6) From DNA to Protein
Activation of an amino acid by an aminoacyl-tRNA synthetase involves the covalent linkage of... A. AMP to the amino acid. B. ADP to the amino acid. C. AMP to the tRNA. D. ADP to the tRNA. E. ADP to the enzyme.
A. AMP to the amino acid. The amino acid is initially activated by an ester linkage, through its carboxyl group, to an AMP molecule resulting from the hydrolysis of an ATP molecule. The amino acid is then transferred onto the appropriate tRNA molecule.
An elongating ribosome is bound to appropriate tRNAs in both the A and the P sites and is ready for peptidyl transfer. What happens next? A. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site tRNA. B. The amino end of the polypeptide chain is released from the P-site tRNA and joined to the free carboxyl group of the amino acid linked to the A-site tRNA. C. The carboxyl end of the amino acid is released from the A-site tRNA and joined to the free amino group of the polypeptide chain linked to the P-site tRNA. D. The amino end of the amino acid is released from the A-site tRNA and joined to the free carboxyl group of the polypeptide chain linked to the P-site tRNA.
A. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site tRNA.
After the first and before the second chemical step of RNA splicing, the intron of the pre-mRNA... A. is still covalently connected to the 3' exon and has an internal branch in the shape of a lariat. B. is still covalently connected to the 3' exon and is linear. C. is still covalently connected to the 5' exon and has an internal branch in the shape of a lariat. D. is still covalently connected to the 5' exon and is linear. E.is still covalently connected to both of its flanking exons and is linear
A. is still covalently connected to the 3' exon and has an internal branch in the shape of a lariat. Due to the first transesterification step, the branch point in the intron is covalently bound to the 5' end of the intron, and the intron is no longer continuous with the upstream (5') exon.
No matter where translation begins, only one polypeptide sequence pattern can be obtained from the translation of an mRNA chain with the sequence (AC)n; that is, an mRNA with the sequence 5'...ACACACACAC...3'. In contrast, an mRNA chain with the sequence (ACG)n can be translated to polypeptide sequences of three distinct patterns, depending on the choice of the reading frame. Using similar judgment, how many different polypeptide sequence patterns can be obtained from an mRNA chain with the sequence (ACGU)n? Write down your answer in digits, e.g. 12.
Answer: 1 Even though the polypeptide sequence pattern for the (ACGU)n RNA polymer is more complicated than for the other two examples, there is only one such pattern regardless of the choice of the reading frame: (Thr.Tyr.Val.Arg)n. For the (AC)n RNA, also only one polypeptide sequence pattern is possible: (Thr.His)n.
For the bacterial transcription machinery, which of the following mRNA sequences would you expect to constitute a potent transcriptional termination signal? Note that the two underlined regions in each sequence are complementary to each other. A. 5'... UGGCCCAGUCGGAAGACUGGGCCUUUUGUUUU...3' B. 5'... UGGCCCAGUCGGAAGACUGGGCCCGCGGAGCU...3' C. 5'... UUUUGUUUUAGGCCCAGUCGGAAGACUGGGCCA...3' D. 5'... CGCGGAGCUAGGCCCAGUCGGAAGACUGGGCCA...3'
Answer: A Feedback: Many bacterial transcription terminators consist of a stable hairpin followed by a string of U nucleotides (that form weak A-U base pairs with the template), which together help disengage the mRNA transcript from RNA polymerase
You have identified an RNA-binding protein that can bind directly (with varying affinities) to mRNA molecules that bear the following sequences near the 3' splice sites of their exons. Based on these results, what is the five-nucleotide consensus sequence for the binding site of this protein? Your answer would be a five-letter string composed of letters A, U, C, or G only; e.g. UUUAG. 5'...AACGG...3' 5'...AUCGG...3' 5'...GACGC...3' 5'...AACCC...3' 5'...AUCGC...3' 5'...AACAC...3' 5'...AACCC...3'
Answer: AACGC The most common nucleotide at each position of the alignment is chosen for the consensus sequence. Note that none of the examined mRNAs has the consensus sequence.
