BIOLOGY
5th Edition
ISBN: 9781264104680
Author: BROOKER
Publisher: MCG
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Chapter 13, Problem 5TY
Summary Introduction
Introduction: Protein sorting and the function of secreting is carried out by SRP RNA – protein complex. Polypeptides are directed to the plasma membrane in bacteria and to the ER in eukaryotes. In bacteria, the SRP is made up of one ncRNA and one protein and in eukaryotes; it has six proteins and one ncRNA.
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The addition of the poly-A tail adds more than 200 units of adenine to the strand of mRNA, yet no protein has a continuous line of more than 200 phenylalanines in its structure. Why is this so?
A. RNA splicing occurs and removes the noncoding sequences.
B. The poly-A tail and 5' GTP cap do not translate into proteins.
C. The adenines are reabsorbed by the cell and help to form other forms of RNA.
D. The tRNA does not recognize the poly-A tail.
A scientist mutates eIF-2 to eliminate its GTP hydrolysis capability. How would this mutated form of eIF-2 alter translation? a. Initiation factors would not be able to bind to mRNA. b. The large ribosomal subunit would not be able to interact with mRNA transcripts. c. tRNAi-Met would not scan mRNA transcripts for the start codon. d. eIF-2 would not be able to interact with the small ribosomal subunit.
A strain of bacteria possesses a temperature-sensitive mutation in the gene that encodes the rho subunit. At high temperatures, rho is not functional. When these bacteria are raised at elevated temperatures, which of the following effects would you expect to see? Explain your reasoning for accepting or rejecting each of these five options. a. Transcription does not take place. b. All RNA molecules are shorter than normal. c. All RNA molecules are longer than normal. d. Some RNA molecules are longer than normal. e. RNA is copied from both DNA strands.
Chapter 13 Solutions
BIOLOGY
Ch. 13.1 - Prob. 1CCCh. 13.2 - Prob. 1CCCh. 13.3 - Prob. 1EQCh. 13.3 - Prob. 2EQCh. 13.3 - Prob. 3EQCh. 13.3 - Effects of Non-coding RNAs on Translation and mRNA...Ch. 13.4 - Non-coding RNAs and Protein Sorting Core Skill:...Ch. 13.5 - Core Skill: Modeling The goal of this modeling...Ch. 13 - Prob. 1TYCh. 13 - Prob. 2TY
Ch. 13 - Prob. 3TYCh. 13 - Prob. 4TYCh. 13 - Prob. 5TYCh. 13 - Prob. 6TYCh. 13 - With regard to miRNAs and siRNAs, which of the...Ch. 13 - Cas1 and Cas2 proteins play a role during which of...Ch. 13 - Which of the following components bind to...Ch. 13 - Abnormalities in the expression of ncRNAs are...Ch. 13 - An ncRNA may have one or more of the following...Ch. 13 - What is RNA interference (RNAi)? Explain how the...Ch. 13 - Prob. 3CQCh. 13 - Prob. 1COQCh. 13 - Go to the PubMed website and search for non-coding...
