Question

In a Mendel's dihybrid experiment, a homozygous pea plant with round yellow seeds was crossed with a homozygous plant with wrinkled green seeds. F\(_1\) intercross produced 560 F\(_2\) progeny. The number of F\(_2\) progeny having both dominant traits (round and yellow) is \(\underline{\hspace{1cm}}\).

Hint:

A classic Mendelian dihybrid cross always yields a 9:3:3:1 phenotypic ratio.

Answer 1

Correct Answer: 315

In a dihybrid cross, the F\(_2\) phenotypic ratio is:
\[ 9 : 3 : 3 : 1 \]
The fraction showing both dominant traits (round and yellow) is \( \frac{9}{16} \).
Thus, number of round yellow progeny:
\[ \frac{9}{16} \times 560 = 315 \]