Q.In Mendel's experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY x rryy?

Step 1. Set up the cross. RRYY (round, yellow) × rryy (wrinkled, green). Since round (R) and yellow (Y) are both dominant, every F1 individual is RrYy — round-seeded with yellow cotyledons, uniform.

Step 2. Self the F1 to obtain F2. Crossing RrYy × RrYy and applying the Law of Independent Assortment (the two genes assort independently, exactly as this chapter's own summary states) gives the classic dihybrid ratio 9 R_Y_ : 3 R_yy : 3 rrY_ : 1 rryy, i.e. out of 16: 9 round-yellow, 3 round-green, 3 wrinkled-yellow, 1 wrinkled-green.

Step 3. Note the recombinant classes. Two of these four F2 phenotype classes — round-green and wrinkled-yellow — are brand-new combinations that neither parent showed (the parents were only round-yellow and wrinkled-green). This is precisely the point this chapter's summary makes: independent assortment produces 'two new recombinations...round green peas or wrinkled yellow peas.' …

Work out the full dihybrid cross (P → F1 → F2) using the branch/Punnett-square method and check each option against the actual 9:3:3:1 segregation, rej …

• Assuming F2 shows only the parental phenotypes and missing that independent assortment creates new recombinant combinations
• Stopping the analysis at F1 instead of continuing to self F1 to reach F2 …