mrlucky0 Posted February 24, 2008 Share Posted February 24, 2008 My first post so please bear with me if my question is a bit unclear. My professor was lecturing about enzyme kinetics and I am a little unclear about the meaning of Km. Specifically, Km is supposed to measure how effectively the enzyme binds to substrate and is defined: km = ((K-1)+(K2)) / (K1) Where K-1 is the rate of the reverse reaction: ES--> E+S, K2 is the rate of the product forming reaction: ES-->P, and K1 is the rate forward reaction: E+S-->ES My question: is K2 dependent on K-1 and K1 or independent? If independent, it would seem that if K1 was very large (good enzyme-substrate binding), and K-1 small, K2 could either be very large or very small. Thus, we wouldn't really be able estimate the magnitude of Km. Alternatively my hunch is that the rates are dependent but can someone explain how this could be so? Link to comment Share on other sites More sharing options...
Psycho Posted February 26, 2008 Share Posted February 26, 2008 Your professor who I guess isn't actually a professor, is wrong km isn't a measurement of the affinity of an enzyme for its substrate, that is only true when K-1 is much greater than K2. Km is the substrate concentration that gives half Vmax. Link to comment Share on other sites More sharing options...
iNow Posted February 26, 2008 Share Posted February 26, 2008 Your professor who I guess isn't actually a professor... A much more likely scenario is that Mr.Lucky didn't accurately quote the professor, or misunderstood his/her words immediately upon hearing them. Link to comment Share on other sites More sharing options...
Psycho Posted February 26, 2008 Share Posted February 26, 2008 A much more likely scenario is that Mr.Lucky didn't accurately quote the professor, or misunderstood his/her words immediately upon hearing them. I was going on the fact that lots of people in some countries call there teachers professor even though they aren't. Not many people are professors lots are doctors but few in relative terms professors. Link to comment Share on other sites More sharing options...
iNow Posted February 26, 2008 Share Posted February 26, 2008 Thanks for the clarification. Link to comment Share on other sites More sharing options...
Gyrase Posted February 27, 2008 Share Posted February 27, 2008 mrlucky0, to answer your question K1 & k-1 have nothing to do with k2 but as others suggested km can be interpreted as enzyme-substrate affinity when k2<<k-1. regards, Link to comment Share on other sites More sharing options...
Hobbz Posted February 28, 2008 Share Posted February 28, 2008 I'd like to hop on this because I am very confused about what Km is and would appreciate some clarification. So, Km is mathematically defined as "Km=(k-1 + k2)/k1" in a simple enzyme model. But it is graphically defined on a Michaelis-Menten plot as the substrate concentration at which the rate equals 1/2 of Vmax. These two definitions appear very different to me. Why?... Because if Km can be "graphically defined on a Michaelis-Menten plot as the substrate concentration at which the rate equals 1/2 of Vmax" therefore it will change with enzyme concentration because each Michaelis-Menten plot is specific for a known enzyme concentration. Too put it more simply, the Michaelis-Menton plot will change with enzyme concentration and therefore change the graphically calculated Km. Something must be wrong here and I don't know what? Link to comment Share on other sites More sharing options...
Gyrase Posted February 29, 2008 Share Posted February 29, 2008 greetings, you can find mathematical proof in every biochemistry textbook, known as michaelis menten formula that proves V = Vmax ([s]/([s]+km)) you can see if you place instead of Km, V will be 1/2Vmax regards, Link to comment Share on other sites More sharing options...
Hobbz Posted February 29, 2008 Share Posted February 29, 2008 I know that is why you can take 1/2 Vmax to find Km on the curve but won't the curve from a Michaelis-Menten plot change if you change the enzyme concentration. Therefore, Km would be dependent on enzyme concentration. Another way to look at is that Km and Vmax are related. Basically if one was to use less enzyme to make the Michaelis-Menten plot, wouldn't it take less substrate to reach Vmax because that is the point at which the enzyme is saturated with substrate? Link to comment Share on other sites More sharing options...
Psycho Posted March 1, 2008 Share Posted March 1, 2008 I know that is why you can take 1/2 Vmax to find Km on the curve but won't the curve from a Michaelis-Menten plot change if you change the enzyme concentration. Therefore, Km would be dependent on enzyme concentration. Another way to look at is that Km and Vmax are related. Basically if one was to use less enzyme to make the Michaelis-Menten plot, wouldn't it take less substrate to reach Vmax because that is the point at which the enzyme is saturated with substrate? Yes but if you have a higher concentration then K1 will be faster anyway as will K2 and most like K-1 as well, everything changes. Link to comment Share on other sites More sharing options...
Gyrase Posted March 1, 2008 Share Posted March 1, 2008 First of all Ks never change! Even if all bears of the world suddenly start belly dancing still Ks(k1, k2, k-1, km,...) remain constant hence the name K. Second Vmax depends on concentration of enzyme no argue. If you increase enzyme concentration, Vmax increases and also Vmax/2 increases but if you plot michaelis-menten plots at different enzyme concentrations and draw a perpendicular line from Km it will pass from Vmax/2 points of all plots. In michaelis menten plot we plot V against , it is possible to write an equation to relate V to [E] but it is not useful. Link to comment Share on other sites More sharing options...
Hobbz Posted March 3, 2008 Share Posted March 3, 2008 I understand that k1 k2 and k-1 won't change but I don't understand how Km can't change if you change enzyme concentration. For example, if one were to use 1 pM of enzyme and found the Km to be 100 uM but if one were to change the enzyme concentration to 1 fM that the substrate concentration to half saturate the enzyme would still be 100 uM? Link to comment Share on other sites More sharing options...
Gyrase Posted March 6, 2008 Share Posted March 6, 2008 Hello Hobbz, if your are looking for an intuitive proof rather than mathematical, then I should tell you something; when I was student I remember that I exactly had this question in my mind and whenever I asked professors this question they were giving me mathematical explanations and were accusing me that "I'm shaking and questioning foundation of logic and math" with such questions They didn't help but the conclusion that I reached to it myself is: The speed is defined as [ES]*k2, so in a fact more [ES] means faster operation so if [ES] reaches to half of its maximum concentration then speed is Vmax/2. Now to the contrary of chemical reactions there is a concentration of substrate that increasing it from that amount will not change V, we call it max at which the plot reaches a plateau, now in a chemical reaction we would expect that having the concentration of the other component [E] constant then [ES]max/2 to be reached at max/2 so km should be max/2 so we should not see km to be the same for all enzyme concentrations but the trick is in enzymes we also have k2 that removes [ES] from equation so in higher concentration of enzyme the speed is faster and removes [ES] faster so to reach [ES]max we need more concentration of more than km*2. Think about it for a while. this explanation helped me to sleep more comfortable at nights, hope it helps you too. cheers, DG 1 Link to comment Share on other sites More sharing options...
Hobbz Posted March 6, 2008 Share Posted March 6, 2008 That makes sense. It's easy to get lost conceptually in enzyme kinetics, thanks for the help. Another perspective that helped reassure me Km does not change with enzyme concentration was when I tried to visualize the difference between one enzyme in the system and then two enzymes. If one gets half saturated at a specific substrate concentration than two enzymes will get saturated at the same substrate concentration. Link to comment Share on other sites More sharing options...
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