Exam 4
What protein in the E. coli replication fork is 1) responsible for unwinding the DNA double helix and 2) what is the mechanism by which this strand separation is accomplished? A. 1) DNA polymerase 2) synthesis of DNA on the leading and lagging strand leads to unwinding the DNA B. 1) Helicase 2) strand separation requires ATP for energy for DNA translocation and disrupts H-bonds. C. 1) DNA gyrase 2) relieves torsional stress front of the fork by DNA cleavage, which also unwinds DNA. D. 1) Primase 2) synthesizes RNA primers front of the Okazaki fragment and this results in DNA unwinding. E. 1) Beta clamp 2) prevents DNA Pol III dissociating from the DNA and this provides energy for unwinding.
1) Helicase 2) strand separation requires ATP for energy for DNA translocation and disrupts H-bonds
Briefly describe the structure and function of the one key 1) human protein and the one key 2) E. coli protein that are each needed for high fidelity termination of DNA replication. A. 1) Telomerase prevents leading strand deletion 2) Ter protein prevents fork replication past termination B. 1) Ter protein works in humans to prevent over-replication, 2) Telomerase binds to E. coli Tus sequence C. 1) Telomerase prevents lagging strand deletion 2) Tus protein prevents fork replication past termination D. 1) Gyrase functions in humans to prevent early termination, 2) Primase functions in E. coli. E. None of these answers are correct.
1) Telomerase prevents lagging strand deletion 2) Tus protein prevents fork replication past terminatio
What happens to the rate of lac operon transcription when 1) lactose is added to an E. coli culture containing glucose? and 2) glucose is added to an E. coli culture containing lactose? A. 1) big increase 2) small decrease B. 1) small decrease 2) big decrease C. No change in 1) or 2) D. 1) small increase 2) big decrease E. 1) big increase 2) big decrease 20. (5 pts)
1) small increase 2) big decrease
Name two reasons why the same mode of regulation of trp biosynthetic enzymes used in bacteria (transcriptional repression and attenuation) are not found in yeast. A. 1) hairpin loops do not form in eukaryotic mRNA 2) bacterial ribosomes are 70S eukaryotic are 80S B. 1) attenuation would not work because of RNA splicing 2) trp is an essential amino acid:no enzymes C. 1) transcription/translation are uncoupled:no attenuation 2) trp is an essential amino acid:no enzymes D. 1) chromatin packaging of mRNA blocks hairpin loops 2) trp synthesis in yeast is blocked by glyphosate E. 1) yeast have more than one trp biosynthetic pathway 2) bacterial genes are in operons: not so in yeast
1) transcription/translation are uncoupled:no attenuation 2) trp is an essential amino acid:no enzymes
Where in an anticodon are you likely to find inosine? Choose one: A. 5' B. middle C. 3'
5'
From the following sequence, locate the sites of potential photoproduct formation. Indicate what photoproducts will most likely form and what the potential effect on the DNA would be if left unrepaired. 5′-ACGTCAGTTACGTACTGACGT
5′-ACGTCAGTTACGTACTGACGT. The TC photoproduct is likely to be a (6-4) photoproduct, which can stall replication forks if not repaired. The TT and CT photoproducts would most likely be pyrimidine dimers and, if unrepaired, would most likely result in adenine on the daughter strand. The TT photoproduct would not result in a mutation, but CT would result in a mismatch, with high potential for mutation, depending upon when repair occurs. If the daughter strand were used to correct the lesion, then the C would be converted to a T.
In the plaque, lysogens are often infected by other λ viruses present in the plaque. Why don't these newly infecting viruses grow lytically and kill the cells?
A lysogen in a λ plaque can be superinfected by other λ virions, but the lysogen makes CI, which will bind to the lytic promoters of the incoming viral DNA, repressing them and preventing them from growing lytically. This is termed "immunity."
rocessing in the Golgi apparatus? Choose one: A. N-linked glycosylated protein B. Phosphorylated protein C. Unmodified protein D. Acetylated protein
A. N-linked glycosylated protein
Which mechanisms are similar in prokaryotic and eukaryotic transcription of mRNA? Choose one or more: A. recruitment of RNA polymerase to DNA B. RNA synthesis C. transcription termination D. processing of mRNA
A. recruitment of RNA polymerase to DNA B. RNA synthesis
What is the role of the -35 and -10 boxes in bacterial transcription? Choose one: A. to recruit RNA polymerase to the promoter B. to signal the start of the coding region C. to signal the end of the coding region D. to form secondary structure to regulate transcription
A. to recruit RNA polymerase to the promoter
Which of these RNAs are a noncoding RNA if consider mRNA as the only known coding RNA? A. ribosomal RNA B. snoRNA C. RNase P RNA D. miRNA E. All of these
All of these
Why is it important that all tRNA molecules adopt a similar tertiary structure?
