Mapping the Bacterial Chromosome Help (page 3)
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.
Establishing Gene Order
Mapping small regions in microorganisms has revealed that multiple crossovers often occur with much greater than random frequency, a phenomenon called "localized negative interference." One unambiguous method for determining the order of very closely linked sites is by means of three-factor reciprocal crosses. Suppose that the location of gene a is known to be to the left of gene b but that the order of two mutants with in the adjacent b gene is unknown. Reciprocal crosses will yield different results, depending upon the order of the mutant sites.
EXAMPLE 10.17 Assume the order of sites is a-b1-b2.
In the original cross, prototrophs (+ + +) can be produced by crossovers in regions (1) and (3). In the reciprocal cross, prototrophs arise by crossovers in regions (3) and (4). The numbers of prototrophs should be approximately equivalent in the two crosses.
An F particle that carries another bacterial gene other than the sex factor produces a relatively stable F+ merozygote. These partial diploids can be used for complementation tests of mutants affecting the same trait.
EXAMPLE 10.18 An Hfr strain of E. coli is unable to ferment lactose (lacZ–1) and can transfer the (lacZ–1) gene through conjugation to a mutant (lacZ–2) recipient, forming the heterogenote (lacZ–2)/(F-lacZ–1). If (lacZ–1) and (lacZ–2) are mutations in the same gene (i.e., functional alleles), then complementation does not occur and only mutant phenotypes are produced. If (lacZ–1) and (lacZ–2) are mutations in different genes, complementation could produce wild types able to ferment lactose.
Intragenic complementation may sometimes be possible when the enzyme product is composed of two or more identical polypeptide chains. Experimental evidence has shown that an in vitro mixture of inactive enzymes from some complementing mutants can "hybridize" to produce an enzyme with up to 25% normal activity. Mutants that fail to complement with some but not all other mutants are assumed to overlap in function. A complementation map can be constructed from the experimental results of testing all possible pairs of mutants for complementary action in bacterial merozygotes or in fungal heterokaryons. A complementation map cannot be equated in any way with a crossover map, since the gene is defined by different criteria. A complementation map tells us nothing of the structure or location of the mutations involved. Complementation maps are deduced from merozygotes or heterokaryons; crossover maps are deduced from recombination experiments.
This indicates that mutants 1 and 2 are complementary and do not overlap in function. Hence, 1 and 2 are nonallelic mutations by this criterion. Mutant 3 fails to complement with either 1 or 2 and hence must overlap (to some degree) with both 1 and 2. Hence, 3 is functionally allelic with both 1 and 2.
Mapping By Deletion Mutants
A deletion in some segment of a functional gene cannot recombine with point mutations in that same region even though two point mutations at different sites within this region may recombine to produce wild type. Another distinctive property of deletion mutants is their stability; they are unable to mutate back to wild type. The use of overlapping deletions can considerably reduce the work in fine structure analysis of a gene.
EXAMPLE 10.20 Determining the limits of a deletion. Suppose that a series of single mutants (l, 2, 3, 4) has already been mapped as shown below:
A deletion that fails to recombine with point mutants 1 and 2 but does produce wild type with 3 and 4 extends over region X. A deletion that yields no recombinants with 3 and 4 has the boundaries diagrammed as Y. A deletion mutant that produces wild type only with point mutants 1 or 4 has the limits of Z.
EXAMPLE 10.21 Assigning point mutations 1–4 to deletion regions R, S, and T.
Given the deletions R, S, and T as shown above, the point mutation that recombines to give wild type with deletions S and T, but not with R, is 1. Number 3 is the only one of the four mutants that fails to recombine with one of the three deletions.
Practice problems for these concepts can be found at:
Today on Education.com
- Coats and Car Seats: A Lethal Combination?
- Kindergarten Sight Words List
- Child Development Theories
- Signs Your Child Might Have Asperger's Syndrome
- 10 Fun Activities for Children with Autism
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development
- Social Cognitive Theory
- GED Math Practice Test 1
- The Homework Debate
- First Grade Sight Words List