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Yes, Vmax has a linear relationship with the amount of enzyme. This in turn deceases the Km of the reaction.
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∙ 12y ago
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Is Vmax a threshold of substrate concentration for initiation of an enzymatic reaction?
Oddly phased question in my opinion. Vmax is only effected by the amount of enzyme present in the reaction. Substrate concentration has zero effect on Vmax. There for I believe the answer in no. {Enzyme concentration is responsible for this}
What is the type of enzyme inhibition in which the Km changes but the Vmax does not?
competitive
What happens to the vmax when a competitive reversible inhibitor is added to an enzyme?
The vmax stays the same as the competitive reversible inhibitor does not affect catalysis in the enzyme-substrate.
The initial rate of an enzyme catalysed reaction depend on?
Based on Michaelis-Menten enzyme kinetics, the initial rate of reaction, vi, is dependent on maximum rate Vmax, substrate concentration [S], and the enzyme's Michaelis constant Km, which represents the the tendency of the substrate/enzyme complex to dissociate. The dependence on enzyme concentration is factored into the maximum rate. The equation to describe this is: vi = Vmax([S]/(Km+[S])) Follow the link below for details.
When the active site of an enzyme is continuously filled and the maximum enzyme rate has been reached what happens if more substrates are added to the reaction?
Nothing. The Vmax remains constant and the graph produced at this point would show your curve flattening at a plateau.
What does increased Vmax suggest?
An increase in Vmax suggest an increase in the amount of enzyme in the reaction. Also this increase in Vmax deceases the Km vaule, which means less substrate is needed.
How do you estimate the value of Vmax for two different concentrations of the substrate?
The Vmax would be the highest rate, when the enzyme is fully saturated. So as you increase substrate the Vmax will increase to a certain point (Vmax). Beyond that point, no matter how much substrate you add the Vmax will not increase.
Is Vmax a threshold of substrate concentration for initiation of an enzymatic reaction?
Oddly phased question in my opinion. Vmax is only effected by the amount of enzyme present in the reaction. Substrate concentration has zero effect on Vmax. There for I believe the answer in no. {Enzyme concentration is responsible for this}
Suppose that the enzyme concentration was increased by a factor of four what would be the value of km and vmax?
KM would not change, since it is a constant. Vmax would half, because Vmax depends on the concentration of the enzyme.
How do you determine Vmax in enzyme kinetics?
Vmax is the maxim initial velocity (Vo) that an enzyme can achieve. Initial velocity is defined as the catalytic rate when substrate concentration is high, enough to saturate the enzyme, and the product concentration is low enough to neglect the rate of the reverse reaction. Therefore, the Vmax is the maximum catalytic rate that can be achieved by a particular enzyme. Km is determined as the substrate concentration at which 1/2 Vmax is achieved. This kinetic parameter therefore importantly defines the affinity of the substrate for the enzyme. These two parameters for a specific enzyme defines: Vmax - the rate at which a substrate will be converted to product once bound to the enzyme. Km - how effectively the enzyme would bind he substrate, hence affinity.
How might amino acid change affect the enzyme?
The pKA of enzyme affects its ionization which could alter enzyme activity. For pH < pKa, the value of vmax is constant and that for pH > pKa, vmax decreases; ie. enzyme activity starts to decline.
What is the type of enzyme inhibition in which the Km changes but the Vmax does not?
competitive
What happens to the vmax when a competitive reversible inhibitor is added to an enzyme?
The vmax stays the same as the competitive reversible inhibitor does not affect catalysis in the enzyme-substrate.
What are k-series and v-series enzymes?
A K-Series enzyme is an enzyme that changes the Km of an enzyme on the michaelis menten or lineweaver burk graphs A V-series enzyme affects the Vmax
What is the meaning and the unit of Vmax?
The units of Vmax and V are amount of product over time, typically µmol/min or similar. The source is linked.
The initial rate of an enzyme catalysed reaction depend on?
Based on Michaelis-Menten enzyme kinetics, the initial rate of reaction, vi, is dependent on maximum rate Vmax, substrate concentration [S], and the enzyme's Michaelis constant Km, which represents the the tendency of the substrate/enzyme complex to dissociate. The dependence on enzyme concentration is factored into the maximum rate. The equation to describe this is: vi = Vmax([S]/(Km+[S])) Follow the link below for details.
How do you calculate turnover number for an enzyme?
As enzyme concentration increases the more active sites there are avalible, so the rate of reaction increases. therefore the turnover number increases.Hope it helped!TashaThe above it not true. The turn over number is Vmax/Et so if the enzyme concentration is doubled the velocity will also be doubled. Therefore the turn over number will remain constnat.
