15. Enzyme catalysed reaction is first order kinetics so so please concentration of substrate X is doubled then rate of reaction will be doubled.
16. The rate determining step is substrate + enzyme gives ES.
So reaction is first order kinetics with respect to enzyme also so if the concentration of enzyme is doubled then the rate of reaction will be two times and if the enzyme concentration is reduced to half then the rate of reaction will become half.
The concentration of substrate X is low. What happens to the rate of the enzyme- catalyzed reaction if the concentr...
The rate of an enzyme-catalyzed reaction initially increases with an increase in the substrate concentration, but eventually reaches a maximum value, even though the concentration of substrate continues to increase. Which of the following best explains why? O As substrate concentration increases, the substrates preferentially bind with each other instead of the active site of the enzyme, and no additional catalysis occurs. As substrate concentration increases, the active sites of all the enzyme molecules become occupied with substrate molecules, and...
2-In an enzyme-catalyzed reaction, the rate of the reaction depends on which of the following at very low substrate concentrations? Select one: Neither enzyme concentration nor substrate concentration Enzyme concentration but not substrate concentration Substrate concentration but not enzyme concentration Both substrate concentration and enzyme concentration
The initial rate, V, of an enzyme catalyzed reaction varies with substrate concentration as follows: 106 x Initial rate, Ms SJ, M 0.020 0.585 0.004 0.495 0.002 0.392 0.001 0.312 0.250 0.00066 Determine Vmax and Km for this reaction
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
Under what circumstances does an enzyme catalyzed reaction rate resemble a non-enzyme catalyzed reaction? At very low concentrations of substrate (Km is greater than S) the Michaelis-Menton equation can be simplified to? At very high concentrations of substrate, the Michaelis-Menton equation can be simplified to? How do you determine the initial rate of reaction
Uncatalyzed Catalyzed Enzyme-substrate Complex In the above reaction, the lower curve is an enzyme-catalyzed reaction where the activation energy is notably lower than the uncatalyzed reaction. Suppose the enzyme in the diagram was mutated in such a way that its affinity for the substrate increased 100 fold, thereby affecting the enzyme-substrate complex portion of the curve. Assume that there was no other effect. Would you expect the reaction rate catalyzed by the altered enzyme to be faster, slower, or equal...
An enzyme catalyzed reaction is studied at substrate concentration that is equal to 8Km. What will be the value of V0 in terms of fraction of Vmax?
Vmax of an enzyme-catalyzed reaction is A. the rate observed when the enzyme active sites are saturated with substrate B. independent of the amount of enzyme present C. the rate observed at the highest substrate concentration that can be experimentally obtained D. the initial rate observed at very low substrate concentrations
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
What is the rate of an enzyme catalyzed reaction if the Vmax is 100µmol S→P/min and the Km is 7 mM and the substrate concentration is 11mM? Is the enzyme working at Vmax? What if the substrate concentration is raised to 25mM?