Question

2a. (4 pts) A potent mutagen was added to an E. coli culture growing in rich broth. Prior experiments established that this F cell line, designated WWU113, could not synthesize leucine. You are interested in measuring the rate of induced reverse mutation. You plan to grow a liquid culture of the leu strain and then plate various dilutions of the culture onto selective media. What media should you use to grow up the liquid culture? This will be Media A in the chart a. below. b. Describe the media in the plates you should use to select for reverse mutations. This will be Media B in the chart below. Let's say you already know the rate of spontaneous leu+ to leu- mutations. Do you expect the c. rate of spontaneous reverse mutations to be greater or less than the rate of leu+ to leu- mutation? Why? 2b. (7 pts) The WWU113 cells were allowed to grow overnight after which a series of dilutions were made. The dilutions were leucine. onto either minimal media, or onto minimal media with plated From the results given here, determine the original concentration of the cell culture and the frequency of leu leu' mutant cells. Show your work. Dilution Colonies Culture Condition 10-8 142 Media A 10-2 115 Media B Original concentration of cell culture: a. b. What is the mutation frequency? Please state it in both the number mutations/cell and the number of cells you would have to screen to find one mutant cell. Culture Condition Dilution Colonies Media A 10-8 142 Media B 10-2 115

Answer 1

A. The mutant is unable tot synthesis leucine. So media should have leucine suppliment.

B. The point mutation is successful can result in leu+ revertant. So the media in the plates should be without leucine suppliment.

C. Spontaneous rate of reverse mutations are less than the rate of leu+ to leu- mutation. There are several ways to creat loss of function for example, point mutations, misses each mutation, nonsense mutation...etc. but there are very few ways to reverse mutations back to normals or gain of function. So forward rate of mutation are more than ten rivers rate of mutations.

A. Original concentration of cell culture= no. Of colonies x dilution / volume plated.

The media A is the one with leucine suppliment . So all cells will grow on this.

= (142*10^8)/0.1 ml

= 142*10^9 cells per ml.

B. The mutant cell number = (115*10^2)/0.1ml

= 115*10^3 cells per ml.

= 1.15*10^5 cells per ml.

This means to screen one mutant cell 10^5 cells will have to be screened.

Mutation frequency = no. Of mutants / total cells.

= 1.15*10^5 / 142*10^9

= 1.15*10^5 / 1.42*10^11

= 0.809*10^-6

= 8 * 10^-6

In other words, we need to screen 10^6 cells to get 8 mutant cells.