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Title: Hardy Weinberg principle? Post by: Mike146 on Mar 21, 2012 Using the Hardy-Weinberg principle, find q if the frequency of recessive homozygote Albinos is 1 in ever 17,000 individuals.
What is the frequency of the dominant allele in the population? What is the frequency of heterozygous carriers of Albinism? Title: Hardy Weinberg principle? Post by: zzz1090 on Mar 21, 2012 In the Hardy-Weinberg formula, p + q = 1, where p is the frequency of one allele and q is the frequency of the other allele (often p is used as the frequency of the dominant allele and q is used as the frequency of the recessive allele). Since there are only two, then the total has to add to 1.
If you square both sides of the above equation, you get: p^2 + 2pq + q^2 = 1, where p^2 is the frequency of homozygotes for one allele, 2pq is the frequency of heterozygotes and q^2 is the frequency of homozygotes for the other allele. So, if q^2 = 1/17000 = 0.0000588, then q = 0.00767 so p = 1 - 0.00767 = 0.99233 (the frequency of the dominant allele). As an aside, note that the frequency of homozygous dominant individuals would be p^2 = 0.99233 * 0.99233 = 0.9847 and finally, the frequency of heterozygotes (carriers of Albinism) is: 2pq = 2 * 0.99233 * 0.00767 = 0.0152 (or 1.52%) |