C&M Quiz 2
What does "export ready" mRNA mean, and what distinguishes an "export ready" mRNA from a bit of excised intron that needs to be degraded?
"Export ready" means that an mRNA is bound by the appropriate set of proteins. Proteins such as the cap-binding complex, the exon junction complex, and the poly-A-binding protein must be present, while proteins such as spliceosome components must be absent. RNA fragments from excised introns do not acquire the necessary set of proteins and are thus doomed to degradation.
The intron-exon structure of eukaryotic genes came as a shock. In the early, skeptical days, the most convincing demostration was visual, as shown, for example, by the electron micrograph in FIg. 6-15A. This image was obtained by hybridizing ovalbumin mRNA to a long segment of DNA that contained the gene. To those used to looking at single- and double-stranded nucleic acids in the electron microscope, the structure was clear: a set of single-stranded tails and loops emanating from a central duplex segment whose ends corresponded to the ends of the mRNA. To the extent possible, describe the intron-exon structure of this gene.
A schematic diagram of the structure of the mRNA-DNA hybrid and the intron-exon structure of the gene
The antibiotic edeine inhibits protein synthesis but has no effect on either DNA synthesis or RNA synthesis. When added to a reticulocyte lysate, edeine stops protein synthesis after a short lag, as shown in FIg. 6-23. By contrast, cycloheximide stops protein synthesis immediately. Analysis of the edeine-inhibited lysate by density gradient centrifugation showed that no polyribosomes remained at the time protein synthesis had stopped. Instead, all the globin mRNA accumulated in an abnormal 40s peak, which contained equimolar amounts of the small ribosomal subunit and initiator tRNA. A. What step in protein synthesis does edeine inhibit? B. Why is there a lag between addition of edeine and cessation of protein synthesis? What determines the length of lag? C. Would you expect the polyribosomes to disappear if you added cycloheximide at the same time as ediene?
A. Edeine specifically inhibits initiation of protein synthesis by preventing the joining of the 60s ribosomal subunit to the 40s subunit/mRNA/initiator tRNA complex. Since elongation is not blocked, ribosomes that have already begun synthesis complete their individual chains and fall off the mRNA, leaving attached only the small subunit and the initiator tRNA. Edeine is an antibiotic produced by certain strains of Bacillus brevis. B. A lag occurs before protein synthesis shuts off because edeine inhibits initiation but has no effect on elongation. Thus, a ribosome that has just started making a new polypeptide is free to complete it. Incorporation of label continues for just the length of time it takes to complete the protein which takes about a minute. C. If cycloheximide (or any other elongation inhibitor) is added at the same time as an initiation inhibitor, the polyribosomes are frozen. Polyribosome breakdown by initiation inhibitors requires ribosome movement, which is blocked by elongation inhibitors.
In the electron micrograph in Fig. 6-1, are the RNA polymerase molecules moving from right to left or from left to right? How can you tell? Why do the RNA transcripts appear so much shorter than the length of DNA that encodes them?
Actually, the RNA polymerases are not moving at all because they have been fixed and coated with metal to prepare the sample for viewing in the electron microscope. Before they were fixed, they were moving from left to right, as indicated by the gradual lengthening of the RNA transcripts. The RNA transcripts are shorter than the DNA that encodes them because they begin to fold up as soon as they are synthesized, whereas the DNA is an extended double helix.
Enzyme that attaches the correct amino acid to a tRNA molecule to form the activated intermediate used in protein synthesis.
Aminoacyl-tRNA synthetase
Sequence of three nucleotides in a tRNA that is complementary to a three-nucleotide sequence in an mRNA molecule.
Anticodon
In a reticulocyte lysate, the polynucleotide 5'-AUGUUUUUUUUU directs the synthesis of Met-Phe-Phe-Phe. In the presence of farsomycin, a new antibiotic perfected by Fluhardy Pharmaceuticals, this polymer directs synthesis of Met-Phe only. From this information, which of the follwoing deductions could you make about farsomycin?
D. The formation of one peptide bond, but no more, eliminates all choices except D. If farsomycin inhibited formation of the 80s initiation complex (A), inhibited binding of aminoacyl-tRNAs to the A site (B), or inactivated peptidyl transferase (C), no peptide would have been formed. If it interfered with chain termination and release (E), the entire peptide would have been made
Segment of a eukaryotic gene consisting of a sequence of nucleotides that will be represented in mRNA or other functional RNAs.
