Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 13, Problem 22PDQ
Present an overview of various forms of posttranscriptional RNA processing in eukaryotes. For each, provide an example.
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Chapter 13 Solutions
Concepts of Genetics (12th Edition)
Ch. 13 - In a mixed heteropolymer experiment using...Ch. 13 - When repeating copolymers are used to form...Ch. 13 - The following represent deoxyribonucleotide...Ch. 13 - Prob. 1CSCh. 13 - A 30-year-old woman was undergoing therapy for...Ch. 13 - A 30-year-old woman was undergoing therapy for...Ch. 13 - HOW DO WE KNOW? In this chapter, we focused on the...Ch. 13 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 13 - Assuming the genetic code is a triplet, what...Ch. 13 - The mRNA formed from the repeating tetranucleotide...
Ch. 13 - In studies using repeating copolymers, AC ......Ch. 13 - In a coding experiment using repeating copolymers...Ch. 13 - Prob. 7PDQCh. 13 - When the amino acid sequences of insulin isolated...Ch. 13 - Prob. 9PDQCh. 13 - Why doesnt polynucleotide phosphorylase (Ochoas...Ch. 13 - Refer to Table 13.1. Can you hypothesize why a...Ch. 13 - Predict the amino acid sequence produced during...Ch. 13 - A short RNA molecule was isolated that...Ch. 13 - A glycine residue is in position 210 of the...Ch. 13 - Refer to Figure 13.7 to respond to the following:...Ch. 13 - Most proteins have more leucine than histidine...Ch. 13 - Define the process of transcription. Where does...Ch. 13 - Prob. 18PDQCh. 13 - Describe the structure of RNA polymerase in...Ch. 13 - Prob. 20PDQCh. 13 - Messenger RNA molecules are very difficult to...Ch. 13 - Present an overview of various forms of...Ch. 13 - One form of posttranscriptional modification of...Ch. 13 - Describe the role of two forms of RNA editing that...Ch. 13 - Substitution RNA editing is known to involve...Ch. 13 - Prob. 26ESPCh. 13 - Prob. 27ESPCh. 13 - Prob. 28ESPCh. 13 - Shown here are the amino acid sequences of the...Ch. 13 - The genetic code is degenerate. Amino acids are...Ch. 13 - M. Klemke et al. (2001) discovered an interesting...Ch. 13 - Recent observations indicate that alternative...Ch. 13 - Isoginkgetin is a cell-permeable chemical isolated...
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- For each of the five short mRNA nucleotide sequences given in the table below: 3. Translate the original sequence (for these short sequences start translation at the first nucleotide) 4. Identify (and highlight or underline) the one nucleotide difference between the original (left) and altered (right) sequences 5. For each altered nucleotide sequence give the type of mutation (effect at the DNA/nucleotide level; see #1 above) 6. Translate each changed sequence. Does the mutation result in a change in the amino acid sequence? If so, what is the effect of the mutation on protein structure (amino acid sequence; see #2 above)arrow_forwardThe entire genome of the yeast Saccharomyces cerevisiae has been sequenced. This sequencing has led to the identification of all the open reading frames (ORFs, gene-size sequences with appropriate translational initiation and termination signals) in the genome. Some of these ORFs are previously known genes with established functions; however, the remainder are unassigned reading frames (URFs). To deduce the possible functions of the URFs, they are being systematically, one at a time, converted into null alleles by in vitro knockout techniques. The results are as follows:15 percent are lethal when knocked out.25 percent show some mutant phenotype (altered morphology, altered nutrition, and so forth).60 percent show no detectable mutant phenotype at all and resemble wild type.Explain the possible molecular-genetic basis of these three mutant categories, inventing examples where possible.arrow_forwardWhy Noncoding RNAs is important for Posttranscriptional Regulation ?arrow_forward
- The following RNA sequence represents a small messenger which can be translated in a prokaryotic cell: 5'-ACGAAUGCACAGUAAAACUGGCUAGCGUAGGCUGA-3 Assume that the messenger RNA is translated in the cell, using the correct machinery and signals required for accurate protein synthesis. Using this RNA sequence and the Genetic Code Dictionary (see your textbook for the dictionary), solve the following problems A. Write the sequence of a protein that would be translated from this mRNA, using the appropriate stop and start signals, and indicating the correct termini of the protein product. B. Suppose that the underlined A in the sequence is changed to a U. Write the expected protein product of this mRNA.arrow_forwardMany uridine molecules are inserted into some mitochondrial mRNAs in trypanosomes. The uridine residues come from the poly(U) tail of a donor strand. Nucleoside triphosphates do not participate in this reaction. Propose a reaction mechanism that accounts for these findings. (Hint: Relate RNA editing to RNA splicing.)arrow_forwardWhich of the following statements regarding splicing in eukaryotes is correct?a) Several reactions in the splicing process involve hydrolysis of ATPb) Exons are spliced out and introns are retained in the mature mRNA transcriptc) Splicing takes place in the cytosold) Small nuclear RNAs are retained in the mature mRNA transcriptarrow_forward
- Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?arrow_forwardThere are several instances which challenge the “one gene, one polypeptide” hypothesis. Describe TWO alternative processing pathways which lead to the production of multiple polypeptides from a single gene.arrow_forwardBy comparing Figures 8-16 and 8-17 with Figure 8-18,speculate what features of RNA permit self-splicing (thatis, in the absence of proteins)arrow_forward
- The entire genome of the yeast Saccharomyces cerevisiaehas been sequenced. This sequencing has led to the identification of all the open reading frames (ORFs, gene-sizesequences with appropriate translational initiation andtermination signals) in the genome. Some of these ORFsare previously known genes with established functions;however, the remainder are unassigned reading frames(URFs). To deduce the possible functions of the URFs,they are being systematically, one at a time, convertedinto null alleles by in vitro knockout techniques. The results are as follows:15 percent are lethal when knocked out.25 percent show some mutant phenotype (alteredmorphology, altered nutrition, and so forth).60 percent show no detectable mutant phenotype at alland resemble wild type.Explain the possible molecular-genetic basis of thesethree mutant categories, inventing examples wherepossible.arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forward
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