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Modern Genetic Analysis Chapter 5. Recombination of Genes
Linkage Maps

Geneticists are frequently interested in knowing where a gene is located within the genome. There are many reasons for this interest. For example, in many well-studied genetic systems, including humans, an accurate genetic location makes it feasible to proceed with a molecular biological analysis of the gene and to eventually learn what protein the gene encodes. Also, knowledge of gene location is useful in constructing special strains for specific experimental purposes. Maps of the gene loci on chromosomes can be constructed by recombination analysis; mapping techniques will be discussed below.

The recombinant frequency (RF) is different in crosses involving different linked heterozygous genes. However, for any one case the RF value is characteristic and experimentally reproducible. The RF for two loci is proportional to the physical distance between them on the chromosome. This makes sense because if two loci are far apart on a chromosome, we would expect that the proportion of meiocytes in which a crossover occurs between the loci would be greater than in cases in which two linked loci are very close together. (If we think of chiasmata as pellets fired from a shotgun, it is intuitive that there will be a greater chance of hitting a large targeta large chromosomal distancethan a small one.) Thus, by determining the frequency of recombinants, we should be able to obtain a measure of the map distance between the genes. In fact, one genetic map unit (m.u.) is defined as that distance between genes for which one product of meiosis out of 100 is recombinant. Put another way, a recombinant frequency of 1 percent is defined as 1 m.u., and an RF of 6 percent is defined as 6 m.u., and so on. A map unit is sometimes referred to as a centimorgan (cM) in honor of Thomas Hunt Morgan, a pioneer in the field of chromosome mapping.

A direct consequence of the way map distance is measured is that if 5 map units (5 m.u.) separate the genes A and B whereas 3 m.u. separate the genes A and C, then B and C should be either 8 or 2 m.u. apart (Figure 5-10). Generally this is found to be the case; geneticists say that map distances are roughly "additive." In other words, the RF analysis is compatible with the fact that genes are arranged in linear order on the chromosome (see Genetics in Process 5-2).

Note that the simple map of the type developed in Figure 5-10 does not place the gene loci at specific places on the chromosome; all the RF analysis does is to determine the positions of genes relative to one another. In fact such a map is only an abstract construct. The small cluster of three genes could in theory be anywhere on the actual chromosome. However, as more and more recombination analyses are done with many more genes, the entire chromosome becomes "fleshed out." Only after the genes close to each chromosome end are placed in the map, and gene positions are determined by chromosome rearrangement analysis (Chapter 8), can absolute chromosome position be assigned.

As mentioned earlier, genetic analysis works in two directions. In the case of linkage analysis, we have seen that by using RF measurements we can draw a genetic map; however, working in the other direction, if we have the map we can predict frequencies of progeny in different genotypic classes. For example, if we know that the Drosophila autosomal genes for eye color (P = red, p = purple) and wing size (V = long, v = vestigial) are linked on the same chromosome 11 m.u. apart, and if we testcross a trans dihybrid P v/p V to p v/p v, we can predict that there will be 11 percent recombinants. Furthermore, there should be 5.5 percent each of P V/p v and p v/p v because the two recombinant types are reciprocal products of the same type of crossover event. The remaining 89 percent will be nonrecombinant, 44.5 percent each of P v/p v and p V/p v.

MESSAGE
Recombination between linked genes can be used to map their distance apart on the chromosome. The map unit (1 m.u.) is defined as a recombinant frequency of 1 percent.

In a dihybrid for linked loci the recombinant frequency never exceeds 50 percent no matter how far apart the loci are. This is nonintuitive because one imagines that as chromosomal distance increases then more crossovers occur and more recombinants will be produced. However, it can be shown simply that in meiocytes with any number of crossovers other than zero, the average proportion of recombinants is 50 percent. Hence the determining factor of recombinant frequency is the relative proportion of the zero crossover class. For long chromosomal regions there will be no zero class, and so the RF will be 50 percent but can never exceed this.

MESSAGE
In a dihybrid of linked genes the RF will be between 0 percent and 50 percent.

