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Enzyme Kinetics: Turnover Number

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2. Carbonic anhydrase of erythrocytes (Mr 30,000) is among the most active of know enzymes. It catalyzes the reversible hydration of CO2:
H2O + CO2 H2CO3

which is important in the transport of CO2 from the tissues to the lungs. If 10g of pure carbonic anhydrase catalyzes the hydration of 0.30 g of CO2 in 1 min at 37oC under optimal conditions, what is the turnover number (kcat) of carbonic anhydrase (in units of min-1)?

3. The following data describe the catalysis of cleavage of peptide bonds in small peptides by the enzyme elastase.

Substrate Km (mM) kcat (s-1)
PAPAG 4.0 26
PAPAA 1.5 37
PAPAF 0.64 18

The arrow indicates the peptide bond cleaved in each case.
(a) If a mixture of these three substrates were presented to elastase with the concentration of each peptide equal to 0.5 mM, which would be digested most rapidly? Which most slowly? (Assume the elastase is present in excess.)
(b) On the basis of these data, what molecular property of an amino acid sequence appears to dictate the specificity of cleavage by elastase. Your answer should go beyond listing the presence of one or more amino acids to a conclusion as to what about the amino acid(s) is causing the specificity.

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Question 1: Enzyme Kinetics: Turnover Number

Carbonic anhydrase of erythrocytes (Mr 30,000) is among the most active of know enzymes. It catalyzes the reversible hydration of CO2...

H2O + CO2 <-------> H2CO3

...which is important in the transport of CO2 from the tissues to the lungs.

If 10 mg of pure carbonic anhydrase catalyzes the hydration of 0.30 g of CO2 in 1 min at 37 degrees C under optimal conditions, what is the turnover number (kcat) of carbonic anhydrase (in units of min-1)?

Response:

The constant kcat is a first-order rate constant with units of reciprocal time. It is also called the "turnover number." In other words, it is equal to the number of substrate molecules that one molecule of enzyme can convert into product in a given amount of time, assuming that the enzyme is saturated with substrate.

First, we need to find out the number of molecules of enzyme in 10 mg of the enzyme.

(6.023x10^23 molecules/mol)(1 mol/30,000 g)(10 mg)(1 g/1000 mg) = 2.01x10^17 molecules of ...

Solution Summary

Solution provides detailed lengthly description of active enzymes and how they catalyze in a variety of settings.

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Understanding the Michaelis-Menten Equation, Enzyme Kinetics

For an enzyme which obeys Henri-Michaelis-Menten kinetics,
(1) At what substrate concentration will an enzyme characterized by a kcat of 30 s-1 and Km of 0.005 M show one quarter of its maximal rate?

(2) Calculate the fraction of Vmax that would be found when [S]=0.5 Km, [S]=3Km, and [S]=10Km

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