Question

3.Your task is to determine the concentration of dopamine in a solution of homogenized brain tissue; i.e. determine the concentration of dopamine in an extremely complex matrix of brain tissue run through a blender that you are unable to reproduce such that you could construct a traditional calibration curve –thus you will utilize the method of standard additions. You have verified that your method and instrument yield a linear electrical current response that is directly proportional to the concentration of dopamine in the overall solution. You have a 50 mL aliquot of the brain solution to which you spike witha 0.279 nM dopamine solution. Utilize the table of data and the graphical procedure for standard addition to determine the concentration of dopamine in the initial brain solution.

Total spike volume added [mL]	Detector Signal [uA]
0.000	1.78
0.050	2.00
0.250	2.81
0.400	3.35
0.550	3.88
0.700	4.37
0.850	4.86
1.000	5.33
1.150	5.82

Answer 1

From the data in table, when no exogenous dopamine has been added, the detector signal was 1.78uA. As Dopamine is increased detector signal also increased in linear fashion (graph1). So if we subtract the initial detector signal (when 0.0 ml of dopamine was spiked), we will get detector signal only from external dopamine added. From the subtracted values, calibration curve can be plotted with number of moles of dopamine and their respected detector signals (graph 2). Thus from equation of calibration curve, we can find the number of moles of dopamine molecule which corresponds to detector signal 1.78 uA. Hence, molarity of dopamine in 50 ml of brain tissue solution which is 2.68 pico molar.

Calculations

0.05 ml of 0.279 nM dopamine solution contains 13.95 femto moles of Dopamine. Similarly, we can calculate number of moles for each volume added to the solution as shown table below with their respective detector signal.

Spiked dopamine (ml)	Moles of Dopamine (fmole)	Detector signal (uA)
0.0	0	0
0.05	13.95	0.22
0.25	69.75	1.03
0.4	111.6	1.57
0.55	153.45	2.1
0.7	195.3	2.59
0.85	237.15	3.08
1	279	3.55
1.15	320.85	4.04

For detector signal 1.78 uA,

Putting y=1.78 value in equation y=0.0125x + 0.0993, getting value of x =134.456 fmol. Hence calculate the molarity for 134.456 fmol in 50 ml of solution, which comes around 2.68 pico Molar.