EXAM 2 Flashcards
Match the subunit of the RNA polymerase of E. coli in the left column with its putative function during catalysis from the right column.
(a) α
(b) β
(c) β′ sequences
(d) σ⁷⁰
(1) binds the DNA template
(2) binds regulatory proteins and
(3) binds NTPs and catalyzes bond formation
(4) recognizes the promoter and initiates synthesis
(a) 2 (b) 3 (c) 1 (d) 4
Which of the following statements about E. coli promoters are correct?
(a) They may be associated with different transcription efficiencies.
(b) For most genes they include variants of consensus sequences.
(c) They specify the start sites for transcription on the DNA template.
(d) They have identical and defining sequences.
(e) They are activated when C or G residues are substituted into their −10 regions by mutation.
(f) Those that have sequences that correspond closely to the consensus sequences and are separated by 17 base pairs are very efficient.
a, b, c, f. The promoters of most E. coli genes include variants of defining consensus sequences that are centered at about the −35 and −10 positions. The nearer the sequences of a promoter are to the consensus sequence and the nearer the separation between them is to the optimal 17-bp spacing, the more efficient the promoter. The −10 consensus sequence is TATAAT. The substitution of a C or G into the sequence would likely lower the efficiency of a promoter.
The sequence of a duplex DNA segment in a longer DNA molecule is
5′-ATCGCTTGTTCGGA-3′
3′-TAGCGAACAAGCCT-5′
When this segment serves as a template for E. coli RNA polymerase, it gives rise to a segment of RNA with the sequence 5′-UCCGAACAAGCGAU-3′. Which of the following statements about the DNA segment are correct?
(a) The top strand is the coding strand.
(b) The bottom strand is the sense strand.
(c) The top strand is the template strand.
(d) The bottom strand is the antisense strand.
b, c. The sense (bottom) strand of the template DNA has the same sequence as the mRNA.
It is said that RNA polymerases backtrack for proofreading. Why would this be energetically unfavorable?
Backtracking involves breaking hydrogen bonds between an RNA-DNA base pair, but if the base pair is incorrect there will often be fewer hydrogen bonds to break.
Which of the following statements about the σ subunit of RNA polymerase are correct?
(a) It enables the enzyme to transcribe asymmetrically.
(b) It confers on the core enzyme the ability to initiate transcription at promoters.
(c) It decreases the affinity of RNA polymerase for regions of DNA that lack promoter sequences.
(d) It facilitates the termination of transcription by recognizing hairpins in the transcript.
a, b, c. The σ subunit recognizes promoter sites, decreases the affinity of the enzyme for regions of DNA lacking promoter sequences, and facilitates the specific, oriented initiation of transcription. Orienting the binding of the enzyme to the DNA results in only one of the two DNA strands functioning as a template for RNA transcription; that is, it gives rise to asymmetric transcription.
When growing E. coli are subjected to a rapid increase in temperature, a new and characteristic set of genes is expressed. Explain how this alteration in gene expression occurs.
The temperature increase induces the synthesis of a new σ factor, σ³², which directs RNA polymerase to promoters that have −10 and −35 sequences different from those recognized by σ⁷⁰. Transcription from these promoters gives rise to characteristic heat-shock proteins.
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Which of the following statements about the ρ protein of E. coli are correct?
(a) It is an ATPase that is activated by binding to single-strand DNA.
(b) It recognizes specific sequences in single-strand RNA.
(c) It recognizes sequences in the DNA template strand.
(d) It causes RNA polymerase to terminate transcription at template sites that are different from those that lead to ρ-independent termination.
(e) It acts as a RNA-DNA helicase.
d, e. The ρ protein recognizes and binds stretches of RNA that are devoid of hairpins and are at least 72 nucleotides long. It acts to hydrolyze ATP and to unwind the RNA-DNA hybrid in the transcription bubble.
Match the functions in the right column with the appropriate antibiotic inhibitor of E. coli transcription in the left column.
(a) rifampicin
(b) actinomycin D
(1) interacts with the template
(2) interacts with nascent mRNA polymerase
(3) prevents initiation
(4) prevents elongation
(5) intercalates into mRNA hairpins
(a) 2, 3 (b) 1, 4. Actinomycin D intercalates only into duplex DNA.
Explain how a mutation might give rise to an E. coli that is resistant to the antibiotic rifampicin.
Rifampicin must bind to the β subunit of RNA polymerase to inhibit the enzyme. A mutation in the gene encoding this subunit that would interfere with the binding of the antibiotic but not with polymerization would produce a rifampicin-resistant cell.
The rate constant for the binding of RNA polymerase holoenzyme to a promoter on a long DNA molecule is greater than that for the collision of two small molecules in solution. Since small molecules diffuse through solutions more rapidly than large ones, how can this be true?
RNA polymerase holoenzyme has lower affinity for nonspecific DNA sequences than for promoter sequences. The nonspecific affinity, however, allows the enzyme to bind to “random-sequence” DNA and then “slide” along the molecule in a unidimensional random walk until it encounters a promoter sequence for which its binding affinity is higher. Diffusion in one dimension is much faster than diffusion in three dimensions, thereby explaining the observed rapid rate constant for the binding of RNA polymerase holoenzyme to promoter sequences. If one measured the encounter of the polymerase with the nonspecific regions of the DNA rather than with promoter sequences, the value of the rate constant would be much lower and would fit our expectations for a three-dimensional, diffusion-limited reaction between macromolecules.
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The ciliated protozoan Tetrahymena contains an enzyme that can synthesize 5′-pseudouridine monophosphate from a mixture of PRPP and uracil. For a time it was thought that this enzyme was instrumental in the synthesis of transfer RNAs in Tetrahymena. Explain why this is not the case.
Nascent polynucleotides formed by RNA polymerases contain only the four usual bases. Subsequently, some of the bases are chemically modified. Were unusual nucleotides to be incorporated into a growing RNA chain, this would in turn require the presence of unusual bases on DNA. The pseudouridine found in transfer RNAs is formed by breaking the nitrogen–carbon bond linking uracil to ribose and forming a carbon–carbon bond instead. Only certain uracils are modified in this manner, owing to their position in the three-dimensional structure of the RNA and to the specificity of the enzymes that carry out the modification.
