Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b.What is the probability of producing an A_B_ phenotype from a cross AaBb × AaBb?
A)1/4
B)1/2
C)1/16
D)9/16
E)1 (100%)

Question

Quizplus · Accepted Answer

- The genotype of the parent AaBb can be separated into the possible gamete combinations: AB, Ab, aB, and ab.
- We can use the product rule to determine the probability of each possible gamete combination from the first parent, and the probability of each possible gamete combination from the second parent.
- For example, the probability of producing an A gamete from the first parent is 1/2, and the probability of producing an A gamete from the second parent is also 1/2. Therefore, the probability of producing an AA genotype is (1/2) x (1/2) = 1/4.
- We can use this approach to calculate the probability of each possible genotype for the offspring. The possible genotypes are: AABB, AABb, AaBB, AaBb, AAbb, aaBB, aaBb, and aabb.
- The only genotype that corresponds to an A_B_ phenotype is AaBb. We can see that there are four possible ways to produce an AaBb genotype: AaBb, aAbB, aaBB, and AAbb. Each of these possibilities has a probability of 1/16 (1/4 x 1/4), so the total probability of producing an A_B_ phenotype is 4/16 or 9/16.

Two Gene Loci, a and B, Assort Independently, and Alleles