Transcript Hardy Weinberg Equilibrium

Hardy Weinberg Equilibrium
Gregor Mendel
(1822-1884)
Wilhem Weinberg
G. H. Hardy
(1877 - 1947)
(1862 – 1937)
Lectures 4-11: Mechanisms of Evolution
(Microevolution)
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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
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
Darwin’s Observation
HOWEVER, Darwin did not
understand how genetic
variation was passed on from
generation to generation
Gregor Mendel, “Father of Modern
Genetics”
http://www.biography.com/people/gregor-mendel-39282#synopsis
Gregor Mendel
• Mendel presented a mechanism
for how traits got passed on
“Individuals pass alleles on to
their offspring intact”
(the idea of particulate (genes)
inheritance)
(1822-1884)
Gregor Mendel, “Father of Modern
Genetics”
http://www.biography.com/people/gregor-mendel-39282#synopsis
Mendel’s Laws of Inheritance
• Law of Segregation
Gregor Mendel
– 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
(1822-1884)
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
• Mathematical description of Mendelian
inheritance
Godfrey Hardy
(1877-1947)
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
(will put population out of HW Equilibrium):
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•
•
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)
Requirements of HW
Violation
Evolution
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
•What is a “population?”
A group of individuals within
a species that is capable of
interbreeding and producing
fertile offspring
Beaufort Sea
Porcupine
herd range
(definition for sexual species)
Fortymile
herd range
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?
Generation 1
Generation 2
Generation 3
AA
0.25
0.20
0.10
Aa
0.50
0.60
0.80
aa
0.25
0.20
0.10
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
A2
Eggs
A1
A3
A1
A2
A1
A4
A2
Sperm
A1
A3
A4
A1
A1
Random Mating (Sex)
Zygotes
A1A1
A1A3
• Genotypes
A1A1
A1A1
A2A4
A3A1
A2
Eggs
A1
A3
A1
A2
A1
A4
So then can we
predict the % of
alleles and genotypes
in the population at
each generation?
A2
Sperm
A1
A3
A4
A1
Zygotes
A1
A1A1
A1A3
A1A1
A1A1
A2A4
A3A1
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
p = frequency of
CR allele
= 0.8
q = frequency of
CW allele
= 0.2
Gametes produced
Each egg:
80%
chance
20%
chance
Each sperm:
80%
chance
Hardy-Weinberg proportions indicate the
expected allele and genotype frequencies,
given the starting frequencies
20%
chance
• 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
ALLELES
Probability of A = p
Probability of a = q
p+q=1
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
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
Allele frequencies remain the same at
next generation
p2 + 2pq + q2
= 0.64 + 0.32 + 0.04 = 1
Expected Allele Frequencies at 2nd Generation
p = AA + Aa/2 = 0.64 + (0.32/2) = 0.8
q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2
Hardy Weinberg Theorem
ALLELE Frequency
Frequency of A = p = 0.8
Frequency of a = q = 0.2
p+q=1
Expected GENOTYPE Frequency
AA:
Aa:
aa :
pxp=
p2 = 0.8 x 0.8 = 0.64
p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32
qxq=
q2 = 0.2 x 0.2 = 0.04
p2 + 2pq + q2 = 0.64 + 0.32 + 0.04 = 1
Expected Allele Frequency at 2nd Generation
p = AA + Aa/2 = 0.64 + (0.32/2) = 0.8
q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2
Similar example,
But with different starting allele frequencies
p
q
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
120
Aa
60
aa
35 (# of individuals)
#A = (2 x AA) + Aa = 240 + 60 = 300
#a = (2 x aa) + Aa = 70 + 60 = 130
Proportion A = 300/total = 300/430 = 0.70
Proportion a = 130/total = 130/430 = 0.30
A + a = 0.70 + 0.30 = 1
Proportion AA = 120/215 = 0.56
Proportion Aa = 60/215 = 0.28
Proportion aa = 35/215 = 0.16
AA + Aa + aa = 0.56 + 0.28 +0.16 = 1
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
So, let’s calculate HW frequencies
• The occurrence of PKU is 1 per 10,000 births
(frequency of the disease allele):
q2 = 0.0001
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 = 0.0198
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
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
Example: Does this population remain in
Hardy Weinberg Equilibrium across
Generations?
