chapter 6 n.
Skip this Video
Download Presentation
Chapter 6

Loading in 2 Seconds...

play fullscreen
1 / 93

Chapter 6 - PowerPoint PPT Presentation

  • Uploaded on

Chapter 6. Selection and Mutation. Adding Selection to the Hardy-Weinberg Analysis. If either of the following occurs then the population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age . -or-

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'Chapter 6' - tamra

Download Now An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter 6

Chapter 6

Selection and Mutation

adding selection to the hardy weinberg analysis
Adding Selection to the Hardy-Weinberg Analysis
  • If either of the following occurs then the population is responding to selection.

1. Some phenotypes allow greater survival to reproductive age.


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.
we will look at two possible effects of natural selection on the gene pool
We will look at two possible effects of natural selection on the gene pool.
  • Selection may alter allele frequencies or violate conclusion #1
  • 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.

FIRST: An example of what we might normally see happen to allele frequencies when natural selection is at work
Let B1 and B2 = the allele frequencies of the initial population with frequencies of B1 = .6 and B2 = .4
  • After random mating which produces 1000 zygotes we get:
sample analysis of a population for the effects of selection
Sample analysis of a population for theeffects of selection
  • 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 fitnessas the survival rates of individuals which survive to reproduce.
fitness formulas
Fitness formulas


  • 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

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

for our previous example
For our previous example
  • 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.
for our previous example1
For our previous example
  • 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.
  • ŵ= (.6)2(1)+
for our previous example2
For our previous example
  • 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.
  • ŵ= (.6)2(1)+(2(.6)(.4)(.75)) +
for our previous example3
For our previous example
  • 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.

ŵ= (.6)2(1)+(2(.6)(.4)(.75)) + (.4)2 (.5) = .80


We can also use the fitness to calculate the expected frequency of each genotype in the next generation. We can do it the long way , if we know actual numbers OR…….

B1B1 = P2w11ŵ

B1B2 = 2pqw12ŵ

B2B2 = q2w22ŵ

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.


We can also use the fitness values and current allele frequencies to calculate the expected frequency of each allele in the next generation.

B1 = p2w11+pqw12B2 = pqw12+q2w22ŵ ŵ

finally we can calculate the change in the frequency of alleles b 1 or b 2 directly as follows
Finally, we can calculate the change () in the frequency of alleles B1 or B2 directly as follows:

Δ B1 = Δp = p (pw11+qw12 – ŵ)ŵ

Δ B2 = Δq = q (pw12+qw22 – ŵ)ŵ

let s do a sample problem
Let’s do a sample problem.
  • 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 genotype and allele frequencies 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.

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 B1B1; B1B2; and B2B2 values for the next generation now

ŵ = p2w11 + 2pqw12 + q2w22

ŵ= (.82) 2 (.8) + 2(.82)(.18)(1.0) + (.18)2 (.6)

ŵ = .537 + .295 + .019 = .85

B1B1 = P2w11ŵ

B1B2 = 2pqw12ŵ

B2B2 = q2w22ŵ

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

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

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

Hint: Do they add up to 1.0?


We an also calculate the new allele frequencies as well

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

B1 = p2w11+pqw12 ŵ

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

B2 = pqw12+q2w22ŵ


So…… B1B1 = .63 B1B2 = .35 B2B2 = .02


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
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 picked at random.
  • Took random samples of flies every few generations and calculated the allele frequencies for AdhF and AdhS
can we identify which assumption is being violated
Can we identify which assumption is being violated?
  • 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.
Cavener and Clegg demonstrated that selection pressure can lead to changes in allele frequencies in just a few generations
a second selection scenario
A second selection scenario
  • Selection may upset therelationship 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.
empirical research on selection and genotype frequency
Empirical research on Selection and Genotype frequency

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.
compare to older women who participated in the feasts but did not get sick
Compare to older women who participated in the feasts but did not get sick:
  • 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.
There are TWO conditions necessary for a favorable allele to increase in a population due to selection.
  • 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.

what pattern of allele frequency changes might be caused by selection
What pattern of allele frequency changes might be caused by selection
  • If selection is acting, does the rate of evolution of a particular allele depend on whether it is….

dominant or recessive?

heterozygote or homozygote?


Natural Selection is most potent as an evolutionary force against recessive alleles when recessive alleles are common and the dominant form is relatively rare.

  • 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.



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

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

  • Dawson showed that dominanceand allele frequencyinteractto determine the rate of evolution when acted on by selection
  • If a recessive allele is commonevolution 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.
selection on heterozygotes and homozygotes
Selection on Heterozygotes and Homozygotes
  • 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
scenario 1 from mukai and burdick fruit fly experiment
Scenario #1 from Mukai and BurdickFruit fly experiment
  • 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
the experiment
The experiment
  • Started with all heterozygotes to establish a new population (each allele =.5)

After several generations equilibrium was reached at .79 frequency for the viable allele

This means that the lethal allele was maintainedat frequency of 0.21! How could this 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 .....


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

Figure 6.18 pg 200

this is a case of heterozygote superiority or overdominance also called heterosis
This is a case of Heterozygote superiority or overdominance (also called heterosis)
  • 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 polymorphism
can also have heterozygote inferiority or underdominance
Can also have Heterozygote inferiority or underdominance
  • Where the heterozygote condition is inferior to either of the homozygotes
  • What do you predict would happen 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.
what is the evolutionary impact
What is the evolutionary impact?
  • Leads to a loss of genetic diversity
  • Although if differentalleles are fixed in different populations can help maintain genetic diversity amongpopulations
  • 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.
frequency dependent selection
Frequency-dependent selection
  • 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.
the favored allele fluctuates because
The favored allele fluctuates because
  • 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.
effects of mutation
Effects of Mutation
  • Mutation is the source of all new alleles
  • Mutation provides the raw material on which selection can act
hardy weinberg and mutation
Hardy Weinberg and Mutation
  • Mutation alone is a weakor 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 Aa at a rate of 1 in 10,000 per generation.
  • then the rates are very slow as shown in figure 6.23
mutation and selection

Mutation and Selection

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

richard lenski and colleagues working with e coli
Richard Lenski and colleagues working with E. coli
  • 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
mutation selection balance
Mutation –Selection Balance
  • 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
intuition tells us that
Intuition tells us that ...
  • If the mutation is only mildlydeleterious and therefore selection against it is weak; andMutation rate is high then
    • The equilibrium frequency of the mutated allele will be relatively in the population.
  • If, on the other hand, there is strong selection against a mutation (the mutation is highly deleterious) and the mutation rate is low then
    • Equilibrium ratio of the mutated allele will be in the population



  • 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.
cystic fibrosis
Cystic Fibrosis
  • 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
huntingtons disease new research 2007
Huntingtons Diseasenew research 2007.
  • 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.