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POPULATION GENETICS

POPULATION GENETICS. Predicting inheritance in a population. © 2008 Paul Billiet ODWS. Predictable patterns of inheritance in a population so long as…. the population is large enough not to show the effects of a random loss of genes by chance events i.e. there is no genetic drift

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POPULATION GENETICS

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  1. POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWS

  2. Predictable patterns of inheritance in a population so long as… • the population is large enough not to show the effects of a random loss of genes by chance events i.e. there is no genetic drift • the mutation rate at the locus of the gene being studied is not significantly high • mating between individuals is random (a panmictic population) • new individuals are not gained by immigration or lost be emigrationi.e. there is nogene flow between neighbouring populations • the gene’s allele has noselective advantage or disadvantage © 2008 Paul Billiet ODWS

  3. SUMMARY All can affect the transmission of genes from generation to generation • Genetic drift • Mutation • Mating choice • Migration • Natural selection Genetic Equilibrium If none of these factors is operating then the relative proportions of the alleles (the GENE FREQUENCIES) will be constant © 2008 Paul Billiet ODWS

  4. THE HARDY WEINBERG PRINCIPLE Step 1 • Calculating the gene frequencies from the genotype frequencies • Easily done for codominant alleles (each genotype has a different phenotype) © 2008 Paul Billiet ODWS

  5. Iceland Population • 313 337 (2007 est) Area • 103 000 km2 Distance from mainland Europe • 970 km Google Earth © 2008 Paul Billiet ODWS

  6. Sample Population Phenotypes Type M Type MN Type N Genotypes MmMm MmMn MnMn 747 Numbers 233 385 129 Contribution to gene pool 2 Mm alleles per person 1 Mm allele per person 1 Mn allele per person 2 Mn alleles per person Example Icelandic population: The MN blood group © 2008 Paul Billiet ODWS

  7. MN blood group in Iceland Total Mm alleles = (2 x 233) + (1 x 385) = 851 Total Mnalleles = (2 x 129) + (1 x 385) = 643 Total of both alleles =1494 = 2 x 747 (humans are diploid organisms) Frequency of the Mm allele = 851/1494 = 0.57 or 57% Frequency of the Mn allele = 643/1494 = 0.43 or 43% © 2008 Paul Billiet ODWS

  8. In general for a diallellic gene A and a (or Ax and Ay) If the frequency of the A allele = p and the frequency of the a allele = q Then p+q = 1 © 2008 Paul Billiet ODWS

  9. Step 2 • Using the calculated gene frequency to predict the EXPECTED genotypic frequencies in the NEXT generation OR • to verify that the PRESENT population is in genetic equilibrium © 2008 Paul Billiet ODWS

  10. SPERMS Mm 0.57 Mn 0.43 EGGS Mm 0.57 Mn 0.43 Assuming all the individuals mate randomly NOTE the gene frequencies are the gamete frequencies too MmMm0.32 MmMn 0.25 MmMn 0.25 MnMn0.18 © 2008 Paul Billiet ODWS

  11. Close enough for us to assume genetic equilibrium © 2008 Paul Billiet ODWS

  12. In general for a diallellic gene A and a (or Ax and Ay) Where the gene frequencies are p and q Then p + q = 1 and © 2008 Paul Billiet ODWS

  13. THE HARDY WEINBERG EQUATION So the genotype frequencies are: AA = p2 Aa = 2pq aa = q2 orp2 + 2pq + q2 = 1 © 2008 Paul Billiet ODWS

  14. DEMONSTRATING GENETIC EQUILIBRIUM Using the Hardy Weinberg Equation to determine the genotype frequencies from the gene frequencies may seem a circular argument © 2008 Paul Billiet ODWS

  15. Only one of the populations below is in genetic equilibrium. Which one? © 2008 Paul Billiet ODWS

  16. Population sample Genotypes Gene frequencies AA Aa aa A a 100 20 80 0 0.6 0.4 100 36 48 16 0.6 0.4 100 50 20 30 0.6 0.4 100 60 0 40 0.6 0.4 Only one of the populations below is in genetic equilibrium. Which one? © 2008 Paul Billiet ODWS

  17. Only one of the populations below is in genetic equilibrium. Which one? © 2008 Paul Billiet ODWS

  18. SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM  haemoglobin gene • Normal allele HbN • Sickle allele HbS © 2008 Paul Billiet ODWS

  19. Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbNHbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.56 0.4 0.04 0.76 0.24 Expected frequencies 0.58 0.36 0.06 SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM  haemoglobin gene • Normal allele HbN • Sickle allele HbS © 2008 Paul Billiet ODWS

  20. SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION © 2008 Paul Billiet ODWS

  21. Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbNHbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.9075 0.09 0.0025 0.91 0.09 Expected frequencies 0.8281 0.16 0.0081 SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION © 2008 Paul Billiet ODWS

  22. RECESSIVE ALLELES EXAMPLE ALBINISM IN THE BRITISH POPULATION Frequency of the albino phenotype = 1 in 20 000 or 0.00005 © 2008 Paul Billiet ODWS

  23. A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWS

  24. A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWS

  25. Albinism gene frequencies Normal allele = A = p = ? Albino allele = q = ? © 2008 Paul Billiet ODWS

  26. Albinism gene frequencies Normal allele = A = p = ? Albino allele = q =  (0.00005) = 0.007 or 7% © 2008 Paul Billiet ODWS

  27. HOW MANY PEOPLE IN BRITAIN ARE CARRIERS FOR THE ALBINO ALLELE (Aa)? a allele = 0.007 = q A allele = p But p + q = 1 Therefore p = 1- q = 1 – 0.007 = 0.993 or 99.3% The frequency of heterozygotes (Aa) = 2pq = 2 x 0.993 x 0.007 = 0.014 or 1.4% © 2008 Paul Billiet ODWS

  28. Heterozygotes for rare recessive alleles can be quite common • Genetic inbreeding leads to rare recessive mutant alleles coming together more frequently • Therefore outbreeding is better • Outbreeding leads to hybrid vigour © 2008 Paul Billiet ODWS

  29. Example: Rhesus blood group in Europe What is the probability of a woman who knows she is rhesus negative (rhrh) marrying a man who may put her child at risk (rhesus incompatibility Rh– mother and a Rh+ fetus)? © 2008 Paul Billiet ODWS

  30. Rhesus blood group A rhesus positive foetus is possible if the father is rhesus positive RhRh x rhrh  100% chance Rhrh x rhrh  50% chance © 2008 Paul Billiet ODWS

  31. Rhesus blood group Rhesus positive allele is dominant Rh Frequency = p Rhesus negative allele is recessive rh Frequency = q Frequency of rh allele = 0.4 = q If p + q = 1 Therefore Rh allele = p = 1 – q = 1 – 0.4 = 0.6 © 2008 Paul Billiet ODWS

  32. Rhesus blood group • Frequency of the rhesus positive phenotype = RhRh + Rhrh • = p2 + 2pq • = (0.6)2 + (2 x 0.6 x 0.4) • = 0.84 or 84% © 2008 Paul Billiet ODWS

  33. Rhesus blood group • Therefore, a rhesus negative, European woman in Europe has an 84% chance of having husband who is rhesus positive… • of which 36% will only produce rhesus positive children and 48% will produce rhesus positive child one birth in two © 2008 Paul Billiet ODWS

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