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EVOLUTION AT MULTIPLE LOCI
Linkage Equilibrium / Disequilibrium
THE SETTING AND TERMINOLOGY
Deals with the consideration of two loci
simultaneously
 The loci are physically linked on the same
chromosome
 Locus A with alleles “A”, “a” and locus B
with alleles ”B” and “b”
 We track not only frequencies of alleles
but also frequencies of chromosomes

MORE TERMINOLOGY
Possible chromosome genotypes for this
example are:
AB; Ab; aB; ab
 These multi-locus genotypes of
chromosomes (or gametes) are called
haplotypes ( for haploid genotype)
 These haplotypes may occur in either
Linkage Equilibrium or Linkage
Disequilibrium

Loci which are linked together in Linkage
Equilibrium:



Have genotypes that are independent of one
another.
If you know the genotype at one locus (A) you
cannot predict what the genotype will be at the
other locus (B).
Example: Suppose that the gene which controls
the length of toes in frogs (A) is linked to the gene
that controls the amount of webbing between the
toes (B).

Populations that are in linkage equilibrium will show no
correlation between toe length and the degree of
webbing between them.
Loci which are in Linkage Disequilibrium



Genotypes of the chromosomes (Haplotypes) exhibit a
non random association between the linked genes.
If you know the genotype at one locus (A) you have a clue
about the genotype at the other locus (B).
Example: back to the gene which controls the length of
toes in frogs which is linked to the gene that controls the
amount of webbing between the toes.
 Populations that are in linkage disequilibrium will show
a correlation between toe length and the degree of
webbing between them.
 For instance we might observe that the shorter the
toes the more webbing and the longer the toes the less
webbing that occurs.
Comparing linkage equilibrium with linkage
disequilibrium
PREDICTING HAPLOTYPE FREQUENCIES
LINKAGE EQUILIBRIUM
If the frequencies of the haplotypes can
be calculated by multiplying the
frequencies of the two alleles involved,
then they are in linkage equilibrium.
 Also, if the occurrence of “B” allele is
equally likely on either the A or the a
chromosome the alleles are in linkage
equilibrium

Figure 8.2a
LINKAGE DISEQUILIBRIUM
If the frequencies of the haplotypes
cannot be calculated by multiplying the
frequencies of the two alleles involved,
then they are in linkage disequilibrium
 The occurrence of “B” allele is not equal
on the A and the a chromosomes

EFFECTS OF SELECTION
IN LINKAGE EQUILIBRIUM
 If selection acts on one locus only....
 Selection for the “A” allele has no effect on the
“B” allele frequency. See Figure 8.8b
A= 5/25 = .2; a= .8
B = 20/25 = .8; b= .2
A = 20/25 = .8; a= .2
B= 20/25 =.8; b =.2
IN LINKAGE DISEQUILIBRIUM


If selection acts on one locus only....
Selection for the “A” allele changes the B allele frequencies
also. As a chromosomes are lost they drag “B” alleles along
in a disproportionate fashion. See Figure 8.8a
4
16
A= 5/25 = .2; a =.8
B= 17/25 = .68; b= .32
4
A = 20/25 = .8; a = .2
B= 8/25 = .32; b= .68
CHROMOSOME FREQUENCIES
LINKAGE EQUILIBRIUM

In linkage equilibrium chromosome (haplotype)
frequencies do not change, they can still be predicted
(calculated) from allele frequencies.
“B”=20/25=.8 “b”5/25 = .2
“A”20/25= .8
“a”5/25= .2
Calculated haplotype frequencies
AB= .64 Ab = .16 aB= .16 ab = .04
AB= 16/25 =.64 Ab= 4/25 = .16
aB = 4/25 = .16 ab =1/25 = .04
Actual haplotype
frequencies
LINKAGE DISEQUILIBRIUM

