Question

According to the law of Independent Assortment in a dihybrid cross

• A. There are four genotypes in $F_2$
• B. $F_2$ contains 16 phenotypes
• C. There is a single individual which is homozygous recessive for both the characters
• D. It is not possible to forecast the different phenotypes

Answer 1

The correct option is C There is a single individual which is homozygous recessive for both the characters

Mendel conducted dihybrid cross on pea plant and postulated the third law of principle of inheritance. It is Law of Independent assortment. According to this, in the F₂ generation 16 progeny are obtained. In these progeny, four types of phenotypes, 9 types of genotypes are appear. Out of 16, 6 progeny shows genetical recombinations, 1 homozygous dominant for both characters(RRYY), 8 are heterozygous dominant for round and yellow, and 1 progeny is homozygous recessive for both characters(rryy).

So, the correct option is ‘There is a single individual which is homozygous recessive for both the characters’.