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A little knowledgeis a dangerousthing.

So is a lot.

Albert Einstein

Distribution of grades: Exam I

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geGenetics

• If Huntington’s disease is a dominanttrait, shouldn’t most people haveHuntington’s?

Genetics

• How can O be the most common bloodtype if it is a recessive trait?

100%

70-79%80-89%90-99%

Percent type O bloodin Native Americans

Genetics

• Why don’t all traits have 3:1distributionin populations?– Why would we expect a 3:1 ratio?

Genetics

• Mendelian crosses• Population• Reproduction

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General model population genetics Gene pool: f(A)=0.6, f(a)=0.4

Gene pool:f(A)=0.6,f(a)=0.4

Hardy-Weinberg: Derivation

• For 1 locus, 2 alleles:f(A)=pf(a)=q =1-p

• Genotype frequencies next generation:f’(AA)=pXp (=p2)f’(Aa)=pXq + qXp (=2pq)f’(aa)=qXq (=q2)

• (p+q)2 = p2 + 2pq + q2 = 1

Hardy-Weinberg: continued

• Allele frequencies next generation:p’=p2 + 0.5(2pq)

=p2+ pq=p(p+q)=p

– No change in allele frequencies

Hardy-Weinberg Equilibrium

• A way to predict the genotype frequencies in futuregenerations

• Assumptions– Diploid– Sexual– Non-overlapping generations– Random mating– Allele frequencies identical in males and females– Infinite population size– No migration– No mutation– No selection

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What bloodygood is it???

Implications of Hardy-Weinberg

• No evolution, no change• Recessive traits not wiped out• Genotype frequencies can be predicted

from allele frequencies– (p2, 2pq and q2)

Using Hardy-Weinberg:

• Can determine frequency of rarerecessive alleles in population

• e.g. PKU (phenylketonuria)– can’t metabolize phenylalanine– frequency in population: 1/10,000-assuming HWE, what proportion of

population are asymptomatic carriers?

HaWeE

• Why do we use it?– It provides us with a representation of how alleles should

behave in the absence of evolutionary forces.– No evolutionary forces

• Allele frequencies stay constant.• Genotype frequencies stay constant.

– Can you maintain allele frequencies through generations but still havethe genotype frequencies changing

• If frequencies are changing evolution is occurring.• Mendelian inheritance preserves genetic variation

HaWeE

• Populations may sometimes leave HaWeE by chance. How long does ittake to return?

• 1 generation of random mating– Allele frequencies in a population of hamsters with allele frequencies for agouti

allele (p=A1) and non-agouti allele (q=A2). Number of genotypes are A1 A1=110,A1 A2=60, and A2 A2=30.

– HaWeE allele frequencies are p=0.7 and q=0.3.– The next generation we find genotypes of are A1 A1=70, A1 A2=95, and A2 A2=35,

with p = 0.5875 and q = 0.4125.– This population is not in HaWee. If random mating occurs and there is

no selection the allele genotype frequencies in the next generation willbe

• A1 A1 = 0.58752 = 0.345, A1 A2=2*0.5875*0.4125 = 0.485, A2 A2= 0.17• Note: if the generations overlap return to HaWee is more gradual

Using Hardy-Weinberg:

• Null model (no change)– χ2 test for deviations from HWE– deviation indicates assumptions violated

• e.g. random mating• no population structure• equal allele frequencies between sexes• no selection• no migration• no genetic drift…

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Testing Hardy-Weinberg

• e.g. scarlet tiger moth (N=1612)– SS white spots (1469)– Ss intermediate (138)– ss little spotting (5)

Calculate allele frequencies:p=q=

Testing Hardy-Weinberg

• e.g. scarlet tiger moth (N=1612)– SS white spots (1469)– Ss intermediate (138)– ss little spotting (5)

Calculate allele frequencies:p= (2*1469+138)/(2*1612) = 0.954q=1-0.954

Testing Hardy-Weinberg (cont)

• Calculate genotype frequenciesexpected under HWE– SS (p2) =0.9542 = 0.91– Ss (2pq) =2*0.954*0.046 = 0.088– ss (q2) = 0.0462 = 0.002

• Calculate expected genotype numbers– SS =1612*0.954– Ss =1612*0.088– ss =1612*0.002

Testing Hardy-Weinberg (cont)

• Calculate genotype frequenciesexpected under HWE– SS (p2) =0.9542 = 0.91– Ss (2pq) =2*0.954*0.046 = 0.088– ss (q2) = 0.0462 = 0.002

• Calculate expected genotype #s– SS =1612*0.91=1466.9 1469– Ss =1612*0.088=141.8 138– ss =1612*0.002=3.22 5

Testing Hardy-Weinberg (cont)χ2 test compares observed, expected numbersχ2=sum((obs-exp)2/exp)χ2

crit=3.84

Accept nullhypothesis

The populationis in HWE

€

1469 −1466.9( )2

1466.9+138 −141.8( )2

141.8+5 − 3.2( )2

3.2=1.18

Assumptions of Hardy-Weinberg

A) Random mating– Inbreeding/assort. Mating

• heterozygote deficit– Outbreeding/negative assortative mating

• heterozygote excess

– How could you distinguish betweenassortative mating at a certain locus,vs inbreeding?

