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

Copyright © 2004 Pearson Prentice Hall, Inc. 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

Copyright © 2004 Pearson Prentice Hall, Inc. 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.

Copyright © 2004 Pearson Prentice Hall, Inc. Defining a two locus measure for LE and LD: D = g AB g ab -g Ab g aB 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

Copyright © 2004 Pearson Prentice Hall, Inc. The disequilibrium example yields D = (11/25)(1/25) – (4/25)(9/25) = (11/625) – (36/625) = -(25/625) = At LE D = 0 LD D = (1/2)(1/2) – 0 =.25 = 0 – (1/4) = -.25 as upper and lower limits

Copyright © 2004 Pearson Prentice Hall, Inc. 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 which, curiously, is our calculated D

Copyright © 2004 Pearson Prentice Hall, Inc. 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.

Copyright © 2004 Pearson Prentice Hall, Inc.

We may use the haplotype frequencies g ?? to predict or estimate two locus genotypes by multiplying the male and female haplotype frequency vectors. (g AB + g Ab + g aB + g ab ) 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.

Copyright © 2004 Pearson Prentice Hall, Inc. 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 A  A’ where A’ is far fitter than A. A’ drags all the hitch hiking neighbors along.

Copyright © 2004 Pearson Prentice Hall, Inc. 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 ?

Copyright © 2004 Pearson Prentice Hall, Inc. Expected A1.535 A2.465 G1.624 G2.376 A1G A1G A2G A2G g A1G1 = = D= r = 0.10

Copyright © 2004 Pearson Prentice Hall, Inc. Is this population in LE ? Calculate D D= g A1G1 g A2G2 -g A1G2 g A2G1 = What is D if you use expected haplotype frequencies to calculate D? Which diploid genotypes contribute to the breakdown of LD ?

Copyright © 2004 Pearson Prentice Hall, Inc. APPROACH TO LE g A1G1 ’ = g A1G1 – 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

Copyright © 2004 Pearson Prentice Hall, Inc. 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?

Copyright © 2004 Pearson Prentice Hall, Inc. 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.

Copyright © 2004 Pearson Prentice Hall, Inc. g AB ’ = g AB – rD g ab ’ = g ab - rD g Ab ’ = g Ab + rD g aB ’ = g aB + rD Can you estimate the time to origin of a haplotype ?

Copyright © 2004 Pearson Prentice Hall, Inc. 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.

Copyright © 2004 Pearson Prentice Hall, Inc. 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

Copyright © 2004 Pearson Prentice Hall, Inc. P g is the probability that any given haplotype ( say delta32—197—215) has remained unchanged. P g = (1 – c –  ) g c = probability that c.o. will create a different haplotype (.0021* *0.48) 0.36 CCR 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=

Copyright © 2004 Pearson Prentice Hall, Inc. P g = ( ) g = g Since we know that of the delta 32 chromosomes are at the linked STR loci, we may use as our estimate of P g.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 = g log[0.848] = glog[0.994] = g g = 27.5 generation (25 yrs.) about 700 years

Copyright © 2004 Pearson Prentice Hall, Inc. Muller’s