Principles of Biology
2nd Edition
ISBN: 9781259875120
Author: Robert Brooker, Eric P. Widmaier Dr., Linda Graham Dr. Ph.D., Peter Stiling Dr. Ph.D.
Publisher: McGraw-Hill Education
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Chapter 12.4, Problem 2TYK
Summary Introduction
Introduction:
Usually eukaryotic genes contain nucleosome-free regions (NFR). At these regions a preinitiation complex forms. Since NFR is not sufficient for gene activation by itself, an activator binds to it which recruits chromatin remodeling complexes and histone modifying enzymes in order to allow the formation of a preinitiation complex.
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Chapter 12 Solutions
Principles of Biology
Ch. 12.1 - How does gene regulation underlie the different...Ch. 12.1 - Prob. 1TYKCh. 12.1 - The most common point of gene regulation in...Ch. 12.2 - Which genes are under the control of the lac...Ch. 12.2 - Prob. 2CCCh. 12.2 - Prob. 1BCCh. 12.2 - Prob. 1TYKCh. 12.3 - Prob. 1CCCh. 12.3 - Prob. 1TYKCh. 12.3 - Prob. 2TYK
Ch. 12.4 - Prob. 1CCCh. 12.4 - Prob. 1BCCh. 12.4 - A chromatin-remodeling complex may change the...Ch. 12.4 - Prob. 2TYKCh. 12.5 - What is the biological advantage of alternative...Ch. 12.5 - Prob. 2CCCh. 12.5 - Prob. 1TYKCh. 12.5 - Prob. 2TYKCh. 12 - Genes that are expressed at all times at...Ch. 12 - Prob. 2TYCh. 12 - Transcription factors that bind to DNA and...Ch. 12 - Prob. 4TYCh. 12 - Prob. 5TYCh. 12 - Prob. 6TYCh. 12 - Prob. 7TYCh. 12 - Prob. 8TYCh. 12 - Prob. 9TYCh. 12 - Prob. 10TYCh. 12 - What is the difference between inducible and...Ch. 12 - Transcriptional regulation often involves a...Ch. 12 - PRINCIPLES A principle of biology is that the...Ch. 12 - Discuss the advantages and disadvantages of...Ch. 12 - Prob. 2CBQ
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- Gene X codes for a protein in eukaryotes. A mutated eukaryotic cell contains an altered base-pair in an intron of gene X. Which would be the most likely effect of this mutation on the biomolecules in the cell? The amount of pre-mRNA transcribed from gene X would be less than normal. The amount of functional protein corresponding to gene X would be less than normal. The ability of snRNAs to form a spliceosome would be diminished. The breakdown of mature mRNA corresponding to gene X would be fasterarrow_forward3′-->5′ Exonuclease activity allows DNA polymerase III (Pol III) to back-up and fix a mismatched base pair that was just incorporated into a growing strand of new DNA. True Or False In one of the four ways to regulate gene expression, positive control with repression indicates that transcription is activated in the presence of a co-repressor. True Or Falsearrow_forwardThrough alternative splicing, eukaryotes (a) reinforce gene inactivation (b) prevent transcription of heterochromatin (c) produce related but different proteins in different tissues (d) amplify genes to meet the requirement of high levels of a gene product (e) bind transcription factors to enhancers to activate transcriptionarrow_forward
- IS. Alternative splicing has been estimated to occur in more than 95% of multi-exon genes. Which of the following is not an evolutionary advantage of alternative splicing? Alternative splicing increases diversity without increasing genome size Different gene isoforms can be expressed in different tissues Alternative splicing creates shorter mRNA transcripts Different gene isoforms can be expressed during different stages of development.arrow_forwardThe following is a DNA sequence of gene Z. The underlined sequence represents the promoter for gene Z and the underlined and italicized sequence encodes the gene Z ribosome binding (RBS) site. Transcription begins at and includes the T/A base pair at position 60 (bold)arrow_forwardConsider 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.arrow_forward
- ennar region of gene X, which determines the length of the tail in mice, is mutated so that transcription factors bind it at a much higher affinity compared to the wild-type sequence. What is the most likely phenotypic outcome? Tail length will not change because the enhancer is a non-coding sequence Tail length will increased due to increased activity of the gene's promoter Tail length will decreased because any mutation will cause a loss-of-function of these regulatory regions Not just the tail will be enlarged because increased activity of the enhancer will impact many genesarrow_forwardA TATA box is a short, A-T rich sequence of nucleotides located just upstream of genes and is used to recruit transcription factors and RNA polymerase to begin transcription. Based on this description, a TATA box is most likely a(n): promoter terminator transcription factor RNA polymerasearrow_forwardConsider Figure 3, which shows some features of a eukaryotic gene. A, B, C are exons while 1, 2 are introns. E marks an enhancer. The 5’ UTR and 3’ UTR are also marked. Which one of the following features would you expect to NOT be included in the primary RNA (before any processing has occurred) that results from transcription of this gene? Region marked E Region marked 5’ UTR Regions marked A, B and C Regions marked 1 and 2arrow_forward
- Regarding eukaryotic genes, it is correct to state that: a) Distal enhancer-like elements decrease the intensity of gene transcription activation b) Mutations in intronic regions of a gene can alter the levels of its corresponding protein c) They are regulated only by promoter regions, being activated or repressed by the presence of transcription factors d) The junctions of exons and introns are recognized by splicing factors, which guarantee the production of the same mRNA regardless of cell type. e) Activator and repressor proteins bind to the coding region of genes, regulating the intensity of their transcriptionarrow_forwardWhat role does an operator sequence serve in bacterial gene expression regulation? Describe one change in bacterial cells that can cause a repressor protein to go from inactive to active or from active to inactive.arrow_forwardThe MAT locus allows yeast to switch mating type through a very complex mechanism. However, it has informed us a great deal about what aspects of gene expression typical to all organisms? Options: higher order changes in chromatin affect transcriptional efficiency that general transcription factors must first bind directly to histone tails and only then can they interact with their cognate binding sites that DNA methylation is involved in this silencing mechanism that SIR2 is required for all types of transcriptional repression that expression of Pol III genes provides a means of identifying active chromatinarrow_forward
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