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- Imagine that mutations occurred in one of the inverted repeat sequences within the rho-independent terminator sequence of a bacterium. What would likely be the consequence of this mutation? Select one: a. Transcription may not be initiated at all. b. Transcription may end prematurely. c. Transcription may be delayed. d. The rho protein won't be able to bind and transcription may not be terminate. e. Transcription may not be terminated and result in much longer RNA.arrow_forwardThe NFAT family is a ubiquitous family of transcription factors. a. Under resting conditions, where is NFAT localized in a cell? b. Under activated conditions, where is NFAT localized in a cell? c. How is it released from its resting condition and permitted to relocalize? d. Immunosuppressant drugs such as cyclosporin act via inhibition of the calcineurin phosphatase. If NFAT is ubiqitous, how do you think these drugs might act with so few side effects on other signaling processes within the body?arrow_forwardYou design a protein with an N-terminal signal sequence and a single transmembrane domain. You also decide to tag your protein with GFP. If you want the GFP to be outside the cell, where should the GFP appear in the amino acid sequence? а. Between the signal sequence and the transmembrane domain b. At the C-terminal end of the protein (i.e., the last part of the protein to be translated с. Between the transmembrane domain and the C-terminus d. The GFP can be anywhere in the amino acid sequencearrow_forward
- If there is a mutation in importin that prevents it from interacting with the NLS of cargo proteins, what affect, if any, would this have on the UPR (unfolded protein response). A. Ran-GEF would bind to the importin promoting cargo entry into the nucleus and transcription of UPR response genes. B. Gene regulatory protein would not be imported into the nucleus thus preventing transcription of UPR response genes C. MRNA would not be imported into the nucleus thus preventing transcription of UPR response genes D. Ran-GTP would not be imported to the nucleus thus preventing transcription of UPR response genes E. No effect, these are completely unrelated processes.arrow_forwardWhich of the following best explains how the prokaryotic expression of a metabolic protein can be regulated when the protein is already present at a high concentration? a.Repressor proteins can be activated and bind to regulatory sequences to block transcription. b.Regulatory proteins can be inactivated to increase gene expression. c.Transcription factors can bind to regulatory sequences to increase RNA polymerase binding. d.Histone modification can prevent transcription of the gene.arrow_forwardIn which of the following does nitrogenous base pairing (base complementarity via hydrogen bonds) NOT play a key, direct role? A. transcription of RNA from a DNA template B. binding of the TATA binding protein to the TATA box C. the RNAi pathway of gene regulation D. Codon-Anticodon recognition in translationarrow_forward
- What are two reasons that a eukaryotic gene (e.g. Green Fluorescent Protein in jellyfish) will not be expressed if it is inserted into a bacterial genome with no alterations? a. Bacterial genes do not contain introns b. Bacterial promoters have different consensus sequences than eukaryotic ones c. Transcription of this gene is affected by the simultaneous processes of transcription and translation d.arrow_forwardArrange the statements in their proper order by writing the corresponding letter (e.g. A) for each statement in the space provided below. A. The single-stranded RNA would complement the target RNA. B. Gene expression is inactivated once the mRNA is no longer accessible for translation. C. The risk-induced silencing complex which is composed of RNA and protein subunits is formed. D. Double-stranded, non-coding RNA is cleaved by Dicer. E. The mRNA can be cleaved or remain bound by the RISC. 1. 2. 3. 4. 5.arrow_forwardWhat is one function of TFIIH during transcription? a. Recruiting the TATA-box binding protein to the TATA-box. b. Terminating transcription via factor-dependent termination. c. Capping the 5' end of mRNA with 7-methylguanylate. d. Phosphorylating the CTD of RNA polymerase.arrow_forward
- Hypertrophic cardiomyopathy is caused by mutations in the β-myosin heavy chain gene. One common mutation involves a change of an arginine (Arg) to a glutamine (Gln) at position 403. Using the genetic code shown above, what could the sequence of the mutant codon be? a. CAA b. CGA c. CGG d. CACarrow_forwardWhich of the following statements about trp operon regulation is TRUE? A. Low trp levels result in trp repressor binding. O B. High trp levels result in transcriptional antitermination. C. Low trp levels cause the ribosome to fall off at the trp codons in the upstream ORF. O D. High trp levels cause the ribosome to fall off at the trp codons in the upstream ORF. O E. Low trp levels result in transcriptional antitermination. QUESTION 34 Consider the following partial diploid for the lac operon system. lacl+ O+ lac Z+Y-A- lacls O+ 1. 2. lac Z-Y+ A+ In this configuration, which of the following statements is TRUE? A. With no lac present, ZYA will all be expressed. B. with lac present none of the genes (ZYA) will be expressed. O C. With no lac present, only YA will be expressed. D. None of these answers is correct. E. With no lac present, only Z will be expressed.arrow_forwardE. coli are grown on a medium containing lactose. Once glucose is added to the medium, the bacteria stop fermenting lactose. Which of the following BEST explains the observed effect? A. Cellular levels of cAMP are lowB.Glucose is bound to the promoter C. Repressor protein is bound to the operator D. Repressor protein is bound to the promoterarrow_forward
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