Lecture 1 Introduction to Quantitative Genetics

Transcription

1 Lecture Introduction to Quantitative Genetics Population Genetics Foundation Quantitative Genetics Quantitative Traits Continuous variation Varies by amount rather than kind Height, weight, IQ, etc What is the Basis for Quantitative Variation?

2 Mendelian bases for Quantitative Genetics Early experiments by Nilsson-Ehle (98) Wheat color Parents. F Dark red med DR medium red pale red white Dark red med DR medium red pale red white F Blending? Dark red med DR medium red What Mode of Inheritance Would Explain This?.5 pale red.65 white Would Blending Explain this? 4

3 Hypothesis: loci acting independently and cumulatively on one trait? Dark Red AABB X White aabb Medium Red AaBb X Medium Red AaBb Dark Red AABB Medium Dark Red AABb AaBB Medium Red AaBb AAbb aabb Pale Red Aabb aabb White aabb /6 4/6 6/6 4/6 /6 5 Gene Effects Usual Mendelian Concept Gene Trait Gene Trait Simple Traits Gene Trait Trait Pleiotropy Genetic Correlation Between Traits Gene Gene Trait Polygenic Trait 6 3

4 What happens to the distribution as the number of loci increases? A continuous distribution emerges 7 Stability of Distribution From Previous Example with Wheat F Dark red med DR medium red.5 pale red.65 white What is the expected Distribution of the F3? F4? Need to know concepts of probability 8 4

5 Probability Important Concepts Independence Mutual Exclusivity 9 Compound Events Two Events are Independent if Knowledge of one event tells us nothing about the probability of occurrence of the other event Pr(This AND That) Pr(A and B)=Pr(A B)xPr(B) With Independence Pr(A B)=Pr(A) Thus Pr(A and B)=Pr(A)xPr(B) Pr(This given that) Pr(A B)=Pr(A and B)/Pr(B) With Independence Pr(A and B)=Pr(A)xPr(B) Thus Pr(A B)=Pr(A) Only for the case of Independence 5

6 Independence By Design Factor A Mutli-factor experiment Want to estimate effects INDEPENDENT of the other Factorial experiment Each level of each factor occurs equally in each other factor The Effect of A can be estimated independent of the effect of B B B B A x x A x x A3 x x Independence By Nature Factors are naturally independent Loci inherited on DIFFERENT chromosomes The probability of an allele being inherited at the first locus is independent of what is inherited at the other locus What is the Pr(A B ) gamete Gametes produced by double heterozygote A B /A B Unlinked Loci Locus B B B Locus A A A A B A B A B A B 6

7 Independence Pr(This AND That) Pr(A and B)=Pr(A B)xPr(B) Pr(A and B )= Pr(A B )xpr(b ) Pr(A B )=Pr(A )=/ Gametes produced by double heterozygote A B /A B Unlinked Loci Locus A A A ½ ½ Pr(A and B )= Pr(A )xpr(b ) Pr(A and B )= ½x ½=/4 Locus B B ½ A B A B ¼ B ½ A B ¼ A B ¼ 3 Non-Independence By Poor Design or missing data The Effect of Factor A for some combinations of B is dependent of the level of B A3 is confounded with the effect of B A is confounded with the effect of B If B were an unknown, unaccounted for factor, could lead to spurious correlation Factor A may not have any effect, all results could be due to unseen Factor B effects Example: Factor A genotypic frequencies at A locus, Factor B proportion of different subpopulations B and B Factor B B B A x A A x x A3 x 4 7

8 Non-Independence By Nature Factors are naturally dependent Loci inherited on SAME chromosomes perfectly linked If the allele inherited at the A locus was A then the allele at the second locus must be B and vice versa What is the Pr(A B ) gamete? Gametes produced by double heterozygote A B /A B Linked Loci Locus B B ½ B ½ Locus A A A ½ ½ A B A B A B A B 5 Non-Independence Pr(This AND That) Pr(A and B)=Pr(A B)xPr(B) Pr(A and B )= Pr(A B )xpr(b ) Pr(A B )= Pr(A and B )= Pr(A )x Pr(A and B )= ½ x =/ Pr(A B )= Gametes produced by double heterozygote A B /A B Linked Loci Locus B B ½ B ½ Locus A A ½ A B A B A ½ A B A B ½ 6 8

9 Multiple Events This OR That Requires knowledge of Exclusivity and Independence Pr(A or B)=Pr(A)+Pr(B)-P(A and B) General Formula Mutually Exclusive Events The occurrence of one event excludes the other P(A and B)= Pr(A or B)=Pr(A)+Pr(B) Not mutually exclusive but independent Pr(A or B)=Pr(A)+Pr(B)-P(A)P(B) Not mutually exclusive nor independent Pr(A or B)=Pr(A)+Pr(B)-P(A)P(B A) 7 There were 4 birds of each of 4 comb types in a pen of 96 birds, ½ of each comb type are black, the other white. Chose bird. What is the probability of choosing either a Single or Rose comb bird? Single Rose Walnut Pea Single 4 Rose 4 P(Single OR Rose)=Pr(Single)+Pr(Rose)-Pr(Single and Rose) P(Single or Rose)=(4/96)+(4/96)-(/96)=48/96 Mutually exclusive 8 9