Exon
T or F Since introns are largely genetic "junk", they do not have to be removed precisely from the primary transcript during RNA splicing.
False Although intron sequences are mostly dispensable, they must be removed precisely. An error of even one nucleotide during removal would shift the reading frame in the spliced mRNA molecule and produce an aberrant protein
T or F During protein synthesis, the thermodynamics of base-pairing between tRNAs and mRNAs sets the upper limit for the accuracy with which protein molecules are made.
False Because correct base-pairing is only about 10-100 fold more stable than incorrect matches, additional mechanisms, beyond the simply thermodynamics of base-pairing, must be used to reach the accuracy of protein synthesis routinely achieved in the cell.
T or F The sigma subunit is a permanent component of the RNA polymerase holoenzyme from E. coli, allowing it to initiate at appropriate promoters in the bacterial genome.
False The sigma subunit associates with the bacterial RNA polymerase core enzyme to form the RNA polymerase holoenzyme only during the initiation phase of RNA synthesis.
T or F Wobble pairing occurs between the first position in the codon and the third position in the anticodon.
False Wobble pairing occurs between the third position in the codon and the first position in the anticodon
T or F RNA polymerase II generates the end of a pre-mRNA transcript when it ceases transcription and releases the transcript; a poly-A tail is then quickly added to the free 3' end.
Flase The 3' ends of most pre-mRNA transcripts produced by RNA polymerase II are defined not by the termination point of transcription, but by cleavage of the RNA chain 10-30 nucleotides downstream of the sequence AAUAAA
Helps to position the RNA polymerase correctly at the promoter, to aid in pulling apart the two strands of DNA to allow transcription to begin, and to release RNA polymerase from the promoter into the elongation mode once transcription has begun.
General transcription factor
Whaty are the roles of general transcription factors in RNA polymerase II mediated transcription, and why are they referred to as "general"?
General transcription factors play several roles in promoting transcription by RNA polymerase II. They help position the RNA polymerase correctly at the promoter, they aid in pulling apart the two strands of DNA to allow transcription to begin, and they release RNA polymerase from the promoter once transcription has begun. THey are called "general" becasue they assemble on all promoters used by RNA polymerase II; they are identified by names beginning with TFII. Labeling them "general" transcription factors also serves to distinguish them from more specialized gene regulatory proteins that enhance transcription at selected promoters in certain cell types.
Set of rules specifying the correspondence between the nucleotide triplets in DNA or RNA and amino acids in proteins.
Genetic code
An RNA polymerase is transcribing a segment of DNA that contains the sequence 5'-GTAACGGATG-3' 3'-CATTGCCTAC-5' If the polymerase transcribes this sequence from left to right, what will the sequence of the RNA be? What will the RNA sequence be if the polymerase moves right to left?
If the polymerase transcribes the the sequence from left to right, it will use the bottom strand as a template to make the sequence 5'-GUAACGGAUG (the RNA sequence corresponding to the top strand of the DNA). If the polymerase moves from right to left, it will use the top strand as a template to make the sequence 5'-CAUCCGUUAC (the RNA sequence corresponding to the bottom strand of the DNA written 5'-->3').
Both hsp60-like and hsp70 molecular chaperones share an affinity for exposed hydrophobic patches on proteins, using them as indicators of incomplete folding. Why do you suppose hydrophobic patches serve as critical signals for the folding status of a protein?
In a well folded protein, the majority of hydrophobic amino acids will be sequestered in the interior, away from water. Exposed hydrophobic patches thus indicate that a protein is abnormal in some way. Some proteins initially fold with exposed hydrophobic patches that are used in binding to other proteins, ultimately burying those hydrophobic amino acids as well. As a result, hydrophobic amino acids are usually not exposed on the surface of a protein, and any significant patch is a good indicator that something has gone awry. The protein may have failed to fold properly after leaving the ribosome, it may have suffered an accident that partly unfolded it at a later time, or it may have failed to find its normal partner subunit in a larger protein complex.
Given the wobble rules for codon-anticodon pairing in bacteria, the minimum number of different tRNAs that would be required to recognize all 61 codons is 31. What is the minimum number of different tRNAs that is consistent with the wobble rules used in eukaryotes?
