LEHNINGER PRINCIPLES OF BIOCHEMISTRY - VOL 3. INFORMATION PATHWAYS - 2017

PART III. INFORMATION PATHWAYS

26. RNA METABOLISM

Questions and Problems

1. RNA polymerase.

(a) How long will it take for E. coli RNA polymerase to synthesize the primary transcript of the E. coli genes encoding the lactose METABOLISM Enzymes (the lac Operon, which is 5,300 bp long, as discussed in Chapter 28)? (b) How far along the DNA chain can the Transcription "bubble" generated by RNA polymerase advance in 10 seconds?

2. Proofreading by RNA polymerase.

DNA polymerases can proofread, whereas the proofreading ability of RNA polymerase is quite limited. Given that the substitution of a single base during Replication or transcription can lead to an error in Protein Synthesis, provide a plausible biological explanation for this striking difference.

3. Posttranscriptional RNA Processing.

Predict the likely consequences of a mutation in the (5')-AAUAAA sequence of a eukaryotic mRNA transcript.

4. Coding and template sequences.

The RNA genome of phage Qβ is a nontemplate, or coding, strand, and upon entering a Cell, it Functions as an mRNA. Suppose that phage Qβ RNA replicase synthesizes primarily the template strand of RNA and packages this strand, rather than the coding strand, into Viral Particles. What will happen to the template strands when they enter a new cell? What enzyme must be included in the viral particles for successful infection of the host cell?

5. Chemistry of Nucleic acid Biosynthesis.

Describe three General Properties of Reactions Catalyzed by DNA polymerase, RNA polymerase, Reverse Transcriptase, and RNA replicase. In what ways is the enzyme polynucleotide phosphorylase similar to these three enzymes, and in what ways does it differ from them?

6. RNA splicing.

What is the minimum number of transesterification reactions required to splice an intron from an mRNA transcript? Explain your answer.

7. RNA processing.

If mRNA splicing is blocked in vertebrate Cells, rRNA modification reactions are also blocked. Explain this observation.

8. RNA genomes.

Introduction/7.html">RNA-containing Viruses have relatively small genomes. For example, single-stranded RNA molecules of Retroviruses consist of approximately 10,000 NUCLEOTIDES, and phage Qβ RNA contains only 4,220 nucleotides. Knowing The properties of reverse transcriptase and RNA replicase described in this chapter, explain why this virus has such a small genome.

9. Screening RNA molecules by SELEX.

The maximum number of different RNA sequences that can be screened using SELEX is 10¹⁵. (a) Suppose you are working with oligonucleotides 32 nucleotides in length. How many variants of these molecules can be present in a random pool containing all possible sequences? (b) What fraction of them (in %) can be analyzed by SELEX? (c) Suppose you want to select an RNA molecule that catalyzes the Hydrolysis of a specific ester. Based on what you know about catalysis (Chapter 6 in Vol. 1), propose a strategy that will allow you to select a suitable catalyst.

10. Death cap Mushroom poisoning.

The death cap mushroom (Amanita phalloides) contains several dangerous compounds, including the lethal poison α-amanitin. This toxin blocks RNA elongation by binding to eukaryotic RNA polymerase II with very high affinity; it is fatal even at concentrations as low as 10^-8 M. Initially, a person who has eaten this mushroom experiences gastrointestinal distress (caused by Other toxins). These symptoms then subside, but after 48 hours the person dies, typically from Liver failure. Explain why α-amanitin takes so long to prove fatal to a human.

11. Detection of rifampicin-resistant strains of the tuberculosis pathogen.

Rifampicin is an essential antibiotic used to combat tuberculosis and other mycobacterial infections. Certain strains of Mycobacterium tuberculosis (the CAUSATIVE AGENT OF tuberculosis) are resistant to rifampicin. These strains have acquired resistance As a result of a mutation in the rpoB Gene, which encodes the β subunit of RNA polymerase. Rifampicin is unable to bind to the mutant RNA polymerase and therefore fails to block Transcription initiation. It has been found that DNA sequences from A large number of rifampicin-resistant M. tuberculosis strains carry Mutations within a specific 69-base-pair region of the rpoB gene. In one well-characterized rifampicin-resistant strain, a single base pair in the rpoB gene is substituted, leading to a single amino acid substitution in the encoded β subunit: a His residue is replaced by an Asp residue.

a) Based on your knowledge of Protein Chemistry (Chapters 3 and 4 in Vol. 1), propose a method to detect the rifampicin-resistant strain containing this specific mutant protein.

b) Based on your knowledge of nucleic acid chemistry (Chapter 8 in Vol. 1), propose a method to identify the mutant rpoB gene.