10 There were 4 birds of each comb type in a pen of 96 birds, ½ of each comb type are black, the other white. What is the probability of choosing either a Black bird or one with a Single comb? Single Rose Walnut Pea Black 48 Single 4 Independence by Design P(Black OR Single)=Pr(black)+Pr(single)-Pr(black and single) Total 96 P(Black OR Single)=(48/96)+(4/96)-(48/96)(4/96)=6/96 Example of not mutually exclusive but independent events 9 There were 4 birds of each of 4 comb types in a pen of 96 birds, ½ of each comb type are black, the other white. Chose birds with replacement. What is the probability of choosing at least one rose comb bird? Single Rose Walnut Pea Easy way and hard way P(B=Rose or B=Rose) -P(B=not Rose and B=not Rose) P(Rose)+P(Rose)-P(Rose and Rose) -¾ x ¾ = -9/6 = 7/6 /4 +/4-(/4)(/4)= 7/6 Example of not mutually exclusive but independence events

11 There were 4 birds of each comb type in a pen of 96 birds, ½ of each comb type are black, the other white. All black birds with pea combs died. What is the probability of choosing either a Black bird or one with a Single comb? Single Rose Walnut Pea White 48 Black 36 Single 4 Independence by Design P(Black OR Single)=Pr(black)+Pr(single)-Pr(black and single) Total 84 P(Black OR Single)=(36/84)+(4/84)-(36/84)(/36)=3/6=48/84 Example of not mutually exclusive nor independent events Application of Probabilities to Populations: Mating Assume 3 genotypes at the A locus AA, Aa, aa Put a female of genotype AA in a large population of males with frequencies of the genotypes P(AA)=x P(Aa)=y P(aa)=z x+y+z= What is the probability she will mate with a given genotype of male

12 Mating Preferences Independence (no preference) P(AA male AA female)=x P(Aa male AA female)=y P(aa male AA female)=z Extreme negative assortative P(AA male AA female)= P(Aa male AA female)= P(aa male AA female)= Extreme positive assortative P(AA male AA female)= P(Aa male AA female)= P(aa male AA female)= 3 What is the Consequence of Random Mating on Genotypic Frequencies Assume a Perfect World No Forces Changing Gene Frequency Equal Gene Frequencies in the Sexes Autosomal Inheritance Random Mating 4

13 GENERATION Allow The Genotypic Frequencies To Be Any Arbitrary Values genotypic frequency P( AA ) = X P( Aa ) = Y P( aa ) = Z such that X + Y + Z = 5 Allele Frequencies P( A ) = P(AA) + P(Aa) P( A ) = X + Y = p P( a ) = P(aa) + P(Aa) P( a ) = Z + = q p + q = Y 6 3

14 Frequency of Mating male genotype female genotype AA Aa aa frequency ( X ) ( Y ) ( Z ) AA ( X ) X XY XZ Aa ( Y ) XY Y YZ aa ( Z ) XZ YZ Z Random Mating=independence 7 Expected genotypic frequencies that result from matings (Gen ). Possible Matings Frequency of Mating Expected Frequency of Offspring AA Aa aa AA x AA X AA x Aa XY / / AA x aa XZ Aa x Aa Y /4 / /4 Aa x aa YZ / / aa x aa Z Conditional Probabilities given genotypes of parents 8 4

15 GENERATION P( AA offsping ) =( X = X = ) + ( + XY + ( X + Y ) = XY ) + 4 Y [ P( A ) ] BECAUSE P( AA parents P( A ) = X + offspring 4 ) = p ( Y Y = p Thereforefor generation ) 9 P(Aa offsping )= =XY+XZ+ = ( XY)+(XZ)+ ( X+ Y )( Z+ Y) =pq Y +YZ (Y )+ (YZ) P( aa offsping ) = 4 = = ( Y 4 ( Z + Y ) = q Y ) + ( YZ ) +( Z + YZ + Z ) 3 5

16 Generation Frequency of Matings male genotype female genotype AA Aa aa frequency ( p ) ( pq ) ( q ) AA ( p ) p 4 p 3 q p q Aa ( pq ) p 3 q 4p q pq 3 aa ( q ) p q pq 3 q 4 3 Expected genotypic frequencies that result from matings (Gen ). Possible Matings Frequency of Mating Expected Frequency of Offspring AA Aa aa AA x AA p 4 AA x Aa 4p 3 q / / AA x aa p q Aa x Aa 4p q /4 / /4 Aa x aa 4pq 3 / / aa x aa q 4 3 6

17 Overall Genotypic Frequencies P(AA offsping 4 3 ) =( p ) + ( 4p q) = p +p q + p q = p = p = p ( p +pq+ q ) ( p + q ) P( Aa ) = offsping = p () pq ( 4 4 p q ) P( aa ) = offsping q 33 Summary of genotypic frequencies by Generation genotype gen gen gen P( AA ) X p p P( Aa ) Y pq pq P( aa ) Z q q 34 7