Which of the following types of noncoding RNA chiefly functions in the processing and chemical modification of ribosomal RNAs (rRNAs)? A. Small nuclear RNAs (snRNAs) B. Small nucleolar RNAs (snoRNAs) C. Small interfering RNAs (siRNAs) D. Transfer RNAs (tRNAs) E. MicroRNAs (miRNAs)
Answer: B Feedback: In the nucleolus, the snoRNAs help process and chemically modify the rRNAs in various ways.
Which of the following better describes a typical, actively translated mRNA in its journey from the nucleus to the cytosol? A. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (5' end first), and remains linear. B. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (3' end first), and remains linear. C. Initially linear, the mRNA enters the cytosol (5' end first), and adopts a circular conformation. D. Initially linear, the mRNA enters the cytosol (3' end first), and adopts a circular conformation. E. Initially linear, the mRNA enters the cytosol (3' end first), and remains linear.
Answer: C Feedback: Normally, the export of an mRNA to the cytosol occurs with the 5' cap proceeding first. The mRNA can then circularize through the interaction of the cytosolic cap-binding proteins with the cytosolic poly-A-binding proteins to facilitate rapid translation.
In humans, nearly 80% of proteins are acetylated on their N-terminal residue, a modification known to be recognized by a specific E3 enzyme, which directs the ubiquitylation of the protein for rapid degradation. Does this mean that all of these proteins are actively degraded at any given time? A. Yes; this high turnover rate ensures that their activity is under tight control. B. Yes; but they are not fully degraded in this way and can still function as protein fragments. C. No; the destruction signal can be buried in the interior of the protein or bound to other proteins. D. No; the E3 enzyme recognizing this mark is inactive most of the time.
Answer: C Feedback: The modification is not necessarily exposed for the E3 enzymes to recognize. Exposure of the modification can occur if, for example, the protein becomes misfolded.
As an mRNA molecule is processed in the nucleus, it loses some proteins and binds to new ones, some of which are used in mRNA surveillance pathways. The presence of which of the following molecules on an mRNA is a signal that the mRNA is still NOT ready for nuclear export? A. Cap-binding complex B. Exon junction complex C. snRNPs used in splicing D. poly-A-binding proteins E. SR proteins
Answer: C Feedback: The snRNPs remain associated with the pre-mRNA until it is fully spliced. Their presence indicates that the mRNA is still being processed.
Sort the following events in the order that they occur during transcription initiation in Escherichia coli. Your answer would be a four-letter string composed of letters A to D only; e.g. DCAB. (A) Abortive initiation trials (B) σ Factor release from the RNA polymerase holoenzyme (C) Binding of the holoenzyme to the promoter in the "closed" complex (D) Formation of the transcription bubble
Answer: CDAB In bacterial transcription initiation, binding of the RNA polymerase holoenzyme to the promoter region in the closed complex is followed by the formation of the open complex involving the formation of the transcription bubble. After a number of abortive initiations, RNA polymerase clears the promoter and σ factor is released.
Several mechanisms contribute to the diversity of the mRNAs and proteins encoded by a single gene in our genome. Which of the following is normally NOT one of them? A. Alternative choice of polyadenylation sites B. Alternative choice of translation initiation sites C. Alternative choice of transcription initiation sites D. Alternative choice of the reading frames E. Alternative choice of splice sites
Answer: D Feedback: Eukaryotic cells take advantage of various mechanisms such as alternative splicing and alternative choices of polyadenylation sites, transcription start sites, and translation start sites to create a highly complex pool of gene products in the cell. Normally, however, a cellular gene is interpreted in only one of the possible reading frames
Cajal bodies in the eukaryotic cell nucleus... A. are stockpiles of fully mature snRNPs and other RNA processing components. B. can only be observed by electron microscopy. C. are absolutely required in all cell types. D. speed up the maturation and assembly of snRNPs and snoRNPs. E. are the main sites of pre-mRNA splicing.
