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

NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health.

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

## An Introduction to Genetic Analysis. 7th edition.

Show detailsWhen RF values are close to 50 percent, the χ^{2} 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):

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 χ^{2} 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 + *b*110)/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 χ^{2} is then calculated as follows:

The obtained value of χ^{2} is converted into a probability by using a χ^{2}
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.

- Chi-square test for linkage - An Introduction to Genetic AnalysisChi-square test for linkage - An Introduction to Genetic Analysis

Your browsing activity is empty.

Activity recording is turned off.

See more...