2. Carbonic anhydrase catalyzes the reversible hydration of carbon dioxide to fo

Question

2. Carbonic anhydrase catalyzes the reversible hydration of carbon dioxide to form bicarbonate ion: This reaction is important in the transport of carbon dioxide from body tissues to the lungs by red blood cells. Carbonic anhydrase h as a molecular weight of 27,000 and a turnover number (kcat value) of 1x105 that you are given 1 mL of a solution containing 2.0 g of pure carbonic anhydrase. Part A: At what rate (in millimoles of COz consumed per second) will you expect this reaction to proceed under optimal (maximal) conditions? (Show all your calculations with units and express your answer to four decimal places.) (5 pts) Part B: Assuming standard temperature and pressure, how much CO2 is all your calculations with units and express your answer to one decimal place. COz is a gas.) (3 pts) mL per "Hint-Remember that second? (Show

Explanation / Answer

part A) kcat=turnover number=Vmax /[ET] (kcat refers to number of catalytic cycles for each active site per unit time of the enzyme)

where Vmax=maximum velocity of enzyme-substrate rxn

[ET]=total enzyme concentration=2.0 ug/1ml=2.0 ug/ml=(2.0*10^-6 g/27000 g/mol) per ml

[mol=mass/molar mass]

So, [ET]=7.407*10^-11 mol/ml=7.407*10^-11 mol/ml *(1000 ml/L)=(7.407*10^-8 mol/L)

So Vmax=kcat*[ET]=(1*10^5 s^-1) *(7.407*10^-8 mol/L)=7.407*10^-3 mol L^-1 s^-1=7.407 mmol L^-1 s^-1

Vmax=7.407*10^-3 mol L^-1 s^-1

part B) maximal velocity =Vmax=7.407 mmol s^-1 per L

So in 1ml, Vmax=7.407 mmol s^-1 per L *(1000ml/L)=7.407 mol s^-1 per 1ml

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