When mammalian genes are cloned, a strategy that is frequently followed involves the isolation of mRNA rather than DNA from a cell and the preparation of a complementary DNA (cDNA) by the enzyme reverse transcriptase. Suppose that mRNA isolated from a cell specialized for the production of protein X is used as a template for the production of cDNA. What major difference or differences would you expect to find between the structure of that cDNA and genomic DNA for protein X?
The cDNA prepared from mRNA would have a long poly(T) tail, unlike genomic DNA. Remember that the poly(A) tail is added to the 3′ end of mammalian mRNA and that there is no counterpart on DNA. A second striking difference would be that the cDNA would contain no intervening sequences (introns) and would therefore be much shorter than the corresponding sections of genomic DNA. (Remember that most mammalian genes are mosaics of introns and exons.) A third difference would be found if any RNA editing were involved. An edited mRNA could generate a cDNA with nucleotides that did not correspond to those in genomic DNA.
Rifampicin specifically inhibits the initiation of transcription in prokaryotes and may therefore be used in humans as a therapeutic antibacterial agent. Would you expect actinomycin D to be useful in antibacterial therapy? Why or why not?
In order to be useful as a therapeutic antibacterial agent, a compound must selectively inhibit processes in prokaryotes but leave the corresponding processes in eukaryotes (including those in mitochondria) largely unaffected. Because rifampicin selectively inhibits the initiation of transcription in prokaryotes but not in eukaryotes, it is useful as an antibacterial agent. Actinomycin D is an intercalating agent that binds to DNA duplexes and inhibits both DNA replication and transcription, although it has a greater inhibiting effect on transcription than on replication. It cannot discriminate between the duplex DNA of bacteria and that of humans, however, and will therefore bind to both. Because it disrupts eukaryotic as well as prokaryotic processes, it is not very useful as an antibacterial agent. It is sometimes used as an anticancer agent, however, because of its ability to slow the replication rate of human DNA.
The mRNAs produced by mammalian viruses undergo modification at the 5′ and 3′ ends in a fashion similar to that of eukaryotic mRNA. Why do you think this is the case?
Viruses use the host’s enzyme system to replicate their DNA and to synthesize their proteins. Since eukaryotic translation systems must synthesize viral protein, the structure of viral mRNAs must mimic that of the host mRNA.
Suppose that human DNA is cleaved into fragments approximately the size of a given mature human messenger RNA and that mRNA–DNA hybrids are then prepared. The corresponding procedure is then carried out for E. coli. When the mRNA–DNA hybrids from each species are examined with an electron microscope, which will show the greater degree of hybridization? Explain.
The mRNA–DNA hybrid of E. coli will show greater hybridization because it is produced continuously from a DNA template without processing. Because of the presence of intervening sequences in human DNA, there will be regions in the human RNA–DNA hybrids where no base pairing occurs.
What are snRNPs, and how are they involved in the eukaryotic mRNA splicing reaction?
The snRNPs are small ribonucleoprotein particles that occur in the nucleus. Each is composed of a small RNA molecule and several characteristic proteins, some of which are common to different snRNPs. Distinct snRNPs recognize and bind to splice junctions and the branch site and are involved in assembling the spliceosome in an ATP-dependent manner. They are requisite components of the splicing apparatus, and the RNA components of some of them are probably catalytically active. The RNAs of some snRNPs form hydrogen bonds with sequences within introns and exons to help to juxtapose properly the reacting splice junctions.
In an attempt to determine whether a given RNA was catalytically active in the cleavage of a synthetic oligonucleotide, the following experimental results were obtained. When the RNA and the oligonucleotide were incubated together, cleavage of the oligonucleotide occurred. When either the RNA or the oligonucleotide was incubated alone, there was no cleavage. When the RNA was incubated with higher concentrations of the oligonucleotide, saturation kinetics of the Michaelis-Menten type were observed. Do these results demonstrate that the RNA has catalytic activity? Explain.
These results alone do not establish that the RNA has catalytic activity, because a catalyst must be regenerated. It is entirely possible that the results observed could be accounted for by a stoichiometric, as opposed to a catalytic, interaction between RNA and the oligonucleotide, in which the RNA may “commit suicide” as the oligonucleotide is cleaved. In such an interaction, a portion of the RNA would participate chemically in the cleavage of the oligonucleotide, but it would also be cleaved itself as a part of the reaction. Four reaction products would accumulate, two resulting from the cleavage of RNA and two from the cleavage of the oligonucleotide. To show that this particular RNA was catalytic, it would be necessary to demonstrate that it turns over and is regenerated in the course of the reaction. You could perform experiments in which the putative catalytic RNA was present in much lower molar concentrations than those of the substrate oligonucleotide. If more substrate oligonucleotide molecules were hydrolyzed than the total number of the presumed catalytic RNA, you would have evidence that true turnover had occurred.
Negative supercoiling affects transcription at many promoters in E. coli. In addition to facilitating the unwinding of the DNA helix necessary to form a transcription bubble, how might negative supercoiling affect transcription either positively or negatively?
The interaction of the σ subunit of RNA polymerase with a promoter requires the precise positioning of the atomic determinants on the surface of the DNA at the −10 and −35 regions to which it binds. Particular hydrogen bond donors and acceptors in the grooves of the DNA in these regions must be positioned so that the amino acids on the protein can interact optimally with them. The degree of supercoiling of a DNA molecule affects the precise positioning of these groups because the base pairs are moved with respect to one another by twisting the DNA. Thus, supercoiling could either promote or hinder the interaction of the polymerase with the DNA by bringing the groups in the −10 region and in −35 regions of the DNA of particular promoters into a configuration that was better or worse for interaction.
Which of the following statements about functional tRNAs are correct?
(a) They contain many modified nucleotides.
(b) About half their nucleotides are in base-paired helical regions.
(c) They contain fewer than 100 ribonucleotides.
(d) Their anticodons and amino acid accepting regions are within 5 Å of each other.
(e) They consist of two helical stems that are joined by loops to form a U-shaped structure.
(f) They have a terminal AAC sequence at their amino acid accepting end.
a, b, c. The molecules consist of two helical stems, each of which is made of two stacked helical segments. However, the molecules are L-shaped, and the anticodon and amino acid accepting regions are some 80 Å from each other. Functional tRNAs have a CCA sequence, not an AAC sequence at their 3’ termini.
Explain why tRNA molecules must have both unique and common structural features.
Transfer RNAs need common features for their interactions with ribosomes and elongation factors but unique features for their interactions with the activating enzymes.
