LSM1102 Population Genetics

8/6/2019 LSM1102 Population Genetics

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LSM 1102POPULATION GENETICS


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Hardy-Weinberg Equilibrium


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D. S. Falconer & T.F.C. Mackay.

(1996)

Introduction to Quantitative Genetics4th Edition, Longman

RECOMMENDED TEXT


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Population Genetics

Deals with Frequencies/distribution of genes

(alleles), genotypes, and phenotypes in

populations.

Factors that maintain equilibrium or

change gene, genotype and phenotypefrequencies over time.


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Definition:

Large group of inter-breedingindividuals which shares a commongene pool.

A group of sexually reproducingorganisms.

A Genetic Population


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The Hardy-Weinberg Principle

Proposed independently in 1908 byEnglish mathematician G.H. Hardy andGerman physician W. Weinberg.

Relates gene/genotype frequencies at asingle Mendelian locus to phenotypefrequencies in a population.


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The H-W Principle Relative frequency (proportion) of genes

(alleles), genotypes and phenotypes remainsunchanged in the 2 nd generation.

No matter how many generations are studied,the relative frequencies will remain constant.

Actual numbers of individuals with eachgenotype (phenotype) will change as

population size changes, but relativefrequencies will remain constant.


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H-W Equilibrium

If a gene has 2 alleles, D and d, withpopulation frequencies of p and q,

respectively.p+q = 1

With 2 different alleles, there are 3 possible

genotypes: DD, Dd, & dd. Given allele frequencies of p and q, when a

population is in Hardy-Weinberg equilibrium,

these genotypes will have frequencies of p2,2pq, and q 2, respectively.

p 2 + 2pq + q 2 = 1

(A binomial distribution)


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H-W Equilibrium

E.g. for a gene with 2

alleles D and d, withpopulation frequencies of pand q respectively.

Genotype/phenotypefrequencies after 1generation:DD (D homozygotes) = p 2

Dd (Dd heterozygotes) = 2pqdd (d homozygotes) = q 2

dd(q2)

Dd(pq)

d(q)

Dd(pq)

DD(p2)

D(p)

d(q)

D(p)

Male Gametes

F e m a

l e G a m

e t e s

Punnett Square


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H-W Equilibrium

q42pq 3p2q2dd(q2)

2pq 34p 2q22p 3qDd

(2pq)

p2q22p 3qp4DD(p2)

dd(q2)

Dd(2pq)

DD(p2)

Genotype (and frequency) of male parent

G e n o

t y p e

( a n

d f r e q u e n c y

)

o f f e m a l e p a r e n

t


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H-W Equilibrium

q 22pqp 2freqRelative

q 2(p 2+2pq+q 2)2pq(p 2+2pq+q 2)p 2(p 2+2pq+q 2)Totalq 4----q 4dd x dd

2pq 32pq 3--4pq 3Dd x dd--2p 2q 2--2p 2q 2DD x dd

p 2q 22p 2q 2p 2q 24p 2q 2Dd x Dd--2p 3q2p 3q4p 3qDD x Dd----p 4p 4DD x DD

ddDdDDMatingFrequency

Mating type

Frequency of Offspring

Allele and genotype frequencies remain the same after each successive generation


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H-W Equilibrium

0.5

0.0 0.5 1.0Allele Freq of A 2 (q)

A1 A2

A2

A2

A1 A1

1.0

Female gametes

M a

l e g a m e

t e s

A1 A2(pq)

A2 A1(qp)

A1 A1(p2)

A2 A2(q2)


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Assumptions of H-W Principle Large population size small populations are more

susceptible to random fluctuations of allelefrequencies (due to transmission of any one allele bychance), a phenomenon called genetic drift.

Random mating population stratification (e.g. basedon ethnicity or religion), assortative mating (choice ofmates based on desired traits), or inbreeding aredifferent forms of non-random mating, which can leadto an in the frequency of homozygotes in the sub-

population or pedigree. No migration movement of individuals between

populations can gradually or gene frequencies, or

introduce new alleles gene flow.


