5) (10 pts) Dopamine exocytosis can be monitored with amperometry using microele

Question

5) (10 pts) Dopamine exocytosis can be monitored with amperometry using microelectrodes. You are performurones experiment monitoring dopamine release from single grown in culture. Your microscopy measurements sahs dopamine extracellular volume is 15.3 nanoLiters (nL) where released and the working electrode (and reference) is ps Each dopamine treatment with a dru over a total of 1 second. Consider that: 1 Amp Faraday's Constant = 9 85 x 104 C/mol (charge of one suggest the the dopamine is molecule produces 2 electrons when oxidized. After g solution, you measure a current of 52.7 nA and a) Based on the detected current, how event? of W many moles of dopamine are released during the exocytosis b) How many molecules of dopamine are released? c) What is the local concentration (M) of dopamine after the vesicle releases the dopamine into the extracellular volume?

Explanation / Answer

A)

Current of 52.7nA (5.27e-8Amp) was measured over 1second. Hence the charge measured was 5.27e-8C.

Now 96485C is the charge of 1 mole electrons.

Hence, 5.27e-8C is the charge of 14.325 moles of electrons i.e 14.3 X 6.023e23 electrons(Avogadro's number).

Now 2 electrons are produced by each dopamine molecule (as given in the question).

Hence 14.3 X 6.023e23 electrons will be produced by 7.15 X 6.023e23 dopamine molecules.

Dividing by Avogadro's number, 7.15moles of dopamine are released.

B)

Available from the above calculation, 7.15 X 6.023e23 dopamine molecules are released.

C)

15.3nL (15.3e-9L) is the extracellular volume.

Now molarity is defined as moles per litre of solution.

Thus local concentraion of dopamine is 5e-5M

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