How many errors would be expected to occur in a 1000-amino acid protein given an error frequency of 10⁻³?
One error would be expected. An e value of 10⁻³ means 1 error per 1000 amino acids.
Which of the following answers completes the sentence correctly? The wobble hypothesis
(a) accounts for the conformational looseness of the amino acid acceptor stem of tRNAs that allows sufficient flexibility for the peptidyl-tRNA and aminoacyl-tRNA to be brought together for peptide-bond formation.
(b) accounts for the ability of some anticodons to recognize more than one codon.
(c) explains the occasional errors made by the aminoacyl-tRNA synthetases.
(d) explains the oscillation of the peptidyl-tRNAs between the A and P sites on the ribosome.
(e) assumes steric freedom in the pairing of the first (5’) nucleotide of the codon and the third (3’) nucleotide of the anticodon.
The wobble hypothesis (b) accounts for the ability of some anticodons to recognize more than one codon.
Which of the following statements about the aminoacyl-tRNA synthetase reaction are correct?
(a) ATP binds to the synthetase.
(b) GTP binds to the synthetase.
(c) The amino acid is attached to the 2’- or 3’-hydroxyl of the nucleotide cofactor (ATP).
(d) The amino group of the amino acid is activated.
(e) A mixed anhydride bond is formed.
(f) An acyl ester bond is formed.
(g) An acyl thioester bond is formed.
(h) A phosphoamide (P-N) bond is formed.
a, e, f. The carboxyl group of the amino acid is activated in a two-step reaction via the formation of an intermediate containing a mixed anhydride linkage to AMP. The amino acid is ultimately linked by an ester bond to the 2’- or 3’-hydroxyl of the tRNA.
The ΔG°′ of the reaction catalyzed by the aminoacyl-tRNA synthetases is
(a) ~0 kcal/mol.
(b) <0 kcal/mol.
(c) >0 kcal/mol.
a. Since the standard free energy of the hydrolysis of an aminoacyl-tRNA is nearly equal to that of the hydrolysis of ATP, the reaction has a ΔG°′ ~ 0; that is, it has an equilibrium constant near 1.
Match the class I and class II aminoacyl-tRNA synthetases with the appropriate items in the right column.
(a) class I
(b) class II
(1) are generally dimeric
(2) are generally monomeric
(3) acylate the 2’-hydroxyl of the tRNA
(4) generally acylate the 3’-hydroxyl of the tRNA
(5) contain β strands at the activation domain
(a) 2, 3, 5 (b) 1, 4, 5. Both classes of enzymes contain β sheet domains at their activation domains.
Indicate the possible ways in which different aminoacyl-tRNA synthetases may recognize their corresponding tRNAs.
(a) by recognizing the anticodon
(b) by recognizing specific base pairs in the acceptor stem
(c) by recognizing the 3’ CCA sequence of the tRNA
(d) by recognizing both the anticodon and acceptor stem region
(e) by recognizing extended regions of the L-shaped molecules
a, b, d, e. There are many different ways in which aminoacyl-tRNA synthetases recognize their specific tRNAs. Answer (c) is incorrect because the 3’ CCA sequence is common to all tRNAs and cannot be used to distinguish among them.
In an experiment, it was found that Cys-tRNAᶜʸˢ can be converted to Ala-tRNAᶜʸˢ and used in an in vitro system that is capable of synthesizing proteins.
(a) If the Ala-tRNAᶜʸˢ were labeled with ¹⁴C in the amino acid, would the labeled Ala be incorporated in the protein in the places where Ala residues are expected to occur? Explain.
(b) What does the experiment indicate about the importance of the accuracy of the aminoacyl-tRNA synthetase reaction to the overall accuracy of protein synthesis?
(a) No. The tRNA, acting as an adaptor between the amino acid and mRNA, would associate with Cys codons in the mRNA through base-pairing between codon and anticodon. The labeled alanine would incorporate only at sites encoded by the Cys codons and not at those encoded by Ala codons.
(b) The experiment demonstrates that the tRNA and not the amino acid reads the mRNA. Thus, if the activating enzyme mistakenly attaches an incorrect amino acid to a tRNA, that amino acid will be incorporated erroneously into the protein.
Which of the following statements about an E. coli ribosome are correct?
(a) It is composed of two spherically symmetrical subunits.
(b) It has a large subunit comprising 34 kinds of proteins and two different rRNA molecules.
(c) It has a sedimentation coefficient of 70S.
(d) It has two small subunits, one housing the A site and the other the P site.
(e) It has an average diameter of approximately 200 Å.
(f) It has a mass of approximately 270 kDa, one-third of which is RNA.
b, c, e. Answer (f) is incorrect because two-thirds of the 2700-kd mass of a ribosome is rRNA.
What is the significance of the reconstitution of a functional ribosome from its separated components?
It shows that the components themselves contain all the information necessary to form the structure and that neither a template nor any other factors are involved. Thus, the ribosome serves as model from which we might learn the general principles involved in self-assembly. Reassembly allows systematic study of the roles of the individual components through the determination of the effects of substitutions of mutant or altered individual proteins or rRNAs.
Which of the following statements about translation are correct?
(a) Amino acids are added to the amino terminus of the growing polypeptide chain.
(b) Amino acids are activated by attachment to tRNA molecules.
(c) A specific initiator tRNA along with specific sequences of the mRNA ensures that translation begins at the correct codon.
(d) Peptide bonds form between an aminoacyl-tRNA and a peptidyl-tRNA positioned in the A and P sites, respectively, of the ribosome.
(e) Termination involves the binding of a terminator tRNA to a stop codon on the mRNA.
b, c, d. Answer (a) is incorrect because the incoming activated aminoacyl-tRNA, in the A site of the ribosome, adds its free amino group to the activated carboxyl of the growing polypeptide on a peptidyl-tRNA in the P site. Answer (e) is incorrect because termination does not involve tRNAs that recognize translation stop codons but rather protein release factors that recognize these and cause peptidyl transferase to donate the growing polypeptide chain to H₂O rather than to another aminoacyl-tRNA.
What experiment showed that each initiator region displays an AUG (or GUG or UUG) codon?
The initiator regions from a number of mRNAs were isolated using pancreatic ribonuclease to digest mRNA–ribosome complexes (formed under conditions in which protein synthesis could begin but elongation could not take place).
What sequence on the 16s rRNA binds to the Shine-Delgarno sequence on mRNA?
