Sequentially number the following steps of Chymotrypsin's cata...

Question

Sequentially number the following steps of Chymotrypsin's catalytic mechanism in the correct order from 1-8:

a) Released newly formed amine portion of the substrate diffuses away _______

b) His-57 deprotonates the Ser-195 hydroxyl group, generating a stronger nucleophile ______

c) Tetrahedral intermediate collapses & His-57 donates a H+ to N of scissile bond (cleaved peptide bond) ______

d) Via general-base-catalysis, His-57 deprotonates a water molecule, generating OH- ______

e) Tetrahedral intermediate collapses & His-57 donates a proton to Ser-195 (cleaving ester bond) _______

f) Released newly formed carboxylic acid portion of the substrate diffuses away & enzyme is restored _______

g) Hydroxide ion attacks the carbonyl group of the substrate, forming another tetrahedral intermediate ______

h) Nucleophilic Ser-195 attacks the carbonyl C of the substrate forming a tetrahedral intermediate______

Accepted Answer

So this practice problem wants us to sequentially number the following steps of chymotrypsin's catalytic mechanism in the correct order from 1 to 8 and so we've got these 8 steps listed down below. And so what we need to recall from our previous lesson videos is that chymotrypsin's catalytic mechanism is broken up into 2 phases, the acylation phase and the deacylation phase. And, really, their steps can be broken up into similar categories. And so notice we have the acylation phase over here on the left, which has substrate binding, nucleophilic attack, removal of the leaving group, and the end of the phase. And then on the right, what we have is the deacylation phase, which also has substrate binding, nucleophilic attack, removal of leaving group, and end of phase. And so by remembering, these general classes of steps here, that can help us correctly order the steps here of chymotrypsin's catalytic mechanism. And so after reading through each of these steps, the very first step is going to be this off option here, b. So we can go ahead and number it number 1 and so the reason it's number 1 is because it says histidine 57 deprotonates the serine 195 hydroxyl group generating a stronger nucleophile. And, of course, we know that this is going to occur in chymotrypsin's active site even before the acylation phase even begins. And so, this is actually, not going to be part of the substrate binding, but, it's something that takes place before even substrate binding occurs. So, it's step 0 if you will. But for this, here, it's gonna be, the first step. So, after reading through each of these problems, the next one is going to be h And that's because, after the stronger nucleophile has been generated, after searing 195 has been generate, has been established as a stronger nucleophile. The next step is for serine 195 to perform a nucleophilic attack on the carbonyl carbon of the substrate forming a tetrahedral intermediate. And so this is really going to be, step 2 here, the nucleophilic attack of the acylation phase. And of course, we're expecting, removal of the leaving group to come next and so that's going to correspond with, this step here c. And so, what happens with removal of the leaving group is the tetrahedral intermediate that was generated in step number 2 is going to collapse. Histidine 57, is going to donate its acidic hydrogen here to the nitrogen of the sessile bond or, the, peptide bond that's going to be cleaved. And so again, this is gonna be step number 3 and that corresponded with the removal of leaving group. So, the next step that we expect to be is the end of the phase. And so the end of the acylation phase is gonna correspond with a, which is the released newly formed amine portion of the substrate is going to diffuse away. And so, that will correspond with the end of the phase. So then, now that the isolation phase is complete, we expect to shift over into the deacylation phase with sub straight binding. And so this is going to correspond with option d. So it says, via general base catalysis, histidine 57 is going to deprotonate a water molecule that enters the active site. So the water molecule would be, the by the substrate that's going to bind. And so this deprotonation is going to generate a hydroxide ion. So, this will be step 5, which correlates with this step 5 here. And then, of course, we expect nucleophilic attack to occur next. And so, after this hydroxide ion has been, generated, the hydroxide ion is going to perform a nucleophilic attack on the carbonyl group of the substrate forming another tetrahedral intermediate. And so this is gonna correspond with step 6, which again was this nucleophilic attack here. So we expect removal of the leaving group to come next. And so, of course, that's going to correspond here with step E here which is the tetrahedral intermediate that was formed here in step 6 is going to collapse. Histidine 57 is going to donate its acidic proton to serine 195 and that's going to cleave the ester bond. And so again, this is step 7. And so, of course, after removal of the leaving group, the end of the phase is next, which corresponds with this last step that we see here. So the released newly formed carboxylic acid portion of the substrate diffuses away and the chymotrypsin enzyme is restored back to its original state where it can again, begin the isolation phase all over again. And so this will be, step 8. And so, this here concludes our practice problem and, really the best way to test yourself and see if you really know chymotrypsin's catalytic mechanism is to, come back to this practice problem and just try to order these steps correctly. And so that concludes this problem and I'll see you guys in our next video.

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