Download Hardy Weinberg Equilibrium

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Hardy Weinberg Equilibrium Lectures 4-­‐11: Mechanisms of Evolu8on (Microevolu8on) (Mendelian Inheritance) •  Hardy Weinberg Principle • 
• 
• 
• 
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Gregor Mendel
Wilhem Weinberg
(1822-1884)
G. H. Hardy
(1877 - 1947)
(1862 – 1937)
Recall from Previous Lectures
Darwin’s Observa<on Gene<c Dri> Muta<on Recombina<on Epigene<c Inheritance Natural Selec<on These are mechanisms ac<ng WITHIN popula<ons, hence called “popula<on gene<cs”—EXCEPT for epigene<c modifica<ons, which act on individuals in a Lamarckian manner Recall from Lecture on History of
Evolutionary Thought
Darwin’s Observa<on Evolu8on acts through changes in allele frequency at each genera8on Leads to average change in characteris8c of the popula8on
HOWEVER, Darwin did not understand how gene8c varia8on was passed on from genera8on to genera8on
Gregor Mendel, “Father of Modern Gene8cs” http://www.biography.com/people/gregor-mendel-39282#synopsis
Gregor Mendel
•  Mendel presented a mechanism
for how traits got passed on
Gregor Mendel, “Father of Modern Gene8cs” http://www.biography.com/people/gregor-mendel-39282#synopsis
Mendel’s Laws of Inheritance •  Law of Segrega8on Gregor Mendel
–  only one allele passes from each parent on to an offspring “Individuals pass alleles on to
their offspring intact”
•  Law of Independent Assortment –  different pairs of alleles are passed to offspring independently of each other (the idea of particulate (genes)
inheritance)
(1822-1884)
(1822-1884)
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Using 29,000 pea plants, Mendel discovered the 1:3 ratio
of phenotypes, due to dominant vs. recessive alleles
•  In cross-­‐pollina<ng plants with either yellow or green peas, Mendel found that the first genera<on (f1) always had yellow seeds (dominance). However, the following genera<on (f2) consistently had a 3:1 ra<o of yellow to green. •  Mendel uncovered the underlying mechanism, that there are dominant and recessive alleles Hardy-Weinberg Principle
•  Mathematical description of Mendelian
inheritance
Tes<ng for Hardy-­‐Weinberg equilibrium can be used to assess whether a popula<on is evolving Godfrey Hardy
(1877-1947)
Wilhem Weinberg
(1862 – 1937)
The Hardy-­‐Weinberg Principle •  A popula<on that is not evolving shows allele and genotypic frequencies that are in Hardy Weinberg equilibrium •  If a popula<on is not in Hardy-­‐Weinberg equilibrium, it can be concluded that the popula<on 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)
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Violation Genetic drift
Random Mating
Inbreeding & other
No Mutations
Mutations
No Natural Selection
Natural Selection
No Migration
Migration
Beaufort Sea • What is a “population?”
A group of individuals within
a species that is capable of
interbreeding and producing
fertile offspring
Porcupine herd range (definition for sexual species)
Fortymile herd range An evolving population is one that
violates Hardy-Weinberg Assumptions
In the absence of Evolution…
PaVerns of inheritance should always be in “Hardy Weinberg Equilibrium” Following the transmission rules of Mendel T ES S HW IE
RT OR
NO RRIT
TE
Large population size
MAP AREA CANADA Evolution
ALASKA YUKON Requirements of HW
ALASKA Fig. 23-­‐5a Hardy-­‐Weinberg Equilibrium •  According to the Hardy-­‐Weinberg principle, frequencies of alleles and genotypes in a popula<on remain constant from genera<on to genera<on •  Also, the genotype frequencies you see in a popula<on should be the Hardy-­‐Weinberg expecta<ons, given the allele frequencies “Null Model” •  No Evolu8on: Null Model to test if no evolu<on is happening should simply be a popula<on in Hardy-­‐Weinberg Equilibrium •  No Selec8on: Null Model to test whether Natural Selec<on is occurring should have no selec<on, but should include Gene<c Dri> Example: Is this population in Hardy
Weinberg Equilibrium?
