Question

A pea plant that has round seeds has the genotype Rr. It is crossed with a pea plant that has wrinkled seeds and the genotype rr. What is the probability that the offspring will have wrinkled seeds?
(A) 0 percent
(B) 25 percent
(C) 50 percent
(D) 75 percent

Answer 1

Answer: The correct answer is- (C) 50 percent.

Genotype of parents are- Rr ( representing heterozygous dominant ) and rr ( homozygous recessive).

Dominant trait ( depicted by capital letter allele such as 'R' for round seeds)  masks the expression of recessive trait ( depicted by small letter allele such as 'r for wrinkled seeds').

As per the information in the question, round seeds is a dominant trait ( Rr, expressed in heterozygous dominant form) over wrinkled seeds (rr), which is recessive trait.

When the given parents are crossed, they produce offspring with the genotypes  rr ( showing wrinkled seeds) and Rr ( round seeds) in the ratio 1:1.

Thus, the probability that the offspring will have wrinkled seeds is 50 percent.

Refer punnet square.

Answer 2

In order to answer this, you will need to set up a Punett Square (I have attached a picture). 

Since Rr is the genotype of round seeds, we know that round must be dominant to wrinkled, and therefore represented by the allele, R.

Since rr is the genotype of wrinkled seeds, we know that wrinkled must be the recessive trait, and there represented by the allele, r.

Since the Punnett Square shows that the genotype ratio is 2Rr : 2rr, or simplified, 1Rr : 1rr, we now know that there is a 50% chance the offspring will have wrinkled seeds.