Vmax & Enzyme concentration

[cgbb]

I understand that Vmax increases with enzyme concentration, so on the MM graph it would be shifted upwards. However, does the graph also shift to the right as well? Wouldn't there need to be a higher substrate concentration to reach the new Vmax since there is more enzyme present?

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[jakernewt]

Found this on an old post, it is the easy way to think about it:

Km is a measure of the enzyme's affinity for a particular substrate. Think of it as the probablity that an enzyme will "stick" to its substrate if it happens to bump into it. This "stickiness" doesn't have anything to do with concentration.

 

[krogers21]

I came here for this exact problem. If Km is not related to concentration, then why is it on the x-axis of a RR v Substrate Conc. graph?

 

[gettheleadout]

Hah, I had this same question a while ago and wrote out a lengthy explanation for myself. I'll post in here:

Why doesn't change in enzyme concentration result in a change in Km?

First of all, assumptions of M-M kinetics must be clarified. Reaction rates (V, Vmax) are taken to be initial rates, that is, the concentration of substrate used in calculating these is considered constant, either to ignore the inevitable decrease resulting from enzymatic conversion or as an implication of constant influx of substrate. Second, the saturation concentration of substrate is not going to be equal to the concentration of enzyme. While intuitively, one would think 5 substrate molecules + 5 enzyme molecules would = saturation and thus Vmax, this ignores several factors: at low enzyme and substrate concentrations, the likelihood that all substrate will encounter all enzyme at the same time is incredibly low, so the V in such a condition would likely be a tiny fraction of [E](Kcat); as substrate concentration increases relative to enzyme concentration, the likelihood of simultaneous binding to all enzyme increases, such that not only the saturation , but the Km as well, are much, much higher than the [E].

Because change in [E] is literally a change in concentration of enzyme and not just molar amount of enzyme, increasing the [E] (at constant ) also increases the likelihood of a substrate molecule encountering an enzyme molecule at any given moment. Increasing [E] does indeed increase the number of active sites which must be occupied to achieve a given V (whether Vmax/2, Vmax, or any other V), but it also decreases the positional variability of substrate molecules with respect to enzymes. Essentially, adding enzyme decreases the ease with which substrate can exist in solution and be any given distance from enzyme. This effectively increased likelihood of substrate encountering enzyme counteracts the increased occupied-active-site number, resulting in no change of required substrate concentration to achieve a given V.

Fish in a fish tank analogy: If a single fish in a fish tank can consume food pellets at a maximum rate of 1 pellet per second (Vmax), then with very low concentration of food pellets in the tank (low ), the fish will only be able to encounter and consume them at a rate much lower than 1 pellet/sec. At some food concentration, however, food pellets will be abundant enough in the tank that the fish will have no trouble encountering 1 pellet each second, and will consume at it's maximum rate of 1 pellet/sec (Vmax). If food concentration increases further, the fish will encounter more pellets while swimming, but will be occupied with chewing its current pellet, and thus its rate of consumption cannot increase further; the fish, as an agent of consumption, is saturated with food. At some food concentration below the point of saturation, the fish will only be able to encounter enough pellets to consume one every two seconds (1 pellet/2 sec = Vmax/2) and thus this food concentration is Km.

Consider the same fish tank with 100 fish, each of which is identical to our previous fish, and can individually consume food at a maximum rate of 1 pellet per second (Vmax). At a similarly very low concentration of food pellets in the tank (low ) there will be more fish to encounter the pellets, and thus collectively they will consume the food at a higher rate than the single fish at the same food concentration. This makes sense intuitively, as more fish available to eat logically means more food collectively eaten in any given period of time. However, more fish also means a greater probability that any given food pellet in the tank will be in position to be consumed by a fish at any given time. Because of this, the concentration of food at which the lone fish can consume at his maximum rate is such that each of our 100 fish would have a food pellet directly in mouth-range as well.

