### 2.2 Calculation of Chi-square test for deviation from Hardy-Weinberg equilibrium

- Put your observed numbers and click the **Calculate** button.

The number of degrees of freedom for the chi-square is equal to one for
two allele systems; and three for three allele systems; and six for four allele systems.
The results are rounded (four decimals).

Example:

In a population was fund the following number of the three genotypes (obs),
and there is calculated the gene frequencies for the alleles A and B
and the corresponding expected (exp) under Hardy-Weinberg equilibrium.
---------
Genotype AA AB BB Total
Number, obs 36 47 23 = 106 = N
Frequency, exp p^{2} 2pq q^{2} = 1,00
Number, exp 33,4 52,2 20,4 = 106
Deviation 2,6 -5.2 2,6
Chi-square 0.20 0,52 0,33 = 1,05
----------
Frequency of A is calculated as p = (2*36 + 47)/(2*106) = 0,561
do B do q = (2*23 + 47)/(2*106) = 0,439
-----
1.00

Try to put the observed numbers in the program and press
the **Calculation** button.

Questions:

Calculate the gene frequency in the genetic system shown below
and a Chi-square test for H-W equilibrium.

Genotype Observed number
AA 31
AB 51
BB 23
CA 55
CB 71
CC 83

Is there statistically significant deviation from H-W equilibrium ?

In the Danish Landrace are found four hæmopexin alleles
with codominat inheritance.
In 1969 were typed 716 animals and the results of
hæmopexin type are shown below:

Hæmopexin type Number
0/0 2
0/1 16
1/1 75
0/2 10
l/2 67
2/2 20
0/3 21
1/3 229
2/3 116
3/3 160

a) Calculate the gene frequencies for the four alleles.

b) Is the population in Hardy-Weinberg equilibrium ?

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