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Chapter 4: Problem 12

Draw a mechanism for the formation of \(N^{5}, N^{10}-\) methylenetetrahydrofolate and the transfer of a hydroxymethyl group to uridylate.

Short Answer

Expert verified
The formation of \(N^{5}, N^{10}-\)methylenetetrahydrofolate involves the attachment of a methylene group to THF in the presence of a serine hydroxymethyltransferase enzyme. This one-carbon form of tetrahydrofolate then donates the carbon to uridylate in a process catalyzed by the enzyme thymidylate synthase, resulting in the formation of thymidylate and dihydrofolate.

Step by step solution

01

Understanding the components

Firstly, it's necessary to understand the basic structures of methylenetetrahydrofolate and uridylate. Draw structural representations of \(N^{5}, N^{10}-\)methylenetetrahydrofolate (which is a form of folate that carries a single carbon group) and uridylate (a monophosphate form of uridine, which is a pyrimidine nucleoside).
02

Formation of methylenetetrahydrofolate

Congratulations on mastering the basic structures! Now let’s move on: The formation of \(N^{5}, N^{10}-\)methylenetetrahydrofolate from tetrahydrofolate (THF) involves the attachment of a methylene group (\(CH_2\)) to THF. You should represent this with a reaction arrow from a \(CH_2\) (which is coming from serine via a serine hydroxymethyltransferase enzyme) to THF, creating methylenetetrahydrofolate. This reaction simultaneously turns serine into glycine.
03

Transfer of hydroxymethyl group

The next step is the transfer of the one-carbon group (hydroxymethyl) from \(N^{5}, N^{10}-\)methylenetetrahydrofolate to uridylate. Draw another reaction arrow from the methylenetetrahydrofolate to uridylate, indicating the transfer. This reaction results in the formation of thymidylate, a crucial DNA component, with the simultaneous conversion of \(N^{5}, N^{10}-\)methylenetetrahydrofolate back to dihydrofolate. A cofactor in this reaction is the enzyme thymidylate synthase.

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