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Assumptions of H-W Principle No new mutation new mutations constantly

occur and will gene frequency, but these are

offset by those lost by death. However,changes in mutation rates can or genefrequency.

No selection selection against a mutantallele will gene frequency, but this is offsetby new mutations. However, increasedreproductive fitness will gene frequency.Selection for a mutant allele, e.g. heterozygoteadvantage of sickle-cell trait, will theproportion of the advantaged genotype.


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H-W equilibrium is attained after 1 generation if allassumptions hold true.

This equilibrium will be maintained fromgeneration to generation as long as conditions do

not change.

H-W Principle


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Stability of Genes Frequencies

Computer simulatedgraph showing stablegene frequencies in alarge randomlymating population inH-W equilibrium.

Audesirk, Fig 15-2. pg 291

A freq (p) = 0.5, a freq (q) = 0.5


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Genotype A 1A1 A1A2 A2A2 TotalGenotype freq. P H Q 1(Probability)

Frequency of A 1 allele : p = (P + H)/Totalp = P + H

Frequency of A 2 allele : q = (Q + H)/Total

q = Q + H

CALCULATION OF GENE FREQUENCIES


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The M-N blood group in man is determined by 2

alleles at a single autosomal locus. Test whether this population is in H-W

equilibrium.

Genotype MM MN NN TOTAL

Obs. Frequency P H Q 1

Obs. number 50 86 45 181

EXAMPLE: H-W EQUILIBRIUM


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Calculate the values of p and qp = (P + H)/Total

= (50 + 86/2) = (50 + 43) = 93 = 0.51181 181 181

q = (Q + H)/Total= (45+ 86/2) = (45+43) = 88 = 0.49

181 181 181

CALCULATION OF M & N ALLELE FREQUENCIES


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Use the 2 test to test hypothesesHo : Population is in H-W equilibrium.

H1 : Population is not in H-W equilibrium.

Genotype MM MN NN TOTAL

Exp. freq. p 2 2pq q 2 1Exp. No. (p 2181) (2pq 181) (q 2181) 181

= 47.1 = 90.5 = 43.4 181

Where p = 0.51q = 0.49

CALCULATIONS OF ALLELE FREQ. OF M & N


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EXPECTED FREQUENCIES OF THE 3 GENOTYPES

Exp. freq. of MM genotype = p 2 = 0.51 2 = 0.2601

Exp. freq. of MN genotype = 2pq = 2 x 0.51 x 0.49 = 0.4998

Exp. freq. of NN genotype = q 2 = 0.490 2 = 0.2401

EXPECTED NUMBERS OF THE 3 GENOTYPES

Exp. number of MM = p2 x 181 = 0.2601 x 181 = 47.1

Exp. number of MN = 2pq x 181 = 0.4998 x 181 = 90.5Exp. number of NN = q2 x 181 = 0.2401 x 181 = 43.4

CALCULATIONS OF 2


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2 = (Obs. No. Exp. No.) 2Exp. No.

= (50 47.1) 2 + (86 90.5) 2 + (45 43.4) 247.1 90.5 43.4

= 2.9 2 + -4.5 2 + 1.6 2

47.1 90.5 43.4= 8.41 + 20.25 + 2.56

47.1 90.5 43.4

= 0.179 + 0.224 + 0.059

= 0.462

Compare calculated 2 with theoretical 2

CALCULATIONS OF 2


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Check critical 2 value from the table

Theoretical 2 = 2[]

= 20.05(3-2)

= 3.841

Since the calculated 2 (0.462) < theoretical 2

Conclusion: This population did not deviatesignificantly from expected H-W genotype proportionsfor the MN locus

CALCULATIONS OF 2


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(A1 gene) p from 0 1

(A2 gene) q from 0 1

p + q = 1

Gene frequencies around 1 or 0 are considered extreme

Gene frequencies around 0.5 considered intermediate

At intermediate gene frequencies, heterozygotes in a

population would be near the max. frequency of 0.5 (50%)