(a) GEUGCG
(b) CAGGU
(c) GGAGG
(d) There is no specific sequence, it is merely a purine-rich region of the 16s rRNA.
a. Answers b and c are actual Shine-Delgarno sequences or parts of one
What is the role of the vitamin folate in prokaryotic translation?
After folic acid is converted to N¹⁰-formyltetrahydrofolate, it acts as a carrier of formyl groups and is a substrate for a transformylase reaction that converts Met-tRNAᶠ to fMet-tRNAᶠ—the initiator tRNA.
Which of the following statements about occurrences during translation are correct?
(a) The carboxyl group of the growing polypeptide chain is transferred to the amino group of an aminoacyl-tRNA.
(b) The carboxyl group of the amino acid on the aminoacyl-tRNA is transferred to the amino group of a peptidyl-tRNA.
(c) Peptidyl-tRNA may reside in either the A or the P site.
(d) Aminoacyl-tRNAs are shuttled from the A to the P site by EF-G.
a, c. The aminoacyl-tRNA in the A site becomes a peptidyl-tRNA when it receives the carboxyl group of the growing polypeptide chain from the peptidyl-tRNA in the P site. After the free tRNA leaves, the extended polypeptide on its new tRNA is then moved to the P site by EF-G. Answer (d) is incorrect because transfer RNAs bearing aminoacyl derivatives with free amino groups are never found in the P site.
For each of the following steps of translation, give the cofactor involved and the number of high-energy phosphate bonds consumed.
(a) amino acid activation
(b) formation of the 70S initiation complex
(c) delivery of aminoacyl-tRNA to the ribosome
(d) formation of a peptide bond
(e) translocation
(a) ATP, 2 (b) GTP, 1 (c) GTP, 1 (d) none (e) GTP, 1. With regard to the answer for (d), the formation of a peptide bond per se does not require a cofactor. The energy for the exergonic reaction is supplied by the activated aminoacyl-tRNA.
Which of the following statements about release factors are correct?
(a) They recognize terminator tRNAs.
(b) They recognize translation stop codons.
(c) They cause peptidyl transferase to use H₂O as a substrate.
(d) There are two proteins in E. coli, each of which recognizes two mRNA triplet sequences.
b, c, d. Each of the two release factors of E. coli recognizes two of the three translation stop codons and interacts with the synthesis machinery such that peptidyl transferase donates the polypeptide chain to H₂O and thus terminates synthesis by hydrolyzing the ester linkage of the protein to the tRNA.
Which of the following statements about eukaryotic translation are correct?
(a) A formylmethionyl-tRNA initiates each protein chain.
(b) It occurs on ribosomes containing one copy each of the 5S, 5.8S, 18S, and 28S rRNA molecules.
(c) The correct AUG codon for initiation is selected by the base-pairing of a region on the rRNA of the small ribosomal subunit with an mRNA sequence upstream from the translation start site.
(d) It is terminated by a release factor that recognizes stop codons and hydrolyzes GTP.
(e) It involves proteins that bind to the 5’ ends of mRNAs.
(f) It can be regulated by protein kinases.
b, d, e, f. Eukaryotic ribosomes usually scan the mRNA from the 5’ end for the first AUG codon, which then serves to initiate the synthesis. Answer (e) is correct because proteins that bind to the cap of the mRNA are involved in the association of the ribosome with the mRNA.
Many antibiotics act by inhibiting protein synthesis. How can some of these be used in humans to counteract microbial infections without causing toxic side effects due to the inhibition of eukaryotic protein synthesis?
The inhibition of the prokaryotic translation and not that of the eukaryote can result from differences between their respective ribosomes. Some antibiotics interact with the RNA components that are unique to bacterial ribosomes and, consequently, can inhibit bacterial growth without affecting the human cells.
Increasing the concentration of which of the following would most effectively antagonize the inhibition of protein synthesis by puromycin?
(a) ATP
(b) GTP
(c) aminoacyl-tRNAs
(d) peptidyl-tRNAs
(e) eIF3
c. Puromycin is an analog of aminoacyl-tRNA. It inhibits protein synthesis by binding to the A site of the ribosome and accepting the growing polypeptide chain from the peptidyl-tRNA in the P site and thus terminating polymer growth. Because aminoacyl-tRNAs compete with the puromycin for the A site, increasing their concentration would lessen the extent of inhibition.
Which of the following statements about the diphtheria toxin are correct?
(a) It is cleaved on the surface of susceptible eukaryotic cells into two fragments, one of which enters the cytosol.
(b) It binds to peptidyl transferase and inhibits protein synthesis.
(c) It reacts with ATP to phosphorylate eIF2 and prevent the insertion of the Met-tRNA into the P site.
(d) It reacts with NAD⁺ to add ADP-ribose to eEF2 and prevent movement of the peptidyl-tRNA from the A to the P site.
(e) One toxin molecule is required for each translation factor inactivated, that is, it acts stoichiometrically.
d. Answer (e) is incorrect because the toxin acts catalytically and is thus extremely deadly; one toxin molecule can inactivate many translocase molecules by modifying them covalently.
Which of the following proteins are usually synthesized by free ribosomes?
(a) lysosomal proteins
(b) cytoplasmic proteins
(c) secretory proteins
(d) integral membrane proteins
b. Free ribosomes synthesize proteins that remain in the cell, either in the cytoplasm or in organelles such as the nucleus or mitochondria.
Place the following in the order in which a newly synthesized protein is transported out of the lumen of the ER.
(a) Cis-Golgi
(b) Secretory granules
(c) Trans-Golgi
(d) Transport vesicles
d, a, c, b
(a) The template strand of DNA known to encode the N-terminal region of an E. coli protein has the following nucleotide sequence: GTAGCGTTCCATCAGATTT. Give the sequence for the first four amino acids of the protein.
(b) Suppose that the sense strand of the DNA known to encode the amino acid sequence of the N-terminal region of a mammalian protein has the following nucleotide sequence: CCTGTGGATGCTCATGTTT. Give the amino acid sequence that would result.
