Two individuals of genotype AABBCCDD and aabbccdd are mated. The F1 generation is interbred self- fertilized) to yield 5000 progeny. What proportion of the F2 is expected to be AabbCcDD?

Question

Quizplus · Accepted Answer

When the two individuals are crossed, they produce heterozygous (AaBbCcDd) offspring. Therefore, the F1 generation is AaBbCcDd x AaBbCcDd. 
When these F1 individuals are self-fertilized, the resulting genotype proportions can be found using a Punnett square. The possible gametes that can be produced by the F1 generation are AB, Ab, aB, and ab for the A and B loci (and similarly for the C and D loci). 
Therefore, the expected genotype frequencies in the F2 generation are:
- AABBCCDD: (1/4) x (1/4) x (1/4) x (1/4) = 1/256
- AABBCcDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AABBCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AABbCCDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AABbCcDD: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AABbCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AAbbCCDD: (1/4) x (1/2) x (1/4) x (1/2) = 1/64
- AAbbCcDD: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- AAbbCcDd: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- AabbCCDD: (1/4) x (1/4) x (1/4) x (1/2) = 1/128
- AabbCcDD: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AabbCcDd: (1/4) x (1/4) x (1/2) x (1/2) = 1/64
- AabbccDD: (1/4) x (1/2) x (1/4) x (1/2) = 1/64
- AabbccDd: (1/4) x (1/2) x (1/2) x (1/2) = 1/32
- aabbCCDD: (1/2) x (1/4) x (1/4) x (1/2) = 1/64
- aabbCcDD: (1/2) x (1/4) x (1/2) x (1/2) = 1/32
- aabbCcDd: (1/2) x (1/4) x (1/2) x (1/2) = 1/32
- aabbccDD: (1/2) x (1/2) x (1/4) x (1/2) = 1/32
- aabbccDd: (1/2) x (1/2) x (1/2) x (1/2) = 1/16

Therefore, the proportion of the F2 that is AabbCcDD is the sum of the frequencies of AabbCcDD, AabbCcDd, aabbCcDD, and aabbCcDd:

1/64 + 1/64 + 1/64 + 1/32 = 3/64

This is closest to choice C, 1/64.

Two Individuals of Genotype AABBCCDD and Aabbccdd Are Mated