In: Biology
Peas heterozygous for three independently assorting genes were intercrossed.
(example cross: Aa Bb Cc x Aa Bb Cc )
What proportion of the offspring will be homozygous for one gene and heterozygous for the other two?
(hint: apply binomial expansion equation)
1/4
3/8
1/8
3/16
1/2
already answered 1/4 and 1/2 and they were incorrect
As per the Mendelian Laws if Inheritance (Segregation,
Independent Assortment, and Dominance):
A cross of AaBbCc x AaBbCc, has A, B, and C as 3 different genes.
here, the alleles A, B, and C are dominant over/to alleles are a,
b, and c. Going for the crosses on an allele-by-allele basis, we
see that:
Aa x Aa = AA, Aa, aA, aa, hence, 1/2 of the progeny will be
homozygous and 1/2 heterozygous.
Bb x Bb = BB, Bb, bB, bb, hence, 1/2 of the progeny will be
homozygous and 1/2 heterozygous.
Cc x Cc = CC, Cc, cC, cc, hence, 1/2 of the progeny will be
homozygous and 1/2 heterozygous.
To obtain proportion of the offspring that will be homozygous for
one gene and heterozygous for the other two, we see that we
have:
For the Aa * Aa pair: when we have AA or aa we need BbCc alone to
set up this condition.
For one of AA alone we end up with 4 of AABbCc.
Hence, for aa we have 4 of aaBbCc.
Similarly, we will have 8 such for the B allele
and 8 such for the C allele.
A AaBbCc x AaBbCc gives us 8 allelic combinations from each parent
and as such 64 progeny.
We obtained a sum of 4+4+8+8 = 24 such occasions wherein offspring
is homozygous for one gene and heterozygous for the other
two.
Hence proportion = 24/64 = 3/8 ANSWER