Question

Consider five independently assorting gene pairs, A/a, B/b, C/c, D/d, and E/e, where each demonstrates complete dominance (A_, B_, C_, D_, and E_) and recessiveness (aa, bb, cc, dd, and ee). An individual with genotype AaBbCCddEe is crossed to an individual with genotype AABbCcDdEe. What is the probability of obtaining an offspring that is dominant for all five traits?

A. 1/4
B. 3/4
C. 3/16
D. 1/8

Answer 1

The probability of obtaining an offspring with all dominant phenotypes from the given genotypes is calculated per gene pair, using the product rule, resulting in a 3/16 chance, making option C the correct answer.

Step-by-step explanation:

The question is about the probability of obtaining offspring with dominant phenotypes for all five traits from a cross between individuals with genotypes AaBbCCddEe and AABbCcDdEe. We calculate this by looking at each gene pair independently and determining the probability of a dominant phenotype for each, and then we multiply these probabilities together using the product rule.

For the A gene, the offspring will always have a dominant phenotype since one parent is AA (homozygous dominant). For genes B, C, and E, where one parent is heterozygous, the probability of a dominant phenotype in the offspring is 3/4. However, for the D gene, one parent is heterozygous and the other is homozygous recessive, so the probability of a dominant phenotype is 1/2.

Therefore, using the product rule, we multiply the probabilities of obtaining a dominant phenotype for each independently assorting gene pair: 1 (for A) × 3/4 (for B) × 3/4 (for C) × 1/2 (for D) × 3/4 (for E), which equals 3/16. Hence, the correct option in the final answer is C. 3/16.