·Use the average initial rate of your IEX-LDH for the following calculation Init

Question

·Use the average initial rate of your IEX-LDH for the following calculation

Initial average rate of IEX-LDH f is 1.90 x 10^-3 OD/sec

Convert from sec to min: (OD/sec) (60 sec/min) = 0.114 OD/min
(1.90 x 10^-3 OD/sec) * (60 sec/min) = 0.114 OD/min

· Use Beer’s Law (A = l c; see Lab03) to convert OD into concentration (M): [OD/min]/[ x l] = M/min [Remember: M stands for molar = mol/liter]

Convert from concentration to amount by including volume of substrate cocktail (2 x 10-3 l) in the cuvette in your calculation: (M/min) x (vol) = ______ mol/min

Enzyme units (U) are in units of mol/min, so you need to convert from mol to mol for the final answer in U: (mol/min) x (106 mol/mol) = ______ U. This is the number of U of LDH in your cuvette.

Explanation / Answer

According to beer's law A = l c

A=  0.114 OD/min

mol/min = (M/min) x (vol)

0.114*2 x 10-3 mol/min

0.228*10-3

U = (mol/min) x (106 mol/mol)

0.228*10-3 *106

0.228*103

228  mol/min

So the Unit is 228  mol/min

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