All tRNA molecules must have similar structure in the region where they bind to elongation factors and the A, P, and E sites on the ribosome. These interactions are not dictated by the particular amino acid charged to the tRNA, so all tRNAs have similar structures.
Describe the basic aminoacylation reaction. How does this reaction differ in class I and class II aminoacyl-tRNA synthetases?
Aminoacylation is a two-stage reaction: First the amino acid is activated through the formation of an aminoacyl-AMP moiety; second, the amino acid is transferred to the acceptor stem of tRNA. Class I aminoacyl-tRNA synthetases transfer the amino acid to the 2′-hydroxyl on the last conserved adenosine residue in the acceptor stem. Class II enzymes transfer the amino acid to the 3′-hydroxyl of this nucleotide residue.
Your colleague is studying the regulation of an E. coli gene, and she purifies a protein that stimulates expression of the gene in an in vitro transcription system. In this system, the gene is expressed at a very low level in the absence of this protein and at a high level in its presence. She interprets these data to mean that it is an activator protein. Can you suggest another possibility, along with one or more experiments that would distinguish between the two models?
Another possible explanation is that the purified protein is an alternate σ factor. Sigma determines the promoter binding specificity, and there are multiple σ factors. To distinguish these models, the in vitro transcription system could use an RNA polymerase without the usual σ70 factor, called the "core" RNA polymerase. The new-sigma model predicts that the target gene would be transcribed; the activator model predicts no transcription.
What would be the effect of a mutation in the HIV-1 reverse transcriptase gene that decreased the rate of mismatches?
Any decrease in the error rate of HIVRT would result in an overall decrease in the rate of viral mutations. This would probably not be an advantage for the virus, as the relatively high error rate often helps to thwart the use of antiviral drugs.
RNA is typically more susceptible to backbone hydrolysis than DNA because of a chemical property. Choose the one correct answer from the right. A. The presence of simpler nucleotides. B. The presence of a 2' OH group. C. The presence of a 3' OH group. D. The lack of thymine nucleotides. E. None of these answers are correct.
B. The presence of a 2' OH group.
Many antibacterial agents work by inhibiting RNA polymerase. These agents are typically molecules that interact with various regions of the catalytic center to prevent DNA binding or transcript synthesis. Why are many of these considered "broad-spectrum antibiotics"; that is, antibacterial agents that act against a large number of different bacterial species? Why are these broad-spectrum antibiotics advantageous for both physicians and their patients? These agents, however, do not affect eukaryotic RNA Pol I, II, or III. What does this tell you about the sequence conservation between the bacterial and eukaryotic enzymes?
Bacterial RNA polymerase is highly conserved; therefore, an inhibitor of its activity would likely affect all bacteria, meaning that the antibiotic can be used without knowing the species of bacteria causing the infection, an advantage because most bacterial infections require treatment as soon as possible. The structure of RNA polymerase is conserved from bacteria to eukaryotes, but these enzymes do not share a high degree of sequence conservation. The interaction between the antibacterial molecule and RNA polymerase depends on specific base contacts, and differences at the primary sequence level means that the eukaryotic RNA polymerases do not form the same contacts, and therefore the inhibitor has no effect.
What would happen if a mutation in the λ phage Xis gene occurred such that the resulting protein was not functional?
Because Xis is required for λ phage DNA excision, a mutation rendering Xis nonfunctional would prevent excision and prevent initiation of a lytic cycle.
Approximately how many ATP are required by helicase for the replication of the E. coli chromosome?
Because helicase requires 1 ATP per two bases, it requires 2.3 × 106 molecules of ATP to replicate 4.6 × 106 bases.
In bacteria, the same DNA strand is often a template for both replication and transcription. Why are codirectional collisions a common occurrence?
Both DNA and RNA polymerase read 3′ to 5′, resulting in a 5′ to 3′ polymerization reaction. When both replication and transcription complexes proceed down the same region of DNA, the DNA polymerase can overtake RNA polymerase because transcription is much slower than replication; this results in a codirectional collision.
Which of the following components of eukaryotic mRNA listed at the right need to be removed before translation? A. 7-Methylguanosine cap sequence B. Poly adenylation sequences C. Intronic sequences D. Exonic sequences E. None of these answers are correct.
C. Intronic sequences
Which of these steps in the RNAi pathway is the final step required for mRNA degradation to occur? Choose one: A. assembly of siRNA into a RISC B. cleavage of dsRNA by Dicer C. siRNA-mRNA duplex formation in a RISC D. siRNA-mRNA duplex formation
C. siRNA-mRNA duplex formation in a RISC
Use the nine nucleotide DNA base pairs below to determine the mRNA and protein sequences in Q12 and Q13. The section of DNA sequence shown is located in the middle of the gene (...), i.e., it is not at start site. (Leftside: termination, rightside: promotor) 5'....... CCA TAC CGG .......3' 3'....... GGT ATG GCC........5' The correct order for codons 1, 2, and 3 written 5' to 3' on the mRNA strand is: A. CCA, UAC, CGG B. GGU, AUG, GCC C. GGC, CAU, ACC D. CCG, GUA, UGG E. None of these. What is the amino acid sequence (start at N-term) encoded by these three mRNA codons? A. Pro-His-Arg B. Gly-Met-Glu C. Pro-Val-Trp D. Gly-His-Ile E. Pro-Val-Gln
CCG, GUA, UGG Pro-Val-Trp
Mutants of λ can arise that form clear plaques. Which virus genes are likely to be mutated?
