Solve a Dihybrid Cross and Test It With Chi-Square

Genetics problems in a biology course usually have two halves: predict the offspring of a cross, then decide whether real results fit the prediction. This guide works through one classic problem from start to finish, with every number checked by the calculators, so that each step can be reproduced by hand in an exam.

The problem

In guinea pigs, black coat (B) is dominant to brown (b) and rough coat (R) is dominant to smooth (r); the genes are on different chromosomes. A black, rough guinea pig with genotype BbRr is crossed with a black, smooth one with genotype Bbrr. What offspring are expected? The cross produces 160 offspring: 64 black rough, 58 black smooth, 22 brown rough and 16 brown smooth. Do the results fit?

Step 1: the gametes

Each parent passes one allele of each gene. Because the genes assort independently, BbRr makes four gamete types in equal proportions: BR, Br, bR and br. Bbrr can only pass r for the second gene, so it makes Br and br, each half the time; the Punnett square calculator lists them as Br, Br, br, br so that the square stays 4 × 4. Writing the gametes out first is the step most students skip, and most errors start there.

Step 2: the square and the ratios

The 16 cells give six genotypes: 4 BbRr, 4 Bbrr, 2 BBRr, 2 BBrr, 2 bbRr and 2 bbrr. Grouping by phenotype: black rough (B_R_) 6 of 16, black smooth (B_rr) 6, brown rough (bbR_) 2 and brown smooth (bbrr) 2, a ratio of 3 : 3 : 1 : 1. The same answer comes from multiplying the single-gene ratios: Bb × Bb gives 3 black : 1 brown, and Rr × rr gives 1 rough : 1 smooth, so (3 : 1) × (1 : 1) = 3 : 3 : 1 : 1. Checking a dihybrid answer by the product rule is quick and catches counting mistakes.

Step 3: expected counts

With 160 offspring, the expected counts are 160 × 3/8 = 60 black rough, 60 black smooth, 160 × 1/8 = 20 brown rough and 20 brown smooth. The observed 64, 58, 22 and 16 are close but not identical, as real data never are.

Step 4: the chi-square test

The chi-square calculator works out each class’s contribution, (O − E)² ÷ E: (64 − 60)² ÷ 60 = 0.267, (58 − 60)² ÷ 60 = 0.067, (22 − 20)² ÷ 20 = 0.200 and (16 − 20)² ÷ 20 = 0.800. The total is χ² = 1.333. There are four classes, so 3 degrees of freedom; the critical value at the 5% level is 7.815. With p = 0.72, the deviations are well within what chance produces, and the 3 : 3 : 1 : 1 hypothesis is not rejected. For contrast, testing the same counts against a 9 : 3 : 3 : 1 ratio, which a student might wrongly assume for any dihybrid cross, gives χ² = 39.4 and p below 0.0001: the data clearly reject it.

Common mistakes

• Assuming 9 : 3 : 3 : 1. That ratio applies only when both parents are heterozygous for both genes.
• Wrong gametes. List them before drawing the square; a homozygous gene contributes only one allele.
• Using percentages in χ². The test needs counts.
• Wrong degrees of freedom. Classes minus one, here 3.
• Reading p > 0.05 as proof. It means the data are consistent with the hypothesis, nothing more.

Cross problem checklist

1. Write each parent’s genotype and list all its gametes.
2. Draw the Punnett square, or multiply the single-gene ratios.
3. Group the offspring by phenotype and simplify the ratio.
4. Convert the ratio to expected counts for the observed total.
5. Run a chi-square test with classes minus one degrees of freedom.
6. State the conclusion in terms of the hypothesis, not proof.