Generation 1
Generation 2
Generation 3
AA
0.25
0.20
0.10
Aa
0.50
0.60
0.80
aa
0.25
0.20
0.10
Generation 1
Generation 2
Generation 3
AA
0.25
0.20
0.10
Aa
0.50
0.60
0.80
aa
0.25
0.20
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)
Testing for Deviaton from HardyWeinberg 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
AA
Aa
aa
Total
Observed Expected (O-E)2/E
30
33
0.27
55
49
0.73
15
18
0.50
100
100
1.50
• Since c2 = 1.50 < 3.841 (from Chi-square table), 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.
• One generation of Random Mating could put a
population back into Hardy Weinberg
Equilibrium
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
Examples of Deviation from
Hardy-Weinberg Equilibrium
Generation 1
Generation 2
Generation 3
Generation 4
AA
0.64
0.63
0.64
0.65
Aa
0.32
0.33
0.315
0.31
aa
0.04
0.04
0.045
0.04
Is this population in HW equilibrium?
If not, how does it deviate?
What could be the reason?
Examples of Deviation from
Hardy-Weinberg Equilibrium
Generation 1
Generation 2
Generation 3
Generation 4
AA
0.64
0.63
0.64
0.65
Aa
0.32
0.33
0.315
0.31
aa
0.04
0.04
0.045
0.04
This is a case of Genetic Drift, where
allele frequencies are fluctuating
randomly across generations
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.64
Aa
0.32
aa
0
Is this population in HW equilibrium?
If not, how does it deviate?
What could be the reason?
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.64
Aa
0.32
aa
0
Here this appears to be Directional
Selection favoring AA
Or… Negative Selection disfavoring aa
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.25
Aa
0.70
aa
0.05
Is this population in HW equilibrium?
If not, how does it deviate?
What could be the reason?
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.25
Aa
0.70
aa
0.05
This appears to be a case of Heterozygote
Advantage (or Overdominance)
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.10
Aa
0.10
aa
0.80
Is this population in HW equilibrium?
If not, how does it deviate?
What could be the reason?
Examples of Deviation from
Hardy-Weinberg Equilibrium
AA
0.10
Aa
0.10
aa
0.80
Selection appears to be 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
(1822-1884)
Wilhem Weinberg
G. H. Hardy
(1877 - 1947)
(1862 – 1937)
Fig. 23-7-4
80% CR ( p = 0.8)
20% CW (q = 0.2)
Sperm
CR
(80%)
64% ( p2)
CR CR
16% (qp)
CR CW
CW
(20%)
16% ( pq)
CR CW
4% (q2)
CW CW
64% CR CR, 32% CR CW, and 4% CW CW
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
Perform the same
calculations using
percentages
Fig. 23-7-1
80% CR (p = 0.8)
20% CW (q = 0.2)
Sperm
CR
(80%)
CW
(20%)
64% (p2)
CRCR
16% (pq)
CRCW
16% (qp)
CRCW
4% (q2)
CW CW
Fig. 23-7-2
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
Fig. 23-7-3
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:
64% CRCR, 32% CRCW, and 4% CWCW plants
Gregor Mendel
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.
Generation 1:
Generation 2:
Generation 3:
Pink
901
1204
1510
White
302
403
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.
Generation 1:
Generation 2:
Generation 3:
(A1A1)
Purple
360
100
0
(A1A2)
Purple
480
200
100
(A2A2)
White
160
200
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, 0.40
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 2: 0.20, 0.40, 0.40
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
Generation2: Dominant allele = 0.5, Recessive allele = 0.5
Generation3: 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 = A
2. A
3. C
4. A
5. D