In cats, being awesome (A) is dominant to being average (a), and being bashful (B) is dominant to being boring (b). A male cat that is homozygous dominant for the A trait and heterozygous for the B trait is mated with a female cat that is homozygous recessive for both. Both the A and B traits assort independently from one another. What is the probability that they will have an offspring who is heterozygous for both traits? Choose 1

(Choice A)

50%

(Choice B)

100%

(Choice C)

0%

(Choice D)

25%

The probability that both cats described will have an offspring who is heterozygous for both traits is 50%.

Explanation:

When a cat awesome and bashful is crossed with a cat average and boring, it is a crossing where two traits are taken into account that will be segregated independently by the parents, and the result will be produced by the analysing the genotypes and making crossing in a punnett square.

The genotypes are given as for male cats AA Bb and for female cats aa bb. The crossing is done based on punnett square and it is denoted as

 Alleles       AB    Ab

  ab             AaBb  Aabb

  ab             AaBb  Aabb

The offspring is AaBb awesome and bashful(it is heterozygous for two traits 50%) whereas in Aabb awesome and blashful( it is heterozygous for trait A and homozygous for trait B).

Thus The probability that both cats described will have an offspring who is heterozygous for both traits is 50%.