natural selection dna encodes information that interacts with the environment to influence phenotype...

Natural Selection

DNA encodes information that interacts with the environment to influence phenotype

Among The Traits That Can Be Influenced By Genetically Determined Responses to the Environment Are:

1. The Viability in the Environment2. Given Alive, the Mating Success in the

Environment3. Given Alive and Mated, Fertility or

Fecundity in the Environment.

Viability

Hb- Locus In Africa:

A/A A/S S/S

Non-Malarial Area

No Anemia No Anemia Anemia

Viability: High High Low

Malarial AreaNot Resistant

to MalariaResistant Anemia

Viability: Low High Low

p/p fetus develops in Low

Phenylalanine in utereo

Environment

Mentally Retarded

Institutionalized

Low Chance of Mating

p/p BabyBorn With

Normal Brain

NormalDiet

LowPhenylalanine

Diet

Mating Success

Normal Intelligence

High Chance of Mating

Fecundity/Fertility

H/+ In A Society

With No Birth

Control, No

Genetic Literacy, and Low Expected Lifespan:Normal

Fecundity

H/+ In A Society

With Birth Control, Genetic

Literacy, and High Expected Lifespan:

Low Fecundity

Why Are Viability, Mating Success, and Fecundity/Fertility

Important Phenotypes in Evolution?

Because All Of These Phenotypes Influence The Chances For

Successful DNA Replication

Physical Basis of Evolution

• DNA can replicate

• DNA can mutate and recombine

• DNA encodes information that interacts with the environment to influence phenotype





ViabilityMating Success

Fecundity/Fertility





ViabilityMating Success

Fecundity/Fertility

These Are Combined Into A SinglePhenotype of Reproductive Success

Or FITNESS

DNA can replicateDNA can mutate and recombine

Genotypic VariationIn Demes and Species

Environment

PhenotypicVariationIn Fitness

HeritableVariationIn Fitness

Natural Selection Is Heritable Variation In Fitness

That Is, The Genes Borne By A Gamete Influence The

Probability of That Gamete Being Passed On To The Next

Generation.

THINK LIKE A GAMETE!

DNA can replicateDNA can mutate and recombine

Genotypic VariationIn Demes and Species

It’s the Environment, stupid!

PhenotypicVariationIn Fitness

HeritableVariationIn Fitness

NATURAL SELECTION IS NOT CIRCULAR

Natural Selection At A Single Locus in A Randomly Mating Deme

AA

p2

Aa

2pq

aa

q2

Zygotic Frequencies

VAA VAa VaaViabilities

aa

q2 Vaa

Aa

2pqVAa

AA

p2 VAA

Environment

Adult Frequencies

CAA CAa CaaMating Prob.Environment

aa

q2 VaaCaa

Aa

2pqVAaCAa

AAp2 VAACAA

Mated Adult Frequencies

bAA bAa baaAve. No. OffspringEnvironment

aaq2 VaaCaabaa

Aa

2pqVAaCAabAa

AAp2 VAACAAbAA

Mated Adult FrequenciesWeighted By No. of Off.

Let WAA = VAACAAbAA; WAa = VAaCAabAa; Waa = VaaCaabaa

AA

p2

Aa

2pq

aa

q2

WAA

Zygotic Frequencies

WAa WaaFitness

AA

p2 WAA

Aa

2pqWAa

aa

q2 Waa

Environment

Mated Adult FrequenciesWeighted By No. of Off.

aa

q2Waa/W

Aa

2pqWAa/WAA

p2 WAA/WMated Adult Frequencies

Convert to Freq. By Dividing by = W = p2WAA+2pqWAa+q2Waa

1 1/21/2

aq’= q2 Waa/W + pqWAa/W

Ap’= p2 WAA/W + pqWAa/W

1Meiosis

Gene Pool

Ap’= p2 WAA/W + pqWAa/W

aq’= q2 Waa/W + pqWAa/W

Gene Pool

p’= p2 WAA/W + pqWAa/W

=( p2 WAA+ pqWAa)/W

p’ = p(pWAA+ qWAa)/W

Does Evolution Occur?

