MATH 786: Mathematical Biology Population Geneticsguindon/math764/... · MATH 786: Mathematical Biology Population Genetics St´ephane Guindon Department of Statistics, UoA. Room

MATH 786: Mathematical Biology

Population Genetics

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Stephane Guindon

Department of Statistics, UoA.Room 211.

[email protected]

Introduction Molecular essentials The Hardy-Weinberg law

Outline

Introduction

Molecular essentials

The Hardy-Weinberg law


Why bother ?

• Population genetics is the bridge between Darwin’s naturalselection and Mendel’s laws of heredity.

• Population genetics explains the genetic basis of evolution

(origin of the “Modern Evolutionary Synthesis”).

• Population geneticists study the genetic constitution ofpopulations and how this constitution changes fromgeneration to generation.


Why bother ?


A theoretical approach to biology

• The genetic constitution of a population can be estimated,

• but it is more difficult to assess the variations of itsconstitution during the course of evolution (time-scaleproblem).

• Progress is made in population genetics by constructingmathematical models of evolution.

• An example : genetic drift.

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Time

DECREASE

INCREASE


Darwin’s theory of evolution

• Variations are observed among individuals in a givenpopulation or, at a higher level, across populations belongingto the same species.

• Some individuals are fitter thanothers (“Survival of thefittest”).

• Natural selection gets rid ofthe less fit individuals.

• According to Darwin, evolution is a continuous

phenomenon.


Mendel’s theory of heredity

• Mendel’s work seems to conflict with Darwin’s continuousevolution...


Founders

R.A. Fisher S. Wright J.B.S. Haldane


Outline

Introduction




Genetic and molecular essentials

• Gene: the physical entity transmitted from parent to offspringduring reproduction.

• Genes exist in different forms or states which are called alleles.

• Genotype: the allelic makeup of an individual (with referenceto a specific character)

• Phenotype: physical expression of the genotype (e.g., eye’scolour, sickle cell anemia)


DNA double-strand


Transcription (DNA → mRNA)


Transcription (DNA → mRNA)


The genetic code


The central dogma in molecular biology


Genomes and chromosomes

• Genomes: the whole hereditary information encoded in DNA.

• Chromosomes: discrete physical sections of the genome.

23 chromosome pairs (humans are diploids)20,000-25,000 genes – 3 billion base pairs


Alleles in dipoid organisms

• The position of a gene along a chromosome is called thelocus of the gene.

• If the two alleles at a locus are indistinguishable in theireffects on the organism, then the individual is said to behomozygote.

• If the two alleles at a locus are distinguishable because of theirdiffering effects on the organism, then the individual is said tobe heterozygote at the locus.

• Let’s consider a gene with two alleles A and a:• Genotypes AA and aa are found among homozygotes.• Genotypes Aa (or Aa) are found among heterozygotes.


From individuals to populations

• Population genetics focuses on allele and genotypes

frequencies in evolving populations.

• e.g., frequency of alleles a and A, or genotype aA in apopulation monitored over a certain period of time.

• Note: with n alleles, if the n(1+n)2 − 1 genotype frequencies are

known, the n allele frequencies can easily be worked out. Howabout the opposite...


Outline

Introduction





• Relationship between the allele and genotype frequencies.

• Applies to diploid organisms with sexual reproduction.

• Life cycle:

Aa

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Aa

AA

Aa

A a A a

aa



Mating Freq. of mating Genotype freq. (AA, Aa, aa)

AA x AA P2AA

1, 0, 0AA x Aa PAAPAa 1/2, 1/2, 0AA x aa PAAPaa 0, 1, 0Aa x AA PAAPAa 1/2, 1/2, 0Aa x Aa P2

Aa1/4, 1/2, 1/4

Aa x aa PAaPaa 0, 1/2, 1/2aa x AA PAAPaa 0, 1, 0aa x Aa PaaPAa 0, 1/2, 1/2aa x aa P2

aa 0, 0, 1

• P ′

AA= P2

AA+ PAAPAa + 1

4P2Aa

= (PAA + 12PAa)

2 = p2.

• P ′

aa = P2aa +PAaPaa + 1

4P2Aa

= (Paa + 12PAa)

2 = (1−p)2 = q2.

• P ′

Aa= 1 − p2

− q2 = 2pq.

• p′ = P ′

AA+ 1

2P ′

Aa= p2 + pq = p, and q′ = q.



• p and q are assumed to be the same among males andfemales.

p q

p P ′

AA= p2 P

′

Aa

2 = pq

qP′

Aa

2 = pq P ′

aa = q2

• p′ = P ′

AA+ 1

2P ′

Aa= p and q′ = q.

• PAA = p2, PAa = 2pq, Paa = q2.



• Assumptions:

1 Randomly mating population.2 No mutation.3 No migration.4 No natural selection.5 Infinite population size.6 Discrete generations.


Overlapping generations

• Assuming non-overlapping generations is reasonable only fororganisms which breed synchronously and only once in theirlifetime.

• During an amount δt of time, a fraction δt of the populationdies and is replaced.

PAA(t + δt) = PAA(t)(1 − δt) + pA(t)2δt (1)

PAa(t + δt) = PAa(t)(1 − δt) + 2pA(t)pa(t)δt (2)

• Show that pA is constant...

• Solving (1) gives: PAA(t) = PAA(0)(e−t) + p2A(1 − e−t).

• A fraction e−t of the population consists of survivors of theoriginal population.

• Individuals born later are in Hardy-Weinberg proportions.


Overlapping generations

PAA(0) = 0.3, pA = 0.6 (p2A

= 0.36).

0 1 2 3 4 5 6

0.3

00

.31

0.3

20

.33

0.3

40

.35

0.3

6

t

PA

A(t

)


Different gene frequencies in the two sexes

• Let pf = p + δ and pm = p − δ

• We have:

P ′

AA= (p + δ)(p − δ) = p2

− δ2

P ′

Aa = (p + δ)(1 − p + δ) + (p − δ)(1 − p − δ) = 2pq + 2δ2

P ′

aa= (1 − p − δ)(1 − p + δ) = q2

− δ2

• We also have p′

m = p′

f= 1

2 (pf + pm).

• We reach Hardy-Weinberg at the ??? generation...


Sex linked locus

• Locus linked to the X chromosome (female are XX, males areYX)

• pf , pm: frequencies of A among females and malesrespectively.

• For female offspring :

P ′

AA = pf pm

P ′

Aa= pf (1 − pm) + pm(1 − pf )

P ′

aa = (1 − pf )(1 − pm)

• For male offspring :

P ′

AY = pf

P ′

aY = (1 − pf )

• Hence, p′

f= 1

2(pf + pm) and p′

m = pf .


Sex linked locus

5 10 15 20

0.1

00

.12

0.1

40

.16

0.1

80

.20

Generation

Alle

le f

req

ue

ncy


Sex linked locus

• Analytical solution ?

• Let pf = p and pm = q.

• pn = 12(pn−1 + pn−2)

• Solution is of the form pn = rn. Find r .

• rn = 12(rn−1 + rn−2) → r2

−12 r − 1

2 = 0.

• Solutions: r1 = −12 and r2 = 1.

• Hence, pn = −12

nC + 1nD .

• p0 = C + D. Hence, D = 23(p0 + 1

2q0) and C = p0 − D.


Documents

MATH 786: Mathematical Biology Population Geneticsguindon/math764/... · MATH 786: Mathematical Biology Population Genetics St´ephane Guindon Department of Statistics, UoA. Room