# Hardy Weinberg Equilibrium

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Hardy Weinberg Equilibrium
Population Genetics Hardy Weinberg Equilibrium

6.1 Mendelian Genetics in Populations:
The Hardy-Weinberg Equilibrium Principle

Population Genetics 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?

Predicting Allele Frequencies
Populations in Hardy-Weinberg equilibrium

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.

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

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? _____ (36*(2)+48) 120 (16*(2) +48) 80

What percent of the alleles are A or a ?
120 / 200 = .6 or 60% A ; or .6 = frequency of allele A 80 / 200 = .4 or 40% a ; or .4 = frequency of allele a

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

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

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

Calculating Genotype Frequencies
We can predict the genotype frequencies by multiplying probabilities.

Hardy-Weinburg equation
Genotype Frequencies

Genotype frequencies described by p2+2pq+q2=1.0
Zygotes Allelic frequency Genotype frequency AA (p)(p) p2 Aa (p)(q) 2pq aA (q)(p) aa (q)(q) q2 Genotype frequencies described by p2+2pq+q2=1.0

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

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 0.2 Determine the genotype frequencies of this population? AA Aa aa 0.64 0.32 0.04 AA = p2 ; Aa = 2pq ; and aa = q2 as follows…

The allele frequency can therefore be calculated as follows.
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 +q2 The allele frequency can therefore be calculated as follows. A = p2 + ½(2pq) and a = q2 + ½(2pq)

Examining our example again we see that if we use the frequencies we calculated for each genotype….
p pq q2 0.64 AA Aa aa A = p2 + ½ (2pq) A=.64 + ½ (.32) A = 0.8 and since q = 1-p ; then a = 1-(0.8 ) a = 0.2

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.

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

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 another There 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

Exceptions to Hardy Weinberg cont.
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.

Problem #6 on page 219 Go to your text page 219 and answer question number 6.

In humans, the COL1A1 locus codes for a certain collagen protein found in bone. The normal allele at the locus is denoted with S. A recessive allele s is associated with reduced bone mineral density and increased risk of fractures in both Ss and ss women. A recent study of 1,778 women showed that 1,194 were SS, 526 were Ss, and 58 were ss. Are these two alleles in Hardy-Weinberg equilibrium in this population? What information would you need to determine whether the alleles will be in Hardy-Weinberg equilibrium in the next generation?

Problem approach Check that conclusion #2 holds
First figure genotype frequencies from the data (percentages) Then from the data, count the actual A alleles in the population and the actual a alleles in the population. What are their frequencies? Then calculate the predicted genotype frequencies of Hardy Weinberg and compare to actual numbers.

The genotype frequencies are:
SS =1194/1778 = Ss= 526/1778 = ss= 58/1778 = .03 Calculate allele frequencies from genotype frequencies 2914/ 3556 S alleles = S frequency or .67+1/2(.30) = 0.82 642 / s alleles = s frequency or ½ (.30)=0.18 If the population is in Hardy-Weinberg equilibrium, the allele frequencies should predict the genotype frequencies. SS genotype frequency would be (0.82)2 , = 0.67; Ss frequency would be 2 (.82) (.18), = 0.30; ss frequency would be (0.18)2 , = 0.03. These numbers almost exactly match the measured genotype frequencies - so this population may be in Hardy-Weinberg equilibrium

However, what also must we know to be sure?
We would need to check in future generations to make sure that the allele frequencies are not changing. So here we confirmed conclusion #2 but have not yet verified conclusion#1.

Example Initial frequencies 15 B1B1 50 B1B2 15 B2B2 80 total

Example Initial frequencies 15 B1B1 50 B1B2 15 B2B2 80 total Calculate genotype frequencies 15/80 = .1875 50/80 = .625 15/80 = .188

Example Initial frequencies 15 B1B1 50 B1B2 15 B2B2 80 total Calculate genotype frequencies 15/80 = .1875 50/80 = .625 15/80 = .188 Calculate allele frequencies in the population B1= 15+1/2(50)/80 = .5 B2=

Example Initial frequencies 15 B1B1 50 B1B2 15 B2B2 80 total Calculate genotype frequencies 15/80 = .1875 50/80 = .625 15/80 = .188 Calculate allele frequencies in the population B1= 15+1/2(50)/80 = .5 B2= Can we predict the genotype frequency from the allele frequency? (Frequency of B1)2 (0.5) 2 = .25 2(B1 B2) 2(.5)(.5) = .5

