In mammalian cells, genes that are expressed in a particular cell are reported to undergo replication during the first half of S phase, and genes not ex- pressed in that cell are replicated in the latter half of S phase. Briefly describe an experiment that could lead to this conclusion. You might consider ap- proaches that involve 5-bromouracil incorporation.
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- In a bacterial culture in which all cells are unable to synthesizeleucine (leu-), a potent mutagen is added, and the cells areallowed to undergo one round of replication. At that point, samplesare taken, a series of dilutions are made, and the cells areplated on either minimal medium or minimal medium containingleucine. The first culture condition (minimal medium) allowsthe growth of only leu+ cells, while the second culture condition(minimal medium with leucine added) allows growth of all cells.The results of the experiment are as follows: Culture Condition Dilution ColoniesMinimal medium 10-1 18Minimal medium + leucine 10-7 6What is the rate of mutation at the locus associated with leucinebiosynthesis?In addition to Tc1, the C. elegans genome contains otherfamilies of DNA transposons such as Tc2, Tc3, Tc4, andTc5. Like Tc1, their transposition is repressed in thegerm line but not in somatic cells. Predict the behaviorof these elements in the mutant strains where Tc1 is nolonger repressed due to mutations in the RNAi pathway.Justify your answer.The E coli cell contains 107 non-specific DNA binding sites for R, 10 R molecules, and 1 Operator. And the mechanism of specific binding of R to O works well with these concentrations. Would it work if R = 1? Explain with a calculation.
- PolyADP-ribose polymerase (PARP) plays a keyrole in the repair of DNA single-strand breaks. In the pres-ence of the PARP inhibitor olaparib, single-strand breaksaccumulate. When a replication fork encounters a sin-gle-strand break, it converts it to a double-strand break,which in normal cells is then repaired by homologousrecombination. In cells defective for homologous recom-bination, however, inhibition of PARP triggers cell death.Patients who have only one functional copy of theBrca1 gene, which is required for homologous recombina-tion, are at much higher risk for cancer of the breast andovary. Cancers that arise in these tissues in these patientscan be treated successfully with olaparib. Explain how it isthat treatment with olaparib kills the cancer cells in thesepatients, but does not harm their normal cells.In Figure 12-26, provide a biochemical mechanism forwhy HP-1 can bind to the DNA only on the left side of thebarrier insulator. Similarly, why can HMTase bind onlyto the DNA on the left of the barrier insulator?Assume that there is a double stranded break on DNA double helix of an eukaryotic cell due to X-ray radiation and it is not repaired. In addition, the cell’s Apaf-1 protein is not expressed due to a null mutation in the Apaf-1 gene. Please discuss the effect of not having Apaf-1 expression in the cell with non-repaired double stranded break.
- Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.Telomerase activity has been found to be 10 to 20 times more active in cancer cells than in normal somatic cells. What is the significance of this circumstance?Consider the following sequence of DNA: 3'-TTA CGG-5'What dipeptide is formed from this DNA after transcription and translation? b. If a mutation converts CGG to CGT in DNA, what dipeptide is formed? c. If a mutation converts CGG to CCG in DNA, what dipeptide is formed? d. If a mutation converts CGG to AGG in DNA, what dipeptide is formed?
- The 3′ → 5′ exonuclease activity of Pol I excises only unpaired 3′-terminal nucleotides from DNA, whereas this enzyme’s pyrophosphorolysis activity removes only properly paired 3′-terminal nucleotides. Discuss the mechanistic signifi cance of this phenomenon in terms of the polymerase reaction.A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. Incorporation of radioactivity, cpm Excess (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?