Answer: D Feedback: In Cajal bodies, snRNPs and snoRNPs undergo maturation and assembly at enhanced rates compared to the rates estimated for cells in which these aggregates are disrupted. However, this enhancement is not crucial for all cells. Pre-mRNA splicing is thought to happen in mRNA production "factories" dispersed throughout the nucleus.
Comparing mRNA molecules from human and Escherichia coli cells, which of the following is typically NOT true? A. A human mRNA has a special 5' cap, while a bacterial mRNA does not. B. A human mRNA has a poly-A tail, while a bacterial mRNA does not. C. A human mRNA undergoes alternative splicing, while a bacterial mRNA does not. D. A human mRNA contains noncoding sequences, while a bacterial mRNA does not. E. A typical human mRNA encodes one protein, while many bacterial mRNAs encode several different proteins.
Answer: D Noncoding regions are also found in E. coli mRNAs, although these regions are typically short compared to those in human cells.
The enzyme poly-A polymerase is responsible for adding 3' poly-A tails to eukaryotic mRNAs. This enzyme... A. cuts the mRNA after recognition of the cleavage/polyadenylation signal by CstF and CPSF proteins. B. polymerizes the tail using an RNA template that is part of the enzyme. C. is extremely processive. D. normally adds about 1000 A nucleotides to the mRNA. E. uses ATP as a substrate.
Answer: E Feedback: After the mRNA is cleaved, this enzyme adds about 200 A nucleotides (from ATP) one by one without using a template.
The nucleolus is a dynamic subcompartment within the nucleus and its size varies depending on the circumstances. In which of the following cells would you NOT expect to see the nucleoli? A. A yeast cell undergoing DNA replication B. A human neuron in a quiescent (G0) state C. A human macrophage active in phagocytosis D. A fruit fly embryonic nucleus in the G2 phase of the cell cycle E. A mouse embryonic cell in the metaphase stage of mitosis
Answer: E Feedback: During the M phase, the chromosomes condense and the nucleoli disappear. In telophase, the chromosomal regions contributing their rRNA genes to the subcompartment coalesce again and their transcription resumes.
Which of the following processes takes place in the nucleoli within the eukaryotic nucleus? A. Ribosome assembly B. rRNA gene transcription C. Telomerase assembly D. tRNA processing E. All of the above
Answer: E Feedback: The nucleolus is a factory for the processing of various noncoding RNAs and their assembly into larger complexes
Due to their high transcription rate, active ribosomal RNA (rRNA) genes can be easily distinguished in electron micrographs of chromatin spreads. They have a characteristic "Christmas tree" appearance, where the DNA template is the "trunk" of the tree and the nascent RNA transcripts form closely packed "branches." At the base of each branch is an RNA polymerase extending that branch, while RNA processing complexes at the tip of the branch form terminal "ornaments." The top of the tree represents the ... of the rRNA gene, and the "ornaments" are at the ... end of the nascent rRNA molecules. A. end; 3' B. end; 5' C. beginning; either 3' or 5' D. beginning; 3' E. beginning; 5'
Answer: E Feedback: The shortest rRNA "branches" at the "top" of the tree emanate from the beginning of the gene since their transcription has just started. All branches extend at their 3′ end at their base while their 5′ end starts to associate with RNA processing complexes.
To ensure the fidelity of splicing, the spliceosome... A. hydrolyzes ATP to undergo complex rearrangements. B. examines the splicing signals on the pre-mRNA several times. C. assembles on the pre-mRNA co-transcriptionally. D. takes advantage of "exon definition." E. All of the above.
Answer: E Feedback: The spliceosome uses ATP hydrolysis to produce a series of rearrangements, allowing the splicing signals to be checked and rechecked to ensure the accuracy of the process. The fact that splicing is coupled to transcription provides an advantage and helps the cell keep track of the exons, as does exon definition by SR proteins and other factors that mark the splice sites around each exon.
How is tRNA splicing different from mRNA splicing in eukaryotic cells? A. tRNA splicing does not proceed via transesterification reactions. B. tRNA splicing is carried out by proteins only. C. tRNA splicing does not create a lariat intermediate. D. tRNA splicing involves RNA endonuclease and RNA ligase activities. E. All of the above.
