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Principles of Biochemistry

David L. Nelson, Michael M. Cox

Chapter 26

RNA Metabolism - all with Video Answers

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Chapter Questions

01:22

Problem 1

RNA Polymerase (a) How long would it take for the $E .$ coli RNA polymerase to synthesize the primary transcript for the $E$. coli genes encoding the enzymes for lactose metabolism, the 5,300 bp lac operon (considered in Chapter 28 )? (b) How far along the DNA would the transcription "bubble" formed by RNA polymerase move in 10 seconds?

Prashant Bana
Prashant Bana
Numerade Educator
06:18

Problem 2

Error Correction by RNA Polymerases DNA polymerases are capable of editing and crror correction, whereas the capacity for error correction in RNA polymerases seems to be limited. Given that a single base error in either replication or transcription can lead to an error in protein synthesis, suggest a possible biological explanation for this difference.

Rashmi Gondi
Rashmi Gondi
Numerade Educator
01:51

Problem 3

RNA Posttranscriptional Processing Predict the likely effects of a mutation in the sequence (5')AAUAAA in a cukaryotic mRNA transcript.

Rashmi Gondi
Rashmi Gondi
Numerade Educator
03:05

Problem 4

Coding versus Template Strands The RNA genome of phage $Q \beta$ is the nontemplate strand, or coding strand, and when introduced into the cell, it functions as an mRNA. Suppose the RNA replicase of phage $Q \beta$ synthesized primarily template-strand RNA and uniquely incorporated this, rather than nontemplate strands, into the viral particles. What would be the fate of the template strands when they entered a new cell? What enzyme would have to be included in the viral particles for successful invasion of a host cell?

Rashmi Gondi
Rashmi Gondi
Numerade Educator
04:15

Problem 5

Transcription The gene encoding the $E .$ coli enzyme $\beta$-galactosidase begins with the sequence ATGACCATGATTACG. What is the sequence of the RNA transcript specified by this part of the gene?

Jennifer Hudspeth
Jennifer Hudspeth
Numerade Educator
09:28

Problem 6

The Chemistry of Nucleic Acid Biosynthesis Describe three properties common to the reactions catalyzed by DNA polymerase, RNA polymerase, reverse transcriptase, and RNA replicase. How is the enzyme polynucleotide phosphorylase similar to and different from these four enzymes?

Rashmi Gondi
Rashmi Gondi
Numerade Educator
04:59

Problem 7

RNA Splicing What is the minimum number of transesterification reactions needed to splice an intron from an mRNA transcript? Explain.

Rashmi Gondi
Rashmi Gondi
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05:24

Problem 8

RNA Processing If the splicing of mRNA in a vertebrate cell is blocked, the rRNA modification reactions are also blocked. Suggest a reason for this.

Rashmi Gondi
Rashmi Gondi
Numerade Educator
02:29

Problem 9

RNA Genomes The RNA viruses have relatively small genomes. For example, the single-stranded RNAs of retroviruses have about 10,000 nucleotides, and the Q $\beta$ RNA is only 4,220 nucleotides long. Given the properties of reverse transcriptase and RNA replicase described in this chapter, can you suggest a reason for the small size of these viral genomes?

Rashmi Gondi
Rashmi Gondi
Numerade Educator
05:18

Problem 10

Screening of RNAs by SELEX The practical limit for the number of different RNA sequences that can be screened in a SELEX experiment is $10^{15}$. (a) Suppose you are working with oligonucleotides 32 nucleotides long. How many sequences exist in a randomized pool containing every sequence possible? (b) What percentage of these can be screened in a SELEX experiment? (c) Suppose you wish to select an RNA molecule that catalyzes the hydrolysis of a particular ester. From what you know about catalysis, propose a SELEX strategy that might allow you to select the appropriate catalyst.

