Question

The key intermediate for the biosynthesis of C6-C3 units is:

• A. Pyruvic acid
• B. Shikimic acid
• C. Dehydroquinic acid
• D. Mevalonic acid

Hint:

Shikimic acid is central to the biosynthesis of aromatic compounds, including the C6-C3 units, which are essential for the production of several biologically active molecules.

Answer 1

The Correct Option is B

Shikimic acid is the key intermediate for the biosynthesis of C6-C3 units in the pathway that leads to the synthesis of aromatic compounds in plants and microorganisms. This acid plays a crucial role in the shikimate pathway, which is responsible for the production of many essential metabolites, including aromatic amino acids.

Answer 2

The question is about the biosynthetic pathway that builds C6-C3 units (a benzene ring attached to a three-carbon side chain, as seen in cinnamic acid derivatives). Let's trace which of the given compounds actually sits on this pathway.

1. Pyruvic acid: Pyruvic acid is a simple three-carbon compound formed from glycolysis. It feeds into the shikimate pathway as a starting material along with erythrose-4-phosphate, but it is not itself the intermediate that carries the C6-C3 skeleton.
2. Shikimic acid: Shikimic acid is the seven-carbon intermediate formed partway through the shikimate pathway, after pyruvic acid and erythrose-4-phosphate have combined and cyclised. From shikimic acid, the pathway goes on to chorismic acid and then to the aromatic amino acid phenylalanine, whose side chain gives rise to C6-C3 phenylpropanoid units such as cinnamic acid and its derivatives. This makes it the key intermediate for C6-C3 biosynthesis.
3. Dehydroquinic acid: This is also an intermediate of the shikimate pathway, but it appears at an earlier stage, before shikimic acid is formed, and is not the compound directly associated with C6-C3 unit formation.
4. Mevalonic acid: Mevalonic acid belongs to a completely different pathway, the mevalonate pathway, which builds isoprenoid (terpenoid) units, not aromatic C6-C3 phenylpropanoids.

Since the C6-C3 phenylpropanoid skeleton is built by way of the shikimate pathway, with shikimic acid as its defining intermediate, this is the compound the question is looking for.

Therefore, the correct answer is Shikimic acid.