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This was a question in a test in my school today.

Q) Which intermediate of Krebs cycle further can form chlorophyll?

A) Oxaloacetic acid B) Citric acid C) Succinyl CoA D) Fumarate

Given answer was (C).

How can Succinyl CoA form chlorophyll? All I could find was related to porphyrine ring. Can anyone explain it or recommend a good source?

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  • $\begingroup$ I think you will find my answer indicates what your teacher wanted. I would explain politely that the question was wrong as regards the porphyrin ring of chlorophyl for the reason explained by March Ho, and illustrated in my answer. $\endgroup$ – David Jan 15 '18 at 17:04
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The standard treatment of this topic is covered in Section 24.4.3 of Berg et al., available freely online:

Succinyl CoA is a precursor for porphyrin in mammalian cells by condensing with glycine to form δ-aminolevulinate as shown in this diagram from that book:

δ-aminolevulinate synthesis

This reaction is catalyzed by mitochondrial δ-aminolevulinate synthase.

To quote from that reference:

Two molecules of δ-aminolevulinate condense to form porphobilinogen, the next intermediate. Four molecules of porphobilinogen then condense head to tail to form a linear tetrapyrrole in a reaction catalyzed by porphobilinogen deaminase. The enzyme-bound linear tetrapyrrole then cyclizes to form uroporphyrinogen III, which has an asymmetric arrangement of side chains. This reaction requires a cosynthase. In the presence of synthase alone, uroporphyrinogen I, the nonphysiologic symmetric isomer, is produced.

This is illustrated in Fig. 24.35 of that section, which also shows that uroporphyrinogen I gives rise to protoporphyrin IX, the precursor of haem:

porphyrin synthesis fro aminolevulinate (Berg Fig. 24.35)

As regards the synthesis of plant chlorophyl (which also contains a porphyrin ring), the review quoted by March Ho contains the following statement:

The porphyrin ring with its conjugated double bonds is assembled in the chloroplast from eight molecules of 5-aminolevulinic acid…

(5-aminolevulinic acid is the chemically preferred name for δ-aminolevulinic acid.)

However the review goes on to explain that although 5-aminolevulinic acid synthesized in plant mitochondria uses the same pathway as mammals, that produced in the chloroplast and used for the synthesis of haem that gives rise to chlorophyl uses a different pathway, termed the C5 pathway, in which — as March Ho stated — glutamate is the precursor:

C5 pathway

I suspect that the person who set the question is an animal or biochemist who thought he would try a variant of the standard question “which intermediate is a precursor of haem”, and ventured into waters where he was out of his depth.

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  • $\begingroup$ Thanks for the explanation of what the question was intending to ask. Given that, it makes a lot more sense that the question referred to a specific molecule, since it was the molecule stated in the textbook. $\endgroup$ – March Ho Jan 15 '18 at 19:20
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This is a poorly set question. If any one member of the Krebs Cycle is used as the raw materials in a biosynthetic pathway, it will follow that all of the other members are also part of the same biosynthetic pathway, since they are part of the same cycle.

The full biosynthetic pathway of chlorophyll is a very complex process, and there are dozens if not hundreds of different steps and enzymes involved. This is one paper that attempts to cover the issue.

In short, the porphyrin ring region of chlorophyll is generated from glutamate, and the phytyl tail region of chlorophyll is generated via the phytol synthesis pathway, via the geranyl phosphate pathway

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  • $\begingroup$ I think it is a poorly set question, but not for the reasons you state. It is fairly clear what is meant. I do feel a complete answer needs to point out the animal/microbial pathway for synthesis of δ-aminolevulinate, which, is what is causing the confusion. I feel a little bad about using your info on the C5 pathway of synthesis in chloroplasts in my answer, but I have acknowledged you and, again, I feel it needs spelling out. $\endgroup$ – David Jan 15 '18 at 16:49

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