Hardy Weinberg Equilibrium

Hardy Weinberg Equilibrium

Wilhem Weinberg(1862 – 1937)

Gregor Mendel

G. H. Hardy(1877 - 1947)

(1822-1884)

Evolution acts through changes in allele frequency at each generation

Leads to average change in characteristic of the population

Recall from Previous Lectures

Darwin’s Observation

Lectures 4-11: Mechanisms of Evolution (Microevolution)

• Hardy Weinberg Principle (Mendelian Inheritance)

• Genetic Drift• Mutation• Recombination• 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

4 Major Evolutionary Mechanisms acting at the population level, changing allele frequencies:

• Genetic Drift• Natural Selection• Mutation• Migration

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

Fig. 23-5a

Porcupineherd range

Beaufort Sea NORTHWEST

TERRITORIESMAPAREA

ALAS

KA

CAN

ADA

Fortymileherd range

ALAS

KAYU

KON

• What is a “population?”A group of individuals within a species that is capable of interbreeding and producing fertile offspring

(definition for sexual species)

Fig. 23-5 Porcupine herd

Porcupineherd range

Beaufort Sea NORTHWEST

TERRITORIES

MAPAREA

ALAS

KA

CAN

ADA

Fortymileherd range

Fortymile herd

ALAS

KAYU

KON

• Mathematical description of Mendelian inheritance

Hardy-Weinberg Principle

Hardy-Weinberg Equilibrium

• According to the Hardy-Weinberg principle, frequencies of alleles and genotypes in a population remain constant from generation to generation

• In a given population where gametes contribute to the next generation randomly, allele frequencies will not change

• Allelic and genotypic frequencies follow the transmission rules of Mendelian inheritance, which maintains constant proportions in a population across generations

Patterns of inheritance should always be in “Hardy Weinberg Equilibrium”

Following the transmission rules of Mendel

In the absence of Evolution…

Requirements of HWEvolution

Large population size Genetic drift

Random MatingInbreeding & other

No MutationsMutations

No Natural Selection Natural Selection

No Migration Migration

An evolving population is one that violates Hardy-Weinberg Assumptions

Violation

“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

Hardy-Weinberg Theorem

In a non-evolving population, frequency of alleles and genotypes remain constant over generations

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

A1

A2

A3

A4A1

A1

A2

Sperm

Eggs

• Genotypes

Random Mating (Sex)

Zygotes

A1A3

A1A1 A1A1

A2A4

A3A1

A1A1

A1

A2A1

A1

A3A4

So then can we predict the % of alleles and genotypes in the population at each generation?

A1

A2

A3

A4A1

A1

A2

Sperm

Eggs

Zygotes

A1A3

A1A1 A1A1

A2A4

A3A1

A1A1

A1

A2A1

A1

A3

A4

Hardy-Weinberg Theorem

In a non-evolving population, frequency of alleles and genotypes remain constant over generations

Fig. 23-6

Frequencies of allelesAlleles in the population

Gametes producedEach egg: Each sperm:

80%chance

80%chance

20%chance

20%chance

q = frequency of

p = frequency ofCR allele = 0.8

CW allele = 0.2

Hardy-Weinberg proportions indicate the expected allele and genotype frequencies, given the starting frequencies

• 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

ALLELESProbability of A = p p + q = 1Probability of a = q

GENOTYPESAA: p x p = p2

Aa: p x q + q x p = 2pqaa: 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 + p2

2 + p3

2 + p42… …+ pn

2 + 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

ALLELE FrequenciesFrequency of A = p = 0.8Frequency of a = q = 0.2

p + q = 1

Expected GENOTYPE FrequenciesAA: p x p = p2 = 0.8 x 0.8 = 0.64Aa: p x q + q x p = 2pq

= 2 x (0.8 x 0.2) = 0.32aa: q x q = q2 = 0.2 x 0.2 = 0.04

p2 + 2pq + q2 = 0.64 + 0.32 + 0.04 = 1

Expected Allele Frequencies at 2nd Generationp = AA + Aa/2 = 0.64 + (0.32/2) = 0.8q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2

Allele frequencies remain the same at next generation

Hardy Weinberg Theorem

ALLELE FrequencyFrequency of A = p = 0.8 p + q = 1Frequency of a = q = 0.2

Expected GENOTYPE FrequencyAA: p x p = p2 = 0.8 x 0.8 = 0.64Aa: p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32aa : q x q = q2 = 0.2 x 0.2 = 0.04

p2 + 2pq + q2 = 0.64 + 0.32 + 0.04 = 1

Expected Allele Frequency at 2nd Generationp = AA + Aa/2 = 0.64 + (0.32/2) = 0.8q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2

Similar example,But with different starting allele frequencies

p q

p2

2pqq2

Fig. 23-7-4

Gametes of this generation:

64% CR CR, 32% CR CW, and 4% CW CW

64% CR + 16% CR = 80% CR = 0.8 = p

4% CW + 16% CW = 20% CW = 0.2 = q

64% CR CR, 32% CR CW, and 4% CW CW plants

Genotypes in the next generation:

SpermCR

(80%)

CW

(2

0 %)

80% CR ( p = 0.8)

CW (20%)

20% CW (q = 0.2)

16% ( pq) CR CW

4% (q2) CW CW

CR

(80%

)

64% ( p2) CR CR

16% (qp) CR CW

Eggs

Perform the same calculations using percentages

Fig. 23-7-1

SpermCR

(80%)

CW

(2

0 %)

80% CR ( p = 0.8)

