Announcements

1. Reminder: bring your lab 9/10 “data” to lab this week

2. Look over problems in Ch. 25: 2, 3, 4, 6, 16, 19.

3. Most of Friday lecture will be for final review/questions

4. We will not cover chapter 26; ie. do not read it. Focus on other chapters.

5. Sun. Dec. 1 was World AIDS day; 6000 people in NY city were diagnosed with AIDS last year; 42 million people world-wide are HIV-positive - thus interest in understanding role of CC-CKR-5 gene and HIV resistance.

Review of lecture 38

1. Cancer overview

- Viruses and Carcinogens

- Genetic testing for cancer

2. Population genetics

- Calculating allele frequencies

Overview of lectures 39/40

I. Allele frequencies

II. Hardy-Weinberg

- assumptions- example to demonstrate H-W law- testing for H-W equilibrium- applications of H-W

III. Natural selection, mutation, migration, genetic drift, nonrandom mating and effects on allele frequencies

Real-life example - calculating allele frequencies CCR5 Function, Genotypes and Phenotypes

A small number of individuals seem to be resistant to acquiring HIV, even after repeated exposure. How?

Breakthrough 1996 - all have mutations in CC-CKR-5 gene

• CC-CKR-5 gene encodes chemokine receptor, CCR5.• Chemokines are signaling molecules used by the immune

system.• HIV-1 uses CCR5 receptors to enter host immune cells.

• 32/32 genotype associated with resistance to HIV-1 infection.

• +/32 genotype is susceptible, but may progress to AIDS slowly.

• +/+ genotype is susceptible to HIV-1.

Allelic variation in the CCR5 gene RFLP analysis

32 bp deletion in exon of CCR5 gene results in non-functional protein, and therefore resistance to HIV infection

Determine Allele Frequencies from GenotypesHow common is ∆32 allele and where is it present?

A sample of 100 French individuals in Brittany revealed the following genotypes. Genotype: +/+ +/32 32/32 Total

No. of individuals 79 20 1 100

1) Determining the allele frequencies by counting alleles:

No. of + alleles 158 20 0 178

No. of 32 alleles 0 20 2 22

200

Frequency of CCR5+ in sample = 178 / 200 = 0.89 = 89%

Frequency of CCR32 in sample = 22 / 200 = 0.11 = 11%

2) Determining the allele frequencies from genotype frequencies:

No. of individuals 79 20 1 100

Genotype frequency 79/100 20/100 1/100 1.00

(0.79) (0.20) (0.01)

Frequency of CCR5+ in sample = 0.79 + (1/2) 0.20 = 0.89 = 89%

Frequency of CCR 32 in sample = (1/2) 0.20 + 0.01 = 0.11= 11%

Conclusions and more questions

• Highest frequency of ∆32 allele is in Northern Europe; populations without European ancestry = no ∆32

• Why is the 32 allele present in this distribution? Where did it originate?

• Would we expect the allele to become more common where it is presently rare?

• Use tools developed to model answers to such questions:

Godfrey H. Hardy, a mathematician, and Wilhelm Weinberg, a physician, independently proposed a simple algebraic equation for analyzing alleles in populations.

– Under certain conditions, one can predict what will happen to genotype and allele frequencies

Assumptions of Hardy-Weinberg

1. No natural selection; equal rates of survival, equal reproductive success.

2. No mutation to create new alleles.

3. No migration in or out of population.

4. Population size is infinitely large.

5. Random mating.

If these assumptions are true, then:

1. The allele frequencies in the population will not change from generation to generation.

2. After one generation of random mating, the genotype frequencies can be predicted from the allele frequencies.

How does such a strict law, where there is NO change from generation to generation, help in

studying evolution?

KEY POINT: By specifying ideal conditions when allele frequencies do NOT change, H-W law identifies forces of evolution (forces that cause allele frequencies to change). Know these five forces of evolution and H-W law.

Demonstration of H-W Law

• Suppose the gene pool for a population for two alleles is fr(A) = 0.7 and fr(a) = 0.3 in eggs and sperm. (Note freq. of dominant allele plus freq. of recessive allele (0.7 + 0.3) = 1)

• If random mating occurs, then what are the probabilities that each of the following genotypes will occur? AA, Aa, aa.

• You can solve using a Punnet square:

Calculating Genotype Frequencies from Allele Frequencies

fr(A) = 0.7 fr(a) = 0.3

fr(A) = 0.7 fr(AA) = 0.7 X 0.7 =0.49

fr(Aa) = 0.7 X 0.3 =0.21

fr(a) = 0.3fr(Aa) = 0.7 X 0.3 =

0.21fr(aa) = 0.3 X 0.3 =

0.09

Egg

s

Sperm

Total fr(Genotypes): 0.49 AA + 0.42 Aa + 0.09 aa = 1

What are the allele frequencies in the next generation?

• Determine allele frequencies from genotype frequencies:

Genotype: AA Aa aa Total

Frequency 0.49 0.42 0.09 1.00

Frequency of A in sample = 0.49 + 1/2 (0.42) = 0.7

Frequency of a in sample = 1/2 (0.42) + 0.09 = 0.3

• So after one generation of random mating, the allele frequencies can be predicted and have not changed. We’re back where we started. No evolution of population.

General Allele and Genotype Frequencies under H-W Assumptions

Total fr(Genotypes): p2 + 2pq + q2 = 1

Summing up H-W Equations

• Gene Pool Equation: p + q = 1 where p = frequency of the dominant allele in the population, q = frequency of the recessive allele in the population.

• Genotype Equation: p2 + 2pq + q2 = 1 where p2 = frequency of dominant homozygotes, 2pq = frequency of heterozygotes, q2 = frequency of recessive homozygotes.

KEY POINT: When population has constant allele frequencies from generation to generation, and when genotype frequencies can be predicted from allele frequencies, then population is in Hardy - Weinberg equilibrium.

Three important consequences of H-W law

1. Dominant traits do NOT automatically increase in frequency from generation to generation

2. Genetic variation can be maintained

3. Knowing the frequency of one genotype can allow for calculation of other genotypes

Testing for Equilibrium

1. Determine genotypes, either directly from phenotypes or by DNA or protein analysis.

2. Calculate allele frequencies from genotype frequencies.

3. Predict genotype frequencies in next generation.

4. Test by Chi-square analysis.

Example: CCR5+ and CCR532

• Note: the textbook example (p. 689) is in error! You must do the X2 calculation on actual data (numbers of genotypes), not frequencies!

• Sample: 283 English (Table 25.3)

• (1) Observed data: 223 +/+, 57 +/32, 3 32/32.

• (2) Allele frequencies (566 total alleles):

– fr(+) = (2 X 223 + 57)/566 = 0.89 = p– fr(32) = q = 1 - p = 0.11

Predict genotype frequencies and numbers in next generation from H-W Law

(3) Expected genotype frequencies and numbers:

fr(+/+) = p2 = (0.89)2 = 0.792; No. +/+ = (0.792)(283) = 224.1

fr(+/32) = 2pq = 2(0.89)(.011) = 0.196; No. +/32 = (0.196)(283) = 55.5

fr(32/32) = q2 = (0.11)2 = 0.012; No. 32/32 = (0.012)(283) = 3.4

Total = 224.1 + 55.5 + 3.4 = 283

(4) Chi-square analysis:

o e o-e (o-e)2 (o-e)2/e

+/+ 223 224.1 -1.1 1.21 0.0054

+/32 57 55.5 1.5 2.25 0.0405

32/32 3 3.4 -0.4 0.16 0.0471

X2 0.0930

(in book, incorrect chi-square value is 0.00023)

p Value Calculation

Degrees of Freedom = k - 1 - m where k = # categories /genotypes and m = # of independent allele freq. estimated

( m= 1 since p was estimated from the data, assuming sample was representative of the population, and q was

determined directly from p).

With df = 3 - 1 - 1 = 1, 0.5 < p < 0.9

Chance alone could account for this much deviation from expected values between 50-90% of the time, so H-W

equilibrium is not rejected.

Conclusions

• Population is in H-W equilibrium (null hypothesis not rejected).

• So there is random mating, no natural selection*, no mutation, no migration, no gene flow in sampled population.

• If significant difference was shown, then something is happening (selection, mutation, migration, gene flow, or no random mating).

Application of H-W Law: Determining frequency of heterozygotes in population

• Cystic Fibrosis, an autosomal recessive trait, occurs at 1/2500 = 0.0004 in people of northern European descent. So q2 = 0.0004.

• Frequency of recessive CF allele is q = 0.0004 = 0.02 (we have now estimated q).

• Frequency of dominant wt allele is p = 1 - q = 1 - 0.02 = 0.98.

• Frequency of heterozygotes is 2pq = 2(0.98)(0.2) = 0.04 = 4% or 1/25.

Learning Check

Suppose the frequency of sickle-cell anemia in a population is 20%. What are the allele frequencies? Estimate the frequencies of heterozygotes and dominant homozygotes.

Natural Selection is a strong force of change in Allele Frequencies

• Natural Selection is a force driving differential rates of survival and/or reproduction among individuals in a population.

• Suppose in population of 100 there are 25 AA, 50 Aa, and 25 aa, so fr(A) = 0.5 and fr(a) = 0.5.

• Suppose different rates of survival: all AA survive, 90% Aa survive, 80% aa survive.

• In next generation, 2(25) + 2(45) + 2(20) = 180 gametes.

Fr(A) = (50 + 45)/180 = 0.53

Fr(a) = (45 + 40)/180 = 0.47

• Now we have a change in allele frequencies! Natural selection is one of the most important factors in evolutionary change.

Mutation is a weak force of change in Allele Frequencies

Migration (Gene Flow) Homogenizes Allele Frequencies across Populations.

Map of fr(B allele) of ABO locus parallels Mongol migration into Europe after end of Roman Empire.

Genetic Drift

• Random change in allele frequencies.

• Important in small populations.

• Kerr and Wright (1954) set up 100 lines of flies, each founded by 4 males and 4 females.

• In each line, at forked locus both fr(f) and fr(f+) = 0.5.

• After 16 generations, complete loss of one allele and fixation of the other occurred in 70 lines; remainder still segregating or extinct.

Nonrandom Mating changes Genotype Frequencies but not Allele Frequencies

Homozygotes increase,heterozygotes decrease.

Checkfr(alleles)for yourself!

H-W Problem 2

• Are the following genotypes in equilibrium?– (1) 35 AA, 50 Aa, 15 aa– (2) 42 AA, 36 Aa 22 aa

H-W Problem 3

• What are the allele frequencies for each of the following?– Generation 1: 25 AA, 50 Aa, 25 aa– Generation 2: 36 AA, 48 Aa, 16 aa– Generation 3: 49 AA, 42 Aa, 9 aa