Assignment_3_ANSWERS - Chem 237 Assignment #3 ...

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Unformatted text preview: Chem 237 Assignment #3 ANSWERS 1. This is a reaction mechanism for an enzyme that converts carbamoyl phosphate (on the far left) and aspartate (you know that one!) to carbamoylaspartate (on the right) plus inorganic phosphate (Pi) O C - O P + NH2 O- O O- O O H3N CH C CH2 + C O- O O- C H2N HN CH C O- CH2 O C O O+ Pi Using the hints in Quiz #4 to help you, write out a chemical (meaning with structures!), enzyme (meaning you show how the enzyme participates in the reaction) mechanism for this enzyme. The equation is not necessarily balanced; if not, what is needed to balance the equation? ANSWER: For the mechanism of aspartate transcarbamoylase (I’ve also seen “transcarbamylase”), I have based my answer on the detailed studies that have been done on the E. coli enzyme over the years. The properties of this enzyme have been studied in many respects, mechanism, structure, allosteric mechanism, genetic regulation, etc. Now, as for your answers, there are many possibilities, depending on which base(s) you use in the reaction. The main error here on your papers was that many of you did not show how the enzyme participates in the reaction. You treated it as a simple organic reaction. Anyway………… General comments: the “Arg”, “His”and “Lys” are amino acid residues on the enzyme and are part of the primary structure of the protein, that are participating in the reaction. The substrates bind, and as that happens, the Asp loses a proton to the water solution. This is due to the very electropositive nature of the active site (lots of + charges there already). Next, nucleophilic attack of the deprotonated Asp on the CP Arg+ Arg+ O H O OH -O NH N H O OH + H2N N NH H N H +HN NH2 +H N 3 COO- COO- COO- NH2 His P P -O O- O His O L ys COOLys O O- O O His P -O His P H O OH -O N O OH NH H H N HN NH H +HN Arg+ Arg+ NH2 +HN NH3+ H2N COO- COO- COO- Lys Lys COO- Arg+ (go to top of page) O H -O N P -O OOH NH HN His H +HN NH3+ COO- COO- L ys Release of N-carbamoylAsp product Arg+ O His P -O NH +HN OH OH NH3+ L ys Release of phosphate and the H+ bound to the His This returns the active site to the configuration for another catalytic cycle. The Lys in the active site is thought to be the base that deprotonates the tetrahedral intermediate. Note how the Arg and His stabilize the ( ­) charge on the oxygen. Some side group will protonate the leaving phosphate, and the structure suggests that it is the Lys. Products are released. Other suggestions about the mechanism of this enzyme have been made. One suggestion is that the phosphate leaving group itself deprotonates the intermediate. The case for the Lys as base is very strong, bases on experiments where the Lys was changed in the protein structure to Asn, which can not donate a proton. Other possibilities: (i) other side chains could be the bases involved in the reaction, such as His, Glu or Asp. (ii) instead of being deprotonated as it enters the active site, the Asp could be deprotonated by an active site base one of the usual suspects, His, Glu, Asp, Lys), in which case your mechanism would need to include two bases, not just one. Main error made in your answers: The main error was that many of you did not involve the enzyme in any way. Or you mentioned the enzyme (“the enzyme binds the substrate in a binding pocket”) and that was about it. 2. One of the many antibiotic resistance mechanisms found in bacteria is that of modification of the antibiotic. One such modification is acetylation. Given the following reaction (I have simplified H NH2 the antibiotic to a simple sugar, the structure of which we will look at in O more detail in a later module); this particular enzyme has been shown to HO CoA have a Cys at the active site and shows ping ­pong kinetics. Using the HO S H H3C HO same approach as above and in Quiz 4, draw a chemical enzyme reaction H OH mechanism for the reaction, suggesting how the enzyme participates. H OH (Hint: the ping ­pong kinetics gives you an important hint as to how the enzyme works!!) Also, “CoA” is Coenzyme A, a large ­ish molecule that is often involved in these types of reactions. You do not need to know the O structure of Coenzyme A to do this problem—the part of the molecule I show in the figure is enough to do this problem. CH3 H HN HO HO HO H H H OH OH CoA-SH (free coenzyme A) ANSWER: The ping pong kinetics tells us that the enzyme binds to one of the substrates, the first product is released, then the enzyme binds the second substrate, and then the second product is released, and the enzyme is returned to the initial state. A very complete answer would consider that the amino group would be protonated in this case, and so would need to be deprotonated to start. The Cys is the nucleophile, and as such, would need to be deprotonated. Arg+ Arg+ O Cys O S SCoA CH3 Cys CH3 S H -O Glu Glu COO- N N O His His NH NH Sugar NH3+ Arg+ CoASH O Cys O- CH3 +H N 2 SCoA S Sugar -O N Glu His Glu COO- NH O NH+ H CH3 His S Arg+ Cys NH Arg+ Arg+ O O- S Cys N+ H2 CH3 S- Sugar Cys N H Sugar CH3 NH+ His HO N NH Glu O HO His Glu NH O Now, the product, the acetylated sugar, is released. In order for this to bring the enzyme back to the initial state, the Glu must lose a proton, so as the product leaves, the Glu releases the proton. This sounds like a coincidence, but it is how the process works. It has been shown for a number of enzymes that the pKa of groups can change when the active site opens or closes. I used Glu and His here, but you could have used other amino acids instead, like Lys or Asp. Arg has a pKa that is too high to be a good choice for an acid or base. Again, many of you ignored the enzyme and/or had the amino sugar directly attack the acetylCoA. No one drew the structure below, so that is good. A couple of hints: In the reactions with a carbonyl group, the intermediate is has tetrahedral geometry. Also, students often draw intermediates like the following: O +H 3N C+ This is NOT correct. I do not know where you get this. But it is not correct, so if your mechanism has a structure like this in it, it is wrong. Do it over. ...
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This note was uploaded on 07/18/2011 for the course CHEM 237DE taught by Professor Daub during the Spring '11 term at Waterloo.

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