at what substrate concentration is v= 5.0 mM s^-1 for the enzyme catalyzed hydrolysis of trehalose?

If Vmax for a reaction is 10 μM ⋅ s-1 and the KM  is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?

You are working on an enzyme that obeys standard Michaelis-Menten kinetics. What variable is the V, dependant on if the concentration of the substrate is substantially higher than the concentration of the enzyme? [S] [E] [ES] O [P] O not enough information provided

An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s

what concentration of a competitive inhibitor is required to yield 60% inhibition of an enzyme at a substance concentration of 1.0x10^-4 M. The Km for the enzyme is 22 uM and the Ki for the inhibitor is 59 uM. To what concentration must the substrate be increased to reestablish velocity at the uninhibited value?

The Ki for a competitive inhibitor pyrrole-2-carboxylate is 0.05M.Calculate the initial velocity if the substrate concentration is 0.25M and 1.0 mole/liter of the inhibitor is added to thereaction.

. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins of bacterial wall) are 37 C - temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. Concidor NH3: 5.

Chymotrypsin is an enzyme that is optimized to function in the small intestine (at a pH of ~7.4). Explain, given what you know about the chymotrypsin mechanism and what we learned about acid/base chemistry, why chymotrypsin would not be able to initiate catalysis if it were secreted into the stomach (pH ~2.5).

When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

Calculcate Kcat for PNP substrate for both enzyme concentrations.  enzyme volume: 20 ul  Bovine Intensince Alkaline phosphatase molecular weight: 140,000 Bovine intenstine Alkaline phosphatase activity: 300 units/ml and 14 units/mg extinction coefficient PNP: 18.5 abs (mM-1 cm-1) Vmax: 0.332 moles/sec a) enzyme 1 concentration: undiluted b) enzyme 2 concentration: 1:1 dilution