1. Announcements -Pick up Problem Set 3 from the front -Ave.- 41.7 SD- 4.1 Group work -Amy is lecturing on Thursday on Methods in Quantitative Genetics -Test Oct. 22

2. Introduction to Quantitative Genetics 10/6/09 Quantitative Genetics is the statistic relationship between genotype and phenotype. It does not concern itself with the mechanistic theory of how genotype (interacting with environment) produces phenotype (developmental genetics). Only statistical relationships are important, because evolution can only act on genotypes that are associated with particular phenotypes. Population parameters such as allele frequency and mating factors can strongly effect these statistical associations!

3. Physical Basis of Evolution DNA can replicate DNA can mutate and recombine DNA encodes information that interacts with the environment to influence phenotype

4. Goal of Today Derive a measure to describe the phenotype that each individual gamete contributes on to the next generation relative to the average phenotype. Two of these measures (Average Excess and Average Effect)

5. Mendelism After 1908 Hardy-Weinberg Helped Establish That Many Traits Were Mendelian Still, the Majority of Quantitative Traits Could Not Be Put Into A Mendelian Framework Most Traits Were Quantitative Therefore, Many Believed That An Alternative and More Important Mechanism of Heredity Existed in Addition to Mendelism

6. Ronald A. Fisher By Age 22 (1912) was laying the basis for much of modern statistics, but could not get an academic position In 1913 Became Convinced of Mendelism In 1916 submitted a paper that explained the inheritance of quantitative traits through Mendelian factors

7. Fisher’s 1916 Paper: The Correlation Between Relatives On The Supposition of Mendelian Inheritance Was rejected by several journals Fisher personally paid to have it published by the Royal Society of Edinburgh in 1918 When published, it ended all serious opposition to Mendelism It established the genetic basis for natural selection It laid the foundation for modern plant and animal breeding, and genetic epidemiology It presented new statistical techniques (e.g., ANOVA -- ANalysis Of VAriance) soon used by virtually all empirical sciences.

8. Two Basic, Non-Mutually Exclusive Ways of Having Discrete Genotypes Yield Continuous Phenotypes 1.Polygenes 2.Environmental Variation

9. Polygenes Fewer Loci More Loci

10. Environmental Variation

11. Most Traits Are Influenced By Both Many Genes and Environmental Variation: Frequently Results in a Normal Distribution. E.g. Cholesterol in Framingham, MA Relative Frequency in Population Total Serum Cholesterol in mg/dl 150220 290

12. The Normal Distribution Can Be Completely Described by Just 2 Numbers: The Mean (  ) and Variance (   )

13. Let x be an observed trait value The mean (  ) is the average or expected value of x. The mean measures where the distribution is centered If you have a sample of n observations, x 1, x 2, …, x n, Then  is estimated by:

14. The variance (   ) is the average or expected value of the squared deviation of x from the mean; that is, (x  ) 2 The variance measures the amount of dispersion in the distribution (how “fat” the distribution is) If you have a sample of n observations, x 1, x 2, …, x n, Then   given  is estimated by: s 2 = [(x 1 -  ) 2 + (x 2 -  ) 2 + … + (x n -  ) 2 ]/n If you do not know  then   is estimated by:

15. By 1916, Fisher Realized 1.Could Examine Causes of Variation, but not cause and effect of quantitative phenotypes. 2.Therefore, what is important about an individual’s phenotype is not its value, but how much it deviates from the average of the population; That is, focus is on variation. 3.Quantitative inheritance could not be studied in individuals, but only among individuals in a population.

16. Fisher’s Model P ij =  + g i + e j The mean (average) phenotype for The entire population:  i  j P ij /n Where n is the number of individuals sampled.

17. Fisher’s Model P ij =  + g i + e j The genotypic deviation for genotype i is the Average phenotype of genotype i minus the Average phenotype of the entire population: g i =  j P ij /n i -  Where n i is the number of individuals with genotype i.

18. Fisher’s Model P ij =  + g i + e j The environmental deviation is the deviation Of an individual’s phenotype from the Average Phenotype of his/her Genotype: e j =P ij -  j P ij /n i = P ij -(g i +  )=P ij -  -g i

19. P ij =  + g i + e j Fisher’s Model Although called the “environmental” deviation, e j is really all the aspects of an individual’s Phenotype that is not explained by genotype in This simple, additive genetic model.

