Question

5-Fluorouracil, an anti-metabolite used in cancer treatment, is activated to:

• A. 5-fluoro-2-oxyuridylic acid
• B. 3-fluoro-3-deoxyuridylic acid
• C. 3-fluoro-3-oxyuridylic acid
• D. 5-fluoro-2-deoxyuridylic acid

Answer 1

The Correct Option is D

The correct option is (D): 5-fluoro-2-deoxyuridylic acid.

Answer 2

5-Fluorouracil is given as an inactive-looking pyrimidine analogue that must be converted inside the cell to its active form before it can block DNA synthesis. Let's check each option for what that active form actually is.

1. 5-fluoro-2-oxyuridylic acid: This name does not describe a real intermediate in 5-FU metabolism; the activation pathway does not pass through an "oxy" derivative of this kind.
2. 3-fluoro-3-deoxyuridylic acid: The fluorine substitution and the deoxy modification in the real active metabolite both sit at the 5 and 2 positions of the pyrimidine/sugar system, not at position 3, so this naming does not match the true structure.
3. 3-fluoro-3-oxyuridylic acid: Like the second option, the position given here is wrong, and there is no "oxy" step in this pathway.
4. 5-fluoro-2-deoxyuridylic acid: Inside the cell, 5-FU is converted through a series of enzymatic steps into 5-fluoro-2-deoxyuridine-5-monophosphate (FdUMP), commonly named 5-fluoro-2-deoxyuridylic acid. This metabolite binds tightly to thymidylate synthase and blocks the enzyme, which stops the cell from making thymine needed for DNA synthesis.

The correctly named and correctly positioned active metabolite of 5-fluorouracil is the deoxyuridylic acid derivative fluorinated at position 5.

So the correct answer is 5-fluoro-2-deoxyuridylic acid.