2009年7月24日 · Km and Vmax are related to enzyme kinetics in a biological system. Km is the substrate concentration that is required for the reaction to occur at 1/2 Vmax. In other words, it is how much substrate is needed for the reaction to occur at 1/2 its max possible rate. Obviously Vmax is the maximum rate that the reaction can proceed at.

2008年6月15日 · If Vmax is dependent on enzyme concentration and Km is the substrate concentration = 1/2Vmax. If you increase the enzyme concentration, Vmax increases, then Km must also increase to fufill this 1/2 Vmax requirement no? I found a book that support my theory, it says that Km and Vmax are constants for a given enzyme concentration.

2018年8月27日 · "Vmax is the maximum velocity of the enzyme. Competitive inhibitors can only bind to E and not to ES. They increase Km by interfering with the binding of the substrate, but they do not affect Vmax because the inhibitor does not change the catalysis in ES because it cannot bind to ES."

2012年1月4日 · Hey guys - I know this is a very old thread, but it still appeared in the top few google searches when I was trying to figure out why Vmax decreases in noncompetitive inhibition, but stays the same in allosteric inhibition, since both involve an …

2014年12月2日 · Km = the [substrate] at (1/2)Vmax ^based on that definition, increasing the transporter protein shouldn't have any effect. (equation version: Km = (Vmax[substrate]) / Vo) Other important equation that does involve [Enzyme]: Kcat = Vmax / [Enzyme] In words, Kcat is the number of cycles catalyzed per second.

2009年5月7日 · Km is equal to the substrate concentration at 1/2 Vmax. Vo = (Vmax*)/(Km + ) 1/2 Vmax = (Vmax*)/(Km + ) Km + = 2 Km = is the substrate concentration. So you can see that Km is NOT dependent on enzyme concentration; otherwise, a change in Vmax (which IS affected by enzyme concentration) would affect Km as well. Hope this helps.

2016年7月19日 · Uncompetitive lowers Km because uncomp. inhibitors only bind to Enzyme-Substrate complexes effectively rendering some enzyme useless: The potential Vmax goes down and Km appears lower in uncomp. inhibition because active sites are taken up and sequestered --> higher affinity. That was a mouthful! Let me know if any of this does not make sense.

2012年10月9日 · Km is the concentration of substrate at 1/2 Vmax and is an intrinsic property of the enzyme. Therefore with irreversible binding you decrease the Vmax because there are less enzymes but you haven't changed the intrinsic affinity that the substrate has for those enzymes that remain (so the slope of the graph is the same).

2007年2月26日 · So I just need some confirmation. In competitive inhibition, Vmax does not change but Km does. Km most likely shifts to the right? which means a decrease in substrate affinity? Correct? In non-competitive inhibition, Vmax …

2006年12月26日 · If Km is large, that means it has a lower affinity for the substrate, and likewise, if Km is small, it means it has a higher affinity for the substrate. If you look at the graph, Km is when velocity is 1/2 of Vmax. So if the enzyme binds very specifically to the substrate, you're going to reach Km faster. Ergo, Km is lower in that case.