Copyright © 2004 Pearson Prentice Hall, Inc. Chapter 7 Multiple Loci & Sex=recombination

Copyright © 2004 Pearson Prentice Hall, Inc.

Chapter 7

• Multiple Loci & Sex=recombination


• Many genes are linked together on a chromosome; they are physically joined by being on the same DNA strand.

• Does this matter ?

• interactions

• association/distance

• origin and longevity

• Utility


• The approach is similar to the approach under H-W equilibrium models. We calculate ideal expectations and look for differences which we, then, try to explain as violations of the assumptions.

• The LE assumption is independence;• LE is LINKAGE EQUILIBRIUM.


• Defining a two locus measure for LE and LD:

• D = gABgab-gAbgaB

D = (12/25)(2/25)-(3/25)(8/25)

= (24/625) – (24/625) as expected the

example we just examined is, by this

metric in LE

p(A),q(a) & s(B)t(b)

D = psqt – ptqs

is the expectation


• The disequilibrium example yields• D = (11/25)(1/25) – (4/25)(9/25)

• = (11/625) – (36/625) = -(25/625)• = - 0.04

• At LE D = 0• LD D = (1/2)(1/2) – 0 =.25• = 0 – (1/4) = - .25• as upper and lower limits


• At linkage equilibrium the complete independence means that knowing the allele present at the A locus gives no information about the likelihood of B or b being present at the B locus.

• p= 0.6 and t = 0.2 f(Ab)=0.12• In LD, on the other hand, knowing one locus

DOES provide insight into the more likely allele at the other locus.

• f(Ab) = 0.16 b with A more than expected and knowing A tells you that it will be b at .12 + .04

• which, curiously, is our calculated D


• Soooooooooooooooooo ?• From an associative point of view we know that all

new mutations are in strong LD when they arise:• A………….B• a…………..B new B’ occurs• a…………..B’ f(aB’)= 1/2N• Any new mutation of interest will arise on a

particular genetic background a…. and will remain associated with a until recombination breaks up the association.




• We may use the haplotype frequencies g?? to predict or estimate two locus genotypes by multiplying the male and female haplotype frequency vectors.

• (gAB + gAb + gaB + gab)2 if male=female

• This is clearly the formal equivalent of a single locus four allele system, BUT recombination = mutation and the AaBb double heterozygote can generate new haplotypes. The other two locus genotypes do not alter frequencies through recombination.


• Selection, mutation & migration, or real N may alter haplotype frequencies and generate D non zero.

• It is very difficult to formulate realistic models where selection acts to maintain polymorphism, but directional selection is easy and even a single locus with an advantage will “carry” the rest of the chromosome along in a selective sweep.

• A………B and AA’ where A’ is far fitter than A. A’ drags all the hitch hiking neighbors along.


• LDH has two alleles A1 and A2• GYP has two alleles G1 and G2• Haplotype frequencies• A1G1 306, A1G2 229• A2G1 318, A2G2 147• What are the allele frequencies ?• Expected haplotype frequencies ?


• Expected• A1 .535 A2 .465 G1 .624 G2 .376

• A1G1 333.84• A1G2 201.6• A2G1 290.16• A2G2 174.84• gA1G1 = 0.306 + .0028= 0.3088• @ D=-0.028 r = 0.10


• Is this population in LE ?• Calculate D• D= gA1G1gA2G2-gA1G2gA2G1

• = -0.028

• What is D if you use expected• haplotype frequencies to• calculate D?• Which diploid genotypes contribute• to the breakdown of LD ?


• APPROACH TO LE• gA1G1’ = gA1G1 – rD• What is r?• What is D?• What are the other three equations?• What are the haplotype frequencies• in the next generation?

Calculate haplo. freq.Calculate D


• New mutants generate LD, because they must occur on a genetic background and their association is only broken down by recombination.

Why will this NOT work in an infinite population size?


• Population admixture will produce LD unless the populations have similar or identical haplotype frequencies.

• Population 1 is AB• Population 2 is ab• A 1:1 admixture generates complete LD

which is D = . D breaks down at a rate determined by r which means that even loci on different chromosomes(r=1/2) will be in LD for hundreds of years.


• gAB’ = gAB – rD

• gab’ = gab - rD

• gAb’ = gAb + rD

• gaB’ = gaB + rD

• Can you estimate the time

• to origin of a haplotype ?


• 1) New mutations occur on specific and random genetic(haplotypic) background.

• 2) If we determine the map location for a new mutation of interest and linked loci, we may ask: How old is the new mutation ?

• 3) How ? Knowing that linked loci are in LE, we may ask: Is the new mutation in LD ? If it is in LD, we may estimate how far it has progressed towards LE, because we know r and the allele frequencies. AND we know that the new mutation must have occurred on a specific chromosome.


Identification of Bckgrnd

• 1- c - • probability that recombination• puts the new mutation on to a• different chromosome• - probability that mutation converts• the chromosome with the new• mutation into a new haplotype


• Pg is the probability that any given haplotype ( say delta32—197—215) has remained unchanged.

• Pg = (1 – c – )g

• c = probability that c.o. will create a• different haplotype• (.0021*0.64 +.0072*0.48)• 0.36 CCR5+ 197 215• 0.64 of CCR5 haplotypes are not• normal--GAAT(197) –AFMB(215)• delta32—GAAT(197)-AFMB(215)• arose on this haplotype and now• it’s frequency is 0.36• = 0.005=0.001344 + 0.003456


• Pg = (1-0.005-0.001)g

• = 0.994g

• Since we know that 0.848 of the delta 32 chromosomes are 197-215 at the linked STR loci, we may use 0.848 as our estimate of Pg.That is we use the current frequency of delta32—197—215 as our estimate of the percentage of haplotypes unchanged from the original by recombination or mutation.

• 0.848 = 0.994g

• log[0.848] = glog[0.994]• -0.0716 = -0.0026g• g = 27.5 generation (25 yrs.)• about 700 years


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Copyright © 2004 Pearson Prentice Hall, Inc. Chapter 7 Multiple Loci & Sex=recombination