20. Fisher’s Model  2 p = Phenotypic Variance  2 p = Average(P ij -    2 p = Average(g i + e j ) 2

21. Fisher’s Model  2 p = Average(g i + e j ) 2  2 p = Average(g i 2 + 2g i e j + e j 2 )  2 p = Average(g i 2 ) + Average(2g i e j ) + Average(e j 2 )

22. Fisher’s Model  2 p = Average(g i 2 ) + Average(2g i e j ) + Average(e j 2 ) Because the “environmental” deviation is really all the aspects of an individual’s Phenotype that is not explained by genotype, This cross-product by definition has an average Value of 0.

23. Fisher’s Model  2 p = Average(g i 2 ) + Average(e j 2 )  2 p =  2 g +  2 e Phenotypic Variance

24. Fisher’s Model  2 p = Average(g i 2 ) + Average(e j 2 )  2 p =  2 g +  2 e Genetic Variance

25. Fisher’s Model  2 p = Average(g i 2 ) + Average(e j 2 )  2 p =  2 g +  2 e Environmental Variance (Really, the variance not Explained by the Genetic model)

26. Fisher’s Model  2 p =  2 g +  2 e Phenotypic Variance = Genetic Variance + Unexplained Variance In this manner, Fisher partitioned the causes Of phenotypic variation into a portion explained By genetic factors and an unexplained portion.

27. Fisher’s Model  2 p =  2 g +  2 e Phenotypic Variance = Genetic Variance + Unexplained Variance This partitioning of causes of variation can only be Performed at the level of a population. An individual’s phenotype is an inseparable Interaction of genotype and environment.

28. ApoE and Cholesterol in a Canadian Population Total Serum Cholesterol (mg/dl) 3/4 3/3 2/3 4/4 2/2 2/4 Relative Frequency  = 174.6  2 p = 732.5

29.  2 0.078  3 0.770  4 0.152 Random Mating Geno- type 3/33/23/42/22/44/4 H-W Freq. 0.5920.1210.2340.0060.0240.023 Mean Pheno. 173.8161.4183.5136.0178.1180.3

30. Geno- type 3/33/23/42/22/44/4 H-W Freq. 0.5920.1210.2340.0060.0240.023 Mean Pheno. 173.8161.4183.5136.0178.1180.3 Step 1: Calculate the Mean Phenotype of the Population  = (0.592)(173.8)+(0.121)(161.4)+(0.234)(183.5)+(0.006)(136.0)+(0.024)(178.1)+(0.023)(180.3)  = 174.6

31. Geno- type 3/33/23/42/22/44/4 Mean Pheno. 173.8161.4183.5136.0178.1180.3 gigi 173.8-174.6 -0.8 161.4-174.6 -13.2 183.5-174.6 8.9 136.0-174.6 -38.6 178.1-174.6 3.5 180.3-174.6 5.7 Step 2: Calculate the genotypic deviations  = 174.6

32. Geno- type 3/33/23/42/22/44/4 H-W Freq. 0.5920.1210.2340.0060.0240.023 gigi -0.8-13.28.9-38.63.55.7 Step 3: Calculate the Genetic Variance  2 g = 50.1  2 g = (0.592)(-0.8) 2 +(0.121)(-13.2) 2 +(0.234)(8.9) 2 +(0.006)(-38.6) 2 +(0.024)(3.5) 2 +(0.023)(5.7) 2

33. Step 4: Partition the Phenotypic Variance into Genetic and “Environmental” Variance  2 g 50.1  2 e 682.4  2 p = 732.5

34. Broad-Sense Heritability h 2 B is the proportion of the phenotypic variation that can be explained by the modeled genetic variation among individuals.

35. Broad-Sense Heritability For example, in the Canadian Population for Cholesterol Level h 2 B = 50.1/732.5 = 0.07 That is, 7% of the variation in cholesterol levels in this population is explained by genetic variation at the ApoE locus.

36. Broad-Sense Heritability Genetic Variation at the ApoE locus is therefore a cause of variation in cholesterol levels in this population. ApoE does not “cause” an individual’s cholesterol level. An individual’s phenotype cannot be partitioned into genetic and unexplained factors.

37. Broad-Sense Heritability Measures the importance of genetic Variation as a Contributor to Phenotypic Variation Within a Generation The more important (and difficult) question Is how Phenotypic Variation is Passed on to The Next Generation.

