Question

The number of glycosidic bonds in 3 full turn of B-DNA are
30
10
60
120

Answer 1

3 turns have 60 glycosidic bonds…. Since there are 60 nitrogenous bases in 3 turns of DNA … ^.^

Answer 2

Answer:

In B-DNA, there will be 10.5 base pairs per turn of DNA of 2 phosphodiester bonds which are present between 2 adjacent base pairs. Therefore, there must be a 9 phosphodiester bond on the other side. So, the total number of bonds is 18 in one turn of DNA. One base pair has 2 glycosidic bonds, the total number of glycosidic bonds associated with the DNA of human cell is 9.2 x 10^6. So, in full 3-turns there would be 60 glycosidic bonds in B-DNA.

Explanation:

The nitrogen-carbon coupling between both the 9' nitrogen of purine bases or the 1' nitrogen of pyrimidine bases and the 1' carbon of the sugar group is known as a glycosidic bond. Glycosidic bonds were divided into two types: 1,4 alpha and 1,4 beta glycosidic linkages.