p = p’ - p

= p(pWAA+ qWAa)/W - p

= p[pWAA+ qWAa)/W - 1]

p = p[pWAA+ qWAa- W]/W

Does Evolution Occur?Note, W = W(p+q)=pW+qW

p = p[pWAA+ qWAa- W]/W

=p[p(WAA-W)+ q(WAa-W)]/W

Since p and W are always > 0,This is the only part of the equationThat Can Change Sign and Hence

Determine the Direction of EvolutionUnder Natural Selection.

Does Evolution Occur?What is:

p(WAA-W)+ q(WAa-W)?

Mean Phenotype of Fitness


p(WAA-W)+ q(WAa-W)?

Genotypic Deviations for thePhenotype of Fitness


p(WAA-W)+ q(WAa-W)?

This is the Average Excess of the A Allelefor the Phenotype of Fitness


p = paA/W


p = paA/W

Natural Selection is An Evolutionary ForceWhenever p ≠ 0 or p ≠ 1 (that is, there isGenetic variation) and when aA ≠ 0 (that is,When there is heritable variation in thePhenotype of fitness).

To Understand Natural Selection

THINK LIKE A GAMETE!

Sickle Cell Anemia In Africa

An Example of Natural Selection

The Sickle Cell Mutation

Infection of a Red Blood Cell By a Malarial

Parasite• Sickle-Cells Are Filtered Out

Preferentially by the Spleen• Malaria Infected Cells Are Often Filtered

Out Because of Sickling Before the Parasite Can Complete Its Life Cycle

• The Sickle Cell Allele is Therefore an Autosomal, Dominant Allele for Malarial Resistance.

The SickleCell Anemia Phenotype

Most Deaths Due to Sickle Cell Anemiaand Due to Malaria Occur Before

Adulthood. Viability Is The Phenotypeof Living To Adulthood

• In a non-Malarial Environment, The S Allele is a Recessive Allele For Viability Because Only the Homozygotes Get Sickle Cell Anemia.

• In a Malarial Environment, The S Allele is an Overdominant Allele For Viability Because Only the Heterozygotes Are Resistant to Malaria And Do Not Get Sickle Cell Anemia.

Two Complications to This Simple Story in Africa:

1. Epidemic Malaria is Recent to Most of Wet, Tropical Africa and the Process of Adaptation to Malaria in Africa Is Still Not in Equilibrium.

2. There is a Third Allele, Hemoglobin C, Involved in the Adaptation to Malaria in Africa.

Epidemic Malaria in Africa

MADAGASCAR

ICELAND

About 2000 years ago, A Malayo-Indonesian

Colony Was Established on Madagasgar


This Colony Introduced The

Malaysian Agricultural

Complex into This Region


This Agricultural Complex Was

Taken Up By Bantu-Speaking Peoples,

Followed by A Large Expansion of the Bantu In Africa About 1500 years

Ago.

The Malaysian Agricultural Complex In Africa

• Is associated with slash-and-burn agriculture: Provides habitat and breeding sites for Anopheles gambiae, the primary mosquito vector for falciparum malaria.

• Results in the high local densities of human populations that are necessary to establish and maintain malaria as a common disease.

The Hemoglobin C Mutation

GAGGlutamic Acid

Hb-A GTGValine

Hb-S

AAGLysine

Hb-C

6th Codon

The Hemoglobin C Mutation

Hb-C Is A “Recessive”Allele for Malarial

Resistance

Hb-A, S and CGenotypes AA AS SS AC CS CC

Anemia No NoYes

(Severe)No

Yes

(Mild)NO

Malarial Resistance

No Yes Yes No Yes Yes

Viability No Malaria

1 1 0.2 1 0.7 1


Anemia No NoYes

(Severe)No

Yes

(Mild)NO

Malarial Resistance



1 1 0.2 1 0.7 1

The A and S Alleles Define An Autosomal Recessive Genetic Disease: Selection Will Insure it is Rare But Difficult to

Eliminate in a Random Mating Population.


