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Transcript
MCB142/IB163
Thomson
Mendelian and population genetics
Fall 2002
Handout 3
• Non-random mating, selection (ch. 17: 568-579)
deviations from Hardy Weinberg assumptions: the strength of the Hardy Weinberg (HW) law is that one can
deviate from the assumptions quite a bit and the data will still approximate Hardy Weinberg proportions
(HWP). The weakness of the HW test is that the deviation from the assumptions has to be very strong in order
to detect the effect of this evolutionary force, e.g., selection. Deviations from HW assumptions involve:
(1) non-random mating, e.g., inbreeding,
(2) effects of mutation are negligible as mutation rates are low—but mutation is an important force in creating
new variation,
(3) migration is important if the migration rate is high and the two populations are very distinct genetically,
(4) genetic drift due to small population size (chance effects)—genetic drift effects are important in both small
and large (but finite) populations in terms of long term effects of changes in allele frequencies over
generations due solely to drift effects, however the finite size of a sample taken from a population is taken
into account in the statistical tests for HWP and finite population size itself does not cause significantly
detectable deviations from HWP,
(5) selection has to be strong to cause deviations from HWP, e.g., it can be detected with sickle cell anemia, see
below.
non-random mating: individuals with certain genotypes sometimes mate with one another more commonly than
would be expected on a random basis.
When like mates more often with like we term this positive assortative mating, e.g., height, IQ. Positive
assortative mating increases the proportion of homozygous individuals but does not alter the allele
frequencies.
With self-fertilizing plants the level of heterozygosity is reduced by 1/2 each generation (see Figure 17.7 of
text.)
Self-fertilizing plants have more homozygotes than expected under Hardy-Weinberg and often show
significant deviations from HWP.
inbreeding: mating with close relatives is another form of non-random mating.
Relatives are more likely to carry the same recessive allele for a rare recessive trait—inbreeding increases the
number of affected individuals with rare recessive traits. Marriages between first cousins have about twice the
rate of birth defects as random matings.
mutation: in and of itself does not change allele frequencies to a noticeable extent as mutation rates are low.
However, mutations are the raw material of evolution, the ultimate source of genetic variation. Although the
frequencies of mutants are initially rare, and most are lost from the population, nevertheless some increase in
frequency due to genetic drift effects and selection (see below).
migration: is the movement of individuals from one population into another, which can alter allele frequencies,
and if there are large genetic differences cause a statistically significant deficiency of heterozygotes from
Hardy-Weinberg expectations.
relative fitness: the average number of offspring produced by individuals with a certain genotype, relative to the
number produced by individuals with other genotypes.
heterozygote advantage: the heterozygote has a higher relative fitness than both homozygotes, this leads to a
balanced polymorphism. If we denote the relative fitness values of the 3 genotypes in a 2-allele system as
follows:
A1A1
A1A2
A2A2
relative fitness
1 - s1
1
1 - s2
then the equilibrium frequency for allele A1, denoted p1* = s2/(s1 + s2), and that for allele A2 is
p2* = s1/(s1 + s2), with p1* + p2* = 1. The allele A1 is more frequent if the selection is stronger against the
A2 homozygote than the A1 homozygote and vice versa.
1
sickle cell anemia: Individuals with sickle cell trait (AS heterozygotes) are more resistant to malaria and have a
higher fitness than individuals who are homozygous AA or SS (these individuals are also more resistant to
malaria than AA individuals, but they suffer the deleterious effects of sickle cell anemia). The relative fitness
values in a malarial environment have been estimated as:
genotype
AA
AS
SS
relative fitness
0.88 1.0
0.14
The predicted equilibrium frequencies with this selection regime (s1 = 0.12, s2 = 0.86) are f(A) = 0.88, and
f(S) = 0.12 using the formula above. The observed allele frequencies are very close to these values. The
selection in this case is very strong and causes detectable deviations from HWP.
