91Ó°ÊÓ

Streptomyces coelicolor has a linear chromosome. Interestingly, there are no genes that encode essential proteins near the ends of the chromosome in this bacterium. Why do you think this is the case?

1 has a point mutation in the \(-10\) region of the promoter of a structural gene encoding an enzyme needed for synthesi… # You have isolated several E. coli mutants: Mutant #1 has a point mutation in the \(-10\) region of the promoter of a structural gene encoding an enzyme needed for synthesis of the amino acid serine. Mutant #2 has a mutation in the \(-35\) region in the promoter of the same gene. Mutant #3 is a double mutant with mutations in both the \(-10\) and \(-35\) region of the promoter of the same gene. Only Mutant #3 is unable to make serine. Why do you think this is so?

DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an \(\mathrm{N}\)-terminal domain with \(5^{\prime}\) to \(3^{\prime}\) exonuclease activity required for removal of the RNA primer, a central domain responsible for \(3^{\prime}\) to \(5^{\prime}\) exonuclease proofreading, and a C-terminal domain with polymerase activity. Pol I is thought to simultaneously remove RNA primers and fill in the gaps that result (figure 13.14). A group of proteins known as RNaseH also have \(5^{\prime}\) to \(3^{\prime}\) exonuclease activity and can thus remove RNA primers. However, they lack the other two functions observed for Pol I. Predict the ability of the following mutants to replicate DNA: (1) a strain with a mutant gene encoding Pol I such that it no longer has polymerase activity (but retains both types of nuclease activities); (2) a strain without RNaseH proteins; (3) a strain with a mutant gene encoding Pol I such that it no longer has \(5^{\prime}\) to \(3^{\prime}\) exonuclease activity (but retains \(3^{\prime}\) to \(5^{\prime}\) nuclease and polymerase activities); (4) a strain with the mutant Pol I described in (3) and a strain lacking all RNaseH proteins. Explain your reasoning for each.