Biochemistry on the Internet

12. The Ribonuclease Gene.

Human pancreatic ribonuclease consists of 128 amino acid residues.

a) What is the minimum number of nucleotide pairs required to encode this protein?

b) mRNA from human pancreatic cells was reverse-transcribed to generate a human DNA library. The mRNA sequence encoding human pancreatic ribonuclease was determined by sequencing the complementary DNA (cDNA) from this library, which includes the Open Reading Frame for this protein. Use the Entrez database system (www.ncbi.nlm.nih.gov/Entrez) to find the published sequence of this mRNA (search the CoreNucleotide database for accession number D26129). What is the length of this mRNA?

c) How can you account for the discrepancy between the size you calculated (part a) and the actual size of the mRNA?

Analysis of Experimental Data

13. An Example of RNA Editing.

The AMPA receptor (α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid receptor) is a vital component of the human Nervous system. It exists in multiple forms across different Neurons, with this diversity partly arising from post-transcriptional modifications. This problem explores the mechanisms underlying such RNA editing.

In 1991, Sommer et al. analyzed the sequence encoding a crucial Arg residue in the AMPA receptor molecule. In the cDNA sequence (see Fig. 9-14, Vol. 1) of the AMPA receptor, the Arginine residue corresponded to the CGG codon (see Fig. 27-7). However, in the genomic DNA, this position harbored a CAG (Gln) codon.

a) Explain how this result is consistent with the occurrence of post-transcriptional modification of AMPA receptor mRNA. Rueter and coworkers (1995) investigated this question in detail. First, they developed a method to distinguish between modified and unmodified transcripts using Sanger DNA Sequencing (see Fig. 8-33, Vol. 1). They modified the Procedure to determine whether this base was A (as in CAG) or something else. They synthesized two DNA primers based on the genomic DNA sequence in this region of the AMPA receptor gene. The primers, along with the noncoding strand genomic DNA sequence for the corresponding AMPA receptor gene region, are shown below, with the modified base A highlighted in red.

To determine whether the A base was retained or replaced by another base, Rueter et al. used the procedure described below.

1. Prepare cDNA complementary to the mRNA using primer 1, reverse transcriptase, dATP, dGTP, dCTP, and dTTP.

2. Remove the mRNA.

3. Anneal ³²P-labeled primer 2 to the cDNA and perform a reaction with DNA polymerase, dGTP, dCTP, dTTP, and ddATP (dideoxy-ATP; see Fig. 8-33).

4. Denature the resulting duplexes and separate them by Polyacrylamide gel Electrophoresis (p. 136, Vol. 1).

5. Detect the ³²P-labeled DNA by autoradiography.

The researchers found that the modified mRNA yielded a 22-nucleotide-long [³²P]DNA, whereas the unmodified mRNA yielded a 19-nucleotide-long [³²P]DNA.

b) Using the sequences presented below, explain why the modified and unmodified mRNAs produced different products.

Using the same approach to measure the fraction of transcripts undergoing editing under various conditions, the researchers found that extracts from cultured epithelial cells (so-called HeLa cells) actively mediate mRNA editing. To investigate the mechanisms of this process, the scientists prepared an active HeLa cell extract as described in the table and tested its ability to modify AMPA receptor mRNA. Proteinase K digests only Proteins, whereas micrococcal nuclease digests only DNA.

c) Use these data to demonstrate that proteins are involved in mRNA editing processes. What is the main weakness of this evidence?

To precisely identify the edited base, Rüter and colleagues used the procedure described below.

1. Synthesized mRNA by adding [α-³²P] ATP to the reaction mixture.

2. Modified the labeled mRNA by incubation with HeLa cell extract.

3. Hydrolyzed the modified mRNA down to individual nucleotide monophosphates using nuclease P1.

4. Separated the nucleotide monophosphates by Thin-Layer Chromatography (see Fig. 10-24, Vol. 1).

5. Identified the resulting ³²P-labeled nucleotide monophosphates by autoradiography.

In the unmodified mRNA, the researchers detected only [³²P] AMP, whereas the modified mRNA contained predominantly [³²P] AMP along with a certain amount of [³²P] IMP (inosine monophosphate, see Fig. 22-34, Vol. 2).

c) Why did this experiment use [α-³²P] ATP rather than [β-³²P] ATP or [γ-³²P] ATP?

d) Why was [α-³²P] ATP used instead of [α-³²P] GTP, [α-³²P] CTP, or [α-³²P] UTP?

e) How do these results rule out the possibility that the entire nucleotide A (sugar, base, and phosphate) was removed and replaced by nucleotide I?

Next, the researchers modified mRNA labeled with [2,8-³H] ATP and repeated the procedure described above. The ³H label was found exclusively in the AMP and IMP mononucleotides.

f) How does this result rule out the possibility that the base A was removed (while preserving the sugar-phosphate backbone) during replacement with base I? What is the most likely mechanism of editing in this case?

g) How does the replacement of an A residue with I in the mRNA account for the Gln-to-Arg substitution in the protein sequence of the two AMPA receptor isoforms? (Hint: See Fig. 27-8.)