In a mating between AaBBcc x AaBbCC, what fraction of offspring will have the genotype: AaBbCc?

Question

In a mating between AaBBcc x AaBbCC, what fraction of offspring will have the genotype: AaBbCc?


a. 0

b. 1/2

c. 1/4

d. 1/6

e. 1/8

Answer 1

Try simplifying it. Just look at the Aa at first. There are four possibilities here: AA, Aa, aA, and aa.But the two in the middle are equivalent, so there's a 2 in 4 possibility that the organism is heterozygous for this trait. That simplifies to 1 in 2.

Now look at Bb. Since the first parent has to pass on a B, there are only two possibilities: Bb and BB. So there's a 1 in 2 possibility for Bb.

Now look at Cc. The first parent has to pass down c, and the second has to pass down C, so the child will definitely be Cc.

So the probability is 1/2 * 1/2 * 1 or 1/4. C.

Answer 2

Should be 1/4

Answer 3

c. 1/4
Using Punnett Square, the AaBBcc parent can have 2 gametes (ABc & aBc), the AaBbCC parent can have 4 gametes (ABC, AbC, aBC, abC):

ABc aBc
ABC AABBCc AaBBCc
AbC AABbCc AaBbCc
aBC AaBBCc aaBBCc
abC AaBbCc aaBbCc

So 2 out of 8 reproductions would resuld in AaBbCc = 2/8 = 1/4