Final answer:
Possible genotypes from the cross with at least one heterozygous parent include Rr and rr for round or wrinkled peas, respectively. A 4x4 Punnett square is used for the RrYY-rrYy cross and shows possible offspring genotypes. Mendel's F2 dihybrid cross yields a 9:3:3:1 phenotypic ratio.
Step-by-step explanation:
The genotypes you would expect in future offspring from a marriage between a pea plant with wrinkled peas (genotype rr) and a plant with round peas of unknown genotype (either RR or Rr) will depend on the genotype of the round pea parent. If the round pea parent is heterozygous Rr, then the possible genotypes of the offspring could be Rr (round peas) and rr (wrinkled peas). If the parent is homozygous RR, then all offspring would have the genotype RR and exhibit round peas. Given that three offspring plants all have round peas, it is more likely that the unknown parent is heterozygous Rr. As such, the option (b) Rr would definitely be expected. If we assume the parent is Rr, the probability of having only round pea plants from a random sample of 3 progeny would be 1/2 * 1/2 * 1/2 = 1/8 since each offspring has a 1/2 chance of inheriting the dominant allele R necessary for round peas.
When considering the cross between RrYy and rrYy pea plants, the possible genotypes for the offspring would be RrYy, Rryy, rrYy, and rryy. You would need a 4x4 Punnett square analysis for this cross to account for all possible gametes.
In the example of the four-o'clock flowers, the correct genotype of the pink offspring from a cross between a true-breeding red (RR) and true-breeding white (rr) plants would be Rr, indicating incomplete dominance.
The F2 generation ratio observed by Mendel for a dihybrid cross would be 9 round/yellow:3 round/green:3 wrinkled/yellow:1 wrinkled/green. For a self-cross of heterozygous Yy offspring, the genotype ratio would be YY:Yy:yy in a 1:2:1 ratio, with the phenotypic ratio being 3 yellow:1 green.