Example Initial frequencies 15 B1B1 50 B1B2 15 B2B2 80 total Calculate genotype frequencies 15/80 = .1875 50/80 = .625 15/80 = .188 Calculate allele frequencies in the population B1= 15+1/2(50)/80 = .5 B2= Can we predict the genotype frequency from the allele frequency? (Frequency of B1)2 (0.5) 2 = .25 2(B1 B2) 2(.5)(.5) = .5 Allele frequency does not predict genotype frequency. Population is not in Hardy-Weinberg equilibrium because it violates conclusion 2

Using the Hardy-Weinberg equilibrium with more than two alleles
In a population of mice, coat color is determined by 1 locus with 4 alleles: A, B, C, and D. The possession of an A allele confers black coat color with another A allele, or a D allele. If a B allele is present with an A allele then coat color is brown, and if C is present with and A allele, coat color is grey. All other phenotypes are light tan. Given that the frequencie of the A, B, and C alleles are .05, .4, and .3, respectivelly, what are the phenotpic frequencies of black, brown, grey and light tan mice when the population is in Hardy-Weinberg equilibrium?

Adding Selection to the Hardy-Weinberg Analysis
How do you know if a population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age. -or- 2. Equal numbers of individuals from each genotype reach reproductive age but some genotypes are able to produce more viable (reproductively successful) offspring. If these differences are heritable then evolution may occur over time.

Caution most phenotypes are not strictly the result of their genotypes. Environmental plasticity and interaction with other genes may also be involved. not as simple as we are making it here.

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.

An example of what we might 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:

Selection Example 360 B1B1 480 B1B2 160 B2B2 1000 total
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10 Using the genotype frequencies, calculate the allelic frequencies in the new population B1 = B2 =

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10 Using the genotype frequencies, calculate the allelic frequencies in the new population B1 = .45+1/2(.45) = 0.675

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10 Using the genotype frequencies, calculate the allelic frequencies in the new population B1 = .45+1/2(.45) = 0.675 B2 = 1/2(.45)+0.10 = 0.325

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10 Using the genotype frequencies, calculate the allelic frequencies in the new population B1 = .45+1/2(.45) = 0.675 B2 = 1/2(.45)+0.10 = 0.325 an increase of .075 a decrease of .075

Selection Example 360 B1B1 480 B1B2 160 B2B2 100% survive 75 %
Initial frequencies B1= 0.6; B2 = 0.4 360 B1B1 480 B1B2 160 B2B2 1000 total differential survival of offspring leads to reduced numbers of some genotypes 100% survive 75 % 50 % survive number surviving 360 80 800 total The genotype frequencies of mating individuals which survive is .45 .10 Using the genotype frequencies, calculate the allelic frequencies in the new population 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 because the allele frequencies are changed; we are not in equilibrium. The population is evolving!

Creating an equation that allows for selection
First, we analyze the population on the basis of the fitness of the offspring. Fitness (w) is defined as the survival rates, or percentage of individuals which survive to reproduce. We can use the fitness (w) of each genotype to calculate the average fitness of the population

Fitness formulas If : w11 = fitness of allele #1 homozygote
MEAN FITNESS If : w11 = fitness of allele #1 homozygote w12 = fitness of the heterozygote w22 = fitness of allele #2 homozygote mean fitness of the population will be described by the formula: ŵ = p2w11 + 2pqw12 + q2w22

For our previous example
B1= 0.6 and B2 = 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 example
B1= 0.6 and B2 = 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 example
B1= 0.6 and B2 = 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 example
B1= 0.6 and B2 = 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 = P2w ŵ B1B2 = 2pqw ŵ B2B2 = q2w ŵ

We can also use the fitness to calculate the expected frequency of each allele in the next generation. B1 = p2w11+pqw B2 = pqw12+q2w ŵ ŵ

Finally, we can calculate the change () in the frequency of B1 or B2 directly as follows:
Δ B1 = Δp = p (pw11+qw12 – ŵ) ŵ Δ B2 = Δq = q (pw12+qw22 – ŵ) ŵ

In class problem Go back to problem # 6 on page 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 0.7 ; Ss individuals 1.0 and the ss individuals 0.8.

