C2005/F2402 '06 -- Key to Recitation Problems #1

1A. HINTS:

(i) Where will the N atoms go? Will they replace old N atoms already in molecules? 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?)

Answers: Amino acid, phospholipids, polypeptides, and the cell membrane (in proteins and phospholipids there).

1B. HINT: How many bacteria will be present at the end of 5 hrs? See approach #1.

Answer: Approach #1 -- Figure out the number of cells at the start and the # at the end. 6 mg of N was consumed, which is enough to build,
at 2 X 10^-9 mg N/cell, 6/(2 X 10^-9) = 3 X 10⁹ cells. You started with 1 X 10^9, so you have 4 X 10⁹ cells at the end.

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.

There are now two ways to view the calculation of doubling time:
1) Simple way: You make 3 X 10⁹ new cells (or use up 6 mg N), so a 4-fold increase overall: 1 x 10⁹ -> 4 x 10⁹ cells, or 2 mg N --> 8 mg N. This means 2 generations (2 mg --> 4 mg --> 8 mg, or 1 x 10⁹ cells --> 2 x 10⁹ --> 4 x 10⁹) in 5 hours. So doubling time = 5/2= 2.5 hrs/gen, or t_D = 2.5h.
2) Using equations: 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. 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

1C. HINT: How much N is left in the medium? Enough for the bacteria to keep doubling?

Answer: Stationary phase. 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.

1D. HINT: What is the limiting factor? Time or raw materials?

Answer: 6 X 10⁹ cells. 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.)

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