Sana Riaz
Sana Riaz
Numerade Educator
02:58

Problem 11

Slow Death The death cap mushroom, Amanita phalloides, contains several dangerous substances, including the lethal $a$ -amanitin. This toxin blocks RNA elongation in consumers of the mushroom by binding to eukaryotic RNA polymerase II with very high affinity; it is deadly in concentrations as low as $10^{-8} \mathrm{M}$. The initial reaction to ingestion of the mushroom is gastrointestinal distress (caused by some of the other toxins). These symptoms disappear, but about 48 hours later, the mushroom-eater dies, usually from liver dysfunction. Speculate on why it takes this long for $\alpha$ -amanitin to kill.

Caroline Jones
Caroline Jones
Numerade Educator
04:47

Problem 12

Detection of Rifampicin-Resistant Strains of Tuberculosis Rifampicin is an important antibiotic used to treat tuberculosis and other mycobacterial diseases. Some strains of Mycobacterium tuberculosis, the causative agent of tuberculosis, are resistant to rifampicin. These strains become resistant through mutations that alter the $r p o B$ gene, which encodes the $\beta$ subunit of the RNA polymerase. Rifampicin cannot bind to the mutant RNA polymerase and so is unable to block the initiation of transcription. DNA sequences from a large number of rifampicin-resistant $M$. tuberculosis strains have been found to have mutations in a specific 69 bp region of rpoB. One well-characterized rifampicin-resistant strain has a single base pair alteration in $r p o B$ that results in a His residue being replaced by an Asp residue in the $\beta$ subunit.
(a) Based on your knowledge of protein chemistry, suggest a technique that would allow detection of the rifampicin-resistant strain containing this particular mutant protein.
(b) Based on your knowledge of nucleic acid chemistry, suggest a technique to identify the mutant form of rpoB.

Sana Riaz
Sana Riaz
Numerade Educator
05:47

Problem 13

The Ribonuclease Gene Human pancreatic ribonuclease has 128 amino acid residues.
(a) What is the minimum number of nucleotide pairs required to code for this protein?
(b) The mRNA expressed in human pancreatic cells was copied with reverse transcriptase to create a "library" of human DNA. The sequence of the mRNA coding for human pancreatic ribonuclease was determined by sequencing the complementary DNA (cDNA) from this library that included an open reading frame for the protein. Use the nucleotide database at NCBI (www.ncbi.nlm.nih.gov/nucleotide) to find the published sequence of this mRNA. (Search for accession number D26129.) What is the length of this mRNA?
(c) How can you account for the discrepancy between the size you calculated in (a) and the actual length of the mRNA?

Rashmi Sinha
Rashmi Sinha
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04:59

