To calculate the internal concentration of [14C]-isoleucine within the filter examined we need to:
- Calculate the number of moles of [14C]-isoleucine needed to give a reading of xCPM.
- Calculate the number of moles required to give the readings we observed in each experiment (nN).
- Calculate the internal cell volume.
- Divide nN by the internal cell volume.
1. A 10μl sample of 0.26mM [14C]-isoleucine registered 576 CPM.
We'll initially subtract the background radiation, approximately 10 CPM ; 576-10 = 566 CPM
We'll then convert the mM and μl figures into M and l ; 0.26/1000 = 0.00026M ; 10/1,000,000 = 0.00001l
We'll then calculate the number of moles in the reference sample; 0.00026M * 0.00001l = 0.0000000026 moles
Hence, we require 0.0000000026 moles of [14C]-isoleucine to deliver a CPM reading of 566 (minus background radiation).
From this we can calculate the number of moles required to give a reading of x CPM.
Moles required for x CPM = 0.0000000026/(566/x)
2.
e.g Moles required for 736 CPM (experiment 1, cpm1, at t=15, -background radiation)
Moles required for 736 CPM = 0.0000000026/(566/736) = 0.000000003381 moles
3. We now need to calculate the internal cell volume of the bacteria on the filter - While the cells were resuspended before the experiment took place to 1 mg dried weight / ml-1 - The addition of [14C]-isoleucine occurred AFTER resuspension. This will change the mg dried weight / ml-1 of the flask, so this needs to be recalculated.
The cells were resuspended to 1 mg dried weight / ml-1, therefore in the 10ml added to the conical flask each ml will contain 1 mg dried weight / ml -1.
250μl of [14C]-isoleucine was then added, increasing the overall volume of the liquid in the flask, decreasing the mg dried weight / ml -1.
1ml was extracted at t=1, t=5, t=10 and t=15.
1000/10250 = 0.09756098
0.09756098*10 = 0.9756098
0.9756098*1.55μl = 1.512μl
For experiments 2, and 4-6, 0.25ml of 21mM glucose was also added. The addition of this requires us to recalculate internal cell volume for these experiments also.
1000/10500 = 0.0952381
0.0952381*10 = 0.952381
0.952381*1.55μl = 1.47619μl
For experiment 3, 5μl of CCCP was also added, as well as 0.25ml of 21mM glucose. The addition of this requires us to recalculate internal cell volume for this experiment.
1000/10505 = 0.09519277
0.09519277 * 10 = 0.9519277
0.9519277*1.55μl = 1.4754879μl
4. We can now calculate the concentration of [14C]-isoleucine on the filter:
e.g Concentration of [14C]-isoleucine within experiment 1, cpm1, at t=15, -background radiation
0.000000003381 moles / 0.000001512l = 0.002236 M
0.002236 * 1000 = 2.236mM
A series of two tailed t.tests, assuming equal variances were performed on samples to see if there were statistically significant difference between experiments, with and without uncouplers present.
All three uncouplers; CCCP, valinomycin and nigericin, are all significantly different to the experiment just performed with glucose (ρ = 0.00000025905 ; ρ = 0.002236 ; ρ = 0.0000007601 respectively). In comparison to each other, the difference between uncouplers in isolation are (CCCP:val ρ = 0.0009410 ; CCCP:nig ρ = 0.00002315 ; val:nig ρ = 0.007665 respectively).
In fact the null hypothesis was rejected for every experiment in relation to all others, except one between +Glucose, +CCCP and +Glucose, +Nigericin, +Valinomycin. This t. test did not reject the null hypothesis (ρ= 0.2191)and implied there's no significant difference between the two effects of CCCP; and nigericin and valinomycin in combination.
ii) The calculated external concentration of isoleucine.
TO MARKER: When running these calculations I kept getting negative results (which have been included), the intracellular concentration of isoleucine for all cells seems to be greater than the isoleucine added at the beginning - I'm honestly not sure where I'm going wrong on this, and would really appreciate feedback on where I went wrong!