(a) The sequence of the first four amino acids of the protein is (formyl)Met-Glu-Arg-Tyr. As the name implies, the template (antisense) strand of DNA serves as the template for the synthesis of a complementary mRNA molecule. (Remember that by convention nucleotide sequences are always written in the 5’ to 3’ direction unless otherwise specified.) The template strand of DNA and the mRNA synthesized are as follows:
DNA template strand: 5’-GTAGCGTTCCATCAGATTT-3’
mRNA: 3’-CAUCGCAAGGUAGUCUAAA-5’
Remember that the codons of an mRNA molecule are read in the 5’-to-3’ direction. Because this particular nucleotide sequence specifies the N-terminal region of an E. coli protein, the first amino acid must be (formyl)-methionine, which may be encoded by either AUG or GUG. Because there is no GUG and only a single AUG in the mRNA sequence, the location of the initiation codon can be established unambiguously. The portion of the mRNA sequence encoding protein and the first four amino acids it encodes are
mRNA: 5’-AUG-GAA-CGC-UAC-3’
Amino acid sequence: (Formyl)Met-Glu-Arg-Tyr
(b) The expected amino acid sequence is Met-Leu-Met-Phe. The nucleotide sequences on DNA and mRNA are
Sense strand of DNA: 5’-CCTGTGGATGCTCATGTTT-3’
mRNA: 5’-CCUGUGGAUGCUCAUGUUUU-3’
In eukaryotes the first triplet specifying an amino acid is almost always the AUG that is closest to the 5’ end of the mRNA molecule. In this example there are two AUGs, so there will be two Met residues in the polypeptide that is produced. The reading frame and the resulting amino acids are as follows:
mRNA: 5’-CCUGUGG-AUG-CUC-AUG-UUU-3’
Amino acid sequence: Met-Leu-Met-Phe
The nucleotide sequence on the sense strand of the DNA known to encode the amino acid sequence of the N-terminal region of a long protein of E. coli has the following nucleotide sequence: CCATGCAAAGTAATAGGT. Give the resulting amino acid sequence.
The sequence is His-Ala-Lys. The DNA and mRNA sequences are
Sense strand of DNA: 5’-CCATGCAAAGTAATAGGT-3’
mRNA: 5’-CCAUGCAAAGUAAUAGGU-3’
Since this sequence specifies the carboxyl end of the peptide chain, it must contain one or more of the chain-termination codons: UAA, UAG, or UGA. UAA and UAG occur in tandem in the sequence, so we can infer the reading frame. The mapping of the amino acid residues to the mRNA is as follows:
mRNA: 5’-C-CAU-GCA-AAG-UAA-UAG-GU-3’
Amino acid sequence: His-Ala-Lys
Suppose that a particular aminoacyl-tRNA synthetase has a 10% error rate in the formation of aminoacyl-adenylates and a 99% success rate in the hydrolysis of incorrect aminoacyl-adenylates. What percentage of the aminoacyl-tRNAs produced by this aminoacyl-tRNA synthetase will be faulty?
The percentage of tRNAs that will be faulty is 0.11%. For every 1000 aminoacyl-adenylates that are produced, 100 are faulty and 900 are correct. The 900 correct intermediates will be converted to correct aminoacyl tRNAs because the intermediates are tightly bound to the active site of the aminoacyl-tRNA synthetase. Of the 100 incorrect aminoacyl-adenylates, 99 will be hydrolyzed and will therefore not form aminoacyl tRNAs. Only one will survive to become an incorrect aminoacyl tRNA. The value 0.11% is obtained by dividing the number of incorrect aminoacyl tRNAs (i.e., 1) by the total number of aminoacyl tRNAs that are synthesized without subsequent hydrolysis (i.e., 1000 - 99 = 901).
Students of biochemistry are frequently distressed by “Svedberg arithmetic,” that is, for instance, by the fact that the 30S and 50S ribosomal subunits form a 70S particle rather than an 80S particle. Why don’t the numbers add up to 80?
The Svedberg unit (S) is a sedimentation coefficient, which is a measure of the velocity with which a particle moves in a centrifugal field. It represents a hydrodynamic property of a particle, a property that depends on, among other factors, the size and shape of the particle. When two particles come together, the sedimentation coefficient of the resulting particle should be less than the sum of the individual coefficients because there is no frictional resistance between the contact surfaces of the particles and the centrifugal medium.
According to the wobble principle, what is the minimum number of tRNAs required to decode the six leucine codons—UUA, UUG, CUU, CUC, CUA, and CUG? Explain.
A minimum of three tRNAs would be required. One tRNA having the anticodon UAA could decode both UUA and UUG. For the other four codons, which have C in the first position and U in the second, there are two different combinations of two tRNAs each that could decode them. The first combination would be two tRNAs that have anticodons with A in the second position and G in the third, one with I in the first position to decode CUU, CUC, and CUA, and the other with U or C in the first position to decode CUG. The second combination would be two tRNAs that have anticodons with A in the second position and G in the third, one with G in the first position to decode CUU and CUC, and the other with U in the first position to decode CUA and CUG.
Coordination of the threonine hydroxyl by an active site Zn in the threonyl-tRNA synthetase allows discrimination between threonine and the isosteric valine (Sankaranarayanan et al., Nat. Struct. Biol. 7[2000]:461–465). Given the similarity of serine and threonine (Ser lacks only the methyl group of Thr), if this is the only mechanism for amino acid discrimination available, threonyl-tRNA synthetase mistakenly couples Ser to threonyl-tRNA frequently. Since this would lead to unacceptably high error rates in translation, how it is it avoided?
Threonyl-tRNA synthetase has a proofreading mechanism. Any Ser-tRNAᵀʰʳ that is mistakenly formed is hydrolyzed by an editing site 20 Å from the activation site. The “decision” to hydrolyze the aminoacyl-tRNA appears to depend on the size of the amino acid substituent. If it is smaller than the correct amino acid, the amino acid fits into the hydrolytic site and is cleaved. If it is the same size as the correct amino acid, it does not fit and is not destroyed. Discrimination between amino acids that are larger than the correct one or are not isoelectronic with it occurs at the aminoacylation step.
Mutations from codons specifying amino acid incorporation to one of the chain-terminating codons, UAA, UAG, or UGA, so-called nonsense mutations, result in the synthesis of shorter, usually nonfunctional, polypeptide chains. It was discovered that some strains of bacteria can protect themselves against such mutations by having mutant tRNAs that can recognize a chain-terminating codon and insert an amino acid instead. The result would be a protein of normal length that may be functional, even though it may contain an altered amino acid residue. How can bacteria with such mutant tRNA molecules ever manage to terminate their polypeptide chains successfully?
