selection and mutation. if either of the following occurs then the population is responding to...

Selection and Mutation

If either of the following occurs then the population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age.

-or-2. All individuals reach reproductive age but some individuals are able to produce more viable (reproductively successful) offspring. If these differences are heritable then evolution may occur over time.

It needs to be mentioned that most phenotypes are not strictly the result of their genotypes.

Environmental plasticity and interaction with other genes may

also be involved. In other words it is not as simple as we

are making it here but we have to start somewhere.

CautionCaution

1. Selection may alter allele frequencies or violate conclusion #1

2. Selection may upset the relationship between allele frequencies and genotype frequencies.

Conclusion #1 is not violated but conclusion #2 is violated.

In other words the allele frequencies remain stable but genotype frequencies change and can no longer be predicted accurately from allele frequencies.

After random mating which produces 1000 zygotes we get:

Initial frequencies

B1= 0.6; B2 = 0.4B1B1 B1B2 B2B2 1000 total

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 B1B2 B2B2 1000 total

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 B2B2 1000 total

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2 1000 total

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total

differential survival of offspring leads to reduced numbers of some genotypes

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

number surviving

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

number surviving 360

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

number surviving 360 360

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

number surviving 360 360 80

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive

number surviving 360 360 80 800 total

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive


The genotype frequencies of mating individuals which survive is

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive


The genotype frequencies of mating individuals which survive is .45

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive


The genotype frequencies of mating individuals which survive is .45 .45

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive


The genotype frequencies of mating individuals which survive is .45 .45 .10

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive



The resulting allelic frequencies in the new reproducing population is

B1 = B2 =

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive




B1 =

.45+1/2(.45)

= 0.675

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive




B1 =

.45+1/2(.45)

= 0.675

B2 =

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive




B1 =

.45+1/2(.45)

= 0.675

B2 = 1/2(.45)+0.10

= 0.325

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive




B1 =

.45+1/2(.45)

= 0.675

B2 = 1/2(.45)+0.10

= 0.325

an increase of .075

a decrease of .075

Initial frequencies

B1= 0.6; B2 = 0.4360 B1B1 480 B1B2 160 B2B2

1000 total


100%

survive

75 %

survive

50 % survive




B1 =

.45+1/2(.45)

= 0.675

B2 = 1/2(.45)+0.10

= 0.325

an increase of .075

a decrease of .075

Thus, conclusion #1 is violated and the allele frequencies areare changing; we are not in equilibrium. The population is evolvingevolving!

analyze the population on the basis of the fitness of the offspring produced.

The fittest individuals will survive the selection process and leave offspring of their own.

We are going to define fitness as the survival rates of individuals which survive to reproduce.

If :w11 = fitness of allele #1 homozygote (exp B1B1)w12 = fitness of the heterozygote (exp B1B2)w22 = fitness of allele #2 homozygote exp (B2B2)

mean fitness of the population will be described by the formula: ŵ = p2w11 + 2pqw12 + q2w22

MEAN FITNESSMEAN FITNESS

CAUTION! Use ONLY allele frequencies in these formulas NOT genotype frequencies!

B1= 0.6 and B2 = 0.4 and

fitness of B1B1 = 1.0 (100% survived)

fitness of B1B2 = .75 ( 75% survived)

fitness of B2B2 = .50 (50% survived) Figure the mean fitness now.Figure the mean fitness now.

B1= 0.6 and B2 = 0.4 and



fitness of B2B2 = .50 (50% survived) Figure the mean fitness now.Figure the mean fitness now. ŵ= (.6)2(1)+

B1= 0.6 and B2 = 0.4 and



fitness of B2B2 = .50 (50% survived) Figure the mean fitness now.Figure the mean fitness now. ŵ= (.6)2(1)+(2(.6)(.4)(.75)) +

B1= 0.6 and B2 = 0.4 and



fitness of B2B2 = .50 (50% survived) Figure the mean fitness now.Figure the mean fitness now.ŵ= (.6)2(1)+(2(.6)(.4)(.75)) + (.4)2 (.5) = .80

B1B1 = P2w11

ŵ

B1B2 = 2pqw12

ŵ

B2B2 = q2w22

ŵ

We can use these formulas which can calculate the new expected genotype frequencies based on the fitness of each genotype and the allele frequencies in the current generation.

