Final answer:
In a deoxyribonucleotide, a phosphate can only be esterified to the 5' carbon of the sugar. This position is crucial for the formation of the phosphodiester bonds that create the sugar-phosphate backbone of DNA, with a free 5' phosphate and a 3'-OH group marking the ends of the polynucleotide chain.
Step-by-step explanation:
In a deoxyribonucleotide, a phosphate group can be esterified to the 5' carbon of the sugar deoxyribose. Specifically, a nucleotide is formed by the esterification of phosphoric acid to the -OH group at the 5th position of the pentose sugar in a nucleoside. This allows for the formation of phosphodiester bonds during the polymerization of DNA, where a phosphate group attached to the 5' carbon of one deoxyribonucleotide forms a bond with the hydroxyl group on the 3' carbon of the sugar of the next nucleotide, resulting in a 5'-3' linkage.
Additionally, in a polynucleotide chain, the sugar-phosphate backbone is characterized by having a free 5' phosphate at one end and a free 3'-OH at the other end.