Concept explainers
DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 5' to 3' exonuclease activity required for removal of the RNA primer, a central domain responsible for 3' to 5' 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' to 3' exonuclease activity and can thus remove RNA primers. However, they lack the other two functions observed for Pol 1. 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' to 3' exonuclease activity (but retains 3' to 5' 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.
Read the original paper: Fukushima, S., et al. 2008. Reassessment of the in vivo functions of DNA polymerase I and RNaseH in bacterial cell growth. J. Bacteriol. 189:8575.
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Prescott's Microbiology
- DNA polymerases are capable of editing and error correction, meaning it is able to edit and correct single base error so that the gene is not affected. However, RNA polymerase has a limited capacity for error correction. Given that a single base error in either replication or transcription can lead to error in protein synthesis, suggest a brief explanation for this difference in the capability of error correction between DNA polymerase and RNA polymerase.arrow_forwardIn the following sequence, a cytosine was deaminated and is now a uracil (underlined). 5’-GGTAUTAAGC-3’ a. Which repair pathway(s) could restore this uracil to cytosine? b. If the uracil is not removed before a DNA replication fork passes through, what will be the sequences of the two resulting double helices? Provide the sequences of both strands of both helices. Label the old and new strands and underline the mutation(s). c. Could the mismatch repair pathway fix the mutations you’ve indicated in part b? d. If the cell undergoes mitosis, and the replicated DNAs are distributed into the two daughter cells. Will 0, 1, or 2 daughter cells have a mutation in this sequence?arrow_forwardAs shown, five DnaA boxes are found within the origin of replication in E. coli. Take a look at these five sequences carefully. A. Are the sequences of the five DnaA boxes very similar to each other? (Hint: Remember that DNA is double-stranded; think about these sequences in the forward and reverse directions.) B. What is the most common sequence for a DnaA box? In other words, what is the most common base in the first position, second position, and so on until the ninth position? The most common sequence is called the consensus sequence. C. The E. coli chromosome is about 4.6 million bp long. Based on random chance, is it likely that the consensus sequence for a DnaA box occurs elsewhere in the E. coli chromosome? If so, why aren’t there multiple origins of replication in E. coli?arrow_forward
- As described in Table, what is the difference between a rapidstop and a slow-stop mutant? What are different roles of the proteins that are defective in rapid-stop and slow-stop mutants?arrow_forwardPlease write the sequence of the mRNA transcript transcribed from the given DNA double helix by indicating template and non template strands.(SLO1)5’-ACGGCATGCATGGTTTAAAAGGGGCCCAAAA-3’3’-TGCCGTACGTACCAAATTTTCCCCGGGTTTT-5’arrow_forwardAlthough DNA polymerases require both a template and a primer, the following single-stranded polynucleotide was found to serve as a substrate for DNA polymerase in the absence of any additional DNA.3′ HO-ATGGGCTCATAGCCGGAGCCCTAACCGTAGACCACGAATAGCATTAGG-p 5′What is the structure of the product of this reaction?arrow_forward
- During the process of transcription, the polymerase must handle several distinct polynucleotide chains: the double-stranded DNA ahead of the polymerase, the single-stranded template DNA, the nontemplate DNA strand, and the freshly-synthesized mRNA. How does the polymerase keep each of these strands--as well as the incoming ribonucleotides needed for the polymerase reaction--separate?arrow_forwardMistakes made by DNA polymerase are corrected either by proofreading mechanisms during DNA replication or by DNA repair systems that operate after replication is complete. The overall rate of errors in DNA replication is about 1 × 10−10, that is, one error in 10 million base pairs. RNA polymerase also has some proofreading capability, but the overall error rate for transcription is significantly higher (1 × 10−4) or one error in each 10,000 nucleotides). Why can organisms tolerate higher error rates for transcription than for DNA replication?arrow_forwardAlthough DNA polymerases require both a template and a primer, the following single-stranded polynucleotide was found to serve as a substrate for DNA polymerase in the absence of any additional DNA.arrow_forward
- This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.arrow_forwardUsing the numbered sites on the DNA double helix strands below, where would the DNA Primer ACTTGCGA bind to for DNA Amplification?arrow_forwardGiven the following DNA sequence of the template strand for a given gene: 5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3' Part A ) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends). Part B ) Use the genetic code to write the peptide sequence translated in a cell from the mRNA in part A. Please use the 3 letter abbreviation for each amino acid. Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?arrow_forward
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