Question

A cross is made between true breeding lines of pea plants with axillary and terminal flowers. If 1240 progeny are obtained, how many plants are homozygous with axillary flowers and heterozygous with axillary flowers in the $F_2$ generation?

• A. 310 & 620
• B. 620 & 620
• C. 310 & 310
• D. 620 & 310

Answer 1

The correct option is A

310 & 620

The $F_2$ genotypic ratio of Mendel’s monohybrid cross is 1:2:1. Out of the $F_2$ progeny 1/4th are homozygous plants with axillary flowers. Homozygous plants with axillary flowers $=1240\times \frac{1}{4}=310$ . Heterozygous plants with axillary flowers $=1240\times \frac{1}{2}=620$. Thus option A is correct.