In linkage disequilibrium chromosome
frequencies change, they can not be predicted
(calculated) from allele frequencies.
“B”8/25= .32 “b”17/25 = .68 “A”20/25= .8 “a”5/25= .2
AB=.256 Ab = .544 aB= .064 ab = .136 calculated
AB= 4/25 =.16 Ab= 16/25 = .64
actual
aB = 4/25 = .16 ab =1/25 = .04
Three tests for linkage Equilibrium
1.
2.
3.
The frequency of “B” on chromosomes carrying
allele “A” is equal to the frequency of “B” on
chromosomes carrying allele “a”.
The frequency of any chromosome haplotypes can
be calculated by multiplying the frequencies of the
alleles which compose that haplotype
The quantity D, (coefficient of disequilibrium)=0
D= gABgab - gAbgaB
Verify
g is the frequency of the various haplotypes
WHAT CAUSES LINKAGE DISEQUILIBRIUM
selection
on multilocus
genotypes
genetic drift
population admixture
SELECTION ON MULTILOCUS GENOTYPES
If we use the population from figure 8.2
(p. 283) to provide gametes to the next
generation which is now undergoing
multilocus selection we have a possibility
of the following haplotypes in each
gamete:
AB Ab aB or aB
 The frequencies of the possible zygotes
formed by this population in the next
generation are given by:

View punnett square
AABB (.2034)
AABb (.0576)
AaBB (.1536)
AaBb (.0384)
AABb (.0576)
AAbb (.0144)
AaBb (.0384)
Aabb (.0096)
AaBB (.1536)
AaBb (.0384)
aaBB (.1024)
aaBb (.0256)
AaBb (.0384)
Aabb (.0096)
aaBb (.0256)
aabb (.0064)
This population is in Linkage equilibrium until….
See Figure 8.3 pg 287
Differential selection now acts on this
population such that….
All individuals
which are smaller
than 13 units in size
(indicted by
individuals with
less than 3
dominant alleles)
are eaten by
predators and
eliminated from the
population, Leaving
……
A population that is
now in
disequilibrium
How can we
verify that this
population is in
linkage
disequilibrium?
Test #1 Frequency of B alleles on “A” and “a”
chromosomes is the same

Looking at the last figure we can count the
frequency of B on A and on a
B on A =
.88
B on a =
1.0
Test #2 the frequency of any haplotype can be calculated by
multiplying the frequencies of constituent alleles
a= ½ (.1536+.1536)/.6528 = 0.24
 b= ½ (.0576 + .0576)/.6528 = .09
• ab frequency should be .02 but it is actually 0

Finally we test to determine if the linkage
equilibrium value for D is equal to zero
 D= gABgab - gAbgaB
 gab = 0
 so D = a negative value and D is not = 0
Let’s try another scenario using these
same chromosomes

Let’s look at problem # 3 on page 313. Work with
the people at your table to answer part a.
We have just examined how selection on
multilocus genes can lead to linkage
disequilibrium


Genetic drift and population admixture also
disrupt linkage equilibrium
We will not be doing examples of these. If you
are interested please refer to your text on pages
288-289.
WHY AND WHEN DOES DISEQUILIBRIUM
MATTER

If populations are in linkage disequilibrium,
single locus models (Hardy Weinberg) may
yield inaccurate predictions about the
population. WHY?
Stop here on day one
WE WILL NOW INVESTIGATE THE ROLE
OF SEXUAL REPRODUCTION IN THE
BEHAVIOR OF LINKED GENES
First we will investigate the basic
concepts of sexual reproduction as it
relates to the distribution of alleles in
offspring.
WHY SEXUAL REPRODUCTION LEADS TO
GENETIC DIVERSITY ?



Get genetic recombination due to:
Meiosis and crossing over
Random mating between unrelated individuals
Millions of different gametes produced by each
parent
Billions of possible combinations of gametes
for each mating
In every generation alleles which are part of a
multilocus genotype will appear in different
combinations
An example from a highly simplified example using
eye color and hair color alleles.
Which haplotypes are possible in the gametes from this parent?
Haplotypes possible are rb or RB only
Hair color
Eye color
Now we have all four haplotypes rb; RB; Rb; and rB
Genetic recombination shuffles genotypes for multilocus
genes and will reduce genetic disequilibrium
SEXUAL REPRODUCTION REDUCES
LINKAGE DISEQUILIBRIUM





Because of crossing-over and outbreeding, Sexual
reproduction reduces linkage disequilibrium
Meiosis and sexual reproduction lead to genetic
recombinations of genes linked on the same
chromosome
Genetic recombination tends to randomize
genotypes at one locus with respect to genotypes at
another locus on the same chromosome
The result is a reduction in linkage disequilibrium
The greater the rate of crossing over between two
loci, the faster linkage disequilibrium will be
eliminated by sexual reproduction
An experiment on the effects of sexual
mating and equilibrium at two loci
Fruit fly experiments of Michael Clegg
 Started with two populations both in total
linkage disequilibrium and at the opposite ends
of the disequilibrium scale
 Within 50 generations of sexual reproduction,
all of the populations were approaching linkage
equilibrium

