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Genetics problem set 3.29
Course: BIOLOGY 282, Spring 2008School: UMass (Amherst)
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set Problem 3.29 When a black AA BB CC is crossed with a colorless aa bb cc, a black Aa Bb Cc is produced in F1. Selfing the AaBbCc will give us 27 genotypes 1/64 AABBCC 2/64 AABBCc 1/64 AABBcc 2/64 AABbCC 4/64 AABbCc 2/64 AABbcc 1/64 AAbbCC 2/64 AAbbCc 1/64 AAbbcc 2/64 AaBBCC 4/64 AaBBCc 2/64 AaBBcc 4/64 AaBbCC 8/64 AaBbCc 4/64 AaBbcc 2/64 AabbCC 4/64 AabbCc 2/64 Aabbcc 1/64 aaBBCC 2/64 aaBBCc 1/64 aaBBcc 2/64...
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set Problem 3.29 When a black AA BB CC is crossed with a colorless aa bb cc, a black Aa Bb Cc is produced in F1. Selfing the AaBbCc will give us 27 genotypes 1/64 AABBCC 2/64 AABBCc 1/64 AABBcc 2/64 AABbCC 4/64 AABbCc 2/64 AABbcc 1/64 AAbbCC 2/64 AAbbCc 1/64 AAbbcc 2/64 AaBBCC 4/64 AaBBCc 2/64 AaBBcc 4/64 AaBbCC 8/64 AaBbCc 4/64 AaBbcc 2/64 AabbCC 4/64 AabbCc 2/64 Aabbcc 1/64 aaBBCC 2/64 aaBBCc 1/64 aaBBcc 2/64 aaBbCC 4/64 aaBbCc 2/64 aaBbcc 1/64 aabbCC 2/64 aabbCc 1/64 aabbcc
In order for an individual to be black, it has to be normal at each of the alleles, or A-/B-/C- . That makes up 27/64 genotypes in the list above. Then, for individual an to be colorless it must have a recessive aa, bb, and/or cc genotypes. Looking at the list above, there are 37/64 colorless genotypes. 27 : 37 is the ratio of the colorless individuals in the F2 progeny.
B) In order for an individual to be black, it must be normal at the A and B allele, but not at the C allele, or A-/B-/cc. If the inhibitor allele is normal, it will prevent the production of the black color, therefore, it must be recessive. Looking at the list above, there are 9/64 such genotypes. The rest of the genotypes (1 9/64 = 55/64) must then be colorless. A 9 black: 55 colorless ratio is expected in the F2.
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