 

[krogers21]

Great explanation! That helps tremendously. However, how can Vmax change but the Km stay the same if Km = 0.5Vmax ? Would the slope essentially be steeper? Also, is the slope of any significance? For instance, the slope of a velocity vs time graph is the acceleration at each moment.

 

[gettheleadout]

Great explanation! That helps tremendously. However, how can Vmax change but the Km stay the same if Km = 0.5Vmax ? Would the slope essentially be steeper? Also, is the slope of any significance? For instance, the slope of a velocity vs time graph is the acceleration at each moment.

Because Km isn't 0.5Vmax, it's at 0.5Vmax. The slope of the L-B plot is Km/Vmax, and since Vmax is essentially [E]*Kcat, as you increase the enzyme concentration and thus increase Vmax you get a shallower slope. In fact, this is why competitive inhibition actually makes the slope of the plot steeper; we know a competitive inhibitor just reduces the number of active sites available, effectively lowering [E] without changing Km. Thus, we get a steepened plot with the same x-intercept at -1/Km (remember that Km is based on enzyme properties and temperature, and only noncomp. inhibitors change enzyme properties and thus Km). Relative slopes can give you info about your enzyme concentration or the presence of competitive inhibitors, as we just saw, but I'm afraid I don't know if the instantaneous slope gives you any meaningful information about a particular point.

 

[krogers21]

Because Km isn't 0.5Vmax, it's at 0.5Vmax. The slope of the L-B plot is Km/Vmax, and since Vmax is essentially [E]*Kcat, as you increase the enzyme concentration and thus increase Vmax you get a shallower slope. In fact, this is why competitive inhibition actually makes the slope of the plot steeper; we know a competitive inhibitor just reduces the number of active sites available, effectively lowering [E] without changing Km. Thus, we get a steepened plot with the same x-intercept at -1/Km (remember that Km is based on enzyme properties and temperature, and only noncomp. inhibitors change enzyme properties and thus Km). Relative slopes can give you info about your enzyme concentration or the presence of competitive inhibitors, as we just saw, but I'm afraid I don't know if the instantaneous slope gives you any meaningful information about a particular point.

This makes tremendous sense. Thanks so much! I was unaware that Km is essentially at .5Vmax. After reviewing enzymes through Chad's Videos it has helped a lot.

 

[gettheleadout]

Because Km isn't 0.5Vmax, it's at 0.5Vmax. The slope of the L-B plot is Km/Vmax, and since Vmax is essentially [E]*Kcat, as you increase the enzyme concentration and thus increase Vmax you get a shallower slope. In fact, this is why competitive inhibition actually makes the slope of the plot steeper; we know a competitive inhibitor just reduces the number of active sites available, effectively lowering [E] without changing Km. Thus, we get a steepened plot with the same x-intercept at -1/Km (remember that Km is based on enzyme properties and temperature, and only noncomp. inhibitors change enzyme properties and thus Km). Relative slopes can give you info about your enzyme concentration or the presence of competitive inhibitors, as we just saw, but I'm afraid I don't know if the instantaneous slope gives you any meaningful information about a particular point.

Whoops. I'm wrong here. Competitive inhibitors do result in a steeper slope of the L-B plot, but it's because they increase the apparent Km, not decrease the Vmax. Vmax is equal to [E]Kcat, but the reason [E] doesn't change as a result of competitive inhibitors is that they don't permanently bind to the active sites (that would make them irreversible/suicide inhibitors). Competitive inhibitors don't change Kcat because they don't change the flexibility/activity/anything else of the enzyme itself since they bind at the active site (non-competitive inhibitors do, because they bind to the enzyme elsewhere). Thus, Vmax stays the same, but apparent Km goes up (because it takes a higher to get the same number of enzyme-substrate complexes formed as would be present at 1/2 Vmax without an inhibitor), and thus the slope of the line increases.

 

[drechie]

why is stuff crossed out? what does mean it is factually incorrect