Hardy Weinberg Equilibrium

Hardy Weinberg Equilibrium
Gregor Mendel Wilhem Weinberg (1862 – 1937) ( ) G. H. Hardy ( )

Lectures 4-11: Mechanisms of Evolution (Microevolution)
Hardy Weinberg Principle (Mendelian Inheritance) Genetic Drift Mutation Sex: Recombination and Random Mating Epigenetic Inheritance Natural Selection These are mechanisms acting WITHIN populations, hence called “population genetics”—EXCEPT for epigenetic modifications, which act on individuals in a Lamarckian manner

Recall from Previous Lectures
5/6/2018 Recall from Previous Lectures Darwin’s Observation Evolution acts through changes in allele frequency at each generation Leads to average change in characteristic of the population

Recall from Lecture on History of Evolutionary Thought
5/6/2018 Recall from Lecture on History of Evolutionary Thought Darwin’s Observation HOWEVER, Darwin did not understand how genetic variation was passed on from generation to generation

Gregor Mendel, “Father of Modern Genetics”
Mendel presented a mechanism for how traits got passed on “Individuals pass alleles on to their offspring intact” (the idea of particulate (genes) inheritance) Gregor Mendel ( )

Gregor Mendel, “Father of Modern Genetics”
Mendel’s Laws of Inheritance Law of Segregation only one allele passes from each parent on to an offspring Law of Independent Assortment different pairs of alleles are passed to offspring independently of each other Gregor Mendel ( )

Gregor Mendel

Using 29,000 pea plants, Mendel discovered the 1:3 ratio of phenotypes, due to dominant vs. recessive alleles In cross-pollinating plants with either yellow or green peas, Mendel found that the first generation (f1) always had yellow seeds (dominance). However, the following generation (f2) consistently had a 3:1 ratio of yellow to green.

Mendel uncovered the underlying mechanism, that there are dominant and recessive alleles

Hardy-Weinberg Principle
5/6/2018 Hardy-Weinberg Principle Mathematical description of Mendelian inheritance Godfrey Hardy ( ) Wilhem Weinberg (1862 – 1937)

Testing for Hardy-Weinberg equilibrium can be used to assess whether a population is evolving

The Hardy-Weinberg Principle
A population that is not evolving shows allele and genotypic frequencies that are in Hardy Weinberg equilibrium If a population is not in Hardy-Weinberg equilibrium, it can be concluded that the population is evolving

Evolutionary Mechanisms
5/6/2018 Evolutionary Mechanisms (will put population out of HW Equilibrium): Genetic Drift Natural Selection Mutation Migration *Epigenetic modifications change expression of alleles but not the frequency of alleles themselves, so they won’t affect the actual inheritance of alleles However, if you count the phenotype frequencies, and not the genotype frequencies , you might see phenotypic frequencies out of HW Equilibrium due to epigenetic silencing of alleles. (epigenetic modifications can change phenotype, not genotype) So Evolution acts through Genetic Drift or Natural Selection acting on the genetic variation caused by mutations or recombination, or lack of variation caused by inbreeding

Requirements of HW Evolution
Violation Large population size Genetic drift Random Mating Inbreeding & other No Mutations Mutations No Natural Selection Natural Selection No Migration Migration An evolving population is one that violates Hardy-Weinberg Assumptions

Fig. 23-5a MAP AREA CANADA ALASKA What is a “population?” A group of individuals within a species that is capable of interbreeding and producing fertile offspring (definition for sexual species) Beaufort Sea NORTHWEST TERRITORIES Porcupine herd range Figure 23.5 One species, two populations Fortymile herd range ALASKA YUKON

In the absence of Evolution…
5/6/2018 In the absence of Evolution… Patterns of inheritance should always be in “Hardy Weinberg Equilibrium” Following the transmission rules of Mendel

Hardy-Weinberg Equilibrium
According to the Hardy-Weinberg principle, frequencies of alleles and genotypes in a population remain constant from generation to generation Also, the genotype frequencies you see in a population should be the Hardy-Weinberg expectations, given the allele frequencies

