Required to diagram recombination events that can replace specific genes on the chromosome of a recipient cell with copies of those genes introduced from a donor cell. As seen in the solution to Solved Problem I, only an even number of crossovers can produce viable recombinant chromosomes. Gene mapping is simplified if you remember that progeny classes that result from four crossovers are found much less often than progeny classes that require two crossovers.
One issue with interrupted-mating experiments such as that in Problem 19 is that gene order may be ambiguous if the genes are close together. Another shortcoming is that such experiments do not provide accurate map distances. The reason is that researchers select for the first Hfr marker transferred into the recipient, but the recovery of $\mathrm{F}^{-}$ exconjugants with a later Hfr marker is complex, depending both on transfer of the marker into the cell and on crossovers that transfer the marker into the recipient chromosome. To make more accurate maps, bacterial geneticists often do Hfr $\times \mathrm{F}^{-}$ crosses in a different way: They select for exconjugants that contain a late Hfr marker, and then screen for the presence of the earlier markers. This method ensures that all of the markers have entered the $\mathrm{F}^{-}$ cell, so relative gene distances are now accounted for solely by crossover frequencies. Furthermore, gene order is clarified by considering the crossovers responsible for each class of exconjugants. As an example, suppose you performed the same cross as in Problem $19,$ but you selected for $\mathrm{Arg}^{+}$ exconjugants, and then screened them for the earlier Hfr markers $\mathrm{Mal}^{+} \mathrm{Xyl}^{+}$ and $\mathrm{Pyr}^{+} .$ You obtained the following data:
a. Explain why four of the exconjugant types are much more frequent than the other two.
b. What can you conclude about the relative distances between the four genes?
c. The data allow you to estimate one other relevant genetic distance. Explain.