What is kcat in biochemistry

In situations where k -1 the rate at which substrate unbinds from the enzyme is much greater than k 2 the rate at which substrate converts to product , if the rate of efficiency is: HIGH, k cat is much larger than K M , and the enzyme complex converts a greater proportion of the substrate it binds into product. This increased conversion can be seen in one of two ways -- either substrate binds more firmly to the enzyme , a consequence of relatively low K M , or a greater proportion of the substrate that is bound is converted before it dissociates, due to a large turnover rate k cat.

LOW, k cat is much smaller than K M , and the complex converts a lesser proportion of the substrate it binds into product. References [ edit edit source ] Berg, Jeremy M. Sixth Ed.

New York: W. Category : Book:Structural Biochemistry. Namespaces Book Discussion. Views Read Edit Edit source View history. If you have several experimental conditions, place the first into column A, the second into column B, etc. You can also choose Prism's sample data: Enzyme kinetics -- Michaelis-Menten. After entering data, click Analyze, choose nonlinear regression, choose the panel of enzyme kinetics equations, and choose Kcat.

You must constrain Et to a constant value, based on other experiments. To constrain the value of Et, go to the Constrain tab of the nonlinear regression dialog, make sure that the drop down next to Et is set to "Constant equal to" and enter the value. For the sample data, enter as the value of Et. If you don't know the value of Et, you cannot fit the kcat, but instead should fit the Vmax. It is not possible for Prism to fit both the kcat and Et, as the two parameters are intertwined, and a substrate-velocity curve gives no information about their individual values.

The best fit value of kcat is The product of Kcat times Et the concentration of enzyme sites equals the Vmax. That allosteric model adds an additional parameter: the Hill slope h. When h equals 1. All rights reserved. Introduction Kcat is the turnover number -- the number of substrate molecule each enzyme site converts to product per unit time. K cat , used to describe the limiting rate of any enzyme-catalyzed reaction at saturation.

Secondly K m is HALF of V max so actullay at half saturation but K cat is at full saturation, deviding these two numbers makes no sense to me.

So to summerize I'm asking : What is the "meaning" of the division of these two consants? In fact, when you get into more complex rate equations like inhibitors, pH effects, and kinetic isotope effects there's a decent argument one often made by W.

The reason for this is that if you have a single enzyme in the presence of two different substrates, you have a competitive inhibition setup. This actually makes intuitive sense, with the right mindset - the only stage where the two substrates are competing where you make the decision to do reaction A versus reaction B is when they're binding to free enzyme. With negligible substrate, all you have is free enzyme - there isn't enough substrate to have appreciable amounts of substrate-bound form, and the rate of enzyme-substrate encounter is much lower than the rate of product formation, meaning that in the steady state you don't have appreciable amounts of product-bound form, either.

This ratio is more often called the "catalytic efficiency" and it's a bad measure. In many cases, enzymes are only better when they have a higher kcat , because the substrate concentration is always way above the Km. Using the ratio when this is obviously the case is just misleading imo, but this happens way too often.

In some cases it could be a useful measure to compare to enzymes. For example if you're unable to measure at higher substrate concentrations. You'd just have the initial slope of the [S] vs vmax graph. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.