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Discussion Starter · #2 ·
The best simple explanation that I found

"Vmax is the maxim intial 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." From answer.com

Hence higher the Km lower the affinity such as with Glucokinase and reverse with Hexokinase.
 

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Discussion Starter · #3 ·
While I am at it

"Vmax is the maxim intial 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." From answer.com

Hence higher the Km lower the affinity such as with Glucokinase and reverse with Hexokinase.
Now look at the Lineweaver-Burke plot of 1/Vo vs. 1/[substrate], aka the double reciprocal plot. The important things to remember about Lineweaver-Burke plots are the x and y intercepts. The x-intercept = -1/Km, and the y-intercept = 1/Vmax. Just learn these, and I'll help you make sense of them by discussing inhibition.

InhibitionThe best way to understand these graphs is to look at what happens with different types of inhibition.

"First, think about competitive inhibition. You've got another substrate competing for the same enzyme. So what changes? Well, the enzyme suddenly has something else it can bind to, so its affinity for the substrate is reduced. At the same time, if you cram in enough substrate to overwhelm the competition, you can eventually reach Vmax. So in competitive inhibition, Km increases while Vmax remains the same. Look at your V/ graph, and the curve will stretch, because it takes a lot more substrate to get that Km at 1/2 Vmax. Look at your Lineweaver-Burke plot. The y-intercept stays the same because Vmax doesn't change. But Km has gone up, which means that -1/Km has gotten closer to zero, increasing the slope of the line and rotating it on the y-axis.

Now look at noncompetitive inhibition. In noncompetitive inhibition, you have something binding to another site on the enzyme, changing the structure of the binding site, and thus affecting the amount of enzyme that is able to bind substrate. This means that Vmax is reduced. Km, the affinity of the functional enzyme, remains the same, though. Looking at the V/ curve, you simply squish the maximum down. Looking at Lineweaver-Burke, Km is the same, so your x-intercept doesn't move. Vmax is smaller, so 1/Vmax is larger. This means that your line will have a higher slope and rotate on the x axis."

There are other conditions possible, but that covers the basics.
 
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