Three-Point Linkage with Drosophila

by Carolina Staff

Carolina LabSheets™


In this lab students investigate the effects of gene linkage and crossing-over on the inheritance of three traits of Drosophila. They use the data they collect to construct a linkage map of the gene loci.

172051 Three-Point Linkage: The vial of testcross fruit flies is from a cross between females heterozygous for three chromosome 2 mutants (b+ vg+ bw+/b vg bw) and males homozygous for the mutants (b vg bw/b vg bw). The expected 1:1 phenotype ratio of the testcross flies is modified by crossover events.

The gene loci and phenotypes used in the 172051 culture are as follows:

Gene SymbolMap LocationMutant PhenotypeWild Phenotype
b48.5Black bodyGray body
gg67Vestigial wingsNormal wings
bw104.5Brown eyesRed eyes

Linkage distance from b to vg = 67.0 – 48.5 = 18.5

Linkage distance from vg to bw = 104.5 – 67.1 = 37.5

Linkage distance from b to bw = 18.5 + 37.5 = 56

The phenotypes are easy to recognize by students who have little experience with Drosophila. The loci are far enough apart to give good crossover results in the small sample size offered by one or two cultures. As a drawback, the relatively large distances between the loci means that the linkage calculations will likely vary from accepted values. When the map distance between two loci reaches or exceeds 50 (as is the case for b and bw), the crossover rate cannot be distinguished from the segregation of alleles on different chromosomes, and the loci are unlinked. However, students will be able to generate linkage maps and demonstrate that vg is located between the b locus and the bw locus and that the crossover distance between b and vg is less than the crossover distance between vg and bw.

The activity assumes that students have previously worked with Drosophila and know how to recognize phenotypes, anesthetize flies, and perform other common tasks. If this is not the case, we recommend that students perform monohybrid and/or dihybrid crosses with Drosophila before beginning this more advanced lab activity. The Carolina Drosophila Manual includes a great deal of useful background information.

Since Drosophila crosses must be set up before shipment, please give 2-week notice for delivery of the cultures; otherwise, the crosses may not be available when you need them.

Needed Materials


Ensure that students understand and adhere to safe laboratory practices when performing any activity in the classroom or lab. Demonstrate the protocol for correctly using the instruments and materials necessary to complete the activities, and emphasize the importance of proper usage. Use personal protective equipment such as safety glasses or goggles, gloves, and aprons when appropriate. Model proper laboratory safety practices for your students and require them to adhere to all laboratory safety rules.


Have students work individually or in pairs.

When your cultures arrive, open the package immediately and inspect the cultures to verify that they have arrived in good condition. Examine the label on each vial. The label is divided into four parts. The upper left-hand quadrant gives information on the female flies used for the cross, and the upper right gives information on the male flies. The lower left-hand quadrant is stamped with the date that the parent flies were placed in the vial. F1 flies will likely begin emerging 12–14 days after the date on the label. Each culture will produce approximately 100 flies over a 10-day period. Cultures should not be used beyond 12 days after receipt because offspring of the testcross flies may begin to emerge.

Either prepare the fly morgues with alcohol (one morgue is included in each FlyNap® Kit) or provide materials and instructions so the students can do this.

Each student group will need a three-point linkage culture, an index card, and a stereomicroscope. Provide a FlyNap® kit for every 6–8 groups.

Optional: Since this activity is intended to introduce three-point linkage, it does not include discussion of interference; however, this can be added. A crossover on one section of a chromosome inhibits or interferes with crossing-over in an adjacent area. This means that there are fewer double crossovers than would be expected.

Using the sample data from the table that follows, interference is calculated in this way:

  1. Determine the expected frequency of crossover between b and bw. This is the product of the crossover frequency between b and vgmultiplied by the crossover frequency between vg and bw.
         Expected Frequency = 0.199 x 0.385 = 0.0766
  2. Use the expected crossover frequency to calculate the expected number of double crossovers in the total flies counted, in this case 421. Expected Number of Double Crossovers = 421 x 0.0766 = 32
  3. Since the observed number of double crossovers is 23, which is less than the expected number, there has been interference. Interference is calculated as 1 minus the coefficient of coincidence. The coefficient of coincidence is the observed number of double crossovers/expected number of double crossovers. Thus
          c.o.c = 23/32 = 0.72
          Interference = 1 – 0.72 = 0.28
          This means that 28% of the expected double crossovers have been prevented by interference.

As an aid to less experienced students, the Phenotype Data Table of the student sheet gives the phenotype combinations in their classes and gives the alleles in their proper sequence. You may wish to allow advanced students to construct their own data table. They simply need to know that the parental combinations (non-crossovers) of phenotypes are the most common, the single crossovers are next, and the double combinations are the fewest.

If time permits, students can do the three-point cross beginning with the parental stocks, 172345 black vestigial brown, and 172100 wild type. Since crossing-over does not occur in male Drosophila, it is important that the crosses be made as described in the Student LabSheet.

Answer Key to Questions Asked on the Student LabSheet

The following is sample data. Your class data will differ but should show the general pattern of non-crossovers as most common, single crossovers as next, and double crossovers as the least common.

ClassBodyWingEyePhenotypeCountType cross over
1b+vg+bw+Gray body, normal wings, red eyes103none
2bvgbwBlack body, vestigial, brown eyes95none
3bvg+bw+Black body, normal wings, red eyes29b × vg
4b+vgbwGray body, vestigial wings, brown eyes32b × vg
5b+vg+bwGray body, normal wings, brown eyes71vg × bw
6bvgbw+Black body, vestigial wings, red eyes68vg × bw
7b+vgbw+Gray body, vestigial wings, red eyes10double
8bvg+bwBlack body, normal wings, brown eyes13double
    Total Counted:421

Use the data from the previous table to fill in the following:

Type CrossoverAdd ClassesTotal of ClassesTotal of Classes/
Total Counted

% Crossover
(map distance)

b x vg3, 4, 7, 829 + 32 + 10 + 13 = 8484/421 = 0.19919.9
vg x bw5, 6, 7, 871 + 68 + 10 + 13 = 162162/421 = 0.38538.5

To determine which locus is in the middle, look at the phenotypes of the double crossovers. Which phenotype has switched from the parental association? This identifies the gene locus that is in the middle. Fill in the following, giving the gene locus and the map distances between the loci.

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