Figure 8.7 pg 291
The adaptive significance of sex:
A closer look at the importance of the role
of reproductive strategies in the survival
and evolution of species
•Sexual reproduction:
The cost is too high
Many potential barriers to successful reproduction
What are some of them?
finding a mate • cooperation between
mates •
sexual diseases • mating may prove infertile and result
in no offspring
 Asexual Reproduction
Asexual reproduction is so much more efficient and
produces so many more offspring
The offspring of the original parent are clones so they
may be better adapted to the environment and survive
and reproduce more
Which reproductive mode is better for survival ?


John Maynard Smith (1978) developed a null model to
explore the evolutionary fate of a population under sexual
reproduction versus asexual reproduction.
Involves two assumptions
If both of these assumptions are met then one form of
reproduction will not be favored over the other
1. A female’s reproductive mode does not affect the
number of offspring she can produce.
2. A female’s reproductive mode does not affect
the probability that her offspring will survive
As figure 8.17 shows, assumption # 1 is not met. Asexual parthenogenetic
females will produce larger numbers offspring than sexual reproducers
(16 of 24 are asexual)
Pg 304
The asexuals will constitute an increasingly larger
percentage of the population in each generation and
should completely take over
WHAT ARE THE POTENTIAL CONSEQUENCES?
Just a single mutation in a sexually
reproducing population that produces an
asexual female will lead to inevitable takeover
by asexuals
 This is not what happens in reality and sexual
and asexual forms of many species coexist
just fine
 For sexual species to coexist means they must
confer some benefit for survival
 This benefit could lie in violation of either or
both assumptions

There would be a violation of Assumption #1
if….
A female’s reproductive mode does affect the
number of offspring she can produce
...for instance when paternal care of the young is
required
Sexual populations would leave more young because
asexuals could not take care of their young and not as
many would survive.
Not may species fall into this category.
A VIOLATION OF ASSUMPTION #2 IS MORE LIKELY
This would be violated if a female’s reproductive
mode does affect the probability that her offspring
will survive



A study with flour beetles
Dunbrack and colleagues set up a study that
compared asexual populations and sexual
populations of flour beetles and compared the ability
of the two population to respond to an
environmental stress, namely the application of an
insecticide to their food.
Figure 8.18 shows the results
The control alone, would
supports assumption #1
that if there is an
advantage in the number
of offspring produced
then that type of
reproduction should be
favored
10
Figure 8.18 pg 306
20
30
Looking at this
experimental population
and comparing it to the
control shown above, we
see that there appears to
be a definite advantage to
sexual reproduction. The
sexually reproducing
population eventually
eliminated the asexual
population when exposed
to selection stress.
Why is this?
10
20
30
SEXUAL STRAINS CAN
EVOLVE, ASEXUAL
STRAINS CANNOT
At the level of population genetics, reduction of
linkage disequilibrium is the only consequence of
sex
 Therefore if a population is already in linkage
equilibrium there is no advantage to sexual
reproduction
 Population-genetic Models which propose
evolutionary benefits for sex must include two
things
1. A mechanism to produce linkage
disequilibrium
2. An explanation for why genes that tend to
reduce disequilibrium are favored
Theories dealing with the advantages of
sexual reproduction
There are two categories of models based on the
source of linkage disequilibrium
1. Those that propose genetic drift
2. Those that propose selection on multilocus
genotypes.
Pairs of genes most likely to show
disequilibrium are those that are situated so
closely together on the chromosome that
crossing over between them is rare.
Linkage disequilibrium is most often a problem in
asexual populations since sexual reproduction
tends to eliminate linkage disequilibrium
In freely mating populations most pairs of loci
should be in linkage equilibrium and singlelocus models will work well most of the time
Muller’s Ratchet : Genetic Drift plus Mutation can
make sex beneficial





