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Home Q & A Board Biology-Related Homework Help Genetics and Developmental Biology
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rkkovach rkkovach
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11 years ago
In a cross between a female AaBbccdd and a male AabbCcDd, what proportion of the progeny will be phenotypically identical to the female parent? (Assume independent assortment of all genes and complete dominance).
A. 1/8
B. 3/32
C. 1/64
D. 9/16
The answer is B but could someone explain to me why that is?
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#1
11 years ago
This is not something that someone could explain. Just do the cross and count the offsprings that are identical with the female.
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irinairina
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11 years ago
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#3
11 years ago
The question asks for phenotypically identical, meaning that the same alleles are expressed in the offspring as in the female.
For the first gene, the female expresses the dominant version, A. Both of the parents are heterozygous for this trait. Whenever both parents are heterozygous for a trait, there is a 3/4 probability that an offspring will express the dominant trait. (You can do punnett squares to prove any of the probabilities that I give you)
For the second gene, the female again, expresses the dominant version of the allele, and is heterozygous. This time though, the male is homozygous recessive. This means that there is a 1/2 probability that an offspring will express the dominant allele.
For the second and third genes, the female is homozygous recessive (expresses the recessive trait) while the father is heterozygous. This means that there is a 1/2 probability that the offspring will express the recessive trait.
To arrive at your answer, you need to multiply the probabilities of an offspring expressing each of the traits, so: 3/4*1/2*1/2*1/2=3/32.
The probability of one offspring being phenotypically identical to the female parent is effectively the same as the proportion of the progeny that will be phenotypically identical to the female parent.
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