Mapping the Bacterial Chromosome Help (page 2)
When Hfr and F– cultures are mixed, conjugation can be stopped at any desired time by subjecting the mixture to the shearing forces of a Waring blender, which break the conjugation bridge. The sample is diluted immediately and plated on selective media, incubated, and then scored for recombinants. In addition to the selected marker, an Hfr strain must also carry a distal auxotrophic or sensitivity marker that prevents the growth of Hfr cells on the selective medium and thereby allows only recombinant cells to appear. This technique is called counter-selection. Because of the polarity with which the Hfr chromosome is transferred, the time at which various genetic markers appear in the recipient indicates their linear organization in the donor chromosome. At a given temperature, the transfer of the first half of the Hfr chromosome proceeds at a relatively uniform rate. Therefore, the time of entry of different markers into a recipient (F–) cell is a function of the physical distance between them.
EXAMPLE 10.11 An Hfr strain carrying the prototrophic markers a+, b+, c+ is mixed with an F– strain carrying the auxotrophic alleles a, b, c. Conjugation was interrupted at 5-min intervals and plated on media that revealed the presence of recombinants.
The order of the genes in the Hfr donor strain is ori—b+ —c+ —a+; b is less than 5 time units from the origin (ori); c is less than 5 time units from b; a is less than 5 time units from c.
When conjugation is allowed to proceed normally, the amount of time before the cytoplasmic bridge ruptures is apparently random. The nearer a marker is to the origin (leading end of donor chromosome), the greater its chances of appearing as a recombinant in a recipient cell. Donor and recipient cells are mixed for about an hour in broth and then placed on selective media that allows growth of F recombinants only for a specific marker. Counterselection against Hfr must also be part of the experimental design. The counterselective marker should be located as distally as possible from the selected marker so that unselected recombinants will not be lost by its inclusion. The frequencies with which unselected markers appear in selected recombinants are inversely related to their distances from the selected marker, provided they lie distal to it. Proximal markers more than 3 time units apart exhibit approximately 50% recombination, indicating that the average number of exchanges between them is greater than 1. Just at the point where gross mapping by conjugation becomes ineffective, i.e., for markers less than 2 time units apart, recombination mapping becomes very effective, permitting estimation of distances between closely linked genes or between mutant sites within the same gene. Distances between genes can be expressed in three types of units: (1) time units, (2) recombination units, or (3) chemical units.
EXAMPLE 10.12 If 1 min of conjugation is equivalent to 20 recombination units in E. coli, and the entire chromosome is transferred in 100 min, then the total map length is 2000 recombination units. If 107 nucleotide pairs exist in the chromosome, then 1 recombination unit represents 107 / .2000 = 5000 nucleotide pairs.
Virtually all of the opportunities for recombination in bacteria involve only a partial transfer of genetic material. One or more genes have an opportunity to become integrated into the host chromosome by conjugation, depending upon the length of the donor piece received. Exogenotes (donor DNA) must become integrated if they are to be replicated and distributed to all of the cells in a clone. Only a small segment of DNA is usually integrated during transformation or transduction. Thus, if a cell becomes transformed for two genetic markers by the same transforming piece of DNA (double transformation), the two loci must be closely linked. Similarly, if a cell is simultaneously transduced for two genes by a single transducing phage DNA (cotransduction), the two markers must be closely linked. The degree of linkage between different functional genes (intergenic) or between mutations within the same functional gene (intragenic) may then be estimated from the results of specific crosses.
In merozygotic systems where the genetic contribution of the donor parent is incomplete, an even number of crossovers is required to integrate the donor DNA into the host chromosome (endogenote).
Prototrophic recombinants must integrate the exogenote from somewhere left of the a locus to right of the b locus. Two crossovers (an even number) are required for this integration.
A prototrophic recombinant in this example requires a quadruple (even number) crossover for integration of all wild-type genes.
The total number of progeny is unknown in merozygotic systems so that recombination frequency cannot be expressed relative to this base. Therefore, recombination frequencies must be made relative to some standard that is common to all crosses. For example, the number of prototrophic recombinants produced by crossing two mutant strains can be compared with the number emerging from crossing wild type by mutant type. However, many sources of error are unavoidable when comparing the results of different crosses. This problem can be circumvented by comparing the number of prototrophic recombinants with some other class of recombinants arising from the same cross.
EXAMPLE 10.15 Ratio test for different functional genes. Suppose we have two mutant strains, a and b, where the donor strain (a+b) can grow on minimal medium supplemented with substance B, but the recipient strain (ab+) cannot do so.
Crossing over in regions (1) and (2) produces prototrophic recombinants (a+b+) able to grow on unsupplemented medium. If the medium is supplemented with substance B, then (a+b) recombinants arising by crossing over in regions (1) and (3) can grow in addition to the prototrophs.
EXAMPLE 10.16 Intragenic ratio test. Consider two intragenic mutations, b1, and b2, unable to grow in medium without substance B. The recipient strain contains a mutation in another functionally different gene (a), either linked or unlinked to b, which cannot grow unless supplemented by substances A and B.
On unsupplemented medium, only prototrophs arising through crossovers in regions (1) and (3) appear. On medium supplemented only by substance B, recombinants involving region (1) and any of the other three regions can survive.
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