Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 12, Problem 21PDQ
One form of posttranscriptional modification of most eukaryotic RNA transcripts is the addition of a poly-A sequence at the 3'-end. The absence of a poly-A sequence leads to rapid degradation of the transcript Poly-A sequences of various lengths are also added to many prokaryotic RNA transcripts where, instead of promoting stability, they enhance degradation. In both cases, RNA secondary structures, stabilizing proteins, or degrading enzymes interact with poly-A sequences. Considering the activities of RNAs, what might be the general functions of 3'-polyadenylation?
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As described earlier, DNA damage can cause deletion or insertion of base pairs. If a nucleotide base sequence of a coding region changes by any number of bases other than three base pairs, or multiples of 3, a frameshift mutation occurs. Depending on the location of the sequence change, such mutations can have serious effects. The following synthetic mRNA sequence codes for the beginning of a polypeptide: 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCUUAUC-AUGUUU-3′ First, determine the amino acid sequence of the polypeptide. Then determine the types of mutation that have occurred in the following altered mRNA segments. What effect do these mutations have on the polypeptide products? a. 5′-AUGUCUCCUACUUGCUGACGAGGGAAGGAGGUGGCUUAUCA-UGUUU-3′ b. 5′-AUGUCUCCUACUGCUGACGAGGGAGGAGGUGGCUUAUCAU-GUUU-3′ c. 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCCCUUAUC-AUGUUU-3′ d. 5′-AUGUCUCCUACUGCUGACGGAAGGAGGUGGCUUAUCAU-GUUU-3′
Consider the Rho-dependent terminator sequence
5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription?
Group of answer choices
Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination.
Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds.
Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination.
Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.
The deadenylation pathway is a critical means of maintaining a steady state level of mRNAs and proteins in the cells. How does deadenylation affect protein synthesis if it works on mRNAs?
Options:
loss of the poly A tail results in rapid 3'-5' degradation of transcript leading to reduced protein synthesis
deadenylation can result in an increased exposure of the cap to enzymes involved in decapping, leading to subsequent 5' mediated degradation
it reduces the ability of mRNAs to form 5' 3' circular mRNA structures that protects both termini
a and b only
a, b and c
Chapter 12 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 12 - CASE STUDY | A drug that sometimes works A...Ch. 12 -
CASE STUDY | A drug that sometimes works
A...Ch. 12 -
CASE STUDY | A drug that sometimes works
A...Ch. 12 - HOW DO WE KNOW? In this chapter, we focused on the...Ch. 12 - Review the Chapter Concepts list on p. 215. These...Ch. 12 - In studies of frameshift mutations, Crick,...Ch. 12 -
4. The mRNA formed from the repeating...Ch. 12 - In studies using repeating copolymers, AC......Ch. 12 - Prob. 6PDQCh. 12 - Prob. 7PDQ
Ch. 12 -
8. When the amino acid sequences of insulin...Ch. 12 - Prob. 9PDQCh. 12 - Why doesn't polynucleotide phosphorylase (Ochoa's...Ch. 12 - Refer to Table 12.1. Can you hypothesize why a...Ch. 12 -
12. Predict the amino acid sequence produced...Ch. 12 - A short RNA molecule was isolated that...Ch. 12 - A glycine residue exists at position 210 of the...Ch. 12 - Shown here is a theoretical viral mRNA sequence...Ch. 12 -
16. Most proteins have more leucine than...Ch. 12 - Define the process of transcription. Where does...Ch. 12 - Describe the structure of RNA polymerase in...Ch. 12 - In a written paragraph, describe the abbreviated...Ch. 12 - Messenger RNA molecules are very difficult to...Ch. 12 - One form of posttranscriptional modification of...Ch. 12 - In a mixed copolymer experiment, messages were...Ch. 12 -
23. Shown in this problem are the amino acid...Ch. 12 - Alternative splicing is a common mechanism for...Ch. 12 - The genetic code is degenerate. Amino acids are...
