Nucleotides (dATP, cCTP, dGTP, cTTP) have a -OH at their 3\' position of their d
ID: 176554 • Letter: N
Question
Nucleotides (dATP, cCTP, dGTP, cTTP) have a -OH at their 3' position of their deoxyribose. In the laboratory a nucleotide was created that has a -NH_3 (azido) at this position instead of having an -OH. (The diagrams below show the normal "T" nucleotide: thymine attached to deoxyribose with three phosphates. To the left, the -OH has been replaced by an "azido" group.) If the new nucleotide were incorporated into the growing strand of DNA, would yet another nucleotide be incorporated as well? Explain your answer.Explanation / Answer
Ans. No
DNA polymerization occurs in 5’-3’ direction. 5’ phosphate group of the incoming nucleotide is covalently lined to the 3’-OH group of the terminal nucleotide at growing end. The first abnormal A”TP (or any nucleotide with azido at 3’ end) might got incorporated in the growing chain because it’s 5’phosphate end was needed to form a covalent bond with the 3’-OH of preceding nucleotide on the chain.
3’-OH (preceding nucleotide at 3’ end) + Phosphate group at 5’end of incoming A”TP =
Phosphodiester bond formed
When the second A”TP nucleotide comes at the active site of the enzyme, the formation of phosphodiester bond can’t be catalyzed. Because the 3’end of preceding nucleotide is an azido group, not the normal -OH group. Azido group and -OH groups have different molecular geometry, polarity and orientation of atoms. Therefore, the active site of DNA polymerase can’t catalyze the formation of phosphodiester bond between 3’azido group (first A”TP) and a normal phosphate group on incoming A”TP- because the active site is not specific for these two groups.
3’-Azido (1st alerted nucleotide at 3’ end) + Phosphate group at 5’end of incoming A”TP =
Phosphodiester bond NOT formed
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