Pure and accurately labeled.
Hope this helps!
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
In genetics, when looking for genetic variation, offspring produced by two parents will always be more genetically variable than offspring with one parent.
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