Linkage Mapping. Reading: Mather K (1951) The measurement of linkage in heredity. 2nd Ed. John Wiley and Sons, New York. Chapters 5 and 6.

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1 Linkage Mapping Reading: Mather K (1951) The measurement of linkage in heredity. 2nd Ed. John Wiley and Sons, New York. Chapters 5 and 6. Genetic maps The relative positions of genes on a chromosome can be determined by conducting linkage analysis. As an example, Hedges et al. (1990) crossed the soybean line Peking, which is homozygous for dominant alleles at genes Rj1 (restricts nodulation with certain strains of Bradyrhizobium) and F (confers the fasciated stem and leaf phenotype), to BARC-1, which is homozygous recessive for both genes. The two lines also differed for alleles at the codominant isozyme locus Idh1 (isocitrate dehydrogenase). The triply heterozygous F 1 plant was selfed to form an F 2 population of 250 plants, each of which was scored for the three phenotypes. Development of a genetic map from this type of experiment follows three phases: first, linkage is detected or loci are declared unlinked; second, recombination frequencies between each pair of loci are estimated; and third, the loci are ordered into a linear map. Detection of linkage Are two genes linked? Deviations from expected segregation ratios derived assuming no linkage are evidence for linkage. But in any sampled population, we know that there will be some deviation from expectation simply due to random chance. How do we tell the difference between random deviations from expected segregation ratios and deviations due to linkage? The chi-square (χ 2 ) test provides a statistical basis for making this determination. We know the expected segregation ratio under the null hypothesis of no linkage between two genes, and for an observed segregation ratio, we can compute the probability of observing the data if the null hypothesis were really true. Generally, we consider data that would be observed with 5% or lower probability under the null hypothesis as evidence that the null hypothesis is false. The data collected by Hedges et al. (1990) on the segregation of the F and Rj1 genes are given in Table 1. For now, we ignore the Idh1 gene. Table 1. Observed phenotypic segregation in an F 2 population developed from the soybean cross Peking (FFRj1Rj1) BARC-1 (ffrj1rj1). Phenotypic class Number Observed F_Rj1_ 144 F_rj1rj1 44 ffrj1-39 ffrj1rj1 23 Total 250 1

2 First, we check that the expected single-gene ratios were observed. For a single dominant gene in an F 2 population, what is the expected phenotypic segregation ratio? Pooling data for F over the two R classes, we find 188 F_ and 62 ff phenotypes. The expected numbers based on the null hypothesis are and 62.5; so the observed data are as close to expectation as possible, we won t bother performing a statistical test. Pooling data for R over the two F classes, we find 183 R_ and 67 rr phenotypes. To perform the χ 2 test, we compute the difference between expected and observed numbers in each class, square the difference and divide by the expected number, then sum over the two classes to obtain the χ 2 statistic (Table 2). Table 2. Chi-square test for single-gene segregation of Rj1 phenotype. Phenotypic class Number Observed Number Expected (E-O) 2 /E Rj1_ rj1rj Total In this example, the χ 2 statistic is This statistic has one degree of freedom because there are two classes, so if a plant is not in one class it has freedom to be in only one other class. Looking up the threshold χ 2 value for one d.f. (e.g., Appendix A.5 in Steel and Torrie), we find that a value of 3.84 or greater will occur 5% of the time or less if the null hypothesis is true. If the null hypothesis is true, then the we expect to observe a value of or greater about 50% of the time. Thus, there is no evidence that the null hypothesis is wrong; the Rj1 phenotype segregates as a single gene. Now that we are certain the Rj1 and F phneotypes are each segregating as single genes, we can ask if they are linked. Under the null hypothesis of no linkage, what is the expected segregation ratio for two dominant genes in an F2 population? We compute the χ 2 statistic based on this expectation (Table 3.) Table 3. Chi-square test for linkage of F and Rj1 genes. Phenotypic class Number Observed Number Expected (E-O) 2 /E F_Rj1_ F_rj1rj ffrj1_ ffrj1rj Total What is the threshold χ 2 value for the test of no linkage? It depends on the degrees of freedom associated with the test. This is tricky: the degrees of freedom for the overall deviation from the 9:3:3:1 segregation ratio has d.f. = 3 because there are four classes. But notice that deviations from this expectation could occur for three different reasons: deviation from 3:1 segregation at the F locus, deviation from 3:1 segregation at the Rj1 2

