To examine gene mapping with a three-point testcross, we will consider three recessive mutations in the fruit fly Drosophila melanogaster. In this species, scarlet eyes (st) are recessive to wild-type red eyes (st⁺), ebony body color (e) is recessive to wild-type gray body color (e⁺), and spineless (ss)—that is, the presence of small bristles—is recessive to wild-type normal bristles (ss⁺). The loci encoding these three characteristics are linked and located on chromosome 3.

We will refer to these three loci as st, e, and ss, but keep in mind that either the recessive alleles (st, e, and ss) or the dominant alleles (st⁺, e⁺, and ss⁺) may be present at each locus. So, when we say that there are 10 m.u. between st and ss, we mean that there are 10 m.u. between the loci at which mutations st and ss occur; we could just as easily say that there are 10 m.u. between st⁺ and ss⁺.

To map these genes, we need to determine their order on the chromosome and the genetic distances between them. First, we must set up a three-point testcross: a cross between a fly heterozygous at all three loci and a fly homozygous for recessive alleles at all three loci. To produce flies heterozygous for all three loci, we might cross a stock of flies that are homozygous for wild-type alleles at all three loci with flies that are homozygous for recessive alleles at all three loci:

The order of the genes has been arbitrarily assigned because, at this point, we do not know which one is the middle gene. Additionally, the alleles in these heterozygotes are in coupling configuration (because all the wild-type dominant alleles were inherited from one parent and all the recessive mutations from the other parent), although the testcross can also be done with alleles in repulsion.

In the three-point testcross, we cross the F₁ heterozygotes with flies that are homozygous for all three recessive mutations. In many organisms, it makes no difference whether the heterozygous parent in the testcross is male or female (provided that the genes are autosomal), but in Drosophila, no crossing over takes place in males. Because crossing over in the heterozygous parent is essential for determining recombination frequencies, the heterozygous flies in our testcross must be female. So we mate female F₁ flies that are heterozygous for all three traits with male flies that are homozygous for all the recessive traits:

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The progeny produced by this cross are listed in Figure 5.12. For each locus, two classes of progeny are produced: progeny that are heterozygous, displaying the dominant trait, and progeny that are homozygous, displaying the recessive trait. With two classes of progeny possible for each of the three loci, there will be 2³ = 8 classes of phenotypes possible in the progeny. In this example, all eight phenotypic classes are present, but in some three-point crosses, one or more of the phenotypes may be missing if the number of progeny is limited. Nevertheless, the absence of a particular class can provide important information about which combination of traits is least frequent and, ultimately, about the order of the genes, as we will see.

To map the genes, we need information about where and how often crossing over has taken place. In the homozygous recessive parent, the two alleles at each locus are the same, and so crossing over will have no effect on the types of gametes produced: with or without crossing over, all gametes from this parent have a chromosome with three recessive alleles (st       e       ss). In contrast, the heterozygous parent has different alleles on its two chromosomes, and so crossing over can be detected. The information that we need for mapping, therefore, comes entirely from the gametes produced by the heterozygous parent. Because chromosomes contributed by the homozygous parent carry only recessive alleles, whatever alleles are present on the chromosome contributed by the heterozygous parent will be expressed in the progeny.

As a shortcut, we often do not write out the complete genotypes of the testcross progeny, listing instead only the alleles expressed in the phenotype, which are the alleles inherited from the heterozygous parent. This convention is used in the discussion that follows.

DETERMINING THE GENE ORDER The first task in mapping the genes is to determine their order on the chromosome. In Figure 5.12, we arbitrarily listed the loci in the order st, e, ss, but we had no way of knowing which of the three loci was between the other two. We can now identify the middle locus by examining the double-crossover progeny.

First, determine which progeny are the nonrecombinants; they will be the two most numerous classes of progeny (even if crossing over takes place in every meiosis, the nonrecombinants will constitute at least 50% of the progeny). Among the progeny of the testcross in Figure 5.12, the most numerous are those with all three dominant traits (st⁺       e⁺       ss⁺) and those with all three recessive traits (st       e       ss).

