Does DNA polymerase I require a $3^\prime$ end?

Question

DNA polymerase III adds nucleotides in the $5^\prime \rightarrow 3^\prime$ direction because it can only add nucleotides to the $3^\prime$ end of the previous nucleotide. This is why it requires a primer. However, does DNA polymerase I operate by the same criterion? Does it require a $3^\prime$ end of a previous nucleotide in order to bind successive DNA nucleotides?

If it does, then how can it do so for Okazaki fragments when each Okazaki fragment is unbonded to each other? It is the DNA ligase that finally catalyzes the phosphodiester linkage between the $3^\prime$ end $5^\prime$ beginning of two Okazaki fragments, isn’t it?

If it does not, then what’s the deal with telomeres? After each DNA replication event, the DNA gets shorter and shorter at the very ends because that final primer can be removed but not replaced by DNA via DNA polymerase I, correct? This suggests to me that DNA polymerase I requires a previous nucleotide's $3^\prime$ end to work with, and it has confused me regarding its action on Okazaki fragments in conjunction with DNA ligase.

Answer 1

DNA Pol I does require the 3' end of a previous nucleotide to initiate elongation.

Regarding Okazaki fragments, this is accomplished by the annealing of small RNA primers to the lagging strand part of a replication fork. DNA Pol I extends the lagging strand off of the 3' end of those primers, generating the Okazaki fragments. The RNA primers are later removed, leaving gaps between Okazaki fragments, which are later filled in through the combined actions of DNA PolI and DNA ligase.

In the case of telomeres, the final RNA primer can't be filled by DNA Pol I because polymerase requires a free 3' hydroxyl for the attachment of the first DNA nucleotide. After removing out the RNA primer, the first DNA nucleotide would need to be attached to the DNA nucleotide preceding it, which is not found in the case of the final primer on a telomere, since it's all the way at the end of the linear strand.

In this case then, DNA Pol I removes the RNA primer is removed by DNA pol I but cannot replace it with DNA, leaving a gap (step 6 in the following image).