3 locus, or deviation from independent segregation between the two genes. The contribution of each source of variation to the total χ 2 value is additive. Thus, we can obtain the single degree of freedom χ 2 value for deviation due to linkage alone by subtracting the two χ 2 values for single-locus deviations from the overall χ 2 value: = (0.005 being the χ 2 value for deviations at the F locus). Notice that a value with 3 d.f. minus two values with one d.f. each leaves a value with one d.f. So we use the threshold χ 2 value for one d.f., which is Since the value for linkage is greater than the threshold value, we interpret this to mean that if the two genes were really unlinked, we would observe such a result 5% of the time or less. So, our conclusion is that the genes are not unlinked, thus they are linked. Similar χ 2 tests reveal that the Idh1 locus is linked to both the F and the Rj1 loci. Having determined that the genes are linked, we now must estimate the recombination frequency. Estimating recombination frequency The recombination frequency is the proportion of recombinant gametes divided by the total number of gametes. Consider the data in Tables 1 and 3 on the frequency of F Rj1 phenotypes: which classes represent recombinant gametes, and which represent parental gamete types? Obviously, the ffrj1rj1 class is generated by the union of two nonrecombinant gametes. But what about the F_rj1rj1 class? Obviously, one recombinant Frj1 gamete went into each plant in this class; but notice that these plants could be created from the union of two recombinant gametes (Frj1 + Frj1 = FFrj1rj1) or from the union of one recombinant gamete and one non-recombinant gamete (Frj1 + frj1 = Ffrj1rj1). Without performing progeny tests, we can t distinguish these possibilities, so we cannot simply count the number of recombinant gametes in an F 2 population, as we did in a backcross population. Being unable to count the number of recombinant gametes, how can we possibly estimate the recombination frequency? The most common method for estimating recombination frequency in this case is to use the method of maximum likelihood. This method is a general estimation procedure that is used in many areas of statistics, not only in linkage mapping. The basic idea is that, if we have a model that relates underlying factors to the observed data, then we can compute the probability of observing the data given a specific model. The probability of observing the data given the model is proportional to the likelihood of the model given the observed data. Thus, we can compute likelihoods for different models given the data and choose the model that is more likely as the better model. The idea is quite simple, but there is one major difficulty that is often encountered: there may be an infinite number of possible models, and without being able to compute likelihoods of all possible models, how can you be sure that you choose the most likely model (the maximum likelihood model)? For any particular application of the maximum likelihood method, therefore, batteries of statistical and mathematical tricks have been developed to make it possible to obtain the maximum likelihood estimate with some degree of assurance. For simple models, such as two point recombination frequency estimation, it is quite certain that one will obtain the maximum likelihood result using some simple tricks. For more complex procedures, including multi-point linkage mapping with many loci 3