If two different legitimate stop codons are present in tandem, it would be extremely improbable that mutant tRNAs would exist for both and would simultaneously bind to each of them and thereby prevent proper chain termination.
Change of one base pair to another in DNA coding sequence frequently results in an amino acid substitution. Change of a C-G to a G-U base pair at the 3:70 position of tRNAᶜʸˢ causes that tRNA to be recognized by alanyl-tRNA synthetase.
(a) What amino acid substitution or substitutions would result with the mutated tRNAᶜʸˢ present?
(b) How does the pattern differ from that resulting from base substitutions within a codon?
(a) The tRNAᶜʸˢ will become loaded with Ala rather than Cys, and as a result will insert Ala rather than Cys into polypeptide chains.
(b) In the case of a base change within a codon, only a single amino acid of a single polypeptide is changed. In the case of a tRNA recognition mutation, amino acid substitutions at many positions of many polypeptides would occur.
The methionine codon AUG functions both to initiate a polypeptide chain and to direct methionine incorporation into internal positions in a protein. By what mechanisms are the AUG start codons selected in prokaryotes?
A purine-rich mRNA sequence, three to nine nucleotides long (called the Shine-Dalgarno sequence), which is centered about 10 nucleotides upstream of (to the 5’ side of) the start codon, base-pairs with a sequence of complementary nucleotides near the 3’ end of the 16S rRNA of the 30S ribosomal subunit. This interaction plus the association of [Met-tRNA] with the AUG in the P site of the ribosome sets the mRNA reading frame.
What features of a DNA regulatory element would endow it with properties enabling a homodimeric regulatory protein (one composed of two identical subunits) to bind it specifically?
The regulatory element would likely have a nearly perfect inverted repeat sequence giving it a twofold axis of symmetry that could be matched to the twofold symmetry of the homodimeric binding protein. In addition, the sequence itself would comprise an array of hydrogen-bond donors and acceptors plus hydrophobic methyl groups that were recognizable by the regulatory protein to provide the sequence specificity of binding.
The helix-turn-helix motif
(a) is a protein-folding pattern.
(b) is observed in a variety of prokaryotic DNA-binding proteins.
(c) contains a recognition helix that inserts itself into the minor groove of DNA.
(d) is often observed in proteins that bind DNA as dimers.
a, b, d. Answer (c) is incorrect because the recognition helix inserts into the wider major groove rather than the narrower minor groove.
Which of the following are common mechanisms used by bacteria to regulate their metabolic pathways?
(a) control of the expression of genes
(b) control of enzyme activities through allosteric activators and inhibitors
(c) formation of altered enzymes by the alternative splicing of mRNAs
(d) deletion and elimination of genes that specify enzymes
(e) control of enzyme activities through covalent modifications
a, b, e. Answer (c) is incorrect because the splicing of mRNA is rare in bacteria.
Which of the following statements about β-galactosidase in E. coli are correct?
(a) It is present in varying concentrations depending on the carbon source used for growth.
(b) It is a product of a unit of gene expression called an operon.
(c) It hydrolyzes the β-1,4-linked disaccharide lactose to produce galactose and glucose.
(d) It forms the β-1,6-linked disaccharide allolactose.
(e) It is activated allosterically by the nonmetabolizable compound isopropylthiogalactoside (IPTG).
(f) Its levels rise coordinately with those of galactoside permease and thiogalactoside transacetylase.
a, b, c, d, f. Although IPTG is an inducer for the synthesis of β-galactosidase, it is neither a substrate nor an allosteric activator of the enzyme, so answer (e) is incorrect.
Explain the stoichiometric relationship between the concentrations of β-galactosidase, galactoside permease, and thiogalactoside transacetylase in E. coli.
The genes encoding these three enzymes are transcribed as a polycistronic mRNA, so there are approximately equal numbers of copies of the mRNA sequences specifying each of the enzymes.
When E. coli is added to a culture containing both lactose and glucose, which of the sugars is metabolized preferentially? What is the mechanism underlying this selectivity?
Glucose is metabolized preferentially because it results in a decrease in the synthesis of cAMP by adenylate cyclase. The lack of cAMP prevents the formation of the cAMP–CAP complex, which is necessary for the efficient transcription of the lac operon and other catabolite-repressible operons.
Which of the following statements about the cAMP-CAP complex are correct?
(a) It protects the −87 to −49 sequence of the lac operon from nuclease digestion.
(b) It protects the −48 to +5 sequence of the lac operon from nuclease digestion.
(c) It protects the −3 to +21 sequence of the lac operon from nuclease digestion.
(d) It affects RNA polymerase activity in a number of operons.
(e) Upon binding to the lac operon, it contacts RNA polymerase.
a, d, e. Answers (b) and (c) are incorrect because they correspond to the binding sequences for RNA polymerase and lac repressor, respectively.
For each of the following, indicate whether it is a feature of the lysogenic or lytic pathway of bacteriophage λ.
(a) Most of the genes in the viral genome are transcribed.
(b) The viral genome is incorporated into the bacterial DNA.
(c) Most of the genes in the viral genome are unexpressed.
(d) A large number of viral particles are produced.
Answers a and d are features of the lytic pathway and b and c are features of the lysogenic pathway.
Which of the following events occur in the presence of high levels of λ repressor?
(a) λ repressor binds to O_R1 and O_R2, but not O_R3.
(b) λ repressor binds to all three right operator sites.
(c) The λ repressor bound to O_R1 blocks access to the promoter on the right side of the operator sites, repressing transcription of Cro.
(d) The λ repressor bound to O_R2 stimulates transcription of the λ repressor promoter.
(e) λ repressor transcription is blocked.
b and e. Answers a, c, and d occur when λ repressor is present in moderate levels.
Briefly explain the mechanism of the switch between the two stable states of bacteriophage λ.
When the repressor is high and Cro is low, the phage switches to the lytic state. The reverse causes a switch to the lysogenic state.
Regulation of the trp operon involves which of the following?
(a) controlling the amount of polycistronic mRNA formed at the level of transcription initiation
(b) controlling the amount of polycistronic mRNA at the level of transcription termination
(c) the sequential and coordinate production of five enzymes of tryptophan metabolism from a single mRNA
(d) the sequential and coordinate production of five enzymes of tryptophan metabolism from five different mRNAs produced in equal concentrations
(e) the production of transcripts of different sizes, depending on the level of tryptophan in the cell
a, b, c, e. Answer (a) is correct; although not mentioned in the text, the trp operon contains an operator, and interaction with a repressor, in addition to attenuation, is involved in transcription regulation. Attenuation provides a rapid, sensitive fine-tuning mechanism on top of the control exerted by the repressor-operator interaction. When the RNA is not terminated at the attenuator, a single polycistronic mRNA, which encodes five enzymes, is produced, so answer (d) is incorrect.
