Chapter 36

  1. Transcription is DNA-directed RNA synthesis by RNA polymerase.

  2. The template strand has a sequence complementary to that of the RNA transcript. The coding strand has the same sequence as that of the RNA transcript except for thymine (T) in place of uracil (U).

  3. Beginning, middle, and end, although strictly true, doesn’t count. The three stages are initiation, elongation, and termination.

  4. DNA template, ATP, GTP, UTP, CTP, Mg2+, or Mn2+

  5. The sigma subunit helps the RNA polymerase locate promoter sites. After the promoter is located, the sigma subunit leaves the enzyme and assists another polymerase to find a promoter.

  6. Complete the interactive matching exercise to see answers.

    C37

  7.  

    Transcription

    Replication

    Template

    DNA

    DNA

    Strands copied

    One

    Both

    Enzyme

    RNA polymerase

    DNA polymerase

    Substrates

    Ribonucleotides

    Deoxyribonucleotides

    Primer

    None

    Required

  8. A promoter is a DNA sequence that directs RNA polymerase to the proper initiation site for transcription.

  9. In the closed promoter complex, the DNA is double helical and transcription is not possible. In the open promoter complex, the DNA is unwound, an essential requirement for transcription.

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

    5′-ATGGGGAACAGCAAGAGTGGGGCCCTGTCCAAGGAG-3′

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

    3′-TACCCCTTGTCGTTCTCACCCCGGGACAGGTTCCTC-5′

  11. In RNA synthesis, an error affects 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.

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

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

  14. The core enzyme without σ binds more tightly to the DNA template than does the holoenzyme. The retention of σ after strand initiation would make the mutant RNA polymerase less processive. Hence, RNA synthesis would be much slower than normal.

  15. 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 is synthesized in 54.6 s.

  16. 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.

  17. Riboswitches are special secondary structures formed by some mRNA molecules, capable of directly binding small molecules, that determine whether transcription will continue or cease. For instance, a riboswitch controls the synthesis of an mRNA encoding a protein required for FMN synthesis. If FMN is already present, the riboswitch binds the FMN and traps the RNA transcript in a conformation that favors the termination of further RNA synthesis, preventing the production of functional mRNA. However, if FMN is absent, an alternative conformation forms that allows the production of the full-length mRNA.

    1. Cells will express β-galactosidase, lac permease, and thiogalactoside transacetylase even in the absence of lactose.

    2. Cells will express β-galactosidase, lac permease, and thiogalactoside transacetylase even in the absence of lactose.

    3. The levels of catabolic enzymes such as β-galactosidase will remain low even at low levels of glucose.

  19. Lactose can’t get into the cell, because the permease is missing.

  20. Because of the σ factor, RNA polymerase binds DNA and slides along the DNA in a one-dimensional search for the promoter. Diffusion in one dimension is faster than diffusion in three dimensions, which is how the two small molecules must find each other.

  21. Cleavage of a precursor, addition of CCA to tRNA, and modification of bases

  22. 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 strands are the smallest; the end of transcription is where strand growth stops. Direction is left to right. Many enzymes are actively transcribing each gene.

  23. 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.

  24. A liter is equivalent to 1000 cm3, so 10−12 cm3 is 10−15 l. The concentration is 1/(6 × 1023) mol per 10−15 1 = 1.7 × 10−9 M. Because Kd = 10−13 M, the single molecule should be bound to its specific binding site.

  25. Anti-inducers bind to the conformation of repressors, such as the lac repressor, that is capable of binding DNA. They occupy a site that overlaps that for the inducer and, therefore, compete for binding to the repressor.

  26. 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.

  27. RNAs of different sizes were obtained, designated 10S, 13S, and 17S when boat was added at initiation, a few seconds after initiation, and 2 minutes after initiation, respectively. If no boat was added, transcription yielded a 23S RNA product. Boat, like ρ, is evidently a termination factor. The template that was used for RNA synthesis contained at least three termination sites that respond to boat (yielding 10S, 13S, and 17S RNA) and one termination site that does not (yielding 23S RNA). Thus, specific termination at a site producing 23S RNA can take place in the absence of boat. However, boat detects additional termination signals that are not recognized by RNA polymerase alone. Sadly, your search of the literature reveals that someone has already characterized the factor that you named boat—the termination factor ρ.