4 simultaneously, it is often not guaranteed that one will obtain the maximum likelihood estimate. To illustrate the procedure, we will demonstrate the maximum likelihood procedure for two-point recombination frequency estimation between the F and Rj1 loci. First, we need a general model for how the data could be affected by recombination between the two loci. To do this, we need to compute the probabilities of obtaining the observed phenotypes given some degree of linkage. Starting with the doubly heterozygous F 1 plant, and knowing that the two genes are in coupling phase (the dominant alleles at both loci came from the same parent), we compute the probability of each possible two-locus gamete produced by the F 1 parent. There are two possibilities for recombination events: (1) recombination occurs between the two genes in the F 1, with a probability of r, the recombination frequency that we want to estimate; or (2) recombination does not occur, with probability 1 - r. If recombination does occur, then two types of gametes are produced, F - rj1 and f - Rj1. These occur with equal frequency, so the probability of an F-rj1 gamete is (½)r. And the probability of an f - Rj1 gamete is the same, (½)r. In the same way, the probabilities of the nonrecombinant gametes, F-Rj1 and f - rj1, are both (1/2)(1-r). Knowing the probabilities of the gametes for any arbitrary recombination frequency, r, allows us to compute the probabilities of the 9 possible F 2 genotypic classes, using a Punnet square (Table 4). Table 4. Punnet square used to determine probability of F 2 genotypes based on gamete probabilities from F 1. Male gametes Female gametes F-Rj1 (½)(1- r) F-rj1 (½)r f-rj1 (½)r f-rj1 (½)(1- r) F-Rj1 (½)(1- r) FFRj1Rj1 A FFRjrj1 A FfRj1Rj1 A FfRj1rj1 (1/4)(1- r) 2 (1/4)r(1- r) (1/4)r(1- r) (1/4)(1- r) 2 F-rj1 (½)r FFRjrj1 A FFrj1rj1 B FfRj1rj1 A Ffrj1rj1 B (1/4)r(1- r) (1/4)r 2 (1/4)r 2 (1/4)r(1- r) f-rj1 (½)r FfRj1Rj1 A FfRj1rj1 A ffrj1rj1 C ffrj1rj1 C (1/4)r(1- r) (1/4)r 2 (1/4)r 2 (1/4)r(1- r) f-rj1 (½)(1- r) FfRj1rj1 A Ffrj1rj1 B (1/4)(1- r) 2 (1/4)r(1- r) ffrj1rj1 (1/4)r(1- r) C ffrj1rj1 (1/4)(1- r) 2 Summing over all 16 boxes gives a total probability of one. Since we can only observe four phenotypic classes (A - D), we use the data from Table 4 to determine the probability of each of the four phenotypic classes. For example, the F_rj1rj1 phenotypic class is composed of genotypes FFrj1rj1 (which occurs with probability (1/4)r 2 ) and Ffrj1rj1 (which occurs with probability (1/4)r(1- r) + (1/4)r(1- r) = (½)r(1- r). Therefore, the probability of the F_rj1rj1 class is (1/4)r 2 + (½)r(1- r) = (1/4)r 2 + (½)r - (½)r 2 = (½)r - (1/4)r 2. The probabilities of the other four classes are worked out similarly (Table 5). D A 4

5 Table 5. Probabilities of four F 2 phenotypic classes based on two dominant genes. Phenotypic Class Genotypes Probability A (F_Rj1_) FFRj1Rj1, FfRj1Rj1, FFRj1rj1, FfRj1rj1 3/4 - (½)r + (1/4)r 2 B (F_rj1rj1) FFrj1rj1, Ffrj1rj1 (½)r - (1/4)r 2 C (ffrj1_) ffrj1rj1, ffrj1rj1 (½)r - (1/4)r 2 D (ffrj1rj1) ffrj1rj1 (1/4)(1 - r) 2 Total 1 If we observe one F 2 progeny from this cross, the probability of its being one of the four phenotypic classes is given in Table 5. What if we observe two F 2 progeny? What is the joint probability that both are phenotypic class A? That is the product of the probabilities of each one being phenotypic class A, because the progeny are independent. Similarly, the probability that one progeny is class A and the other is class B is the product of the probabilities that the first is A and the second is B plus the product of the probabilities that the first is B and the second is A. We can compute the joint probability for any sample of n total F 2 progeny with a, b, c, and d numbers in each progeny class using the multinomial probability expression: n! L = + + a! b! c! d! ( 2 ) ( 2 ) ( 2 3 ) ( 2 ) r r a r r b r r c r r d which is simply a special case of the general multinomial likelihood formula: n! a a a L = m m m t t a! a!... a! ( ) 1 ( ) 2...( ) t where n = total number of progeny observed, t = total number of progeny classes possible, a i = number of progeny in class i observed, and m i = expected proportion in class i. This formula can be used for any generation or mating scheme used for linkage analysis. You can see from the formula that n, a, b, c, and d are all observed quantities, and only r is unknown. By plugging in different values for r, one can compute the likelihood of observing the data for different values of r. So, one way to obtain a maximum likelihood estimate of recombination frequency is to compute the likelihood for a range of values of r between 0 and 0.5 and choose the value with the highest likelihood as the maximum likelihood estimate of recombination frequency. This is a numerical solution. Another way to obtain the maximum likelihood estimate is to consider the problem in terms of calculus. One can imagine plotting likelihood values on the Y-axis for recombination values on the X-axis, and obtaining a curve representing likelihood as a function of recombination frequency. Recall from calculus that the derivative of this function with respect to recombination frequency (r) represents the slope of the curve. The slope of the curve is zero where the curve peaks (or where the curve hits a nadir). 5

