A dinucleotide: phosphodiester link between 3′ and 5′ carbonsएक डाइन्यूक्लिओटाइड: 3′ और 5′ कार्बनों के बीच फ़ॉस्फ़ोडाइएस्टर बंधA dinucleotide: phosphodiester link between 3′ and 5′ carbons

NCERT Fig 10.6 · Class 12 Chemistry · Chapter 10 · interactive diagram

Parts and what they mean

• Nucleotide 1 (न्यूक्लिओटाइड 1) — Its 3′-OH has formed an ester with the phosphate.
• Nucleotide 2 (न्यूक्लिओटाइड 2) — Joined through the phosphate at its 5′ carbon.
• Phosphodiester linkage (फ़ॉस्फ़ोडाइएस्टर बंध) — One phosphate joined by ester bonds to the 3′ carbon of one sugar and the 5′ carbon of the next. Repeating it gives the sugar–phosphate backbone: …sugar–phosphate–sugar–phosphate… with a base on each sugar.
• Base of nucleotide 1 (न्यूक्लिओटाइड 1 का क्षारक) — The order of bases along the chain = primary structure of the nucleic acid.
• Base of nucleotide 2 (न्यूक्लिओटाइड 2 का क्षारक) — Attached at C-1′.
• 5′ end (5′ सिरा) — Free 5′-CH₂OH (or phosphate) end of the chain.
• 3′ end (3′ सिरा) — Free 3′-OH; the next nucleotide would join here.

Original diagram drawn by Simple Learning Hub, based on NCERT Fig 10.6. Report a mistake