00:01
In part a, the dry weight of your e.
00:04
Coli is 0 .28 pg, or that would be 0 .3 pg.
00:12
Since the chemical composition of your e.
00:15
Coli is 50 % of carbon, 14 % of nitrogen, therefore the number of carbon atoms per cell required to make up the macromolec contents would be nc is equal to 0 .3 p .g.
00:31
Divided by the molecular weight of your carbon times 50%.
00:39
So we will have 0 .3 pg divided by 12 grams per mole times 0 .5.
00:49
This would be 1 .25 times 10 raised to the power of negative 14 mole is equal to 7 .5 times 10.
01:03
Raised to the power of 9 carbon atoms.
01:10
Next, the number of nitrogen atoms required per cell.
01:13
So we have n.
01:15
That is 0 .3 p .g over mole.
01:22
We have molecular weight.
01:27
So molecular weight of your nitrogen times 14%.
01:33
So 0 .3 divided out 14 grams per mole times 0 .4 .5 .5 .5 .5 .5...