C2005/F2402 '09 -- Key to Recitation Problems #1

   
    1A.  First of all, you have to figure out how many cells you will have at the end. Then you can calculate the doubling time: Here are two approaches to figuring out the # of cells at the end:

Approach #1 -- Figure out the number of cells at the start and the # you can build. Total that up to get the # of cells you will have at the end.  How many cells can you build from the 6 mg of N that was consumed? You know the amount of N per cell =  2 X 10^-9 mg.
 6 mg N /(2 X 10^-9) mg/cell  = 3 X 10⁹ cells. You started with 1 X 10⁹, so you have 4 X 10⁹ cells total at the end. In other words, the initial number of cells was 10⁹,  and the N consumption added 3 X 10⁹ cells to this initial number, so the total at the end of the 5 hours would be  1 + 3 = 4 X 10⁹. 

Approach #2 -- Concentrate on the amount of N in bacteria at start and end, instead of the # of bacteria. You started with the 1 x 10⁹ cells which corresponds to 2 mg of N. You used up 6 mg, so you had 8 mg of N in bacteria at the end.  8 mg N /(2 X 10^-9) mg/cell  = 4 X 10⁹ cells. 

    Now you know you went from 1 X 10⁹ cells to 4 X 10⁹ cells in 5 hrs. There are two ways to view the calculation of doubling time:

        1) Simple way: A 4-fold increase in cell number occurred overall: 1 x 10⁹ → 4 x 10⁹ cells, or 2 mg N → 8 mg N. This means 2 generations occurred in 5 hours (2 mg → 4 mg → 8 mg, or 1 x 10⁹ cells → 2 x 10⁹ → 4 x 10⁹). So doubling time = 5/2= 2.5 hrs/gen, or t_D  = 2.5h.
        2) Using equations:  Using N = No e^kt , ln(N/No)= ln(4/1) = 1.39  = kt., and t = 5h, so k = 1.39/5 = 0.278,  Now k = ln2/t_D = 0.69/t_D.  So 0.69/t_D = 0.278, and so t_D  = (0.69)/0.278  = 2.48.

    1B.  There isn't enough N left in the medium for the bacteria to keep doubling, so they will be in stationary phase. 

Details: There is only 4 mg of N left in the culture at the end of the first 5 h hour growth period, which is only enough for 2 x 10⁹ cells, or a 50% increase over the 4 x 10⁹ cells present at that time. This is only enough for about a 1/2 of a cell division, so the cells will run out of N after a short time, and so for most of the second 5-hour period they will be in stationary phase.

    1C. The limiting factor here is raw materials, not time. So you will end up with 6 X 10⁹ cells.

Details: There were 4 x 10⁹ after the first 5-hour period of growth in part 1 above; to the 4 x 10⁹ present at the start of the second 5-hour growth period must be added the 2 x 10⁹ cells that could be built with the 4 remaining mg of N, so the total will be 4 + 2 = 6 X 10⁹ cells. (Alternatively, calculate how many bacteria you can make if you use up all of the 10 mg N, and add that to the starting number of bacteria.)   
        

    2.  Amino acid, phospholipids, polypeptides, and the cell membrane (in the proteins and phospholipids in the membrane). Keep in mind:

(i) The new N atoms will not replace old N atoms already in molecules; they will be used to make new molecules.

(ii) If you are not sure if some of the complex molecules ever contain N, check out your handouts. (What is the "R" in phospholipids?)