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Population Genetics

Population Genetics . Hardy Weinberg Equilibrium. 6.1 Mendelian Genetics in Populations:. The Hardy-Weinberg Equilibrium Principle. Population Genetics.

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Population Genetics

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  1. Population Genetics Hardy Weinberg Equilibrium

  2. 6.1 Mendelian Genetics in Populations: The Hardy-Weinberg Equilibrium Principle

  3. Population Genetics • Population genetics is concerned with the question of whether a particular allele or genotype will become more common or less common over time in a population, and Why. • Example: • Given that the CCR5-D32 allele confers immunity to HIV, will it become more frequent in the human population over time?

  4. Predicting Allele Frequencies Populations in Hardy-Weinberg equilibrium

  5. Yule vs. Hardy • What are the characteristics of a population that is in equilibrium or another words, not evolving. • Yule thought that allele frequencies had to be 0.5 and 0.5. for a population to be in equilibrium. • Hardy proved him wrong by developing the Hardy-Weinburg equation.

  6. Punnett square • 60 % of the eggs carry allele A and 40% carry allele a • 60% of sperm carry allele A and 40% carry allele a.

  7. Sample problem • In a population of 100 people, we know that 36% are AA , 48%are Aa, and 16% are aa. • Determine how many alleles in the gene pool are A or a. • Each individual makes two gametes.... • How many A alleles are in this population’s gene pool? _____ • How many a alleles? _____ (36*(2)+48) 120 (16*(2) +48) 80

  8. What percent of the alleles are A or a ? 120 / 200 = .6 or 60% A ; or .6 = frequency of allele A 80 / 200 = .4 or 40% a ; or .4 = frequency of allele a

  9. Creating the Hardy-Wienburg equation is a matter of combining probabilities found in the Punnett square.

  10. Combining Probabilities • The combined probability of two independent events will occur together is equal to the product of their individual probabilities. • What is the probability of tossing a nickel and a penny at the same time and having them both come up heads? • ½ x ½ = ¼

  11. Combining Probabilities • The combined probability that either of two mutually exclusive events will occur is the sum of their individual probabilities. When rolling a die we can get a one or a two (among other possibilities), but we cannot get both at once. Thus, the probability of getting either a one or a two is • 1/6 + 1/6 = 1/3

  12. Calculating Genotype Frequencies • We can predict the genotype frequencies by multiplying probabilities.

  13. Hardy-Weinburg equation Genotype Frequencies

  14. Zygotes Allelic frequency Genotype frequency AA (p)(p) p2 Aa (p)(q) 2pq aA (q)(p) aa (q)(q) q2 Genotype frequencies described by p2+2pq+q2=1.0

  15. The relationship between allele and genotype frequency • Let original A frequency be represented by p and original a frequency be represented by q • Since there are only two alleles possible for this gene locus, The frequencies of A and a must equal 1.0 • Therefore, p + q =1.0

  16. Sample: calculating genotype frequencies from allele frequencies? If a given population had the following allele frequencies: allele frequency (p) for A of 0.8 allele frequency (q) for a of 0.2 Determine the genotype frequencies of this population? AAAaaa 0.04 0.64 0.32 AA = p2 ; Aa = 2pq ; and aa = q2 as follows…

  17. We can also calculate the frequency of alleles from the genotype frequencies. When a population is in equilibrium the genotype frequencies are represented as.. P2 + 2pq +q2 The allele frequency can therefore be calculated as follows. A = p2 + ½(2pq) and a = q2 + ½(2pq)

  18. Examining our example again we see that if we use the frequencies we calculated for each genotype…. p2 2pqq2 0.64 AA .32 Aa .04 aa A = p2 + ½ (2pq) A=.64 + ½ (.32) A = 0.8 and since q = 1-p ; then a = 1-(0.8 ) a = 0.2

  19. These rules hold as long as a population is in equilibrium. Hardy Weinberg Equilibrium describes the conclusions and assumptions that must be present to consider a population in equilibrium.

  20. Hardy Weinberg Conclusions • The allele frequencies in a population will not change from generation to generation. You would need at least 2 generations of data to demonstrate this. • If the allele frequencies in a population are given by p and q then the genotype frequencies will be equal top2; 2pq ; q2. Therefore if AA can not be predicted by p2Aa cannot be predicted by 2pq and aa cannot be predicted by q 2 then the population is not in equilibrium

  21. There are 5 assumptions which must be met in order to have a population in equilibrium • There is no selection. In other words there is no survival for one genotype over another • There is no mutation. This means that none of the alleles in a population will change over time. No alleles get converted into other forms already existing and no new alleles are formed • There is no migration (gene flow)New individuals may not enter or leave the population. If movement into or out of the population occurred in a way that certain allele frequencies were changed then the equilibrium would be lost

  22. Exceptions to Hardy Weinberg cont. • There are no chance events(genetic drift) This can only occur if the population is sufficiently large to ensure that the chance of an offspring getting one allele or the other is purely random. When populations are small the principle of genetic drift enters and the equilibrium is not established or will be lost as population size dwindles due to the effects of some outside influence • There is no sexual selection or mate choice Who mates with whom must be totally random with no preferential selection involved.