The external concentration of isoleucine at t=15mins can be calculated by:
- Calculate the total number of moles added to the solution at t=0
- Calculate the number of moles uptaken into all cells at t=1
- Calculate number of moles not in cells at t=1
- Calculate concentration of liquid at t=1
- Calculate the number of non-intracellular moles removed at t=1
- Calculate the total number of cells removed at t=1
- Repeat process for t=5 and t=10
- Calculate number of moles within cells at t=15
- Add number of moles within cells at t=15 to number of moles removed at t=1, t=5 and t=10
- Calculate the number of moles remaining in solution
- Calculate the concentration in solution at t=15
1. The total number of moles added to the solution can be calculated as follows:
250ul * 0.26mM = 0.00025l * 0.00026M = 0.000000065 moles were added ie total number of moles in the solution is 0.000000065
2. Number of moles uptaken into all cells at t=1; can be calculated by using the number of moles calculated from the CPM in the sample, divided by the fraction of cells that are in the sample examined.
e.g Experiment 1, cpm1, t=1; t=5; t=10, -background radiation
Moles required for x CPM = 0.0000000026/(566/x)
t=1
Moles required for 672 CPM = 0.0000000026/(566/672) = 0.000000003087 moles
0.09756098/1 of the cells are on the filter [see Q2)i)3) ]
0.000000003087 / 0.09756098 = 0.00000003164174858 moles
3. Calculate no. of moles not in cells at t=1
This is simply the total number of moles added at t=0, minus the moles present in cells at t=1
0.000000065 - 0.00000003164174858 = 0.00000003335825142 moles not in cells
4. Calculate the concentration in solution at t=1
This can be calculated by the number of moles not present in cells, divided by the (total volume - internal cell volume)
The volume can be calculated as 10250μl - 15.5μl = 10234.5μl = 0.0102345l
0.00000003335825142 moles / 0.0102345l = 0.000003259392391 M
5. Calculate the number of moles removed at t=1, in the 1ml, that isn't present within cells
We can work this out by multiplying the concentration by the volume of liquid removed (1ml - 1.512μl)
1ml - 1.512μl = 0.998488 = 0.000998488l
0.000003259392391 M * 0.000998488l = 0.00000000325446419 moles removed not in cells
6. Calculate number of moles removed at t=1
We calculate this by adding the number of moles removed that weren't in cells from the number of moles removed that were in cells
0.00000000325446419 moles not in cells, but in 1ml extraction + 0.000000003087 moles in extracted cells = 0.00000000634146419 moles removed at t=1
7. We repeat the process for t=5, and t=10, results on the attached sheet, but account for the moles and cells already removed.
Full worked example for t=5 follows (for experiment 1, cpm1):
832 CPM reported = 822 CPM (- background radiation)
0.0000000026/(566/822) = 0.000000003776 moles required for CPM
Total moles in all cells (accounting for cells removed in t=1) = 0.000000003776/(0.09756098/(1-0.09756098)) = 0.00000003492799826 moles
Total moles not in cells at t=5 = 0.000000065 (total added at t=0) - 0.00000003492799826 (total moles in all remaining cells at t=5) - 0.00000000634146419 (moles removed in t=1) = 0.00000002373053755 moles
External concentration at t=5 = 0.00000002373053755 (total moles not in cells) / ( (0.0102345-0.000998488) (Total liquid in flask - liquid removed at t=1) - (0.0000155-0.000001512) (Total internal cell volume - cell volume removed in t=1) ) = 0.000002573246128 M
Moles present in 1ml extraction, not in cells = 0.000002573246128 (molarity of liquid in flask at t=5) * 0.000998488 (Liquid in 1ml extraction - internal cell volume in 1ml extraction) = 0.00000000256935538 moles
Total number of moles removed at t=5 = 0.00000000256935538 (Moles in liquid in 1ml extraction) + 0.000000003776 (Moles in cell volume in 1ml extraction) = 0.00000000634535538 moles
Full worked example for t=10 follows (for experiment 1, cpm1):
851 CPM reported = 841 CPM (- background radiation)
0.0000000026/(566/841) = 0.000000003863 moles required for CPM
Total moles in all cells (accounting for cells removed in t=1 and t=5) = 0.000000003863/(0.09756098/(1- (2*(0.09756098))) = 0.00000003186974822 moles
Total moles not in cells at t=10 = 0.000000065 (total added at t=0) - 0.00000003186974822 (total moles in all remaining cells at t=10) - 0.00000000634535538 (moles removed in t=5) - 0.00000000634146419 (moles removed in t=1) = 0.00000002044343221 moles
External concentration at t=10 = 0.00000002044343221 (total moles not in cells) / ( (0.0102345-(2*0.000998488)) (Total liquid in flask - liquid removed at t=1 and t=5) - (0.0000155-(2*0.000001512)) (Total internal cell volume - cell volume removed in t=1 and t=5) ) = 0.000002485509168 M
Moles present in 1ml extraction, not in cells = 0.000002573246128 (molarity of liquid in flask at t=10) * 0.000998488 (Liquid in 1ml extraction - internal cell volume in 1ml extraction) = 0.000000002481751078 moles