Genetics: From Genes To Genomes (6th International Edition)
Genetics: From Genes To Genomes (6th International Edition)
6th Edition
ISBN: 9781260041217
Author: Leland Hartwell Dr., ? Michael L. Goldberg Professor Dr., ? Janice Fischer, ? Leroy Hood Dr.
Publisher: Mcgraw-Hill
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Chapter 14, Problem 21P

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 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 × 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 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 Arg+ exconjugants, and then screened them for the earlier Hfr markers Mal Xyl and Pyr . You obtained the following data:

Exconjugate type Number of exconjugates

Arg+ Mal+ Xyl+ Pyr+

80
Arg+ Mal+ Xy+l Pyr-
40
Arg+ Mal+ Xyl- Pyr-
20
Arg+ Mal- Xyl-
Pyr-
20
Arg+ Mal- Xyl+ Pyr-
1
Arg+ Mal- Xyl- Pyr+ 1
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.
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A wild-type mouse that is heterozygous for two immunoglobulin heavy chain alleles (IgHa/b) generates the population of B cells shown on the left of the figure below.  A mouse strain, also IgHa/b, carries an inactivating mutation in the VpreB gene. In addition to producing fewer mature B cells than the wild-type mice, the VpreB-deficient mice generate B cells as shown on the right.  What is the explanation of the difference seen between the wild-type and the VpreB-mutant B cells?
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Genetics: From Genes To Genomes (6th International Edition)

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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license