What are the chances that two parents who are heterozygous dominant for two traits of interest would have a child that is dominant for one trait, but not both traits?

a. 1 in 16

b. 6 in 16

c. 3 in 16

d. 9 in 16

Question

Kiersten Vance

Biology

24 February, 22:22

What are the chances that two parents who are heterozygous dominant for two traits of interest would have a child that is dominant for one trait, but not both traits?

a. 1 in 16

b. 6 in 16

c. 3 in 16

d. 9 in 16

+1

Answers (1)

Know the Answer?

Alexander Boyer · Answer 1

c. 3 in 16

Explanation:

Let the two genes be A and B

If the parents are heterozygous dominant for the traits, their genotype will be AaBb

AaBb X AaBb =

A_B_ = 9 = dominant for both traits

aaB_ = 3 = dominant for one trait

A_bb = 3 = dominant for one trait

aabb = 1 = recessive for both traits

Out of the resultant progeny, aaB_ and A_bb will be dominant for one trait but not for both. Their individual ratio will be 3/16

What are the chances that two parents who are heterozygous dominant for two traits of interest would have a child that is dominant for one trait, but not both traits?a. 1 in 16b. 6 in 16c. 3 in 16d. 9 in 16

What are the chances that two parents who are heterozygous dominant for two traits of interest would have a child that is dominant for one trait, but not both traits?

a. 1 in 16

b. 6 in 16

c. 3 in 16

d. 9 in 16