Generation 1
Generation 2
Generation 3
AA
Aa
0.25 0.50
0.20 0.60
0.10 0.80
aa
0.25
0.20
0.10
–  This is because Gene<c Dri> is opera<ng even when there is no Natural Selec<on 3
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important concepts • gene: A region of genome sequence (DNA or RNA), that is Hardy-­‐Weinberg Theorem In a non-­‐evolving popula<on, frequency of alleles and genotypes remain constant over genera<ons the unit of inheritance , the product of which contributes to phenotype • locus: Loca<on in a genome (used interchangeably with “gene,” if the loca<on 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 You should be able to
predict the genotype
frequencies, given the
allele frequencies
colors, BRCA1 breast cancer allele, etc.) • genotype: The combina<on of alleles at a locus (gene) • phenotype: The expression of a trait, as a result of the genotype and regula<on of genes (green eyes, brown hair, body size, finger length, cys<c 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 loca8on in the genome) A2
So then can we predict the % of alleles and genotypes in the popula<on at each genera<on? A4
A2
Sperm
A1
A3
A4
A1
Zygotes
A1A1
A1A3
•  Genotypes A1A1
A1A1
A2A4
A3A1
Hardy-­‐Weinberg Theorem A1
A3
A4
A1
A1
A1A1
A1A3
A2
A1
A2
A1
A2
Zygotes
A1
A3
A1
A3
A4
Sperm
A1
Random Ma<ng (Sex) – 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 popula<on, blue, green, grey, brown, etc. A1
A2
Eggs
A1
•  However, there can be many alleles at a locus in a popula8on. Eggs
•  Alleles in a popula<on of diploid organisms In a non-­‐evolving popula<on, frequency of alleles and genotypes remain constant over genera<ons A1A1
A1A1
A2A4
A3A1
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Fig. 23-­‐6 •  By conven<on, if there are 2 alleles at a locus, p and q are used to represent their frequencies Alleles in the popula8on 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 20% chance •  The frequency of all alleles in a popula<on will add up to 1 –  For example, p + q = 1 Hardy-­‐Weinberg propor8ons indicate the expected allele and genotype frequencies, given the star8ng frequencies What about for a triploid organism? If p and q represent the rela<ve frequencies of the only two possible alleles in a popula<on at a par<cular 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 Hardy Weinberg Theorem
What about for a triploid organism? •  (p + q)3 = 1 = p3 + 3p2q + 3pq2 + q3 = 1 Poten<al offspring: ppp, ppq, pqp, qpp, qqp, pqq, qpq, qqq How about tetraploid? You work it out. 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 5
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ALLELE Frequencies
Frequency of A = p = 0.8
Frequency of a = q = 0.2
p+q=1
More General HW Equa8ons •  One locus three alleles: (p + q + r)2 = p2 + q2 + r2 + 2pq +2pr + 2qr 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
•  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 mul<ple 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 Hardy Weinberg Theorem
ALLELE Frequency Frequency of A = p = 0.8 Frequency of a = q = 0.2 p2 + 2pq + q2
= 0.64 + 0.32 + 0.04 = 1
Allele frequencies remain the same at next genera<on 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
Similar example, But with different star<ng allele frequencies p + q = 1 Expected GENOTYPE Frequency AA: Aa: aa : p x p = p2 = 0.8 x 0.8 = 0.64 p x q + q x p = 2pq = 2 x (0.8 x 0.2) = 0.32 q x q = q2 = 0.2 x 0.2 = 0.04 p2 + 2pq + q2 = 0.64 + 0.32 + 0.04 = 1 p Expected Allele Frequency at 2nd Genera<on q p = AA + Aa/2 = 0.64 + (0.32/2) = 0.8 q = aa + Aa/2 = 0.04 + (0.32/2) = 0.2 p2 2pq q2 6
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Calculating Allele and Genotype Frequencies from
