14. 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 hypothesised 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?
(1) 22.86 (2) 13.33
(3) 10.67 (4) 5.71
Chi-Square Test Analysis of Genetic Duplicate Dominant Gene Interaction
Step 1: Understand the hypothesis
Researcher’s hypothesis:
Two genes with identical (duplicate) functions control the dominant phenotype.
Genes are unlinked.
A classic duplicate dominant gene interaction gives an F2 phenotypic ratio of [translate:15 dominant : 1 recessive] when both loci are segregating (A/a, B/b) and at least one dominant allele at either locus produces the dominant trait.
So expected F2 ratio under the hypothesis = [translate:15:1] (dominant : recessive).
Step 2: Observed data
- Dominant phenotype plants = 140.
- Recessive phenotype plants = 20.
- Total plants = 160.
Step 3: Calculate expected numbers (E)
Using 15:1 ratio:
Expected dominant = 15/16 × 160 = 150.
Expected recessive = 1/16 × 160 = 10.
Step 4: Apply chi-square formula
de>χ² = ∑ (O − E)² / E
For dominant class:
O = 140, E = 150
O − E = −10
(O − E)² / E = 100 / 150 = 0.67.
For recessive class:
O = 20, E = 10
O − E = 10
(O − E)² / E = 100 / 10 = 10.
Total chi-square:
χ² = 0.67 + 10 = 10.67.
Rounded to two decimal places → 10.67.
Step 5: Option-wise explanation
- 22.86 – too large; would require a much bigger deviation from expected values.
- 13.33 – also larger than what we calculated.
- 10.67 – matches the correct chi-square from calculation.
- 5.71 – smaller than the true value.
Therefore, the chi-square value obtained by the researcher is 10.67 (option 3).
