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Transcript
The amino and keto forms represent the most stable forms of the bases, i.e., the equilibria lie far to the left. Note that the tautomeric forms of each base have different H-bonding characteristics- As indicated above, N-1 of adenine is a H-bond acceptor in the amino form, but a H-bond donor in the imino form. Likewise, the keto oxygen on C-4 of thymine can only act as a H-bond acceptor while the enol form can act as a H-bond donor. As we will see later the complementary pairing of bases in double stranded DNA normally involves Hbonding of amino and keto forms as follows- Since the imino form of A (adenine) can’t form H-bonds to the keto form of T (thymine) in the same manner, tautomeric forms have been proposed to contribute to spontaneous mutations that result from the formation of abnormal base pairs such as- Compounds formed from replacement of the OH group of D-ribose or 2-deoxy-D-ribose by a base to form an N-glycosidic bond are called nucleosides. Note that replacement of the OH group by the base, changes the hemiacetal structure of C-1 to an acetal type structure- Fig 21-3(b) shows the structures of the deoxyribonucleosides corresponding to the bases found in DNA. Observe that the N-glycosidic linkage involves C1' of the sugar and N-9 of the purine or N-1 of the pyrimidine. The complete structures, names and abbreviations of the four nucleotides found in DNA (Fig 22-3) are- Observe that the phosphate group is bound to the sugar in the form of an ester linkage, i.e., a phosphate ester of phosphoric acid and an alcohol. Now let's consider the manner in which the nucleotides are linked together to form a linear polymer. The nucleotides are linked by 5'-3' phosphodiester bonds (Fig 22-4) as follows- In the polymer, the terminal nucleotide at one end will have a free 5'-phosphate group whereas the terminal nucleotide at the other end will have a free 3'OH group. The direction of the chain is defined as 5'→3'. Note that when the phosphate group forms a diester linkage there is only one ionizable oxygen that is usually represented in the ionized (charged) form. Consider why the phosphodiester is in the ionized form. Recall that phosphoric acid itself has three acidionization equilibria- The ionization of the phosphodiester group found in a nucleic acid corresponds to- Based on the pKa, we would expect that the unionized (acid) form of such a polymer of nucleotides should be a strong acid, hence the name nucleic acid. We should also recognize that the ionized (basic) form with the negative charge on the phosphate oxygen exists almost exclusively at pH 7.0. We can show this by calculating the ratio of the basic to acidic forms at pH 7.0-