A plant heterozygous for a dominant trait was selfed. The progeny had 140 plants showing the dominant trait and 20 plants showing the recessive trait. A researcher hypothesized that there are two genes with identical functions that control the dominant trait. The researcher also proposed that the two genes are not linked. The researcher carried out a chi-square test to test the hypothesis. Which one of the following options is the correct chi-square value (rounded to second decimal) obtained by the researcher?

Correct option is C

Step 1: Understand the Hypothesis

If the plant is heterozygous for a dominant trait and there are two unlinked genes with identical functions, we are likely dealing with a scenario where the dominant trait is expressed when at least one dominant allele is present at each gene locus. This is similar to a dihybrid cross with complementary gene action or a case where two genes contribute to the same trait (e.g., duplicate dominant epistasis).

For a heterozygous plant (AaBb, where A and B are dominant alleles and a and b are recessive alleles), a self-cross would typically produce a 9:3:3:1 phenotypic ratio in a standard dihybrid cross. However, if the two genes are unlinked and both must contribute to the dominant trait (or if the recessive trait appears only when both genes are homozygous recessive), the expected ratio might be adjusted. Given the problem's hint about identical functions and unlinked genes, we can infer a model where the recessive trait (aa bb) appears only when both genes are homozygous recessive, and the dominant trait appears otherwise.

The expected phenotypic ratio for a dihybrid cross (AaBb × AaBb) is:

However, with duplicate dominant epistasis (where the dominant trait is expressed if at least one dominant allele is present at either gene), the ratio simplifies to 15:1 (15 dominant : 1 recessive), because the recessive phenotype (aa bb) requires both genes to be homozygous recessive.

Step 2: Calculate Expected Values

The total number of plants is 140 (dominant) + 20 (recessive) = 160.

Step 3: Perform Chi-Square Test

The chi-square statistic is calculated as:   $$\chi^2 = \sum \frac{(O - E)^2}{E}$$

Total chi-square value  $$\chi^2$$​ $=0.6667+10=10.6667$

Rounded to the second decimal place, this is 10.67.

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