Lawes and Maloy wanted to map the temperature-sensitive lethal mutation clm-195(TS) in Salmonella typhimurium. Two-factor crosses were done with several linked loci: an insertion mutation with a tetracycline resistant phenotype (zci::Tn10), the nit gene, and the btuC gene. The results are shown in the table below. Calculate the linkage between each of the genes in these two-factor crosses. [Note that Tn10 indicates an insertion mutation with a tetracycline resistant phenotype. Genetic mapping of Tn10 is essentially like any other gene.]
Phage P22 HT was grown on the donor cells shown below. Using these phage lysates, two-factor crosses were done against Salmonella typhimurium strains with mutations in the metA, aceA, and iclR genes. Given the results shown in the table below, draw a genetic linkage map showing the cotransduction frequency and the predicted order of the metA, aceA, and iclR genes. Show your calculations. [Note that Tn10 indicates an insertion mutation with a tetracycline resistant phenotype. Genetic mapping of Tn10 is essentially like any other gene.]
|Donor||Recipient cells||Selected phenotype||Recombinants||Number obtained|
|metA::Tn10 iclR-||metA+ iclR+||TetR||IclR-|
|aceA+ iclR-||aceA- iclR+||Ace+||IclR-
|metA::Tn10 aceA+||metA+ aceA-||TetR||Ace-
ANSWER: Note that you need to determine the frequency of coinheritance -- that is, the fraction of recombinants that inherited both DONOR alleles. The calculations and predicted genetic map are shown below.
Hughes and Roth isolated a mutation (nadD) in a gene required for NAD biosynthesis in Salmonella. They did two-factor crosses with phage P22 to determine the linkage map of nadDrelative to the lip and leuS genes. From the following two-factor cross data, draw a linkage map of the nadD, lip, and leuS genes. [Indicate percent cotransduction and draw appropriate arrowheads.]
ANSWER: Note that the cotransduction frequency is the coinheritance of both donor alleles. The values obtained are shown in the figure below.
In order to map the adenine deaminase (add) gene in E. coli, two-factor crosses were done with the nearby genes aroD, man and uidA [Jochimsen, B., P. Nygaard, and T. Vestergaard. 1975. Mol. Gen. Genet. 143: 85-91.] Given the results shown below, draw a genetic linkage map showing the most likely gene order and the coinheritance frequencies with arrows indicating the selected marker.
ANSWER: The frequency of coinheritance -- that is, the fraction of recombinants that inherited both DONOR alleles -- is shown in the table above and a map of the man, add, and aroD genes is shown in the figure below.
Note that you cannot determine the position of uidA from this data. The order could be either: uidA man add aroD or man uidA add aroD
To confirm the order of the metA, aceA, and iclR genes determined from two-factor crosses, three-factor crosses were done. Given the results shown below, what is the order of these three genes? [Note that metA::Tn10 indicates an insertion mutation in the metA gene with a tetracycline resistant phenotype.]
|Selected phenotype||Recombinants||Number obtained|
ANSWER: The rare class of recombinants would require a double cross-over as shown in the diagram below:
To determine the gene order of the add, man, and uidA genes, three-factor crosses were done. Given the results shown below, what is the order of these three genes? Draw a diagram of each cross to show your rationale for this conclusion.
ANSWER: Note that you can infer the middle marker from the rare class of recombinants (the Man+ Add- UidA- recombinants). This result suggests that the uidA allele is the middle marker in the recipient because four crossovers are required to produce this class of recombinants. The question asked you to draw a diagram of each of the crosses, but only the crossovers yielding the rarest class of recombinants is shown in the figure below.
Hughes and Roth did three factor crosses to confirm the order of the nadD gene relative to the adjacent genes, leuS and lip. Does this data agrees with the linkage map constructed from the two-factor crosses. [For each of the crosses in the table, show a drawing with the relevant crossovers and the inferred gene order.]
ANSWER: Note the rare classes in each cross. In the first cross there is clearly a rare class (suggesting that this class requires 4 X-overs), but in the second cross there is no class that is much rarer than the others (suggesting that they all require 2 X-overs). It is important to indicate an even number of X-overs because an odd number of X-overs would linearize the bacterial chromosome (remember that the donor is linear DNA and the recipient is circular DNA) resulting in cell death.
A three factor cross was done to determine the order of several linked genes: an insertion mutation with a tetracycline resistant phenotype (zci::Tn10), the nit gene, and the btuC gene. The results are shown in the table below. Based on these results, what is the order of the genes nit, btuC, and zci::Tn10?
|nit+ btuC11||nit-9 btuC+ zci::Tn10||Btu-||TetR Nit-||215|
ANSWER: The first thing to look for when analyzing three-factor cross results is a single class of recombinants which is much rarer than the other classes of recombinants -- such rare recombinants typically arise from two pairs of cross-overs. However, these results show two equally rare classes of recombinants, so inferring the gene order is not so simple. The disparate distribution of recombinants in each of the four classes suggests that the selected marker (btuC) is NOT the middle marker. Thus, btuC is probably an end marker in the gene order, but you cannot determine the gene order unambigously.
What three-factor cross could you do to confirm the order of clm, btuC, and zci::Tn10? Clearly indicate the genotypes of the donor and recipient strains you would use.
Any similar combination would work. There need to be three different "factors" in both the donor vs the recipient (i.e. if the donor has a mutant allele the recipient should have the donor allele), otherwise recombinant classes will not be recovered. An addition, one of the markers in the donor needs to be a selectable allele.
Several amber (Am) and temperature sensitive (Ts) mutations were isolated in a bacteriophage. Complementation tests showed that the mutations fall into three complementation groups, designated X, Y, and Z. Three-factor crosses between these mutants gave the results shown in the table below.
|Cross||Donor||Recipient||Selected marker||Number of selected progeny |
that coinherit the non-selected
|1||Z8(Ts) X+||Xl(Am) Z5(Ts)||Z+||X+
|2||Xl(Am) Z8(Ts)||X6(Am) Z+||X+||Z+
|3||Y3(Am) Z+||Y6(Am) Z5(Ts)||Y+||Z+
|4||Y6(Am) Z5(Ts)||Y+ Z8(Ts)||Z+||Y+|