1. Human alcohol dehydrogenase (ADH) is a dimer with a mass of 80 kDa. A prepara
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Question
1. Human alcohol dehydrogenase (ADH) is a dimer with a mass of 80 kDa. A preparation of ADH was assayed in a 1-ml cuvet using UV/VIS spectrophotometer. The assay contained 10 mu l of a 2 mug/ml purified protein solution. alcohol + NAD + --> 4 acetylaldehyde + NADH + H+ The molar extinction coefficient of NADH at 340 nm is 6.2 x 10^4 M^-1 cm^-1 The absorbance at 340 nm as a function of time was: time (sec) 0 30 60 90 120 absorbance (Abs) 0.025 0.055 0.125 0.170 0.260 a) Plot Abs vs. time. b) Determine the velocity in mu moles per minute (units). c) Determine the specific activity in units per mg. d) Determine the turnover number (Kcat). 2. The enzyme fumarase catalyzes the reaction: fumarate + H2O --> malate Km is 2.5 x 10^-5 M and Vmax is 225 mu mol-min^-1-mg^-1. (To answer these questions, please read lecture notes on inhibition kinetics and find out how a presence of inhibitors can change Km and Vmax depending upon the type of inhibition). a. Succinate is a competitive inhibitor with Ki = 2.5 x 10^-5 M. b. Iodate is a noncompetitive inhibitor with Ki = 2.0 x 10^-3 M. a) Calculate the new values of Km and Vmax in the presence of 2.0 x 10^-3 M iodate. b) Calculate the new values of Km and Vmax in the presence of 2.5 x 10^-5 M succinate.Explanation / Answer
Slope = 0.117 Abs./min
Velocity = {0.117/(6.2x104*1x10-4)}*{0.01/1000} = 0.19 micromoles/min
Specific activity = rate of reaction*volume of solution/mass of protein = 0.19/2x10-5 = 9500 micromoles/min-mg
turnover number = 9500/80000 = 0.12 min-1
2.
1) Slope = 0.117 Abs./min
Using Beer-Lambert Law is, rate of reaction = (0.117)/(1*6.2x104) = 1.89 x 10-6 M/min
Velocity = 1.89 mM/min *1x10-5 = 1.89x10-5 mmole per min
Specific activity = rate of reaction / concentration of protein = 0.945 mmole per min per mg
Turnover number = Specific activity*molecular weight = 0.945*80000 = 75600 mmole/min - mmol or 1.26 mmole/s - mmol
2) For competitive enzymes, Vmax is same
Therefore, Vmax = 225 mmol/min-mg
And Km = Km(1-0.444) = 1.39x10-5
For noncompetitive enzyme inhibitors, Km is same = 2.5 x 10-5
And Vmax = 225(1-2x10-3/(2x10-3 + 2x10-3) = 112.5
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