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Sweden population n= 204.
Double mutants = 6
Heterozygote = 46
Homozygotes wild = 204-(6+46)= 152
Each mutant has two mutant alleles and each heterozygote has one mutant alleles.
So frequency of mutant allele= 2*6+1*46= 12+46=58
Total alleles = 2*204=408
Frequency = 58/408= 0.142.
Frequency of wild type allele = 2*152+1*46=304+46=350
Frequency of wild allele = 350/408= 0.857
Yes, the Sweden population is HW equilibrium.
Observed frequencies = frequency of mutant = q^2 = 6/204= 0.029.
q= root of 0.029
So, p = frequency of wild allele = 1-0.17=0.83
So gene frequency of wild p^2= 0.83^2= 0.689
The gene frequency if heterozygote = 2pq= 2*0.83*0.17= 0.282
Expected is (from question 1), p= 0.857, p^2= 0.734
q= 0.142, q^2= 0.02
Heterozygote 2pq= 2(0.857)(0.142)=0.243.
Now, doing chi square analysis:
= [(0.689-0.734)^2/0.734] + [(0.029-0.02)^2/0.02] + [(0.282-0.243)^2/0.243]
= 0.012 this is the X2 value.
As we see that this value is less than 3.84. That means there is no significant difference between the two values. In other words, Sweden population follow HW equilibrium.
The population continues to to be HW equilibrium, since the frequencies have not changed.
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