Microevolution 2 – mutation & migration

A simple model of directional selection


• consider a single locus with two alleles A1 and A2



• let p = frequency of A1 allele



• let p = frequency of A1 allele • let q = frequency of A2 allele




• relative fitnesses are:

A1A1 A1A2 A2A2 w11 w12 w22





A1A1 A1A2 A2A2 w11 w12 w22

• it is also possible to determine relative fitnesses of the A1 and A2 alleles:





A1A1 A1A2 A2A2 w11 w12 w22


let w1 = fitness of the A1 allele





A1A1 A1A2 A2A2 w11 w12 w22




What is the fitness of an allele? Consider fitness from the gamete’s perspective:

♀

A1


♀

A1

A1

♂


♀

A1

A1

♂ A1A1


♀

A1

A1

♂ A1A1 “realized” fitness w11


♀

A1

A1


This route will occur with a probability p, since p is the frequency of the A1 allele


♀

A1

A1

♂ A1A1 “realize” fitness w11

This route will occur with a probability p, since p is the frequency of the A1 allele

p


♀

A1

A1


♂

A2

p


♀

A1

A1


♂

A2

p

A1A2 “realized” fitness w12


♀

A1

A1


♂

A2

p


This route will occur with a probability q, since q is the frequency of the A2 allele

q


♀

A1

A1


♂

A2

p


Therefore, w1 = pw11 + qw12

q


♀

A1

A1

♂ A1A1 “realized” fitness w1w1

♂

A2

p


Therefore, w1 = pw11 + qw12

q

(this is equivalent to a weighted average of the two routes)

What is the fitness of an allele? Similarly for the A2 allele:

♀

A2

A2


♂

A1

q


Therefore, w2 = qw22 + pw12

p

The fitness of the A1 allele = w1 = pw11 + qw12

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define….

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2

The fitness of the A1 allele = w1 = pw11 + qw12 The fitness of the A2 allele = w2 = qw22 + pw12 One final fitness to define…. Mean population fitness = w = pw1 + qw2 (This too is a weighted average of the two allelic fitnesses.)

Let p’ = frequency of A1 allele in the next generation

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2)

Let p’ = frequency of A1 allele in the next generation p’ = pw1/(pw1 + qw2) p’ = p(w1/w) Let q’ = frequency of A2 allele in the next generation q’ = qw2/(pw1 + qw2) q’ = q (w2/w)

An example of directional selection

An example of directional selection Let p = q = 0.5

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975 w2 = qw22 + pw12 = 0.925

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975 w2 = qw22 + pw12 = 0.925 w = pw1 + qw2 = 0.950

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw22 = 0.975 w2 = qw22 + pw11 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487

An example of directional selection Let p = q = 0.5 Genotype: A1A1 A1A2 A2A2 Fitness: 1 0.95 0.90 w1 = pw11 + qw12 = 0.975 w2 = qw22 + pw12 = 0.925 w = pw1 + qw2 = 0.950 p’ = p(w1/w) = 0.513 q’ = q(w2/w) = 0.487 In ~150 generations the A1 allele will be fixed

Conclusion: Natural selection can cause rapid evolutionary change!

Natural selection and mean population fitness


Question: How does w change during the process of directional selection?



Genotype: AA Aa aa Fitness: 1 1 0.50

Here, selection is favoring a dominant allele.

Directional selection always maximizes mean population fitness!



Genotype: AA Aa aa Fitness: 0.4 0.4 1

Here, selection is favoring a recessive allele.

Directional selection always maximizes mean population fitness!


Question: How does w change during the process of balancing selection?



Genotype: AA Aa aa Fitness: 1-s 1 1-t



Genotype: AA Aa aa Fitness: 1-s 1 1-t

^

Results in stable equilibrium point at p = t/(s + t)


Question: How does w change during the process of balancing selection? Example: suppose s = 0.40 and t = 0.60



Genotype: AA Aa aa Fitness: 0.6 1 0.4




^

Stable equilibrium point at p = t/(s + t)




^

Stable equilibrium point at p = t/(s + t) = 0.60

Balancing selection maximizes mean population fitness!

… but is only 0.75!

Balancing selection maximizes mean population fitness!

Conclusion: Natural selection always acts to maximize mean population fitness


• Sewall Wright envisoned populations occupying “adaptive landscapes”.



Wright at the University of Chicago in 1925



Wright in 1965


• Sewall Wright envisoned populations occupying “adaptive landscapes”. • these landscapes were covered with multiple adaptive peaks separated by valleys of reduced fitness

A hypothetical adaptive landscape


• Sewall Wright envisoned populations occupying “adaptive landscapes”. • these landscapes were covered with multiple adaptive peaks separated by valleys of reduced fitness • his shifting balance theory considered how populations could move from one peak to another.

