Population Genetics Population genetics - PowerPoint Presentation

Slide1

Population Genetics

Slide2

Population genetics

is concerned with the question of whether a particular allele or genotype will become more common or less common over time in a population, and Why.Example: Given that the CCR5-D32 allele confers immunity to HIV, will it become more frequent in the human population over time?

Population Genetics

Slide3

Predicting Allele Frequencies

Populations in Hardy-Weinberg equilibrium

Slide4

Yule vs. Hardy

What are the characteristics of a population that is in equilibrium or another words, not evolving.Yule thought that allele frequencies had to be 0.5 and 0.5. for a population to be in equilibrium.Hardy proved him wrong by developing the Hardy-Weinburg equation.

Slide5

Punnett square

60 % of the eggs carry allele A and 40% carry allele a60% of sperm carry allele A and 40% carry allele a.

Slide6

Sample problem

In a population of 100 people, we know that 36% are AA , 48% are Aa, and 16% are aa.Determine how many alleles in the gene pool are A or a.Each individual makes two gametes....How many A alleles are in this population’s gene pool? _____ How many a alleles? _____

120

80

(36*(2)+48)

(16*(2) +48)

Slide7

What percent of the alleles are

A or a ? 120 / 200 = .6 or 60% A ; or .6 = frequency of allele A80 / 200 = .4 or 40% a ; or .4 = frequency of allele a

Slide8

Creating the Hardy-Wienburg equation is a matter of combining probabilities found in the Punnett square.

Slide9

Combining Probabilities

The combined probability of two independent events will occur together is equal to the product of their individual probabilities.What is the probability of tossing a nickel and a penny at the same time and having them both come up heads?½ x ½ = ¼

Slide10

Combining Probabilities

The combined probability that either of two mutually exclusive events will occur is the sum of their individual probabilities. When rolling a die we can get a one or a two (among other possibilities), but we cannot get both at once. Thus, the probability of getting either a one or a two is 1/6 + 1/6 = 1/3

Slide11

Calculating Genotype Frequencies

We can predict the genotype frequencies by multiplying probabilities.

Slide12

Hardy-Weinburg equation

Genotype Frequencies

Slide13

Zygotes

Allelic frequency

Genotype frequency

AA

(p)(p)

p

2

Aa

(p)(q)

2pq

aA

(q)(p)

aa

(q)(q)

q

2

 

 

Genotype frequencies described by

p

2

+2pq+q

2

=1.0

Slide14

The relationship between allele and genotype frequency

Let original A frequency be represented by p and original a frequency be represented by qSince there are only two alleles possible for this gene locus, The frequencies of A and a must equal 1.0Therefore, p + q =1.0

Slide15

Sample: calculating genotype frequencies from allele frequencies?

If a given population had the following allele frequencies: allele frequency (p) for A of 0.8 allele frequency (q) for a of 0.2Determine the genotype frequencies of this population? AA Aa aa

0.64

0.32

0.04

AA

=

p

2

;

Aa

=

2

pq

; and

aa

=

q

2

as follows…

Slide16

We can also calculate the frequency of alleles from the genotype frequencies.

When a population is in equilibrium the genotype frequencies are represented as..P2

+ 2pq +q

2

The allele frequency can therefore be calculated as follows.

A =

p

2

+ ½(2pq) and a =

q

2

+ ½(2pq)

Slide17

Examining our example again we see that if we use the frequencies we calculated for each genotype….

p2 2pq q2

0.64 AA .32 Aa .04 aa

A =

p

2

+ ½ (2

pq)

A=.64 + ½ (.32)

A = 0.8

and since

q

= 1-

p

; then a = 1-(0.8 ) a = 0.2

Slide18

These rules hold as long as a population is in equilibrium.

Hardy Weinberg Equilibrium describes the conclusions and assumptions that must be present to consider a population in equilibrium.

Slide19

Hardy Weinberg Conclusions

The allele frequencies in a population will not change from generation to generation. You would need at least 2 generations of data to demonstrate this.If the allele frequencies in a population are given by p and q then the genotype frequencies will be equal to p2; 2pq ; q2. Therefore if AA can not be predicted by p2

Aa

cannot be predicted by

2pq

and

aa

cannot be predicted by

q

2

then the population is not in equilibrium

Slide20

There are 5 assumptions which must be met in order to have a population in equilibrium

There is no selection. In other words there is no survival for one genotype over anotherThere is no mutation. This means that none of the alleles in a population will change over time. No alleles get converted into other forms already existing and no new alleles are formed There is no migration (gene flow)New individuals may not enter or leave the population. If movement into or out of the population occurred in a way that certain allele frequencies were changed then the equilibrium would be lost

Slide21

There are no chance events

(genetic drift) This can only occur if the population is sufficiently large to ensure that the chance of an offspring getting one allele or the other is purely random. When populations are small the principle of genetic drift enters and the equilibrium is not established or will be lost as population size dwindles due to the effects of some outside influence There is no sexual selection or mate choice Who mates with whom must be totally random with no preferential selection involved.

