6 - Note all alleles can't do as well as other genotypes in the womb but generally assume it's hardy weinburg o P is the amount of frequency A in

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Note, all alleles can’t do as well as other genotypes in the womb but generally assume it’s hardy  weinburg o P’ is the amount of frequency A in the next generation If you have a gene pool of all As, if one mutates into an a: o then p no longer equals 1 o p’ does not equal p o and the population has change A -> then it’s mew ( M) and A <- a then it’s vew (v) So if p=f(A) and q=f(a) dQ= M (creation rate of a by A->a) * p – v (destruction of a by a->) * q if delta Q=0 then o 0=pM-qv (all have hats) o P+q=1 so 0= (1-q)M-qv (hats) o M-qM-qv (hats) o qM + qv = M o q hat (M+v) = M o q hat = M / M+v if q hat = ½ then dq = 0, and it will be straight  if q is ½ then it will be straight o if q is almost at 1, that means there is almost no p, so then it can’t turn into more q, so it goes  down
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This note was uploaded on 02/05/2012 for the course BIEB 150 taught by Professor Chao during the Winter '11 term at San Diego.

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6 - Note all alleles can't do as well as other genotypes in the womb but generally assume it's hardy weinburg o P is the amount of frequency A in

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