1. The sequence of the coding (+, sense) strand is

5′-ATGGGGAACAGCAAGAGTGGGGCCCTGTCCAAGGAG-3′

and the sequence of the template (−, antisense) strand is

3′-TACCCCTTGTCGTTCTCACCCCGGGACAGGTTCCTC-5′

2. An error will affect only one molecule of mRNA of many synthesized from a gene. In addition, the errors do not become a permanent part of the genomic information.

3. At any given instant, only a fraction of the genome (total DNA) is being transcribed. Consequently, speed is not essential.

4. The active sites are related by convergent evolution.

5. Heparin, a glycosaminoglycan, is highly anionic. Its negative charges, like the phosphodiester bridges of DNA templates, allow it to bind to lysine and arginine residues of RNA polymerase.

6. This mutant σ will competitively inhibit the binding of holoenzyme and prevent the specific initiation of RNA chains at promoter sites.

7. The core enzyme without σ binds more tightly to the DNA template than does the holoenzyme. The retention of σ after chain initiation would make the mutant RNA polymerase slower. The mutant enzyme would also be unlikely to bind alternative σ factors.

8. A 100-kDa protein contains about 910 residues, which are encoded by 2730 nucleotides. At a maximal transcription rate of 50 nucleotides per second, the mRNA would be synthesized in 55 s.

9. The RNA polymerase slides along the DNA rapidly rather than simply diffusing through three-dimensional space.

10. The start site is in red:

5′-gccgttgacaccgttcggcgatcgatccgctataatgtgtggatccgctt-3′

11. Initiation at strong promoters takes place every 2 s. In this interval, 100 nucleotides are transcribed. Hence, centers of transcription bubbles are 34 nm (340 Å) apart.

12. (a) The lowest band on the gel will be that of strand 3 alone (i). Band ii will be at the same position as band i because the RNA is not complementary to the nontemplate strand, whereas band iii will be higher because a complex is formed between RNA and the template strand. Band iv will be higher than the others because strand 1 is complexed to 2, and strand 2 is complexed to 3. Band v is the highest because core polymerase associates with the three strands.

(b) None, because rifampicin acts before the formation of the open complex.

(c) RNA polymerase is processive. When the template is bound, heparin cannot enter the DNA-binding site.

(d) When GTP is absent, synthesis stops when the first cytosine residue downstream of the bubble is encountered in the template strand. In contrast, with all four nucleoside triphosphates present, synthesis will continue to the end of the template.

13. Backtracking must occur prior to cleavage leading to dinucleotide products.

14. The base-pairing energy of the di- and trinucleotide DNA–RNA hybrids formed at the very beginning of transcription is not sufficient to prevent strand separation and loss of product.

A43

15. (a) Because cordycepin lacks a 3′-OH group, it cannot participate in 3′ → 5′ bond formation. (b) Because the poly(A) tail is a long stretch of adenosine nucleotides, the likelihood that a molecule of cordycepin would become incorporated is higher than with most RNA. (c) Yes, it must be converted into cordycepin 5′-triphosphate.

16. There are 2⁸ = 256 possible products.

17. The relationship between the −10 and −35 sequences could be affected by torsional strain. Since topoisomerase II introduces negative supercoils in DNA, this prevents this enzyme from overstimulating the expression of its own gene.

18. Ser-Ile-Phe-His-Pro-Stop

19. A mutation that disrupted the normal AAUAAA recognition sequence for the endonuclease could account for this finding. In fact, a change from U to C in this sequence caused this defect in a thalassemic patient. Cleavage occurred at the AAUAAA 900 nucleotides downstream from this mutant AACAAA site.

20. One possibility is that the 3′ end of the poly(U) donor strand cleaves the phosphodiester bond on the 5′ side of the insertion site. The newly formed 3′ terminus of the acceptor strand then cleaves the poly(U) strand on the 5′ side of the nucleotide that initiated the attack. In other words, a uridine residue could be added by two transesterification reactions. This postulated mechanism is similar to the one in RNA splicing.

21. Alternative splicing, RNA editing. Covalent modification of the proteins subsequent to synthesis.

22. Attach an oligo(dT) or oligo(U) sequence to an inert support to create an affinity column. When RNA is passed through the column, only poly(A)-containing RNA will be retained.

23. (a) Different amounts of RNA are present for the various genes.

(b) Although all of the tissues have the same genes, the genes are expressed to different extents in different tissues.

(c) These genes are called housekeeping genes—genes that most tissues express. They might include genes for glycolysis or citric acid cycle enzymes.

(d) The point of the experiment is to determine which genes are initiated in vivo. The initiation inhibitor is added to prevent initiation at start sites that may have been activated during the isolation of the nuclei.

24. DNA is the single strand that forms the trunk of the tree. Strands of increasing length are RNA molecules; the beginning of transcription is where growing chains are the smallest; the end of transcription is where chain growth stops. Direction is left to right. Many enzymes are actively transcribing each gene.