A pure-breeding pea plant with round yellow seeds (RRYY) is crossed with a pure-breeding plant with wrinkled green seeds (rryy).
(i) Write the genotype and phenotype of the F1 offspring.
(ii) In the F2 generation, what fraction of the total offspring would be expected to have round green seeds? Write their possible genotypes.
(iii) Identify one F2 genotype that would breed true for wrinkled yellow seeds and explain how you determined it.
Generated by claude-sonnet-4-6 · 2026-06-26 01:03 · grounding rag
Model Answer
(i) F₁ Genotype: RrYy; Phenotype: Round yellow seeds (R and Y are dominant).
(ii) Round green seeds = R\_yy genotype. From the F₂ Punnett square, this phenotype appears in 3/16 of the offspring.
Possible genotypes: RRyy and Rryy.
(iii) The F₂ genotype rryy breeds true for wrinkled yellow — wait, wrinkled yellow = rrY\_.
The true-breeding genotype is rrYY.
It is homozygous for both recessive (rr = wrinkled) and dominant (YY = yellow) alleles, so it will always produce wrinkled yellow offspring on self-pollination.
Source: Chapter 8, Section 8.2.3
Explanation
- Part (i): Since RRYY × rryy, all F₁ get one R and one Y from each parent → RrYy. Round and yellow are dominant, so phenotype is round yellow.
- Part (ii): Use the 9:3:3:1 ratio from Fig. 8.5. Round green = 3/16. The two genotypes giving round green are RRyy and Rryy (R\_yy).
- Part (iii): "Breeds true" means homozygous. Wrinkled = rr, yellow = YY → rrYY. Examiners want you to state why it breeds true: because both gene pairs are homozygous, self-pollination always gives the same phenotype. Note: rrYy also gives wrinkled yellow but does NOT breed true.