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Hardy-Weinberg

Hardy-Weinberg. Taylor Pruett AP biology 3 rd block. Who are Hardy and Weinberg?. British mathematician Godfery H. Hardy and German physician Wilhelm Weinberg. What did they do?.

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Hardy-Weinberg

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  1. Hardy-Weinberg Taylor Pruett AP biology 3rd block

  2. Who are Hardy and Weinberg? • British mathematician Godfery H. Hardy and German physician Wilhelm Weinberg.

  3. What did they do? • 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.

  4. Hardy-Weinburg Principle: • 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.

  5. Requirements: • 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

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

  7. 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

  8. How to use the equation: • 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

  9. Putting them into frequencies • 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.

  10. Answers to problem 1: • 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%.

  11. Problem 2: • 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.

  12. Answer: • 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

  13. Problem 3:

  14. Answers:

  15. Answers Cont’d:

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