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Stata Technical Bulletin

Methods and formulas

Borrowing the notation from Weir (1996), let A_u, и = {1,..., ∕√} represent к alleles at a locus and A_uA_v represent each
of the possible k(k + 1)/2 distinct genotypes.

Consider a random sample of n individuals. Then the observed alleles counts, n_u, are

n_u = 2n_uu + n_uv
u≠v

where n_uv and n_uu are respectively, the observed number of heterozygotes A_uA_v and homozygotes A_uA_u in the sample.

The population allele frequencies are therefore estimated as

n_u
Pu =
2n

and their variances as

var(⅛) = P(j>_u + P_uu - 2p²_u)
δu

where P_uu is the observed frequency of the A_uA_u genotype.

Each allele variance under Hardy-Weinberg equilibrium simplifies to the variance of a binomial distribution with parameters
ρ_u and 2n:

^var(⅛) = -^~Pu^-Pu")

2n

The expected genotype frequencies under the assumption of Hardy-Weinberg equilibrium are estimated as
for homozygotes, and

for heterozygotes.

The disequilibrium coefficients for heterozygous genotypes are estimated as

D_uv = p_up_v - -P_uv

Cv V       ɪ Cvx V              Cv V

The Pearson’s chi-squared test statistic is computed using the observed and expected genotype counts as

y' (jiuv - 2np_up_v)^2
2nPuPv

ιt≠∙υ

and the likelihood-ratio chi squared test statistic as

^²¹ⁿ(⅛)

where

⅛ = Σ'-n(⅛)²÷ΣΣ-n(⅞⅛)
n        ^{v z}      U U≠V       ^{v      z}

and

iι=∑ """^ln (⅛) ` ∑ ∑ """'n (⅛)
U         '            U u≠v          ”■ ≠

Both Pearson’s and the likelihood-ratio chi-squared test statistics are distributed with k(k — 1)/2 degrees of freedom.

References

Sham, P. 1998. Statistics in Human Genetics. New York: John Wiley & Sons.

Spencer, N., D. A. Hopkinson, and H. Harris. 1964. Quantitative differences and gene dosage in the human red cell acid phosphatase polymorphism.

Nature 201: 299-300.

Weir, B. S. 1996. Genetic Data Analysis II. Sunderland, MA: Sinauer Associates.

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