We know that DNA polymerase III requires free 3′-OH groups for initiating replication. Would pol-I require free 3′-OH groups when filling in the gaps between Okazaki fragments? If yes, how are free 3′-OH groups provided since primers are removed prior to synthesis
DNA polymerase III is responsible for the latest Okazaki fragment's extension (a DNA-dependent DNA polymerase). Deoxynucleotide polymerization proceeds before the 3′ hydroxyl of the primer on the prior Okazaki fragment is reached. DNA polymerase I, which has 5′ to 3′ exonuclease activity, removes the primer on the prior Okazaki fragment one base at a time. All errors associated with the RNA primer are corrected, and each ribonucleotide is replaced with the equivalent deoxyribonucleotide. A separate enzyme, DNA ligase, joins the last deoxyribonucleotide, which uses one ATP to attach the Okazaki fragment into the growing lagging chain.
Okazaki fragment joining requires removal of the RNA primer, DNA replication to complete synthesis, and processing of the ends by nucleases to create a ‘nick’ that can be closed by the action of DNA ligase
The answer is yes, and the 3 -OH groups are provided by an exonuclease which creates a 'nick' for attachment