“Null Model” No Evolution: Null Model to test if no evolution is happening should simply be a population in Hardy-Weinberg Equilibrium No Selection: Null Model to test whether Natural Selection is occurring should have no selection, but should include Genetic Drift This is because Genetic Drift is operating even when there is no Natural Selection

Example: Is this population in Hardy Weinberg Equilibrium?
5/6/2018 Example: Is this population in Hardy Weinberg Equilibrium? AA Aa aa Generation Generation Generation

Hardy-Weinberg Theorem
5/6/2018 Hardy-Weinberg Theorem In a non-evolving population, frequency of alleles and genotypes remain constant over generations You should be able to predict the genotype frequencies, given the allele frequencies

important concepts gene: A region of genome sequence (DNA or RNA), that is the unit of inheritance , the product of which contributes to phenotype locus: Location in a genome (used interchangeably with “gene,” if the location is at a gene… but, locus can be anywhere, so meaning is broader than gene) loci: Plural of locus allele: Variant forms of a gene (e.g. alleles for different eye colors, BRCA1 breast cancer allele, etc.) genotype: The combination of alleles at a locus (gene) phenotype: The expression of a trait, as a result of the genotype and regulation of genes (green eyes, brown hair, body size, finger length, cystic fibrosis, etc.)

important concepts allele: Variant forms of a gene (e.g. alleles for different eye colors, BRCA1 breast cancer allele, etc.) We are diploid (2 chromosomes), so we have 2 alleles at a locus (any location in the genome) However, there can be many alleles at a locus in a population. For example, you might have inherited a blue eye allele from your mom and a brown eye allele from your dad… you can’t have more alleles than that (only 2 chromosomes, one from each parent) BUT, there could be many alleles at this locus in the population, blue, green, grey, brown, etc.

Alleles in a population of diploid organisms
Eggs Alleles in a population of diploid organisms A1 A3 A2 A1 A4 A2 A3 A1 Sperm A4 A1 A1 Random Mating (Sex) A1A1 A1A1 Zygotes A1A3 A1A1 Genotypes A2A4 A3A1

A2 A1 Eggs A1 A3 A2 A1 A4 So then can we predict the % of alleles and genotypes in the population at each generation? A2 A3 A1 Sperm A4 A1 A1 A1A1 A1A1 Zygotes A1A3 A1A1 A2A4 A3A1

Hardy-Weinberg Theorem
5/6/2018 Hardy-Weinberg Theorem In a non-evolving population, frequency of alleles and genotypes remain constant over generations

Fig. 23-6 Alleles in the population Frequencies of alleles Gametes produced p = frequency of Each egg: Each sperm: CR allele = 0.8 q = frequency of 80% chance 20% chance 80% chance 20% chance CW allele = 0.2 Figure 23.6 Selecting alleles at random from a gene pool Hardy-Weinberg proportions indicate the expected allele and genotype frequencies, given the starting frequencies

The frequency of all alleles in a population will add up to 1
By convention, if there are 2 alleles at a locus, p and q are used to represent their frequencies The frequency of all alleles in a population will add up to 1 For example, p + q = 1

If p and q represent the relative frequencies of the only two possible alleles in a population at a particular locus, then for a diploid organism (2 chromosomes), (p + q) 2 = 1 = p2 + 2pq + q2 = 1 where p2 and q2 represent the frequencies of the homozygous genotypes and 2pq represents the frequency of the heterozygous genotype

What about for a triploid organism?

What about for a triploid organism?
(p + q)3 = 1 = p3 + 3p2q + 3pq2 + q3 = 1 Potential offspring: ppp, ppq, pqp, qpp, qqp, pqq, qpq, qqq How about tetraploid? You work it out.

