How to calculate recombination frequency between three genes?
Answers
Explanation:
So far, we have looked at linkage in crosses of double heterozygotes to doubly recessive testers. The next level of complexity is a cross of a triple heterozygote to a triply recessive tester. This kind of cross, called a three-point testcross, illustrates the standard approach used in linkage analysis. We shall consider two examples of such crosses here.
First, we focus on three Drosophila genes that have the non-wild-type alleles sc (short for scute, or loss of certain thoracic bristles), ec (short for echinus, or roughened eye surface), and vg (short for vestigial wing). We can cross sc/sc · ec/ec · vg/vg triply recessive flies with wild-type flies to generate triple heterozygotes, sc/sc+ · ec/ec+ · vg/vg+. We analyze recombination in these heterozygotes by testcrossing heterozygous females with triply recessive tester males. The results of such a testcross follow. The progeny are listed as gametic genotypes derived from the heterozygous females. Eight gametic types are possible, and they were counted in the following numbers in a sample of 1008 progeny flies:
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The systematic way to analyze such crosses is to calculate all possible recombinant frequencies, but it is always worthwhile to inspect the data for obvious patterns before doing so. At first glance, we can note in the preceding data that there is a considerable deviation from the 1:1:1:1:1:1:1:1 ratio that is expected if the genes are all unlinked. So let’s begin to calculate recombinant frequency values, taking the loci a pair at a time. Starting with the sc and ec loci (ignoring the vg locus for the time being), we determine which of the gametic genotypes are recombinant for sc and ec. Because we know that the heterozygotes were established from sc · ec and sc+ · ec+ gametes, we know that the recombinant products of meiosis must be sc · ec+ and sc+ · ec. We note from the list that there are 12 + 14 + 14 + 16 = 56 of these types; therefore, RF = (56/1008) × 100 = 5.5 m.u. This frequency tells us that these loci must be linked on the same chromosome, as follows:
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