Answer: E Feedback: tRNA splicing appears to be simpler compared to the splicing of pre-mRNAs.
What is the function of the 19S cap in the proteasome complex? A. Recognizing the proteins that will be degraded B. Unfolding the proteins that will be degraded and threading them into the central 20S cylinder for digestion C. Removing the polyubiquitin chain from the proteins that will be degraded D. Hydrolyzing ATP to make the digestion process highly efficient E. All of the above
Answer: E The 19S cap in the proteasome is a large complex with various functions including substrate recognition, de-ubiquitylation, unfolding and feeding into the 20S core, as well as regulatory roles
How many ATP molecules must be hydrolyzed by the proteasome for the digestion of a protein that has been tagged for degradation with a polyubiquitin chain? A. None; the digestion does not require ATP hydrolysis and is exergonic B. One C. Two D. Ten E. Many; the number depends on the specific protein
Answer: E The AAA proteins in the cap hydrolyze tens or hundreds of ATP molecules to unfold and thread each protein into the central 20S digestive core.
Indicate true (T) and false (F) statements below regarding mRNA splicing in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Introns must be removed in the order in which they occur along the pre-mRNA. ( ) Nucleosomes tend to be positioned over exons (on average, about one nucleosome per exon). ( ) Exon size tends to be much more variable than intron size. ( ) The exon junction complexes mark the sites on the mRNA where splicing has not been successful.
Answer: FTFF Exon size tends to be much more uniform than intron size and averages at about 150 nucleotide pairs, close to the length of DNA wrapped by one nucleosome. The bound nucleosome is suggested to act as a "speed bump" to allow the assembly of splicing proteins involved in exon definition.
Indicate true (T) and false (F) statements below regarding the RNA WORLD HYPOTHESIS. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) According to this hypothesis, RNA in primitive cells was responsible for storing genetic information, while proteins were responsible for the catalysis of chemical reactions. ( ) The existence of natural ribozymes supports this hypothesis. ( ) In support of this hypothesis, all peptides in present-day cells are made by the ribosome. ( ) All present-day cells use DNA as their hereditary material.
Answer: FTFT The hereditary material in all present-day cells is DNA, but according to the RNA world hypothesis, RNA preceded DNA (as genetic material) as well as proteins (as catalysts). Some short peptides are synthesized without the ribosome, an observation that provides clues as to how the early transition from an RNA world to a protein-dominated world might have initiated.
Indicate true (T) and false (F) statements below regarding the folding of proteins upon their synthesis. Your answer would be a five-letter string composed of letters T and F only, e.g. TFFFF. ( ) A globular protein can fold to its native form while still within the ribosome. ( ) A protein can begin to fold and to bind to other proteins while still being synthesized on the ribosome. ( ) A protein may undergo several cycles of hsp70 binding and release while still being synthesized on the ribosome. ( ) A molten globule can form only after the synthesis of the protein is complete. ( ) A protein typically undergoes hsp60-assisted folding while still being synthesized.
Answer: FTTFF Proteins can start folding as soon as they emerge from the ribosome (or even within the exit tunnel of the ribosome, but only for small helical segments) and one domain can be fully folded and functional before an entire multidomain protein is completely synthesized. However, folding within the hsp60 chaperonin complex typically happens only post-translationally.
The sequence of a region of DNA around the 5′ end of a gene in Escherichia coli is shown below. The -10 hexamer and the transcription start site are highlighted. What would be the sequence of the first 10 nucleotides of the mRNA transcribed from this gene? Write down the sequence from 5′ to 3′, e.g. CGGAUAAACT. ---- 5′...GCGCTTGGTATAATCGCTGGGGGTCAAAGAT...3′
Answer: GGUCAAAGAU Feedback: The mRNA sequence starts from the transcription start site and is the same as the sequence of the coding strand of the DNA (and complementary to the "template strand") except that T is replaced with U.