Mapping Using a Trihybrid Testcross

So far we have considered the analysis of recombination in dihybrids. We have seen that by compiling map distances obtained in many dihybrid testcrosses it is possible to build up entire linkage maps that correspond to maps of loci on chromosomes. However, in some situations multiple heterozygotes (trihybrids, tetrahybrids, etc.) are available to the researcher, and in these cases it is possible to perform several map-distance calculations in one testcross to a multiple recessive. We will follow an example using a testcross of a trihybrid, sometimes called a three-point testcross.

The example analyzes alleles at three loci in Drosophila:

1. v (vermilion eyes) versus v+ (red eyes)
2. cv (crossveinless, or absence of a crossvein on the wing) versus cv+ (presence of a crossvein)
3. ct (cut, or snipped wing edges) versus ct+ (smooth wing edges)

Initially we do not know if these are linked, and the genes must be written in no particular order. The parental stocks available are homozygous doubly recessive flies of genotype v+/v+.cv/cv.ct/ct and homozygous singly recessive flies of genotype v/v.cv+/ cv+.ct+/ct+. From this cross, trihybrid progeny of genotype v/v+.cv/cv+.ct/ct+ are obtained, and females of this genotype are testcrossed to triple recessives of genotype v/v.cv/cv.ct/ct. Because there are three allele pairs, there are 2 × 2 × 2 = 8 gametic genotypes possible from the female. As with simpler testcrosses, these gametic genotypes determine the eight progeny types from this testcross. For simplicity the progeny are written in terms of these gametic genotypes, and the v cv ct contribution from the tester simply omitted. The data are shown below as numbers of progeny out of a total sample of 1448 flies from an actual cross:

The standard recombination approach is called for, but we must be careful in our classification of parental and recombinant types. Note that the parental input genotypes for the triple heterozygotes are v+.cv.ct and v.cv+.ct+; we must take this into consideration when we decide what constitutes a recombinant. The data are analyzed two loci at a time.

Starting with the v and cv loci, we know that the recombinants must be of genotype v.cv and v+.cv+ and that there are 45 + 40 + 89 + 94 = 268 of these. Out of a total of 1448 flies, this gives an RF of 18.5 percent.

For the v and ct loci, the recombinants are v.ct and v+.ct+. There are 89 + 94 + 3 + 5 = 191 of these among 1448 flies, so that RF = 13.2 percent.

For ct and cv, the recombinants are cv.ct+ and cv+.ct. There are 45 + 40 + 3 + 5 = 93 of these among the 1448, so that RF = 6.4 percent.

Obviously, all the loci are linked on the same chromosome, because the RF values are all considerably less than 50 percent. Because the v and cv loci show the largest RF value, they must be farthest apart; therefore, the ct locus must be between them. A map can be drawn as follows:

Note several important points here. The first is that we have deduced a different gene order from that used when listing the parental and progeny genotypes. As the point of the exercise was to determine the linkage relationships of these genes, the original listing was of necessity arbitrary; the order simply was not known before the data were analyzed.

Second, we have definitely established that ct is between v and cv and have determined what the distances are between ct and these other loci in map units. But we have arbitrarily placed v to the left and cv to the right; the map could equally well be inverted.

A third point to note is that the two smaller map distances, 13.2 m.u. and 6.4 m.u., add up to 19.6 m.u., which is greater than 18.5 m.u., the distance calculated for v and cv. Why is this so? The answer to this question lies in the way in which we have analyzed the two rarest classes in our classification of recombination for the v and cv loci. Now that we have the map, we can see that these two rare classes are in fact double recombinants, arising from two crossovers (Figure 5-11). However, we did not count the v ct cv+ and v+ ct+ cv genotypes when we calculated the RF value for v and cv; after all, with regard to v and cv they are parental combinations (v cv+ and v+ cv). In the light of our map, however, we see that this led to an underestimate of the distance between the v and cv loci. Not only should we have counted the two rarest classes, we should have counted each of them twice because each represents a double recombinant class. Hence, we can correct the value by adding the numbers 45 + 40 + 89 + 94 + 3 + 3 + 5 + 5 = 284. Out of the total of 1448, this is exactly 19.6 percent, which is identical with the sum of the two component values.

