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Two interacting genes (independently assorting) were involved in the same pathway. Absence of either gene's function leads to absence of the end product of the pathway. A dihybrid cross involving the two genes is carried out. What fraction of the F2 progeny will show the presence of the end product?

 

Option: 1

\frac{1}{4}


Option: 2

\frac{3}{4}


Option: 3

\frac{9}{16}


Option: 4

\frac{15}{16}


Answers (1)

best_answer

In a dihybrid cross, if two independently assorting genes are involved in the same pathway and the absence of either gene's function leads to the absence of the end product, we can determine the fraction of progeny showing the presence of the end product by using a Punnett square.

Let's assume that the two genes are represented by capital letters (A and B) for the dominant alleles and lowercase letters (a and b) for the recessive alleles. The parental generation (P) can be represented as:

AA BB (both homozygous dominant)

crossed with

aa bb (both homozygous recessive)

The F1 generation will all be heterozygous for both genes: Aa Bb.

When we perform a dihybrid cross between the F1 individuals, we get the following genotypic ratios in the F2 generation:

9/16 will have at least one functional allele for both genes (A- B-, A- bb, aa B-, A- Bb)

1/16 will have both genes nonfunctional (aa bb)

Therefore, the fraction of the F2 progeny that will show the presence of the end product is 9/16.Hence, the correct answer is option 3.

Posted by

Sumit Saini

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