6 Thus, points at which the derivative is zero represent either maximum or minimum points of the likelihood curve. So, another way to find the maximum likelihood point without actually graphing the likelihood curve is to take the derivative of the likelihood function given above with respect to r, setting this derivative equal to zero and solving for values of r. Remember that if multiple solutions are obtained, they may be maxima or minima, in theory, but in practice, often there is only a single solution, or if there are multiple solutions, only one exists within the parameter space between 0 and 0.5. From this point on, it is simply a matter of mathematics to obtain the correct answer. In this example, an additional math trick is used to simplify things. Differentiation of the likelihood function itself is quite difficult because of the powers involved, so usually we take the natural log of the likelihood function and differentiate that instead. The resulting equation is easier to differentiate and we know from calculus that the maximum point of a function is also the maximum point of the log of the function. A further trick is made by substituting P for (1-r) 2, and taking the derivative with respect to P, solving for the maximum likelihood estimate of P, and then converting the estimate of r. The log of the likelihood function is: ln( L n! ) = ln( ) + aln 4 ( 2 + P) + bln 4 ( 1 P) + cln 4 ( 1 P) + d ln 4 P a! b! c! d! The derivative of the log function with respect to P is set equal to zero: ln( L) P = a b c d + = P 1 P 1 P P Multiplying both sides of the equation by the common denominator (2+P)(1-P)P gives: a( 1 P) P ( 2 + P) P( b + c) + ( 2 + P)( 1 P) d = a( P P ) ( 2P + P )( b + c) + ( 2 P P ) d = 0 2 2d + ( a 2b 2c d) P ( a + b + c + d) P = 0 Now, we fill in the numbers for a, b, c, and d from Hedges et al. (1990): 46 + P(-45) + P 2 (-250) = 0 This has the form of a quadratic equation, ax 2 + bx + c = 0, which has roots: b b ac x = ± 2 4 2a Using the quadratic equation, we solve for P = -0.52, P = (1- r) 2, so (1- r) = square root of P, which means that only is a real solution for P. (1- r) = r =