# of Individuals
Calculating Allele Frequencies from # of Individuals
AA Aa aa 120 60 35 (# of individuals) •  The frequency of an allele in a popula<on can be calculated from # of individuals: #A = (2 x AA) + Aa = 240 + 60 = 300 #a = (2 x aa) + Aa = 70 + 60 = 130 Propor<on A = 300/total = 300/430 = 0.70 Propor<on a = 130/total = 130/430 = 0.30 –  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 A + a = 0.70 + 0.30 = 1 Propor<on AA = 120/215 = 0.56 Propor<on Aa = 60/215 = 0.28 Propor<on aa = 35/215 = 0.16 –  plus 1 allele for each heterozygous individual; the same logic applies for recessive alleles AA + Aa + aa = 0.56 + 0.28 +0.16 = 1 Applying the Hardy-­‐Weinberg Principle •  The occurrence of PKU is 1 per 10,000 births •  How many carriers of this disease in the popula<on? •  Example: es<mate frequency of a disease allele in a popula<on •  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 à mental retarda<on –  Rare deleterious recessives o>en remain in a popula<on because they are hidden in the heterozygous state (the “carriers”) –  Natural selec<on can only act on the homozygous individuals where the phenotype is exposed (individuals who show symptoms of PKU) 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: –  We can assume HW equilibrium if: •  There is no migra<on from a popula<on with different allele frequency •  Random ma<ng •  No gene<c dri> •  Etc 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. popula<on 7
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Condi<ons for Hardy-­‐Weinberg Equilibrium •  The Hardy-­‐Weinberg theorem describes a hypothe<cal popula<on •  The five condi<ons for nonevolving popula<ons are rarely met in nature: –  No muta<ons –  Random ma<ng –  No natural selec<on –  Extremely large popula<on size –  No gene flow DEVIATION
from
Hardy-Weinberg Equilibrium
Indicates that
EVOLUTION
Is happening
•  So, in real popula<ons, allele and genotype frequencies do change over <me Hardy-Weinberg across a Genome
Allele A1 Demo •  In natural popula<ons, 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 selec<on and become out of HW equilibrium, while the rest of the genome remains in HW equilibrium How can you tell whether a popula<on is out of HW Equilibrium? •  Perform HW calcula<ons to see if it looks like the popula<on is out of HW equilibrium •  Then apply sta<s<cal tests to see if the devia<on is significantly different from what you would expect by random chance 8
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Generation 1
Generation 2
Generation 3
Example: Does this population remain in
Hardy Weinberg Equilibrium across
Generations?
Generation 1
Generation 2
Generation 3
AA
Aa
0.25 0.50
0.20 0.60
0.10 0.80
aa
0.25
0.20
0.10
How can you tell whether a popula<on 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)
AA
Aa
0.25 0.50
0.20 0.60
0.10 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
Tes<ng for Deviaton from Hardy-­‐
Weinberg Expecta<ons •  A χ2 goodness-­‐of-­‐fit test can be used to determine if a popula<on is significantly different from the expec<ons of Hardy-­‐Weinberg equilibrium. •  If we have a series of genotype counts from a popula<on, 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 χ2 value: Genotype
AA Aa aa Total Observed Expected
30 33 55 49 15 18 100 100 (O-­‐E)2/E 0.27 0.73 0.50 1.50 •  One genera<on of Random Ma<ng could put a popula<on back into Hardy Weinberg Equilibrium •  Since χ2 = 1.50 < 3.841 (from Chi-­‐square table), we conclude that the genotype frequencies in this popula<on are not significantly different than what would be expected if the popula<on is in Hardy-­‐Weinberg equilibrium. 9
