## Info

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:

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 more double-crossover progeny appear than expected, which happens when a crossover increases the probability of another crossover occurring nearby. In this case, the coefficient of coincidence is greater than 1 and the interference will be negative.

Concepts

The coefficient of coincidence equals the number of double crossovers observed, divided by the number of double crossovers expected on the basis of the single-crossover frequencies. The interference equals 1 — the coefficient of coincidence; it indicates the degree to which one crossover interferes with additional crossovers.

1. Write out the phenotypes and numbers of progeny produced in the three-point cross. The progeny phenotypes will be easier to interpret if you use allelic symbols for the traits (such as st+ e+ ss).

2. Write out the genotypes of the original parents used to produce the triply heterozygous individual in the test-cross and, if known, the arrangement of the alleles on their chromosomes (coupling or repulsion).

3. Determine which phenotypic classes among the progeny are the nonrecombinants and which are the double crossovers. The nonrecombinants will be the two most-common phenotypes; the double crossovers will be the two least-common phenotypes.

4. Determine which locus lies in the middle. Compare the alleles present in the double crossovers with those present in the nonrecombinants; each class of double crossovers should be like one of the nonrecombinants for two loci and should differ for one locus. The locus that differs is the middle one.

5. Rewrite the phenotypes with genes in correct order.

6. Determine where crossovers must have taken place to give rise to the progeny phenotypes by comparing each phenotype with the phenotype of the nonrecombinant progeny.

7. Determine the recombination frequencies. Add the numbers of the progeny that possess a chromosome with a crossover between a pair of loci. Add the double crossovers to this number. Divide this sum by the total number of progeny from the cross, and multiply by 100%; the result is the recombination frequency between the loci, which is the same as the map distance.

8. Draw a map of the three loci, indicating which locus lies in the middle, and label the distances between them.

9. Determine the coefficient of coincidence and the interference. The coefficient of coincidence is the number of observed double-crossover progeny divided by the number of expected double-crossover progeny. The expected number can be obtained by multiplying the product of the two single-recombination probabilities by the total number of progeny in the cross.

Connecting Concepts 9

### Stepping Through the Three-Point Cross

We have now examined the three-point cross in considerable detail, seeing how the information derived from the cross can be used to map a series of three linked genes. Let's briefly review the steps required to map genes from a three-point cross.

### Worked Problem

In D. melanogaster, cherub wings (ch), black body (b), and cinnabar eyes (cn) result from recessive alleles that are all located on chromosome 2. A homozygous wild-type fly was mated with a cherub, black, and cinnabar fly, and the resulting F1 females were test-crossed with cherub, black, and cinnabar males. The following progeny were produced from the testcross:

ch |
b+ |
cn |

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