Hardy Weinberg Theorem
5/6/2018 Hardy Weinberg Theorem ALLELES Probability of A = p p + q = 1 Probability of a = q GENOTYPES AA: p x p = p2 Aa: p x q + q x p = 2pq aa: q x q = q2 p2 + 2pq + q2 = 1

More General HW Equations
One locus three alleles: (p + q + r)2 = p2 + q2 + r2 + 2pq +2pr + 2qr One locus n # alleles: (p1 + p2 + p3 + p4 … …+ pn)2 = p12 + p22 + p32 + p42… …+ pn2 + 2p1p2 + 2p1p3 + 2p2p3 + 2p1p4 + 2p1p5 + … … + 2pn-1pn For a polyploid (more than two chromosomes): (p + q)c, where c = number of chromosomes If multiple loci (genes) code for a trait, each locus follows the HW principle independently, and then the alleles at each loci interact to influence the trait

Expected GENOTYPE Frequencies AA: p x p = p2 = 0.8 x 0.8 = 0.64
5/6/2018 ALLELE Frequencies Frequency of A = p = 0.8 Frequency of a = q = 0.2 p + q = 1 Expected GENOTYPE Frequencies AA: p x p = p2 = 0.8 x 0.8 = 0.64 Aa: p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32 aa: q x q = q2 = 0.2 x 0.2 = 0.04 p2 + 2pq + q2 = = 1 Allele frequencies remain the same at next generation Expected Allele Frequencies at 2nd Generation p = AA + Aa/2 = (0.32/2) = 0.8 q = aa + Aa/2 = (0.32/2) = 0.2

Hardy Weinberg Theorem
5/6/2018 Hardy Weinberg Theorem ALLELE Frequency Frequency of A = p = p + q = 1 Frequency of a = q = 0.2 Expected GENOTYPE Frequency AA: p x p = p2 = 0.8 x 0.8 = 0.64 Aa: p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32 aa : q x q = q2 = 0.2 x 0.2 = 0.04 p2 + 2pq + q2 = = 1 Expected Allele Frequency at 2nd Generation p = AA + Aa/2 = (0.32/2) = 0.8 q = aa + Aa/2 = (0.32/2) = 0.2

But with different starting allele frequencies
5/6/2018 Similar example, But with different starting allele frequencies p q

5/6/2018

5/6/2018 p2 2pq q2

Calculating Allele Frequencies from # of Individuals
The frequency of an allele in a population can be calculated from # of individuals: For diploid organisms, the total number of alleles at a locus is the total number of individuals x 2 The total number of dominant alleles at a locus is 2 alleles for each homozygous dominant individual plus 1 allele for each heterozygous individual; the same logic applies for recessive alleles

Calculating Allele and Genotype Frequencies from # of Individuals
AA Aa aa (# of individuals) #A = (2 x AA) + Aa = = 300 #a = (2 x aa) + Aa = = 130 Proportion A = 300/total = 300/430 = 0.70 Proportion a = 130/total = 130/430 = 0.30 A + a = = 1 Proportion AA = 120/215 = 0.56 Proportion Aa = 60/215 = 0.28 Proportion aa = 35/215 = 0.16 AA + Aa + aa = = 1 Roughly in HW equilibrium Genetic Drift

Applying the Hardy-Weinberg Principle
Example: estimate frequency of a disease allele in a population Phenylketonuria (PKU) is a metabolic disorder that results from homozygosity for a recessive allele Individuals that are homozygous for the deleterious recessive allele cannot break down phenylalanine, results in build up  mental retardation

The occurrence of PKU is 1 per 10,000 births
How many carriers of this disease in the population?

