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

- Put your observed numbers and click the Calculate button.

 Three alleles put 3 -> Four alleles, put 4 -> Set digit for print ->

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   p2               2pq             q2    = 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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