Getting to the bottom of Km.

[Lunasly]

Hello everyone,

Spare me the rants – I have read the other threads, but I still don't quite get it conceptually.

As far as I know, Km is used as a means of describing the affinity for the substrate. It's value is at 1/2Vmax. Now if you increase the concentration of enzyme in your solution, it makes sense that your overal rate will increase (double the number of workers the job will get done faster). If Vmax increases as a result of increasing the concentration of enzyme, then won't 1/2Vmax be higher? Therefore, wouldn't the value of Km also increase?

Since this is not the case, I want to come up with an analogy that I hope someone can confirm is correct. It is to my understanding that we assume that the concentration of enzyme within our solution does not change and that the only thing that does change is the concentration of substrate. We know that if we have a set amount of enzyme in our solution, then the more substrate we add, the faster we will produce our product. That is, however, until we saturate the solution with our substrate. As a result, our rate of reaction can no longer increase because we don't have enough enzyme to accommodate oversupply of substrate – our rate evens out.

Now lets say hypothetically we do increase our concentration of enzyme in our solution. It then makes sense that our maximum velocity would increase because we have a larger supply of enzymes which can accommodate the large supply of substrate. With that being said, 1/2Vmax would also increase. So it makes sense that if 1/2Vmax is equal to Km, then Km should also increase. However, this is not the case and its because Km is entirely based on the affinity for the substrate at 1/2Vmax. Regardless of whether we add more enzyme or more substrate, the Km will always be the same because at 1/2Vmax, half of the enzymes are working while the other half are not. So even if we increased the amount of enzyme to accommodate the large supply of substrate, our Vmax would increase and so would our 1/2Vmax, but the ratio of enzymes working to those not working would be the same. For example, lets say I originally have 10 enzymes where 5 enzymes are working and another 5 enzymes not working at 1/2Vmax. Now lets also say that I add more enzyme and Vmax increases and therefore 1/2Vmax also increase. Now that I have done this, lets say I have 20 enzymes in total, where 10 of them are working and another 10 are not working. The ratio, regardless of whether we increase the amount of enzyme or not, is still the same.

Is what I just wrote correct?

Also can someone explain to me why binding affinity is inversely proportional to Km?

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[MedPR]

Km is independent of enzyme concentration. It is a measure of affinity, hence why non-competitive inhibitors do not change the Km.

Km is the concentration of substrate that allows the enzyme to reach 1/2Vmax. If you can achieve 1/2Vmax with very few substrates, that must mean there is very high affinity. In other words, low Km means high affinity.

 

[Lunasly]

Was my analogy correct, however?

 

[Arctangent]

It seems like the analogy you presented is correct. As I follow, an increase in number of enzymes implies an increase in Vmax, and thus 1/2Vmax increases. What I am a bit more unsure of is the point that while Km is not equal to 1/2Vmax or linearly proportional to it, it is the value of substrate concentration that corresponds to it. In other words the literal value of 1/2Vmax can change, but Km will not. So as your last sentence in the paragraph states, the ratio of enzymes working to not working will remain the same regardless of enzyme concentration.

As far as binding affinity and Km, as Km is defined as the concentration of substrate where 1/2 the enzymes are working, a small number suggests a large fraction of those are being catalyzed by enzymes, or in other words there is a large binding affinity. As far as inverse relationships go, a similar concept might be that of LD50, which is the amount of a given chemical that is lethal to half of a population. As LD50 is higher, the chemical is less lethal, as more is required to be fatal to the same population, and vice versa.

In some instances, intuition or analogies may not work with Michaelis-Menten kinetics. As far as I understand the concept of Km, it does break down under certain conditions. Considering a fixed amount of substrate equal to the Km, and an infinitely increasing enzyme concentration, at some point there will be too many enzymes for half of the enzymes to be occupied. Going with the worker analogy,with a fixed number of objects to be made, if you keep increasing workers (from 10, to 20, to 30 etc.), 1/2 the total number of workers will eventually outnumber the number of objects. That is actually one of the assumptions of using this system, which is that the substrate concentration must be substantially greater than enzyme concentration.

 

[Lunasly]

I think that is an important assumption because aren't we always dealing with only a single enzyme concentration when presented with any given curve?

I thought of some other ways to think about it as well in regards to competitive and non-competitive inhibition. Im competitive inhibition, it is to my understanding that Vmax is not effected, while Km is increased. Well, if we make the assumption that we are only ever dealing with a single concentration of enzyme, then it makes sense that Vmax is constant because the competitive inhibitor is only bound to the active site for a small period of time. Therefore, if we increase the [substrate], then we overcome this issue and Vmax will remain unchanged. Km, however, needs to be thought of as binding affinity. Even, though our 1/2Vmax will stay the same (because Vmax stays unchanged), the affinity for the substrate will decrease because there is now more than one substrate competing for that same binding spot. Because binding affinity is inversely proportional to Km, it makes sense that a lower binding affinity means a higher Km.

For non-competitive inhibition, a similar idea holds true. This time Vmax changes, but Km doesn't. Think again that Vmax represents the number of workers. If I have 10 workers, but 5 of them get non-competitivly inhibited (rendering them useless), then it makes sense that Vmax (which is my number of workers) is going to decrease. 5 workers can't get the job done as fast as 10 workers. Now in terms of Km – if we think of it as binding affinity. Because Vmax is decreasing, it makes sense that 1/2Vmax is also going to be a lower number. With that being said, Km doesn't change because if we think of it as binding affinity, the affinity of the substrate for the remaining 5 workers is going to be the same. It's like candy: If I have 10 workers that love candy bars. Their affinity for the candy bars is high because that is all they are exposed to. If I kill 5 of those workers, well their affinity for the candy bars is still the same because they still all love candy bars. It's the same idea here: Even though we lost 5 of our enzymes, their affinity for the substrate stays the same because there is nothing else competing for that active site. Thus, Km stays unchanged.

I hope to God what I just wrote was not a bunch of bull****. Can anyone confirm?

 

[Morsetlis]

You want us to confirm candy bars analogies?

 

[Lunasly]

You want us to confirm candy bars analogies?

Yes, please.

 

[MedPR]

Yes, please.

Your candy bar analogy is correct.

 

[JLeBling]

Answer: http://www.sumarios.org/sites/default/files/pdfs/80-676-3-pb.pdf

 

[MedPR]

Answer: http://www.sumarios.org/sites/default/files/pdfs/80-676-3-pb.pdf

Nice bump.

 

[drechie]

does Km change with [substrate] concentration?

 

[PharmHoppe]

does Km change with [substrate] concentration?

No. Once 1/2Vmax is reached, increase in [substrate] has no effect on km. Km only changes if a competitive inhibitor is present.

From a rudimentary perspective, km=k2/k1 if K2>>>k3. If equilibrium is already established, addition of substrate will affect the forward and reverse equilibrium constants similarly, i.e no net change.