Clear plaques have mutations in the cI gene; they cannot make CI and cannot establish or maintain the lysogenic state. Clear plaques led to the discovery of this gene and two others; hence the name of the gene cI as clear plaque, class I.
An Ames test of a compound suspected to be a mutagen was examined before and after incubation with rat liver extract. Based on the results shown, what do you conclude? A. The control plate shows low mutation rate, which is surprising. B. The Ames test does not work considering the liver is worse. C. Compound is not mutagenic at all because colonies grow meaning that their DNA is undamaged. D. Compound is mutagenic in without liver enzymes and this activity is further enhanced by liver enzymes. E. All three bacterial plates must contain a large amounts of histidine to prevent colonies from forming.
Compound is mutagenic in without liver enzymes and this activity is further enhanced by liver enzymes.
Why was Francis Crick dissatisfied with protein synthesis models that brought amino acids into direct contact with DNA or mRNA?
Crick did not support these models because in eukaryotes, DNA and protein synthesis occur in different cellular compartments. Also, on a physicochemical basis, nucleotide structure seemed insufficiently complex to act directly as a template for 20 amino acids.
The modified base 5-methylcytosine is used in higher eukaryotes as an epigenetic mark of particular chromatin states. But 5-methylcytosine is also known to be a "hot spot" for mutation; that is, it has a higher rate of mutation from C to T than most C residues. Explain
Cytosine residues are deaminated at a very low rate to form uracil, which will lead to a uracil-guanine (U-G) base pair. Because U is not a normal component of DNA, this is quickly repaired to a C-G base pair by uracil N-glycosylase and subsequent events. Deamination of 5-MeC would lead to formation of a T-G base pair, and the cell would not rapidly repair this, as there is no way to be sure which base is the correct base. If DNA replication occurs before the mismatch is repaired, it fixes a C-to-T transition mutation in one of the progeny duplexes.
How are RNA structures different from protein structures? A. RNA is always single-stranded, whereas proteins are not. B. RNA can H-bond with itself, whereas proteins cannot. C. RNA mutations can lead to nonfunctioning proteins, whereas protein mutations do not. D. RNA adopts less defined and more dynamic tertiary structures than proteins. E. None of these answers are correct.
D. RNA adopts less defined and more dynamic tertiary structures than proteins.
DNA ??? causes supercoils to develop, and DNA ??? alleviates those supercoils.
DNA helicase causes supercoils to develop, and DNA gyrase alleviates those supercoils.
Which DNA polymerase is responsible for most DNA replication in E. coli? Choose one: A. DNA polymerase I B. DNA polymerase II C. DNA polymerase III
DNA polymerase III
During DNA replication, the ??? strand is synthesized continuously.
During DNA replication, the leading strand is synthesized continuously.
Which of these is a difference in the total RNA content of prokaryotes and eukaryotes? Choose one: A. Prokaryotic mRNA is generally longer and contains more introns than eukaryotic mRNA. B. Eukaryotes transcribe several different types of rRNA, while prokaryotes transcribe only one rRNA. C. Eukaryotes encode a greater variety of ncRNA than prokaryotes. D. Transcription and translation are coupled in prokaryotes but not in eukaryotes.
Eukaryotes encode a greater variety of ncRNA than prokaryotes.
A scientist wishes to express a eukaryotic protein in bacterial cells. The gene is cloned along with its promoter region and is inserted into a plasmid. After transforming the plasmid into bacterial cells, protein expression is initiated, but no protein is observed after the cells are lysed. Why? How could this problem be fixed?
Eukaryotic and prokaryotic promoters have different sequences for RNA polymerase, and so the bacterial σ factor would most likely not bind the eukaryotic promoter sequence, thus RNA polymerase would not associate with the DNA, preventing transcription and thus protein expression. Cloning the gene behind a bacterial promoter sequence could solve this problem.
For the polymerase involved in leading strand synthesis, its level of processivity is expected to be
For the polymerase involved in leading strand synthesis, its level of processivity is expected to be hih
Which protein, involved in regulating GAL1 and GAL10, binds to galactose? Choose one: A. GAL4 B. GAL2 C. GAL80 D. GAL3
GAL3
In the galactose regulatory system of yeast, what would be the behavior of mutants that lack one of the following components: GAL3, GAL4, or GAL80? Consider the expression of the target genes with and without the presence of galactose. What would be the behavior of a GAL3-GAL80 double mutant?
