In a dihybrid cross involving two autosomal traits on different chromosomes in which the parents are purebred for the opposite forms of both traits, how many of the offspring would be expected to be homozygous recessive for both traits?

The question lacks options, the options are:

A) 1 out of 16

B) 3 out of 16

C) 6 out of 16

D) 9 out of 16

The answer is 1 out of 16

Explanation:

This is a DIHYBRID cross because it involves two different genes coding for distinct traits. One of the traits will be dominant while the other recessive. Hence, parents that are purebred for opposite forms of the trait means that one parent is homozygous dominant while the other is homozygous recessive. When these two parents cross, they produce F1 offsprings that all possess the dominant trait but heterozygous/hybrids.

When these hybrids are self-crossed, they produce four different combinations of gametes which when crossed using a punnet square will result in F2 offsprings with a 9:3:3:1 phenotypic ratio according to Mendel's observation.

9 represents offsprings that are dominant for both traits

The two 3's represents offsprings that are recessive for one trait and dominant for the other respectively.

1 represents offsprings that are homozygous recessive for both traits.

Hence, 1 out of 16 offsprings will be homozygous recessive for both traits.

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