Anemia No NoYes

(Severe)No

Yes

(Mild)NO

Malarial Resistance



1 1 0.2 1 0.7 1

The A and C Alleles Define A Set of Neutral Alleles in a Non-malarial Environment: Their Frequencies Are

Determined by Genetic Drift and Mutation.


Anemia No NoYes

(Severe)No

Yes

(Mild)NO

Malarial Resistance



1 1 0.2 1 0.7 1

Viability Malaria

0.9 1 0.2 0.9 0.7 1.3

Observed Relative Viabilities In Western Tropical Africa

Hb-A, S and C• CC is the Fittest Genotype By Far• If Natural Selection is “Survival of the

Fittest”, Then Natural Selection Should Increase the Frequency of the C allele and the CC Genotype.

• Contrary to Rumor, Natural Selection is Not “Survival of the Fittest.”

• Natural Selection Is Heritable Variation in Fitness, so Think Like A Gamete: Which Gamete Has the Highest Average Excess of Fitness?

Initial Gene Pool Before Malaria

A

pA = 0.99

pS=.005 pC=.005

Initial Ave. Fitness After Transition to Malaysian Agricultural Complex

A

pA = 0.99

pS=.005 pC=.005

Under Random Mating, theMean Phenotype = W = 0.901

Initial Phenotypes After Transition to Malaysian Agricultural Complex

Genotypes AA AS SS AC CS CC

Viability Malaria 0.9 1 0.2 0.9 0.7 1.3

Genotypic Deviation (W = 0.901)

-.001 .099 -.701 -.001 -.201 .399

A

pA = 0.99

pS=.005 pC=.005




-.001 .099 -.701 -.001 -.201 .399


aA = -0.0005aS = 0.0935aC = 0.0000


The Initial AdaptiveResponse To A MalarialEnvironment MediatedBy Natural Selection IsTo Decrease A, IncreaseS, and Leave C The Same px = px(ax)/W)

aA = -0.0005aS = 0.0935aC = 0.0000

Gene Pool After Several Generations of Selection Under

A Malarial EnvironmentA

pA = 0.95S

pC = 0.005

W = 0.907pS = 0.045


A Malarial Environment pC=.005

A

pA = 0.95S

.045




-.007 .093 -.707 -.007 -.207 .393


A Malarial EnvironmentAfter the Initial AdaptiveResponse To A MalarialEnvironment, NaturalSelection Continues to Decrease A, Increase S, but Now It Also Decreases C Because aC= -0.014.

aA = -0.003aS = 0.055aC = -0.014

As pS increases in frequency, W increases and theseGenotypic Deviations Become Increasingly Negative.Therefore, Natural Selection Eliminates the C Allele.



pA 1-pS

S

pS

pC 0



A Selective Equilibrium Will Only Occur When p = 0 Under Natural Selection For All Alleles.

Genotypes AA AS SS

Viability Malaria 0.9 1 0.2

aA = (1-pS)(0.9-W)+pS(1-W) = 0 = aS = (1-pS)(1-W)+pS(0.2-W)

A

pA = 1-pS

S

pS

A Selective Equilibrium Will Only Occur When p = 0 Under Natural Selection For All Alleles.