We can assume HW equilibrium if:
Rare deleterious recessives often remain in a population because they are hidden in the heterozygous state (the “carriers”) Natural selection can only act on the homozygous individuals where the phenotype is exposed (individuals who show symptoms of PKU) We can assume HW equilibrium if: There is no migration from a population with different allele frequency Random mating No genetic drift Etc

So, let’s calculate HW frequencies
The occurrence of PKU is 1 per 10,000 births (frequency of the disease allele): q2 = q = sqrt(q2 ) = sqrt(0.0001) = 0.01 The frequency of normal alleles is: p = 1 – q = 1 – 0.01 = 0.99 The frequency of carriers (heterozygotes) of the deleterious allele is: 2pq = 2 x 0.99 x 0.01 = or approximately 2% of the U.S. population

Conditions for Hardy-Weinberg Equilibrium
The Hardy-Weinberg theorem describes a hypothetical population The five conditions for nonevolving populations are rarely met in nature: No mutations Random mating No natural selection Extremely large population size No gene flow So, in real populations, allele and genotype frequencies do change over time

Hardy-Weinberg Equilibrium
DEVIATION from Hardy-Weinberg Equilibrium Indicates that EVOLUTION Is happening

Hardy-Weinberg across a Genome
In natural populations, some loci might be out of HW equilibrium, while being in Hardy-Weinberg equilibrium at other loci For example, some loci might be undergoing natural selection and become out of HW equilibrium, while the rest of the genome remains in HW equilibrium

Allele A1 Demo

How can you tell whether a population is out of HW Equilibrium?

Perform HW calculations to see if it looks like the population is out of HW equilibrium
Then apply statistical tests to see if the deviation is significantly different from what you would expect by random chance

5/6/2018 Example: Does this population remain in Hardy Weinberg Equilibrium across Generations? AA Aa aa Generation Generation Generation

In this case, allele frequencies (of A and a) did not change.
5/6/2018 AA Aa aa Generation Generation Generation In this case, allele frequencies (of A and a) did not change. ***However, the population did go out of HW equilibrium because you can no longer predict genotypic frequencies from allele frequencies For example, p = 0.5, p2 = 0.25, but in Generation 3, the observe p2 = 0.10

How can you tell whether a population is out of HW Equilibrium?
When allele frequencies are changing across generations When you cannot predict genotype frequencies from allele frequencies (means there is an excess or deficit of genotypes than what would be expected given the allele frequencies)

Testing for Deviaton from Hardy-Weinberg Expectations
A c2 goodness-of-fit test can be used to determine if a population is significantly different from the expections of Hardy-Weinberg equilibrium. If we have a series of genotype counts from a population, then we can compare these counts to the ones predicted by the Hardy-Weinberg model. O = observed counts, E = expected counts, sum across genotypes

Example Genotype Count: AA 30 Aa 55 aa 15 Calculate the c2 value:
Genotype Observed Expected (O-E)2/E AA Aa aa Total Since c2 = 1.50 < (from Chi-square table, alpha = 0.05), we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population is in Hardy-Weinberg equilibrium.

A c2 goodness-of-fit test can be used to determine if a population is significantly different from the expections of Hardy-Weinberg equilibrium. If we have a series of genotype counts from a population, then we can compare these counts to the ones predicted by the Hardy-Weinberg model. O = observed counts, E = expected counts, sum across genotypes

O = observed counts, E = expected counts, sum across genotypes We test our c2 value against the Chi-square distribution (sum of square of a normal distribution), which represents the theoretical distribution of sample values under HW equilibrium And determine how likely it is to get our result simply by chance (e.g. due to sampling error); i.e., do our Observed values differ from our Expected values more than what we would expect by chance (= significantly different)? ? Less likely to get these values by chance

Test for Deviation from HW equilibrium
Genotype Count Generation 4: AA 65 Aa 31 aa 4 Calculate the c2 value: Genotype Observed Expected (O-E)2/E AA Aa aa Total Since c2 = < (from Chi-square table for critical values, alpha = 0.05), we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population were in Hardy-Weinberg equilibrium.