GENE FREQUENCIES


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SINGLE GENE LOCUSAUTOSOMAL TWO ALLELES, CODOMINANT

GENES (ALLELES) A 1 A2

GENE FREQUENCY p q

GENOTYPES A 1A1 A1A2 A2A2 TOTALExp. genotype freq. p 2 2pq q 2 1

Obs. genotype freq. P H Q

Able to distinguish the 3 genotypes

p = P + H q = Q + HTOTAL TOTAL p + q = 1

EXAMPLE: M-N BLOOD GROUP IN HUMANS

GENOTYPES MM MN NN


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Phenotypes Normal Albino Total

Genotypes AA Aa aa

Exp. freq. p 2 2pq q 2 1

Obs. freq. ? ? 1/20,000 1

Assume Exp. freq. = Obs. freq.

Therefore q 2 = 1/20,000 , q = 1/141

Freq. of carrier (Aa) = 2pq = 2 1/141 140/141

= 1/70

About one in 70 persons is a carrier.

SINGLE GENE LOCUS, AUTOSOMAL TWOALLELES, COMPLETE DOMINANCE


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1 single locus with than 2 alleles

Genotype frequencies can be found byexpanding equations

2 Alleles with frequencies p & q

Expand : (p + q)2

(binomial equation) 3 Alleles with frequencies p 1 q 1 & r

Expand : (p + q + r) 2

n Alleles with freq p 1, p 2, p 3, .. P n Expand (p 1 + p 2 + p 3 + + p n)2

MULTIPLE ALLELESSINGLE LOCUS AUTOSOMAL


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In a Scottish population, the observed

numbers of individuals in the A,B, AB and Oblood groups are given.

What are the gene frequencies in this pop?

(evaluate p, q and r)

MULTIPLE ALLELES: HUMAN A-B-O BLOOD GROUPS


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HUMAN A-B-O BLOOD GROUPSGENOTYPIC AND PHENOTYPIC FREQUENCIES

894

894/2610= 0.342

AIAIA IAIO

p2 2pr

Observed No.

Of Phenotypes

Observed freq. OfPhenotypes

PhenotypeGenotypeExp. genotypeFrequencies

70

70/2610= 0.027

309

309/2610= 0.119

ABIAIB

2pq

BIBIB IAIO

q2 2qr

2610

1

1337

1337/2610= 0.512

Total1

1

OIOIO

r2

Data of ABO Blood Types in Scottish Population


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Assume that the pop. is in H-W equilibrium.

Freq. of Io

gene is r Given that r 2 = 0.512, therefore r = 0.715

Freq of IA

(p) and IB

(q) genes can be found indirectly Freq. of A and O phenotypes = p 2 + 2pr + r 2

= (p + r)2

since p + q + r = 1, therefore (p + r) = (1 q)

HUMAN A-B-O BLOOD GROUPSCALCULATION OF GENE FREQUENCIES (p, q & r)

substitute


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Freq. of A & O phenotypes = (1 q) 2

0.342 + 0.512 = (1 q)2

1 q = 0.854

q = 0.076 (frequency of the IB

allele)p = 1 q r

= 1 0.076 0.715= 0.209 (frequency of I A allele)

HUMAN A-B-O BLOOD GROUPSCALCULATION OF GENE FREQUENCIES (p, q & r)

http://www umds ac uk/physiology/daveb/brainday/colourblindness/cblind htm


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Color blindness afflicts 8% of males and 0.04 % of human females.

Color perception depends on three genes, each producing chemicals sensitiveto different parts of the visible light spectrum.

Red and green detecting genes are on the X-chromosome, while the blue

detection is on an autosome.

http://www.umds.ac.uk/physiology/daveb/brainday/colourblindness/cblind.htm


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In Humans is XX (homogametic) is XY (heterogametic)

Single locus (sex-linked)2 alleles : A 1 A2Frequency : p q

SINGLE GENE LOCUS, SEX-LINKEDTWO ALLELES

R SP H QOBS. NO.