Indicate whether each of the following is located (or takes place) in the large (L) or the small (S) ribosomal subunit. Your answer would be a four-letter string composed of letters L and S only, e.g. LLLS. ( ) Peptide bond formation ( ) Codon-anticodon interaction ( ) The path of the mRNA ( ) The polypeptide exit tunnel
Answer: LSSL The RNA components of the ribosome in the large and small subunit are responsible for peptide bond formation and the verification of the codon-anticodon match, respectively. mRNA is threaded through the small subunit, while the nascent protein emerges from the large subunit.
Indicate true (T) and false (F) statements below regarding the accuracy of translation by the ribosome. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) The difference in the binding affinity of a codon to a correct versus incorrect anticodon CANNOT by itself account for the high accuracy of translation. ( ) GTP hydrolysis by EF-Tu both speeds up translation and increases its accuracy. ( ) The ribosome can detect correct Watson-Crick base-pairing between the codon and anticodon in the A site, and consequently trigger GTP hydrolysis by EF-Tu. ( ) Even after GTP hydrolysis by EF-Tu, the ribosome can prevent wrong amino acid incorporation by kinetic proofreading.
Answer: TTTT The difference in affinity between correct and incorrect codon-anticodon matches cannot by itself account for the high accuracy of translation. The ribosome further assesses the correctness of the match; in the case of a correct match, EF-Tu hydrolyzes its bound GTP. This contributes to the speed and fidelity of translation. Even after EF-Tu release, there is a second proofreading opportunity for the ribosome to prevent an incorrect amino acid from being added to the growing chain.
Three consecutive nucleotides in RNA (such as AUC or GUA) constitute a triplet called a ..., which can specify an amino acid or a stop signal for translation.
Answer: codon Three consecutive nucleotides in a protein-coding sequence of mRNA constitute a codon. Each codon specifies either one amino acid or a stop in the translation process.
Fill in the blank. These RNA molecules can catalyze a wide variety of biochemical reactions, including peptide bond formation and RNA splicing. They can be found in nature or selected in vitro for a desired function. Such a molecule is called a(n) ...
Answer: ribozyme Ribozymes are RNAs capable of catalyzing many different biochemical reactions. The ribosome provides an example in which the rRNA of the large subunit bears peptidyl transferase activity. Many other functions not found naturally can be screened for in the laboratory.
On the ribosome, the mRNA is read from ..., and the polypeptide chain is synthesized from... A. 5' to 3'; C- to N-terminus. B. 5' to 3'; N- to C-terminus. C. 3' to 5'; C- to N-terminus. D. 3' to 5'; N- to C-terminus.
B. 5' to 3'; N- to C-terminus. During translation elongation, the ribosome translocates on the mRNA from the 5′ end to the 3′ end in consecutive sets of three nucleotides. The polypeptide is concomitantly elongated at its C-terminus by addition of individual amino acids.
Which of the following nucleotides is hydrolyzed in both transcription and in translation elongation? A. ATP B. GTP C. TTP D. UTP E. CTP
B. GTP
This general transcription factor recognizes the TATA box in RNA polymerase II promoters. It is... A. the only single-subunit general transcription factor. B. able to introduce a rather sharp kink in the double helix upon binding to DNA. C. responsible for the phosphorylation of the RNA polymerase CTD during transcription initiation. D. also responsible for the recognition of the BRE element in the promoter. E. All of the above.
B. able to introduce a rather sharp kink in the double helix upon binding to DNA. Feedback: It is the multisubunit TFIID, containing a TATA-binding protein, which, upon binding to the TATA box, creates a sharp kink in the double helix that serves as a landmark to attract other transcription factors.
The transcript for which of the following noncoding RNA in our cells is expected to undergo 5' cap addition after transcription? A. 5S rRNA B. miR-21 (a microRNA) C. tRNAPhe D. 5.8S rRNA E. 18S rRNA
B. miR-21 (a microRNA) Feedback: MicroRNAs are mainly transcribed by the RNA polymerase II machinery and, similar to mRNAs, their primary transcripts normally undergo capping and polyadenylation.
