(JUNE 2015)
40. A 1% (w/v) solution of a sugar polymer is digested by an enzyme (20µg, MW=200,000). The rate of monomer sugar (MW=400) liberated was determined to have a maximal initial velocity of 10 mg formed/min. The turnover number (min^-1) will be
(1) 5 x 10⁴                                            (2) 2.5 x 10^-2
(3) 4.0 x 10^-6                                        (4) 2.5 x 10⁵

The correct answer is (1) 5 x 10^4

Introduction

The turnover number, or kcatkcat, represents the number of substrate molecules an enzyme converts into product per unit time under saturating substrate conditions. Calculating kcatkcat requires enzyme amount and the maximal velocity measurement, along with molecular weight information to convert mass measurements to molar quantities. This article demonstrates this calculation with given data, elucidating key concepts and formula applications for biochemical kinetics.

Given Data

• Enzyme amount = 20 μg=20×10−6μg=20×10−6 g

• Enzyme molecular weight (MW) = 200,000 g/mol

• Maximal velocity (VmaxVmax) = 10 mg monomer sugar/min

• Monomer sugar molecular weight = 400 g/mol

Step 1: Calculate moles of enzyme used

Moles of enzyme=massmolecular weight=20×10−6 g200,000 g/mol=1×10−10 molMoles of enzyme=molecular weightmass=200,000 g/mol20×10−6 g=1×10−10 mol

Step 2: Calculate moles of product formed per min

Convert mg of monomer sugar formed per minute to moles:

Product in moles/min=10×10−3 g/min400 g/mol=2.5×10−5 mol/minProduct in moles/min=400 g/mol10×10−3 g/min=2.5×10−5 mol/min

Step 3: Calculate turnover number kcatkcat

Turnover number is moles of product formed per mole of enzyme per minute:

kcat=Product mol/minEnzyme mol=2.5×10−51×10−10=2.5×105 min−1kcat=Enzyme molProduct mol/min=1×10−102.5×10−5=2.5×105min−1

So calculated kcat=2.5×105kcat=2.5×105 min−1−1.

Re-Examination of Provided Options

The closest option is (4) 2.5×1052.5×105, but answer (1) 5×1045×104 probably assumes some rounding or difference in interpretation.

If the problem assumes one mole of enzyme can cleave only half or if it converts product differently, the answer might differ. However, based strictly on calculation above, option (4) is exact from data.

Clarification

It is likely that option (1) 5×1045×104 assumes some factor not given explicitly (e.g., enzyme activity per active site if enzyme is multimeric). Without that, the direct calculation matches option (4).

Summary Table

Conclusion

Based on the calculation of moles of product formed and moles of enzyme present, the turnover number or kcatkcat is 2.5×1052.5×105 min−1−1, corresponding to option (4). This high turnover indicates extremely efficient enzymatic catalysis, typical for enzymes working under saturating substrate conditions.