2. Linkage Equilibrium / Disequilibrium

3.  Deals with the consideration of two loci simultaneously  The loci are physically linked on the same chromosome Aa Bb  Locus A with alleles “A”, “a” and locus B with alleles ”B” and “b” not  We track not only frequencies of alleles but also frequencies of chromosomes

4. ABAbaBab  Possible chromosome genotypes for this example are: AB; Ab; aB; ab genotypes of chromosomes haplotypes  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

5.  Have genotypes that are independent of one another. A B  If you know the genotype at one locus (A) you cannot predict what the genotype will be at the other locus (B). A 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.

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

9.  If the frequencies of the haplotypes can be calculated by multiplying the frequencies of the two alleles involved, then they are in linkage equilibrium. B Aa  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

10.  If the frequencies of the haplotypes cannot be calculated by multiplying the frequencies of the two alleles involved, then they are in linkage disequilibrium B  The occurrence of “B” allele is not equal on the A and the a chromosomes

12.  If selection acts on one locus only.... A B  Selection for the “A” allele has no effect on the “B” allele frequency. See Figure 8.8b AaAa BbBb A= 5/25 =.2; a=.8 A = 20/25 =.8; a=.2 B = 20/25 =.8; b=.2 B= 20/25 =.8; b =.2

13.  If selection acts on one locus only.... AB aB  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 16 4 4 A= 5/25 =.2; a =.8 A = 20/25 =.8; a =.2 B= 17/25 =.68; b=.32 B= 8/25 =.32; b=.68

15.  In linkage equilibrium chromosome (haplotype) frequencies do not change, they can still be predicted (calculated) from allele frequencies. “B”==bAa”= “B”=20/25=.8 “b”5/25 =.2 “A”20/25=.8 “a”5/25=.2 ABAbaB16 ab AB=.64 Ab =.16 aB=.16 ab =.04 ABAb AB= 16/25 =.64 Ab= 4/25 =.16 aBab aB = 4/25 =.16 ab =1/25 =.04 Actual haplotype frequencies Calculated haplotype frequencies

16.  In linkage disequilibrium chromosome frequencies change, they can not be predicted (calculated) from allele frequencies. BbAa “B”8/25=.32 “b”17/25 =.68 “A”20/25=.8 “a”5/25=.2 ABAbaBab AB=.256 Ab =.544 aB=.064 ab =.136 ABAb AB= 4/25 =.16 Ab= 16/25 =.64 aBab aB = 4/25 =.16 ab =1/25 =.04 calculated actual

17. B AB a 1. The frequency of “ B ” on chromosomes carrying allele “ A ” is equal to the frequency of “ B ” on chromosomes carrying allele “ a ”. 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= g AB g ab - g Ab g aB g is the frequency of the various haplotypes Verify

18.  selection on multilocus genotypes  genetic drift  population admixture

19.  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: ABAbaBaB AB Ab aB or aB  The frequencies of the possible zygotes formed by this population in the next generation are given by:

20. 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

21. 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….

22. A population that is now in disequilibrium How can we verify that this population is in linkage disequilibrium?

23. Aa  Looking at the last figure we can count the frequency of B on A and on a BA B on A = Ba B on a = 1.0.88

24. aa=bb=aa=bb= ½ (.1536+.1536)/.6528 = 0.24 ½ (.0576 +.0576)/.6528 =.09 ab ab frequency should be.02 but it is actually 0

25.  D= g AB g ab - g Ab g aB  g ab = 0  so D = a negative value and D is not = 0

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

27.  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.

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

29. Stop here on day one

32.  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.

33. Haplotypes possible are rb or RB only Which haplotypes are possible in the gametes from this parent?

35. reduce genetic disequilibrium Genetic recombination shuffles genotypes for multilocus genes and will reduce genetic disequilibrium Eye color Hair color rbRBRbrB Now we have all four haplotypes rb ; RB ; Rb ; and rB

36.  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

37.  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

38. Figure 8.7 pg 291

40. 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 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 Sexual reproduction :

41.  John Maynard Smith  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

42. 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

43.  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

44. ...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

45.  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

46. 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? 10 20 30

48. We can conclude that assumption #2 is incorrect Same experiment but with the black beetle being given the sexual role. We see essentially the same results 10 20 30

49.  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

50. source of linkage disequilibrium 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.

51. 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.

52.  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

53.  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

54. 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.

55.  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

56.  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

57.  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

58.  In general constant environments favor asexual reproducers that are well adapted to the current environment  When the environment changes they are at a disadvantage  The changing environment theories for the advantage of sex assume trade-offs such that …  genotypes that do well in one type of environment necessarily do more poorly in others

59.  Basically involves alternating between a selection regime which favors selection of particular multilocus genotypes at one point in time and then later selection is for a different set of multilocus genotypes  Sex can recreate the genotypes that were recently eliminated but have now become favored  Differences in fitness may be caused by either physical changes in the environment or by changes in biotic interactions ( Elderflower example)

60.  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.

61.  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

62. I

63.  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

64.  Those that credit genetic drift with introducing disequilibrium by creating high fitness genotypes that can be lost by drift selection patterns  natural selection patterns which continuously alter the currently best-adapted genotype. Sex allows lost genotypes to be reclaimed that were formerly selected against

66. Fig. 8.2a pg 283

67. Figure 8.2b pg 283 Go to conditions

68. Figure 8.3a pg 287

69. 4992 % of chromosomes that are A =.2304 +.0576 +.0576 +.1536 =.4992/.6528 = 76% 1536 % of chromosomes that are a =.1536/.6528 = 24% B on A =.2304 +.0576 +.1536 /.4992 = 88% b on A =.0576 /.4992 = 12% B on a =.1536/.1536 = 1.0

70. 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

72. .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