Exceptions to Hardy Weinberg cont.

Slide22

Genetic Distance: Definitions

Allele: Different forms of a gene.

Genotype:

The specific allele in an individual.

Phenotype:

The expression of a genotype.

Allele

Genotype

Phenotype

Homozygote

Homozygote

Heterozygote

Slide23

Genetic Distance: Definitions

Microsatellite: Short consecutive repeats:

Single nucleotide polymorphism (SNP):

Variation in a single nucleotide of a genome between two individuals.

Slide24

Linkage disequilibrium (LD):

Correlation between alleles at two different position.

Haplotype:

Combination of alleles at multiple linked loci which are transmitted together.

Genetic Distance: Definitions

Slide25

Evolution

Evolutionary forces:

Natural selection

: Probability of survival and reproduction

- Genetic drift:

Change in allele frequencies entirely by chance.

Slide26

Drift

Change in allele frequencies due to samplinga ‘stochastic’ processNeutral variation is subject to driftSelection

Change in allele frequencies due to

function

‘

deterministic

’

Functional variation may be subject to selection (more later)

The two forces that determine the fate of alleles in a population

Selection

vs

Drift

Slide27

Genetic Drift 1

Slide28

4 populations

2 at N=25

2 at N=250

Genetic Drift 2: Population Size Matters

Slide29

Effective population size N

eSewall Wright (1931, 1938)“The number of breeding individuals in an idealized population that would show the same amount of dispersion of allele frequencies under random genetic drift or the same amount of inbreeding as the population under consideration".Usually, N

e

< N

(absolute population size)

N

e

!= N can be due to:

fluctuations in population size

unequal numbers of males/females

skewed distributions in family size

age structure in population

Slide30

Selection vs Drift 1: |s| and Pop Size

If |s| < 1/Ne,then selection is ineffective and the alleles are solely subject to drift: the alleles are “effectively neutral”What is the probability of fixation?

If |s| < 1/N

e

, then P(fix) = q

N

e

= effective pop size

s = selection coefficient

q = allele frequency

If |s| > 1/N

e

, then P(fix) =

1 - e-4 N sq

1 - e-4 N se

e

Source: A. Sidow, BIOSCI 203

Slide31

Mutation

: Change in nucleotide sequence of genes caused by copying error or exposure to radiation, chemical substance, viruses,...

- Migration

Slide32

Fixation Index (

Fst): Measure of population differentiation.Π

Between

(

Π

Within

): Average number of pairwise difference between two individuals sampled from different (the same) population.

Π

Between

Π

Within

Slide33

NON-RANDOM MATING

Inbreeding:

mating between close relatives leads to deviations from H-W equilibrium by causing a deficit of heterozygotes.

In the extreme case of self-fertilization:

Generation AA Aa aa

0 p

2

2pq q

2

1 p

2

+ (pq/2) pq q

2

+ (pq/2) 2 p2 + (3pq/4) pq/2 q2 + (3pq/4)

Slide34

HOW CAN WE QUANTIFY THE AMOUNT OF INBREEDING IN A POPULATION?

The inbreeding coefficient,

F

The probability that a randomly chosen individual caries two copies of an allele that are

identical

by descent from a recent ancestor.

The probability that an individual is

autozygous

Slide35

Consider two pedigrees:

A

1

*

A

2

A

1

A

2

A

1

A

2

A

1

*

A

2

A

1

*

A

2

A

1

*

A

1

*

A

1

*

A

2

A

1

*

A

1

A

1

*

A

1

*

IBD

AVERAGE

F

FROM EACH MATING IS 0.25

Full-sib mating

Backcross

IBD

Slide36

LOSS OF HETEROZYGOSITY IN LINE OF SELFERS

Population Size (N) = 1

Heterozygosity after one generation, H

1

= (1/2) x H

0

Heterozygosity after two generations, H

2

= (1/2)

2

x H

0

After

t generations of selfing, Ht = (1/2)t x H0Example: After t = 10 generations of selfing, only 0.098% of the loci that were heterozygous in the original individual will still be so. The inbred line is then essentially completely homozygous.

Slide37

DECLINE IN HETEROZYGOSITY DUE TO INBREEDING

Slide38

HETEROZYGOSITY IN A POPULATION THAT IS PARTIALLY INBRED

In an inbred population the frequencies of homozygous individuals are higher than expected under HWE. Thus, the observed heterozygosity will be lower that expected under HWE.

H

obs

= 2pq(1-F) = H

exp

(1-F).

F

ranges from

0

(no inbreeding) to

1

(completely inbred population)

Slide39

F

CALCULATED FROM HETEROZYGOTE DEFICIT

Where,

H

exp

= frequency of heterozygotes if all matings were

random

F

= (H

exp

– H

obs

) / H

exp

Slide40

INBREEDING COEFFICIENT,

F

As the inbreeding coefficient (

F

) increases, fitness often decreases.

