Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations.

Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations in H-W assumptions cause changes in the distribution of alleles within/across populations Unit 2: Evolution and Pop. Structure (a.k.a. violations in H-W assumptions) Unit 2.1: genetic drift Unit 2.2: natural selection Unit 2.3: mutation Unit 2.5: assortative mating Unit 2.6: inbreeding Unit 2.4: migration

Assortative mating Feb. 18, 2015 HUGEN 2022: Population Genetics J. Shaffer Dept. Human Genetics University of Pittsburgh

Objectives After this lecture you should be able to: 1.Define positive and negative assortative mating, provide examples of each, and describe the qualitative effects on genotype and allele frequencies 2.Calculate the effect of assortative mating on genotype and allele frequencies given mating-type probabilities 3.Make inferences from mating-type frequency tables (i.e. type of mating scheme, e.g., random, assortative mating, etc.)

Hardy-Weinberg assumptions diploid organism sexual reproduction nonoverlapping generations random mating large population size equal allele frequencies in the sexes no migration no mutation no selection

Human mating systems Random –Assumed for HWE –Realistic? Non-random –Assortative mating –Population substructure –Inbreeding Note: these notions are imprecisely defined

Assortative mating Definition –Choice of mate is dependent on a particular phenotype or genotype Examples –People more often mate with those of similar height / skin tone –Deaf people more often mate with deaf people Types –Positive assortative mating: mating between people with like phenotypes/genotypes –Negative assortative mating: mating between people with unlike phenotypes/genotypes Additonal comments –Degree of assortative mating can vary from a slight tendency to near complete –Assortative mating on a simple characteristic usually will affect only one or a limited number of genes (e.g. height, skin tone, hearing)

Assortative mating: key qualitative concepts Additional comments –amount of increase or decrease depends on the specific scheme (examples to follow) –assortative mating does NOT change allele frequencies but it DOES affect the rate at which natural selection changes allele frequencies type of assortative mating effect on freq. homozygotes effect on freq. of heterozygotes positiveincreasesdecreases negativedecreasesincreases

Assortative mating: human populations Humans tend to have positive assortative mating on characteristics they observe in their mates –height –weight –disability –ethnicity Negative assortative mating may (??) be rare in humans –HLA and scent –recessive mutation testing

Assortative mating: human populations Humans tend to have positive assortative mating on characteristics they observe in their mates –height –weight –disability –ethnicity Negative assortative mating may (??) be rare in humans –HLA and scent –recessive mutation testing Dor Yeshorim

Modeling assortative mating We can mathematically model the effects of assortative mating based on knowledge of … 1.how often people with given genotypes mate 2.starting genotype frequencies

Mating type frequency table mating-typeFrequencyP(child is AA)P(child is Aa)P(child is aa) AA x AAU100 AA x AaV0.5 0 AA x aaW010 Aa x AaX0.250.50.25 Aa x aaY00.5 aa x aaZ001

Mating type frequency table mating-typeFrequencyP(child is AA)P(child is Aa)P(child is aa) AA x AAU100 AA x AaV0.5 0 AA x aaW010 Aa x AaX0.250.50.25 Aa x aaY00.5 aa x aaZ001 after one-generation: P(AA) = 1U + 0.5V + 0W + 0.25X + 0Y + 0Z P(Aa) = 0U + 0.5V + 1W + 0.5X + 0.5Y + 0Z P(aa) = 0U + 0V + 0W + 0.25X + 0.5Y + 1Z

Mating type frequency table mating-typeFrequencyP(child is AA)P(child is Aa)P(child is aa) AA x AAU100 AA x AaV0.5 0 AA x aaW010 Aa x AaX0.250.50.25 Aa x aaY00.5 aa x aaZ001 NOTE: Assortative mating leads to changes in genotype frequencies. Therefore, except under specific circumstances, mating frequencies (U-Z) will also change after every generation.

