Population Genetics AQA


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Population Genetics AQA

  2. 2. Lesson Objective To be able to use the Hardy-Weinberg equation to calculate allele frequencies in a population.
  3. 3. Population definition A POPULATION is a group of individuals of the same species that can interbreed. Populations are dynamic - they can expand or contract due to changes in their birth or death rates or migration. The set of genetic information carried by a population is the gene pool.
  6. 6. Populations rather than individuals are functional units of variation The goal of our previous discussions in this class has been to understand the inheritance of a single trait, a trait that may be controlled by one, a few, or many genes. The goal of population genetics is different. Rather than studying the inheritance of a trait, population genetics attempts to describe how the frequency of the alleles which control the trait change over time. To study frequency changes, we analyse populations rather than individuals. Furthermore, because changes in gene frequencies are at the heart of evolution and speciation, population and evolutionary genetics are often studied together.
  7. 7. We observe the phenotype (and not the genotype) of individuals. To measure the frequency of an allele we need to know: 1. the mechanism of inheritance of a particular trait 2. how many different alleles of that gene there are in the population. For traits that show codominance, the frequency of the heterozygous phenotype is the same as the frequency for the heterozygous genotype. Work through blood group example.
  8. 8. The Hardy-Weinberg Law of Genetic Equilibrium In 1908 G. Hardy and W. Weinberg independently proposed that the frequency of alleles and genotypes in a population will remain constant from generation to generation if the population is stable and in genetic equilibrium. Five conditions are required in order for a population to remain at Hardy-Weinberg equilibrium: 1.A large breeding population * 4. No immigration or emigration 2. Random mating *5. No natural selection 3. No mutations
  9. 9. A large breeding population A large breeding population helps to ensure that chance alone does not disrupt genetic equilibrium. In a small population, only a few copies of a certain allele may exist. If for some chance reason the organisms with that allele do not reproduce successfully, the allelic frequency will change. This random, non selective change is what happens in genetic drift or a bottleneck event.
  10. 10. Large breeding population
  11. 11. Random Mating In a population at equilibrium, mating must be random. In assortative mating, individuals tend to choose mates similar to themselves; for example, large blister beetles tend to choose mates of large size and small blister beetles tend to choose small mates. Though this does not alter allelic frequencies, it results in fewer heterozygous individuals than you would expect in a population where mating is random.
  12. 12. Random Mating
  13. 13. No Change in Allelic Frequency Due to Mutation For a population to be at Hardy-Weinberg equilibrium, there can be no change in allelic frequency due to mutation. Any mutation in a particular gene would change the balance of alleles in the gene pool. Mutations may remain hidden in large populations for a number of generations, but may show more quickly in a small population.
  14. 14. No mutations
  15. 15. No Immigration or Emigration For the allelic frequency to remain constant in a population at equilibrium, no new alleles can come into the population, and no alleles can be lost. Both immigration and emigration can alter allelic frequency.
  16. 16. No Migration or Emigration
  17. 17. No Natural Selection In a population at equilibrium, no alleles are selected over other alleles. If selection occurs, those alleles that are selected for will become more common. For example, if resistance to a particular herbicide allows weeds to live in an environment that has been sprayed with that herbicide, the allele for resistance may become more frequent in the population
  18. 18. No natural selection
  19. 19. Estimating allelic frequency If a trait is controlled by two alternate alleles, how can we calculate the frequency of each allele? For example, let us look at a sample population of pigs. The allele for black coat is recessive to the allele for white coat. Can you count the number of recessive alleles in this population?
  20. 20. Estimating allelic frequency of pigs Answer: There are 4 individuals with black coat, so it might seem that there are 8 copies of the recessive allele. In fact, some of the individuals with white coat may be heterozygous for the trait. So you cannot estimate the number of recessive alleles simply by looking at the phenotypes in the population – unless, that is, you know that the population is at Hardy-Weinberg equilibrium. If that is the case, then you can determine the frequencies of alleles and genotypes by using what is called the Hardy-Weinberg equation.
  21. 21. The Hardy-Weinberg equation To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation. According to this equation: p = the frequency of the dominant allele (represented here by A) q = the frequency of the recessive allele (represented here by a) For a population in genetic equilibrium: p + q = 1.0 (The sum of the frequencies of both alleles is 100%.) (p + q)2 = 1 so p2 + 2pq + q2 = 1 The three terms of this binomial expansion indicate the frequencies of the three genotypes: p2 = frequency of AA (homozygous dominant) 2pq = frequency of Aa (heterozygous) q2 = frequency of aa (homozygous recessive)
  22. 22. Sample Problem 1 Let's return to our population of pigs. Remember that the allele for black coat is recessive. We can use the Hardy-Weinberg equation to determine the percent of the pig population that is heterozygous for white coat. Calculate q2 Count the individuals that are homozygous recessive in the illustration above. Calculate the percent of the total population they represent. This is q2.