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What would Gene<c Dri> look like? Examples of Devia8on from Hardy-­‐Weinberg Equilibrium •  Most popula<ons are experiencing some level of gene<c dri>, unless they are incredibly large Examples of Devia<on from Hardy-­‐Weinberg Equilibrium Genera<on 1 Genera<on 2 Genera<on 3 Genera<on 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 Examples of Devia<on from Hardy-­‐Weinberg Equilibrium Genera<on 1 Genera<on 2 Genera<on 3 Genera<on 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 popula<on in HW equilibrium? If not, how does it deviate? What could be the reason? This is a case of Gene8c Dri`, where allele frequencies are fluctua<ng randomly across genera<ons Examples of Devia<on from Hardy-­‐Weinberg Equilibrium Examples of Devia<on from Hardy-­‐Weinberg Equilibrium AA Aa aa 0.64 0.32 0 Is this popula<on in HW equilibrium? If not, how does it deviate? What could be the reason? AA Aa aa 0.64 0.32 0 Here this appears to be Direc8onal Selec8on favoring AA Or… Nega8ve Selec8on disfavoring aa 10
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Examples of Devia<on from Hardy-­‐Weinberg Equilibrium AA Aa aa 0.25 0.70 0.05 Is this popula<on in HW equilibrium? If not, how does it deviate? What could be the reason? Examples of Devia<on from Hardy-­‐Weinberg Equilibrium AA Aa aa 0.25 0.70 0.05 This appears to be a case of Heterozygote Advantage (or Overdominance) Examples of Devia<on from Hardy-­‐Weinberg Equilibrium AA Aa aa 0.10 0.10 0.80 Is this popula<on in HW equilibrium? If not, how does it deviate? What could be the reason? Summary
(1) A nonevolving popula8on is in HW Equilibrium (2) Evolu8on occurs when the requirements for HW Equilibrium are not met (3) HW Equilibrium is violated when there is Gene8c Dri`, Migra8on, Muta8ons, Natural Selec8on, and Nonrandom Ma8ng Examples of Devia<on from Hardy-­‐Weinberg Equilibrium AA Aa aa 0.10 0.10 0.80 Selec<on appears to be favoring aa Hardy Weinberg Equilibrium Gregor Mendel
(1822-1884)
Wilhem Weinberg
G. H. Hardy
(1877 - 1947)
(1862 – 1937)
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80% CR ( p = 0.8) Eggs CR (80%)
Sperm CR (80%)
Perform the same
calculations using
percentages
16% ( pq) CR CW
CR CR
16% (qp) CR CW
80% CR ( p = 0.8) 20% CW (q = 0.2) CW (20%)
64% ( p2) CW (20%)
Fig. 23-­‐7-­‐1 20% CW (q = 0.2) Sperm 4% (q2) CW CW
CR (80%)
CW (20%)
64% ( p2) CRCR
16% ( pq) CRCW
CR (80%)
Fig. 23-­‐7-­‐4 Eggs 64% CR CR, 32% CR CW, and 4% CW CW
Gametes of this genera<on: 64% CR + 16% CR = 80% CR = 0.8 = p 4% CW + 16% CW = 20% CW = 0.2 = q CW (20%)
Genotypes in the next genera<on: 4% (q2) CW CW
16% (qp) CRCW
64% CR CR, 32% CR CW, and 4% CW CW plants Fig. 23-­‐7-­‐2 Fig. 23-­‐7-­‐3 64% CRCR, 32% CRCW, and 4% CWCW
Gametes of this genera8on: 64% CRCR, 32% CRCW, and 4% CWCW
64% CR + 16% CR = 80% CR
Gametes of this genera8on: = 0.8 = p 4% CW + 16% CW = 20% CW = 0.2 = q 64% CR + 16% CR = 80% CR = 0.8 = p
Genotypes in the next genera8on: 4% CW + 16% CW = 20% CW = 0.2 = q
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
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The following are numbers of purple and white peas in a population.
The following are numbers of pink and white flowers in a population.
Generation 1:
Generation 2:
Pink
901
1204
White
302
403
Generation 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
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
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
(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
Answers: 1. Parents: Aa x Aa = Offspring: AA (25%), Aa (50%), aa (25%) Answer = A 2. A 3. C 4. A 5. D 13