Calculate Chi-squared test statistic Figure out degrees of freedom
The chi-squared distribution is used because it is the sum of squared normal distributions Calculate Chi-squared test statistic Figure out degrees of freedom Select confidence interval (P-value) Compare your Chi-squared value to the theoretical distribution (from the table), and accept or reject the null hypothesis. If the test statistic > than the critical value, the null hypothesis (H0 = there is no difference between the distributions) can be rejected with the selected level of confidence, and the alternative hypothesis (H1 = there is a difference between the distributions) can be accepted. If the test statistic < than the critical value, the null hypothesis cannot be rejected

Test for Significance of Deviation from HW Equilibrium
Degrees of Freedom is n – 1 = 2 alleles (p, q) -1 = 1

Testing for significance
The results come out not significantly different from HW equilibrium This does not necessarily mean that genetic drift is not happening, but that we cannot conclude that genetic drift is happening Either we do not have enough power (not enough data, small sample size), or genetic drift is not happening Sometimes it is difficult to test whether evolution is happening, even when it is happening... The signal needs to be sufficiently large to be sure that you can’t get the results by chance (like by sampling error)

Test for Deviation from HW equilibrium
5/6/2018 Genotype Count Generation 4  increase sample size AA Aa aa 4000 Calculate the c2 value: Genotype Observed Expected (O-E)2/E AA Aa aa Total , , Since c2 = > (from Chi-square table for critical values, alpha = 0.05), we conclude that the genotype frequencies in this population ARE significantly different than what would be expected if the population were in Hardy-Weinberg equilibrium.

Test for Significance of Deviation from HW Equilibrium
Degrees of Freedom is n – 1 = 2 alleles (p, q) -1 = 1

One generation of Random Mating could put a population back into Hardy Weinberg Equilibrium

Examples of Deviation from Hardy-Weinberg Equilibrium
Lethal homozygote of the recessive allele

What would Genetic Drift look like?
Most populations are experiencing some level of genetic drift, unless they are incredibly large

AA Aa aa Generation Generation Generation Generation Is this population in HW equilibrium? If not, how does it deviate? What could be the reason? Roughly in HW equilibrium Genetic Drift

AA Aa aa Generation Generation Generation Generation This is a case of Genetic Drift, where allele frequencies are fluctuating randomly across generations Roughly in HW equilibrium Genetic Drift

AA Aa aa Is this population in HW equilibrium? If not, how does it deviate? What could be the reason? Lethal homozygote of the recessive allele

AA Aa aa Here this appears to be Directional Selection favoring AA Or… Negative Selection disfavoring aa Lethal homozygote of the recessive allele

AA Aa aa Is this population in HW equilibrium? If not, how does it deviate? What could be the reason? Heterozygote advantage

AA Aa aa This appears to be a case of Heterozygote Advantage (or Overdominance) Heterozygote advantage

AA Aa aa Is this population in HW equilibrium? If not, how does it deviate? What could be the reason? Selection favoring aa

AA Aa aa Selection appears to be favoring aa Selection favoring aa

Summary (1) A nonevolving population is in HW Equilibrium
(2) Evolution occurs when the requirements for HW Equilibrium are not met (3) HW Equilibrium is violated when there is Genetic Drift, Migration, Mutations, Natural Selection, and Nonrandom Mating

Hardy Weinberg Equilibrium
Gregor Mendel Wilhem Weinberg (1862 – 1937) ( ) G. H. Hardy ( )

Perform the same calculations using percentages
Fig 80% CR ( p = 0.8) 20% CW (q = 0.2) Sperm CR (80%) CW (20%) Perform the same calculations using percentages CR (80%) Eggs 64% ( p2) CR CR 16% ( pq) CR CW 16% (qp) CR CW 4% (q2) CW CW CW (20%) 64% CR CR, 32% CR CW, and 4% CW CW Figure 23.7 The Hardy-Weinberg principle Gametes of this generation: 64% CR + 16% CR = 80% CR = 0.8 = p 4% CW + 16% CW = 20% CW = 0.2 = q Genotypes in the next generation: 64% CR CR, 32% CR CW, and 4% CW CW plants