18 Hardy-Weinberg Equilibrium or the Squared Law If a population starts with any arbitrary distribution of genotypes, provided they are equally frequent in the two sexes, the proportions of genotypes (AA, Aa, aa) with initial gene frequencies p and q will be in the proportion ( ) p + q = p + ( pq ) + q A a AA Aa aa after one generation of random mating and will remain in that distribution until acted upon by other forces 35 Limitations If genotypic frequencies conform to the squared law, does that mean that all the assumptions hold? NO Selection can be occurring after the time the population was counted 36 8

19 What if Organisms Do not Mate But Release Gametes to Search Out Each Other? Do the Same Properties Hold? 37 Fertilization What is the probability that a male gamete (sperm) carrying the A allele Will fertilize an egg given that the egg is carrying the a allele If Independence P(A male a female)=p(a male) If not Independence P(A male a female)= to Incompatibility 38 9

20 What is the Consequence of Random Union of Gametes Assume a Perfect World No Forces Changing Gene Frequency Equal Genotypic Frequencies in the Sexes Autosomal Inheritance 39 GENERATION lets allow the genotypic frequencies to be any arbitrary value and the gene frequencies to be the appropriate function of those values. genotypic frequency P( AA ) = X P( Aa ) = Y P( aa ) = Z such that X + Y + Z = 4

21 Allele Frequency P( A ) = X + = p P( a ) = Z + = q Y Y p + q = 4 With Independence male gamete/frequency A a ( p ) ( q ) female gamete/ A AA Aa frequency ( p ) ( p ) ( pq ) a Aa aa ( q ) ( pq ) ( q ) Random Union of Gametes Produces the Same Outcome as Random Mating 4

22 Random Mating Assumptions Mates Chose Partners Independent of Genotype Random Union of Gametes Gametes Pair Independent of the Alleles Which they Carry Then Random Mating= Random Union of Gametes 43 What Happens If The Allele Frequencies Are Not Equal Between The Sexes? Generation Gametic Frequencies A Male Gamete Frequency a p m q m Female Gamete frequency A a p f q f p p p q m f f m AA Aa Genotypic Frequencies in Generation p m q f Aa q m q f aa 44

23 Gametic Frequencies Produced by Adults of First Generation Frequency of Homozygous Class + ½ frequency of heterozygous class p p m f = = p p m m p p f f + + ( p q + q p ) m ( p q + q p ) Note, gametic frequencies are now equal in sexes p m = p f = m p f f m m f f 45 Generation Generation Gametic Frequencies Female Gamete frequency A a q A Male Gamete Frequency a p p p p = ( p ) AA p q q = q Aa q Genotypic Frequencies in Generation p q Aa aa ( q ) 46 3

24 Example Cross Between Populations G Population Males ¾ AA ¼ aa Population Females ¼ AA ¾ aa G Random Mate G 47 Generation Population Males ¾ AA ¼ aa Population Females ¼ AA ¾ aa ¼A ¾a ¾A ¼a 3/6 AA 9/6 Aa /6 Aa 3/6 aa p =3/6 + ½(9/6 + /6)=8/6= ½ G Genotypic Freq G Gene Freq 48 4

25 Generation GMales 3/6 AA /6 Aa 3/6 aa GFemales 3/6 AA /6 Aa 3/6 aa ½ A ½ a ½ A ½a ¼AA ¼ Aa ¼ Aa ¼ aa G Genotypic Freq p = ¼+ ½( ¼+ ¼)= ½ G Gene Freq 49 Summary G Population Males ¾ AA ¼ aa Population Females ¼ AA ¾ aa G 3/6 AA /6 Aa 3/6aa p = ½ Genotypic Frequencies Different G 4/6 AA 8/6 Aa 4/6aa p = ½ Gametic Frequencies Same G3? 5 5

26 Summary Random Mating Autosomal Loci Equal Allele Frequency in Sexes Equilibrium Established After Generation Unequal Allele Frequencies in Sexes One Generation Required to Establish Equal Allele Frequencies in Sexes Second Generation Required to Establish Equilibrium Excess of Heterozygotes Produced in First Generation Indicative of Crossing between populations Useful in Plant and Animal Breeding May result in Heterosis 5 Lecture Problems. Two separate populations of equal size are in equilibrium for the same pair of alleles because of random mating within each. In population I, p A =.6, while in population II, p A =., with q = - p in each population. (a) If a random sample of females from one population is crossed to a random sample of males from the other population, what would be the expected genotypic frequencies among the progeny? If these progeny are then allowed to mate at random, what would be the expected gene and genotypic frequencies in the next-generation? What happens to heterozygote frequencies between the F and F generations? (b) If equal numbers of both sexes from each population are combined and allowed to mate at random, what would be the expected gene and genotypic frequencies in the next-generation? (c) Compare results in part a and b, what conclusions can you draw from this. 5 6