Question

In: Biology

Peas heterozygous for three independently assorting genes were intercrossed. (example cross: Aa Bb Cc x Aa...

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

Solutions

Expert Solution

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


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