How does a eukaryotic cell deal with the superhelical tension in its genomic DNA resulting from the activity of RNA polymerases? A. DNA gyrase introduces negative supercoils, keeping the DNA under constant tension. B. The RNA polymerases are allowed to rotate freely around their templates during transcription, leading to the relaxation of the tension. C. DNA topoisomerases rapidly remove the superhelical tension caused by transcription. D. The nucleosomes adjust the tension by binding to positively supercoiled regions behind a moving RNA polymerase. E. All of the above.
C. DNA topoisomerases rapidly remove the superhelical tension caused by transcription. Unlike bacteria, which maintain a fairly constant level of negative supercoiling in their circular DNA, eukaryotic topoisomerases rapidly remove the superhelical tension caused by DNA translocating machines such as the RNA polymerases.
Each aminoacyl-tRNA synthetase activates a certain amino acid by attaching it to its corresponding tRNA(s) through a high-energy linkage... A. between the amino group of the amino acid and the 3' hydroxyl group of the tRNA. B. between the amino group of the amino acid and the 5' hydroxyl group of the tRNA. C. between the carboxyl group of the amino acid and the 3' or 2' hydroxyl group of the tRNA. D. between the carboxyl group of the amino acid and the 5' hydroxyl group of the tRNA. E. between the amino group of the amino acid and the 3' or 2' hydroxyl group of the tRNA.
C. between the carboxyl group of the amino acid and the 3' or 2' hydroxyl group of the tRNA. The amino group of the amino acid is free, and will serve as the attacking group in the formation of the peptide bond on the ribosome. The carboxyl group, on the other hand, forms a high-energy and susceptible ester bond with the tRNA in preparation for that same reaction.
Polysomes... A. are large cytoplasmic assemblies made of several ribosomes each translating their exclusive mRNA. B. are only found in the eukaryotic cytoplasm. C. can take advantage of the circularization of eukaryotic mRNA (by interactions between the 5' and 3' ends of the mRNA) to further speed up the rate of protein synthesis. D. are mostly translationally inactive and are normally used by the cell to store the ribosomes and their associated mRNAs for future use. E. All of the above.
C. can take advantage of the circularization of eukaryotic mRNA (by interactions between the 5' and 3' ends of the mRNA) to further speed up the rate of protein synthesis. Polysomes form as a result of frequent initiation events leading to several ribosomes bound to and translating each mRNA simultaneously. If the mRNA is circularized, the ribosomes that have finished the synthesis of the protein can reinitiate immediately for another round.
Eukaryotic pre-mRNAs undergo a number of modifications such as capping at the 5' end. A 5' cap... A. consists of a modified terminal adenine nucleotide. B. has a 3'-to-5' linkage between the terminal nucleotide and the 5' end of the pre-mRNA. D. carries a negative charge in the terminal base due to methylation. E. is identical for all mRNAs that are transcribed by RNA polymerase II.
C. contains a triphosphate bridge between the terminal base and the 5' end of the pre-mRNA. The cap is a 7-methyl G residue linked to the 5' end of the transcript via a 5'-to-5' triphosphate bridge. Methyl transferase enzymes methylate the terminal guanine base (giving it a positive charge) and sometimes other nucleotides, creating different 5' caps.
Of the following proteins or protein complexes, which one does NOT typically interact with an elongating RNA polymerase II? A. Histone-modifying enzymes B. Capping enzymes C. Chromatin remodeling complexes D. Mediator complex E. Histone chaperones
D. Mediator complex In the initiation phase of transcription, Mediator mediates the interaction of RNA polymerase with the transcriptional activators and is required for the activation of transcription by enhancers.