  23. 23. q2 is: Four of the sixteen individuals show the recessive phenotype, so the correct answer is 25% or 0.25.
  24. 24. Find q Find q Take the square root of q2 to obtain q, the frequency of the recessive allele.
  25. 25. q is: q = 0.5
  26. 26. Find p The sum of the frequencies of both alleles = 100%, p + q = l. You know q, so what is p, the frequency of the dominant allele?
  27. 27. p is: p = 1 - q, so p = 0.5
  28. 28. Find 2pq Find 2pq The frequency of the heterozygotes is represented by 2pq. This gives you the percent of the population that is heterozygous for white coat:
  29. 29. 2pq is: 2pq = 2(0.5) (0.5) = 0.5 , so 50% of the population is heterozygous.
  30. 30. Sample Problem 2 In a certain population of 1000 fruit flies, 640 have red eyes while the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye colour? Hint: The first step is always to calculate q2! Start by determining the number of fruit flies that are homozygous recessive. If you need help doing the calculation, look back at the Hardy-Weinberg equation.
  31. 31. Solution You should expect 160 to be homozygous dominant. Calculations: q2 for this population is 360/1000 = 0.36 q = = 0.6 p = 1 - q = 1 - 0.6 = 0.4 The homozygous dominant frequency = p2 = (0.4)(0.4) = 0.16. Therefore, you can expect 16% of 1000, or 160 individuals, to be homozygous dominant.
  32. 32. Sample Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a human population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic disorder that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10,000 babies is born with the disorder. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?
  33. 33. Solution Answer:Approximately 2% of the U.S. population carries the PKU allele. Calculation: q2= 1/10,000 = 0.0001 q = = 0.01 p = 1 - q = 1 - 0.01 = 0.99 The carriers are heterozygous. Therefore, 2pq = 2 (0.99) (0.01) = 0.0198= 1.98%
  34. 34. Allelic frequency vs. Genotypic frequency Allelic Frequency If you are told that the frequency of a recessive allele in a population is 10%, you are directly given q, since by definition q is the frequency of the recessive allele. This comprises all the copies of the recessive allele that are present in heterozygotes as well as all the copies of the allele in individuals that show the recessive phenotype. What is q for this population?
  35. 35. Answer q = 0.1
  36. 36. Allelic frequency vs. Genotypic frequency Genotypic Frequency Genotypic frequency is the frequency of a genotype — homozygous recessive, homozygous dominant, or heterozygous — in a population. If you don't know the frequency of the recessive allele, you can calculate it if you know the frequency of individuals with the recessive phenotype (their genotype must be homozygous recessive).
  37. 37. Sample Problem If you observe a population and find that 16% show the recessive trait, you know the frequency of the aa genotype. This means you know q2. What is q for this population?
  38. 38. Answer q is the square root of 0.16 = 0.4
  39. 39. Class Quiz 1(c)The question tells you that p = 0.9 and q = 0.1. From this, you can calculate the heterozygotes: 2pq = 2 (0.9) (0.1) = 0.18. If you selected e as your response, you may have confused the allelic frequency with genotypic frequency. This problem gives you the allelic frequency of a, which is 10%. 2(b) The conditions described all contribute to genetic equilibrium, where it would be expected for initial gene frequencies to remain constant generation after generation. If you chose e, remember that genetic equilibrium does not mean that the frequency of A = the frequency of a. 3(d) Like question 2, this question is intended to emphasise the point that the initial frequency of alleles has nothing to do with genetic equilibrium. 4(d) Where q2 = 0.09, so q = 0.3. p = 1 - q, so p = 1 - 0.3 = 0.7 AA = q2 = 0.49 5(d) Where q2 = 0.16; q = 0.4 p = 1 - q, so p = 0.6 = 60%
  40. 40. Cohen Syndrome is a developmental disorder inherited as an autosomal recessive trait. http://www.cbsnews.com/video/watch/? id=700552n&tag=related;photovideo
  41. 41. Ellis-Van Creveld Syndrome Ellis-van Creveld is passed down through families (inherited). It is caused by defects in one of two Ellis van Creveld syndrome genes (EVC and EVC2) that are next to each other. The disease is autosomal-recessive The severity of the disease varies from person to person. The highest rate of the condition is seen among the Old Order Amish population of Lancaster County, Pennsylvania. It is fairly rare in the general population.
  42. 42. Consanguinity Consanguinity means descent from a common ancestor; a consanguineous couple is usually defined as being related as second cousins or closer. The word derives from ‘con’+ ‘sanguine’ – from the Latin, meaning ‘of the same blood’. Consanguinuous marriage today is most prevalent in communities originating from North Africa, the Middle East, and large parts of Asia. In the British Pakistani community it is estimated that 50-60% of marriages are consanguineous, and there is evidence that this proportion is rising. Geographical or social isolation of migrant groups may play a part in this. http://www.youtube.com/watch?v=Swadss8D8zw