GAL3 and GAL4 mutants will not express the target genes in the presence of galactose. A GAL80 mutant and a GAL3-GAL80 double mutant will express the target genes in the absence of galactose.
Why did George Gamow state that at least three nucleotides must be required to code for each amino acid?
Gamow concluded that three nucleotides were the minimum number required for each codon because a two-nucleotide codon would result in only 42, or 16, possible combinations. Because there are 20 standard amino acids, a two-nucleotide codon would have required more than one amino acid to be associated with the same codon. However, a codon of three nucleotides gives 43, or 64, possible combinations.
In the experiment by Jacob and Meselson, bacteria were grown for several generations in a medium containing heavy isotopes (15NH4Cl and 13C-glucose). The bacteria used these isotopes to synthesize all cellular components, resulting in their incorporation into all of the bacterial nucleic acids and proteins. The bacteria were switched to a normal medium after infection with bacteriophage, and the ribosomes were isolated. Jacob and Meselson discovered that the isolated ribosomes contained the heavy isotopes. Why did this result indicate that ribosomes are not carriers of genetic information?
If ribosomes were the source of genetic material, the bacteriophage would synthesize its own ribosomes soon after infection as a way of propagating itself. This would occur in the normal medium, and so these ribosomes would not contain heavy isotopes. However, if ribosomes were only sites of protein synthesis, then the viral mRNA would be translated by bacterial ribosomes, which were synthesized while the heavy isotopes were present. The presence of isotopes in ribosomes enabled Jacob and Meselson to confirm that ribosomes were not the viral genetic material.
How do the subunits required for elongation by bacterial RNA polymerase compare to the subunits required for transcription initiation? Choose one: A. Initiation requires σ factor only, while elongation requires the addition of the α2ββ′ω holoenzyme to σ factor. B. Initiation requires the α2ββ′ω holoenzyme plus σ factor, while elongation requires the α2ββ′ω holoenzyme without the presence of σ factor. C. Initiation requires the α2 subunits plus σ factor, while elongation requires these subunits with the addition of the ββ′ω subunits. D. Initiation requires the α2ββ′ω holoenzyme, while elongation requires the addition of σ factor to this complex.
Initiation requires the α2ββ′ω holoenzyme plus σ factor, while elongation requires the α2ββ′ω holoenzyme without the presence of σ factor.
What is the role of the mRNA 5' cap in eukaryotic translation? Choose one: A. It is the start of translation. B. It recruits the 60S ribosome. C. It forms a complex with the poly(A) tail, which orients the ribosome on the mRNA. D. It protects the mRNA from degradation.
It forms a complex with the poly(A) tail, which orients the ribosome on the mRNA.
Why is the ribosome considered a ribozyme
It is a ribozyme because it contains a catalytic RNA (23S rRNA in E. coli) that catalyzes the formation of a peptide bond between amino acids.
Assume that you have identified all the cis-acting sites in a prokaryotic genome for a particular trans-acting transcription factor. What is the function of this transcription factor when binding to these sites? A. If the site has the sequence GATC-CTAG then it is an activator, if GATC-GATC then it is a repressor B. It is most likely a ligand-dependent de-repressor since the lac repressor uses this mode of regulation. C. It is most likely a ligand-dependent repressor since the trp repressor uses this mode of regulation. D. It is not possible to predict the function from the sites since could activate or repress at different sites. E. Assume that it is a repressor unless it requires cAMP and then assume it is an activator like CRP-cAMP.
It is not possible to predict the function from the sites since could activate or repress at different sites.
Why does the initiator fMet-tRNAfMet bind to the P site rather than the A site of the ribosome? A. It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed. B. It needs to bind to the P site because the fMet chemical structure only fits the P site ribosomal active site. C. It binds to the A site as do all incoming AA-tRNAAA molecules so that a peptide bond can be formed. D. It needs to bind to the P site first and then it moves to the A site so that a peptide bond can be formed. E. None of these answers are correct.
It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed.
Consider the regulation of the trp operon in E. coli. If this system were present in yeast instead of E. coli, would you expect it to function properly? Explain.
It would not work in eukaryotes because this mechanism depends on transcription and translation occurring simultaneously; in eukaryotes these processes occur in different cellular compartments.
Why was Khorana's method of nucleotide synthesis so essential to the discovery of the genetic code?
Khorana's method of nucleotide synthesis was the first procedure that allowed directed synthesis, not simply random combinations of nucleotides. This was essential to cracking the genetic code, as it allowed Khorana and others to test specific triplet combinations in filter-binding and in vitro translation experiments.
Which of the following genes in the lac operon encodes lac permease? Choose one: A. lacZ B. lacA C. lacY D. lacI
LacY
Why are multiple lipid modifications often found on membrane-associated proteins?