INBREEDING DEPRESSION

Slide41

INBREEDING DEPRESSION IN HUMAN POPULATIONS

Slide42

INBREEDING VERSUS RANDOM GENETIC DRIFT

Inbreeding

is caused by non-random mating and leads to changes in genotype frequencies but

not

allele frequencies.

Random genetic drift

occurs in finite populations, even with completely random mating, and leads to changes in

both

genotype and allele frequencies.

Both processes cause a decline in heterozygosity.

Slide43

Smith et al.

Why does inbreeding cause a decrease in fitness?

What genetic mechanisms, or type of gene action are responsible?

Slide44

QUANTIFYING POPULATION SUBDIVISION

Vs.

Random Mating Population -

Panmictic

Subdivided Population -

Random mating within but

not

among populations

Slide45

HOW DO WE MEASURE MIGRATION (

GENE FLOW)?

Direct Methods – e.g., mark-recapture studies in natural populations. For many organisms this is

not

a realistic option.

Indirect Methods – e.g., molecular marker variation.

SS

FS

FF

SS

FS

FS

FF

FF

FS

SS

Slide46

CONSIDER TWO COMPLETELY ISOLATED POPULATIONS

Due to random genetic drift, the allele frequencies in the populations diverge.

In an extreme case, they can be fixed for alternate alleles:

A

1

A

1

A

1

A

2

A2A2Population 1 1.0 0 0Population 2 0 0 1.0Overall HWE 0.25 0.50 0.25Individuals in population 1 are clearly more closely related to one another than they are to individuals in population 2.In this context,

the inbreeding coefficient (F) represents the probability that two gene copies within a population are the same, relative to gene copies taken at random from all populations lumped together.

Slide47

QUANTIFYING POPULATION SUBDIVISION WITH F

ST

F

st

measures variation in allele frequencies among populations.

Ranges from 0 to 1

F

st

compares the

average expected heterozygosity

of individual subpopulations (

S

) to the

total expected heterozygosity if the subpopulations are combined (T).

Slide48

F

ST AND POPULATION SUBDIVISION

At

Panmixis,

F

ST

= 0

All subpopulations have the same allele frequencies.

Complete Isolation, F

ST

= 1

All subpopulations are fixed for different alleles.

Slide49

Example:

Consider three subpopulations with 2 alleles at frequencies p and q, p q HS=2pq

Subpop 1: 0.7 0.3 0.42

Subpop 2: 0.5 0.5 0.50

Subpop 3: 0.3 0.7 0.42

Average H

S

= 0.446

Slide50

The

total expected heterozygosity across all subpopulations is calculated from the average allele frequency, p q Subpop 1: 0.7 0.3 Subpop 2: 0.5 0.5 Subpop 3: 0.3 0.7 p = 0.5 q = 0.5Remember that,

H

T

= 2pq = 0.5

F

ST

= (0.50 - 0.466) / (0.50) = 0.11

Slide51

WRIGHT

’

S ISLAND MODEL:

Consider

n

subpopulations that are diverging by drift alone, not by natural selection, and with an equal exchange of migrants between populations each generation at rate

m

……

What is the equilibrium level of population subdivision (

F

ST

)?

m

m

m

m

Slide52

RELATIONSHIP BETWEEN F

ST AND Nm IN THE ISLAND MODEL

N

m

is the absolute number of migrant organisms that enter each subpopulation per generation.

At equilibrium:

And:

When

N

m

= 0, F

ST

= 1

Nm = 0.25 (1 migrant every 4th generation), Fst

= 0.50 Nm = 0.50 (1 migrant every 2nd generation), Fst = 0.33 Nm = 1.00 (1 migrant every generation), Fst = 0.20 Nm = 2.00 (2 migrants every generation), Fst = 0.11

Slide53

If

N

m

>> 1

,

little divergence by drift;

If

N

m

<< 1

,

drift is very important

ROLE OF DRIFT IN POPULATION DIVERGENCE

Slide54

Find Genes which are candidates to have been under selection:

Very low and very high

F

st

distance.

Compare expected and observed values of

F

st

.

Slide55

Detection of Selection in Humans with SNPs

Large-scale SNP-survey looked at:106 Genes in an average of 57 human individuals60,410 base pairs of noncoding sequence (UTRs, introns, some promoters)135,823 base pairs of coding sequence

Some salient points:

Because survey is snapshot of

current

frequencies, evidence for selection or drift is indirect

This is about bulk properties, not about individual genes

Slide56

F

st matrix analysis:

Phylogenetic tree

Based on SNP of 120 genes in 1,915 individuals

Slide57

Based on 783 microsatellites in 1,027 individuals

Principal Component Analysis

Slide58

Mitochondrial DNA (

mtDNA):

In mitochondria (out of nucleus)‏

transmitted along only female lineages.

No recombination.

High mutation rate:

Abundance of polymorphic

Difficult genealogy reconstruction