AP Biology Unit 5: Heredity — Worked Examples

Chi-Square Test of a Dihybrid Cross

Hard

A student crosses two pea plants heterozygous for seed shape (Rr) and seed color (Yy). The student expects a 9:3:3:1 phenotypic ratio. The observed results from 160 offspring are shown below. Using a chi-square test at a significance level of 0.05 with 3 degrees of freedom (critical value = 7.82), what should the student conclude?

The chi-square value is approximately 2.78; fail to reject the null hypothesis — the data fit a 9:3:3:1 ratio ✓ Correct
The chi-square value is approximately 2.78; reject the null hypothesis — the genes are linked
The chi-square value is approximately 8.44; reject the null hypothesis — the genes do not assort independently
The chi-square value is approximately 2.78; reject the null hypothesis — the data show epistasis

Solution

Calculating chi-square: $\\chi^2 = \\sum \\frac{(O-E)^2}{E}$ = (100-90)^2/90 + (28-30)^2/30 + (26-30)^2/30 + (6-10)^2/10 = 100/90 + 4/30 + 16/30 + 16/10 = 1.11 + 0.13 + 0.53 + 1.60 = 2.78. Since 2.78 < 7.82 (critical value), we fail to reject the null hypothesis. The data are consistent with a 9:3:3:1 ratio and independent assortment. Choice B gets the correct chi-square but draws the wrong conclusion (fail to reject means the data fit). Choice C uses an incorrect chi-square value. Choice D correctly calculates but misinterprets what 'fail to reject' means.

Recombination Frequency and Gene Mapping

Hard

In Drosophila, body color (B = gray, b = black) and wing shape (V = normal, v = vestigial) are linked on the same chromosome. A testcross of a heterozygous female (BbVv, with B and V in cis configuration) to a homozygous recessive male (bbvv) produces the offspring shown below. What is the recombination frequency between these two genes?

20%, indicating the genes are 20 map units apart ✓ Correct
50%, indicating the genes assort independently
80%, indicating the genes are tightly linked
10%, because recombination frequency is calculated from the parental classes only

Solution

Recombination frequency = (recombinant offspring / total offspring) x 100. The parental classes (gray-normal and black-vestigial) total 800 and the recombinant classes (gray-vestigial and black-normal) total 200. RF = 200/1000 x 100 = 20%. This means the genes are 20 map units (cM) apart. Choice B (50%) would indicate independent assortment, not linkage — but the parental classes are clearly in excess. Choice C (80%) incorrectly uses the parental class frequency instead of the recombinant class. Choice D (10%) miscalculates by using an incorrect denominator or numerator.