Lipid modifications can be used for strong association of a protein with a membrane. As soluble proteins fold, the hydrophobic regions are mostly sequestered in the interior of the protein, while hydrophilic regions remain exposed to the polar, aqueous environment of the cell. Membranes have hydrophobic fatty acid tails arranged in the center of the membrane. The addition of lipids to the exterior of a protein adds hydrophobic regions that can insert within the membrane and therefore firmly associate with the membrane.
Why do acetylation and methylation not occur simultaneously on the same lysine residue on histones?
Methylation and acetylation are not found on the same regions of histones because they have opposite effects on histones. Methylation leads to tighter compaction of histones, often found in transcriptionally inactive regions. Acetylation is often found in histones that are not as tightly condensed and are therefore more accessible by the transcriptional machinery.
In the trp attenuation system of E. coli, predict the effects of the following mutations on the operation of page 1194 the regulatory system, and give your reasons. To simplify matters, assume that the Trp repressor is not active. Three broad possibilities exist for whether the operon is transcribed in the absence or presence of charged tRNATrp: behavior is normal, transcription is always low ("noninducible"), or transcription is always high ("constitutive"). a. Mutation in region 4 so it cannot pair with region 3. SHOW ANSWER b. Mutation in region 3 so it can pair with region 4 but not with region 2. SHOW ANSWER c. Change of the two Trp codons to glycine codons. SHOW ANSWER d. Change of the AUG codon of the leader peptide to AUA. SHOW ANSWER e. Combination of the first mutation with each of the other three.
Mutation in region 4 so it cannot pair with region 3: would be constitutive, because no terminator is formed. Mutation in region 3 so it can pair with region 4 but not with region 2: would be noninducible; antiterminator cannot form in the absence of Trp-tRNATrp, so terminator would always form. Change of the two Trp codons to glycine codons: noninducible, because the ribosome would not pause at the Gly codons in the presence of low Trp-tRNATrp. It might pause if glycine levels are low, but glycine is rarely low. Change of the AUG codon of the leader peptide to AUA: noninducible, because the ribosome would not be able to stabilize the antiterminator. Combination of the first mutation (a) with each of the other three (b, c, or d). Constitutive, because if the terminator cannot form, it does not matter what happens upstream. For the mutation (a) and (d) combined: pausing at the 1-2 stem-loop is too brief to reduce the rate of transcription significantly. This latter is an "epistasis" test, often used to order the events in a regulatory pathway.
Which transcription factor is NOT used to induce a pluripotent state? Choose one: A. Nanog B. Sox2 C. Oct4 D. Klf4
Nanog
Assume that you have identified all the cis-acting sites in a prokaryotic genome for a particular gene regulatory protein. Can you predict the consequences of the regulatory protein binding to these sites? Give your reasons, with examples.
No, you cannot be certain what the effect of protein binding will be. Numerous examples indicate that the effect of a bound trans-acting factor depends on its context. Bound bacteriophage λ CI protein and CRP-cAMP both can act as either activators or repressors. The mammalian Oct4-Sox2 heterodimer activates several ES-specific genes and works to repress developmental genes.
What accounts for the high fidelity of DNA polymerization? A. DNA repair takes care of all errors so no need for proofreading activity in DNA polymerase III enzyme. B. DNA polymerase fidelity is ensured by the G-C and A-T base pairs forming between two DNA strands. C. Mg2+ in the active site provides proofreading functions through ionic bonds with only correct nucleotides. D. The 5' to 3' exonuclease activity is a proofreading function that removes incorrect nucleotides at the fork. E. None of these answers are correct.
None of the above
The mRNA sequence AUGCACAGU codes for the first three amino acids of a particular protein. Which nucleotides can be changed without modifying the amino acid sequence that will result after translation?
Normally, AUGCACAGU is translated to produce the tripeptide Met-Thr-Gly. The sixth or ninth nucleotides can be changed without changing the amino acid sequence, as there are four codons for Thr and Gly that only differ at the third position. It would not be possible to change the third nucleotide because there is only one Met codon.
What is the molecular biology term that describes where DNA replication commences? Choose one: A. origin B. leader C. starter D. beginner With regard to Part 1, how many of these sequences should exist within a single replication bubble? Within a replication bubble, how many replication forks are expected if bidirectional replication is taking place?
Origin, 1, 2
Which amino acid was coded by the first codon deciphered? Choose one: A. Phe B. Pro C. Lys D. Gly
Phe
Of these eukaryotic polymerases, which is primarily responsible for leading strand synthesis during nuclear DNA replication? Choose one: A. Pol α B. Pol ε C. Pol γ D. Pol δ
Pol ε
Which of the following is correct regarding DNA replication? Choose one: A. Polymerization makes the new DNA strand in the 5'→3' direction. B. Polymerization makes the new DNA strand in the 3'→5' direction. C. Polymerization makes the new DNA strand in either the 5'→3' direction or the 3'→5' direction.
Polymerization makes the new DNA strand in the 5'→3' direction.