Now that we have had some experience with the data of this cross, we can look back at the progeny listing and see that it is usually possible to deduce gene order by inspection, without a recombinant frequency analysis. Only three gene orders are possible, each with a different gene in the middle position. It is generally true that the double recombinant classes are the smallest ones. Only one order should be compatible with the smallest classes having been formed by double crossovers, as shown in Figure 5-12 on the following page. Only one order gives double recombinants of genotype v ct cv+ and v+ ct+ cv. Notice in passing that the ability to detect double crossovers depends on having a heterozygous gene between the two crossovers; if the mothers of these progeny had not been heterozygous (ct/ct+), we could never have identified the double recombinant classes.

Finally, note again that linkage maps merely map the loci in relation to one another, using standard map units. We do not know where the loci are on a chromosomeor even which specific chromosome they are on. The linkage map is essentially an abstract construct that can be correlated with a specific chromosome and with specific chromosome regions only by applying special kinds of cytogenetic analyses, as we shall see in Chapter 8.

MESSAGE
Testcrosses of multiple heterozygotes enable linkage between several loci to be evaluated in one cross.

Interference

The detection of double recombinant classes shows that double crossovers must occur. Knowing this, one of the questions that arises is whether the crossovers in adjacent chromosome regions are independent or whether they interact with each other in some way. For example, we might ask if a crossover in one region affects the likelihood of there being a crossover in an adjacent region. It turns out that often the answer is "yes"; this interaction is called interference.

The analysis can be approached in the following way. If the crossovers in the two regions are independent, then according to the product rule, the frequency of double recombinants would equal the product of the recombinant frequencies in the adjacent regions. In the vctcv recombination data, the vct RF value is 0.132 and the ctcv value is 0.064, so if there is independence, double recombinants might be expected at the frequency 0.132 × 0.064 = 0.0084 (0.84 percent). In the sample of 1448 flies, a total of 0.0084 × 1448 = 12 double recombinants is expected. But the data show that only 8 were actually observed. If this deficiency of double recombinants were consistently observed, it would show us that the two regions are not independent and suggest that the distribution of crossovers favors singles at the expense of doubles. In other words, there is some kind of interference: a crossover reduces the probability of a crossover in an adjacent region.

Interference is quantified by first calculating a term called the coefficient of coincidence (c.o.c.), which is the ratio of observed to expected double recombinants, and then subtracting this value from 1. Hence,

In our example I = 1 (8/12) = 4/12 = 1/3, or 33 percent.

In some regions, there are never any observed double recombinants. In these cases, c.o.c. = 0, so I = 1 and interference is complete. Most of the time, the interference values that are encountered in mapping chromosome loci are between 0 and 1, but in certain special situations, observed double crossovers exceed expected numbers, giving negative interference values.

Interference is commonly encountered in linkage analysis, but it is worth noting that no molecular mechanism for interference is known.

Examples of Linkage Maps

Linkage maps are the prelude to the serious experimental genetic study of any organism. Many organisms have had their chromosomes intensively mapped in this way. The resultant maps represent a vast amount of genetic analysis generally achieved by collaborative efforts of research groups throughout the world.

The stage of analysis that we have reached in our discussion is well illustrated by linkage maps of the screwworm (Cochliomyia hominivorax). The larval stage of this insectthe "worm"is parasitic on mammalian wounds and is a costly pest of livestock in some parts of the world. A genetic system of population control has been proposed, of a type that has been successful against other insects. In order to accomplish this goal, an understanding of the basic genetics of the insect is needed, and one important part of this is to obtain a chromosome map. This animal has six chromosome pairs, and chromosome mapping has begun.

The job of general mapping starts by finding and analyzing as many variant phenotypes as possible. Although the genes controlling most of these phenotypes will not be directly involved in the fly-control project, they are needed as markers to build up the chromosome map. The adult stage of this insect is a fly, and geneticists have found numerous phenotypic variants among screwworm flies. For example, they have found flies of six different eye colors, all different from the brown-eyed wild-type flies, as Figure 5-13a shows. They also have found five variant phenotypes for other characters. Eleven mutant alleles were shown to determine the 11 variant phenotypes, each at a different autosomal locus. Pure lines of each phenotype were intercrossed to generate dihybrid F1s, and then these were testcrossed or selfed according to the methods discussed earlier. The analysis of these mutants has revealed so far the set of four linkage groups shown in Figure 5-13b. Notice that the ye and cw loci are shown tentatively linked, although the recombinant frequency is not significantly different from 50 percent.