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- Once a primary RNA transcript is created from a DNA template, it must be modified in several ways before becoming messenger RNA (mRNA), ribosomal RNA (rRNA) or transfer RNA (tRNA). The following image shows RNA processing of one pre-mRNA into mRNA. Note that pre-RNA is processed in three ways: 1) a 5' methylguanylate cap (G cap) is added, 2) a poly-A tail is added, and 3) the spliceosome removes introns from the pre-mRNA transcript. Please redraw the following diagram and label the following on your diagram: DNA Promoter pre-mRNA (unprocessed) mRNA *5' methylguanylate cap *polyadenylation *Exon (may be more than one) *Intron (may be more than one) Transcription RNA processing AAAAAarrow_forwardDiscuss the types of RNA transcripts and the functional roles theyplay. Why do some RNAs form complexes with protein subunits?arrow_forwardThe most common type of termination signal in E. Coli is a symmetrical inverted repeat of GC rich sequences followed by about 7 As that forms a stable stem-loop structure in the RNA, which disrupts its association with the DNA template and terminates transcription. true or falsearrow_forward
- The following four mutations have been discovered in a gene that has more than 60 exons and encodes a very large protein of 2532 amino acids. Indicate which mutation would likely cause a detectable change in the size of the mRNA and/or the size of the protein product. Consider a detectable change to be >10% of the wild-type size. A table of the genetic code is shown below. First letter 0 00 U O A บบบ UUC UUA UUG U CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu >Leu AUU AUC lle AUA AUG Met >Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Second letter C Ser Pro Thr Ala CAU CAC CAA CAG UAU UGU Tyr UAC UGC UAA Stop UGA UAG Stop UGG AAU AAC AAA AAG A GAU GAC GAA GAG His Gin Asn Lys Asp G Glu CGU CGC CGA CGGJ AGU AGC AGA AGG GGU GGC GGA GGG O AAG576UAG (changes codon 576 from AAG to UAG) Cys Stop Trp O GUG326AUG (changes codon 326 from GUG to AUG) Arg Ser Arg Gly DUAG DUA G DCAG DO AG deletion of codon 779 insertion of 1000 base pairs into the sixth intron (this particular…arrow_forwardA common feature of many eukaryotic mRNAs is the presence of a rather long 3′ UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an important enzyme in cellular metabolism. Certain cells—termed U937D cells—have lots of CK-B mRNA, but no CK-B enzyme is present. In these cells, the 5′ end of the CK-B mRNA is bound to ribosomes, but the mRNA is apparently not translated. Something inhibits the translation of the CK-B mRNA in these cells. Researchers introduced numerous short segments of RNA containing only 3′ UTR sequences into U937D cells. As a result, the U937D cells began to synthesize the CK-B enzyme, but the total amount of CK-B mRNA did not increase. The introduction of short segments of other RNA sequences did not stimulate the synthesis of CK-B; only the 3′ UTR sequences turned on the translation of the enzyme. On the basis of these results, propose a mechanism for the inhibition of CK-B translation in the U937D cells. Explain how the introduction of short…arrow_forward) A normal mRNA that reads 5'- UGCCAUGGUAAUAACACAUGAAGGCCUGAAC-3' was an insertion mutation that changes the sequence to 5'- UGCCAUGGUUAAUAACACAUGAGGCGUGAAC-3'. Translate the original mRNA and the mutated mRNA and explain how insertion mutations can have dramatic effects on proteins. ( Hint; Be sure to find the initiation site).arrow_forward
- Imagine you are going to label a gene associated with apoptosis in Symbiodiniaceae with a Yellow Fluorescent Protein (YFP). To generate the YFP, you know the pre-MRNA looks as follows: Unspliced YFP premature mRNA Сap 5' UTR Exon 1 Intron Exon 2 Intron Exon 3 3' UTR Poly-A tail If Exon 2 is also required for mRNA stability, what can be predicted from the possible spliced alternative isoforms formed? One of the isoforms will not have a poly-A tail O The alternative splicing of YFP pre-MRNA prevents 5'-capping The MRNA isoform without Exon 2 will be degraded faster than the other isoform Exon 2 will be added to isoform B later to correct the mistake in splicing The protein translated from one of the mRNA isoforms will possess an additional functional domainarrow_forwardAll eukaryotes possess a surveillance pathway referred to asnon-sense-mediated mRNA decay (NMD). Its principalfunction is to eliminate mRNA transcripts with prematurestop codons. Such faulty transcripts are detected duringtranslation and subsequently destroyed by removal of the 5′cap followed by degradation by a nuclease. Describe howpremature stop codons are detected and what type of errorcauses them.arrow_forwardWith regard to RNA polymerase proofreading ability, which of the following is true? A 3'5' exonuclease fixes all errors in mRNA. There are specialized mRNA repair pathways that remove noncomplementary nuclec No proofreading occurs. RNA backtracking can remove noncomplementary nucleotides. RNA and DNA polymerase proofreading are essentially identical. OOOOOarrow_forward
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