38. 3/3 0.592 3/2 0.121 3/4 0.234 2/2 0.006 2/4 0.024 4/4 0.023 Deme Development Environment  = 174.6  2 p = 732.5 h2Bh2B  4 0.152  3 0.770  2 0.078 4/4 0.023 2/4 0.024 2/2 0.006 3/4 0.234 3/2 0.121 3/3 0.592 Random Mating Meiosis Gene Pool Deme Environment ? Development

39. Fisher’s Model 1.Assume that the distribution of environmental deviations (e j ’s) is the same every generation 2.Assign a “phenotype” to a gamete

40. Phenotypes of Gametes 1.Average Excess of a Gamete Type 2.Average Effect of a Gamete Type 3.These two measures are identical in a random mating population, so we will consider only the average excess for now.

41. The Average Excess The Average Excess of Allele i Is The Average Genotypic Deviation Caused By A Gamete Bearing Allele i After Fertilization With A Second Gamete Drawn From the Gene Pool According To The Deme’s System of Mating.

42. The Average Excess Where g ij is the genotypic deviation of genotype ij, t ij is the frequency of ij in the population (not necessarily HW), p i is the frequency of allele i, and:

43. The Average Excess Note, under random mating t ii = p i 2 and t ij = 2p i p j, so:

44.  2 0.078  3 0.770  4 0.152 Random Mating Gene Pool Deme 3/3 0.592 3/2 0.121 3/4 0.234 2/2 0.006 2/4 0.024 4/4 0.023 Average Excess of An Allele

45.  2 0.078 Gene Pool Average Excess of An Allele What Genotypes Will an  2 allele find itself in after random mating?

46.  2 0.078 Random Mating Gene Pool Deme 3/22/22/4 Average Excess of An Allele What are the probabilities of these Genotypes after random mating given an  2 allele?

47.  2 0.078  3 0.770  4 0.152 Random Mating Gene Pool Deme 3/2 0.770 2/2 0.078 2/4 0.152 Average Excess of An Allele These are the Conditional Probabilities of the genotypes Given random mating and a gamete with the  2 allele.

48.  2 0.078  3 0.770  4 0.152 Random Mating Gene Pool Deme 3/2 0.770 2/2 0.078 2/4 0.152 Average Excess of An Allele Environment h2Bh2B Development Genotypic Deviations -13.2-38.63.5 Average Genotypic Deviation of a  2 bearing gamete = (0.770)(-13.2)+(0.078)(-38.6)+(0.152)(3.5) = -12.6

49.  2 0.078  3 0.770  4 0.152 Random Mating Gene Pool Deme 3/3 0.770 3/2 0.078 3/4 0.152 Average Excess of Allele  3 Environment h2Bh2B Development Genotypic Deviations -0.8-13.28.9 Average Excess of  3 = (0.770)(-0.8)+(0.078)(-13.2)+(0.152)(8.9) = -0.3

50.  2 0.078  3 0.770  4 0.152 Random Mating Gene Pool Deme 3/4 0.770 2/4 0.078 4/4 0.152 Average Excess of Allele  4 Environment h2Bh2B Development Genotypic Deviations 8.93.55.7 Average Excess of  4 = (0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0

51.  2 0.078 -12.6  3 0.770 -0.3  4 0.152 8.0 Gene Pool Alleles Frequencies “Phenotype” (Average Excess) The critical breakthrough in Fisher’s paper was assigning a “phenotype” to a gamete, the physical basis of the transmission of phenotypes from one generation to the next.

52. Average Excess of  4 = (0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0 The Average Excess Depends Upon the Genotypic Deviations, which in turn Depend Upon the Average Phenotypes of the Genotypes and And the Average Phenotype of the Deme, which in turn Depends Upon The Genotype Frequencies.

53. Average Excess of  4 = (0.770)(8.9)+(0.078)(3.5)+(0.152)(5.7) = 8.0 The Average Excess Depends Upon the Gamete Frequencies in the Gene Pool and Upon the System of Mating.

54. The Average Excess The Portion of Phenotypic Variation That Is Transmissible Through a Gamete Via Conditional Expectations

55. Purposes Population Genetics vs. Quantitative Genetics Populations Genetics -Concerned pop gen forces -Average Excess- understand NS Quantitative Genetics -Concerned with Breeding Values -Measurable values before genetics -Average Effect (of the gene)= Average Effect (Alan) -Average Effect (of allelic substitution)

56. The Average Effect The Portion of Phenotypic Variation That Is Transmissible Through a Gamete Measured At the Level of a Deme and Its Associated Gene Pool via Least-Squares Regression.

