 Taylor Pruett AP biology 3 rd block.  British mathematician Godfery H. Hardy and German physician Wilhelm Weinberg.

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Taylor Pruett AP biology 3 rd block

 British mathematician Godfery H. Hardy and German physician Wilhelm Weinberg.

 In 1908, Hardy and Weinberg came up with a mathematical model to estimate the genotypic frequencies of a population that is in genetic equilibrium.  Genetic Equilibrium: where allele frequencies do not change.

 The Hardy-Weinberg principle states that in a large randomly breeding population, allelic frequencies will remain the same from generation to generation assuming that there is no mutation, gene migration, selection or genetic drift.

 Genetic equilibrium is referred to as Hardy- Weinberg equilibrium.  This describes a stable, nonevolving population; allelic frequencies do not change.  Requirements: ◦ Population must be large ◦ Population must be isolated ◦ No mutations ◦ Mating must be random ◦ No natural selection

 The Hardy-Weinberg principle is illustrated in a mathmatical equation:  p²+2pq+q²=1  p+q=1

 p = frequency of the dominant allele in the population  q = frequency of the recessive allele in the population  p2 = percentage of homozygous dominant individuals  q2 = percentage of homozygous recessive individuals  2pq = percentage of heterozygous individuals

 Example: ◦ D= p ◦ d= q  So, set up your equation like: ◦ D²+2Dd+d²=1 ◦ D²= frequency of DD ◦ 2Dd= frequency of Dd ◦ d²= frequency of dd ◦ D= frequency of the D allele ◦ d= frequency of the d allele

◦ You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: ◦ The frequency of the "aa" genotype. ◦ The frequency of the "a" allele. ◦ The frequency of the "A" allele. ◦ The frequency for the “AA” allele.

 36%, as given in the problem itself  If q² = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the “a” allele, then the frequency is 60%.  Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.  Since p=0.4, to find p², (0.4)²=0.16. So the frequency of the “AA” genotype is 16%.

 A census of albatrosses nesting on a Galapagos Island revealed that 24 of them showed a rare recessive condition that affected beak formation. The other 63 show no beak defect. What is the frequency of the dominant allele? Give your answer to the nearest hundreth.

 24+63= 87 total birds.  24/87=0.28  Take the square root  You get 0.53  1-0.53= 0.47  P= 0.47

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