Every mark here is for using a table or formula — not memorising one. The booklet is open. The skill is knowing what to do with it.
The booklet hands you the formula. The first mark is choosing the right one for the question.
Write the formula, then put the numbers in. Method marks survive a slip in the final number.
Convert carefully (mm → µm), and give the answer to a sensible precision. Units earn marks.
Type an mRNA sequence, or apply a mutation, and watch the table do the work. (Supports Q2.)
Attempt each with the booklet, then open the worked solution to check your method.
In a population of 800 beetles, a wing-colour gene has alleles A (dominant) and a (recessive). The frequency of A is 0.6. Observed genotypes:
| Genotype | AA | Aa | aa |
|---|---|---|---|
| Observed | 270 | 410 | 120 |
(a) p = 0.6, q = 0.4. AA = p²N = 0.36 × 800 = 288; Aa = 2pqN = 0.48 × 800 = 384; aa = q²N = 0.16 × 800 = 128 (total 800).
(b) χ² = Σ(O−E)²/E: AA (270−288)²/288 = 1.125; Aa (410−384)²/384 = 1.760; aa (120−128)²/128 = 0.500 → χ² = 3.39. df = 3 − 1 = 2; critical value (p = 0.05) = 5.99. Since 3.39 < 5.99…
(c) The calculated value is less than the critical value → do not reject the null hypothesis → no significant difference → the population is consistent with Hardy–Weinberg equilibrium for this gene.
An mRNA reads: 5′ – AUG GCA AAU UGG ACU UAA – 3′
Tip: use the decoder above to check each part.
(a) AUG–GCA–AAU–UGG–ACU–UAA → Met–Ala–Asn–Trp–Thr (then stop).
(b) UGG → UAG = a stop codon → nonsense mutation. Translation ends early → a truncated polypeptide (Met–Ala–Asn), likely non-functional.
(c) GCA → GCG, still alanine. The code is degenerate (several codons per amino acid) → a silent mutation, sequence unchanged.
(d) Deleting the G: AUG–CAA–AUU–GGA–CUU–… → Met–Gln–Ile–Gly–Leu–… Every downstream codon changes → a frameshift mutation.
Two woodland sites, same three species, same total (N = 30):
| Species | Site A | Site B |
|---|---|---|
| 1 | 10 | 26 |
| 2 | 10 | 2 |
| 3 | 10 | 2 |
(a) D = Σn(n−1) / N(N−1), then 1/D. N(N−1) = 30 × 29 = 870.
Site A: Σn(n−1) = 3 × (10×9) = 270; D = 0.310; 1/D = 3.22.
Site B: Σn(n−1) = (26×25)+(2×1)+(2×1) = 654; D = 0.752; 1/D = 1.33.
(b) Site A has the higher index → greater biodiversity. Richness is equal, but Site A is far more even; Site B is dominated by one species. Simpson's accounts for evenness, not just the number of species.
To estimate a snail population: 40 caught, marked, released. Second visit: 50 caught, of which 8 were marked.
(a) N = (M × n)/R = (40 × 50)/8 = 250 snails.
(b) Any two: no significant migration / birth / death between samples; marking doesn't affect survival or behaviour; marked individuals mix randomly back in; marks aren't lost.
A mitochondrion in an electron micrograph measures 40 mm long. Magnification is ×2000.
(a) actual = image / magnification = 40 mm / 2000 = 0.02 mm.
(b) 0.02 mm = 20 µm.
A recessive condition affects 1 in 2500 newborns in a population at equilibrium.
(a) q² = 1/2500 = 0.0004 → q = 0.02.
(b) p = 1 − 0.02 = 0.98; carriers = 2pq = 2 × 0.98 × 0.02 = 0.0392 (≈ 3.9%).
(c) 1 / 0.0392 ≈ 1 in 26.
Tap a tool to see what it finds — and how the genetics tools feed each other.
Every formula and table you need is in the booklet — marks are for using them, not memorising them. Tap a tool to see exactly what it finds.
Tap any tool to trace its job · tap the background to reset
Drag the right tool into each scenario. Two tools don't fit any of these — they're the trap.
Single best answer, Paper 1 style. Pick one — you'll see why.