Why would antibiotics that affect the ribosome kill bacteria without harming their host? Choose one: A. Eukaryotes are more complex than prokaryotes. B. Prokaryotic ribosomes have a different structure than eukaryotic ribosomes, so antibiotics that bind to one cannot bind to the other. C. Antibiotics form different types of bonds with prokaryotic proteins and eukaryotic proteins, so only the interactions with prokaryotic proteins affect ribosome function. D. Fungi do not need to kill eukaryotes, only prokaryotes.
Prokaryotic ribosomes have a different structure than eukaryotic ribosomes, so antibiotics that bind to one cannot bind to the other.
Formation of the 5′ m7G cap requires that the 5′-nucleotide be a triphosphate. Explain why only this nucleotide retains all three phosphoryl groups.
RNA synthesis requires NTPs; the first nucleoside added has three phosphoryl groups. The reaction to add each subsequent nucleotide is initiated by a nucleophilic attack on the α phosphate, resulting in cleavage and release of PPi.
Choose 3 reasons why RNA is a dynamic biomolecule, and in that way, distinct from DNA. 1. RNA contains a ribose sugar, a nucleotide base, and phosphate groups. 2. RNA tertiary structure is altered by the binding of protein and small molecule ligands. 3. RNA forms base pairs with complementary nucleotides and predictably transmits information. 4. RNA forms a double helical structure with anti-parallel strands with 5' - 3' and 3' - 5' polarity. 5. RNA is synthesized and degraded continuously to control biochemical processes 6. RNA can be degraded in the 5' to 3' direction or in the 3' to 5' direction, and it can be cleaved internally. 7. RNA is able to participate in transitory functional base pairing interactions with RNA or DNA.
RNA tertiary structure is altered by the binding of protein and small molecule ligands. RNA is synthesized and degraded continuously to control biochemical processes RNA is able to participate in transitory functional base pairing interactions with RNA or DNA.
RNaseP can cleave both pre-tRNA and pre-rRNA sequences. If the primary sequences of these RNAs are not always similar, how the does the enzyme recognize and bind its substrates?
RNaseP recognizes the secondary structure of tRNA and rRNA rather than the primary structure. Such structures form co-transcriptionally, so the primary transcript is cleaved on the basis of the presence of these structures and not any individual sequences.
Why does RNA contain uracil and DNA contain thymine? Choose the one BEST answer. A. Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a C-G base pair to a T-A base pair. B. Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a C-G base pair to an A-T base pair. C. RNA transcription base pairs U-T in the RNA with DNA, whereas this process base pairs A-T in the RNA with DNA. Therefore, if uracil were present in DNA, then U-A and A-U could not be distinguished. D. Uracil deamination to cytosine provides a survival advantage to rapidly evolving species like viruses. E. Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a G-C base pair to a A-T base pair.
Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a C-G base pair to a T-A base pair.
What is the primary difference between strong and weak bacterial promoters? How does transcription differ from these two types of promoters?
Strong bacterial promoters differ only slightly from the conserved sequences at the −35 and −10 boxes, whereas weak promoters can contain several base pair changes. These are the sites of σ-factor binding, so the closer the −35 and −10 boxes are to the conserved sequences, the stronger the association between RNA polymerase and the promoter. Therefore, strong promoters are more likely to be bound by σ factor and transcribed than weak promoters.
What effect is likely from the alteration of the sequence at the 5' end of the RNA component of the U1 snRNP from 5'AUACYYACCU3" to 5'AUACYYGCCU3'? (Y stands for the modified base "psi"). A. The splicing junction will not be recognized by the U2 snRNP because the tertiary structure of the U1 snRNP has altered. B The splicing reaction will not occur because the A nucleotide that was changed to G was required for nucleophilic attack and formation of the lariat structure. C. The 5' splicing junction that will be recognized by this altered spliceosome will be 5'AGGCAGGU3' because the nucleotide sequence of the base pairing portion of U1 snRNP has changed. D. No effect at all because the sequences can bind to each other using noncanonical base pairs. E. None of the above
The 5' splicing junction that will be recognized by this altered spliceosome will be 5'AGGCAGGU3' because the nucleotide sequence of the base pairing portion of U1 snRNP has changed.
What must the rate of nucleotide incorporation by primase be such that it completes a primer in exactly the same time that an Okazaki fragment is completed?
The average Okazaki fragment is 2,000 bases and synthesized in 2 seconds (1,000 bases/second). A typical primer is about 10-12 bases; thus, primase would need a rate of 5-6 nucleotides/second to avoid being the rate-limiting step of replication.
Calculate the number of Okazaki fragments produced during the replication of a single E. coli chromosome.
The average Okazaki fragment is 2,000 bases; the E. coli chromosome is 4,638 kb. Therefore, ∼2,319 Okazaki fragments form per replication.
16. (5 pts) Choose the one correct statements regarding tRNA synthetases below. A. Hydrolysis of ATP drives conformational changes resulting in formation of the tRNA-amino acid complex. B. tRNA synthetase is a ribozyme that uses RNA to catalyze the formation of the tRNA-amino acid complex. C. tRNA synthetases use small RNA molecules to correctly recognize the anticodon of the tRNA molecules. D. The amino acid is linked to the 3'OH of the ATP before formation of the tRNA-amino acid complex. E. The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct.
