Pop gen part 2new

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Pop gen part 2new

  1. 2. 14.2 Hardy-Weinberg Law
  2. 3. Concept review… Population genetic Gene pool concept Hardy-Weinberg Law
  3. 4. At the end of the lesson, you should be able to : <ul><li>a) Hardy-Weinberg law </li></ul><ul><li>State the Hardy-Weinberg law </li></ul><ul><li>Explain five assumptions of Hardy –Weinberg Law for genetic equilibrium: </li></ul><ul><li>Large population size </li></ul><ul><li>Random fertilization </li></ul><ul><li>no net mutation </li></ul><ul><li>no migration (include genetic drift) </li></ul><ul><li>no natural selection </li></ul><ul><li>b) Calculate allele and genotype frequencies </li></ul>
  4. 5. The Hardy-Weinberg equilibrium <ul><li>In 1908, G. H. Hardy (an English mathemathician) & W. Weinberg (a German physician) independently identified a mathematical relationship between alleles and genotypes in </li></ul><ul><li>populations. </li></ul><ul><li>This relationship has been called the Hardy-Weinberg equilibrium and it concerns allele frequency </li></ul>
  5. 6. Hardy-Weinberg Law <ul><li>The principle that frequencies of alleles and genotypes in a population remain constant from generation to generation , </li></ul>
  6. 7. Hardy-Weinberg Law <ul><li>Condition of a population for Hardy-Weinberg equilibrium </li></ul><ul><ul><li>very large population size so that genetic drift can be avoided (chance fluctuation in the gene that can cause phenotype frequencies to change over time). </li></ul></ul><ul><ul><li>no migration that cause gene flow due to immigration into or emigration out from the population </li></ul></ul>
  7. 8. <ul><ul><li>no net mutations because mutation may change an allele into another & this changes allele frequencies </li></ul></ul><ul><ul><li>random fertilization because if individuals choose mates only with certain traits, frequencies of certain alleles may change </li></ul></ul><ul><ul><li>All individuals must be equally fertile and able to pass the alleles to the next generation so that no natural selection is taking place </li></ul></ul>
  8. 9. Hardy-Weinberg Law. <ul><li>Related to allele frequencies are the genotypic frequencies </li></ul><ul><li>Genotype frequency is the ratio of number of individuals with certain genotype in a population </li></ul><ul><li>Hardy-Weinberg equations are used to estimate the frequencies of alleles & genotypes in a population which is in genetic equilibrium </li></ul>Hardy-Weinberg Equations
  9. 10. Hardy-Weinberg Law. <ul><li>The equations: </li></ul><ul><li>p 2 + 2 pq + q 2 = 1 and p + q = 1 </li></ul><ul><li>p 2 = genotypic frequency of homozygous dominant </li></ul><ul><li>2pq = genotypic frequency of heterozygous </li></ul><ul><li>q 2 = genotypic frequency of homozygous recessive </li></ul><ul><li>p = frequency of dominant allele </li></ul><ul><li>q = frequency of recessive allele </li></ul>Hardy-Weinberg Equations
  10. 11. Hardy-Weinberg Law. <ul><li>Question 1 </li></ul><ul><li>Resistance toward a type of pesticide for a population of rats is controlled by dominant allele, R. 64% of the rat population show the resistance. </li></ul><ul><li>a) Calculate the frequency for R allele. </li></ul><ul><li>Assume that the population is in genetic equilibrium and </li></ul><ul><li>p 2 + 2pq + q 2 = 1 while p + q = 1 </li></ul><ul><li>36% of rat population are homozygous recessive (rr). </li></ul><ul><li>Genotypic frequency for homozygous recessive (rr), q 2 = 0.36 </li></ul><ul><li>Frequency for recessive allele (r), q = √0.36 </li></ul><ul><li>= 0.6 </li></ul><ul><li>Frequency for dominant allele (R), p = 1 - q </li></ul><ul><li>= 1 – 0.6 </li></ul><ul><li>= 0.4 </li></ul>Example of calculation
  11. 12. Hardy-Weinberg Law. <ul><li>b) Calculate the number of rats with genotypes RR, Rr and rr for a population of 200 rats. </li></ul><ul><li>It is already known that p = 0.4 and q = 0.6 </li></ul><ul><li>Genotypic frequency for homozygous dominant (RR), p 2 = (0.4) 2 </li></ul><ul><li> = 0.16 </li></ul><ul><li>Number of rats with genotype RR = 0.16 x 200 </li></ul><ul><li>= 32 </li></ul><ul><li>Genotypic frequency for heterozygous (Rr), 2pq = 2(0.4)(0.6) </li></ul><ul><li> = 0.48 </li></ul><ul><li>Number of rats with genotype Rr = 0.48 x 200 </li></ul><ul><li>= 96 </li></ul><ul><li>Genotypic frequency for homozygous recessive (rr), q 2 = 0.36 </li></ul><ul><li>Number of rats with genotype rr = 0.36 x 200 </li></ul><ul><li>= 72 </li></ul>Example of calculation
  12. 13. Hardy-Weinberg Law. <ul><li>Question 2 </li></ul><ul><li>For a population of Shorthorns, the following data was obtained: </li></ul><ul><li>Calculate the frequencies for alleles C M and C P . </li></ul><ul><li>Total number of individuals for the population = 308 </li></ul><ul><li>Frequency for allele C M = 2(110) + 150 </li></ul><ul><li> 2(308) </li></ul><ul><li>= 0.6 </li></ul><ul><li>Frequency for allele C P = 2(48) + 150 </li></ul><ul><li> 2(308) </li></ul><ul><li>= 0.4 </li></ul>Examples of calculation Genotype Phenotype Number of individuals C M C M red 110 C M C P red & white 150 C P C P white 48
  13. 14. Sekilas pandang… Thank you for your attention

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