Mutations in the dnaA gene of E. coli are lethal. Therefore, we can only study such mutations if we are able to generate conditional mutations (mutations whose effect takes place only in certain conditions). One type of conditional mutation manifests the mutant phenotype at high temperatures and is thus called a temperature-sensitive mutation. Temperature-sensitive dnaA mutants grow and replicate their DNA at the permissive temperature of 25°C, but they do not grow or replicate their DNA at the restrictive (non-permissive) temperature of 37°C. Two observations helped to determine the function of the dnaA gene product: • First, in vitro studies using DNA templates that have been unwound do not require the DnaA protein for DNA synthesis. Second, if intact cells are grown at 25°C and then shifted to 37°C, DNA synthesis continues at this temperature until one round of replication is completed, and then DNA synthesis stops. a. What two conclusions can you draw from these observations about the role of the dnaA gene product? b. Explain your answers.

Scientists have observed that individuals with LOW birthweight are MORE likely to become obese as teenagers. To study this observation in more depth, a scientist compared the methylation of the promoter region of the leptin gene in low birthweight (LBW) and normal birthweight (NBW) individuals by treating isolated DNA from these individuals with bisulfite and comparing the results to untreated DNA. The results are shown below. What conclusion can you draw?   Untreated: ATTCGATTCGCGAA Treated NBW: ATTCGATTCGCGAA Treated LBW: ATTTGATTCGTGAA Scientists have observed that individuals with LOW birthweight are MORE likely to become obese as teenagers. To study this observation in more depth, a scientist compared the methylation of the promoter region of the leptin gene in low birthweight (LBW) and normal birthweight (NBW) individuals by treating isolated DNA from these individuals with bisulfite and comparing the results to untreated DNA. The results are shown below. What…

The experiment below is from a seminal set of experiments in the 1960s that illustrated the role of various repair pathways for DNA damage caused by UV radiation. In this experiment, the scientists isolated E coli strains that are mutant in the Rec A gene, the UvrA gene or both. They then irradiated cultures of each strain with increasing doses of UV light and measured the effect on cell viability. Answer the following questions about this data. A. Which DNA repair pathway and repair activity is inhibited by the Rec A mutant? B. Which DNA repair pathway and repair function is inhibited by UvrA mutant? C. Why is the UvrA/RecA double mutant so much more senitive to UV light than either mutant alone?

You are working with a recA- strain of E. coli (cannot produce functional RecA, nor carry out homologous recombination). In which horizontal gene transfer process(es) could cells of this strain successfully undergo genetic change?     In conjugation, as a recipient of a portion of chromosomal DNA from a Hfr donor.     In conjugation, as a recipient of a plasmid from a F+ donor.     In natural transformation, as a recipient of a portion of chromosomal DNA.     There are no HGT processes that could result in genetic change of this strain.     In generalized transduction, as a recipient of a portion of chromosomal DNA.

Describe a typical program used for DNA amplification with the thermal cycler and the purpose of each steps.

A biologist is attempting to clone the gene encoding a particular enzyme (Enz) into a plasmid vector in E.coli. This plasmid has a gene encoding a green fluorescent protein (GFP) as well as a gene for tetracycline antibiotic resistance (TetR). The restriction site (to clone foreign DNA into) is within the GFP sequence. Which of the following would be expected when trying to see which E. coli cells acquired the recombinant plasmid (i.e., carrying the Enz gene)?     Bacteria UNABLE to grow on tetracycline-containing media AND are NOT able to make green fluorescent protein are the ones that contain the recombinant plasmid.     Bacteria able to grow on tetracycline-containing media AND that are NOT able to make green fluorescent protein are the ones that contain the recombinant plasmid.     Bacteria able to grow on tetracycline-containing media AND are able to make green fluorescent protein are the ones that contain the recombinant plasmid.     Bacteria UNABLE to…

In the experiment of Figure shown , a met− bio− thr+ leu+ thi+ cellcould become met+ bio+ thr+ leu+ thi+ by a (rare) double mutationthat converts the met− bio− genes into met+ bio+. Likewise, a met+bio+ thr− leu− thi− cell could become met+ bio+ thr+ leu+ thi+ bythree mutations that convert the thr− leu− thi− genes into thr+ leu+thi+. From the results of Figure shown, how do you know that the occurrenceof 10 met+ bio+ thr+ leu+ thi+ colonies is not due to thesetypes of rare double or triple mutations?

DNA is extracted from the blood cells of two different humans, individuals 1 and 2. In separate experiments, the DNA from each individual is cleaved by restriction endonucleases A, B, and C, and the fragments are separated by electrophoresis. A hypothetical map of a 10,000 bp (base pair) segment of a human chromosome is shown (1 kbp= 1,000 bp). Individual 2 has point mutations that eliminate restriction recognition sites B* and C*. After the 10kbp segment is digested with each restriction endonuclease A, B, and C one by one, samples are loaded onto an agarose gel for electrophoresis for analysis. I would like you to draw the result of gel electrophoresis. To answer the question, first, draw the whole gel image (on the right) on your answer sheet, then indicate where you expect to see the bands on the gel. (Hint1: Same-size bands will appear at the same position. Hint 2: On the gel bands from individuals 1 and 2 might be at different positions.) The left lane (with an M label on the…

You are analyzing the intracellular DNA intermediates formed during the replication of the single-stranded (ss) DNA phage M13. In a replication mutant, you find accumulation of DNA molecules with the following properties:Results of a 0.7% agarose gel electrophoresis run at pH 8.0 – gel is running top tobottom, with the origin at the top of the diagram. The smudges indicate several bandswith slightly different migration rates.   open circle M13closed circle M13 Left lane — markers (wild-type open circle M13 and closed circle M13)2nd lane — DNA extracted from mutant M13 phage cells3rd lane — DNA extracted from mutant M13 phage cells denatured4th lane — DNA extracted from normal M13 phage cellsRight lane — DNA extracted from normal M13 phage cells denaturedProfiles from a 5 – 20% sucrose gradient centrifuging of the mutant M13 phage DNA   The arrow on the neutral (pH 7.4) gradient shows the position of the covalently closed circular DNA. On the alkaline (pH 12.5) gradient, note two peaks…

Refer to Figure, which presents the Meselson–Stahl experiment. DNA molecules can be denatured by high pH, as well as by heat. Suppose that the CsCl gradient centrifugations were run at pH 12, conditions under which DNA strands separate. Sketch the gradient profiles expected for each of the four samples depicted in the figure.