The Central Dogma:
DNA makes RNA makes protein


TOPICS: (Click on the underlined text to directly jump to that section)


DNA Replication


Briefly diagram the salient features of DNA replication in E. coli, including: (a) the direction of DNA synthesis relative to the 5' and 3' ends of the DNA, (b) properties of the leading and lagging strands (including the synthesis and joining of Okazaki fragments), and (c) the enzymes involved in each step.

ANSWER: Critical features of DNA replication are shown in the following diagram. DNA replication by DNA pol III begins at the chromosomal origin (ori) and proceeds bidirectionally (i.e. by theta-replication) around the chromosome until the two replication forks meet at the terminus. Initiation of DNA replication requires DNA primase, the DnaA protein, and specific DNA sequences. Termination occurs when the two DNA replication forks collide and requires a topoisomerase to separate the two strands of DNA. (Termination is facilitated by specific DNA sequences and a protein encoded by the tus gene, but they are not essential because they can be deleted and cells still divide normally.)


In order to understand the biochemical mechanism of DNA replication, Kornberg purified an enzyme from E. coli (called DNA polymerase I or Pol I) that could catalyze DNA synthesis Although this enzyme could catalyze DNA synthesis in vitro, this does not prove that the enzyme is essential for DNA synthesis in vivo. To test this question, De Lucia and Cairns sought to isolate a temperature sensitive mutant in the polA gene which encodes Pol I. In such a mutant, Pol I would be active at 30 C but inactive at 42 C. If Pol I was essential for DNA synthesis in vivo, what would happen to a polA(Ts) mutant if it was shifted to 42 C? [Briefly explain your answer.]

ANSWER: If the polA gene product was the primary DNA polymerase, the cells would grow normally at 30 C but when the cells were shifted to 42 C DNA replication would stop and no further cell division or growth could occur. [Note -- DeLucia and Cairns' isolated a polA mutant that was not conditional, indicating that the polA gene is not essential under certain growth conditions and, thus, DNA Polymerase I is NOT the primary DNA polymerase in E. coli.]


Mutants defective for the proofreading function of DNA polymerase III typically form small, unhealthy looking colonies on rich medium. Why?

ANSWER: Such mutations in the dnaQ gene (called mutD) produce a "mutator" phenotype. Because they are unable to proofread errors that occur during DNA replication, such strains accumulate mutations at a high frequency. The resulting large number of "lethal mutations" that arise during cell division slows the growth of the colonies. The mutation frequency is highest when the cells are growing on rich medium which allows the cells to grow more rapidly than the resulting errors can be corrected by other repair systems.


Why does the lagging DNA strand replicate via fragments while synthesis of the leading strand is continuous?


Transcription


Briefly describe the salient features of transcription in E. coli, including: (a) the features of the promoter required for the specific initiation of transcription, (b) the direction of transcription relative to the 5' and 3' ends of the DNA and how elongation occurs, (c) the features of typical transcription termination sites, (d) and the enzyme required.

ANSWER:

  1. The promoter sequence is recognized by RNA polymerase + a specific sigma factor. The "housekeeping" sigma factor in E. coli is called sigma 70 (s70). The RNA polymerase + s70 protein holoenzyme (Es70) binds to a set of sites centered around upstream of the start of transcription. (The first nucleotide of the mRNA is called +1.) The consensus sequence for Es70 is TATAAT (centered around -10) and TTGACA (centered around -35).
  2. RNA polymerase moves along the DNA in a 5' to 3' direction, adding nucleotides to the 3'OH group of the growing RNA chain. The sigma-factor normally dissociates from the RNA polymerase core soon after transcription is properly initiated and chain elongation has begun. The energy for elongation comes from hydrolysis of NTPs.
  3. Factor independent termination described on Simon Cawley's webpage and Rho dependent termination is described on the Microbial Genetics supplement page (see the supplement on Rho-dependent Polarity).
  4. The enzyme required is RNA polymerase. RNA polymerase core is a large protein with four subunits -- two a subunits, one B-subunit, and one B'-subunit. When a sigma-factor associates with the RNA polymerase core, it is called the RNA polymerase holoenzyme.


