Biology
12th Edition
ISBN: 9781260494570
Author: Raven, Peter
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Textbook Question
Chapter 16, Problem 7A
A mechanism of control in E. coli not discussed in this chapter involves pausing of ribosomes allowing a transcription terminator to form in the mRNA. In eukaryotic fission yeast, this mechanism should
a. be common since they are unicellular.
b. not be common since they are unicellular.
c. not occur as transcription occurs in the nucleus and translation in the cytoplasm.
d. not occur due to possibility of alternative splicing.
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All of the following regarding ribosomes are true EXCEPT:
A. Ribosomes are comprised of protein and RNA.
B. Ribosomes coordinate with initiation and elongation factors to regulate translation.
C. Ribosomes bind transcripts in their promoters.
D. Prokaryotic ribosomes are comprised of a large and a small subunit.
E. Ribosomes in eukaryotic cells bind to complete transcripts.
A.
Individual ribosomal proteins can be isolated that have catalytic activity
B.
The ribosome associates with only three nucleotides of mRNA at a time
C.
There is more RNA mass than protein mass in eukaryotic ribsomes
D.
Only the larger subunits of ribosomes contain ribosomal RNA
E.
rRNA has a catalytic role in translation
1)You add a drug to cells that can pass through hydrophilic or hydrobic environments (e.g., membranes & cytosol). The drug specifically prevents the binding of KDEL (ER retention signal) to any other molecules. In the short term, what would happen to the synthesis and localization of translocon proteins?
a. New complete translocon proteins would be made, but they would end up in the golgi apparatus.
b. New complete translocon proteins would be made, but they would end up at the plasma membrane
c. No new intact translocon proteins would be synthesized
d. New complete translocon proteins would be made, but they would be secreted
e. New complete translocon proteins would be made and localized normally
Chapter 16 Solutions
Biology
Ch. 16.1 - Prob. 1LOCh. 16.1 - Prob. 2LOCh. 16.1 - Prob. 3LOCh. 16.2 - Explain how proteins can interact with base-pairs...Ch. 16.2 - Prob. 2LOCh. 16.3 - Prob. 1LOCh. 16.3 - Prob. 2LOCh. 16.3 - Explain control of gene expression in the trp...Ch. 16.4 - Prob. 1LOCh. 16.4 - Prob. 2LO
Ch. 16.4 - Prob. 3LOCh. 16.5 - Describe at least two kinds of epigenetic mark.Ch. 16.5 - Explain the function of chromatin-remodeling...Ch. 16.6 - Prob. 1LOCh. 16.6 - Prob. 2LOCh. 16.7 - Prob. 1LOCh. 16.7 - Prob. 2LOCh. 16 - Prob. 1DACh. 16 - What advantage might a bacterium gain by linking...Ch. 16 - Prob. 2IQCh. 16 - Prob. 3IQCh. 16 - In prokaryotes, control of gene expression usually...Ch. 16 - Prob. 2UCh. 16 - Prob. 3UCh. 16 - The lac operon is controlled by two main proteins....Ch. 16 - In eukaryotes, binding of RNA polymerase to a...Ch. 16 - In eukaryotes, the regulation of gene expression...Ch. 16 - In the trp operon, the repressor binds to DNA a....Ch. 16 - Prob. 1ACh. 16 - Specific transcription factors in eukaryotes...Ch. 16 - Repression in the trp operon and induction in the...Ch. 16 - Regulation by small RNAs and alternative splicing...Ch. 16 - Eukaryotic mRNAs differ from prokaryotic mRNAs in...Ch. 16 - In the cell cycle, cyclin proteins are produced in...Ch. 16 - A mechanism of control in E. coli not discussed in...Ch. 16 - You have isolated a series of mutants affecting...Ch. 16 - Examples of positive and negative control of...Ch. 16 - What forms of eukaryotic control of gene...Ch. 16 - The number and type of proteins found in a cell...
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- Which of the following statements are true about eukaryotic mRNA?a. The sigma factor is essential for the correct initiationof transcription.b. Processing of the nascent mRNA may begin beforeits transcription is complete.c. Processing takes place in the cytoplasm.d. Termination is accomplished by the use of a hairpinloop or the use of the rho factor.e. Many RNAs can be transcribed simultaneously fromone DNA templatearrow_forwardTermination of a prokaryotic transcript A. is a random process.B. requires the presence of the rho subunit of the holoenzyme. C. does not require rho factor if the end of the gene contains a G-C rich palindrome. D. is most efficient if there is an A-T-rich segment at the end of the gene. E. requires an ATPase in addition to rho factor.arrow_forwardWhich of the following is true of ribosome rescue? (C & E wrong) A. The process produces a protein that is tagged for degradation. B. Proteases enter the A-site of the ribosome to degrade the C-terminus of the synthesized peptide. C. A partially translated mRNA becomes fully degraded before it can be translated further. D. No protein is produced in the process. E. Stalled ribosomes are released in the absence of a stop codon.arrow_forward
- Identify the eukaryotic level of gene regulation. Choose the correct answer below. a. Pre- transcriptional control b. transcriptional control c. translational control d. post-translational control 1. in the fruit fly, genes from rRNA can be replicated more or less often compared to the rest of the chromatin depending on the needs of the cell. 2. in the human beta-globin, two introns are spliced out in order to produce the mature mRNA. 3. DNA methylation can change the degree of condensation of the chromatin. 4. The mouse REST gene is under the control of a promoter region that contains alternative promoters.arrow_forwardOne of the following is CORRECT: a. By each cycle the polypeptide has grown by one residue and consumed four GTP b. Peptidyl transferase mediates the release of amino acids from the growing chain c. Elongation process involves the addition of amino acids to the carboxyl end of the growing polypeptide chain d. Ribosomal RNAs are translated into proteinsarrow_forwardIf a cell's DNA was mutated such that it lost the promoter of a gene, you would expect: a. That ribosomes wouldn’t bind to that gene b. No effect since the promoter doesn't have any coding information for the amino acids in the polypeptide c. Ribosomes would never stop translating that gene d. Transcription factors for that gene would no longer bind to activate expression of that gene e. tRNA would no longer bind to the codons of that genearrow_forward
- 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_forwardGive one example of a prokaryotic system of gene expression that is more complicated (duringtranscription at the promoter) than its corresponding eukaryotic system of gene expression.A. prokaryotes require general transcription factors to initiate transcriptionB. prokaryotes typically have activator and repressor DNA sequences thousands of nucleotides awayC. prokaryotes have three classes of RNA polymerasesD. prokaryotes always have their DNA tightly packed into higher-level structures than nucleosomesE. prokaryotes usually have polycistronic mRNAarrow_forwardWhich of the following gene regulation mechanisms are found ONLY in prokaryotes? a. coordinate regulation b. transcriptional regulation c. alternative splicing d. translational regulation e. DNA modificationsarrow_forward
- Arrange 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_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_forwardMatch the following with what organelle (one per blank) it is designated to go (letters can be used once, twice or not all): A. Lysosome B. Golgi body C. Endoplasmic reticulum mRNA transcript has a KDEL (amino acid code). protein is N- glycosylated protein has a mannose-6-phosphate sugar D. Nucleus MRNA a transcript has a long string of hydrophobic amino acids E. Peroxisome MRNA transcript encodes for a string of basic amino acidsarrow_forward
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