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Unformatted text preview: • STABALIZING SELECTION o SELECTION FOR THE MIDDLE FITNESS, ELIMINATES EXTREMES o Favors heterozygote • AA= p²= 1-S Aa=2pq=1 aa=q²=1-t • P’ = p²(1-S) + 2pq * ½ (1) / p²(1-S) + 2pq + q²(1-t) o Bottom = p² - p²S + 2pq + q² – q²t and p²+2pq+q²=1 o SO p’ = p²(1-s)+pq/ 1-p²s-q²t • dP = p’-p o (p²(1-S)+pq/1-p²s-q²t) - P o multiply p by denominator o add both neumerators (bottom will be “total”) o Cancel out the p²+2pq+q² because it equals one o Factor out the Ps o You’ll get 1-p which will equal q o So you will get –p²qs+pq²t/total Which will equal pq(qt-ps)/total for the change in p • The graph of p in a pure heterozygote world would remain constant over time o If p = ½ then dP = 0 o If p is large then dP = negative • Consequence of stabilizing selection for heterozygotes, the fitness of p will eather lower or increase in...
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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.
- Winter '11