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Griffiths AJF, Miller JH, Suzuki DT, et al. An Introduction to Genetic Analysis. 7th edition. New York: W. H. Freeman; 2000.

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An Introduction to Genetic Analysis. 7th edition.

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Chi-square test for linkage

When RF values are close to 50 percent, the χ2 test can be used as a critical test for linkage. 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):

Image ch5e27.jpg

From these data the recombinant frequency is 225/500 = 45 percent. This seems like a case of linkage because the RF is less than 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 χ2 test.

The problem then is to find the expectations, E, for each class. For linkage testing, it might be supposed that these expectations are simply given by the 1:1:1:1 ratio of the four backcross classes that we expect when there is independent assortment. For the 500 progeny in our example, then, we would expect 500/4 = 125 in each class. But the 1:1:1:1 ratio is not the appropriate test for linkage, because to get such a ratio two things must be true. There must be independent assortment between the A and B locus, but, in addition, there must be an equal chance for the different genotypes formed at fertilization to reach the age at which they are scored for the test, which usually means that the four genotypes must have equal chance of survival from egg to adult. However, it is often the case that mutations used in linkage tests have some deleterious effect in the homozygous condition; so a/a or b/b genotypes have a lower probability of survival than do the wild-type heterozygotes A/a and B/b. We might then be led to reject the hypothesis of independent assortment even when it is correct because the differential survivorship of the genotypes causes deviations from the 1:1:1:1 expected ratio. What we need is a method of calculating the expectations, E, that is insensitive to differences in survivorship.

No matter what the frequencies of a/a or b/b genotypes are among the backcross adults, if there is independent assortment, we expect the frequency of the a · b genotypes to be the product of the frequencies of the a and the b alleles. In our example, the total proportion of a alleles is (135 + 115)/500, which is indeed the expected 50 percent, but the frequency of alleles is only (135 + b110)/500 = 49 percent. Thus we expect the proportion of a · b genotypes to be 0.50 × 0.49 = 0.245 and the number of a · b genotypes in a sample of 500 to be 500 × 0.245 = 122.5. The same kind of calculation can be performed for each of the other genotypes to give all the expected numbers. The comparison is usually done in a contingency table, as shown in Table 5-3. The expected number for an entry in the contingency table is the product of the proportion observed in its row, the proportion observed in its column, and the total sample size. But the row and column proportions are the row and column totals divided by the grand total, so the actual calculation of the expected number in each entry is simply to multiply the appropriate row total by the appropriate column total and then divide by the grand total. The value of χ2 is then calculated as follows:

Image ch5e28.jpg

Table 5-3. Contingency Table Comparing Observed and Expected Results of a Testcross to Examine Linkage Between Loci A/a and B/b.

Table 5-3

Contingency Table Comparing Observed and Expected Results of a Testcross to Examine Linkage Between Loci A/a and B/b.

The obtained value of χ2 is converted into a probability by using a χ2 table (see Table 4-1, page 126). To do so, we need to decide on the degrees of freedom (df) in the test, which, as the name suggests, is the number of independent deviations of observed from expected that have been calculated. We notice that, because of the way that the expectations were calculated in the contingency table from the row and column totals, all deviations are identical in absolute magnitude, 12.5, and that they alternate in sign and so they cancel out when summed in any row or column. Thus, there is really only one independent deviation, so there is only one degree of freedom. Therefore, looking along the one degree of freedom line in Table 4-1, we see that the probability of obtaining a deviation from expectations this large (or larger) by chance alone is 0.025 (2.5 percent). Because 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.

By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.

Copyright © 2000, W. H. Freeman and Company.
Bookshelf ID: NBK22084


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