The following MN blood group data were collected from a popu

The following MN blood group data were collected from a population in southern Pennsylvania: Calculate allele frequencies in this population and hardy-Weinberg expected genotypic frequencies. Use the chi-square test to determine whether observed genotypic frequencies deviate significantly from Hardy-Weinberg expected values.

Solution

According to Hardy-Weinberg Law; if alleles are M and N

(M+N)² = M² +N²+2MN =1

According the data type M has 76 individuals in population of 200 individuals.

The frequency = number of individuals for particular allele/ total population

Frequency for M = 76/200 = 0.38

Applying same formula for others, and frequency for N = 0.18 and for allele MN = 0.44

In a population the expected population always follow rule of independent assortment. In which, M=N =0.25, and MN = 0.50 frequencies occur.

So in the 200 individuals population expected numbers of individuals for M frequency = 0.25 x 200 = 50

For N = 50 and for MN = 100

Chi square (X²) test = (observed- expected)² /expected = (O-E)²/ E

alleles

Observed population

Expected population

(O-E)²

X² values

M

76

50

676

13.52

N

36

50

196

3.92

MN

88

100

144

1.44

Total chi square value

18.88

Therefore, in the results there is not significant deviation for the results. The value 18.88 lies in Chi-Square Distribution table for 0.01 p values which tell there is 1% deviation by chance.      

alleles	Observed population	Expected population	(O-E)2	X2 values
M	76	50	676	13.52
N	36	50	196	3.92
MN	88	100	144	1.44
Total chi square value				18.88