In labrador retrievers, a common breed of dog, 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?
the data given regarding the alleles are as follows black(B)-dominant , chocolate(b) - recessive normal vision - dominant, progressive retinal atrophy - recessive normal hip joint - dominant, hip dysplasia - recessive
lets first consider one trait parents are heterozygous when crossed Bb x Bb offspring BB Bb Bb bb --------black----- chocolate the probability of dogs being black - 3/4 dogs being chocolate - 1/4 likewise for the other two traits probability of normal vision - 3/4 and retinal atrophy - 1/4 probability of normal hip joint - 3/4 and hip dysplasia - 1/4
we have to calculate the probability of 1st pup being born is chocolate - 1/4 normal vision - 3/4 normal hip - 3/4 since all these traits should be acquired the probability of having the above three traits is calculated by getting the product of individual probabilities therefore answer is 1/4 x 3/4 x 3/4 probability = 9/64