Figures 5-14 and 5-15 show two examples of linkage maps: one from Drosophila and one from the tomato. The Drosophila genome is one of the most intensively mapped of all model genetic organisms. The map in Figure 5-15 on page 148 shows only a fraction of the known loci. Tomatoes, also, have been interesting from the perspectives of both basic and applied genetic research, and the tomato genome is one of the best characterized of plants.

The different panels of Figure 5-15 illustrate some of the stages of understanding through which research arrives at a comprehensive map. First, although chromosomes are visible under the microscope, there is initially no way to locate genes on them. However, the chromosomes can be individually identified and numbered, based on their inherent landmarks such as staining patterns and centromere positions, as has been done in parts a and b. Next, analysis of recombinant frequencies generates a set of linkage groups that must correspond to chromosomes, but specific correlations cannot necessarily be made with the numbered chromosomes. At some stage cytogenetic analysis must be performed to enable the linkage groups to be assigned to specific chromosomes (Chapter 8). Part c of Figure 5-15 shows a tomato map made in 1952, with the linkages of the genes known at that time. Each locus is represented by the two alleles used in the original mapping experiments. As more and more loci became known, they were mapped in relation to the loci shown in the figure, so today the map contains hundreds of loci. Some of the chromosome numbers shown in part c are tentative and do not correspond to the modern chromosome numbering system used for tomatoes.

Molecular technology has extended and supplemented the above techniques. This technology has provided molecular "probes" that can detect sites of neutral variation in the DNAvariation that results in no detectable phenotypic variation. There is a wide spectrum of such neutral DNA variation ranging from single base-pair differences to different numbers of repeats of specific DNA segments. Such variation is sometimes called molecular polymorphism. If a pair of homologous chromosomes differs at a site of neutral variationthat is, one chromosome shows variant a1 at the neutral site and its partner shows variant a2, then this heterozygous locus becomes a "molecular marker" that can be used in mapping. Because they are parts of chromosomes, molecular markers segregate and recombine like genes. They form valuable chromosomal landmarks for fleshing out a chromosome map, and can also be used as starting points for gene isolation (Chapter 11).

The c 2 Test

In research generally, it is often necessary to compare experimentally observed numbers of items in several different categories with numbers that are predicted on the basis of some hypothesis. For example, you might want to determine whether or not the sex ratio in some specific population of insects is 1:1 as expected. If there is a close match, then the hypothesis is upheld, while if there is a poor match, the hypothesis is rejected. As part of this process a judgment has to be made about whether the observed numbers are close enough to those expected. Very close matches and blatant mismatches generally present no problem, but inevitably there are gray areas in which the match is not obvious. Genetic analysis often involves the interpretation of numbers in various phenotypic classes. In such cases a statistical procedure called the c 2 (chi square) test is used to help in making the decision to hold on to or reject the hypothesis. For example, the test can be used to analyze Mendelian ratios, answering such questions as "Are these experimental results compatible with the 3:1 ratio expected on the basis of a hypothesis of monohybrid selfing?" Another widespread application is in testing for linkage, which we will use to illustrate the c 2 test.

The genetic criterion for linkage is the absence of independent assortment. It is not possible to test for linkage directly because a priori we do not have a precise linkage distance to use in the hypothesis, so we are forced to test the hypothesis of the absence of linkage. If the observed results cause rejection of the hypothesis of no linkage, then we can infer linkage. This type of null hypothesis is generally useful in c 2 analysis by providing a precise experimental prediction that can be tested.