In eukaryotes, a minimum of 45 tRNAs would be required to recognize all 61 codons, given the rules for wobble bsae-pairing. Pairs of codons that end in U or C always encode the same amino acid. Thus a single tRNA with an I in the wobble position of the anticodon would be required for each such pair of codons. The 16 pairs of such codons would require 16 tRNAs. Each of the remaining 29 codons, which end in either an A or a G, would require a a specific tRNA with a corresponding U or C in the wobble position. Thus, the minimum number of tRNAs is 16 plus 29 or 45.
Polycistronic mRNAs are common in prokaryotes but extremely rare in eukaryotes. Describe the key differences in protein synthesis that underlie this observation.
In eukaryotic cells, protein synthesis is normally initiated by scanning from the 5' end of the mRNA until the first AUG codon is found. This mechanism of initiation ensures that ribosomes will all start translating near the 5' end. By contrast, in prokaryotic cells, protein synthesis is initiated by base-pairing between mRNA sequences adjacent to an initiation AUG codon and sequences in the 16s rRNA of the small ribosomal subunit. The prokarytoic inittiation strategy allows ribosomes to recognize several start sites in the same mRNA. This key difference in mechanism underlies their ability to make several proteins from a single polycistronic mRNA.
Special tRNA that carries methionine and is used to begin translation.
Initiator tRNA
Protein that helps other proteins fold correctly.
Molecular chaperone
Most proteins require molecular chaperones to assist in their correct folding. Hows do you supposed the chaperones themselves manage to fold correctly?
Molecular chaperones fold like any other protein. Molecules in the act of synthesis on ribosomes are bound by hsp70 chaperones. And incorrectly folded molecules are helped by hsp60-like chaperones. That they function as chaperones when they have folded correctly makes no difference to the way they are treated before they reach their final, functional conformation. Of course, properly folded hsp60-like and hsp70 chaperones must already be present to help fold the newly made chaperones. At cell division, each daughter cell inherits a starter set of such chaperones form the parental cell.
Which of the following mutational changes would you predict to be the most deleterious to gene function? Explain. 1. Insertion of a single nucleotide near the end of the coding sequence. 2. Removal of a single nucleotide near the beginning of the coding sequence. 3. Deletion of three consecutive nucleotides in the middle of the coding sequence. 4. Deletion of four consecutive nucleotides in the middle of the coding sequence. 5. Substitution of one nucleotide for another in the middle of the coding sequence.
Mutations of the type described in 2 and 4 are often the most harmful. In both cases, the reading frame would be changed. Because these framseshifts occur early in the coding sequence or in the middle of it, the encoded protein will contain a nonsenical and usually truncated sequence of amino acids. In contrast, a reading frameshift that occurs toward the end of the coding sequence, as described in 1, will result in a largely correct protein that may be functional. 3-slient mutation 5-harmless because it does not change the amino acid
Surveillance system in eukaryotes that eliminates defective mRNAs before they can be translated into protein.
Nonsense-mediated mRNA decay
Large multiprotein structure forming a channel through the nuclear envelope that allows selected molecules to move between the nucleus and cytoplasm.
Nuclear pore complex
Nucleotide sequence in DNA to which RNA polymerase binds to begin transcription.
Promoter
Large protein complex in the cytosol and nucleus with proteolytic activity that is responsible for degrading the proteins marked for destruction.
Proteasome
The enzyme that carries out transcription.
RNA polymerase
Process in which intron sequences are excised from RNA transcripts in the nucleus during the formation of messenger and other RNAs.
RNA splicing
The three-nucleotide phase in which nucleotides in an mRNA are translated into amino acids in a protein.
Reading frame
RNA molecule with catalytic activity.
Ribozyme
Signal in bacterial DNA that halts transcription.
Terminator
For the RNA sequence below, indicate the amino acids that are encoded in the three reading frames. If you were told that this segment of RNA was in the middle of an mRNA that encoded a large protein, would you know which reading frame was used? How so? AGUCUAGGCACUGA
The amino acids encoded in each of the three reading frames are shown in FIg. 6-47. If this segment of RNA encoded part of a larger protein, it would have to be translated in reading frame 1, which is the only one that does not contain a stop codon.