Which of the following statements concerning the trp operon leader RNA, which has 162 nucleotides preceding the initiation codon of the first structural gene of the operon, are correct?
(a) A deletion mutation in the DNA encoding the 3’ region of the leader RNA gives rise to increased levels of the biosynthetic enzymes forming Trp.
(b) A short open reading frame, containing Trp codons among others, exists within the leader RNA.
(c) The leader RNA encodes a “test” peptide whose ability to be synthesized monitors the level of Trp-tRNA in the cell.
(d) The leader RNA may form two alternative and mutually exclusive secondary structures.
(e) The structure of the leader RNA in vivo depends on the position of the ribosomes translating it.
a, b, c, d, e. The discovery of a deletion mutation in front of the first structural gene of the operon, and not in the operator, was the first clue that a control mechanism operating at the level of transcription termination was involved in regulating the expression of the trp biosynthetic enzymes. This deletion changed the potential mRNA structures so that the rho-independent transcription-termination structure could no longer form.
Which of the following is correct about leader sequences in amino acid operons?
(a) An abundance of one amino acid residue in the leader peptide sequence leads to attenuation.
(b) Operons that synthesize amino acids all contain leader sequences.
(c) Leader peptides contain at least 10 of the same amino acids in a row.
(d) Leader sequences are found on the 5’ end of the operon.
a and d. Answer b is incorrect because only some of the operons coding for amino acid biosynthetic genes are regulated by attenuator sites. Answer c is incorrect because some leader peptides (tryptophan and histidine, for example) contain fewer than 10 amino acids.
Describe the mechanism of attenuation in an amino acid operon.
In E. coli, translation begins shortly after transcription of mRNA. A short leader peptide on the N-terminus of the nascent polypeptide chain contains multiple copies of the amino acid whose biosynthetic genes are being transcribed. If the amino acid is plentiful, translation continues at a relatively fast rate compared to transcription and an RNA structure forms that terminates transcription. If levels of the amino acid are low, the ribosome stalls at the leader peptide, and an alternative RNA structure forms that does not terminate transcription.
What property of enzymes makes them more suitable than, say, structural proteins for studies of the genetic regulation of protein synthesis? Explain.
Enzymes are catalysts, and thus small amounts can be readily detected. An enzyme molecule can convert many (perhaps millions) substrate molecules to product, so an enzyme assay can be much more sensitive than an assay for a structural protein, which must be assayed stoichiometrically, that is as a single molecule. Many cellular proteins are produced in amounts that are too small to be detected by direct chemical methods.
Some of the known constitutive mutations of the lactose operon occur in the operator sequence rather than the regulator gene.
(a) Would you expect such an oᶜ mutant to be dominant or recessive to its wild-type o⁺ allele? Explain.
(b) Is a constitutive mutation in an operator cis-acting or trans-acting in its effects? Explain.
(c) Design an experiment involving the genes i⁺, oᶜ, o⁺, and z⁺ that would confirm your answer to part (b). Assume that it is possible to detect whether enzymes are produced in diploid (++) or haploid (+) amounts.
(a) Imagine a partial diploid that has one o⁺ and one oᶜ sequence. The o⁺ gene will bind a repressor, so the structural genes on its chromosome will not be expressed. The oᶜ sequence will not bind a repressor, so the structural genes on its chromosome will always be expressed. Thus, an oᶜ mutant would be dominant to its wild-type o⁺ allele.
(b) Repressors do not bind oᶜ sequences. Only the structural genes on the same chromosome as the oᶜ mutant will be affected, a cis-acting effect.
(c) One could prepare a partial diploid with the following genotype
i⁺ o⁺ z⁺
i⁺ oᶜ z⁻
If the effect of the mutation is cis, the haploid amount of enzyme Z will be produced in the absence of inducer. (Its synthesis will be specified by the chromosome containing i⁺ oᶜ z⁺.) If the effect is trans, the diploid amount of Z will be produced in the absence of inducer.
Since the permease required for the entry of lactose into E. coli cells is itself a product of the lactose operon, how might the first lactose molecules enter uninduced cells? Explain.
Very low levels of lactose operon enzymes are synthesized even in the absence of an inducer (see page 928 of the text, which indicates that few enzymes are produced in the absence of inducer).
Design an experiment to show that lactose stimulates the synthesis of new enzyme molecules in E. coli rather than fostering the activation of preexisting enzyme molecules, for example, by zymogen activation.
Add IPTG to E. coli cells growing in a medium containing a carbon source other than lactose in both the presence and the absence of an inhibitor of prokaryotic protein synthesis, like chloramphenicol. If zymogen activation is involved, chloramphenicol will not inhibit induction. If the synthesis of new protein is involved (as it is), induction will not be observed in the presence of chloramphenicol.
The three enzymes of the lactose operon in E. coli are not produced in precisely equimolar amounts following induction. Rather, more galactosidase than permease is produced, and more permease than transacetylase is produced. Propose a mechanism to account for this that is consistent with known facts about the lactose operon.
Differential expression of the three structural genes in the lactose operon must be at the level of translation and not transcription since a single, polycistronic mRNA molecule is formed. Following induction, mRNA transcripts containing genetic information for all three genes are produced. Some ribosomes might drop off the messenger at the end of the structural genes, with a smaller number reading through the more distal genes.
Assume that the structural genes x and y, which code for the repressible enzymes X and Y, are subject to negative control by a regulator gene r, which produces a substance S whose ability to bind to the operator gene o is modified by cosubstance T. Assume that the following allelic possibilities exist for genes r and o:
o⁺ = wild-type operator gene
oᶜ = operator gene unable to bind S (with or without T)
r⁺ = wild-type regulator gene
rᶜ = inactive regulator gene product (with or without T)
rˢ = regulator gene product always active (with or without T)
In addition, assume that the mutations x⁻ and y⁻ in the two structural genes lead, respectively, to nonfunctional enzymes X and Y.
(a) Are enzymes X and Y likely to be biosynthetic or degradative? Justify your answer.
