What are the Kcat,Km,Kcat/Km enzymology values?

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Kcat, Km, and kcat/Km are key enzymology values that describe enzyme activity and efficiency. Kcat represents the maximum rate of substrate conversion when the enzyme is saturated, while Km indicates the substrate concentration at which the reaction rate is half of kcat, reflecting the enzyme's affinity for the substrate. Kcat/Km is a measure of enzyme efficiency, combining both the turnover rate and substrate affinity. To calculate substrate conversion rates, the Michaelis-Menten equation can be used, and specific activity relates to kcat as it measures enzyme activity per unit mass. Understanding these parameters is crucial for optimizing enzyme reactions in various applications.
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Hello!

Can someone explain in simple terms the enzymology values of kcat, km and kcat/km?

From what I gathered all mean how many substrate molecules are converted by an enzyme molecule per unit of time, is this correct? (let's say the conversion rate)

The kcat is the conversion rate when there is maximum substrate concentration that saturate the enzyme.
The km is the conversion rate when there is half of the saturation concentration of the substrate.
the kcat/km is the conversion rate when there is minimum substrate concentration.

Are these accurate? What do we mean by minimum substrate concentration? How can I find the exact
conversion rate when I know the kcat/km and the substrate concentration of my solution?

I basically want to calculate the substrate conversion rate per mg of enzyme per minute for a specific concentration of the two substrates. What do I need to do that?

Also, there is specific activity as well, is this the same with kcat/km?

Thanks!
 
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cyentist said:
Hello!

Can someone explain in simple terms the enzymology values of kcat, km and kcat/km?

From what I gathered all mean how many substrate molecules are converted by an enzyme molecule per unit of time, is this correct? (let's say the conversion rate)

The kcat is the conversion rate when there is maximum substrate concentration that saturate the enzyme.

This is a correct interpretation of kcat. You can think of kcat as the speed limit of the enzyme. This shows how fast the enzyme can act under optimal conditions (i.e. saturating substrate). kcat is also called the turnover number of the enzyme, since it represents the maximum number of substrate molecule the enyzme can react with per unit time.

The km is the conversion rate when there is half of the saturation concentration of the substrate.

This is incorrect. The units of Km are generally in terms of concentration (e.g. molar), so Km is not a rate. Km is usually interpreted to be a measure of the enzyme's affinity for the substrate. When the substrate concentration is greater than Km, the enzyme is easily able to grab onto the substrate, while when the substrate concentration is less than Km,

the kcat/km is the conversion rate when there is minimum substrate concentration.

This is a good interpretation. kcat/Km is a useful measure of the efficiency of the enzyme because it considers both the maximal rate of the enzyme kcat, as well as the affinity of the enzyme for its substrate (Km). More efficient enzymes will have high kcat values (the enzyme can react with its substrates quickly) and low Km values (the enzyme can bind its substrate at very low concentrations), and both influence the kcat/Km value.

Are these accurate? What do we mean by minimum substrate concentration? How can I find the exact
conversion rate when I know the kcat/km and the substrate concentration of my solution?[

I basically want to calculate the substrate conversion rate per mg of enzyme per minute for a specific concentration of the two substrates. What do I need to do that?

The Michaelis-Menten equation will give you the initial rate of reaction between your enzyme and your substrate:

$$ v = \frac{k_{cat}[E][S]}{K_m+[S]}$$

You can also see how the interpretations of Km and kcat derive from the following equations. For example, when [S] >> Km, then Km +[S] ~ [S], so $$v = \frac{k_{cat}[E][S]}{[S]} = k_{cat}[E]$$

In other words, at saturating substrate concentration, the rate is the turnover rate of the enzyme times the number of enzymes you have.

In contrast, when [S] << Km, then Km+[S] ~ Km, so $$v = \frac{k_{cat}[E][S]}{K_m} = \frac{k_{cat}}{K_m}[E][S]$$

In other words, at low substrate concentration (i.e. substrate concentrations well below the Km), the rate is proportional to the amount of enzyme, the amount of substrate and kcat/Km.

Also, there is specific activity as well, is this the same with kcat/km?
The specific activity of the enzyme is similar to the kcat. The specific activity of the enyzme is Vmax/(mg enzyme), where Vmax is the maximal rate of the enzyme solution at saturating substrate concentration. Specific activity is a useful measurement when your enzyme preparation is not completely pure and there are other proteins present. Enzyme solutions that are more pure will have a higher specific activity. The kcat sets the theoretical maximum value of the specific activity.
 
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Thanks!
What do I put for [E] if I have a porous material or surface where rhe enzyme is immobilised on and put in a solution of the substrate?
Also, what are the units of V and how do I convert them into substrate moles per mg enzyme per minute? It seems v units are mM/sec which is not convenient to convert into moles/sec
 
cyentist said:
Thanks!
What do I put for [E] if I have a porous material or surface where rhe enzyme is immobilised on and put in a solution of the substrate?

In this case, I'd probably fit the kinetics data to this alternate form of the MM equation: $$v = \frac{V_{max}[S]}{K_m+[S]}$$

Also, what are the units of V and how do I convert them into substrate moles per mg enzyme per minute? It seems v units are mM/sec which is not convenient to convert into moles/sec

The units of v are generally units of concentration per time, though moles per time is also used. You can convert between mM/s to moles/s if you know the volume of your reaction. To calculate the specific activity of your immobilized enzyme, you would just calculate Vmax then divide by the mass of enzyme immobilized.
 
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