Compare and contrast RNA polymerase and DNA polymerase III with respect to the following characteristics: (a) ability to initiate polymerization, (b) substrates and chemical bond formed, (c) ability to correct incorporation mistakes, (d) requirement for specific sequences to start polymerization, (e) requirement for a primer to initiate polymerization, (f) rate of elongation of the nucleic acid chain produced.


It is possible to isolate temperature sensitive mutants in genes that are essential for cell growth. Suppose that temperature sensitive mutations inactivate a gene product at high temperature (42 C), but that the gene product remains functional at low temperatures (30 C). If you had a temperature sensitive mutation in the sigma subunit of RNA polymerase, what would happen to transcription when the cells were shifted from 30 C to 42 C?


The antibiotic rifampicin inhibits RNA polymerase. Propose a genetic experiment designed to learn which subunit of the RNA polymerase is the binding site (target) for rifampicin.


Compare and contrast transcription of RNA by RNA polymerase and replication of DNA by DNA pol III.


Translation


Briefly describe the salient features of translation in E. coli, including: (a) the role of ribosomes, mRNA, and tRNA, (b) the ribosome binding site and start codon required for the specific initiation of protein synthesis, and (c) the codons and release factors that signal termination of protein synthesis.

ANSWER:

  1. See the brief cartoon review of transcription and translation. For more review, see your textbook.
  2. In bacteria the ribosome binds to a specific sequence on the mRNA called a ribosome-binding (or Shine-Delgarno) site. This sequence is complementary to a sequence near the 3' end of the 16S rRNA and serves to properly align the ribosome relative to the start codon. A typical ribosome-binding site consists of 3-9 purine bases located about 3-12 nucleotides upstream of the start codon. The start codon is usually AUG, but occasionally GUG, UUG, or AUU is used. In bacteria, at least some Archae, mitochondria, and chloroplasts the start codon is recognized by a distinct tRNA, fMet-tRNA.
  3. Translation termination occurs at stop (or nonsense) codons -- UAG (amber), UAA (ochre), and UGA (opal). There are no cognate tRNAs that recognize these codons, but they are recognized by specific proteins called release factors. When a release factor binds to a stop codon, the ribosome catalyzes hydrolysis of the polypeptide chain. The ribosome then dissociates from the mRNA.


Compare and contrast the termination of translation, termination of transcription, and termination of DNA replication.


An important difference between eukaryotes and prokaryotes is the coupling of transcription and translation in prokaryotes.

  1. What is coupling of transcription and translation?

    ANSWER: Coupling of transcription and translation means that ribosomes bind to the mRNA and protein synthesis (translation) begins while the mRNA is still being synthesized (transcription). A cartoon of coupled transcription and translatio is shown below:

  2. How come transcription and translation are not also coupled in eukaryotes?

    ANSWER: Transcription and translation occur in different cellular compartments in Eukaryotes, separated by a membrane barrier -- transcription occurs inside the nucleus and translation occurs in the cytoplasm.


    The E. coli lac operon encodes three gene products. The three gene products can be assayed independently. Many mutations in the upstream region of the lacZ gene (labeled 1) are strongly polar on expression of the lacY and lacA genes, but mutations near downstream end of the lacZ gene (labeled 2) are much less polar. Propose a molecular explanation for this observation. [Hint: think about the coupling of transcription and translation and the role of Rho factor.]

    ANSWER: The most likely explanation is that when translation termination occurs early in the lacZ gene (due to a nonsense mutation or an out of frame nonsense codon read in a frameshift mutant), Rho factor interacts with the naked mRNA and causes transcription termination or "polarity" (thereby preventing expression of downstream genes). In contrast, although mutations may cause translation termination near the end of the lacZ gene, the translation start site preceding the lacY gene are very close, so the dissociated ribosomes re-bind to the mRNA and prevent Rho dependant transcription termination. (See on the supplement on the web). Missense mutations do not cause translation termination so they would not be polar, but recall that because they often have no observable phenotype missense mutations with a LacZ- phenotype would be expected to be relatively rare compared to other types of mutations.