The specific transfer RNA used for the incorporation of selenocysteine in proteins recognizes the UGA codon, which is normally a translation stop codon. What prevents this tRNA from reading through all the other "legitimate" UGA stop codons in the cells and causing a massive, disastrous recoding? A. The cells that have selenocysteine in their proteins do not normally use the UGA stop codon and always use the other two stop codons (UAA and UAG) instead. B. This tRNA is made in such a small amount that its side effects are negligible. C. This tRNA only interacts with the UGA codon in the P site of the ribosome, and therefore does not interfere with the normal function of the codon in translation termination, which takes place in the A site. D. The mRNAs encoding the selenocysteine-containing proteins also contain additional sequences required for the recoding event.
D. The mRNAs encoding the selenocysteine-containing proteins also contain additional sequences required for the recoding event. The presence of the UGA codon is only one of the signals required for the recoding event. Other sequences (that can form stem-loop structures) in the vicinity of the UGA codon on the mRNA collaborate with a specific protein factor to incorporate the amino acid into the growing protein chain only for particular mRNAs and not any mRNA with a UGA codon.
Which of the following features is common between the bacterial and eukaryotic ribosomes in translation initiation? A. They both use an initiator tRNA that carries formylmethionine. B. They both bind to the 5' end of the mRNA and move forward to find the start codon. C. They both recognize the start codon by interacting with the Shine-Dalgarno sequence. D. They both interact with various translation initiation factors. E. All of the above.
D. They both interact with various translation initiation factors.
This complex uses ATP and forms an "isolation chamber" for its misfolded protein substrates, allowing them to fold to their native conformation. It is called... A. a proteasome. B. a ribosome. C. Mediator. D. a chaperonin. E. an exosome.
D. a chaperonin The hsp60-like proteins form a large barrel-shaped complex, sometimes called a chaperonin, that forms an "isolation chamber" for the folding process of a protein after it has been fully synthesized.
DNA and RNA polymerase differ in ALL of the following EXCEPT... A. the nucleotide substrates they incorporate. B. their requirement for a primer. C. their error rate. D. the type of chemical reaction they catalyze. E. their processivity.
D. the type of chemical reaction they catalyze. Despite the lack of homology, the reactions catalyzed by these two polymerases are analogous, even though the enzymes work on different substrates. DNA polymerases require a primer and have a much lower error rate compared to RNA polymerases. They are also not as processive on their own.
In a gene that normally contains three exons, which of the following changes probably will NOT activate the nonsense-mediated mRNA decay pathway? The sizes of exons 1 to 3 are 100, 150, and 200 nucleotide pairs, respectively. A. A nonsense mutation in exon 1. B. A nonsense mutation in exon 2. C. A mutation in the first intron resulting in the inclusion of a large intronic fragment in the mature mRNA. D. A frameshift mutation in exon 1. E. A mutation in exon 2 leading to its loss through exon skipping.
E. A mutation in exon 2 leading to its loss through exon skipping. Feedback: Nonsense mutations in the upstream exons can directly trigger nonsense-mediated decay (NMD). Frameshift mutations in these exons are likely to introduce premature termination codons and activate NMD as well. Similarly, retention of intronic sequences carries a high risk of premature stop codons. In some cases, exon skipping can lead to NMD if the size of the skipped exon is not a multiple of 3 and therefore its absence changes the reading frame of the downstream exon.
How fast does a bacterial ribosome move on an mRNA? A. At about 2 nucleotides per second, significantly lower than the speed of the RNA polymerase. B. At about 5 nucleotides per second, comparable to the speed of the RNA polymerase. C. At about 10 nucleotides per second, significantly lower than the speed of the RNA polymerase. D. At about 20 nucleotides per second, comparable to the speed of the RNA polymerase. E. At about 60 nucleotides per second, comparable to the speed of the RNA polymerase.
E. At about 60 nucleotides per second, comparable to the speed of the RNA polymerase.
This large and complex general transcription factor has a DNA helicase activity that exposes the template for RNA polymerase II transcription. It also has a kinase activity that phosphorylates the C-terminal domain of the polymerase on Ser5 leading to promoter clearance. It is... A. TFIIB B. TFIID C. TFIIE D. TFIIF E. TFIIH
E. TFIIH Feedback: TFIIH is a large complex with several key functions in transcription initiation.