pob5e_solutions_ch26 - 2608T_ch26sm_S294-S300 12:19 am Page...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
RNA Metabolism S-294 1. RNA Polymerase (a) How long would it take for the E. coli RNA polymerase to synthesize the pri- mary transcript for the E. coli genes encoding the enzymes for lactose metabolism (the 5,300 bp lac operon, considered in Chapter 28)? (b) How far along the DNA would the transcription “bubble” formed by RNA polymerase move in 10 seconds? Answer (a) Elongation of the RNA transcript in E. coli proceeds at about 50 to 90 nucleotides per second. Thus, the time required to produce the primary transcript in this case is ± 60 to 100 s (b) Because elongation occurs at 50 to 90 nucleotides per second, in 10 seconds the bubble travels (10 s)(50 to 90 nucleotides/s) ± 500 to 900 nucleotides 2. Error Correction by RNA Polymerases DNA polymerases are capable of editing and error correc- tion, whereas the capacity for error correction in RNA polymerases seems to be quite limited. Given that a single base error in either replication or transcription can lead to an error in protein synthesis, suggest a possible biological explanation for this difference. Answer For each gene, RNA polymerase produces many RNA transcripts, so an error in any one transcript would result in only a small fraction of protein with an incorrect amino acid residue. This defective protein will probably be degraded quickly. An error in the mRNA does not pass to subsequent generations of cells because the mRNA itself is degraded. For DNA replication, however, errors would be transmitted to the next generation of cells. 3. RNA Posttranscriptional Processing Predict the likely effects of a mutation in the sequence (5 ² )AAUAAA in a eukaryotic mRNA transcript. Answer A key signal for cleavage and 3 ² -polyadenylation of the primary mRNA transcript is the sequence (5 ² )AAUAAA. After RNA polymerase II has transcribed beyond this sequence, an endonuclease cleaves the transcript at a position about 25 to 30 nucleotides 3 ² to the AAUAAA, and polyadenylate polymerase adds a string of 20 to 250 A residues to the 3 ² end, generating the 3 ² poly(A) tail. The mutation would prevent this cleavage and polyadenylation. A transcript that is not polyadenylated is unstable, so the steady-state level of this mRNA would be very low and little or no protein would be produced. 5,300 nucleotides ³³³ 50 to 90 nucleotides/s chapter 26 2608T_ch26sm_S294-S300 02/26/2008 12:19 am Page S-294 pinnacle OS9:Desktop Folder:WHQY028:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Chapter 26 RNA Metabolism S-295 4. Coding versus Template Strands The RNA genome of phage Q b is the nontemplate strand, or cod- ing strand, and when introduced into the cell it functions as an mRNA. Suppose the RNA replicase of phage Q b synthesized primarily template-strand RNA and uniquely incorporated this, rather than nontemplate strands, into the viral particles. What would be the fate of the template strands when they entered a new cell? What enzyme would have to be included in the viral particles for successful invasion of a host cell?
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 7

pob5e_solutions_ch26 - 2608T_ch26sm_S294-S300 12:19 am Page...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online