Respuesta :
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%.
The probability that they will have an offspring who is heterozygous for both traits in this cat with being awesome (A) is dominant to being average (a), and being bashful (B) is dominant to being boring (b) -is - 50%
Cross between two independent assorted allele
In cats, being awesome (A) is dominant to being average (a), and being bashful (B) is dominant to being boring (b). Here the cross is between A male cat that is dominant for the A trait and heterozygous for the B trait is mated with a female cat that is wild type for recessive for both
Male cat - AA Bb
Female cat - aa bb
Cross - AA Bb x aa bb
Gametes: AB, Ab and ab
Punntte:
AB Ab
ab AaBb Aabb
A trait probability for a heterozygous trait is 1 or 100%
For b trait, the probability of heterozygous trait is 50% or 1/2
So the total probability is 50%
= 50%
Learn more about Independent assortment:
https://brainly.com/question/2376592
