In Labrador dogs, black coat is dominant to chocolate, normal vision is dominant to progressive retinal atrophy (PRA), and normal hip joint is dominant to hip dysplasia. All these genes assort independently. Two dogs that are heterozygous for alleles of all three genes are crossed. Using rules of probability (not a Punnett square), what is the chance that the first pup born to these dogs will be chocolate, have normal vision, and have normal hip joints?

In Labrador dogs, black coat is dominant to chocolate, normal vision is dominant to progressive retinal atrophy (PRA), and normal hip joint is dominant to hip dysplasia. All these genes assort independently. Two dogs that are heterozygous for alleles of all three genes are crossed. Using rules of probability (not a Punnett square), what is the chance that the first pup born to these dogs will be chocolate, have normal vision, and have normal hip joints?

BbVvHh x BbVvHh= BBVVHH, BbVvHh, BbVvHh, bbvvhh

Bb= black coat dominant

Vv= Normal vision dominant

Hh= Normal hip join dominant

probability of having a first born of these dogs will be chocolate, have normal vision and have normal hip joint is 0

Explanation:

As the punette square gives 3:1 phenotype having three black coat, normal vision and normal hip joint and one chocolate, progressive retina altropy and hip dysplasia

it's     c       this is a cell (b) with a nucleus (c) containing dna (a). the dna is organized into chromosomes (d). a section of the chromosome that codes for a trait is called a gene (e).

your spelling is a little off : ]