We calculated S = .82 and s = .18 and the fitnesses are w11(SS)=.7; w12(Ss)=1; w22(ss)=.8 We know this to start ŵ = p2w11 + 2pqw12 + q2w22 ŵ= (.82) 2 (.7) + 2(.82)(.18)(1.0) + (.18)2 (.8) ŵ = = .791 B1B1 = P2w ŵ B1B2 = 2pqw ŵ B2B2 = q2w ŵ ; SS = (.82)2(.7) / = .595 ;Ss = 2(.82)(.18)(1.0) / .791 = .373 ;ss = (.18)2(.8) / .791 = .032

S = (.82)2(.7) + (.82)(.18)(1.0) = B1 = p2w11+pqw ŵ s = (.82)(.18)(1.0) + (.18)2(.8) = B2 = pqw12+q2w ŵ

So…… B1B1 = B1B2 = B2B2 = .032 and B1 = B2 = .22 Is this population in equilibrium? Have the allele frequencies changed? Can we predict the genotype frequencies from the allelic frequencies?

Experimental confirmation of loss of Hardy Weinberg equilibrium

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

Figure 5.13 pg 158

Control populations showed no consistent change in the frequency of the alleles
AdhF showed a rapid and consistent increase in the experimental populations

Can we identify which equilibrium assumption is being violated?
only difference is ethanol in food migration? random mating? drift? mutations? If we eliminate the other factors... 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 the 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
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.

Genetic variation for resistance to kuru in a Fore population
Kuru is a fatal neurological disorder that is caused by the misfolding of a prion protein in neurological tissue. Humans contract the disease by eating contaminated tissue. Symptoms start with shivering and trembling and lead to staggering and trouble talking and swallowing to coma and death.

Genetic variation for resistance to kuru in a Fore population
The greatest outbreak of kuru occurred in the 1950’s in the Fore people. They practiced ritual funereal cannibalism. Researchers wanted to know whether certain genotypes were more susceptible than others. They looked at 2 different alleles for the encoded protein that was responsible for the disease. One allele contained Met at position 129 while the other contained a Val at the same position.

Genotypes of the survivors
Researchers surveyed a population of 30 females who had eaten dead relatives and yet survived the kuru epidemic without getting sick. The numbers of individuals with their genotypes were as follows: Met/Met Met/Val Val/Val

Do these numbers deviate significantly from H.W. equallibrium?
Met/Met Met/Val Val/Val 1. Calculate the allele frequencies. Met (8+23) =0.52 60 Val (23+6) =0.48

Calculate the genotype frequencies. expected under the Hardy-Weinberg
Calculate the genotype frequencies expected under the Hardy-Weinberg principle. Met/Met Met/Val Val/Val (.52)2 2(.52)(.48) (.48)2 = =.5 = .23

Calculate the expected number of individuals of each genotype under Hardy-Weinberg equilibrium. Met/Met Met/Val Val/Val (.27)(30) (.5)(30) (.23)(30) = = = 7 These numbers are different from the ones that were actually observed in the population. = 4 = = 3

Is the difference statistically significant?
Is it plausible that the difference between the expected and observed values arose by chance? Our null hypothesis is that the difference is simply due to chance.

Chi-square 4. Calculate the test statistic using Chi-square.
4. Calculate the test statistic using Chi-square. c2 =  (observed – expected)2 c2 = (4-8)2 + (23-15)2 + (3-7)2 = 8.55 Is the value 8.55 statistically significant or could it reasonably have occurred by chance? (expected)

Determine whether the value of the test statistic is significant.
Look up the value of 8.55 in the table of “Critical values of the chi-square distribution”. To use this table we need to calculate the degrees of freedom (df) for the test statistic. df = number of classes – number of independent values. There are three classes: the number of genotypes. We calculated two values in determining the expected values: the total number of individuals, and the frequency of the Val allele df = 1

The critical value for most research studies is P = 0.05.
This means that there is a 5% chance that are null hypothesis is correct. Any P value less than 0.05 means that we reject our null hypothesis. Our c2 value of 8.55 has a P value of Therfore we reject our null hypothesis that the difference between the expected and the observed is simple due to chance events.

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?