The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct.
If the sequence 5′-AACGC-3′ were damaged by reactive oxygen species, what would be the most prevalent product, and what would be the result of replication (show both strands after replication)?
The guanine would become 8-hydroxyguanine and tautomer, 8-oxoguanine. These base pair with adenine, resulting in a G-A mismatch that is not likely to be repaired by DNA repair mechanisms. After a second round of replication, the result is a G→C substitution in one of the daughter strands. First round of replication: 5′-AACGC-3′ 5′-AACGC-3′ (8-hydroxyguanine = G): 3′-TTGAG-5′ 3′-TTGCG-5′ Second round of replication: 5′-AACCC-3′ 5′-AACGC-3′ 3′-TTGAG-5′ 3′-TTGCG-5′ 5′-AACGC-3′ 5′-AACGC-3′ 3′-TTGAG-5′ 3′-TTGCG-5
The presence of the β clamp ??? the processivity of DNA polymerase.
The presence of the β clamp increases the processivity of DNA polymerase.
What happens during protein synthesis if a tRNA is charged with the wrong amino acid? A. Protein synthesis will continue, however the RNA splicing reaction will be blocked by the amino acid. B. The amino acid will be removed by a ribozyme that cleaves tRNA molecules at the CCA end. C. The tRNA will be excluded from the A site of the ribosome but not from the P site. D. The protein will contain the wrong amino acid in the corresponding codon position. E. The protein will not be synthesized because the incorrectly charged tRNA will be stuck in the P site.
The protein will contain the wrong amino acid in the corresponding codon position.
An Ames test of a suspected mutagen was examined both before and after incubation with rat liver extract, giving the following results. What can you conclude about the suspected mutagen?
The results suggest the suspected mutagen is mutagenic without liver metabolism, and even more mutagenic after metabolism by liver enzymes.
RNAi-based modification of amylopectin content in wheat occurs through decreased expression of starch-branching enzymes. If the siRNA used to knock down plant starch-branching enzymes was transmitted to humans, what enzyme might it affect, and why would this be harmful?
The structure of amylopectin is similar to that of glycogen, as both contain an α(1→4) chain of glucose molecules with α(1→6) branches. The spacing between branches varies, so it is reasonable to conclude that starch-branching enzymes and glycogen-branching enzymes that catalyze formation of the same bond are similar. If siRNA against a starch-branching enzyme was transmitted to mammals, it could result in repression of a glycogen-branching enzyme, and the cell would store only straight-chain glycogen, reducing the rate of glycogen breakdown and the capacity to store glycogen. These effects would negatively affect ATP production and could lead to symptoms similar to those observed for glycogen storage disorders.
What are the three types of coatomer proteins that assist with vesicle formation, and with what type of transport is each associated?
The three types are COPI, COPII, and clathrin. COPI vesicles mediate transport within the Golgi apparatus; COPII vesicles mediate transport between the endoplasmic reticulum and the Golgi apparatus; and clathrin vesicles mediate transport between the Golgi apparatus and plasma membrane.
What is the wobble hypothesis, and how does it explain the degeneracy of the genetic code?
The wobble hypothesis states that noncanonical base pairs can exist between the third (3′) position in the codon and the first (5′) position in the anticodon. Wobble base pairing explains how some mRNA codons that differ by the nucleotide at the third position can bind to the same tRNA, facilitated by the presence of modified nucleotides such as inosine at the wobble position. Inosine can form hydrogen bonds with A, C, or U.
What are the three possible amino acids resulting from the translation of the RNA sequence CCAAA?
They are Pro (CCA), Gln (CAA), or Lys (AAA).
Which of the following statements about pluripotent cells is true? Choose one: A. They can develop into many different differentiated cell types. B. They can develop into only one differentiated cell type. C. They cannot become differentiated. D. All cells in an organism are pluripotent.
They can develop into many different differentiated cell types.
mRNA and lncRNA share several common characteristics. Which of the following are not characteristics of both mRNA and lncRNA, however? Choose one: A. Their 5' and 3' ends are chemically modified. B. They are transcribed by RNA polymerase II. C. They function as a scaffold for the assembly of large ribonucleoprotein complexes. D. Their transcript is greater than 200 nt in length.
They function as a scaffold for the assembly of large ribonucleoprotein complexes.
Why does the human mitochondrion have a slightly different genetic code than that found in the cytosol?
This is probably due to its ancestral origin. It is widely believed that mitochondria evolved from bacteria taken up by early nucleated cells. Because mitochondrial DNA has different evolutionary origins from chromosomal DNA, it contains a separate protein translation system, including tRNA genes.
