AP Biology • Genetics • Probability • Chi-Square

Do Seed Color and Seed Shape Assort Independently?

Use a dihybrid cross and a chi-square goodness-of-fit test to determine whether the observed offspring data are consistent with independent assortment.

χ² = Σ ((O − E)² / E)

O = observed result • E = expected result • Σ = sum of all categories

1.Research Question and Trait Information

Research question: Do seed color and seed shape assort independently?

Seed color

R = purple / pigmented
r = yellow / lacking pigment

Seed shape

Su = smooth / starchy
su = wrinkled / sweet

Rewrite the research question in your own words.

The parental cross is RrSusu × RrSusu.

2.Null Hypothesis (H₀)

State what would be expected if the two genes assort independently.

3.Alternative Hypothesis (Hₐ)

State what would be expected if the traits do not assort independently.

4.Count the Observed Values

The corn sample contains 186 seeds across four phenotypic categories.

Phenotype	Observed (O)
Purple, Smooth	118
Purple, Wrinkled	29
Yellow, Smooth	33
Yellow, Wrinkled	6
Total	186

5A.Complete the Dihybrid Cross

Cross: RrSusu × RrSusu. Each parent can produce the gametes RSu, Rsu, rSu, rsu.

×	RSu	Rsu	rSu	rsu
RSu
Rsu
rSu
rsu

From this cross, the expected phenotypic ratio is 9 : 3 : 3 : 1.

5B.Determine the Expected Values

Use the total of 186 offspring and the expected 9:3:3:1 phenotypic ratio.

Phenotype	Observed	Expected Ratio
Purple, Smooth	118	9/16
Purple, Wrinkled	29	3/16
Yellow, Smooth	33	3/16
Yellow, Wrinkled	6	1/16

Show at least one expected-value calculation.

6.Calculate the Chi-Square Value

For each phenotype, calculate (O − E)² / E, then add all four contributions.

Phenotype	O	E
Purple, Smooth	118	104.6
Purple, Wrinkled	29	34.9
Yellow, Smooth	33	34.9
Yellow, Wrinkled	6	11.6

Total χ²

Calculation structure:
(118−104.6)²/104.6 + (29−34.9)²/34.9 + (33−34.9)²/34.9 + (6−11.6)²/11.6

7–9.Degrees of Freedom, p Value, and Critical Value

Degrees of freedom

Significance level

Critical χ² value

p	df 1	df 2	df 3	df 4	df 5	df 6	df 7	df 8
0.05	3.84	5.99	7.81	9.49	11.07	12.59	14.07	15.51
0.01	6.63	9.21	11.34	13.28	15.09	16.81	18.48	20.09

10.Make the Statistical Decision

Compare the calculated χ² value with the critical χ² value.

Reject H₀
The observed deviation from the expected ratio is statistically significant. Fail to reject H₀
The observed deviation can reasonably be attributed to chance.

11.AP Biology Evidence-Based Conclusion

Your conclusion should state χ², df, p, the critical value, whether H₀ is rejected, and what the result means for independent assortment.

Worksheet Progress

AP Extension: Interpret the Result

Why does a χ² value below the critical value support the idea that the genes assort independently?