A

pA = 1-pS

S

pS

aA = (1-pS)(0.9-W)+pS(1-W) = aS = (1-pS)(1-W)+pS(0.2-W) (1-pS)(0.9)+pS(1) = (1-pS)(1)+pS(0.2)

0.9+0.1pS = 1-0.8pS

0.9pS = 0.1pS = 0.1/0.9 = 0.11

So At Equilibrium, pS = 0.11 and pA=0.89

The Equilibrium Allele Frequencies Are Maintained By Natural Selection, Resulting in a

Balanced Polymorphism A

pA = 0.89

S

pS=0.11

The Balance Occurs Because When pS < 0.11, aS > 0(malarial resistance dominates the average excess)

And When pS > 0.11, aS < 0(anemia dominates the average excess)

The EquilibriumA

pA = 0.89

S

pS=0.11

At Equilibrium, There is Genotypic Variation in Fitness (Broad-Sense Heritability), but No

Heritability (Average Excesses = 0).

AA

0.79

AS

0.20

SS

0.01

WAA = 0.9 WAS = 1WSS

=0.2

Adaptation By Natural Selection Depends Upon History:

Which Mutations Are Present and Their Frequencies. The

course of adaptation is always constrained by the available

genetic variation and proceeds until there is no heritability of

fitness.

Two Possible Responses to Malaria

A

pA

pS 0 pC 0

A

pA = 0.89

S

pS=.11

C

pC = 1

1. The Fittest Genotype is Eliminated. 1. The Fittest Genotype is Fixed.

2. Average Fitness goes from .9 to .91. 2. Average Fitness goes from .9 to 1.3.

3. 20% of the individuals have a relative viability of 1 and 80% have either anemia or malarial susceptibility.

3. 100% of the individuals have a relative viability of 1.3 and none have anemia nor malarial susceptibility.

Two Possible Responses to Malaria

A

pA

pS 0 pC 0

A

pA = 0.89

S

pS=.11

C

pC = 1

1. The Fittest Genotype is Eliminated. 1. The Fittest Genotype is Fixed.

2. Average Fitness goes from .9 to .91. 2. Average Fitness goes from .9 to 1.3.

3. 20% of the individuals have a relative viability of 1 and 80% have either anemia or malarial susceptibility.

3. 100% of the individuals have a relative viability of 1.3 and none have anemia nor malarial susceptibility.

With One Exception



1 1 0.2 1 0.7 1

C is a neutral allele relative to A, so

sometimes the C allele will drift to high

frequencies relative to the A allele.

S is a recessive, deleterious allele relative to A, so

natural selection in the pre-Malarial environment will

keep it rare (no h2).

Suppose There Was A Deme With This Gene Pool Before The Malaysian Agricultural Complex

A

pA = 0.95C

.045

pS=.005

Such a gene pool is likely to evolve in the pre-malarial environment because of the neutrality of A and C

relative to each other.




-.002 .098 -.702 -.002 -.202 .398


aA = -0.001aS = 0.081aC = 0.015

The Initial Adaptive Response To A Malarial Environment Is To Increase The Frequency of The S and C Alleles.



pA = 0.78S

.05

C

0.17




-.01 .09 -.71 -.01 -.21 .39


A Malarial EnvironmentAfter the Initial AdaptiveResponse To A MalarialEnvironment, NaturalSelection Continues to Decrease A, Increase C, but Now It Also Decreases S Because aS= -0.005.

aA = -0.009aS = -0.005aC = 0.044

o.o

o.o5

o.15

0.20

0.25

0.10

o.o o.o5 0.10 o.15

S Allele Frequency in 72 West African Populations

CAllele

Frequency

A Negative Correlation Exists Between the Frequencies of the S and C alleles in Malarial

Regions in Africa

Even uniform selective pressures produce divergent adaptive responses because selection

operates upon variation whose creation and initial frequencies are profoundly influenced by

random factors such as mutation and drift.

Although adaptation is often portrayed as “optimizing”

individual or population fitness, only gametic fitness is optimized via natural selection. Individuals or demes with the highest fitness are not necessarily favored and can be actively selected against.