B1 = p2w11+pqw12 B2 = pqw12+q2w22

ŵ ŵ

Δ B1 = Δp = p (pw11+qw12 – ŵ)

ŵ

Δ B2 = Δq = q (pw12+qw22 – ŵ)

ŵ

Go back to the problem we did in class last time. Taking this current population that you have already analyzed, figure out what the new genotypenew genotype and allele allele frequenciesfrequencies will be if the fitness of these individuals is actually as follows:

SS individuals 0.8 ; Ss individuals 1.0 and the ss individuals 0.6.

ŵ = p2w11 + 2pqw12 + q2w22

ŵ= (.82) 2 (.8) + 2(.82)(.18)(1.0) + (.18)2 (.6)ŵ = .537 + .295 + .019 = .85B1B1 = P2w11

ŵ

B1B2 = 2pqw12

ŵ

B2B2 = q2w22

ŵ

; SS = (.82)2(.8) / .85 = .633

;Ss = 2(.82)(.18)(1.0) / .85 = .347

;ss = (.18)2(.6) / .85 = 0.023

Last time we calculated S = .82 and s = .18

Now we set the fitnesses at w11(SS)=.8;w12(Ss)=1;w22(ss)=.6

Calculate the ŵ and BCalculate the ŵ and B11BB1;1; B B11BB2;2; and B and B22BB22 values for the next values for the next

generation nowgeneration now

Hint: Do they add up to 1.0?

B1 = p2w11+pqw12

ŵ

B2 = pqw12+q2w22

ŵ

S = (.82)2(.8) + (.82)(.18)(1.0) = .806 .85

s = (.82)(.18)(1.0) + (.18)2(.6) = .196 .85

We an also calculate the new allele frequencies as well

So…… B1B1 = .63 B1B2 = .35 B2B2 = .02

and

B1 = .80 B2 = .20

Is this population in equilibrium?

Have the allele frequencies changed?

Can we predict the genotype frequencies from the allelic frequencies?

Fruit fly experiments of Cavener and Clegg

Worked with fruit flies having two versions of the ADH (alcohol dehydrogenase) enzyme, F and S. (for fast and slow moving through an electrophoresis gel)

Grew two experimental populations on food spiked with ethanol and two control populations on normal, non-spiked food. Breeders for each generation were Breeders for each generation were picked at randompicked at random.

Took random samples of flies every few generations and calculated the allele frequencies for AdhF and AdhS

Figure 6.13 pg 185

only difference is ethanol in food no migration assured random mating population size, drift? no mutation.

Must be selection for the fast form of gene. Indeed studies show that AdhF form breaks

down alcohol at twice the rate as the AdhS form.

Therefore offspring carrying this allele are more fit and leave more offspring and the make-up of gene pool changes.

Selection may upset the relationship between allele frequencies and genotype frequencies.

Conclusion #1 ( allele frequencies do not change) is not violated but conclusion #2 (that we can predict genotype frequencies from allele

frequencies) is violated.

Initial B1 = 0.5

Initial B2= 0.5 250 B1B1 500 B1B2 250 B2B2 1000 Total

Differential fitness of the genotypes Fitness .50 Fitness 1.0 Fitness .50

Number of survivors to reproductive age 125 500 125 750 total

Initial B1 = 0.5




The genotype frequencies of mating individuals which survive

Initial B1 = 0.5





125 / 750

0.167

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167

The resulting allelic frequencies in new mating population

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167


B1 =

.167+1/2 (0.667)

= 0.5

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167


B1 =

.167+1/2 (0.667)

= 0.5

B2=

½ (.667) + .167

= 0.5

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167


B1 =

.167+1/2 (0.667)

= 0.5

B2=

½ (.667) + .167

= 0.5

No change No change

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167


B1 =

.167+1/2 (0.667)

= 0.5

B2=

½ (.667) + .167

= 0.5

No change No change

Thus conclusion #1 is not violated therefore this population has notnot evolved…..but…..

