Harris 1966 measured the number of genotypes at the alkaline phosphate locus in

Question

Harris 1966 measured the number of genotypes at the alkaline phosphate locus in the English population. There are three alleles, designated F, I, and S (for fast, intermediate,and slow mobility in an electropheritc apparatus)

1. What are the observed frequencies of the six genotypes?

2. What are the observed allele frequencies for the three alleles?

3. What are the Hardy-Weinberg equilibrium frequencies of the genotypes?

4. What are the Hardy-Weinberg equilibrium frequencies of the alleles?

5. What is the total observed heterozygosity (i.e. frequency of all heterozygotes)?

6. What is the total heterozygosity at Hardy-Weinberg equilibrium?

7. Calculate F from the observed and equilibrium heterozygosities. Does this suggest to you any departures from Hardy-Weinber conditions, and, if so, what kind of departure?

Explanation / Answer

The Hardy-Weinberg model, predicts genotype and allele frequencies. The model has five basic assumptions:

The Hardy-Weinberg model can also be applied to the genotype frequency of a single gene.

they are two variables;

P; frequency of me of the two alles

Q. frequency of the pther two alleles

Hardy-Weinberg model two equations:

one that calculates allele frequencies and one that calculates genotype frequencies. The equation

p + q = 1 , this equation describes the allelic frquency for the genotype.

in a diploid organism the equation changes from, possible genotypes can be genotypes: AA, Aa, and aa.

p2+ 2pq + q2 = 1.

the 6 phenotypes are referred to as S, F, I,

SF, SI, and FI. In types S, F, and I most of the alkaline phosphatase activity and hardy weinburg equilibrium;

alkaline phosphatase and their frequencies

genotype S   = r2 = 0.410

genotype SF=2pr =0.346

genotype F =p2=0.073

genotype SI = 2qr =0.115

genotype FI =2pq =0.049

genotype I =q2 =0.008,

from the above caluculation we get

p = 0.27, q = 0.09, r = 0.64

applying the hardy weinburg equation for phenotype ;

p + q = 1

0.27+0.09=0.36 ,which is not equal to one

applying the hardy weinburg equation for genotype ;

p2+ 2pq + q2 = 1.

{0.27}2 + 2.*0.27*0.09 + [0.09]2

0.0729+0.0486+0.0081

0.12 , the genotype frequency

there is a large number of heterozygosity takes place because of the structural changes in the enzyme and the production of isozymes in diffrent populations .more allelic have been found to have frequencies greater than 0.01 , this suggest the hetrozygosity of the enzymes in the diffrent population.

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