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uranmaximum [27]
2 years ago
4

Nucleotide Structure Which positions in the purine ring of a purine nucleotide in DNA have the potential to form hydro-gen bonds

but are not involved in Watson-Crick base pairing?
Biology
2 answers:
DedPeter [7]2 years ago
4 0

Answer:

Purine is referred to as a heterocyclic aromatic organic compound that consists of a pyrimidine ring fused to an imidazole ring. Purines nitrogenous base makes up the nucleotide base in DNA. All the five purine ring nitrogens (N-1, N-3, N-7, and N-9) have the potential to form hydrogen bonds yet only N-1 is involved in Watson-Crick hydrogen bonding with a pyrimidine. The N-9 is actively involved in the the N-glycosyl linkage with deoxyribose and has very little hydrogen bonding capacity.  The N-3 and N-7 are readily available to form further hydrogen bonds.

slamgirl [31]2 years ago
4 0

Answer:

N-3 and N-7 position are available

Explanation:

DNA is a biopolymer of

 deoxyribonucleic acid (a type of nucleic acid) that has four different chemical groups in the category of purine and pyrimidine, called bases: adenine, guanine, cytosine, and thymine .

Purine refers to any of a class of organic heterocyclic compounds, composed of fused pyrimidine and imidazole rings, that constitute one of the two groups of organic nitrogenous bases (the other being the pyrimidines) and are components of nucleic acids.

A pyrimidine is a diazine in which the two nitrogen atoms are in the meta- positions; it is the basis of three of the bases found in DNA and RNA, thymine, uracil and cytosine.

Linking to the above nitrogens of the purine ring at position (N-1, N-3, N-7, and N-9) have the potential to form hydrogen bonds.

According to Watson-Crick model they showed that N-1 is involved in the hydrogen bonding with a pyrimidine base, while N-9 is takes on with the N-glycosyl linkage with sugar (deoxyribose) and has relatively a very limited tendency for hydrogen bonding.

N-3 and N-7 positions are available for to form other sorts of bonds (hydrogen bonding) and are not busy with the Watson-Crick pairing.

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