7 In some cases, one does not end up with a derivative of the log of the likelihood function that can be solved mathematically, as was possible in this case. In such a case, one must use numerical methods to obtain the correct solution. Nevertheless, it is still useful to get the derivative of the likelihood function because that leads to an estimate of the variance of the linkage estimate. The variance of the estimate requires taking the second derivative of the function, which we won t go into here (see Mather (1951) for details). The standard error of the recombination frequency estimate in this example is The important message is that these recombination frequency estimates are just that, estimates, and they are measured always with some error. Standard Errors of Recombination Frequencies and Information Factors that influence the precision of recombination frequency estimates include the sample size (bigger is better always), the gene action of the marker genes (more information is obtained with codominant marker genes than with dominant genes), and the type of population used (F 2 populations with codominance are better than backcross, but backcrosses are better than F 2 populations complete dominance). A way to compare the informativeness of different types of populations is to compare their information value per data point (one individual). This discussion follows Mather (1951). The total amount of information in a set of data is the inverse of the variance of the recombination frequency estimate from those data: I(r) = 1/V(r). The variance of the estimate is the variance of a single data point divided by the number of data points. Turning this around, the variance of a single data point is equal to the variance of the total data set divided by the number of data points. So, we can obtain the information per data point as a function of the variance of the estimator: I(r) = 1/V(r) = ni(r), where n = number of individuals sampled and i(r) = the information per individual: = = c 2 1 mi i( r), where c = total number of classes of progeny in the population. i 1 mi r The information for DH populations was derived as follows: Class Coupling Repulsion m i dm i /dr i m i dm i /dr i AABB (1/2)(1-r) -1/2 1/2(1-r) (1/2)r 1/2 (1/2)r AAbb (1/2)r 1/2 (1/2)r (1/2)(1-r) -1/2 1/2(1-r) aabb (1/2)r 1/2 (1/2)r (1/2)(1-r) -1/2 1/2(1-r) AABB (1/2)(1-r) -1/2 1/2(1-r) (1/2)r 1/2 (1/2)r Total 1 0 1/r(1-r) 1 0 1/r(1-r) 7

8 This is the same as for backcross populations (Mather 1951). (But note that in BC populations with dominant markers, only those loci for which the recurrent parent is recessive will be informative). The information values can be computed similarly for other types of populations, permitting a comparison of the relative informativeness (relative efficiency) of different population types per individual sampled (Figure 1 from Mather, 1951). I added relative efficiencies to this figure for recombinant inbred line (RIL) populations (equation from Liu et al. (1998)) and for doubled haploid (DH) populations. 8

9 Thus, F 2 populations with codominant markers are overall the best choice among these population types for mapping. However, if one only has dominant markers, F 2 populations have VERY poor precision for repulsion-phase linked markers. With dominant markers only, RILs give the best precision for loci linked at r < 0.15, and DH and BC populations give better precision for r > If you can develop them, DH populations are superior to BC populations in the sense that you can map all polymorphic markers in DH populations, but you can map only those polymorphic markers at which the recurrent parent is homozygous recessive in BC populations. Table 6 provides an example of how population type and population size influence the precision of recombination frequency estimates (Allard 1956). Table 6. Standard errors of recombination frequency estimates for different linkage intensities, population types, and population sizes (n). n = 100 n = 200 r = 0.05 r = 0.10 r = 0.20 r = 0.05 r = 0.10 r = 0.20 F2 dom. (coupling) F2 dom. (repulsion) F2 codominant Backcross = DH RIL LOD Scores An alternative statistic for linkage analysis that is used in combination with maximum likelihood methods is the LOD score. The LOD score is the logarithm of odds of a particular linkage arrangement and recombination frequency relative to the likelihood of the null hypothesis of no linkage (Lander and Botstein 1986): LOD = log 10 L( r$ MLE) L( r = 05. ) Using the data from the example of the F and Rj1 genes, the likelihood of the maximum likelihood estimate model, with a recombination frequency of 0.41 is obtained by simply plugging in r = 0.41 to the likelihood equation for the data set: 250! 144 ( ) 23 L( r$ = 041. ) = ( ) ( ) ( ) 144! 44! 39! 23! We won t bother to work this out because the first part with the factorials will cancel out in the LOD score equation. The likelihood of the model with no linkage is computed in the same way, but substituting 0.5 for r in the same equation. The LOD score is: 9