Problem 14

( $\alpha$ -amino-3-hydroxy-5-methyl-4- isoxazolepropionic acid) receptor is an important component of the human nervous system. It is present in several forms, in different neurons, and some of this variety results from posttranscriptional modification. This problem explores research on the mechanism of this RNA editing.
An initial report by Sommer and coauthors (1991) looked at the sequence encoding a key Arg residue in the AMPA receptor. The sequence of the cDNA (see Fig. 9-14) for the AMPA receptor showed a CGG (Arg; see Fig. $27-7$ ) codon for this amino acid. Surprisingly, the genomic DNA showed a CAG (Gln) codon at this position.
(a) Explain how this result is consistent with posttranscriptional modification of the AMPA receptor mRNA.
Rueter and colleagues (1995) explored this mechanism in detail. They first developed an assay to differentiate between edited and unedited transcripts, based on the Sanger method of DNA sequencing (see Fig. $8-34$ ). They modificd the technique to determine whether the base in question was an A (as in CAG) or not. They designed two DNA primers based on the genomic DNA sequence of this region of the AMPA gene. These primers, and the genomic DNA sequence of the nontemplate strand for the relevant region of the AMPA receptor gene, are shown at the bottom of the page; the A residue that is edited is in red.
To detect whether this A was present or had been edited to another base, Rueter and coworkers used the following procedure:
1. Prepared cDNA complementary to the mRNA, using primer 1, reverse transcriptase, dATP, dGTP, dCTP, and dTTP.
2. Removed the mRNA.
3. Annealed 32 P-labeled primer 2 to the cDNA and reacted this with DNA polymerase, dGTP, dCTP, dTTP, and ddATP (dideoxy ATP; see Fig. $8-34$ ).
4. Denatured the resulting duplexes and separated them with polyacrylamide gel electrophoresis (see Fig. $3-18$ ).
5. Detected the $^{32}$ P-labeled DNA species with autoradiography.
They found that edited mRNA produced a 22 nucleotide $\left[^{32} \mathrm{P}\right] \mathrm{DNA},$ whereas unedited mRNA produced a 19 nucleotide $\left[^{32} \mathrm{P}\right]$ DNA.
(b) Using the sequences below, explain how the edited and unedited mRNAs resulted in these different products.
Using the same procedure, this time to measure the fraction of transcripts edited under different conditions, the researchers found that extracts of cultured cpithelial cells (a common cell line called HeLa) could edit the mRNA at a high level. To determine the nature of the editing machinery, they pretreated an active HeLa cell extract as described in the following table and measured its ability to edit AMPA mRNA. Proteinase K degrades only proteins; micrococcal nuclease, only DNA.
(c) Use these data to argue that the editing machinery consists of protein. What is a key weakness in this argument?
To determine the exact nature of the edited base, Rueter and colleagues used the following procedure:
1. Produced mRNA, using $[a-32]$ ATP in the reaction mixture.
2. Edited the labeled mRNA by incubating with HeLa extract.
3. Hydrolyzed the edited mRNA with nuclease P1 to produce single nucleotide monophosphates.
4. Separated the nucleotide monophosphates with thin-layer chromatography (TLC; sce Fig. $10-25 \mathrm{b}$ ).
5. Identified the resulting 32 p-labeled nucleotide monophosphates with autoradiography.
In unedited mRNA, they found only $\left[^{32} \mathrm{P}\right] \mathrm{AMP} ;$ in edited mRNA, they found mostly $\left[^{32} \mathrm{P}\right] \mathrm{AMP}$ with some $\left[^{32} \mathrm{P}\right] \mathrm{IMP}$ (inosine monophosphate; see Fig. $22-36$ ).
(d) Why was it necessary to use $\left[\alpha-^{32} \mathrm{P}\right] \mathrm{ATP}$ rather than $\left[\beta^{32} \mathrm{P}\right] \mathrm{ATP}$ or $\left[\gamma^{32} \mathrm{P}\right] \mathrm{ATP}$ inthis experiment? (e) Why was it necessary to use $\left[a^{32} \mathrm{P}\right] \mathrm{ATP}$ rather than $\left[a-^{32} \mathrm{P}\right] \mathrm{GTP},\left[\alpha-^{32} \mathrm{P}\right] \mathrm{CTP},$ or
$\left[a-^{32} \mathrm{P}\right] \mathrm{UTP} ?$
(f) How does the result exclude the possibility that the entire A nucleotide (sugar, base, and phosphate) was removed and replaced by an I nucleotide during the editing process? The researchers next edited mRNA that was labeled with $\left[2,8-^{3} \mathrm{H}\right] \mathrm{ATP}$ and repeated the above procedure. The only $^{3}$ H-labeled mononucleotides produced were AMP and IMP.
(g) How does this result exclude removal of the A base (leaving the sugar-phosphate backbone intact) followed by replacement with an I base as a mechanism of editing? What, then, is the most likely mechanism of editing in this case?
(h) How does changing an A to an I residue in the mRNA explain the Gln to Arg change in protein sequence in the two forms of AMPA receptor protein? (Hint: See Fig. 27-8.)
TABLE CANT COPY
FIGURE CANT COPY

Sana Riaz
Sana Riaz
Numerade Educator