Genotypes
AA
AS
SS
Observed no. among adults
9,365
2,993
29
Total: 12,387
Observed freq. among adults
0.756
0.242
0.002
Expected no. among adults (HWP)
9,523
2,676
188 Total: 12,387
Expected freq. among adults (HWP)
0.769
0.216
0.015
Also see problem 3 below.
frequency dependent selection: is another cause leading to balanced polymorphism. The reproductive success of
a phenotype depends on its frequency, with higher fitness associated with lower frequency of the phenotype
in the population, e.g., ‘right-mouthed’ and ‘left-mouthed’ species of cichlid fish.
balanced polymorphism: 2 or more alleles are maintained in a population due to selection. Both heterozygote
advantage and frequency dependent selection are examples of balancing selection, they both lead to a stable
polymorphic equilibrium state.
directional selection: leads to fixation of an allele. An example would be relative fitnesses
AA
AB
BB
1.0
0.9
0.8
which would lead to fixation of the A allele.
rates of change: in the table below the number of generations required for the frequency of allele A to change
from 0.01 to 0.99 is given for the fitness model
AA
AB
BB
1
1 - hs
1-s
for values of h = 0 (or 1) (A is dominant or recessive), 0.1 (or 0.9) and 0.5 (additive effect model), and s =
0.005, 0.01, 0.05, 0.1, and 0.2:
h s = 0.2
0.1
0.05
0.01
0.005
0.5
83
175
359
1,830
3,700
0.1
155
317
640
3,225
6,500
0
534
1,075
2,157
10,800
21,600
Weak selection, of course, takes longer to effect an allelic substitution.
Even with fairly strong selection it can take quite a long time to effect a substitution, e.g., with h = 0.5 and s =
0.1, it takes 175 generations to change the allele frequency from 0.01 to 0.99, and taking an average of 25
years per generation for humans this would be 4,375 years. Weaker selection, e.g., h = 0.5, s = 0.01, which
nonetheless is strong selection in an evolutionary sense, takes 1,830 generations (45,750 years for humans).
The time required is approximately proportional to 1/s, e.g., the time length with s = 0.1 is close to twice the
time length with s = 0.2.
The total number of generations is smallest for genic selection (h = 0.5). If either allele is recessive, or nearly
so, the total length of time is greatly extended, this is attributable, for the most part, to the weak effect that
selection has when the recessive allele is rare.
quantitative trait: determined by a large number of genes each of small effect and environmental factors, e.g.,
height and weight.
stabilizing selection: selection acts to eliminate both extremes from an array of phenotypes (this is the
quantitative equivalent of balancing selection for a single gene trait).
directional selection: selection acts to eliminate one extreme from an array of phenotypes.
2
disruptive selection: (also called diversifying selection) selection acts to eliminate the intermediate type,
favoring both extremes (this is the quantitative equivalent of heterozygote disadvantage for a single gene
trait).
Problems
1. Do problem 15 from chapter 17 on page 585 of the textbook. Write out the numbers of males and females of
each genotype in the two separate groups assuming HWP and equal numbers of males and females, and also
test the assumption stated in the problem that in the mixed population the 1,000 females would be in HWP
(answer is given below).
2. In a population of self fertilizing plants of size 100, the genotypes at a codominant locus are 20 AA, 60 AB,
and 20 BB individuals. What will the genotype counts be in the next generation (assume a population size of
100 again)? What would they be if there was random mating?
3. In Africa in an area where malaria is prevalent, 1,000 newborns and 1,000 adults are typed for the sickle-cell
polymorphism:
Genotypes
AA
AS
SS
Newborns (O)
780
204
16
Total: 1,000
Adults (O)
769
229
2
Total: 1,000
For both samples, determine the allele frequencies of the A and S alleles, and test fit to HWP. Comment on
any differences, or not, between the allele frequencies and the test of HWP in the two samples.
4. For the following selection schemes, where the relative fitness values for each genotype are given, state
whether one or the other allele will be fixed (and if so which allele) or whether a balanced polymorphism will
be maintained, in which case determine the allele frequencies at this equilibrium:
AA
AB
BB
(a)
0.8
0.9
1.0
(b)
0.2
1.0
0.6
(c)
1.2
1.1
1.0
3
Answers to Problems
1. Problem 15: In population 1 q1 = 0.10 and in population 2 q2 = 0.30 (the proportion of color-blind males in
each population). After equal mixing of the two populations, the frequency of colorblind males is 0.2.
The numbers in each group before and after mixing are as follows, and the test of HWP on the females:
females X+X+ X+Xc XcXc males X+Y XcY
population 1
405
90
5
450
50
Total: 1,000
population 2
245
210
45
350
150
Total: 1,000
mixed (O)
650
300
50
800
200
Total: 2,000
mixed (E)
640
320
40
Total: 1,000
mixed (O - E)
10
20
10
Total: 1,000
2
mixed (O - E) /E 0.156
1.25
2.50
Total chi-square = 3.90, 1df, P<0.05*
There is evidence of lack of fit to HWP in the females. In fact mixing of two distinct populations with
different allele frequencies is a well known cause of lack of fit to HWP, so the text should not have told you
to make this assumption.