10 LOD = log = log 10 L( r$ = 0. 41) = log L( r = 05. ) ( + ) ( ) ( ) ( ) 144 ( ) 23 ( 9 / 16) ( 3/ 16) ( 1/ 16) = log ( ) = The odds ratio shows that the model of linkage of r = 0.41 is more than eight times more likely than the model of no linkage. In most applications in which many genes are mapped simultaneously, these two loci would probably be considered unlinked, because very stringent thresholds are normally used (e.g., linkage is often declared only when genes are linked with LOD of 3.0 or greater, meaning linkage is 1,000 times more likely than non-linkage). This is because when many genes are being mapped, the chance of declaring linkages falsely at least once in the entire analysis becomes quite large, so a high LOD threshold is used. At the same time, when many genes are being tested for linkage, it is likely that a gene between F and Rj1 would be mapped, and being closer to both of them than they are to each other, it would probably exhibit higher LOD scores with both F and Rj1, and the F and Rj1 genes would end up in the same linkage group anyway. The real reason the LOD score was developed was because human geneticists work with numerous pedigrees, each of which has only a few individuals. From any single family, there is not enough information to determine linkage, but by combining results across pedigrees, it is possible to detect linkage. A LOD score can be computed for each pedigree separately, then the combined LOD score for the whole data set is obtained by multiplying together the individual LOD scores. Plant breeders still need to understand LOD scores because the most commonly used software for linkage mapping is MAPMAKER, which was developed by human geneticists and uses LOD scores. LOD scores also provide a convenient way to compare how much more likely a model is than different alternative models. Ordering Loci Using the methods described above, Hedges et al. (1990), estimated recombination frequency between the three pairs of genes in soybean as follows: Gene Pair Recombination Frequency F - Rj ± 0.05 F - Idh ± 0.03 Rj1 - Idh ± 0.03 What is the order of the three loci? The most common sense ordering is F - Idh1 - Rj1. The linkage map for these three genes is drawn to summarize both the order and recombination distance information: 10

11 F Idh1 Rj Notice that there is something weird here: if this were really like a road map, the distance from F to Rj1 should be the sum of the distances from F to Idh1 and Idh1 to Rj1. But the recombination frequency between F and Rj1 is 0.41, which is less than the sum of 0.22 and Consider the effect of a recombination between F and Idh1 and a recombination between Idh1 and Rj1 in the same meiosis. The alleles at F and Idh1 will be recombined and the alleles at Idh1 and Rj1 will be recombined, but in those gametes, the parental combinations at the F and Rj1 genes will exist. So, when genes get very far apart, double-recombinations are more likely to occur between them. If the recombination frequency between F and Idh1 were 30% and that between Idh1 and Rj1 were 30%, the maximum recombination frequency between F and Rj1 would still be at most 50%. For this reason, mapping functions were developed to linearize maps, meaning that they make the distance between any two loci equal to the sum of the distances of the intervals between them. Mapping Functions Mapping functions are mathematical tools that are used to linearize genetic maps and to put genetic distances on a basis that relates to the number of crossovers that occur between genes, rather than on the basis of recombination. In order to develop a mapping function, one has to make assumptions about whether or not crossovers occur independently along a chromosome, or whether one crossover in a region tends to prevent a second crossover from occurring near it. This latter possibility is called chromosome interference, as we discussed in the previous lecture. We now return to the equation relating recombination frequency between loci A and C as a function of the recombination frequencies in the intervals A B and B C between these loci: r AC = r AB + r BC 2(1 - i)r AB r BC, where i = coefficient of coincidence. As we discussed in the previous lecture, if one assumes no interference, then i = 0 and we have: r AC = r AB + r BC 2r AB r BC. This is the basis for Haldane s mapping function, which we will derive. First, we define the average number of crossovers in an interval to be λ. We next define a map unit to be m = (1/2)λ. Map units are reported in Morgans, or more typically, centimorgans (cm). The idea is that an interval in which an average of one crossover occurs every meiosis is defined to have a genetic map distance 50 cm ( = 0.5 M, thus m = 0.5λ). 11

