9: Hardy-Weinberg Equilibrium

Students calculate allele and genotype frequencies using Hardy-Weinberg relationships and use chi-square goodness-of-fit tests to evaluate whether observed genotype counts fit equilibrium expectations.

Chi-Square Tests for Hardy-Weinberg Equilibrium

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Table of Chi-Squared (χ2) Critical Values
Degrees of Freedom	Probability
	0.95	0.90	0.75	0.50	0.25	0.10	0.05	0.01
1	0.00	0.02	0.10	0.45	1.32	2.71	3.84	6.63
2	0.10	0.21	0.58	1.39	2.77	4.61	5.99	9.21
3	0.35	0.58	1.21	2.37	4.11	6.25	7.81	11.34
4	0.71	1.06	1.92	3.36	5.39	7.78	9.49	13.28

Table 1
Phenotype	Observed	Expected	Calculation	Statistic
Red Flowers	270	279.5	(270-279.5)2⁄ 279.5	0.323
Pink Flowers	208	188.6	(208-188.6)2⁄ 188.6	1.996
White Flowers	22	31.8	(22-31.8)2⁄ 31.8	3.020
(sum) χ2 =				5.339

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You finally have a new competent lab partner that you trust.
This lab partner calculated the allele frequencies of p=0.75 and q=0.25. Then they did a chi-squared (χ²) test for your Hardy-Weinberg data.
They need you to decide whether you reject or accept the null hypothesis using the information provided.

• A. the chi-squared (χ²) sum is GREATER than the critical value and we ACCEPT the null hypothesis
• B. the chi-squared (χ²) sum is GREATER than the critical value and we REJECT the null hypothesis
• C. the chi-squared (χ²) sum is LESS than the critical value and we ACCEPT the null hypothesis
• D. the chi-squared (χ²) sum is LESS than the critical value and we REJECT the null hypothesis

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Hardy-Weinberg Variables from Population Data

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In the grassland population of deer, there is a display of complete dominance in antler shape. Individuals with at least one R allele show the heavily branched antler shape, while those no R alleles show the straight antler shape.
A group of 2500 deer were observed. 2211 exhibited the heavily branched antler shape, while 289 showed the recessive form of straight antler shape.
Which Hardy-Weinberg variable is represented by the fraction 

289

2,500

 ?

• A. p — Frequency of the dominant allele
• B. q — Frequency of the recessive allele
• C. p² — Frequency of homozygous dominant individuals
• D. 2pq — Frequency of heterozygous individuals
• E. q² — Frequency of homozygous recessive individuals
• F. p² + 2pq — Frequency of individuals with the dominant phenotype
• G. 2pq + q² — Frequency of individuals that are NOT homozygous dominant

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Hardy-Weinberg Allele and Genotype Frequencies from Population Data

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beetles
genotype	phenotype	count
homozygous dominant	red	7,911
heterozygous	orange	11,178
homozygous recessive	yellow	5,911
SUM		25,000

In a field there are 7,911 red beetles, 11,178 orange beetles, and 5,911 yellow beetles that show incomplete dominance. What is the frequency of the dominant allele?
Note: Do not enter a percentage on the blank. For example, if the answer is 48.2%, enter 0.482 on the blank. Your answer will be a decimal value between 0 and 1.

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