Chapter 15

Question 15.1

The promoter is a site for RNA polymerase loading and initiation of transcription. To accomplish this, the RNA polymerase must form an open complex in which the two DNA strands are separated over a short distance. Due to the effects of AT-rich versus GC-rich DNA on the overall thermal stability of DNA, the strand separation is more readily accomplished in sequences that are AT-rich (see Chapter 4).

S-15

Question 15.2

The deletion would move the −35 sequence closer to the −10 sequence by half a helical turn of the DNA, putting the two elements on opposite faces of the DNA duplex. This would dramatically reduce binding of sigma factor to the promoter, thereby decreasing transcription efficiency.

Question 15.3

5′-AUGACCAUGAUUACG. The sequence reported for a gene is, by convention, that of the coding strand, and sequences are always written in the 5′→3′ direction.

Question 15.4

At 50 to 90 nucleotides per second, the enzyme would take 34 to 61 seconds to transcribe the gene.

Question 15.5

Assuming the two genes use similar transcription initiation modes, tesA will be more efficiently transcribed. The tesB −10 sequence deviates more from the consensus sequence, and its higher G≡C content will be more difficult to melt.

Question 15.6

The number of transcripts would increase, because Pol III typically generates many more transcripts than Pol II. tRNAs, which are synthesized by Pol III, are required in much greater quantities in the cell than are most mRNAs, which are made by Pol II.

Question 15.7

Initiation (including abortive initiation), the slowest step; elongation, the fastest step but punctuated by pauses; and termination, requiring stalling and subsequent dissociation of the polymerase from the DNA.

Question 15.8

About 8 to 10 phosphodiester bonds can be formed in the initiation phase. RNA synthesis always begins at a unique location (defined as +1) relative to the promoter sequence. Elongation can be prevented by controlling the sequence of the DNA template strand and the rNTPs added to the reaction, and the reaction products are defined by the same parameters. For example, if the first G residue in the DNA template strand does not appear until position +6, and you add ATP, UTP, and GTP, but no CTP, RNA synthesis will be limited to the first five nucleotides, and the dominant reaction product is likely to be pentanucleotides. This strategy has been used in research on RNA polymerase. (See, for example, W. R. McClure, C. L. Cech, and D. E. Johnston, J. Biol. Chem. 253:8941–8948, 1978.)

Question 15.9

Many RNAs are present in the body, synthesized before ingestion of the α-amanitin. The mRNAs support cell and tissue function until they are degraded, and typically last for about two days.

Question 15.10

Intercalation means that the planar portion of a small molecule inserts into the double-helical DNA between successive base pairs, deforming the DNA and preventing movement of RNA polymerase along the template. Actinomycin D and acridine act in a similar way.

Question 15.11

This is difficult to do experimentally. The TATA box and the Inr sequence are often, but not always, found upstream from genes transcribed by Pol II. A significant minority of eukaryotic promoters lack a well-defined TATA box—the so-called TATA-less promoters—and this can confound identification by computer algorithms alone.

Question 15.12

No. The two strands of a DNA molecule are antiparallel and complementary (not identical). When the promoter is inverted, it will direct RNA synthesis that uses what was originally the coding strand as the template strand. The mRNA sequence, derived from a different DNA strand and synthesized in the opposite direction, would be very different from the mRNA produced by the original gene and might not even contain an open reading frame.

Question 15.13

Errors in the genetic information in actively transcribed genes are of greater immediate importance to the cell than errors in silent genes. Transcription-coupled DNA repair focuses repair on those DNA sequences most heavily utilized by the cell.

Question 15.14

There may be a palindromic sequence that allows formation of a hairpin structure during transcription, creating a ρ-independent terminator that is not perfectly efficient, or an imperfect rut sequence to guide the loading of the ρ protein. In both cases, the sequences would have to be imperfect so that some transcripts could be elongated through gene B.

Question 15.15

(a) 10–13. (b) 5–8. (c) 2–13. (d) 5–8. (e) Transcription from the PHS promoter requires a factor present in fractions 5–8. (f) Transcription from P1 and P2 uniquely requires the RNA polymerase reconstituted with the 32 kDa protein, and the designation “σ32” is probably warranted. (g) The use of an alternative sigma factor is an efficient way to coordinate regulation of transcription of a group of genes that express products not always required by the cell.