section bank c/p #77

[taeyeonlover]

What is the most likely reason for decreased Km value observed in D45G variant compared to other two versions of GalK with respect to each substrate?

substrate binding pocket is too crowded
key hydrogen bonding is lost
enzyme is unfolded
Vmax is much lower which means less substrate is needed to reach it.


In here aspartate is replaced with either alanine or glycine.

I guess D makes sense but D45A's Kcat also decreases, which means that V max decreases for D45A since Vmax/[substrate] = kcat. But for D45A the Km increases.

So why is D the correct answer?

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[nostra_damus]

I want to say it's because the Vmax decreases a lot more?

 

[aldol16]

D45A's kcat is on the same order of magnitude as the wild type enzyme's - hardly enough to cause a huge difference in the Vmax. D45G, on the other hand, has a kcat that is an order of magnitude lower than the other enzymes. This would substantially affect Vmax.

 

[nostra_damus]

Following up on this question, how do you make the jump that a decrease in Vmax means a decrease in Km? What am I missing here?

 

[663697]

Following up on this question, how do you make the jump that a decrease in Vmax means a decrease in Km? What am I missing here?

I'm wondering this too, but I think it may be because Km is defined as the concentration of substrate at which v= (1/2)*Vmax. Since Vmax is proportional to kcat and the table shows that kcat decreases, this would cause a decrease in Vmax. Therefore the decrease in Vmax leads to a smaller concentration of substrate required to reach 1/2 of it. This in turn reduces the Km value.

Looking back at this, I don't think the change in the Km value signifies an increased affinity for the substrate in this case, because I think we usually refer to increased/decreased affinities by looking at Km when Vmax is held constant. I could be wrong about this so I'll leave it to @aldol16 to expand if there's something I missed

 

[nostra_damus]

So basically we're not saying anything about the change in affinity and it's all basically due to numbers: decreasing vmax --> 1/2 vmax will go down --> lower km

 

[663697]

So basically we're not saying anything about the change in affinity and it's all basically due to numbers: decreasing vmax --> 1/2 vmax will go down --> lower km

That's the way I see it at least. If it had been something like a competitive inhibitor where the VMAX doesn't change but the Km does, we would conclude there was a change in affinity. But since both variables are changing in this case I think it's just a matter of one affecting the other.