You isolate an enzyme from teddy bear liver. The enzyme catalyzes the following

Question

You isolate an enzyme from teddy bear liver. The enzyme catalyzes the following reaction: A + B rightarrow C + D You determine the Km for A to be 3 mu M, and for B to be 25mM. What are the approximate physiological concentrations of A and B in teddy bear cells? The V max for A is 2.7 mu moles/min/mg enzyme, what is the Vmax for B? The molecular weight of the enzyme is 40 What is the turnover number (K cat) for the enzyme? (Units = mu moles of substrate/s/mu mole of protein). You have an assay mix containing 1 mu M A, 100 mM B, and 0.2 mg of enzyme. What is the rate of formation of product under these conditions (B is saturating, A is not) (in mu moles/min/mg enzyme)?

Explanation / Answer

Km is count for the half of Vmax, at where the half of the total catalytic sites of an enzyme is accomplished by the substrate. Here we have given Km of A is 2um which means 2 uM of substrate A is sufficient to bind the half of the total catalytic sites. It means at least double of the concentration of substrate must to present in the cell.

So the approximate physiological concentration for substrate A is 4 um and B is 50mM.

64 Kdalton protein concentration would be = moles = grams/ molecular weight

= 0.2mg/ 64Kd = 0.2 x 10-3g/ 64 x 103 dalton

= 0.2/64 x 10-6 moles = 0.003125 x 10-6 moles

=3.125 x 10-3 x 10-6 moles = 3.125 nmoles

To calculate Kcat, we do it by Vmax/Et where Et is enzyme concentration in the assay.

So we have given 0.2 mg of enzyme and for substrate A is 2.7umoles/3.125 nmoles per minute

= 2.7 x 103 nmoles/ 3.125 nmoles per minute

= 0.864 x103 per minute

= 0.864 x103 / 60 per second

= 0.0144 x103 sec-1 = 14 sec-1

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