12 Haldane derived his mapping function by assuming that crossovers occur independently across the chromosome, so that the probability that a crossover occurring between any two points on a chromosome follows a Poisson distribution. Based on the Poisson distribution, the probability of no crossover between the loci defining the interval is: P[no crossover] = e -λ = e -2m. The probability of a crossover is: P[one or more crossovers] = 1 - e -2m. Each time one or more crossovers occurs in the interval, 50% recombinant gametes are produced (see last lecture), so the probability of a recombination in the interval is: r 2m 1 e =. 2 The inverse of this allows one to convert map distances into recombination frequencies: ln( 1 2r) m =. 2 Another commonly used mapping function is Kosambi s, which assumes that a moderate amount of interference occurs. Specifically, the Kosambi mapping function is derived under the assumption that the coefficient of coincidence is related to the size of the interval being mapped, C = 1 - i = 2r. This assumption implies that if the interval is large and r is near 0.5, then C approaches 1, i approaches 0, and interference is absent. But if the interval is small, and r is near 0.0 then C approaches 0, i approaches 1, and interference is nearly complete. With this assumption, the Kosambi mapping function is (see Liu, 1998 for derivation): 1 m = ln r. 1 2r The inverse of the Kosambi function is: 4m 1 ( e 1) r =. 4m 2 ( e + 1) The relationship between these two mapping functions and recombination frequency is shown in the figure below, from Ott (1999): 12

13 Notice that at small recombination frequencies (r < 0.10), there is little difference between recombination frequency itself and any of the mapping functions. This is because if recombination occurs at low frequency in an interval, than multiple crossovers must occur at very low frequency in such intervals. If the only reasonably likely events are zero of one crossover at any one meiosis, then the relationship between crossing over and recombination is linear! Thus, if one has a dense genetic map (with markers spaced at intervals no greater than, say, 10% recombination), there will be very little difference among maps created with different mapping functions. In reality, nobody knows what the precise relationship between recombination frequency and crossing-over is, and it probably varies among species, populations, and chromosomal regions, as we saw in the previous lecture (Sherman and Stack 1995). Furthermore, Sherman and Stack s (1995) result suggest that interference is not linearly related to interval length, so neither Kosambi nor Haldane s functions can be correct in all situations. Mapping functions are just a way to simplify the presentation of a linkage map. Multi-point ordering and mapping How does one order large numbers of loci in a common linkage group? As the number of loci increases, the number of possible orders increases geometrically. Worse, the twopoint recombination frequency estimates can not be used as the sole guide to locus ordering, because two-point estimates can lead to conflicting locus orders, particularly for 13

14 sets of genes that are closely linked. Remember that recombination frequencies are estimated with error! Multi-point maximum likelihood linkage mapping is most often used to order large numbers of loci and develop complete linkage maps. This method requires calculating the likelihood of different multiple locus linkage orders and distances, using the same method described above, but augmented by extending the multinomial probability function to include all possible multi-locus genotypes. Clearly, the computations quickly become unwieldy. Furthermore, if one has to attempt different locus orders and compute the maximum likelihood estimates of recombination frequencies and their probabilities for many possible orders, the approach is intractable without computer help. The advantage of multi-point linkage mapping is that it uses the available data most efficiently, which is particularly important when there is missing data at some genes on some individuals. Mapmaker/EXP is the most commonly used linkage mapping program because it can perform multi-point maximum likelihood linkage mapping conveniently. The details of multi-point linkage mapping can be found in Lander and Green (1987) and in the Mapmaker manual, and will be discussed only briefly here. The basic steps involved in making a linkage map are: 1. Assigning loci to linkage groups. The first step is to test each pair of loci for linkage, and if there is evidence for linkage (usually a fairly stringent statistical threshold is used), then the pair of loci is considered to be on a common linkage group. Establishment of linkage groups simplifies the following steps because different locus orders and map distances can be estimated and compared for one linkage group at a time, reducing the number of loci involved in the calculations for any one linkage group. 2. Choosing the order of loci. Various methods have been developed to choose the most likely order of loci in a linkage group (Liu 1998). Mapmaker/EXP chooses orders (when there are many loci) by first selecting a subset of five loci that are well spaced along the linkage group, and then by adding additional loci one at a time to this initial order. 3. Each time a locus is added to the group, it is placed in each possible interval and the likelihood for its position within each interval is computed and the most likely position is selected. Once the locus is tried in an interval, the locus order is taken to be known and the multipoint likelihood is computed for some starting values of recombination frequencies. An iterative procedure is used to find the most likely recombination frequencies given the chosen order and this yields the likelihood of the maximum likelihood estimate of all of the recombination frequencies on the linkage group simultaneously. Actually computing the likelihood for a given locus order and set of recombination frequencies is quite complicated see Liu (1998) and Lander and Green (1987) for details. 4. The best likelihoods of each order of markers are then compared and the most likely order is chosen. 5. Another locus is added to the linkage group and the program returns to step 3. 14