A1Y A2YA1 A1 A1 A2 A2 A2GENOTYPE

MALEFEMALE


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Freq. of A 1 among : p f = (P + H)Total

Freq. of A 1 among : p m = RTotal

(assuming equal population)

Freq. of A 1 in whole populationp = 2/3 p f + 1/3 p m

Freq. of A 2 in whole populationq = 2/3 q f + 1/3 q m

Population at equilibrium: p f = p m and q f = q m

SINGLE GENE LOCUS, SEX-LINKEDTWO ALLELES


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Color blindness in man is due to a recessivesex-linked gene (c)

Freq. of colorblind men is 7.8%

Freq. of colorblind women is 0.65%

What is the freq. of heterozygous women in thispop.?

Genotype CC Cc cc C cObs. freq. ? 0.0065 0.078

Exp. freq. p f2 2p fq f q f2 p m q m

SINGLE GENE LOCUS, SEX-LINKEDEXAMPLE FROM HUMANS (COLOR BLINDNESS)


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p = freq. of normal gene, q = freq. of c.b. gene

Among females : q 2f = 0.0065 therefore q f = 0.0065= 0.081

Among males : q m = 0.078 (given)

Overall in pop: q = 2/3 q f + 1/3 q m= (2/3 0.081) + (1/3 0.076) = 0.08

Therefore p = 1 q = 1- 0.08 = 0.92

Freq. of heterozygotes = 2 p q= 2 0.92 0.08= 0.147 15%

SINGLE GENE LOCUS, SEX-LINKEDEXAMPLE FROM HUMANS (COLOR BLINDNESS)

To calculate if the allele frequency of one


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To calculate if the allele frequency of one population differs from another population:

Population I Population IIAllele frequency p 1 p2

Population size n 1 n2

p1 - p2

z score = p 1(1-p 1) + p 2(1-p 2)n1x2 n 2x2

z score is significantly different if >1.96

p 0.2 0.1 0.05 0.01 0.001

z 1.28 1.65 1.96 2.58 3.29


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Mutation & Selection


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Terminology

Mutation A change in the genetic material, either of a single

gene or in the number of structure of thechromosomes.

A mutation which occurs in the gametes is inheritedbut not otherwise (ie, those in somatic cells)

Polymorphism The occurrence in a population of two of more

genetically determined forms in such frequencies(>1%) that the rarest of them could not be maintainedby mutation alone.


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Genetic drift Changes in genefrequencies when small groups ofindividuals are separated from orleave a larger populations.

Founder effect A type of geneticdrift in human populations in which afew members leave to found a newsettlement, perpetuating a subset ofthe alleles in the original population.

Terminology


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NON-RANDOM MATING

I. ASSORTATIVE MATING (selective mating)Leads to increase in frequencies of homozygotesand decrease in heterozygotes. The pop. becomespartially divided into 2 groups, mating takes placemore frequently within groups. Most extreme form isinbreeding

II. DISSORTATIVE MATING (best case scenario is random mating)Leads to an increase of heterozygotes and a reduction

of homozygotes.


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FORCES THAT CHANGE GENE FREQUENCIES

- Systematic processes1. MUTATION - amount and direction

2. SELECTION of change are predictable


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Changes in genes. Source of new genetic variationMutation rate is low 10 -5 or 10 -6

1. NON-RECURRENT MUTATION Rare event Not important

A1 A22. RECURRENT MUTATION

Mutational event at a specific gene locus. Itoccurs with a characteristic frequency.

A1 A2 = mutation rate of gene A 1 to A 2

TYPES OF MUTATION


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2 Types

1. One-way recurrent mutationA1 A2

(Initial gene freq.) p o q o

Results in: Decrease in freq. of A 1 gene (p 0)

Increase in freq. of A 2 gene (q 1)

p = change in freq. of A 1 gene due to 1generation of one-way recurrent mutation.

