Populationgenetics

609 views

Published on

Published in: Technology, Health & Medicine
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
609
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
20
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Populationgenetics

  1. 1. POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWS
  2. 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 emigration i.e. there is no gene flow between neighbouring populations  the gene’s allele has no selective advantage or disadvantage © 2008 Paul Billiet ODWS
  3. 3. SUMMARY  Genetic drift  Mutation  Mating choice  Migration  Natural selection All can affect the transmission of genes from generation to generation 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. 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. 5. Iceland Population  313 337 (2007 est) Area  103 000 km2 Distance from mainland Europe  970 km © 2008 Paul Billiet ODWS Google Earth
  6. 6. Example Icelandic population: The MN blood group 2 Mn alleles per person 1 Mn allele per person 1 Mm allele per person 2 Mm alleles per person Contribution to gene pool 129385233Numbers747 Mn Mn Mm Mn Mm Mm Genotypes Type NType MNType MPhenotypesSample Population © 2008 Paul Billiet ODWS
  7. 7. MN blood group in Iceland Total Mm alleles = (2 x 233) + (1 x 385) = 851 Total Mn alleles = (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. 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. 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. 10. Mn Mn 0.18 Mm Mn 0.25 Mm Mn 0.25 Mm Mm 0.32 Mn 0.43 Mm 0.57 Mn 0.43Mm 0.57 Assuming all the individuals mate randomly SPERMS EGGS NOTE the gene frequencies are the gamete frequencies too © 2008 Paul Billiet ODWS
  11. 11. Close enough for us to assume genetic equilibrium Genotypes Expected frequencies Observed frequencies Mm Mm 0.32 233 ÷ 747 = 0.31 Mm Mn 0.50 385 ÷ 747 = 0.52 Mn Mn 0.18 129 ÷ 747 = 0.17 © 2008 Paul Billiet ODWS
  12. 12. SPERMS A p a q EGGS A p AA p2 Aa pq a q Aa pq aa q2 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. 13. THE HARDY WEINBERG EQUATION So the genotype frequencies are: AA = p2 Aa = 2pq aa = q2 or p2 + 2pq + q2 = 1 © 2008 Paul Billiet ODWS
  14. 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. 15. Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Gene frequencies AA Aa aa A a 100 20 80 0 100 36 48 16 100 50 20 30 100 60 0 40 © 2008 Paul Billiet ODWS
  16. 16. Only one of the populations below is in genetic equilibrium. Which one? 0.40.6 0.40.6 0.40.6 40060100 302050100 164836100 0.40.608020100 aAaaAaAA Gene frequenciesGenotypesPopulation sample © 2008 Paul Billiet ODWS
  17. 17. Only one of the populations below is in genetic equilibrium. Which one? 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 © 2008 Paul Billiet ODWS
  18. 18. SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM β haemoglobin gene  Normal allele HbN  Sickle allele HbS Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbN HbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.56 0.4 0.04 Expected frequencies © 2008 Paul Billiet ODWS
  19. 19. SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM β haemoglobin gene  Normal allele HbN  Sickle allele HbS 0.060.360.58 0.240.76 Expected frequencies 0.040.40.56Observed frequencies HbS HbN HbS HbS HbN HbS HbN HbN Genotypes AllelesSickle Cell Anaemia Sickle Cell Trait NormalPhenotypes © 2008 Paul Billiet ODWS
  20. 20. SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbN HbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.9075 0.09 0.0025 Expected frequencies © 2008 Paul Billiet ODWS
  21. 21. SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION 0.00810.160.8281 0.090.91 Expected frequencies 0.00250.090.9075Observed frequencies HbS HbN HbS HbS HbN HbS HbN HbN Genotypes AllelesSickle Cell Anaemia Sickle Cell Trait NormalPhenotypes © 2008 Paul Billiet ODWS
  22. 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. 23. Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal AA p2 Normal Aa 2pq Albino aa q2 A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWS
  24. 24. Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal AA p2 0.99995 Normal Aa 2pq Albino aa q2 0.00005 A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWS
  25. 25. Albinism gene frequencies Normal allele = A = p = ? Albino allele = q = ? © 2008 Paul Billiet ODWS
  26. 26. Albinism gene frequencies Normal allele = A = p = ? Albino allele = q = √ (0.00005) = 0.007 or 7% © 2008 Paul Billiet ODWS
  27. 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. 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. 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. 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. 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. 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. 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 Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Rhesus positive RhRh p2 0.84 Rhesus positive Rhrh 2pq Rhesus negative Rhrh q2 0.16 © 2008 Paul Billiet ODWS

×