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January 11, 2005
Bio 107/207
Winter 2005
Lecture 3
Random mating and the Hardy-Weinberg principle
How much DNA polymorphism exists in natural populations?
- in humans, the amount of nucleotide diversity is extremely small.
- on page 39-41 in the textbook, Hedrick describes a large study of human SNP polymorphism (a
SNP is a Single Nucleotide Polymorphism)
- the survey, based on 1.42 million SNPs (!), yields a estimate of π per site is only 0.000751!
- how does this compare to other species?
- once again, most data has been collected for Drosophila.
- here are some estimates of π and θ for a few nuclear genes:
D. melanogaster
D. simulans
Locus
N
Adh
Pgi
Est-6
Rh3
boss
ci
15
11
13
5
5
10
MEAN
π
0.00881
0.00078
0.00721
0.00070
0.00486
0.00000
θ
N
π
0.00655
0.00082
0.00889
0.00084
0.00490
0.00000
5
6
4
5
5
9
0.00677
0.00395
0.02224
0.01203
0.01245
0.00023
0.00373 0.00367
θ
0.00688
0.00471
0.02147
0.01253
0.01226
0.00038
0.00961 0.00971
- compared to humans, both Drosophila taxa show dramatically higher levels of DNA
polymorphism.
- D. simulans is much more polymorphic than D. melanogaster too!
- why would this be?
The Hardy-Weinberg-Castle Equilibrium
- consider a single locus with two alleles A1 and A2.
- let:
p = frequency of A1 allele
q = frequency of A2 allele
- three genotypes are thus possible: A1A1, A1A2, A2A2.
- let:
P = freq. of A1A1 homozygote
H = freq. of A1A2 heterozygote
Q = freq. of A2A2 homozygote
- from the genotype frequencies, we can estimate allele frequencies:
∧
p = P + 1/2 H
∧
q = Q + 1/2 H
- since only two alleles present, p + q = 1
Question: If mating occurs at random in the population, what will be the frequencies
of A1 and A2 in the next generation?
- what does “random mating” mean?
- very simply, that matings take place in the population independently of the genotype of an
individual.
- if mating is “random” the frequency of matings among different genotypes is determined
simply by their frequencies.
- therefore, if random mating occurs in the population the following frequencies of matings will
occur:
Female genotypes
______________________________
Male genotypes
A1A1 (P)
P2
PH
PQ
A1A1 (P)
A1A2 (H)
A2A2 (Q)
A1A2 (H)
A2A2 (Q)
PH
H2
HQ
PQ
HQ
Q2
- let us now examine the progeny produced by this set of matings:
Progeny
______________________________
Mating
A1A1
A1A1
A1A1
A1A2
A1A2
A2A2
x
x
x
x
x
x
Frequency
A1A1
A1A2
A2A2
A1A2
A2A2
A2A2
P2
2PH
2PQ
H2
2HQ
Q2
(P + H + Q)2
=1
A1A1
P2
PH
H2/4
(P + H/2)2
= p2
A1A2
PH
2PQ
H2/2
HQ
2(P + H/2)*
(Q + H/2)
= 2pq
A2A2
H2/4
HQ
Q2
(Q + H/2)2
=q2
- the frequencies of alleles have not changed.
This leads to two important conclusions:
1. Allele frequencies in the population will remain constant indefinitely.
2. Genotypic proportions occur at Hardy-Weinberg proportions in the population as
determined by the “square law”
- for two alleles (p, q) these proportions are given by the expanding the term (p + q)2.
- for three alleles (p, q, r) these proportions are given by the expanding the term (p + q + r)2.
- the reason why the Hardy-Weinberg equilibrium is so important is that for evolutionary change
to occur in a population it is necessary for one, or more, specific assumptions need to be
violated.
- what are these assumptions?
Assumptions of Hardy-Weinberg Equilibrium:
1. Generations are discrete (i.e., non-overlapping)
2. Allele frequencies are the same in males and females
3. Random mating
4. Infinite population size (i.e., no genetic drift)
5. No migration (gene flow)
6. No mutation
7. No selection
- the Hardy-Weinberg principle thus states that allele frequencies in populations will not change
unless some evolutionary process is acting to result in a change of allele frequency.