Initial B1 = 0.5





125 / 750

0.167

500 / 750

0.667

125 / 750

0.167


B1 =

.167+1/2 (0.667)

= 0.5

B2=

½ (.667) + .167

= 0.5

Change in allele frequencyNo change No change

Thus conclusion #1 is not violated therefore this population has not evolved….but…..

Conclusion #2 is violated. We are not in equilibrium.

Frequency of B1B1 =.167 which is not equal to (.5)2

Kuru example among the Foré in New Guinea◦ Pg 188-191◦ Wanted to determine if there was a genetic basis

to the resistance of kuru infection. Ritualistic mortuary feasts, only young women

ate the contaminated nervous system tissue leading to CJD (similar to mad cow disease)

Among young women who never participated he Met allele = 0.48 and the Val allele 0.52; Genotypes were: Met/Met 0.22; Met/Val 0.51 and Val/ Val 0.26 very close to the values expected for H-W.

Met = 0.52 and Val = 0.48 The expected genotypes are Met/Met 0.27 ; Met/ Val 0.5 and Val/Val 0.23 The actual were: Met/Met 0.13 ; Met/ Val 0.77 and Val/Val

0.10 Appears homozygotes are susceptible but

heterozygotes are protected.

HIV example in book. pg 191 Two conditions must be met

1. Need a high enough frequency of the beneficial allele in the population gene pool

2. There must be high selection pressure for the allele in the same area. In this case a high incidence of HIV infection.

If selection is acting, does the rate of evolution of a particular allele depend on whether it is….

heterozygote or homozygote?

dominant or recessive?

Tribolium Beetle example

Dawson’s Flour beetle example Studied a gene locus that had a wild type

(+) allele and a lethal allele. +/+ or +/L are normal L/L is lethal. Two experimental populations composed

of all heterozygotes +/L Therefore started with + = 0.5 and L

=0.5. Expected populations to evolve toward

lower frequency of the L allele.

Results showed that the recessive lethal did drop rapidly at first but slowed down over successive generations.

WHY?

As you go on there are less and less homozygous lethals for selection to act on and the lethal allele hides in the heterozygotes

In each succeeding generation all LL are lost and ++ makes up a greater proportion of the survivors.

Dawson showed that dominancedominance and allele frequencyfrequency interactinteract to determine the rate of evolution when acted on by selection

If a recessive allele is common evolution is rapid because there are a lot of homozygotes that express the phenotype for selection to act on.

If recessive allele is rare, evolution is slow because the rare allele is hidden in the heterozygotes where selection cannot act.

His experiments also demonstrated that ◦ controlled lab situations can accurately predict the

course of evolution ◦ populations do what you would expect if selection is

occurring as predicted by the evolutionary theory.

Normally in a recessive/ dominant gene, the fitness of the heterozygote will be equal to one of the homozygotes

Also, it is possible for the heterozygotes to have a fitness intermediate to the two homozygotes.

Thirdly we may find Heterozygote Superiority or Inferiority

Studied a gene in which Homozygotes for one allele are viable Homozygotes for the other allele are

not viable and are lethal. Heterozygotes have a higher fitness

than either homozygotes

Started with all heterozygotes to establish a new population (each allele =.5)

After several generations equilibrium was reached at .79 frequency for the viable alleleThis means that the lethal allele was maintainedmaintained at frequency of 0.21! How could How could this bethis be?

Started more populations beginning with frequency of .975 of viable allele. Expect the population to eliminate all lethal alleles and fix the viable allele at 1.0.

But .....

Figure 6.18 pg 200

The viable allele droppeddropped in frequency and the same equilibrium around a frequency of .79 was reached for the viable allele!

There is some advantage to the heterozygote condition and the heterozygote actually has a superior fitness to either homozygote.

