA) (1), avoiding HDR's low efficiency and reducing indel byproducts / random insertions (1).

Question 5 (10 marks)

(a) (3) DNA is negatively charged (phosphate backbone) (1), so migrates toward the positive electrode (anode) under an electric field (1); the gel matrix sieves fragments so smaller fragments move faster/further (1).

(b) (4) Linear model: distance $d = m\log_{10}(\text{size}) + c$. Using points (400 bp, 3.0) and (100 bp, 6.0): $m = \dfrac{6.0-3.0}{\log_{10}100 - \log_{10}400} = \dfrac{3.0}{2 - 2.602} = \dfrac{3.0}{-0.602} = -4.983$ (1) $c$: $3.0 = -4.983(2.602) + c \Rightarrow c = 3.0 + 12.97 = 15.97$ (1) For $d = 4.5$: $4.5 = -4.983\log_{10}S + 15.97 \Rightarrow \log_{10}S = \dfrac{15.97-4.5}{4.983} = 2.302$ (1) $S = 10^{2.302} \approx \textbf{200 bp}$ (1).

(c) (3) STR loci are highly polymorphic — repeat number varies greatly between individuals (1); they are non-coding, so under low selective pressure and highly variable (1), giving high discriminating power for individual identification (1).

[
  {"claim":"500*2^25 = 1.6777216e10 target copies after 25 cycles at 100% efficiency","code":"val = 500*2**25\nresult = (val == 16777216000)"},
  {"claim":"Average amplification factor per cycle for 500->1.2e7 in 25 cycles is ~1.497","code":"f = (Rational(12000000,500))**Rational(1,25)\napprox = float(f)\nresult = abs(approx - 1.497) < 0.005"},
  {"claim":"ddCt method: fold change = 2^-(-2) = 4","code":"dd = 2.0 - 4.0\nfold = 2**(-dd)\nresult = (fold == 4)"},
  {"claim":"Fourth fragment at 4.5cm estimated ~200 bp","code":"import sympy as sp\nm = (6.0-3.0)/(sp.log(100,10)-sp.log(400,10))\nc = 3.0 - m*sp.log(400,10)\nlogS = (4.5 - c)/m\nS = 10**logS\nresult = abs(float(S) - 200) < 5"}
]