Consider the expression "central dogma", which refers to the proposition that genetic information flows from DNA to RNA to protein. Is the word "dogma" appropriate in this scientific context?
The answer is best given by Francis Crick himself, who in 1957 coined the terms "the sequence hypothesis," which proposed that genetic information is encoded in the sequence of the DNA bases, and "the central dogma," which stated that DNA makes RNA makes protein.
The interaction of U1 snRNP with the sequences at the 5' ends of introns is usually shown to involve pairing between the nucleotides in the pre-mRNA and those in the RNA component of the U1 snRNP, as illustrated in Fig. 6-16A. But given the relatively small number of bases involved, that arrangement is hardly convincing. A series of experiments tested the hypothesis that base-pairing was critical to the function of U1 snRNP in splicing. As shown in FIg. 6-16B, a mutant pre-mRNA was generated that could not be spliced. Several mutant U1 snRNAs were then tested for their ability to promote splicing of the mutant pre-mRNA, with the results shown in FIg. 6-16C. Do these experiments argue that base-pairing is critical to the role of U1 snRNP in splicing? If so, how?
The results of these experiments argue convincingly that base-pairing occurs between the pre-mRNA and the U1 RNA. The base-pairing between the U1 snRNA and the pre-mRNA is not extensive, but it is better when splicing occurs successfully than when it does not. These experiments illustrate a classic approach for testing the reality of proposed base-pairing schemes.
The rules for wobble base-pairing in bacteria and eukaryotes are shown in Table 6-3. On the left side of the table, the rules are expressed as a wobble codon base and its recognition by possible anticodon bases. Reformulate these rules as particular anticodon bases and their recognition by possible codon bases, as suggested by the partial information on the right side of the table.
The rules for wobble base-pairing between the anticodon and the codon, expressed both ways, are shown in Table 6-5. It is striking that A in the wobble position of the anticodon in eukaryotes does not have a pairing partner in codons. It turns out that A is not used in the wobble position in eukaryotic tRNAs. In many cases an A is encoded in the wobble position, but it is changed to inosine after transcription, giving rise to a mature tRNA that recognizes U or C in the wobble position of the codon
You are studying a DNA virus that makes a set of abundant proteins late in its infectious cycle. AN mRNA for one of these proteins maps to a DNA restriction fragment from the middle oft he linear genome. To determine the precise location of this mRNA you anneal it with the purified restriction fragment under conditions where only DNA-RNA hybrid duplexes are stable and DNA strands do not reanneal. When you examine the reannealed DNA-RNA duplexes by electron microsocopy, you see structures such as that in Fig. 6-4. Why are there single-stranded tails at the ends of the DNA-RNA duplex region?
The tails arise because the ends of the mRNA are not complementary to the ends of the DNA restriction fragment. Thus, one of the single-stranded tails at each end is DNA from the restriction fragment. A single-strand tail at one end corresponds to the 5' end of the mRNA, which must come from an upstream exon that is not present in the restriction fragment. A single-strand tail at the other end corresponds to the 3' end of the mRNA, which may come from a downstream exon not present in the restriction fragment or may simply be the poly-A tail itself. Without additional information you cannot identify which single strand comes from which source.
T or F Eukaryotic mRNA molecules carry 3' ribosyl OH groups at both their 3' and 5' ends.
True At its 3' end, each eukaryotic mRNA has a string of adenine nucleotides, the last of which has a terminal ribose with a free 3'-OH group
T or F The consequences of errors in transcription are less severe than those of errors in DNA replication.
True Errors in DNA replication have the potential to affect future generation of cells, while errors in transcription have no genetic consequence.
RNA molecule that specifies the amino acid sequence of a protein.
mRNA (messenger RNA)
Match the following list of RNAs with their functions mRNA rRNA snoRNA snRNA tRNA piRNA miRNA siRNA
mRNA-codes for proteins rRNA-components of ribosome snoRNA-modification and processing of rRNA snRNA-splicing of RNA transcripts tRNA-adaptor for protein synthesis piRNA-protects germ line from transposable elements miRNA-blocks translation of selected mRNAs siRNA-directs degradation of selected mRNAs
Small RNA molecules that are complexed with proteins to form the ribonucleoprotein particles involved in RNA splicing.
snRNA (small nuclear RNA)
