Competitive inhibition: Vmax remains the same and Km Changes Non-competitive (pure): Vmax changes and Km remain the same
The units of Vmax and V are amount of product over time, typically µmol/min or similar. The source is linked.
The lineweaver burk plot is: y= 1/v x=1/[S] m=Km/Vmax B=1/Vmax This plot is a linear plot and follows the y=mx+b equation
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.
Vmax
very much so.
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.
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.
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Yamaha started making the VMAX in 2008. The VMAX, however, is no longer being manufactured in 2013 though and is out of production at the moment with no plans for future production.
KM would not change, since it is a constant. Vmax would half, because Vmax depends on the concentration of the enzyme.
Lactate Dehydrogenase
Competitive inhibition: Vmax remains the same and Km Changes Non-competitive (pure): Vmax changes and Km remain the same
The units of Vmax and V are amount of product over time, typically µmol/min or similar. The source is linked.
Mass/Synthesised pressure = SHM Vmax
Yes, Vmax has a linear relationship with the amount of enzyme. This in turn deceases the Km of the reaction.
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