RECURRENT MUTATION


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RECURRENT MUTATIONOne-way recurrent mutationp = mut. rate orig. freq. of A 1

p = p op 1 : freq. of A 1 gene after 1 generation of one-way recurrentmut.

p 1 = p o - p op 1 = p o (1 )

p 2, the freq. of A 1 will decrease by the amount p 1p 2 = p 1 p 1p 2 = p 1 (1 )p 2 = p o (1 ) (1 )p 2 = p o (1 )2

Repeating the process for t generationsp

t= p

o(1 )t


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2. Two-way recurrent mutation

Change of frequencies of the A 1 and A 2 alleles

A1 A2

p o q o

Forward mut: Freq. of A 1 will decrease by - p o

Backward mut: Freq. of A 1 will increase by +vq o

Net change in A 1 after 1 gen. of 2-way recurrent

mut.

p = q o - p o

RECURRENT MUTATION

RECURRENT MUTATION


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Two-way Recurrent MutationGene Frequencies At EquilibriumAT EQUILIBRIUM- no net change in gene frequencies- p e , q e are equilibrium gene frequencies- q e = p e therefore p = 0

p = q e - p eq e - p e = 0 Let p e = (1-q e)

q e - (1 q e) = 0

q e - + q e = 0 q e + q e = u Similarly

q e = p e =

+ +

RECURRENT MUTATION


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SELECTION


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Individuals differ in viability and fertility

They therefore contribute different numbers of

offspring to the next generation FITNESS (adaptive value or selective value) is

the contribution of offspring to the next

generation Natural or artificial selection operates on the

gene

Therefore causing changes in the genefrequency

SELECTION

FITNESS


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In a particular environment, the most favoredgenotype (fittest) has optimum fitness of 1

Contributes the most offspring to the nextgeneration.

The fitness (W) of the genotypes selected against

is 1- s s is the coefficient of selection

This is the proportionate reduction in the gameticcontribution of particular genotype

s is also referred to as the selection intensity

FITNESS

FITNESS


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If s = 0:Fitness = (1- s) = 1 (no selection against genotype)

If s = 1:Fitness = 0 (complete selection against genotype)

If s = 0.1: Fitness = 0.9

For every 10 zygotes produced. By the fittestgenotype, the genotype with fitness of 0.9produced only 9.

Examine the degrees of dominance in relation tofitness.

FITNESS

Degrees of Dominance in Relation to Fitness


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Degrees of Dominance in Relation to Fitness

(Falconer, Fig 2.1)

SELECTION AGAINST A RECESIVE GENE


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A Hopi Albino girl.

Weaver & Hedrick, Fig 2.17, pg 36



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EXAMPLE ALBINISM (A > a)Complete selection against the recessive albino gene, a.s = 1 (fitness = 0) if albino are prevented from reproducing

Present frequency of albinos = 1/20,000 in EuropeAfter 100 gen. of selection against albinos (t = 100), what isthe frequencies of albinos in this population?

Phenotype Normal AlbinoGenotype AA Aa aa TOTALExp. Freq. p 2 2pq q 2 1

Fitness 1 1 1-sGametic contrib. p 2 2pq q 2(1-s) 1-sq 2


NO. OF GENERATIONS REQUIRED TO PRODUCE A CHANGE OF


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Initial frequency of a (albino gene) = qFreq. after 1 generation of selection = q 1q 1 = q

2(1-s) + pq since (1-s) = 0, s=1, and p=1-q1-sq 2

q 1 = q (1-q)1 q 2 (substitute 1 q 2 = (1-q)(1+q)

q 1 = q1 + qAfter 2 generations of selection : q 2

q 2 = q 1 (substitute q 1 = q/ 1+q 1 )

(1+ q 1)q 2 = q

1 + 2q

FREQUENCY OF A RECESSIVE GENE (t)

NO OF GENERATIONS REQUIRED TO PRODUCE A CHANGE


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After t generations : q t = q1 + tq

After 100 generations of selection ( t = 100)

Gen. 0: q = 1/20,000 = 1/141 p = 140/141Gen. t = 100 q 100 = q

1+ tq

q 100 = 1/141 / 1 + (100 1/141)q 100 = 1/241

Therefore freq. of albinos (aa) after 100 gens.

q 2100 = (1/241) 2 = 1/58,080One person in 58,080 persons is expected to be an albino.

