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?
a) 1/64
b) 1/16
c) 3/32
d) 9/64

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.

the correct answer is b. possesses the characteristic of compartmentalization of organelles, wherein the organelles are bound by membranes.

explanation: