9.  You are doing a conjugation/recombination experiment with two strains of Escherichia coli; one is a donor strain (Hfr cells) and the other a recipient (F^– cells). Each strain is given to you as a young broth culture in a concentration of 8X10⁸ cells per ml. You mix equal amounts of the two cultures together and then let the mixture sit for a half hour. Then, from dilutions of the mixture, inoculations are made onto a special plating medium which supports the growth of recombinants but not parents, and the plates are incubated. From the plate which received 0.1 ml of a 10^–4 dilution of the mixture, you count 40 colonies.

For this problem and the other recombination problems below, consider CFUs and cells to be the same. (This is done for convenience only.)

a.  What was the total number of recombinant cells per ml of the mixture?

b.  What proportion (percentage or fraction) of the F^– cells in the mixture became recombinants (i.e., what was the recombination frequency)?

	dilution factor  X  no. of colonies  =  no. of recombinant F– cells/ml 105  X  40  =  4.0 X 106 That is, 4.0X106 is the number of F– cells (per ml) which can form colonies on the medium.
	no. of recombinant F– cells/ml  X  100%  =  recombination frequency total no. of F– cells/ml 4X106 / 4X108  X  100%  =  1 %

10.  You are given an Hfr strain of bacteria which is resistant to the antibiotic ampicillin, but not resistant to the antibiotic chloramphenicol. The F^– strain has the opposite phenotype – i.e., chloramphenicol resistant but ampicillin sensitive. You combine 1.0 ml of undiluted Hfr with 1.0 ml of undiluted F^–, allow enough time for conjugation to occur, and then plate dilutions. As you read through this problem, what is not made clear is that the platings (all of which are detailed in the following paragraphs) are all made on a certain day, and the plates are all observed and counted on another day – after a suitable period of incubation.

You plate 0.1 ml of a 1:10,000 (10^–4) dilution of the Hfr strain of bacteria on Nutrient Agar. 300 colonies appear after incubation. Dilution factor = 10⁵. Therefore, the concentration of cells (no. per ml) in the Hfr culture is 300 X 10⁵ or 3.0 X 10⁷.

You plate 0.1 ml of a 1:10,000 dilution of the F^– strain of bacteria on Nutrient Agar. 100 colonies appear after incubation. Dilution factor = 10⁵. Therefore, the concentration of cells (no. per ml) in the F^– culture is 100 X 10⁵ or 1.0 X 10⁷.

So, if one ml of each culture were added to a sterile empty tube, the resulting mixture would contain 4 X 10⁷ cells in the 2 ml amount which would be a concentration of 2 X 10⁷ cells per ml. The Hfr cells in this mixture outnumber the F^– cells 3 to 1, so the number of F^– cells would be one-fourth of the total number, and the concentration of F^– cells would therefore be 5 X 10⁶ per ml of the mixture. Remember that the F^– cells are the only cells which can undergo recombination.

You plate 1.0 ml of a 1:100 (10^–2) dilution of the conjugation of Hfr and F^– (i.e., the mixture) on Nutrient Agar plates containing chloramphenicol and ampicillin. After incubation, you count 45 colonies. What is the recombination frequency?

This is Problem No. 10 in the first 2 editions of the manual.  A conjugation/recombination experiment was run with two strains of Escherichia coli: an Hfr strain (met^–thr⁺trp^–) and an F^– strain (met⁺thr^–trp^–). Broth cultures were used, each at a concentration of 10⁸ CFUs/ml. Equal amounts of both strains were mixed together, and the mixture was incubated for a half hour and then plated on three different media, as were the original strains. (Note: met=methionine, thr=threonine, trp=tryptophan.)

a.  Place a + in each box where you expect growth and a O in each box where no growth is expected.

b.  One-tenth ml of a 10^–4 dilution of the incubated conjugation mixture was plated on MM+trp. After incubation, 50 colonies were counted. Determine the CFU/ml of the mixture for the recombinants growing on the plate.

c.  Calculate the recombination frequency for those recipient cells which gained the ability (through recombination) to grow on MM+trp.

	dilution factor  X  no. of colonies  =  no. of recombinant F– cells/ml 105  X  50  =  5.0 X 106 That is, 5.0X106 is the number of F– cells (per ml) which can grow and form colonies on MM+trp.
	no. of recombinant F– cells/ml  X  100%  =  recombination frequency total no. of F– cells/ml 5X106 / 5X107  X  100%  =  10%

11.  You are given a 10^–2 dilution of a bacteriophage suspension. You then make two, successive 1/10 dilutions of this dilution. From the last dilution you make, 0.1 ml is added to a tube containing 4.8 ml of melted top agar. One-tenth ml of a young culture of host organism is also added to the same tube of top agar. After mixing the tube of top agar is poured onto a plate of bottom agar. After incubation, 120 plaques are counted. Determine the number of plaque-forming units (PFUs) per ml of the original, undiluted bacteriophage suspension.

Do any of these matter?

• Amount of top agar.
  
• Amount of host culture.
  
• No. of phage particles in any plaque.

Note the analogies!

• cell vs. phage
  
• CFU vs. PFU
  
• colony vs. plaque