80% CR ( p = 0.8) 20% CW (q = 0.2) Sperm CR (80%) Eggs CW (20%) CR CW
Fig 80% CR ( p = 0.8) 20% CW (q = 0.2) Sperm CR (80%) CW (20%) CR (80%) Figure 23.7 The Hardy-Weinberg principle Eggs 64% ( p2) CRCR 16% ( pq) CRCW 4% (q2) CW CW 16% (qp) CRCW CW (20%)

Gametes of this generation:
Fig 64% CRCR, 32% CRCW, and 4% CWCW Gametes of this generation: 64% CR + 16% CR = 80% CR = 0.8 = p Figure 23.7 The Hardy-Weinberg principle 4% CW + 16% CW = 20% CW = 0.2 = q

Gametes of this generation:
Fig 64% CRCR, 32% CRCW, and 4% CWCW Gametes of this generation: 64% CR + 16% CR = 80% CR = 0.8 = p 4% CW + 16% CW = 20% CW = 0.2 = q Genotypes in the next generation: Figure 23.7 The Hardy-Weinberg principle 64% CRCR, 32% CRCW, and 4% CWCW plants

1. Nabila is a Saudi Princess who is arranged to marry her first cousin. Many in her family have died of a rare blood disease, which sometimes skips generations, and thus appears to be recessive. Nabila thinks that she is a carrier of this disease. If her fiancé is also a carrier, what is the probability that her offspring will have (be afflicted with) the disease? (A) 1/4 (B) 1/3 (C) 1/2 (D) 3/4 (E) zero

2. Which of the following is most likely to be TRUE?
The following are numbers of pink and white flowers in a population. Pink White Generation 1: Generation 2: Generation 3: 2. Which of the following is most likely to be TRUE? (A) The heterozygotes are probably pink (B) The recessive allele here (probably white) is clearly deleterious (C) Evolution is occurring, as allele frequencies are changing greatly over time (D) Clearly there is a heterozygote advantage (E) The frequencies above violate Hardy-Weinberg expectations

The following are numbers of purple and white peas in a population.
(A1A1) (A1A2) (A2A2) Purple Purple White Generation 1: Generation 2: Generation 3: 3. What are the genotype frequencies at each generation? (A) Generation 1: 0.30, 0.50, 0.20 Generation 2: 0.20, 0.40, 0.40 Generation 3: 0, 0.333, 0.666 (B) Generation 1: 0.36, 0.48, 0.16 Generation 2: 0.10, 0.20, 0.20 Generation 3: 0, 0.10, 0.30 (C) Generation 1: 0.36, 0.48, 0.16 Generation 3: 0, 0.25, 0.75 (D) Generation 1: 0.36, 0.48, 0.16 Generation 2: 0.36, 0.48, 0.16 Generation 3: 0.36, 0.48, 0.16

4. From the example on the previous slide, what are the frequencies of alleles at each generation?
(A) Generation1: Dominant allele (A1) = 0.6, Recessive allele (A2) = 0.4 Generation2: Dominant allele = 0.4, Recessive allele = 0.6 Generation3: Dominant allele = 0.125, Recessive allele = 0.875 (B) Generation1: Dominant allele = 0.6, Recessive allele = 0.4 Generation2: Dominant allele = 0.6, Recessive allele = 0.4 Generation3: Dominant allele = 0.6, Recessive allele = 0.4 (C) Generation1: Dominant allele = 0.6, Recessive allele = 0.4 Generation2: Dominant allele = 0.5, Recessive allele = 0.5 Generation3: Dominant allele = 0.25, Recessive allele = 0.75 (D) Generation1: Dominant allele = 0.4, Recessive allele = 0.6

5. From the example two slides ago, which evolutionary mechanism might be operating across generations? (A) Mutation (B) Selection favoring A1 (C) Heterozygote advantage (D) Selection favoring A2 (E) Inbreeding

Answers: 1. Parents: Aa x Aa = Offspring: AA (25%), Aa (50%), aa (25%)
Answer = A 2. A 3. C 4. A 5. D

Hardy Weinberg Equilibrium

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