- it thus predicts that no evolution will occur unless one of the above assumptions is
violated.
- therefore, the study of microevolution is largely concerned with understanding the conditions
under which the above assumptions are violated!
- in other words, it involves determining the relative importance of random drift, migration,
mutation, and natural selection in affecting the frequency of genetic polymorphism in natural
populations.
- given the large number of assumptions that are required for Hardy-Weinberg equilibrium to
occur, it is reasonable to ask whether it is ever observed in natural populations.
- surprisingly, it is.
- here is an example involving 840 scallops sampled from a natural population off the east coast
of Canada.
- one SNP locus was segregating for two alleles (“A1” and “A2”), three genotypes were found in
the following proportions:
A1A1 = 652
A1A2 = 177
A2A2 = 11
840
Question: Is this population in Hardy-Weinberg equilibrium?
Step 1. Estimate genotype frequencies:
∧
P = NA1A1/N = 652/840
= 0.7762
H = NA1A2/N = 177/840
= 0.2107
Q = NA2A2/N = 11/840
= 0.0131
∧
∧
P+ H+ R
= 0.7762 + 0.2107 + 0.0131 = 1
Step 2. Estimate allele frequencies:
∧
p = frequency of A1 allele
=P+½H
= 0.7762 + ½ (0.2107)
= 0.882
∧
q = frequency of A2 allele
=Q+½H
= 0.0131 + ½ (0.2107)
= 0.118
Step 3. Determine expected number of genotypes
Expected number
of A1A1 homozygotes
Expected number
of A1A2 heterozygotes
Expected number
of A2A2 homozygotes
∧
= p2 x N
= (0.882)2 x 840
= 653.5
∧∧
= 2pq x N
= 2(0.882)(0.118) x 840
= 174.8
∧
= q2 x N
= (0.118)2 x 840
= 11.7
Step 4. Compare observed and expected numbers
Genotype
A1A1
A1A2
A2A2
Observed
Expected
652
177
11
653.5
174.8
11.7
- the observed and expected numbers are very similar!
Allele frequency differences between the sexes
1. Autosomal loci
- if allele frequencies are the same in the two sexes, then H-W equilibrium will be established
after one generation of random mating.
- however, if frequencies differ between males and females then it will take more than one
generation for H-W equilibrium will be established.
- consider first an autosomal locus with two alleles, A1 and A2.
- let:
pf and pm = frequencies of the A1 allele in females and males, respectively
qf and qm = frequencies of the A2 allele in females and males, respectively
- genotype frequencies are now:
P = pfpm
H = pfqm + pmqf
Q = qfqm
- if frequencies differ between the two sexes, there will be an excess of heterozygotes and a
deficiency of homozygotes relative to H-W expectations.
- this can be seen by considering the mean frequencies:
−
p = 1/2 (pf + pm)
−
q = 1/2 (qf + qm)
- this assumes that the sex ratio is 1:1 such that half the genes are in females and half in males.
- the extent of deviation from HW proportions for the A1A1 homozygote is given by:
−
P – p2 = pfpm – [1/2(pf + pm)]2
= pfpm – 1/4(pf2 + 2pfpm + pm2)
= – 1/4(pf2 – 2pfpm + pm2)
= – 1/4(pf – pm)2
- deviations for the other two genotypes can be shown to be:
H – 2pq = 1/2(pf – pm)2
−
Q – q2 = – 1/4(pf – pm)2
- these departures from HW proportions are expected to be small if frequencies are not that
different between males and females.
- they will also persist for only one generation.
2. X-linked loci or loci in haplo-diploid organisms
- there are some interesting dynamics for sex-linked genes.
- unlike the case for an autosomal locus, allele frequency differences between the sexes will
disappear only over several generations
- this is shown in Figure 2.5 of the textbook for an X-linked gene.
- the deviation from the mean allele frequency is halved each generation.
- because of the manner in which X chromosomes are passed between generations, frequencies
oscillate around the mean frequency and slowly disappear over time.