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TheThe
Hardy-WeinbergHardy-Weinberg
PrinciplePrinciple
! The HW principle specifies a null hypothesis as to what
genotype frequencies should be in the absence of evolution
Assumptions of the Hardy-Weinberg Model Assumptions of the Hardy-Weinberg Model
**_- No Natural Selection • No Mutation
- No Migration_** ( No Gene Flow ) • No Genetic Drift
- Hardy-Weinberg equilibrium also makes two additional
assumptions:
**_- Random mating
- Discrete, non-overlapping generations_**
In the absence of evolution, gene and genotype frequencies do not change
Predictions of the Hardy-Weinberg PrinciplePredictions of the Hardy-Weinberg Principle
In the absence of evolution, with random mating and discrete
generations, the proportions of genotypes in one generation
depend directly on the probabilities of union between gametes
derived from the preceding generation
- In other words, if we know the gene or genotype frequencies
in one generation, then in the absence of evolution the
genotype frequencies in the next generation can easily be
predicted.
The HW Principle serves as the basis for testing evolutionary processes
AA p^2 Aa pq aA qp aa q^2 A fr ( A )= p a fr ( a )= q Sperm gametic pool Egg gametic pool A fr ( A )= p a fr ( a )= q
Generation Generation tt GenerationGeneration^ t+1t+
AA Aa aa p^2 + 2 pq + q^2 = ( p + q )( p + q ) = ( p + q )^2 = 1 p^2 2 pq q^2
Hardy-Weinberg
Equilibrium
The Hardy-Weinberg Principle: The Hardy-Weinberg Principle: A Numerical ExampleA Numerical Example The banding patterns of individual fruit flies representing the three possible genotypes at the ADH locus, FF , FS , and S S Direction of protein migration ADH banding patterns FF FS SS Example : Suppose you obtain a sample population of 500 flies and assay their ADH genotypes ( N =500 genotypes and 2 N =1000 alleles). Goal : Determine the expected numbers of the three ADH genotypes assuming HW equilibrium to be true. Is the population in HW Equilibrium? ADH ADH Example Example Genotypes: (^) FF FS SS Calculations summarizing the observed genetic data : Observed genotypic counts 287 126 87 (^ =^ 500) Observed genotypic frequencies 287/500 = 0.574 126/500 = 0.252 = 87/500 0.174 ( = 1) Obser frequenciesved allele^ p^ = fr (^ F^ ) = ! ( 287 * 2 ) + ( 126 * 1 ) 1000 =^ 0. q = fr ( S ) = ! ( 87 * 2 ) + ( 126 1) 1000 =^ 0.^ ( = 1) Calculations assuming HW equilibrium : HW ex genotypicpected frequencies =^ p 0.49^2 =^2 0.42 p q =^ q 0.09^2 ( = 1) HW ex genotypic counpected ts^ 0.49500 = 245 0.42500 = 210 0.09500 = 45 ( = 500)**
ADH ExampleADH Example
Conclusion : The sample data set for ADH appears to contain excess
homozygotes ( FF and SS genotypes) and deficit heterozygotes ( FS
genotypes) relative to their expectations under HW equilibrium.
HW expected
genotypic counts
Obs. genotypic
counts
Genotypes FF FS SS
The ADH Example - Food for Thought The ADH Example - Food for Thought
- What possible violations of the assumptions of the HW model do you think might be responsible for the observed deviations in ADH genotypic frequencies? - What processes could generate an excess of homozygotes and **deficit or heterozygotes relative to HW expectations?
- How would you determine whether the deviation from HW** equilibrium is statistically significant? - How would you test the difference between observed and expected counts? General Notation and Calculations - ObservedGeneral Notation and Calculations - Observed DataData Observed allele frequencies : Let p = fr( A ) = the frequency of A , and q = fr( a ), with p + q = 1. Genotypes : Given alleles A and a , the three possible genotypes are denoted AA , Aa , and aa. Observed genotype frequencies : Let PAA = fr( AA ), PAa = fr( Aa ), and Paa = fr( aa ), with PAA + PAa + Paa = 1. General Notation and Calculations - ObservedGeneral Notation and Calculations - Observed DataData Observed genotypic counts : Let NAA , NAa , and Naa , denote the observed numbers of AA , Aa , and aa genotypes, respectively, with NAA + NAa + Naa = N denoting the total number of sample genotypes. Observed allelic counts : Let NA and Na denote the numbers of A and a alleles, respectively, with NA + Na = 2 N (a sample of N diploid genotypes will contain 2 N alleles). General Notation and CalculationsGeneral Notation and Calculations ! p = NA NA + Na = NA 2 N = 2 NAA + NAa 2 N = NAA N
1 2 NAa N = PAA + 12 PAa q = Na NA + Na = Na 2 N = 2 Naa + NAa 2 N = Naa N
1 2 NAa N = Paa + 12 PAa
An illustrative exampleAn illustrative example
p = PAA + 12 PAa
q = Paa + 12 PAa
Are the genotype frequencies at HW equilibrium?
Pop. PAA PAa Paa p q
Chi-square Goodness of Fit TestChi-square Goodness of Fit Test
Determining degrees of freedom (df) : df = the number of categories in the data minus 1 for each parameter estimated from the data in order to generate the expected counts under the null hypothesis. For testing HW : df = the no. of genotypes – 1 for determining sample size N
And now for the statistical test of the MN locus dataAnd now for the statistical test of the MN locus data
Are the Britishers in HW equilibrium at the MN locus? "^2 = (298 - 294.3)^2 /294.3 + (498 - 496.4)^2 /496.4 + (213 - 209.3)^2 /209.3 = 0. Therefore, in our case we have "^2 = 0.222 with 3 - 1 - 1 = 1 df.
- Since our "^2 = 0.222 is less than 3.84, there is no reason to think that the population of Britishers does not obey the HW law for the MN blood cell locus.
Drawing a conclusionDrawing a conclusion
What does this data set and statistical genetic analysis tell us about the evolutionary behavior of the MN locus in British men and women? ! Since the genotypic frequencies are consistent with Hardy- Weinberg equilibrium, there is no evidence of significant non- random mating (e.g., inbreeding), gene flow, fecundity or viability selection, or non-random mating with respect to the observable variation this locus.
Summary of the Hardy-Weinberg Principle
Assumptions of the HW principle :
**- No natural selection • No mutation
- No genetic drift • No gene flow
- Random mating • Non-overlapping generations** Predictions: - Genotype frequencies in the ratio p^2 , 2 pq , and q^2 **(HWE)
- No evolution at the locus in question**
Important conclusions of the HW model Important conclusions of the HW model
(1) In the absence of evolution, when individuals mate at random, HW genotype frequencies are determined solely by gene frequencies and not by the frequencies of genotypes in the preceding generation. (2) HW genotype frequencies are reached in a single generation and, in the absence of destabilizing evolutionary forces such as mutation, selection, genetic drift, gene flow, and so forth, the HW equilibrium is maintained indefinitely (gene and genotype frequencies are constant over time) and no evolution takes place. (3) When a population's genotype frequencies are related to its gene frequencies in the ratios fr( AA ) = p^2 , fr( Aa ) = 2 pq , and fr( aa ) = q^2 then the population is said to be “in HW equilibrium”. (4) HWE represents a null hypothesis against which we can identify and estimate the forces that cause deviations from that equilibrium - the forces of evolution.
