Answer:
D
Explanation:
Let: black coat=A and chocolate coat=a; normal vision=B and PRA=b; normal hip joint=C and hip dysplasia=c
We are crossing two dogs that are both heterozygous for all traits
AaBbCc x AaBbCc
To determine the probability we will look at each trait separately and then use the multiplication rule and the addition rule.
Aa x Aa = 1/4 AA, 1/2 Aa, 1/4 aa
Bb x Bb = 1/4 BB, 1/2 Bb, 1/4 bb
Cc x Cc = 1/4 CC, 1/2 Cc, 1/4 cc
Now we determine the genotypes that will allow for a pup that will be chocolate, have normal vision and have normal hip joints
aaBBCC, aaBbCC, aaBbCc, aaBBCc
Next we use the multiplication rule to determine the probability of each genotype
aaBBCC: 1/4 aa x 1/4 BB x 1/4 CC = 1/64
aaBbCC: 1/4 aa x 1/2 Bb x 1/4 CC = 1/32
aaBbCc: 1/4 aa x 1/2 BB x 1/2 Cc = 1/16
aaBBCc: 1/4 aax 1/4 BB x 1/2 Cc = 1/32
Lastly, we add all the probabilities together to determine the overall probability
1/64 + 2(1/32) + 1/16 = 9/64
Therefore, the probability that the first pup born to these dogs will be chocolate, have normal vision and have normal hip joints is 9/64.
