Question

Consider a population of Pacific tree frogs (Pseudacris regilla). In this population of frogs, a single locus controls the production of a mating call. This locus has two alleles - S1 and S2. You observe the following genotype counts:

S1S1 = 987

S1S2 = 333

S2S2 = 21

Use these data to calculate the chi-square statistic for a test of whether or not this population is consistent with HWE. Report your calculated chi-square statistic to two decimal places.

Given your calculated chi-square statistic from Question 1, is this population of Pacific tree frogs at HWE assuming a critical value of 6.634897?

NOTE: The answers are 1.40 for the first part, and YES for the second part. For some reason I can't get my math to arrive at these answers. Any help would be appreciated.

Answer 1

Answer:

S1S1 987 = 987*2 = 1964 “S1” alleles

S1S2 333 = 333 “S1” & 333 “S2” alleles

S2S2 21 = 21*2=42 “S2” alleles

Total alleles = 2682

Total S1 alleles = 1974 + 333 = 2307

Frequency of S1 allele = 2307 / 2682 = 0.86

Total S2 alleles = 333+42 = 375

Frequency of S2 allele = 375 / 2682 = 0.14

Total progeny = 987+333+21= 1341

Expected S1S1 individuals = 0.86 * 0.86 * 1341 = 991.81

Expected S2S2 individuals = 0.14 * 0.14 * 1341 = 26.28

Expected S1S2 individuals = 2*0.86*0.14 *1341 = 322.91

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
S1S1	987	991.81	-4.81	23.14	0.023
S1S2	333	322.91	10.09	101.81	0.315
S2S2	21	26.28	-5.28	27.88	1.061
	1341	1341			1.399

Chi-square value = 1.40

Degrees of freedom = number of categories -1

Df = 3-1=2

The chi-square value of 1.40 is less than the critical value of 6.634897. Hence we accept the null hypothesis and the population is consistent with HWE.