57. The Average Effect Derived in Quant. Gen. terms with RM assumption Jim’s Definition- The mean phenotypic deviation of a gamete from the population mean when paired with another allele at random. AND The additive contribution of a gamete to the next generation

58. The Average Effect Templeton (1987) showed:

59. Fisher’s Model The Next Step Is To Assign a “Phenotypic” Value To a Diploid Individual That Measures Those Aspects of Phenotypic Variation That Can be Transmitted Through the Individual’s Gametes. Breeding Value or Additive Genotypic Deviation Is The Sum of the Average Effects (=Average Excesses Under Random Mating) of Both Gametes Borne By An Individual.

60. Additive Genotypic Deviation Let k and l be two alleles (possibly the same) at a locus of interest. Let  k be the Average Effect of allele k, and  l the Average Effect of allele l. Let g akl be the additive genotypic deviation of genotype k/l. Then: g akl =  k +  l

61. Geno- type 3/33/23/42/22/44/4 H-W Freq. 0.5920.1210.2340.0060.0240.023 gigi -0.8-13.28.9-38.63.55.7  2 0.078 -12.6  3 0.770 -0.3  4 0.152 8.0 Alleles Frequencies Average Excess =Effect (rm) g ai -0.3+(-0.3) -0.6 - 0.3+(-12.6) -12.9 -0.3+8.0 7.7 -12.6 -25.4 -12.6+8.0 -4.6 8.0 + 8.0 16.0

62. Geno- type 3/33/23/42/22/44/4 H-W Freq. 0.5920.1210.2340.0060.0240.023 gigi -0.8-13.28.9-38.63.55.7 g ai -0.6-12.97.7-25.4-4.616.0 The Additive Genetic Variance  2 a = 44.7  2 a =(0.592)(-0.6) 2 +(0.121)(-12.9) 2 +(0.234)(7.7) 2 +(0.006)(-25.4) 2 +(0.024)(-4.6) 2 +(0.023)(16.0) 2

63. The Additive Genetic Variance Note that  2 g = 50.1 >  2 a = 44.7 It is always true that  2 g >  2 a Have now subdivided the genetic variance into a component that is transmissible to the next generation and a component that is not:  2 g =  2 a +  2 d

64. The Additive Genetic Variance  2 g =  2 a +  2 d The non-additive variance,  2 d, is called the “Dominance Variance” in 1-locus models. Mendelian dominance is necessary but not sufficient for  2 d > 0.  2 d depends upon dominance, genotype frequencies, allele frequencies and system of mating.

65. The Additive Genetic Variance For the Canadian Population,  2 g = 50.1 and  2 a = 44.7 Since  2 g =  2 a +  2 d 50.1 = 44.7 +  2 d  2 d = 50.1 - 44.7 = 5.4

66. Partition the Phenotypic Variance into Additive Genetic, non-Additive Genetic and “Environmental” Variance  2 g 50.1  2 p = 732.5  2 e 682.4  2 a 44.7  2 e 682.4  2 d 5.4

67. The Additive Genetic Variance  2 g =  2 a +  2 d +  2 i In multi-locus models, the non-additive variance is divided into the Dominance Variance and the Interaction (Epistatic) Variance,  2 i. Mendelian epistasis is necessary but not sufficient for  2 i > 0.  2 i depends upon epistasis, genotype frequencies, allele frequencies and system of mating.

68. The Partitioning of Variance  2 p =  2 a +  2 d +  2 i +  2 e As more loci are added to the model,  2 e goes down relative to  2 g such that h B 2 = 0.65 for the phenotype of total serum cholesterol in this population. Hence, ApoE explains about 10% of the heritability of cholesterol levels, making it the largest single locus contributor.

69. (Narrow-Sense) Heritability h 2 is the proportion of the phenotypic variance that can be explained by the additive genetic variance among individuals.

70. (Narrow-Sense) Heritability For example, in the Canadian Population for Cholesterol Level h 2 = 44.7/732.5 = 0.06 That is, 6% of the variation in cholesterol levels in this population is transmissible through gametes to the next generation from genetic variation at the ApoE locus.