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

Glossary of biochemical terms | Enzyme Problems Set1. | Frequently asked questions | C2005/F2401 ’06– Recitation Problems #3 -- Answers | A. Hint: Can you calculate the Vmax? (What value of [S] was used here? Does that help you get the Vmax?) How do you get from Vmax to the turnover number? | C2005/F2401 '06 -- Answers to Recitation Problems #4 | C2005/F2401 '06 -- Key to Recitation Problems #5 | C2005/F2401 '06 -- Answers to Recitation Problems #6 | C2005/F2401 '06 -- Recitation Problems #7 -- Answers | T A C T C A T C G A 3’ ......................… Promotor; transcription right to left. |


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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 109, 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 tD = 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 ekt, ln(N/No)= ln(4/1) = 1.39 = kt., and t = 5h, so k = 1.39/5 = 0.278, Now k = ln2/tD = 0.69/tD. So 0.69/tD = 0.278, and so tD = (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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Typical Examination Questions| C2005/F2401 '06 Answers to Recitation Problems #2 1. Hint: Consider which molecules (or parts thereof) are hydrophobic or hydrophilic.

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