NO. OF GENERATIONS REQUIRED TO PRODUCE A CHANGEOF FREQUENCY OF A RECESSIVE GENE (t)

Overdominance


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Selection favouring heterozygotesGenotype A 1A1 A1A2 A2A2 TotalExp. freq p 2 2pq q 2 1

Rel. fitness 1-s 1 1 1-s 2Gametic p 2(1-s 1) 2pq q

2(1-s 2) 1-s 1p2-s 2q

2

contribution

Frequency of the A 2 allele after 1 generation of selection:q 1 = No. of A 2 genes

New totalq 1 = q 2(1-s 2) + pq

1 s 1p 2 s 2q 2

Overdominance

O d i


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Change in gene frequency due to one generation

of selection:

q = q 1 q q = pq (s 1p s 2q)

1 s 1p2

s 2q2

At equilibrium - q = 0

then s 1p = s 2q

Therefore at equilibrium s 1p e = s 2q e

Overdominance


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Since delta q = q1-q and q1 is given in the previous slide, I tried to solve q1-q by:

To keep it simple, I am just solving the numerator,

q2 (1-s2) + pq -q(1-s1p 2-s2q 2)

=q2-s2q 2 + pq - q + s1p 2q - s2q 3

=q ( q-s2q + p -1 + s1p 2 +s2q 2)

Try and make p common for all terms within the bracket by subst q = 1-p, for the last term, I subst q = 1-p for one of

the q in q2

=q (1-p-s2q + p -1 +s1p 2 + s2q -s2qp)

Some of the terms like 1, p and s2q cancel out one another, leaving

= q (s1p 2-s2qp)

Take out p

= pq (s1p-s2q) Hoola! which is the same as the numerator in the delta q.

O d i


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Gene frequencies at equilibrium:

s 1p e = s 2q e s 1p e = s 2q es 1p e = s 2 (1- p e) s 1(1-q e) = s 2q e

p e = s 2 q e = s 1

s 1 + s 2 s 1 + s 2 Equilibrium gene frequencies are dependent on the

relative fitness of the two homozygote (s 1 and s 2)

The 3 genotypes are always present in the pop at fairly

high frequencies.

Overdominance

Overdominance


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GENETIC POLYMORPHISM More than one genetic form is maintained; A 1A1,

A1A2, and A 2A2 . Genetic variability is maintained. AT EQUILIBRIUM, GENE FREQUENCES ARE

MAINTAINED AT INTERMEDIATE LEVELS.

Arbitrary range 0.01 to 0.99 The commonest allele is not > 0.99. The rarest allele is not < 0.01

Equilibrium gene frequencies (p e , q e) must be toohigh to be maintained by balance between mutationand selection.

Overdominance

Overdominance


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Normal red blood cells Sickle-shaped red blood cells

(Lewis, pg 183)

Example: Sickle-cell anaemia

Overdominance

Sickle-cell anaemia


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Such cells occur in people carrying a mutant gene,

which causes crystalloid aggregates resulting in

functional abnormality along with distortedmorphology.

The sickle-shaped cells clog small blood vessels,interfere with oxygen transport to various tissues,

and cause anemia.

Sickle-cell anaemia: Distribution of Malaria correlates with thedistribution of sickle cell gene ( Hb s )


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distribution of sickle cell gene ( Hb )

Starr & Taggart Pg.291,

Fig.18.11

Distributionof Malaria

Distribution ofsickle cellgene (Hb s )

Sickle-cell anaemia in Humans: Example


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Hb s gene is present among Africans and theirdescendants in America.

Sickle-cell Sickle-cellPhenotype Normal trait anaemia

Genotype Hb AHb A Hb AHb s Hb s Hb s

Initial freq p 2 2pq q 2

Fitness 1-s 1 1 1-s 2

Sickle-cell anaemia in Humans: Example


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In West Africa, the equilibrium freq of the Hb s gene,q e = 0.1 (rather high freq)