Example in humans is sickle cell anemia Leads to the maintenance of genetic

diversity = balanced polymorphismbalanced polymorphism

Where the heterozygote condition is inferior to either of the homozygotes

What do you predict would happen What do you predict would happen to the allele frequencies here?to the allele frequencies here?

Leads to fixation of one allele in the population, while the other is lost.

Either allele may be fixed depending on conditions and beginning frequencies of each allele in the gene pool.

Leads to a loss of genetic diversity Although if different alleles are fixed in

different populations can help maintain genetic diversity among populations

When one allele is consistently favored it will be driven to fixation

When heterozygote is favored both alleles are maintained and at a stable equilibrium (balanced polymorphism) even though one of the alleles may be lethal in the homozygous state.

The Elderflower orchid example in book

Population’s allele frequencies remain at or near an equilibrium but it is due to the direction of selection fluctuating.

First one allele is favored and then the other.

The population fluctuates around an equilibrium point.

Bumblebees visit yellow and purple flowers alternately

The least frequent phenotype is visited more often and receives more pollination events.

In subsequent generations this color becomes more and more frequent until it becomes the dominant color.

Once this happens then the same color becomes less frequently visited and the other color becomes favored.

Oscillation between the two colors continues and the favored allele alternates over time around some mean equilibrium value.

Mutation is the source of all new alleles Mutation provides the raw material on

which selection can act

Mutation alone is a weak or nonexistent evolutionary force

If all mutations that happened, occurred in gametes so that they would be immediately passed on to their offspring and ….

the rate of mutation were high, say Aa at a rate of 1 in 10,000 per generation.

then the rates are very slow as shown in figure 6.23

Figure 6.23 pg. 211

In concert with selection, mutation becomes a potent evolutionary force.

Used a strain of E. coli that cannot exchange DNA (conjugation) so the only possible source of genetic variation is mutation.

Showed steady increases in fitness and size over 10,000 generations in response to a demanding environment. (little over 4 years)

However, increases in fitness occurred in jumps when a beneficial mutation occurred and then spread rapidly through the population

Figure 6.25 pg 213

When mutations are deleterious Selection acts to eliminate them Deleterious Mutations persist because they

are created anew over and over again When the rate at which deleterious

mutations are formed exactly equals the rate at which they are eliminated by selection the allele is in equilibrium. = mutation-selection balance

If the mutation is only mildlymildly deleteriousdeleterious and therefore selectionselection against it is weakweak; andand MutationMutation rate is highhigh then ◦The equilibrium frequency of the mutated allelemutated allele will be relatively in in the populationthe population.

If, on the other hand, there is strong strong selection againstselection against a mutation (the mutation is highly deleterious) and the mutation rate is mutation rate is low low then ◦Equilibrium ratio of the mutated allele mutated allele will be will be in the population

highhigh

lowlow

Spinal muscular atrophy, second most common lethal autosomal recessive disease in humans. Selection coefficient is .9 against the disease mutations.

However, among Caucasians 1 in 100 people carry the disease causing allele.

Research shows that the mutation rate for this disease is quite high

Mutation selection balance is proposed explanation for persistence of mutant alleles.

http://www.smafoundation.org

Cystic fibrosis is the most common lethal autosomal recessive disease in Caucasians

Mutation-selection balance alone cannot account for the high frequency of the allele = .02

Appears to also be some heterozygote superiority involved

Heterozygotes are resistant to typhoid fever bacteria and have superior fitness during typhoid fever epidemic.

At the current time it is believed that CF is an example of heterosis and not mutation-selection balance

An autosomal dominant allele Is actually increasing in the human

population. Any ideas why? May be because it increases the tumor

supressor activity in cells dramatically lowering the incidence of cancer in those with the defective allele.

They survive through the reproductive years and leave more offspring than their unaffected siblings.

selection and mutation. if either of the following occurs then the population is responding to...

Documents

reduced numbers

surviveinitial frequenciesb1

genotypesinitial frequenciesb1

greater survival

successful offspring

allele frequencies

genotype frequencies

reproductive age