MBG 2400 Lecture Notes - Lecture 2: Reca, Bundesautobahn 45, Allele

• School: University of Guelph
• Department: Molecular Biology and Genetics
• Course: MBG 2400
• Professor: Barry Micallef

Document Summary

Hardy-weinberg equilibrium in a large random mating population in the absence of (utopian state, most populations break some or all of rules) Mutation selection (cid:862)allele f(cid:396)e(cid:395)ue(cid:374)(cid:272)ies a(cid:374)d ge(cid:374)otypi(cid:272) f(cid:396)e(cid:395)ue(cid:374)(cid:272)ies remain constant and genotypic frequencies are dete(cid:396)(cid:373)i(cid:374)ed (cid:271)y allele f(cid:396)e(cid:395)ue(cid:374)(cid:272)ies(cid:863) Start with f(a) = p = 0. 6 and f(a) = q = 0. 4. F(aa) = p2 = 0. 36, f(aa) = 2pq = 0. 48, f(aa) = q2 =0. 16 p2 of aa + 2pq of aa + q2 of aa = 1. 0. 36 + 0. 48 + 0. 16 = 1 p = p + h q = q + h. P a(cid:374)d (cid:395) do(cid:374)"t (cid:272)ha(cid:374)ge (cid:449)ith (cid:396)a(cid:374)do(cid:373) (cid:373)ati(cid:374)g. Using a punnet square is the same as random mating. H = f(aa) = pq + pq = 2pq = 0. 28. Suppose we know a recessive condition, aa, has a frequency of 1% in the population. If we assume f(aa) = 0. 01 is due to the hardy-weinberg equilibrium, we can do it.