Frequency the Hb A gene (normal gene), p e = 0.9

Assume Hb s Hb s homozygotes do not leave offspring,i.e. s 2 = 1 and fitness = 0

What is the fitness of normal homozygotescompared with heterozygotes? (evaluate 1-s 1)

q e = s 1 q e = 0.1s 1 + s 2 s 2 = 1

s 1 = 1/9

Therefore - fitness of Hb AHb A (W) = 1-s 1 = 1-1/9 =8/9

Possible example of overdominanceABO blood group in humans


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A possible example of heterozygote

advantage in AB0 human blood groups.Why are 3 alleles (I A, IB and I o) maintained at

intermediate levels in human populations?There may be a correlation with resistance to

certain human diseases

ABO blood group in humans

Overdominance - Possible correlation betweeni id f ABO ith di


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0

0.5

1

1.5

2

duo ulcer gas ulcer stom ulcer anaemia

disease

Relative Incidence of Diseases in Relation to ABOBlood Types

O

A,B,AB

A

incidence of ABO genes with diseases


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Balanced PolymorphismFactors Maintaining

polymorphism

Mechanisms That May Be Responsible


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1. Heterozygote advantage

(overdominance)2. Frequency-dependent selection3. Heterogeneous environment4. Transition5. Neutral mutation

For Polymorphism

Heterozygote Advantage


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Overdominance for fitness maintains anequilibrium gene frequency at intermediatelevels

Heterozygote advantage is an attractiveexplanation for polymorphism

But only few examples to support this Haploid organisms have been found tohave as much variability as diploid

organisms this argues strongly againstheterozygote advantages importance

Frequency-dependent selection


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Having a phenotype that is rare may itself be anadvantage

The rare phenotype is favoured.

The direction of selection is dependent on the genefrequency - an allele at low frequency is favoured but thesame allele at high frequency is selected against.

This leads to stable equilibrium gene frequency Birds and fish have been shown to take disproportionately

more of the more common type of food when offered achoice

This exerts frequency-dependent selection on polymorphicprey, eg snails, giving advantage to individuals with a rarepattern of colouration

Heterogeneous environment


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The environment experienced by individuals of apopulation is not constant: differs from place to place andfrom time to time

If one allele is advantageous in one environment andanother in a different environment, stable polymorphismcan result without heterozygotes necessarily being onaverage superior.

Selection is complex in such situation: depends on Dominance relations Individuals choose to breed in the environment to which they are

adapted Mating is preferentially between individuals from the same

environment or random Etc



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A relatively simple form of selection in a heterogeneous environmentresults in a cline

A cline is a gradient of gene frequency between one locality andanother

Clines are known to be maintained by selection favouring one allele inone locality and another allele in another locality, with limitedmigration, which allows mating only between individuals fromneighbouring parts of the cline

Location A Location B Location C Location D



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0

0.1

0.2

0.3

0.4

0.5

0.6

F i n n s

N o r w e g

i a n s

D a n e s

S w e d e s

B r i

t o n s

F r e n c

h

I r i s h

R u s s i a n s

C a n a d

i a n s

W h i t e s

( U S A )

B l a c k s

( U S A )

I t a l

i a n s

J e w s

S o u

t h A s i a n s

S ' p o r e

I n d i a n s

C h i n e s e

( C h i n a

)

T a i w a n e s e

S ' p o r e

C h i n e s e

J a p a n e s e

J a v a n e s e

Populations

R a r e

A l l e l e F r e q u e n c y

sp24 c

d X+ h z

North Europeans North

AmericansSouth

Europeans Asians

Allele frequencies (weighted averages) of theapolipoproteinB gene in the various populations of theworld clustered according to geographical locations.


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Transition


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Melanic and typical forms of the peppered moth

a) On a lichen-covered tree b) On a soot covered trunk

Biston betularia (peppered moth)

Dark variety(mutant carbonaria)

Lightvariety

Transition


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A comparison of the prevalence of the carbonariamutant and the typical morph of the moth, B .betularia in polluted and unpolluted area

Birmingham-polluted Dorset-unpolluted

Brooker Pg 726

Neutral mutation


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Selection force not responsible for polymorphism(unlike previous 4 mechanisms)

Polymorphism merely results from balancebetween mutation and loss by chance, whichdepends on population size

LSM1102 Population Genetics

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