For part d I keep on getting 27/64. I get this answer by doing: 3((1/4)((3/4)^2)

Question

For part d I keep on getting 27/64. I get this answer by doing: 3((1/4)((3/4)^2)). Am I doing this right? If not, please describe why my answer is wrong.

3.26 Go Peas heterozygous for three independently assorting (a) What proportion of the offspring will be homozygous for (b) What proportion of the offspring will be homozygous for (c) What proportion of the offspring will be homozygous for (d) What proportion of the offspring will be homozygous for genes were intercrossed. all three recessive alleles? all three genes? one gene and heterozygous for the other two? the recessive allele of at least one gene?

Explanation / Answer

Let there be 3 independently assorting genes A,B and C. Thus,

AaBbCc x AaBbCc,

splitting this trihybrid cross into 3 crosses separately we get-

Aa x Aa

Homozygous recessive aa- 1/4

Bb x Bb

Homozygous recessive bb -1/4

Cc x Cc

homozygous recessive cc-1/4

product rule is very important in genetics which states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the events. Thus multiplying the probabilities of all the recessives obtained in all the crosses are-

1/4 x 1/4 x 1/4 = 1/64

Thus the proportion of the offspring that will be homozygous for the recessive allele of at least one gene would be 1/4.

A a A AA Aa a Aa aa

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