Dawson’s Flour beetle example
Natural Selection is most potent as an evolutionary force when selection acts on recessive alleles which 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; (+ = 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?

In each succeeding generation all LL are lost and ++ makes up a greater proportion of the survivors.
With each new generation there are less and less homozygous lethals for selection to act on and the lethal allele hides in the heterozygotes

Summary Dawson showed that dominance and allele frequency interact to determine the rate of evolution when acted on by selection when a recessive allele is common and there is a great difference in fitness between the phenotypes then evolution is rapid because both the recessive and dominant phenotype are well represented for selection to act on. Example if A= .05 a= .95 AA Aa aa (0.05)2 2(0.05)(0.95) (0.95)2 = =.095 = .9025 Almost 10 % in the population have the dominant phenotype, while 90 % have the recessive phenotype Thus if the two phenotypes differ in fitness there will be a change in allele frequency

Summary Example if A= .95 a= .05
If recessive allele is rare and the dominant allele is common, evolution is slow. Example if A= .95 a= .05 AA Aa aa (0.95)2 2(0.95)(0.05) (0.05)2 = =.095 = .0025 Approximately 100% of the population has the dominant phenotype. Even if the phenotypes differ greatly in fitness, there are so few of the minority phenotype that there will be little change in allele frequencies in the next generation.

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 that of the dominant 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 (incomplete dominance)

Scenario #1 from Mukai and Burdick Fruit fly experiment
Studied a gene in which ·  Homozygotes for one allele are viable (VV) · Homozygotes for the other allele are not viable and lethal. (LL) Started with all heterozygotes to establish a new population (each allele =.5) Predict the frequency of V after 15 generations.

The experiment -After several generations equilibrium was reached at .79 frequency for the viable allele -This means that the lethal allele was at a frequency of 0.21! How could this be?

To further test these results, Mukai and Burdick did a second experiment...
-This time the population began with a frequency of of V allele. - Expected the population to eliminate all lethal alleles and fix the viable allele at 1.0. -However, Same equilibrium around a frequency of .79 was reached for viable allele!

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
Where the heterozygote condition is inferior to either of the homozygotes What do you predict would happen 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.

Example from G.G. Foster’s work with Fruit flies
Looked at chromosome differences where different chromosome forms behave like single alleles. In meiosis compound chromosomes may or may not segregate.

Only certain chromosomal; combinations will lead to viable zygotes
+ Which of these combinations would give viable offspring? B + C + + D

What is the evolutionary impact?
Leads to a loss of genetic diversity Although if different alleles are fixed in different populations can help maintain genetic diversity among populations

SUMMARY 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
Evolution can be effected by the frequency of a particular phenotype in the population.

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 Population fluctuates around an equilibrium point

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.

End of Selection Effects

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
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 Aa at a rate of 1 in 10,000 per generation. then the rates are very slow as shown in figure 5.22

Figure pg. 183

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

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

Figure 5.25 pg 185

Mutation –Selection Balance
Most 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 they 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 ...
If the mutation is only mildly deleterious and therefore selection against it is weak; and Mutation rate is high then The equilibrium frequency of the mutated allele will be relatively high 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 low

Example Spinal muscular atrophy, second most common lethal autosomal recessive disease. Selection coefficient is .9 against the disease mutations. However, 1 in 100 carry a disease causing allele in Caucasians 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 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.

Cystic Fibrosis Pier and his colleagues have found, in 11 European countries, an association between the severity of typhoid outbreaks and the frequency of the delta-F508 allele (the most common loss-of-function mutation) a generation later. Salmonella typhi bacteria manipulate their host cells, causing them to express more CFTR protein on their membranes. Pier et al. engineered cells homozygous for functional CFTR alleles, homozygous for a common loss-of-function allele, and heterozygous for the two. The loss-of-function homozygotes were virtually impervious to invasion by typhus-causing bacteria; heterozygotes were more vulnerable, but accumulated 86% fewer bacteria than did the dominant homozygotes.

Quiz 1. Multiply Choice question: Mark all the choices that are correct. 1. Which of the follow describes a population that is at equilibrium yet has an allele that is lethal. The selective advantage enjoyed by the lethal allele when it is in heterozygotes exactly balances the obvious disadvantage it suffers when it is in homozygotes. A homogenization across populations B haplodiploidy C heterozygote superiority D overdominance