Problem 3. A child with acute lymphocytic leukemia (ALL) has approximately 1012

Question

Problem 3. A child with acute lymphocytic leukemia (ALL) has approximately 1012 leukemic cells when the disease is clinically apparent.

(a) If a cell is about 8 m in diameter, estimate the total mass of leukemic cells.

(b) Cure requires killing every single cell. The doubling time for the cells is about 5 days. If all cells were killed except for one, how long would it take for the disease to become apparent again?

(c) Suppose that chemotherapy reduces the number of cells to 109 and there are no changes of ALL cell proper- ties (no mutations). How long a remission would you expect? What if the number were reduced to 106?

Explanation / Answer

DENSITY OF CELLS IS APPROX 1/10 OF WATER.=1/10

VOLUME OF CELLS = 4/0X3.14X4X4X4=803.84 m3

MASS OF 1CELL= DENSITY X VO;UME= 1/10X803.84=80.384g.

MASS OF 1012 CELLS=81348.608 MICROGRAM.

B. dOUBLING TIME = 5 DAYS

DISEASE WILL APPARENT AFTER IT WILL BECOME 1012 IN NO.

THEREFORE

IT WILL TAKE 50 DAYS.

c.IT WILL NOT REMIT IF THE CHEMOTHERAPY IS REGULAR. AND THETRE WILL BE NO CHANGE IF CELL COUNT WILL BE 106.

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