Mating type freq. under different mating systems mating typeHWE (random) complete positive complete negative general AA x AAD2D2 D0U AA x Aa2DH0 (DH)/(DH+DR+HR) V AA x aa2DR0 (DR)/(DH+DR+HR) W Aa x AaH2H2 H0X Aa x aa2HR0 (HR)/(DH+DR+HR) Y aa x aaR2R2 R0Z P(AA) = D P(Aa) = H P(aa) = R mating type frequencies always sum to 1 many different (potentially complex) mating schemes are possible

Example Multiple choice 1: mating-typefrequencyP(child is AA)P(child is Aa)P(child is aa) AA x AAU100 AA x AaV0.5 0 AA x aaW010 Aa x AaX0.250.50.25 Aa x aaY00.5 aa x aaZ001 Given the following genotype frequencies at generation 0 P(AA) = D P(Aa) = H P(aa) = R (assume D, H, and R are non-zero) Which of the following values for U-Z would be most consistent with positive assortative mating? (a) U < D 2 ; X = 0; and Z = 0 (b) U = D 2 ; X = H 2 ; and Z = R 2 (c) V < 2DH; W < 2DR; and Y < 2HR

Example Multiple choice 2: mating-typefrequencyP(child is AA)P(child is Aa)P(child is aa) AA x AAU100 AA x AaV0.5 0 AA x aaW010 Aa x AaX0.250.50.25 Aa x aaY00.5 aa x aaZ001 Given the following genotype frequencies at generation 0 P(AA) = D P(Aa) = H P(aa) = R (assume D, H, and R are non-zero) Which of the following values for U-Z would be most consistent with negative assortative mating? (a) U < D 2 ; X < H 2 ; and Z < R 2 (b) V < 2DH; W < 2DR; and Y < 2HR (c) U = (DH)/(DH+DR+HR); X = (DR)/(DH+DR+HR); and Z = (HR)/(DH+DR+HR)

Math problem: complete assortative mating In every generation, people mate only with others of their own genotype Genotype frequencies in generation 0: –P(AA) = D –P(Aa) = H –P(aa) = R What are genotype frequencies in the next generation? What happens to the genotype frequencies in the long run? Example

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AA AA x Aa AA x aa Aa x Aa Aa x aa aa x aa

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AA AA x Aa0 AA x aa0 Aa x Aa Aa x aa0 aa x aa

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AA AA x Aa0000 AA x aa0000 Aa x Aa Aa x aa0000 aa x aa

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AA Aa x Aa aa x aa

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD Aa x AaH aa x aaR

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD100 Aa x AaH aa x aaR

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD100 Aa x AaH0.250.50.25 aa x aaR

Example Mating type probability table mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD100 Aa x AaH0.250.50.25 aa x aaR001

Example Example (cont.) Genotype frequencies in generation 1: –P(AA) = 1D + (1/4)H + 0R= D + H/4 –P(Aa) = 0D + (1/2)H + 0D= H/2 –P(aa) = 0D + (1/4)H + 1R= R + H/4 Homozygote frequencies increased by H/4 each Heterozygote frequency decreased by H/2 mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD100 Aa x AaH0.250.50.25 aa x aaR001

Example Example (cont.) What will happen in the next generation? mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AA?100 Aa x Aa?0.250.50.25 aa x aa?001 generation 1: – –P(AA) = D + H/4 – –P(Aa) = H/2 – –P(aa) = R + H/4 generation 0: – –P(AA) = D – –P(Aa) = H – –P(aa) = R

Example Example (cont.) What will happen in the next generation? mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD + H/4100 Aa x AaH/20.250.50.25 aa x aaR + H/4001 generation 1: – –P(AA) = D + H/4 – –P(Aa) = H/2 – –P(aa) = R + H/4 generation 0: – –P(AA) = D – –P(Aa) = H – –P(aa) = R

