Q1A) Fig. 6.10, the blue curve suggests high level of transcription initiation for that set of samples. Why did it increase?Q1B) In the same figure, the data for the green curve was obtained with a so-called rifampicin-resistant core. You would expect that the green curve should overlap with the blue one, but it doesn’t. What is the reason? RNA chain initiation5.04.03.02.01.0Add core1020Time (min)3040Figure 6.10 Sigma can be reused. Travers and Burgess allowedRNA polymerase holoenzyme to initiate and elongate RNA chains on aT4 DNA template at low ionic strength, so the polymerases could notdissociate from the template to start new RNA chains. The red curveshows the initiation of RNA chains, measured by [y-³2PJATP and[y-³2P]GTP incorporation, under these conditions. After 10 min (arrow),when most chain initiation had ceased, the investigators added new,rifampicin-resistant core polymerase in the presence (green) orabsence (blue) of rifampicin. The immediate rise of both curvesshowed that addition of core polymerase can restart RNA synthesis,which implied that the new core associated with o that had beenassociated with the original core. In other words, the σ was recycled.The fact that transcription occurred even in the presence of rifampicinshowed that the new core, which was from rifampicin-resistant cells,together with the old σ, which was from rifampicin-sensitive cells,could carry out rifampicin-resistant transcription. Thus, the core, notthe σ, determines rifampicin resistance or sensitivity. (Source: Adaptedfrom Travers, A.A. and R.R. Burgess, "Cyclic re-use of the RNA polymerase sigmafactor." Nature 222:537-40, 1969.)

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Description: Q1A) Fig. 6.10, the blue curve suggests high level of transcription initiation for that set of samples. Why did it increase?Q1B) In the same figure, the data for the green curve was obtained with a so-called rifampicin-resistant core. You would expec

The enzyme RNA polymerase has following subunits in its core enzyme- 1 beta (β), 1 beta prime (β'),…

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) Fig. 6.10, the blue curve suggests high level of transcription initiation for that set of samples. Why did it increase? Q1B) In the same figure, the data for the green curve was obtained with a so-called rifampicin-resistant core. You would expect that the green curve should overlap with the blue one, but it doesn’t. What is the reason?

) Fig. 6.10, the blue curve suggests high level of transcription initiation for that set of samples. Why did it increase? Q1B) In the same figure, the data for the green curve was obtained with a so-called rifampicin-resistant core. You would expect that the green curve should overlap with the blue one, but it doesn’t. What is the reason?

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 16P

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A fragment of bacterial DNA reads: 3’ -TACCTATAATCTCAATTGATAGAAGCACTCTAC- 5’ Assuming that this fragment is the template strand, what is the sequence of mRNA that would he transcribed? (Hint: Be sure to identify the initiation site.)
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34. Arrange the following list of eukaryotic gene elements in the order in which they would appear in thegenome and in the direction traveled by RNA polymerase along the gene. Assume the gene’s singleintron interrupts the open reading frame. Note thatsome of these names are abbreviated and thus do notdistinguish between elements in DNA versus RNA.For example, splice-donor site is an abbreviation forDNA sequences transcribed into the splice-donor sitebecause splicing takes place on the gene’s RNA transcript, not on the gene itself. Geneticists often usethis kind of shorthand for simplicity, even thoughit is imprecise. (a) splice-donor site; (b) 3′ UTR;(c) promoter; (d) stop codon; (e) nucleotide towhich methylated cap is added; (f) initiation codon;(g) transcription terminator; (h) splice-acceptorsite; (i) 5′ UTR; (j) poly-A addition site; (k) splicebranch site. Concerning the list of eukaryotic gene elements inProblem 34:a. Which of the element names in the list are abbreviated? (That…
BIOC 385 Deciphering the Genetic Code Q11.3: Describe the THREE experimental approaches used to decipher the Genetic Code
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DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 5´ to 3´ exonuclease activities 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. 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 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´ to 3´ exonuclease activities (but retains 3´ to 5´ nuclease and polymerase activities); (4) a strain with…
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