There are many other ways in which human populations have adapted to malaria; e.g. G-6-PD Deficiency:

Plasmodium oxidizes RBC NADPH from the Pentose Phosphate pathway for its metabolism. This results in a deficiency of RBC GSH, most severe in G6PD deficient individuals, leading to peroxide-induced hemolysis which curtails the development of Plasmodium.

There are many other ways in which human populations have adapted to malaria; e.g.

Thalassemia:

Adaptation generally involves many loci with different biochemical, cellular or

developmental functions.

Therefore, we also need to model natural selection as a polygenic

process.

The Fundamental Theorem of Natural Selection

• Fisher was one of the first to model natural selection as a polygenic process.

• Although there are many aspects of his models, the most important results are found in what he termed the “fundamental theorem of natural selection.”

x = phenotypic value of some trait for an individual in a population

f(x) = the probability distribution that describes the frequencies of x in the population.

The mean phenotype is then:

€

= xf(x)dxx∫

w(x) = the fitness of those individuals sharing a common phenotypic value x.

The mean or average fitness of the population is:

€

w = w(x)f(x)dxx∫

w(x)f(x) does not in general define a probability distribution, but w(x)f(x)/ does integrate to one and defines the probability distribution of the selected individuals.

Hence, the mean phenotype of the selected individuals is:

Let h2 = the heritability of the trait. The response to selection is given by R=h2S where S=(s–), R = (o–), and o is the phenotypic mean of the offspring of the selected parents.

€

s =xw(x)f(x)dx

x∫w

€

w

When x = w, w(w) = w by definition, and = .

€

w

€

s =w × wf(w)dw

w∫w

=w2f(w)dw

w∫w

€

s =w2 − w 2( ) + w 2[ ]f(w)dw

w∫w

=w2 − w 2( )f(w)dw + w 2 f(w)dw

w∫w∫w

€

s =w − w ( )

2f(w)dw + w 2

w∫w

=σ 2 + w 2

w

€

S = μ s − μ =σ 2 + w 2

w − w =

σ 2 + w 2 − w 2

w =

σ 2

w

When x = w, the response to selection, R, is . Hence,

€

w

€

R = h2S

Δw =σ a

2

σ 2

⎛

⎝ ⎜

⎞

⎠ ⎟σ 2

w

⎛

⎝ ⎜

⎞

⎠ ⎟

Δw =σ a

2

w

Fundamental Theorem of Natural Selection

Some Implications of FFTNS

• FIRST, natural selection can only operate when there is genetic variation associated with phenotypic variation for fitness in the population.

• SECOND, the only fitness effects that influence the response to natural selection are those transmissible through a gamete.

• THIRD, the adaptive outcome represents an interaction of fitness variation with population structure.


• FOURTH, selective equilibria can only occur when all the average excesses and all the average effects are zero; that is, when all gamete’s have the same average fitness impact. Evolution due to natural selection stops only when there is no heritability for fitness. This in turn means that at a selective equilibrium there is no correlation between the fitness of parents and the fitness of their offspring even when there is genetic variance in the phenotype of fitness.

The EquilibriumA

pA = 0.89

S

pS=0.11

At Equilibrium, There is Genotypic Variation in Fitness (Broad-Sense Heritability), but No

Heritability (Average Effects = 0).

AA

0.79

AS

0.20

SS

0.01

WAA = 0.9 WAS = 1WSS

=0.2


• FIFTH, natural selection acts to increase the average fitness of a population on a per generational basis. Because the additive genetic variation must be greater than or equal to zero, w ≥ 0 under natural selection. Because average fitness can only increase or stay the same under natural selection, the selective equilibria discussed under point four must always correspond to an average fitness local optimum.

Wright’s Concept of An Adaptive Surface or Landscape

AA AS SS

WAA = 0.9 WAS = 1 WSS=0.2

0.0 0.2 0.4 0.6 0.8 1.0p

0.2

0.4

0.6

0.8

1.0Average Fitness

Frequency of S Frequency of S

0.00 0.05 0.10 0.15 0.20p

0.89

0.90

0.91

0.92Average Fitness


• SIXTH, natural selection only takes populations to local adaptive solutions and not necessarily to the adaptive state with the highest average fitness, and indeed may operate to prevent an adaptive state with higher average fitness from evolving.

