Download Quantitative genetics

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.
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
Determine Allele Frequencies from Genotypes
How 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%
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
Sperm
fr(A) = 0.7
Eggs
fr(A) = 0.7
fr(a) = 0.3
fr(a) = 0.3
fr(AA) = 0.7 X 0.7 = fr(Aa) = 0.7 X 0.3 =
0.49
0.21
fr(Aa) = 0.7 X 0.3 = fr(aa) = 0.3 X 0.3 =
0.21
0.09
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:
Frequency
AA
0.49
Aa
0.42
aa
0.09
Total
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 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
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:
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
o-e
(o-e)2 (o-e)2/e
o
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.
• In each line, at forked locus
both fr(f) and fr(f+) = 0.5.
• Important in small
populations.
• After 16 generations,
complete loss of one allele
and fixation of the other
occurred in 70 lines;
remainder still segregating
or extinct.
• Kerr and Wright (1954)
set up 100 lines of flies,
each founded by 4
males and 4 females.
Nonrandom Mating changes Genotype
Frequencies but not Allele Frequencies
Homozygotes increase,
heterozygotes decrease.
Check
fr(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