Works in populations which are small, where drift is a potent
mechanism
As mutations occur in asexual populations, they are passed on to
all offspring of the asexual parent
Over time several mutations can be accumulated in a population
(the frequency of each individual mutant allele is a balance
between mutation rate, the strength of selection and genetic drift)
Asexual populations are doomed to accumulate deleterious
mutations which are passed on to all offspring
Asexual populations cannot get rid of the mutations which are
accumulating until the population is eliminated
Asexual Sub-populations are separate and
reproductively isolated from one another. These subpopulations will have different mutations and differing
numbers of mutations
The fittest of the sub-populations are those
with the fewest mutations
 However, drift can eliminate any of these
populations by chance
 Figure 8.20 pg 308 shows how this works

Each bar represents an asexual sub-population.
Sub populations will differ in the number of
mutations they contain. The sub-population
with the fewest deleterious mutations will be
the fittest.
If the 0 mutation group is lost by drift then
the fittest group now becomes the population
with only one mutation
If drift then takes the1-mutation subpopulation, the fittest is the one with 2
mutations etc
•Over time as the populations age the shift is
toward the accumulation of more and more
mutations

Genetic load increases, the populations are
less and less fit and ultimately the
population becomes extinct

Genetic Load = the accumulation of
deleterious alleles, the more harmful
mutations there are in a population the
greater the genetic load.
Summary of Muller’s Ratchet
The milder the deleterious mutations, the
quicker the ratchet works. If mutations are
too serious, selection will eliminate them
before drift can carry them to fixation
 There are examples from laboratory
experiments and in nature that show that
mutation and drift could indeed be a
mechanism to favor sexual reproduction
 However this mechanism works very slowly
over a long period of time

Sexual reproduction breaks the ratchet



In the case of sexually reproducing species, groups
which are lost by chance can be reconstituted by
outcrossing and recombination
Example: if the 0 mutation group has been lost and
two individuals each with just 1 mutation mate, then
1/4 of their offspring will be mutation-free
Sex reduces linkage disequilibrium by recreating the
missing genotypes
The Red Queen Hypothesis


Red Queen hypothesis, refers to the huffy chess piece
in Lewis Carroll's Through the Looking Glass. In
Looking Glass Land, the Queen tells Alice, "It takes all
the running you can do, to keep in the same place."
According to the Red Queen hypothesis, sexual
reproduction persists because it enables many
species to rapidly evolve new genetic defenses against
parasites that attempt to live off of them.
PBS EVOLUTION VIDEO SEGMENT
You may click the button below to review the
main points of the video.
 As the parasites adapt to new genotypes In the
fish, if they are asexual they are susceptible
 Meanwhile the sexuals can continue to
recombine and present resistant genotypes on
a regular basis

I
FIGURE8.22 PAGE 311
SUMMARY OF THE ADVANTAGE OF SEX
In the context of population genetics, the
advantage of sex is to reduce linkage
disequilibrium
 population-genetic model for the adaptive
value of sex has two parts
1. A mechanism for the creation of
linkage disequilibrium
2. a reason why selection favors traits
that tend to reduce linkage
disequilibrium

THERE ARE TWO CLASSES OF MODELS
Those that credit genetic drift with introducing
disequilibrium by creating high fitness
genotypes that can be lost by drift
 natural selection patterns which continuously
alter the currently best-adapted genotype. Sex
allows lost genotypes to be reclaimed that were
formerly selected against

THE END
Fig. 8.2a pg 283
Figure 8.2b
pg 283
Go to conditions
Figure 8.3a pg 287
% of chromosomes that are A = .2304 + .0576 + .0576 + .1536 =
.4992/.6528 = 76%
% of chromosomes that are a = .1536/.6528 = 24%
B on A = .2304 + .0576 +.1536 / .4992 = 88%
B on a = .1536/.1536 = 1.0
b on A = .0576 / .4992 = 12%
View punnett square
AABB (.2034)
AABb (.0576)
AaBB (.1536)
AaBb (.0384)
AABb (.0576)
AAbb (.0144)
AaBb (.0384)
Aabb (.0096)
AaBB (.1536)
AaBb (.0384)
aaBB (.1024)
aaBb (.0256)
AaBb (.0384)
Aabb (.0096)
aaBb (.0256)
aabb (.0064)
.2304 + ½ (.0576) + ½ (.0576) + ½ (.1536) + ½ (.1536)
.2304 + .0576 + .0576 + ½ (.1536) + ½ (.1536)
.2304+.0576+.1536
= .4416
= 0.88
.2304+.0576+.0576+.1536
.4992