2. With selfing, the frequency of heterozygotes is reduced by 1/2 each generation, with the remaining 1/2
divided equally between the two homozygous classes. The genotype counts in the next generation will be 35
AA, 30 AB, and 35 BB. If there had been random mating, the genotype counts would be 25 AA, 50 AB, and
25 BB.
3. Among the newborns, f(A) = p = 0.8820, f(S) = q = 0.1220, and in the adults f(A) = 0.8835, f(S) = 0.1165.
Newborns (O) 780
204
16
Total: 1,000
Newborns (E) 777.9
208.2
13.9
Total: 1,000
(O - E)
2.1
4.2
2.1
2
(O - E)
0.006
0.085
0.317
Total chi-square: 0.408, df =1, ns
Adults (O)
Adults (E)
(O - E)
(O - E)2
769
780.6
11.6
0.17
229
205.9
23.1
2.59
2
13.5
11.5
9.80
Total: 1,000
Total: 1,000
Total chi-square: 12.56, df =1, P<0.001***
The adult sample shows very significant deviation from HWP, while the newborn sample does not. This
reflects the fact that at birth (pre-selection) the genotype frequencies are expected to be in HWP, whereas
after selection has acted (malaria and sickle-cell anemia), the adult population shows significant deviation
from HWP. The fact that the allele frequencies are very similar in newborns and adults may indicate that the
population is close to equilibrium.
4. (a) fixation of the B allele, loss of the A allele
(b) stable polymorphism, s1 = 0.8, s2 = 0.4, thus p1* = 1/3 (= f(A)), p2* = 2/3 (=f(B))
(c) fixation of the A allele, loss of the B allele
4
• Genetic drift (ch. 17: 580-582)
genetic drift: (chance effects) random change in the frequency of alleles at a locus.
short term genetic drift effects: cause changes in allele frequencies, both in small and large populations. The
change in allele frequency due to genetic drift in a small population appears larger, statistical testing can
determine whether changes are larger than expected by chance.
As an example, an allele frequency change in a population of size 50 from p = 0.5 to 0.56 in 1 generation is
within the range expected by drift, whereas in a population of size 5,000 such a change would be much too
large to be due solely to drift effects.
Think in terms of tossing a coin, if you tossed a coin 50 times you would expect 25 heads and 25 tails, but due
to the finite number of tosses would not be surprised to observe 28 heads and 22 tails (the change given for
the allele frequency above, 56% heads). In fact, 95% of the results of tossing a coin 50 times would fall
within the range from 30 heads (20 tails) to 20 heads (30 tails).
If you toss a coin 5,000 times 95% of the results would fall within the range from 2,550 heads (2,450 tails) to
2,450 heads (2,550 tails), and an observed outcome of 2,800 heads (2,200 tails) (56% heads when 50%
expected) is well outside the range expected by chance (this outcome would occur by chance less than 1 in a
million times).
long term genetic drift effects: are loss or ‘fixation’ of an allele.
It is easy to understand how genetic drift in small populations can lead to loss of alleles (and hence ‘fixation’
of the other allele), e.g., suppose we have a constant population size of 10 haploid individuals, with an initial
allele frequency of p = f(A) = 0.5, q = f(B) = 0.5. We assume at the moment that no new mutations arise.
After the first generation reproduces, the allele frequencies in the 2nd generation are p = 0.6, q = 0.4, and we
now reweight our ‘coin’ so that the probability of an A in the next (3rd) generation is 0.6 and of a B is 0.4, the
observed frequencies in the current (2nd) generation. Due to drift effects the observed frequencies in the 3rd
generation may differ from these expected values. Continuing this process, the genetic drift (random) effects
result in eventual loss of one of the alleles, and ‘fixation’ of the other.
alleles
A
B
5
5
6
4
8
2
7
3
9
1
10
0
In this example, the B allele is lost, and the A allele is ‘fixed’ in the population. From the starting point of
equal allele frequencies it is equally likely that B would be fixed and A lost in a subsequent random sampling
of the genetic material.