0.0 0.2 0.4 0.6 0.8 1.0p

0.7

0.8

0.9

1.0Average Fitness

Wright’s Concept of An Adaptive Surface or Landscape

AA Aa aa

WAA = 1 WAa = 0.5 Waa=0.9

Frequency of A



w

A

S

C

A

C

pC

pS pA

S

pC

pS pA

0.89

0.895

0.9

0.905

0.91

A

C

S

0.25

0.5

0.75

1

1.25

A

S

C


• SEVENTH, natural selection generally does not optimize, even in a local sense, any individual trait other than fitness itself, even if the trait contributes to fitness in a positive fashion.

Given a selective equilibrium ( = 0) at a local peak, let feq(x) = the phenotypic distribution of the trait at equilibrium. Then, the average fitness and average trait at equilibrium is:

€

w

€

w eq = w(x)feq(x)dxx∫

€

x eq = xfeq(x)dxx∫

is the “optimal” value of trait X only if

€

x eq

€

w eq = w(xeq )

€

w(x) ≈ w(xeq ) + w'(x eq )(x − x eq ) + 12 w' '(x eq )(x − x eq )2

Use Taylor’s theorem to expand w(x) around :

€

x eq

Take the average value of both sides of the Taylor’s Series approximation by integrating across the equilibrium probability distribution of the trait:

€

w eq = w(x eq ) feq(x)dxx∫ + ′ w (x eq ) x − x eq( )feq(x)dx + 1

2x∫ ′ ′ w (x eq ) x − x eq( )2feq(x)dx

x∫

€

w eq = w(xeq )+ 12 ′ ′ w (xeq )σ eq

2 (x)

€

x eq is an optimal value of trait X that maximizes w(x) when:

1. the trait has no phenotypic variance at equilibrium [2

eq(x) = 0], or 2. the trait is related to fitness in a strictly linear fashion at equilibrium [w”( ) = 0]

€

x eq


• EIGHTH, the process of adaptation can result in the evolution of some seemingly non-adaptive traits. In general, many traits contribute to fitness, not just one. Consider the case in which two traits, say X and Y, contribute to fitness such that w(x,y) is the fitness of those individuals with trait values x and y for the two traits respectively. Then, the two-dimensional requirement for optimality of both traits is:

€

∂2w(xeq ,yeq )

∂x 2σ eq

2 (x) + 2∂ 2w(xeq , yeq )

∂x∂yCoveq (x,y) +

∂ 2w(xeq , yeq )

∂y 2σ eq

2 (y) = 0

E.g., many human populations have adapted to malaria by increasing the frequency of the trait of hemolytic anemia. Here, natural selection favors the increase of a highly deleterious trait.

Such cases are common because of pleiotropy, and indeed most of the people who die or suffer from genetic disease do so because natural selection favored the genes despite one or more pleiotropic deleterious traits.


• NINTH, the course of adaptive evolution is strongly influenced by genetic architecture.

wA

C

pC

pS pA

S

0.89

0.895

0.9

0.905

0.91

A

C

S

C recessive to A for malarial resistance

w

A

C

pC

pS pA

S

0.89

0.895

0.9

0.905

0.91

A

C

S

C with 4% dominance to A for malarial resistance

Common theme?

p = paA/W

€

w =σ a

2

w

The Course of Adaptive Evolution Is Determined By the Phenotypic Effects

Assigned to GAMETES, not individuals!

natural selection dna encodes information that interacts with the environment to influence phenotype...

Documents

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pqv aa c aa b aa aa

pqv aa c aa aa p

pqv aa aa p

pqw aa aa q

w aa environment

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offspring environment