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Nonrandom mating in humans

• Assortative mating:– Height, IQ, ethnicity, dwarfism…

• Disassortative mating (rare):– MHC (T-shirt studies)

• Inbreeding avoidance

Assumptions of Hardy-WeinbergB) Sexual reproduction

Send in the clones

C) DiploidyCan you do it for others?

D) Discrete generations

Clonaid has been able tofind living cells in a bodythat has been dead for 4months - there is hope!

Assumptions of Hardy-Weinberg

E) Infinite population size-finite pop’s = sampling error (genetic drift)-random fluctuations in allele and genotype frequencies-drift strength proportional to 1/(2N)-eventually one allele fixed, others lost

Assumptions of Hardy-Weinberg

F) Equal allele freqs between sexesG) No selectionH) No migrationI) No mutation

HaWee• Why do we use it?

– It provides us with a representation of how alleles should behave in theabsence of evolutionary forces.

• Null model for evolution– If frequencies are changing evolution is occurring.

» Forces of evolution?– No evolutionary forces

• Allele frequencies stay constant.• Genotype frequencies stay constant.

– Can you maintain allele frequencies through generations but still have the genotype frequencieschanging?

– Prove it yes or no

– In conservation, provides a basis for the following:• detecting deviations from random mating• tests for selection• modeling the effects of inbreeding and selection

• Mendelian inheritance preserves genetic variation

Mechanisms of evolution

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Revision: Hardy Weinberg

p2 + 2pq + q2 = 1,p + q = 1

a AAlleles

q pAllelefrequency

Genotypefrequency

Genotypes

q22pqp2

aaAaAA

What do you do if you have3 allele or are a triploid?

Sources

• The ultimate sourceof all new geneticvariation at a locusis MUTATION

• Within a populationgenetic variationcan be added indifferent ways– Mutation– Dispersal– Transposable

elements

Mutations

• Mutations are rare– 10-4 - 10-7 for most

markers• New mutant alleles

are by definitionrare– 1/2N

• Increasing frequency ofmutant allele– Drift

• Removes low frequencyalleles

– Recurrent mutation• Slow• 1/2N, 2/2N, 3/2N

– Selection• Depends upon the

strength of selection

Recurrent mutation

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1 49 97 145 193 241 289 337 385 433 481 529 577 625 673 721 769 817 865 913 961Generation

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Mutation Irreversible

• HaWeE with µ• Allele frequency with

mutation– pt= pt-1(1- µ)– Only allele which didn’t

mutate– pt-1= pt-2(1- µ)Thus– pt = p0(1- µ)t

• As t ⇒ infinity• p0 ⇒ 0.0

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1 169 337 505 673 841 1009 1177 1345 1513 1681 1849 2017 2185 2353 2521 2689

Generation

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y m=0.001m=0.0001m=0.00001m=0.000001

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Mutation Irreversable

• HaWeE with µ• Allele frequency with

mutation– pt= pt-1(1- µ)– Only allele which didn’t

mutate– pt-1= pt-2(1- µ)Thus– pt = p0(1- µ)t

• Mutation half life– pt = 0.5p0

• Thus– 0.5 = (1- µ)t

– ln(0.5) = t ln(1- µ)– t =ln 0.5/ln(1- µ)– t = -0.693/ln(1- µ)

• ln(1- µ) ⇒ - µ– t = 0.693/µ

Half-life

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Half life

69.3 6931

Mutations

• Reversible– A -> a– A -> A

• Probabilityof– A -> a = µ– a -> A = ν

• HaWeE with reversiblemutation

• Pt = pt-1(1- µ)+qt-1 ν

• Do the math• p = ν/(µ+ν)• Reversible mutations result

in an equilibrium of the two(or more) allleles

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Generations

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A(m=0.1,p=0.9)a(n=0.1)A(m=0.1,p=0.70)a(n=0.05)A(m=0.1,p=0.5)a(n=0.1)

Starting allele frequencydoesn’t matter forthe equilibrium pointonly the two mutationrates

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Infinite Alleles Model (IAM)• Mutations

– Unlikely that only 2 allelesexist.

– How many alleles exist?• Lots

– a protein with 300 AA– 900 nucleotides occupied by

1 of 4 nucs (A,T,C,g)– 4900 = 10542!!!– Typical microsatelite locus

has 10 - 20 alleles– Typical allozyme locus has

2-3– Typical RAPD or AFLP?

– Each mutation shouldcreate a novel allele thusthere are an infinite numberof possible alleles

• Mutation– Always creates a novel

allele• Any two identical alleles must

be Identical By Descent (IBD)– Calculating homozygosity

• Two alleles that must be IBDjoin in an individual

– Autozygous