CW (20%)

20% CW (q = 0.2)

16% ( pq) CRCW

4% (q2) CW CW

CR

(80%

)

64% ( p2) CRCR

16% (qp) CRCW

Eggs

Fig. 23-7-2


64% CRCR, 32% CRCW, and 4% CWCW

64% CR + 16% CR = 80% CR = 0.8 = p

4% CW + 16% CW = 20% CW = 0.2 = q

Fig. 23-7-3


64% CRCR, 32% CRCW, and 4% CWCW

64% CR + 16% CR = 80% CR = 0.8 = p

4% CW + 16% CW = 20% CW = 0.2 = q

64% CRCR, 32% CRCW, and 4% CWCW plants

Genotypes in the next generation:

• 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 Frequencies from # of Individuals

AA Aa aa120 60 35 (# of individuals)

#A = (2 x AA) + Aa = 240 + 60 = 300#a = (2 x aa) + Aa = 70 + 60 = 130Proportion A = 300/total = 300/430 = 0.70Proportion a = 130/total = 130/430 = 0.30

A + a = 0.70 + 0.30 = 1

Proportion AA = 120/215 = 0.56Proportion Aa = 60/215 = 0.28Proportion aa = 35/215 = 0.16

AA + Aa + aa = 0.56 + 0.28 +0.16 = 1

Calculating Allele and Genotype Frequencies from # of Individuals

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?

– 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

• The occurrence of PKU is 1 per 10,000 births(frequency of the disease allele):

q2 = 0.0001q = 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 = 0.0198or approximately 2% of the U.S. population

So, let’s calculate HW frequencies

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

DEVIATION from

Hardy-Weinberg EquilibriumIndicates that

EVOLUTIONIs happening

4 Major Evolutionary Mechanisms:

• Genetic Drift• Natural Selection• Mutation• Migration

• 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

Hardy-Weinberg across a Genome

Allele A1 Demo

Examples of Deviation from Hardy-Weinberg Equilibrium

What would Genetic Drift look like?

• Most populations are experiencing some level of genetic drift, unless they are incredibly large


AAAa aa

Generation 1 0.640.32 0.04Generation 2 0.630.33 0.04Generation 3 0.640.315 0.045Generation 4 0.650.31 0.04

Is this population in HW equilibrium?If not, how does it deviate?What could be the reason?


AAAa aa

Generation 1 0.640.32 0.04Generation 2 0.630.33 0.04Generation 3 0.640.315 0.045Generation 4 0.650.31 0.04

This is a case of Genetic Drift, where allele frequencies are fluctuating randomly across generations


AA Aa aa0.64 0.32 0



AA Aa aa0.64 0.32 0

Here this appears to be Directional Selection favoring AA

Or… Negative Selection disfavoring aa


AA Aa aa0.25 0.70 0.05



AA Aa aa0.25 0.70 0.05

This appears to be a case of Heterozygote Advantage (or Overdominance)


AA Aa aa0.10 0.10 0.80



AA Aa aa0.10 0.10 0.80

Selection appears to be favoring aa

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

Example: Does this population remain in Hardy Weinberg Equilibrium across Generations?

AAAa aa

Generation 1 0.25 0.50 0.25Generation 2 0.20 0.60 0.20Generation 3 0.10 0.80 0.10

AAAa aa

Generation 1 0.25 0.50 0.25Generation 2 0.20 0.60 0.20Generation 3 0.10 0.80 0.10

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?

1. When allele frequencies are changing across generations

2. 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)

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

(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

Summary

Hardy Weinberg Equilibrium

Wilhem Weinberg(1862 – 1937)

Gregor Mendel

G. H. Hardy(1877 - 1947)

(1822-1884)

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

The following are numbers of pink and white flowers in a population.

Pink WhiteGeneration 1: 901 302Generation 2: 1204 403Generation 3: 1510 504

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

WhiteGeneration 1: 360 480

160Generation 2: 100 200

200Generation 3: 0 100

300

3. What are the genotype frequencies at each generation?(A) Generation 1: 0.30, 0.50, 0.20

Generation 2: 0.20, 0.40, 0.40Generation 3: 0, 0.333, 0.666

(B) Generation 1: 0.36, 0.48, 0.16Generation 2: 0.10, 0.20, 0.20Generation 3: 0, 0.10, 0.30

(C) Generation 1: 0.36, 0.48, 0.16Generation 2: 0.20, 0.40, 0.40Generation 3: 0, 0.25, 0.75

(D) Generation 1: 0.36, 0.48, 0.16Generation 2: 0.36, 0.48, 0.16Generation 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) = 0.4Generation2: Dominant allele = 0.4, Recessive allele = 0.6Generation3: Dominant allele = 0.125, Recessive allele = 0.875

(B) Generation1: Dominant allele = 0.6, Recessive allele = 0.4Generation2: Dominant allele = 0.6, Recessive allele = 0.4Generation3: Dominant allele = 0.6, Recessive allele = 0.4

(C) Generation1: Dominant allele = 0.6, Recessive allele = 0.4Generation2: Dominant allele = 0.5, Recessive allele = 0.5Generation3: Dominant allele = 0.25, Recessive allele = 0.75

(D) Generation1: Dominant allele = 0.4, Recessive allele = 0.6Generation2: Dominant allele = 0.5, Recessive allele = 0.5Generation3: Dominant allele = 0.25, Recessive allele = 0.75

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 = A2. A3. C4. A5. D

Hardy Weinberg Equilibrium

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