Question

In a cross between a pea plant heterozygous for round seeds (Rr) and a plant with wrinkled seeds (rr), what is the expected phenotypic ratio of the offspring?

• A. 1 Round : 1 Wrinkled
• B. 3 Round : 1 Wrinkled
• C. All Round
• D. All Wrinkled

Hint:

In a monohybrid cross involving a heterozygous parent (Rr) and a homozygous recessive parent (rr), the phenotypic ratio is always 1:1, as half the offspring inherit the dominant allele and half inherit the recessive allele.

Answer 1

The Correct Option is A

In pea plants, the allele for round seeds (R) is dominant over the allele for wrinkled seeds (r), which is recessive. This cross involves a heterozygous plant (Rr) and a homozygous recessive plant (rr). A Punnett square is utilized to determine the phenotypic ratio of the progeny.
The parental genotypes are:
- Rr (heterozygous, resulting in round seeds), producing gametes R and r.
- rr (homozygous recessive, resulting in wrinkled seeds), producing only r gametes.
The Punnett square for the Rr × rr cross is depicted below:
\[\begin{array}{c|cc} & R & r
\hliner & Rr & rr
\end{array}\] The resulting offspring genotypes are:
- Rr (round seeds due to the dominance of R): 50%
- rr (wrinkled seeds): 50%
Consequently, the phenotypic ratio is 1 Round : 1 Wrinkled (1:1).