  23. Problem #6 on page 219 • Go to your text page 219 and answer question number 6.

  24. In humans, the COL1A1 locus codes for a certain collagen protein found in bone. The normal allele at the locus is denoted with S. A recessive allele s is associated with reduced bone mineral density and increased risk of fractures in both Ss and ss women. A recent study of 1,778 women showed that 1,194 were SS, 526 were Ss, and 58 were ss. • Are these two alleles in Hardy-Weinberg equilibrium in this population? • What information would you need to determine whether the alleles will be in Hardy-Weinberg equilibrium in the next generation?

  25. Problem approach • Check that conclusion #2 holds • First figure genotype frequencies from the data (percentages) • Then from the data, count the actual A alleles in the population and the actual a alleles in the population. What are their frequencies? • Then calculate the predicted genotype frequencies of Hardy Weinberg and compare to actual numbers.

  26. The genotype frequencies are: SS =1194/1778 = .67 Ss=526/1778 = .30 ss= 58/1778 = .03 Calculate allele frequencies from genotype frequencies 2914/ 3556 S alleles = 0.82S frequency or .67+1/2(.30) = 0.82 642 / 3556 s alleles = 0.18s frequency or .03 + ½ (.30)=0.18 If the population is in Hardy-Weinberg equilibrium, the allele frequencies should predict the genotype frequencies. SS genotypefrequency would be (0.82)2,= 0.67; Ss frequency would be 2 (.82) (.18), = 0.30; ss frequency would be (0.18)2,= 0.03. These numbers almost exactly match the measured genotype frequencies - so this population may be in Hardy-Weinberg equilibrium

  27. However, what also must we know to be sure? • We would need to check in future generations to make sure that the allele frequencies are not changing. • So here we confirmed conclusion #2 but have not yet verified conclusion#1.

  28. Example

  29. Example

  30. Example

  31. Example

  32. Example

  33. 0 Using the Hardy-Weinberg equilibrium with more than two alleles • In a population of mice, coat color is determined by 1 locus with 4 alleles: A, B, C, and D. The possession of an A allele confers black coat color with another A allele, or a D allele. If a B allele is present with an A allele then coat color is brown, and if C is present with and A allele, coat color is grey. All other phenotypes are light tan. Given that the frequencie of the A, B, and C alleles are .05, .4, and .3, respectivelly, what are the phenotpic frequencies of black, brown, grey and light tan mice when the population is in Hardy-Weinberg equilibrium?

  34. 0 Adding Selection to the Hardy-Weinberg Analysis • How do you know if a population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age. -or- 2. Equal numbers of individuals from each genotype reach reproductive age but some genotypes are able to produce more viable (reproductively successful) offspring. If these differences are heritable then evolution may occur over time.

  35. Caution • most phenotypes are not strictly the result of their genotypes. • Environmental plasticity and • interaction with other genes may also be involved. • not as simple as we are making it here.

  36. We will look at two possible effects of natural selection on the gene pool • Selection may alter allele frequencies or violate conclusion #1 • Selection may upset the relationship between allele frequencies and genotype frequencies. Conclusion #1 is not violated but conclusion #2 is violated.

  37. An example of what we might see happen to allele frequencies when natural selection is at work

  38. Let B1 and B2 = the allele frequencies of the initial populationwith frequencies of B1 = .6 and B2 = .4 • After random mating which produces 1000 zygotes we get:

  39. Selection Example

  40. Selection Example

  41. Selection Example

  42. Selection Example

  43. Selection Example

  44. Selection Example

  45. Selection Example

  46. Selection Example

  47. Selection Example

  48. Creating an equation that allows for selection • First, weanalyze the population on the basis of the fitness of the offspring. • Fitness (w) is defined as the survival rates, or percentage of individuals which survive to reproduce. • We can use the fitness (w) of each genotype to calculate the average fitness of the population

  49. Fitness formulas MEAN FITNESS • If : w11 = fitness of allele #1 homozygote w12 = fitness of the heterozygote w22 = fitness of allele #2 homozygote mean fitness of the population will be described by the formula: ŵ = p2w11 + 2pqw12 + q2w22

  50. For our previous example • B1= 0.6 and B2 = 0.4 and fitness of B1B1 = 1.0 (100% survived) fitness of B1B2 = .75 ( 75% survived) fitness of B2B2 = .50 (50% survived) • Figure the mean fitness now.

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