Total number of moles removed at t=10 = 0.000000002481751078 (Moles in liquid in 1ml extraction) + 0.000000003863 (Moles in cell volume in 1ml extraction) = 0.000000006344751078 moles
8. We then calculate the number of moles present within cells at t=15 (using the same method as in step 1 & 2), except account for the cells already removed: 
 = 736/ (0.09756098/((1-0.09756098*3))) = 0.00000002451165922 moles
9. We then add the number of moles present within cells at t=15, to the moles already removed at t=1, t=5 and t=10.
0.00000002451165922 (moles present in cells at t=15) + 0.00000000634146419 (moles removed at t=1) + 0.00000000634535538 (moles removed at t=5) + 0.000000006344751078 (moles removed at t=10) = 0.00000004354322987 moles
10. We then calculate the number of moles remaining in solution
0.000000065 - 0.00000004354322987 = 0.00000002145677013 moles
11. We can then calculate the concentration at t=15
Remaining volume inside cells =  10250μl - 3ml - internal volume of remaining cells) = 10250μl - 3000μl - (15.5-3(1.512))μl =7239.036μl
 7239.036μl = 0.007239036l
0.00000002145677013 moles / 0.007239036l = 0.000002964036943 M
These are the calculations carried out for each experiment. The results are on the attached Figure. 3

iii) The factor by which [14C]-isoleucine has been concentrated within the cells.

The factor of concentration within the cell at t=15 can be calculated from the concentration within the cell at t=15, and the concentration outside the cell at t=0. 
e.g  experiment 1, cpm1, t=1; t=5; t=10, -background radiation
2.236mM within the cell at t=15
0.000000065 moles / 0.01024845l = 0.000006342422513M concentration in flask at t=0
0.006342mM outside the cells at t=0
2.236/0.006342 = 352.56
Hence it has been concentrated within the cell by a factor of 352.56 for experiment 1, cpm1. The results for all experiments are on the attached Figure. 3.