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Step-by-Step Solution
Step 1: Identify the Parent Genotypes
The cross given is between plants with genotypes $RRYY$ (round, yellow) and $rryy$ (wrinkled, green). Here:
$R$ = round seed shape (dominant), $r$ = wrinkled seed shape (recessive)
$Y$ = yellow cotyledon (dominant), $y$ = green cotyledon (recessive)
Step 2: Determine the Gametes
Since one parent is $RRYY$, the only gametes it produces are $RY$.
Since the other parent is $rryy$, the only gametes it produces are $ry$.
Step 3: Form the F1 Generation
Crossing these gametes ($RY \times ry$) gives $RrYy$ in the F1 generation.
Thus, the F1 plants have the genotype $RrYy$, which displays round seeds with yellow cotyledons (because both round and yellow traits are dominant).
Step 4: Selfing the F1 Generation
When $RrYy$ plants are self-pollinated, each parent can produce four types of gametes: $RY$, $Ry$, $rY$, $ry$. These combine in a $4 \times 4$ Punnett square.
Step 5: Determine the F2 Phenotypes
The phenotypic ratio for a dihybrid cross (with complete dominance) is typically $9:3:3:1$. The four possible phenotypes are:
Round seeds, Yellow cotyledons (dominant for both traits)
Round seeds, Green cotyledons
Wrinkled seeds, Yellow cotyledons
Wrinkled seeds, Green cotyledons
However, from the provided options and the correct answer given, the question focuses only on the combinations with yellow cotyledons. Hence, the expected phenotypic categories mentioned in the correct option are:
Round seeds with yellow cotyledons
Wrinkled seeds with yellow cotyledons
Thus, the final outcome in the F2 generation includes round-yellow and wrinkled-yellow seeds (along with other combinations if all categories were listed).
Step 6: Conclusion
Because round is dominant over wrinkled, and yellow is dominant over green, the dihybrid cross ensures that some F2 offspring will show round-yellow as well as wrinkled-yellow phenotypes. Therefore, the correct answer is: “Round seeds with yellow cotyledons, and wrinkled seeds with yellow cotyledons.”
