Question

Which carbon atom of ribose sugar is joined to nitrogen base to form nucleoside?

• A. C $-$ 2'
• B. C $-$ 5'
• C. C $-$ 3'
• D. C $-$ 1'

Hint:

To easily remember nucleoside/nucleotide structure, use the "1-3-5" rule for the sugar attachments:
The Base is always at position 1', the next nucleotide connects at position 3', and the Phosphate group is at position 5'.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to identify the specific carbon position on the pentose (ribose) sugar ring that directly bonds with a nitrogenous base to create a nucleoside structure.

Step 2: Key Formula or Approach:
A nucleoside is fundamentally composed of a five-carbon sugar (ribose in RNA, deoxyribose in DNA) covalently bonded to a nitrogenous base (purine or pyrimidine).
This specific covalent connection is known as a $\beta$-N-glycosidic bond.

Step 3: Detailed Explanation:
In the standard nomenclature of a nucleotide/nucleoside, the carbon atoms of the sugar ring are numbered with primes (1', 2', 3', 4', 5') to clearly distinguish them from the unprimed numbered atoms in the nitrogenous base.
The anomeric carbon of the furanose (sugar) ring—which is the most reactive carbon and was part of the carbonyl group in the straight-chain form—is always the C-1' carbon.
The nitrogenous base exclusively attaches to this anomeric C-1' carbon. (Specifically, it binds to the N-1 position of pyrimidines or the N-9 position of purines).

Step 4: Final Answer:
The nitrogenous base joins at the C-1' carbon of the ribose sugar, which corresponds to option (D).