A hypothetical adaptive landscape

How does a population move among peaks??

Mutation

Mutation • Let p = frequency of A1 allele


• Let q = frequency of A2 allele



• Let p = q = 0.50



• Let p = q = 0.50 µ

A1 A2

• Let µ = 1 x 10-5



• Let p = q = 0.50 µ A1 A2

υ

• Let µ = 1 x 10-5 (and ignore υ)

Mutation • denoting the change in A2 in one generation as Δq,


Δq = µ x p


Δq = µ x p

= (1 x 10-5)(0.5)


Δq = µ x p

= (1 x 10-5)(0.5)

= 0.000005


Δq = µ x p

= (1 x 10-5)(0.5)

= 0.000005 • the A2 allele has increased in frequency to 0.500005.


Δq = µ x p

= (1 x 10-5)(0.5)

= 0.000005 • the A2 allele has increased in frequency to 0.500005. • it would take another 140,000 generations to reach 0.875


Δq = µ x p

= (1 x 10-5)(0.5)

= 0.000005 • the A2 allele has increased in frequency to 0.500005. • it would take another 140,000 generations to reach 0.875 Conclusion: The rate of change due to mutation pressure is extremely small!

Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change.

Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly!

Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change.

Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change. 4. Most mutations are deleterious and experience “purifying selection”.

Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change. 4. Most mutations are deleterious and experience “purifying selection”. 5. A small (but unknown) proportion of mutations are beneficial and lead to adaptation.

Migration (gene flow)

Migration (gene flow) • gene flow is simply the movement of genes among populations.

Migration (gene flow) • gene flow is simply the movement of genes among populations. • it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals.

Migration (gene flow) • gene flow is simply the movement of genes among populations. • it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals. • its magnitude is determined by m

Migration (gene flow) • gene flow is simply the movement of genes among populations. • it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals. • its magnitude is determined by m m = the proportion of genes entering a population in individuals (genes) immigrating from a different population.

Sewall Wright’s Continent-Island model

A simple model of migration

A simple model of migration • let pI = frequency of A1 allele on island

A simple model of migration • let pI = frequency of A1 allele on island • let pC = frequency of A1 allele on continent

A simple model of migration • let pI = frequency of A1 allele on island • let pC = frequency of A1 allele on continent • let m = proportion of A1 alleles moving to island each generation

A simple model of migration • let pI = frequency of A1 allele on island • let pC = frequency of A1 allele on continent • let m = proportion of A1 alleles moving to island each generation • let p’I = frequency of A1 allele on island in the next generation


p’I = (1-m)(pI) + m(pc)


p’I = (1-m)(pI) + m(pc) ↑ ↑ “resident” “immigrant”

A simple model of migration • let ΔpI = change in frequency of A1 allele on island

from one generation to the next



ΔpI = p’I - pI



ΔpI = p’I - pI

= (1-m)(pI) + m(pc) - pI



ΔpI = p’I - pI

= (1-m)(pI) + m(pc) - pI

= m(pc – pI)

An example:

An example:

let pc = 0.75

An example:

let pI = 0.25 let pc = 0.75

An example:

let pI = 0.25 let pc = 0.75

let m = 0.10

An example:

let pI = 0.25 let pc = 0.75

let m = 0.10

let’s ignore back-migration

An example:

let pI = 0.25 let pc = 0.75

let m = 0.10

ΔpI = m(pc – pI)

An example:

let pI = 0.25 let pc = 0.75

let m = 0.10

ΔpI = m(pc – pI) = 0.10(0.75-0.25)

An example:

let pI = 0.25 let pc = 0.75

let m = 0.10

ΔpI = m(pc – pI) = 0.10(0.75-0.25) = 0.050

Now:

pI = 0.30 pc = 0.75

let m = 0.10

Now:

pI = 0.30 pc = 0.75

let m = 0.10

ΔpI = m(pc – pI) = 0.10(0.75-0.30) = 0.045

Conclusions

Conclusions 1. Gene flow can cause rapid evolutionary change.

Conclusions 1. Gene flow can cause rapid evolutionary change. 2. The long-term outcome will be the elimination of genetic differences between populations!

Conclusions 1. Gene flow can cause rapid evolutionary change. 2. The long-term outcome will be the elimination of genetic differences between populations! But… natural selection act to oppose gene flow

Example: Lake Erie water snakes (Nerodia sipedon)

Microevolution 2 – mutation & migration · 10/9/2015 · 1 allele = w 1 = pw 11 + qw 12 The...

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Transcript of Microevolution 2 – mutation & migration · 10/9/2015 · 1 allele = w 1 = pw 11 + qw 12 The...