. Dominant alleles of three independently assorting genes A , B , and C control

Question

. Dominant alleles of three independently assorting genes A, B, and C control production of a black pigment. The alternative alleles a, b, and c function abnormally so that an individual homozygous for one or more of these recessive alleles is colorless. The pathway for black pigment production is shown below. Each gene controls a separate step.

                      A          B        C

Colorless                                 black

A black A/AI; B/B; C/C is crossed with a colorless a/a; b/b; c/c, giving a black F1. If the F1 were intercrossed, what proportion of the F2 would be colorless? Show your work to explain your answer.

Explanation / Answer

AABBCC(black) is crossed with aabbcc(colorless)

So the gametes produced from each respectively are ABC and abc.

Thus all F1 progeny will be of AaBbCc genotype and will have a black phenotype.

Now when 2 such F1 progeny are crossed:

Gametes produced by each are: ABC,ABc,AbC,aBC,abc.

So the F2 progeny will be:

Thus out of the 25 probable F2 progenies, 10 are homozygous recessive for 1 or more genes(indicated by *).

Therefore proportion of F2 that is colorless= 10/25=2/5=0.4

Cross ABC ABc AbC aBC abc ABC AABBCC AABBCc AABbCC AaBBCC AaBbCc ABc AABBCc AABBcc* AABbCc AaBBCc AaBbcc* AbC AABbCC AABbCc AAbbCC* AaBbCC AabbCc* aBC AaBBCC AaBBCc AaBbCC aaBBCC* aaBbCc* abc AaBbCc AaBbcc* AabbCc* aaBbCc* aabbcc*

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