Question

A pure breeding pea plant with round yellow seeds was crossed with pea plant having wrinkled green seeds. On selfing of $F_1$ hybrid of his cross $64$ progenies were obtained in $F_2$ generation. Find out the number of $F_2$ progenies showing non-parental characters.

• A. 12
• B. 24
• C. 36
• D. 4

Answer 1

Answer: (B)

24

Answer 2

The $F _{2}$ phenotypic ratio in a dihybrid cross is $9 : 3 : 3 : 1$. The first and last term in ratio i.e, 9 and $1$ represent the parental genotypes while the second and third term i.e, $3$ and $3$ represent the recombinant genotypes. Therefore, the number of $F _{2}$ progenies out of $64$ progenies showing non-parental characters will be = $(6 / 16) \times 64=24 .$