Question

A SNP locus is genotyped for a total of 1000 individuals, among which 485 is AA, 421 is Aa, and 94 is aa. Compute the X2 statistic for Hardy-Weinberg Equilibrium.

Answer 1

here total alleles =(2*485+2*421+2*94)=2000

p² =((2*485+421)/2000)²=0.4837

q² =((2*94+421)/2000)²=0.0927

2*p*q =2*((2*485+421)/2000)*((2*94+421)/2000)=0.4236

hence from Chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
AA	0.4837	485.000	483.72	0.06	0.003
Aa	0.4236	421.000	423.56	-0.12	0.015
aa	0.0927	94.000	92.72	0.13	0.018
total	1.000	1000	1000		0.037

X² test statistic =0.0365~ 0.037