    Two classes of polar mutations can be isolated in the lac operon. One class of polar mutations maps upstream of the lacZ gene and no nonsense mutations are found in this class. A second class of polar mutations that maps within the lacZ gene includes nonsense mutations.

    1. What is polarity? [Give an example using the lac operon.]

      ANSWER: Polarity is failure to express downstream genes in an operon due to a mutation located upstream. For example, a mutation in the lacZ gene which prevents expression of the lacY and lacA genes.

    2. What is the most likely cause of each of the two classes of mutations.?

      ANSWER: The first class of mutation is probably a promoter mutation that prevents RNA polymerase binding and thus prevents expression of the entire lac operon. The second class of mutation includes mutations in the lacZ gene that result in premature translation termination -- this may be due to a nonsense mutation, a frameshift mutation, or an out-of-frame deletion.

    3. How would you determine if a mutation was a nonsense mutation without directly sequencing the DNA?

      ANSWER: Testing for suppression with a collection of nonsense suppressors (i.e. suppressors caused by mutations in tRNA genes that result in insertion of different amino acids instead of translation termination).

    4. Why are no nonsense mutations found among the first class of mutations?

      ANSWER: The first class of mutations affect a DNA-binding site for RNA polymerase. Because this DNA sequence is not transcribed or translated, any mutation in the sequence will not result in translation termination and thus is not a nonsense mutation.

    5. What causes the polarity of the two different classes of mutations?

      ANSWER: The first class of mutations result in polarity because they directly prevent transcription of the lac operon. The second class of mutations result in polarity because they cause translation termination which, because Rho binds to the untranslated sequences, results in transcription termination.


    Many antibiotics act by binding to bacterial ribosomes and inhibiting translation. Resistance to such antibiotics is often due to mutations that produce specific nucleotide changes in one of the ribosomal subunits which prevent the antibiotic from binding to the ribosome. In such cases, the antibiotic resistant alllele is often recessive to the antibiotic sensitive allele. Why?

    ANSWER: If the antibiotic stops the ribosomes in their tracks, it will prevent subsequent ribosomes from transcribing that message. If cells have both resistant and sensitive ribosomes, both the resistant and sensitive ribosomes will bind to the mRNA -- however, the sensitive ribosomes will become stuck and block resistant ribosomes from translating the message. Thus, the sensitive phenotype is dominant to the resistant phenotype.


    The organization of the his operon in Salmonella is shown below:

    The his operon encodes the enzymes for the biosynthesis of the amino acid histidine. Thus, His- mutants are histidine auxotrophes (that is, they require histidine for growth). The hisD gene encodes the enzyme which catalyzes the last step of histidine biosynthesis, the conversion of histidinol to histidine.

    Missense mutations anywhere in the hisG gene do not prevent cells from using histidinol to make histidine. Likewise, strains with nonsense mutations in region "ii" of the hisG gene can use histidinol. However, nonsense mutations in region "i" of the hisG gene prevent cells from using histidinol.

    Suggest an explanation that can account for these results.

    ANSWER: The simplest explanation for these results is based upon Rho-dependent polarity. Recall that Rho-dependent polarity requires a region of untranslated ("naked") mRNA. Missense mutations do not terminate translation, so the mRNA remails coated with ribosomes which prevents Rho-dependent polarity. Termination of translation by a nonsense mutation in region "i" of hisG results in Rho-dependent polarity and subsequent termination of transcription, preventing the expression of hisD. However, when translation is terminated by a nonsense mutation in region "ii" of , the translation start site of hisD is sufficiently close that Rho-dependent polarity does not terminate transcription, and the downstream hisD gene is expressed. [For a more detailed description of Rho-dependent polarity, see the supplement on Rho-dependent Polarity].


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    Last modified July 2, 2004