You are studying the regulation of a Drosophila gene, and you have available the genome sequences of several related Drosophila species. By comparing these sequences in the vicinity of your gene, you identify several regions that appear to be conserved. Moreover, you identify potential binding sites for known regulatory proteins. You speculate that this region is an enhancer. Suggest how you might test this idea.
To validate your hypothesis, you would use recombinant DNA to place the proposed enhancer upstream of a reporter gene in Drosophila embryos. Several constructs should be prepared, with the enhancer in either orientation and at a variable distance from the promoter for the reporter. If the DNA region includes an enhancer, you should see expression of the reporter. See Levine and colleagues' study to analyze the eve stripe 2 enhancer (see Figure 23.54).
Which of the following statements regarding the transcription and mRNA processing in eukaryotes is NOT correct? Choose one: A. Capping of the mRNA 5' end occurs co-transcriptionally. B. Transcription factors (TFIIA, TFIIB, TFIIE, TFIIH) assemble on RNA polymerase II after it binds to TFIID/TBP. C. Splicing of the mRNA occurs co-transcriptionally. D. Hyperphosphorylation of the RNA polymerase II CTD regulates the interaction of mRNA processing enzymes with the newly transcribed regions of mRNA.
Transcription factors (TFIIA, TFIIB, TFIIE, TFIIH) assemble on RNA polymerase II after it binds to TFIID/TBP.
Which of the following is NOT a part of translational termination? Choose one: A. Uncharged tRNA recognition of stop codon in the mRNA B. Ribosome subunit dissociation C. GTP hydrolysis D. Release factor recognition of stop codon in the mRNA
Uncharged tRNA recognition of stop codon in the mRNA
For the polymerase involved in leading strand synthesis, would you expect to observe 3′→5′ exonuclease activity
Yes
When is the 5′ m7G cap added to an mRNA transcript? Choose one: A. near the end of elongation B. after transcription termination C. after transcription initiation but early in elongation D. during transcription initiation E. during transcription termination
after transcription initiation but early in elongation
When does RNA-mediated gene silencing occur? Choose one: A. during transcription initiation B. after transcription of an mRNA but before translation C. during replication D. during translation
after transcription of an mRNA but before translation
DNA is polymerized Choose one: A. always in the 5' to 3' direction. B. usually in the 5' to 3' direction. C. never in the 5' to 3' direction. D. in either 5' to 3' or 3' to 5' directions.
always in the 5' to 3' direction.
Posttranscriptional mRNA decay can occur from both the 5' and 3' ends. What primary process is common to both of these decay pathways? Choose one: A. exonucleolytic decay B. decapping C. m7G cap scavenging D. deadenylation
deadenylation
Which replication fork enzyme relieves torsional stress in front of the fork? Choose one: A. primase B. polymerase C. gyrase D. helicase
gyrase
As of 2017, what is the next step in stem cell differentiation experiment planned for Parkinson's patients in the study being conducted at the Scripps Institute in San Diego? A. isolate fibroblasts from all patients B. transcription factor genes into fibroblasts C. Diagnose Parkinson's D. isolate differentiated DOPA+ cells E. inject differentiated DOPA+ cells into the brains of patients
inject differentiated DOPA+ cells into the brains of patients
Which type of interaction is not a functional mechanism by which lncRNA regulates gene expression? Choose one: A. lncRNA can form base pairs with an mRNA and prevent translation. B. lncRNA can form base pairs with a region of ssDNA and alter transcription of the associated region. C. lncRNA can form secondary structures that provide an assembly platform for factors that regulate gene expression. D. lncRNA can be translated to produce a ribonucleoprotein complex that binds to DNA and recruits other protein complexes that regulate transcription.
lncRNA can be translated to produce a ribonucleoprotein complex that binds to DNA and recruits other protein complexes that regulate transcription.
Which of the following models of DNA replication was verified by the Meselson-Stahl experiments? Choose one: A. conservative B. semiconservative C. dispersive D. none of the above
semiconservative
Describe the secondary and tertiary structure of tRNA.
tRNA forms a cloverleaf secondary structure with three arms: D, TΨC, and anticodon. The top stem of tRNA is the acceptor site for amino acid attachment. The D and TΨC loops further interact to form the L-shaped tertiary structure.
Which of the following properties are associated with the origin of replication in bacterial DNA? Choose one: A. three 13-bp sequences high in G-C content followed by four 9-bp sequences high in A-T content B. three 13-bp sequences high in A-T content followed by four 9-bp sequences high in G-C content C. three 13-bp sequences followed by four 9-bp sequences, both of which are high in A-T content D. three 13-bp sequences followed by four 9-bp sequences, both of which are high in G-C content
three 13-bp sequences followed by four 9-bp sequences, both of which are high in A-T content
What is the role of σ factor in prokaryotic transcription? Choose one: A. to stabilize the RNA polymerase complex B. to ensure transcription begins in the correct location C. to promote elongation by RNA polymerase D. to signal for termination of transcription
to ensure transcription begins in the correct location