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Light Absorbance

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Here we give a worked solution for a particular set of numbers:
Note: $N_A$ refers to Avogadro's number.



An initial statement of the Beer-Lambert Law:

$A = \epsilon cl$


Differentiating this gives:

$\frac{dA}{dt} = \epsilon l \frac{dc}{dt}$

$\therefore \frac{dc}{dt} = \frac {1}{\epsilon l} \frac{dA}{dt} = \frac{1}{6220 \times 2} \times 0.6 = 4.82 \times 10^{-5} M\ min^{-1}$



In the cuvette:
Rate of change of NADH = $4.82 \times 10^{-5} M\ min^{-1}$

$=\ 4.82 \times 10^{-5} \times V_{cuvette}\ mol\ min^{-1}$

$=\ 4.82 \times 10^{-5} \times \frac{1.1 +0.6 + 0.9 +1.57 + 0.03}{1000}\ mol\ min^{-1}$

$= 2.03 \times 10^{-7}\ mol\ min^{-1}$


Scaling up to extract:
Rate of change of NADH = $2.03 \times 10^{-7}\times \frac{V_{extract}}{V_{taken}} \ mol\ min^{-1}$

$= 2.03 \times 10^{-7}\times \frac{5}{0.03} mol\ min^{-1}$

$ = 3.38 \times 10^{-5} mol\ min^{-1}$

$= 5.63 \times 10^{-7} mol\ s^{-1}$

$ = 5.63 \times 10^{-7} katal$

$\mathbf{ \therefore Activity = 0.563\ \mu katal}$


Concentration of Malate Dehydrogenase:
$Activity = 3.38 \times 10^{-5} \times N_A = 2.03 \times 10^{19} molecules\ min^{-1}$
$Activity = Turnover \times [E]$
$\therefore [E] = \frac{Activity}{Turnover}$
$[E] = \frac{2.03 \times 10^{19}}{100000} = 2.03 \times 10^{14} molecules_{enzyme}\ extract^{-1}$
$= \frac{2.03 \times 10^{14}}{N_A}\ mol_{enzyme}\ extract^{-1} = 3.38 \times 10^{-10} mol_{enzyme}\ extract^{-1}.$
$ = 3.38 \times 10^{-10} \times 70000 = 2.36 \times 10^{-5}\ g_{enzyme}\ extract^{-1}$

$=2.36 \times 10^{-5}\ g_{enzyme}\ (5ml)^{-1}$

$\mathbf{=4.73 \times 10^{-3} mg\ ml^{-1}}$