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Next, identify the double-crossover progeny. These progeny should always have the two least numerous phenotypes because the probability of a double crossover is always less than the probability of a single crossover. The least common progeny among those listed in Figure 5.12 are progeny with spineless bristles (st⁺       e⁺       ss) and progeny with scarlet eyes and ebony body (st       e       ss⁺) , so they are the double-crossover progeny.

Three orders of genes on the chromosome are possible: the eye-color locus could be in the middle (e       st       ss) , the body-color locus could be in the middle (st       e       ss), or the bristle locus could be in the middle (st       ss       e). To determine which gene is in the middle, we can draw the chromosomes of the heterozygous parent with all three possible gene orders and then see if a double crossover produces the combination of genes observed in the double-crossover progeny. The three possible gene orders and the types of progeny produced by their double crossovers are

The only gene order that produces chromosomes with the set of alleles observed in the least numerous progeny—the double crossovers (st⁺       e⁺       ss and st       e       ss⁺ in Figure 5.12)—is the one in which the ss locus for bristles lies in the middle (gene order 3). Therefore, this order (st       ss       e) must be the correct sequence of genes on the chromosome.

With a little practice, we can quickly determine which locus is in the middle without writing out all the gene orders. The phenotypes of the progeny are expressions of the alleles inherited from the heterozygous parent. Recall that when we looked at the results of double crossovers (see Figure 5.11), only the alleles at the middle locus differed from those in the nonrecombinants. If we compare the nonrecombinant progeny with the double-crossover progeny, they should differ only in alleles at the middle locus.

Let’s compare the alleles in the double-crossover progeny st⁺       e⁺       ss with those in the nonrecombinant progeny st⁺       e⁺       ss⁺. We see that both have an allele for red eyes (st⁺) and both have an allele for gray body (e⁺), but the nonrecombinants have an allele for normal bristles (ss⁺), whereas the double crossovers have an allele for spineless bristles (ss). Because the bristle locus is the only one that differs, it must lie in the middle. We would obtain the same results if we compared the other class of double-crossover progeny (st       e       ss⁺) with the other class of nonrecombinant progeny (st       e       ss). Again, the only locus that differs is the one for bristles. Don’t forget that the nonrecombinants and the double crossovers should differ at only one locus; if they differ at two loci, the wrong classes of progeny are being compared. Animation 5.1 illustrates how to determine the order of three linked genes.

DETERMINING THE LOCATIONS OF CROSSOVERS When we know the correct order of the loci on the chromosome, we can rewrite the phenotypes of the testcross progeny in Figure 5.12 with the alleles in the correct order so that we can determine where crossovers have taken place (Figure 5.13).

Among the eight classes of progeny, we have already identified two classes as nonrecombinants (st⁺       ss⁺       e⁺ and st       ss       e) and two classes as double crossovers (st⁺       ss       e⁺ and st       ss⁺       e). The other four classes include progeny that resulted from a chromosome that underwent a single crossover: two underwent single crossovers between st and ss, and two underwent single crossovers between ss and e.

To determine where the crossovers took place in these progeny, we can compare the alleles found in the single-crossover progeny with those found in the nonrecombinants, just as we did for the double crossovers. For example, consider the progeny with chromosome st⁺       ss       e. The first allele (st⁺) came from the nonrecombinant chromosome st⁺       ss⁺       e⁺, and the other two alleles (ss and e) must have come from the other nonrecombinant chromosome st       ss       e through crossing over:

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This same crossover also produces st       ss⁺       e⁺ progeny.

This method can also be used to determine the location of crossing over in the other two types of single-­crossover progeny. Crossing over between ss and e produces st⁺       ss⁺       e and st       ss       e⁺ and chromosomes:

We now know the locations of all the crossovers. Their locations are marked with red slashes in Figure 5.13.

CALCULATING THE RECOMBINATION FREQUENCIES Next, we can determine the frequencies of recombination, which we will use to determine the map distances. We can calculate recombination frequency by adding up all of the recombinant progeny, dividing this number by the total number of progeny from the cross, and multiplying the number obtained by 100%. To determine the map distances accurately, we must include all crossovers (both single and double) that take place between two genes.