(b) Is substance S active or inactive in the presence of T? Explain.
(c) For each of the following, predict whether active enzymes X and Y will or will not be produced under the specified conditions. For partially diploid cells, assume semidominance; that is, the enzyme activity in a diploid cell will be twice that found in a haploid cell. Use the following answer code:
0 = active enzyme absent
+ = active enzyme present in haploid amounts
++ = active enzyme present in diploid amounts
(a) The enzymes will be biosynthetic. The clue to this is provided by the statement that the enzymes are repressible. (Degradative enzymes are often inducible, whereas biosynthetic enzymes are often repressible.)
(b) Because this operon is under negative control, substance S must act as a repressor. Substance S is active as a repressor in the presence of T, so T functions as a corepressor.
(c) (1) + + 0 0
(2) 0 + 0 0
(3) 0 0 0 0
(4) + + + +
(5) + + + +
(6) + + + +
(7) + + + 0 0
(8) + + + 0 0
(9) 0 0 0 0
(10) + + + 0
Assume that the dissociation constant K for the repressor-operator complex is 10⁻¹³ M and that the rate constant for association of operator and repressor is 10¹⁰ M⁻¹ s⁻¹. Calculate the rate constant k_diss for the dissociation of the repressor–operator complex. What is the t_1/2 (half-time of dissociation, or half-life) of the repressor–operator complex?
Remembering that the dissociation constant K is equal to the ratio of the off rate to the on rate for a reaction,
[RO] ⇌ [R] + [O]
[R][O]/[RO] = K = 10⁻¹³ M = k_diss/10¹⁰ M⁻¹ s⁻¹
k_diss = 10⁻³ s⁻¹
t_1/2 = 0.693/k_diss = 0.693/10⁻³ s⁻¹ = 693 s ≡ 11.6 min
Since the three enzymes of the lactose operon in E. coli are not produced in precisely equimolar amounts following induction. Rather, more galactosidase than permease is produced, and more permease than transacetylase is produced. Propose a mechanism to account for this that is consistent with known facts about the lactose operon.
Differential expression of the three structural genes in the lactose operon must be at the level of translation and not transcription since a single, polycistronic mRNA molecule is formed. Following induction, mRNA transcripts containing genetic information for all three genes are produced. Some ribosomes might drop off the messenger at the end of the structural genes, with a smaller number reading through the more distal genes.
An operon for the biosynthesis of amino acid X in a certain bacterium is known to be regulated by a mechanism involving attenuation. What can one confidently predict about the amino acid sequence in the leader peptide for that operon? Explain.
The leader sequence should contain codons for amino acid X. If sufficient X is present in the cell, there will be sufficient X-tRNAX for the synthesis of the leader peptide (as well as for the synthesis of other X-containing proteins in the cell). Therefore, there will be no need to biosynthesize the enzymes needed to produce X.
In order to prove that regulation by attenuation occurs in vivo, Charles Yanofsky and others studied tryptophan synthesis regulation in a series of E. coli mutants. For each mutant described below, predict the expression of tryptophan synthesis genes in the presence or absence of tryptophan:
Mutant A: Mutations with decreased, but detectable, tryptophan-tRNA synthetase activity
Mutant B: Mutations in AUG or Shine-Dalgarno sequence of leader peptide sequence
Mutant C: Same as mutant B but with the leader peptide fully expressed on a plasmid
Mutant D: Mutations replacing the two Trp codons with Leu codons
Mutant E: Mutant D with a mutation in the Leu-tRNA synthetase gene
Mutant F: Mutant D that constitutively synthesizes leucine
Mutant A: There will be increased transcription of the Trp synthesis genes in both the presence and absence of tryptophan. The low levels of Trp-tRNA synthetase will slow down the rate of translation regardless of the levels of tryptophan. It is the stalling or slowing of the ribosome in segment 1 that is important for regulation, not just the inability to synthesize the leader peptide.
Mutant B: Since the ribosome will never start translation, transcription will always terminate regardless of the levels of tryptophan.
Mutant C: The results will be the same as in mutant B. Attenuation relies on the coupling of transcription and translation. Providing the leader peptide in trans would have no effect on the level of attenuation.
Mutant D: Removing the Trp codons would eliminate all regulation by tryptophan, and Trp synthesis would instead be regulated by the levels of leucine. In this mutant, the Trp synthesis genes would not be transcribed even in the absence of tryptophan, and this strain, like mutants B and C, would always require tryptophan to grow, unless the strain was grown in the absence of leucine.
Mutant E: There would be constitutive expression of the Trp synthesis genes, even in the presence of tryptophan or leucine.
Mutant F: This mutant would behave like mutants B-E and never express the genes for Trp synthesis.
Suppose that a system regulating the expression of a single copy DNA leads to the synthesis of an enzyme having a turnover number (k_cat) of 10⁴ s⁻¹. Each DNA copy is transcribed into 10³ molecules of mRNA and each of the mRNA molecules is translated into 10⁵ molecules of enzyme protein. How many molecules of substrate are converted into product per second for each wave of transcription that sweeps over the DNA?
For each transcription, the number of molecules of substrate that are converted into product per second is given by 10⁴ s⁻¹ × 10³ × 10⁵ = 10¹² s⁻¹
Antibiotic resistance occurs when bacteria that are not resistant to the antibiotic are killed, leaving behind bacteria that have developed resistance to the antibiotic. Eventually only the antibiotic-resistant bacteria are left. Speculate on how antibiotics could be developed that target quorum sensing, especially in bacteria that form biofilms.
Bacteria that form biofilms are of particular medical importance because organisms that form them are often resistant to the immune response as well as to antibiotics. Quorum sensing appears to play a major role in the formation of biofilms in that cells are able to sense other cells in their environments and to promote the formation of communities with particular compositions. For example, quorum sensing provides the opportunistic bacterium Pseudomonas aeruginosa with a mechanism with which to coordinate the formation of biofilms. If a molecule could be designed that binds to the autoinducer receptor, but blocks dimerization, and therefore blocks activation of transcription, quorum sensing would also be blocked. This should prevent the bacteria from forming biofilms, and give the body’s immune system time to remove the infection through normal immune pathways. This process is called quorum quenching. An advantage to this approach compared to traditional antibiotic design is that there would be few evolutionary forces that select for resistance. Non-resistant strains could still multiply, and resistant strains would have to compete with them. There would be no strong survival advantage to resistant mutations.