15 As an example, if ten loci (A J) are assigned to a linkage group, the program starts by choosing an informative set of five loci and puts them in their most likely order: A B C D E Next, locus F is placed in each possible interval on the map: F A B C D E A F B C D E A B F C D E A B C F D E A B C D F E A B C D E F For each possibility, the maximum likelihood positions of the loci are computed, given the locus order. Thus, for order FABCDE, we estimate the maximum likelihood values for recombination frequencies r FA, r AB, r BC, r CD, and r DE. The likelihood of the maximum likelihood estimates of this multipoint map is remembered. Then, for order AFBCDE we estimate the maximum likelihood values for recombination frequencies r AF, r FB, r BC, r CD, and r DE. The likelihood of the maximum likelihood estimates of this multipoint map is remembered, and so on for each order, and finally the likelihoods for the best maps for each of the six locus orders are compared and the most likely one is chosen. Let s say that order ABCFDE is chosen as most likely. The program then adds the next locus (say, G) to the map, and the procedure repeats, this time comparing likelihoods of maps for orders GABCFDE, AGBCFDE, ABGCFDE, etc. If the best order differs from the next best order by a LOD score of 2 or less, usually, the locus is assigned to the most likely interval without specifying its position (to minimize the chance of establishing an incorrect order during this process). This happens regularly with very tightly linked markers. The critical point is that when one attempts to map numerous loci simultaneously, it is simply not possible to compute likelihoods for all possible locus orders, or even a fraction of all possible locus orders, so users must approach the problem by first eliminating the least likely possible orders quickly as a first step (this is accomplished by establishing linkage groups and when the program chooses the initial set of five informative loci with which to start building the map of a linkage group). It is very possible that the same person starting with the same data set on Mapmaker will end up with at least slightly different linkage maps at the end. It is best to re-analyze the data set several times and choose the most likely order that occurs most frequently. Furthermore, if the genetic data were collected in a different sample of the same 15

16 population, the linkage map developed would probably differ, because of the sampling variance inherent in the recombination frequency estimation procedure. In any case, it is possible to assign loci to linkage groups, and using some cytogenetics tools we will discuss in the next lecture, to assign the loci to actual chromosomes. References Allard RW (1956) Formulas and tables to facilitate the calculation of recombination values in heredity. Hilgardia 24: Hedges BR, Sellner JM, Devine TE, Palmer RG (1990) Assigning isocitrate dehydrogenase to linkage group 11 in soyean. Crop Sci 30: Lander ES, Botstein D (1986) Mapping complex genetic traits in humans: New methods using a complete RFLP linkage map. Cold Spring Harbor Symposia on Quantitative Biology. Vol. LL. Cold Spring Harbor Laboratory, pp Lander ES, Green P (1987) Construction of multilocus genetic linkage maps in humans. Proceedings of the National Academy of Science USA 84: Liu BH (1998) Statistical genomics: Linkage, mapping, and QTL analysis. CRC Press, Boca Raton Mather K (1951) The measurement of linkage in heredity. 2nd Ed. John Wiley and Sons, New York Ott J (1999) Analysis of human genetic linkage. 3rd Ed. Johns Hopkins University Press, Baltimore Sherman JD, Stack SM (1995) Two-dimensional spreads of synaptonemal complexes from solanaceous plants. VI. High-resolution recombination nodule map for tomato (Lycopersicon esculentum). Genetics 141: