Announcements

1
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.

2
Review of lecture 38

• 1. Cancer overview
• - Viruses and Carcinogens
• - Genetic testing for cancer

2. Population genetics - Calculating allele
frequencies
3
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

4
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.

5
Allelic variation in the CCR5 gene RFLP analysis

• ?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.

32 bp deletion in exon of CCR5 gene results in
non-functional protein, and therefore resistance
to HIV infection
6
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 CCR?32 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
7
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

8
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.

9
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.
10
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

11
Calculating Genotype Frequencies from Allele
Frequencies
Sperm
Eggs
Total fr(Genotypes) 0.49 AA 0.42 Aa 0.09 aa
1
12
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. Were back where we started. No
  evolution of population.

13
General Allele and Genotype Frequencies under H-W
Assumptions
Total fr(Genotypes) p2 2pq q2 1
14
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.
15
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

16
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.

17
Example CCR5 and CCR5?32

• 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

18
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

19
(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 0.04
71 X2 0.0930 (in book, incorrect
chi-square value is 0.00023)
20
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 lt p lt 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.
21
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).

22
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.

23
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.
24
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.

25
Mutation is a weak force of change in Allele
Frequencies
26
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.
27
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.

28
Nonrandom Mating changes Genotype Frequencies but
not Allele Frequencies
Homozygotes increase, heterozygotes decrease.
Check fr(alleles) for yourself!
29
H-W Problem 2

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

30
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