Question

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?

Answer 1

Answer:

Probability that both cats described will have an offspring who is heterozygous for both traits is 50%, according to the scenario given.

Explanation:

When a cat amazing 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 can be obtained by analyzing their genotypes and making the corresponding crossing in a Punnett square.

Genotypes:

• Male cat  AABb
• Female cat aabb

Crossing :

Punnett's square (simplified)

Alelles       AB      Ab

ab            AaBb  Aabb

ab            AaBb  Aabb

Offspring:

• AaBb amazing and bashful (heterozygous for two traits) 50%
• Aabb amazing and boring (heterozygous for trait A ans homozygous recessive for B) 50%

Probability that they will have an offspring who is heterozygous for both traits is 50%.