iga11e_ch05

SOLVED PROBLEMS

SOLVED PROBLEM 1. Suppose that a cell were unable to carry out generalized recombination (rec⁻). How would this cell behave as a recipient in generalized and in specialized transduction? First, compare each type of transduction, and then determine the effect of the rec⁻ mutation on the inheritance of genes by each process.

Solution

Generalized transduction entails the incorporation of chromosomal fragments into phage heads, which then infect recipient strains. Fragments of the chromosome are incorporated randomly into phage heads, and so any marker on the bacterial host chromosome can be transduced to another strain by generalized transduction. In contrast, specialized transduction entails the integration of the phage at a specific point on the chromosome and the rare incorporation of chromosomal markers near the integration site into the phage genome. Therefore, only those markers that are near the specific integration site of the phage on the host chromosome can be transduced.

Markers are inherited by different routes in generalized and specialized transduction. A generalized transducing phage injects a fragment of the donor chromosome into the recipient. This fragment must be incorporated into the recipient’s chromosome by recombination, with the use of the recipient’s recombination system. Therefore, a rec⁻ recipient will not be able to incorporate fragments of DNA and cannot inherit markers by generalized transduction. On the other hand, the major route for the inheritance of markers by specialized transduction is by integration of the specialized transducing particle into the host chromosome at the specific phage integration site. This integration, which sometimes requires an additional wild-type (helper) phage, is mediated by a phage-specific enzyme system that is independent of the normal recombination enzymes. Therefore, a rec⁻ recipient can still inherit genetic markers by specialized transduction.

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SOLVED PROBLEM 2. In E. coli, four Hfr strains donate the following genetic markers, shown in the order donated:

Strain 1:	Q	W	D	M	T
Strain 2:	A	X	P	T	M
Strain 3:	B	N	C	A	X
Strain 4:	B	Q	W	D	M

All these Hfr strains are derived from the same F⁺ strain. What is the order of these markers on the circular chromosome of the original F⁺?

Solution

A two-step approach works well: (1) determine the underlying principle and (2) draw a diagram. Here the principle is clearly that each Hfr strain donates genetic markers from a fixed point on the circular chromosome and that the earliest markers are donated with the highest frequency. Because not all markers are donated by each Hfr, only the early markers must be donated for each Hfr. Each strain allows us to draw the following circles:

From this information, we can consolidate each circle into one circular linkage map of the order Q, W, D, M, T, P, X, A, C, N, B, Q.

SOLVED PROBLEM 3. In an Hfr × F⁻ cross, leu⁺ enters as the first marker, but the order of the other markers is unknown. If the Hfr is wild type and the F⁻ is auxotrophic for each marker in question, what is the order of the markers in a cross where leu⁺ recombinants are selected if 27 percent are ile⁺, 13 percent are mal⁺, 82 percent are thr⁺, and 1 percent are trp⁺?

Solution

Recall that spontaneous breakage creates a natural gradient of transfer, which makes it less and less likely for a recipient to receive later and later markers. Because we have selected for the earliest marker in this cross, the frequency of recombinants is a function of the order of entry for each marker. Therefore, we can immediately determine the order of the genetic markers simply by looking at the percentage of recombinants for any marker among the leu⁺ recombinants. Because the inheritance of thr⁺ is the highest, thr⁺ must be the first marker to enter after leu. The complete order is leu, thr, ile, mal, trp.

SOLVED PROBLEM 4. A cross is made between an Hfr that is met⁺ thi⁺pur⁺ and an F⁻ that is met⁻ thi⁻ pur⁻. Interrupted-mating studies show that met⁺ enters the recipient last, and so met⁺ recombinants are selected on a medium containing supplements that satisfy only the pur and thi requirements. These recombinants are tested for the presence of the thi⁺ and pur⁺ alleles. The following numbers of individuals are found for each genotype:

met⁺ thi⁺ pur⁺	280
met⁺ thi⁺ pur⁻	0
met⁺ thi⁻ pur⁺	6
met⁺ thi⁻ pur⁻	52

Why was methionine (Met) left out of the selection medium?
What is the gene order?
What are the map distances in recombination units?

Solution

Methionine was left out of the medium to allow selection for met⁺ recombinants because met⁺ is the last marker to enter the recipient. The selection for met⁺ ensures that all the loci that we are considering in the cross will have already entered each recombinant that we analyze.
Here, a diagram of the possible gene orders is helpful. Because we know that met enters the recipient last, there are only two possible gene orders if the first marker enters on the right: met, thi, pur or met, pur, thi. How can we distinguish between these two orders? Fortunately, one of the four possible classes of recombinants requires two additional crossovers. Each possible order predicts a different class that arises by four crossovers rather than two. For instance, if the order were met, thi, pur, then met⁺ thi⁻ pur⁺ recombinants would be very rare. On the other hand, if the order were met, pur, thi, then the four-crossover class would be met⁺ pur⁻ thi⁺. From the information given in the table, the met⁺ pur⁻ thi⁺ class is clearly the four-crossover class and therefore the gene order met, pur, thi is correct.
Refer to the following diagram:

To compute the distance between met and pur, we compute the percentage of met⁺ pur⁻ thi⁻, which is 52/338 = 15.4 m.u. Similarly, the distance between pur and thi is 6/338 =1.8 m.u.

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SOLVED PROBLEM 5. Compare the mechanism of transfer and inheritance of the lac⁺ genes in crosses with Hfr, F⁺, and F′ lac⁺ strains. How would an F⁻ cell that cannot undergo normal homologous recombination (rec⁻) behave in crosses with each of these three strains? Would the cell be able to inherit the lac⁺ gene?

Solution

Each of these three strains donates genes by conjugation. In the Hfr and F⁺ strains, the lac⁺ genes on the host chromosome are donated. In the Hfr strain, the F factor is integrated into the chromosome in every cell, and so chromosomal markers can be efficiently donated, particularly if a marker is near the integration site of F and is donated early. The F⁺ cell population contains a small percentage of Hfr cells, in which F is integrated into the chromosome. These cells are responsible for the gene transfer displayed by cultures of F⁺ cells. In the Hfr⁻- and F⁺-mediated gene transfer, inheritance requires the incorporation of a transferred fragment by recombination (recall that two crossovers are needed) into the F⁻ chromosome. Therefore, an F⁻ strain that cannot undergo recombination cannot inherit donor chromosomal markers even though they are transferred by Hfr strains or Hfr cells in F⁺ strains. The fragment cannot be incorporated into the chromosome by recombination. Because these fragments do not possess the ability to replicate within the F⁻ cell, they are rapidly diluted out during cell division.

Unlike Hfr cells, F′ cells transfer genes carried on the F′ factor, a process that does not require chromosome transfer. In this case, the lac⁺ genes are linked to the F′ factor and are transferred with it at a high efficiency. In the F⁻ cell, no recombination is required because the F′ lac⁺ strain can replicate and be maintained in the dividing F⁻ cell population. Therefore, the lac⁺ genes are inherited even in a rec⁻ strain.