The exact same process goes on in large populations, it just takes longer.
alleles
A
B
500
500
495
505
510
490
.............
1000
0, or A could be lost, and B ‘fixed’ in the population.
The important point with regard to genetic drift is that because all populations are of finite size, evolution (in
the sense of amino acid or nucleotide (allelic) changes) will always occur. One allelic type will eventually
replace another.
common ancestor: the consequence of finite population sizes and hence genetic drift effects is that we have a
common ancestor for all our genes. It is a different common ancestor for each gene due to independent
assortment of genes on different chromosomes (Mendel’s 2nd law), and recombination between genes on the
same chromosome. Very closely linked genes could share a common ancestor.
For a neutral allele at an autosomal locus in a diploid organism the time to trace back to the common ancestor
is on average 4Ne generations where Ne is the effective population size.
5
The female ancestor of our mitochondrial DNA, which is inherited maternally and without recombination, has
been traced to Africa about 200,000 years ago. Note that this woman is only the common ancestor for our
mitochondrial DNA, and further this observation does not tell us what the population size was at that time, it
certainly does not mean it was just this woman and one man.
Our nuclear genes trace back to many other common ancestors, some presumably from this time period, some
more recent, and others which are older. Because of genetic recombination of nuclear genes, it is much more
difficult than with mitochondrial DNA to trace back to the common ancestor.
The Y chromosome will trace back to a male common ancestor and studies on Y chromosome variation are in
progress in a number of laboratories.
founder effect: the change in allele frequencies when a new colony is formed by a very small number of
founding individuals from a larger population.
Alleles found in the general population may be absent from the founder population, e.g., of an island
population, or a religious isolate, such as the Amish and Hutterite populations in the United States.
On the other hand, by chance rare alleles from the general population may be more frequent in the founder
population, e.g., a form of dwarfism in the Amish.
bottleneck: drastic reduction in population size due, e.g., to over fishing or a natural disaster such as a hurricane,
will lead to changes in allele frequencies.
Problems
1. Consider five different genes in humans, one of which is mitochondrial (inherited maternally and
independently of the nuclear chromosomes); the other four genes are on four separate nuclear chromosomes
(autosomes) and have evolved independently. These five genes trace back to how many different ancestors?
2. Explain how it is possible for humans and chimps to share some polymorphisms, e.g., the ABO blood group
alleles.
6
Answers to Problems: Genetic drift
1. One female common ancestor for the mitochondrial gene, four different common ancestors of either sex for
the four nuclear genes.
2. We need to distinguish between gene trees and species trees. Closely related species may share
polymorphisms, assuming both groups were polymorphic for the same variation when they split from their
common ancestor, and have maintained the polymorphism since then. In the case of ABO blood groups,
almost many primate species have the 3 alleles present, and it is most likely that there is some form of
selection maintaining this polymorphism. When a new species forms, the population does not necessarily go
through a bottleneck, and variation can be maintained.
• Evolutionary genetics (ch. 19: 619-648)
evolutionary trees from DNA data: A series of evolutionary changes involves accumulation of genetic change
in the DNA.
Even if there were no selection operating, i.e., mutations which arose were selectively neutral, because of the
finite size of populations and consequent chance events, alleles always eventually become lost from a
population, so there is eventual replacement of allelic types by another. The more distant two species the
more genetic differences (amino acid changes or nucleotide changes) that will have accumulated between
them. So, the longer the time since the organisms diverged, the greater the number of differences in the
nucleotide sequence of the gene for, e.g., cytochrome c.
Evolutionary trees drawn from DNA data agree well with those drawn from the fossil record, and can be
important where convergent evolution of similar characteristics can cause confusion in drawing evolutionary
trees based on the characteristics of organisms, and/or when the fossil record is poor.
Human
Gorilla
Pig
Rabbit
Human
0
Gorilla
1
0
Pig
19
20
0
Rabbit
26
27
27
0
substitution times: the time required for substitution of different amino acids is long, in the millions of years.
For example, with hemoglobin we expect on average 1 amino acid substitution every 5 million years.
molecular clock: there is a regularity with time in the rate of accumulation of genetic changes at the nucleotide
and amino acid levels.
Different proteins show different rates of accumulation of changes. Some evolve very slowly, such as
histones, others at a medium rate such as hemoglobin, and others faster such as fibrinopeptide. But for each
protein, the molecular clock generally applies.
7