Question

You are culturing fibroblasts and observe that a spherical cell 1 μm in diameter divides into two spherical cells. (a) Calculate the work necessary to increase the surface area if its surface tension is 12.3 x 10-3 N/m. Assume that the volume of the daughter cells taken together is the same as that of the parent. (b) If the hydrolysis of ATP in this cell yields 25 kJ/mol of free energy, how many molecules of ATP are needed for the division above?

Answer 1

SOLUTION:-

First of all the surface area of parent cell = pi x d2 = 3.14 x lum = 3.14 x 10"-12m’

then volume of parent cell= pi x d’l6 = 3.14 x(10"-12)’/ 6= 0.524 x 10“-18m’

the volume of each daughter cell is = 0.524 x10*-18 12 (because 2 daughter cells are present} = 0.262 x
1oA-18m’

now the equation for area and surface of sphere = (36 x pix volume of daughter cells)‘/: = (36 x 3.14 x
.262 x 10“-18 M = 1980 x lot-12 m’

now for two daughter cells = 2 x 1980 x lot-12 m’

for carrying out this we need to calculate = surface tension x difference of area and surface (i.e 3.96-3.14=
0.82) = 123 x 10"-3 Nmt-l x 0.82 x 10“-12 rn2 = 1.0 x 1oA-14Nm.

if the hydrolysis of ATP in the cell is 25 KJ / mol of free energy then about 25000 ATP are needed for the
division.