Recombinant progeny that possess a chromosome that underwent crossing over between the eye-color locus (st) and the bristle locus (ss) include the single crossovers (st⁺   /   ss       e and st   /   ss⁺       e⁺) and the two double crossovers (st⁺   /   ss   /   e⁺ and st   /   ss⁺   /   e) (see Figure 5.13). There are a total of 755 progeny; so the recombination frequency between ss and st is

The distance between the st and ss loci can be expressed as 14.6 m.u.

The map distance between the bristle locus (ss) and the body locus (e) is determined in the same manner. The recombinant progeny that possess a crossover between ss and e are the single crossovers st⁺       ss⁺   /   e and st       ss   /   e⁺ and the double crossovers st⁺   /   ss   /   e⁺ and st   /   ss⁺   /   e. The recombination frequency is

Thus, the map distance between ss and e is 12.2 m.u.

Finally, calculate the map distance between the outer two loci, st and e. This map distance can be obtained by summing the map distances between st and ss and between ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.). We can now use the map distances to draw a map of the three genes on the chromosome:

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A genetic map of D. melanogaster is illustrated in Figure 5.14.

INTERFERENCE AND THE COEFFICIENT OF COINCIDENCE Map distances give us information not only about the distances that separate genes, but also about the proportions of recombinant and nonrecombinant gametes that will be produced in a cross. For example, knowing that genes st and ss on the third chromosome of D. melanogaster are separated by a distance of 14.6 m.u. tells us that 14.6% of the gametes produced by a fly heterozygous at these two loci will be recombinants. Similarly, 12.2% of the gametes from a fly heterozygous for ss and e will be recombinants.

Theoretically, we should be able to calculate the proportion of double-recombinant gametes by using the multiplication rule of probability (see Chapter 3). Applying this rule, we should find that the proportion (probability) of gametes with double crossovers between st and e is equal to the probability of recombination between st and ss multiplied by the probability of recombination between ss and e, or 0.146 × 0.122 = 0.0178. Multiplying this probability by the total number of progeny gives us the expected number of double-crossover progeny from the cross: 0.0178 × 755 = 13.4. But in our example, only 8 double crossovers—considerably fewer than the 13 expected—were observed in the progeny of the cross (see Figure 5.13).

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This phenomenon is common in eukaryotic organisms. The calculation assumes that each crossover event is independent and that the occurrence of one crossover does not influence the occurrence of another. But crossovers are frequently not independent events: the occurrence of one crossover tends to inhibit additional crossovers in the same region of the chromosome, so double crossovers are less frequent than expected.

The degree to which one crossover interferes with additional crossovers in the same region is termed the interference. To calculate the interference, we first determine the coefficient of coincidence, which is the ratio of observed double crossovers to expected double crossovers:

For the loci that we mapped on the third chromosome of D. melanogaster (see Figure 5.13), we find

which indicates that we are actually observing only 60% of the double crossovers that we expected on the basis of the single-crossover frequencies. The interference is calculated as

interference = 1 − coefficient of coincidence

So the interference for our three-point cross is

interference = 1 − 0.6 = 0.4

This value of interference tells us that 40% of the double-crossover progeny expected will not be observed because of interference. When interference is complete and no double-crossover progeny are observed, the coefficient of coincidence is 0 and the interference is 1.

Sometimes a crossover increases the probability of another crossover taking place nearby and we see more double-crossover progeny than expected. In this case, the coefficient of coincidence is greater than 1 and the interference is negative.

Most eukaryotic organisms exhibit interference, which causes crossovers to be more widely spaced along the chromosome than would be expected on a random basis. Interference was first observed in crosses of Drosophila in the early 1900s, yet despite years of study, the mechanism by which interference occurs is still not well understood. One proposed model of interference suggests that crossovers occur when stress builds up along the chromosome. Under this model, a crossover releases stress for some distance along the chromosome. Because a crossover relieves the stress that causes crossovers, additional crossovers are less likely to occur in the same area. TRY PROBLEM 23