Example Example (cont.) What will happen in the next generation? Genotype frequencies in generation 2: –P(AA) = 1(D + H/4) + (1/4)(H/2) + 0(R + H/4)= D + (3/8)H –P(Aa) = 0(D + H/4) + (1/2)(H/2) + 0(R + H/4)= H/4 –P(aa) = 0(D + H/4) + (1/4)(H/2) + 1(R + H/4)= R + (3/8)H mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAD + H/4100 Aa x AaH/20.250.50.25 aa x aaR + H/4001 generation 1: – –P(AA) = D + H/4 – –P(Aa) = H/2 – –P(aa) = R + H/4 generation 0: – –P(AA) = D – –P(Aa) = H – –P(aa) = R

Example Example (cont.) What will happen in the future? What if this mating system keeps going forever? –P(AA) = D + H/2 –P(Aa) = 0 –P (aa) = R + H/2 mating typefrequencyP(child AA)P(child Aa)P(child aa) AA x AAIncreasing100 Aa x AaDecreasing0.250.50.25 aa x aaIncreasing001 generation 1: – –P(AA) = D + H/4 – –P(Aa) = H/2 – –P(aa) = R + H/4 generation 0: – –P(AA) = D – –P(Aa) = H – –P(aa) = R generation 2: – –P(AA) = D + (3/8)H – –P(Aa) = H/4 – –P(aa) = R + (3/8)H

Example Example (cont.) Under complete assortative mating genotype frequencies quickly change What about allele frequencies? genotype freq.allele freq. generation 0 P(AA) = D P(Aa) = H P(aa) = R P(A) = D + H/2 P(a) = R + H/2 generation 1 P(AA) = D + H/4 P(Aa) = H/2 P(aa) = R + H/4 P(A) = P(a) =

Example Example (cont.) Under complete assortative mating genotype frequencies quickly change What about allele frequencies? genotype freq.allele freq. generation 0 P(AA) = D P(Aa) = H P(aa) = R P(A) = D + H/2 P(a) = R + H/2 generation 1 P(AA) = D + H/4 P(Aa) = H/2 P(aa) = R + H/4 P(A) = (D + H/4) + (1/2)(H/2) P(a) = (R + H/4) + (1/2)(H/2)

Example Example (cont.) Under complete assortative mating genotype frequencies quickly change What about allele frequencies? genotype freq.allele freq. generation 0 P(AA) = D P(Aa) = H P(aa) = R P(A) = D + H/2 P(a) = R + H/2 generation 1 P(AA) = D + H/4 P(Aa) = H/2 P(aa) = R + H/4 P(A) = (D + H/4) + (1/2)(H/2) = D + H/2 P(a) = (R + H/4) + (1/2)(H/2) = R + H/2

Example Example (cont.) Under complete assortative mating genotype frequencies quickly change What about allele frequencies? do not change genotype freq.allele freq. generation 0 P(AA) = D P(Aa) = H P(aa) = R P(A) = D + H/2 P(a) = R + H/2 generation 1 P(AA) = D + H/4 P(Aa) = H/2 P(aa) = R + H/4 P(A) = (D + H/4) + (1/2)(H/2) = D + H/2 P(a) = (R + H/4) + (1/2)(H/2) = R + H/2

Genotype frequencies after one generation of assortative mating α = positive assortative mating fraction i.e., proportion not randomly mating, 0 to 1

Genotype frequencies after one generation of assortative mating α = positive assortative mating fraction i.e., proportion not randomly mating, 0 to 1 Aa from randomly mating Aa from positively assorting all genotypes (AA, Aa, and aa) in next generation

Assortative mating: Recap Positive and negative assortative mating –examples of each –qualitative effects on homozygotes, heterozygotes, genotype frequencies, and allele frequencies. Identify type of mating scheme (random, assortative mating, etc.) from mating-type frequencies Calculate next generation genotype frequencies given a mating scheme

Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations.

Similar presentations

Presentation on theme: "Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations.

Similar presentations

Presentation on theme: "Course outline Evolution: When violations in H-W assumptions cause changes in the genetic composition of a population Population Structure: When violations."— Presentation transcript:

Similar presentations

About project

Feedback