Linkage Equilibrium / Disequilibrium

Deals with the consideration of two loci simultaneously

The loci are physically linked on the same chromosome

Locus A with alleles “AA”, “aa” and locus B with alleles ”BB” and “bb”

We track not not only frequencies of alleles but also frequencies of chromosomes

Possible chromosome genotypes for this example are:

ABAB; AbAb; aBaB; abab These multi-locus genotypes of genotypes of

chromosomeschromosomes (or gametes) are called haplotypeshaplotypes ( for haploid genotype)

These haplotypes may occur in either Linkage Equilibrium or Linkage Disequilibrium

Have genotypes that are independent of one another.

If you know the genotype at one locus (AA) you cannot predict what the genotype will be at the other locus (BB).

Example: Suppose that the gene which controls the length of toes in frogs (AA) is linked to the gene that controls the amount of webbing between the toes (BB). Populations that are in linkage equilibrium will

show no correlation between toe length and the degree of webbing between them.

Genotypes of the chromosomes (Haplotypes) exhibit a non random association between the linked genes.

If you know the genotype at one locus (AA) you have a clue about the genotype at the other locus (BB).

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.

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 “BB” allele is equally likely on either the AA or the aa chromosome the alleles are in linkage equilibrium

Figure 8.2a

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 “BB” allele is not equal on the A and the a chromosomes

If selection acts on one locus only.... Selection for the “AA” allele has no effect on

the “BB” allele frequency. See Figure 8.8b

AA= 5/25 = .2; aa= .8 AA = 20/25 = .8; aa= .2BB = 20/25 = .8; bb= .2 BB= 20/25 =.8; bb =.2

If selection acts on one locus only.... Selection for the “AA” allele changes the BB allele

frequencies also. As aa chromosomes are lost they drag “BB” alleles along in a disproportionate fashion. See Figure 8.8a

16

4

4

A= 5/25 = .2; a =.8 A = 20/25 = .8; a = .2B= 17/25 = .68; b= .32 B= 8/25 = .32; b= .68

In linkage equilibrium chromosome (haplotype) frequencies do not change, they can still be predicted (calculated) from allele frequencies.““B”=B”=20/25==.8 “bb”5/25 = .2 “AA”20/25= .8 “a”a”5/25== .2

ABAB= .64 AbAb = .16 aBaB= .16 16 abab = .04

ABAB= 16/25 =.64 AbAb= 4/25 = .16

aBaB = 4/25 = .16 abab =1/25 = .04

Actual haplotype frequencies

Calculated haplotype frequencies

In linkage disequilibrium chromosome frequencies change, they can not be predicted (calculated) from allele frequencies.“BB”8/25= .32 “bb”17/25 = .68 “AA”20/25= .8 “aa”5/25= .2

ABAB=.256 AbAb = .544 aBaB= .064 abab = .136

ABAB= 4/25 =.16 AbAb= 16/25 = .64

aBaB = 4/25 = .16 abab =1/25 = .04

calculated

actual

1. The frequency of “BB” on chromosomes carrying allele “AA” is equal to the frequency of “BB” on chromosomes carrying allele “aa”.

2. The frequency of any chromosome haplotypes can be calculated by multiplying the frequencies of the alleles which compose that haplotype

3. The quantity D, (coefficient of disequilibrium)=0

D= gABgab - gAbgaB

g is the frequency of the various haplotypes

Verify

selection on multilocus genotypes

genetic drift population admixture

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:

ABAB AbAb aBaB or aBaB The frequencies of the possible

zygotes formed by this population in the next generation are given by:

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)

View punnett square

This population is in Linkage equilibrium until….

See Figure 8.3 pg 287

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 ……

Differential selection now acts on this population such that….

A population that is now in disequilibrium

How can we How can we verify that this verify that this population is in population is in linkage linkage disequilibrium?disequilibrium?

Looking at the last figure we can count the frequency of B on AA and on aa

BB on AA =

BB on aa =1.0

.88

aa= bb=

½ (.1536+.1536)/.6528 = 0.24½ (.0576 + .0576)/.6528 = .09

• abab frequency should be .02 but it is actually 0

D= gABgab - gAbgaB

gab = 0 so D = a negative value and D is not = 0

Let’s look at problem # 3 on page 313. Work with the people at your table to answer part a.

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.

If populations are in linkage disequilibrium, single locus models (Hardy Weinberg) may yield inaccurate predictions about the population. WHY?WHY?

Stop here on day one

First we will investigate the First we will investigate the basic concepts of sexual basic concepts of sexual reproduction as it relates to reproduction as it relates to the distribution of alleles in the distribution of alleles in offspring. offspring.

Get genetic recombination due to: Meiosis and crossing over Random mating between unrelated

individuals Millions of different gametes produced by

each parentBillions 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.

Haplotypes possible are rb or RB only

Which haplotypes are possible in the gametes from this parent?

Genetic recombination shuffles genotypes for multilocus genes and will reduce genetic disequilibriumreduce genetic disequilibrium

Eye

color

Hair color

Now we have all four haplotypes rbrb; RBRB; RbRb; and rBrB

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

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 cost is too high Many potential barriers to successful reproduction

What are some of them?What are some of them?finding a mate • cooperation between

mates • sexual diseases • mating may prove infertile and result in no offspring

Asexual ReproductionAsexual 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

•Sexual reproduction:

John Maynard SmithJohn 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

The asexuals will constitute an increasingly larger percentage of the population in each generation and should completely take over

(16 of 24 are asexual)

As figure 8.17 shows, assumption # 1 is not met. Asexual

parthenogenetic females will produce larger numbers offspring than sexual reproducers

Pg 304

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

...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 female’s reproductive mode does affect the number of offspring she can produce

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

This would be violated if a female’s reproductive mode does affect the probability that her offspring will survive

Figure 8.18 pg 306

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.

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

Why is this? Why is this?

10 20 30

10 20 30

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

There are two categories of models based on the source of linkage disequilibriumsource of linkage disequilibrium

1. Those that propose genetic drift

2. Those that propose selection on multilocus genotypes.

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 single-locus models will work well most of the time

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.

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

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

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 sub- population, the fittest is the one with 2 mutations etc

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.

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.

•Over time as the populations age the shift is toward the accumulation of more and more mutations

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

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

Red Queen hypothesisRed 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.

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

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

Those that credit genetic drift with introducing disequilibrium by creating high fitness genotypes that can be lost by drift

natural selection patterns selection patterns which continuously alter the currently best-adapted genotype. Sex allows lost genotypes to be reclaimed that were formerly selected against

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 = .49924992/.6528 = 76%

% of chromosomes that are a = .15361536/.6528 = 24%

B on A = .2304 + .0576 +.1536 / .4992 = 88%

b on A = .0576 / .4992 = 12%

B on a = .1536/.1536 = 1.0

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)

View punnett square

.2304+.0576+.1536

.2304+.0576+.0576+.1536

= .4416

.4992

.2304 + ½ (.0576) + ½ (.0576) + ½ (.1536) + ½ (.1536)

.2304 + .0576 + .0576 + ½ (.1536) + ½ (.1536)

= 0.88