The c 2 test is simply a way of quantifying the various deviations expected by chance if a hypothesis is true. Take a simple hypothesis predicting a 1:1 ratio, for example. Even if the hypothesis is true, we would not always expect an exact 1:1 ratio. We can model this with a barrel full of equal numbers of red and blue marbles. If we blindly remove samples of 100 marbles, on the basis of chance we would expect samples to show small deviations such as 52 red:48 blue quite commonly, and larger deviations such as 60 red:40 blue less commonly. We have seen examples of such deviations in these chapters. The c 2 test allows us to calculate the probability of such chance deviations from expectations if the hypothesis is true. But if all levels of deviation are expected with different probabilities even if the hypothesis is true, how can we ever reject a hypothesis? It has become a general scientific convention that a probability value of less than 5 percent is to be taken as the criterion for rejecting the null hypothesis. The hypothesis might still be true, but we have to make a decision somewhere and the 5 percent level is the conventional decision line. The logic is that although results this far from expectations are anticipated 5 percent of the time even when the null hypothesis is true, we will mistakenly reject the null hypothesis in only 5 percent of cases and we are willing to take this chance of error.

Let's test a specific set of data for linkage using c 2 analysis. Assume that we have crossed pure-breeding parents of genotypes A/A.B/B and a/a.b/b and obtained a dihybrid A/a.B/b, which we have testcrossed to a/a.b/b. A total of 500 progeny are classified as follows (written as gametes from the dihybrid):

From these data the recombinant frequency is 225/500 = 45 percent. On the face of it this seems like a case of linkage because the RF is less than the 50 percent expected from independent assortment. However, it is possible that the two recombinant classes are in the minority merely on the basis of chance; therefore, we need to perform a c 2 test. The statistic c 2 is always calculated from actual numbers, not from percentages, proportions, or fractions. Sample size is therefore very important in the c 2 test, as it is in most considerations of chance phenomena. Samples to be tested generally contain several classes. The letter O is used to represent the observed number in a class, and E the expected number for the same class based on the predictions of the hypothesis. The general formula for calculating c 2 is as follows:

The problem, then, is to calculate the expectations E for each class. The hypothesis that must be tested in this case is one of independent assortment of the two loci, because the hypothesis of linkage cannot provide a precise set of expected proportions. We might assert that if the allele pairs of the dihybrid are assorting randomly (that is, there is no linkage), there should be a 1:1:1:1 ratio of gametic types. Therefore it seems reasonable to use 1/4 as the expected proportion of each gametic class. However, this expectation assumes equal viability of all genotypes, and this is not necessarily the case. A better test is whether or not the two allele ratios are combined randomly in the meiotic products, because this makes no assumptions about equal viability. In this approach, the phenotypic classes are laid out as a contingency table, as shown in Table 5-1. The grid of the table is simply a graphic representation of combining the allele ratios of the two loci. The c 2 analysis tests whether or not the numbers of individuals in the four phenotypic classes of progeny are simply the two allelic ratios combined at random, as expected under the hypothesis of independent assortment. The expected values are calculated simply by subdividing the column totals proportionately to the row totals. The value of c 2 is calculated as follows:

The obtained value of c 2 (5.02) is converted into a probability value using a c 2 table (Table 5-2). To do this we need to compute the number of degrees of freedom (df) used in the c 2 calculation, which is the number of independent deviations, O E. But all the deviations in the contingency table turn out to be equal in magnitude (12.5). This is because the total expected numbers in each row or column have to be equal to the total observed number in each row and column, so there is only one degree of freedom. In general, in contingency tables the degrees of freedom are equal to the number of classes represented in the rows minus one, times the number of classes represented in the columns minus one. In the present example

Therefore, using Table 5-2 we look along the one degree of freedom line until we find our c 2 value of 5.02. We see that the probability of obtaining a deviation from expectations this large (or larger) by chance alone is 0.025 (2.5 percent). Since this probability is less than 5 percent, the hypothesis of independent assortment must be rejected. Thus, having rejected the hypothesis of no linkage, we are left with the inference that the loci must be linked.

MESSAGE
The C 2 test is used to test observed numbers in a class against the expectations derived from a predictive hypothesis. The test generates the probability of obtaining by chance a specific deviation at least as great as the one observed, assuming that the hypothesis is correct.

© 1999 by W. H. Freeman and Company. All rights reserved.