Question

okay consider a cross of beetles with exoskeleton allies B and b and leg size alleles N and n. the male parent is heterozygous in both traits. the females parent is homozygous recessive in both traits

Answer 1

The male parents traits would be Bb and Nn
The female parents traits would be bb and nn

Answer 2

This cross will generate puppies with the genotypes of 25% BbNN, 25% Bbnn, 25% bbNn and 25% bbnn.

Step-by-step explanation:

In this case the parent beetle has genotype BbNn and the mother beetle is bbnn. To solve this intersection, we have to remember Mendel's law of independent segregation and perform the traits intersections separately and multiply the percentages later.

1 - Characteristic of the exoskeleton: Bb x bb that gives us 50% Bb, 50% bb.

2 - Leg feature: Nn vs nn which gives us 50% Nn and 50% nn as well.

Finally, that means the cross of beetles have a chance to origin puppies with (0,5 * 0,5) in each different trait, resulting in the genotypes of 25% BbNN, 25% Bbnn, 25% bbNn and 25% bbnn