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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Textbook Question
Chapter 15, Problem 22PDQ
Many eukaryotic promoter regions contain CAAT boxes with consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?
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Many eukaryotic promoter regions contain CAAT boxes with consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?
Many promoter regions contain CAAT boxes containing consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?
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.
Chapter 15 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 15 -
CASE STUDY | A mysterious muscular dystrophy
A...Ch. 15 -
CASE STUDY |A mysterious muscular dystrophy
A...Ch. 15 -
CASE STUDY |A mysterious muscular dystrophy
A...Ch. 15 -
HOW DO WE KNOW?
1. In this chapter, we have...Ch. 15 -
2. Review the Chapter Concepts list on p. 280....Ch. 15 - Describe which enzymes are required for lactose...Ch. 15 - Contrast positive versus negative regulation of...Ch. 15 -
5. Both attenuation and riboswitches rely on...Ch. 15 - For the lac genotypes shown in the accompanying...Ch. 15 -
7. For the genotypes and conditions (lactose...
Ch. 15 -
8. The locations of numerous lacI– and lacIs...Ch. 15 - Explain why catabolite repression is used in...Ch. 15 - Describe experiments that would confirm whether or...Ch. 15 - Predict the level of genetic activity of the lac...Ch. 15 - Predict the effect on the inducibility of the lac...Ch. 15 -
13. Describe the role of attenuation in the...Ch. 15 -
14. In a theoretical operon, genes A, B, C, and D...Ch. 15 - A bacterial operon is responsible for production...Ch. 15 - A marine bacterium is isolated and is shown to...Ch. 15 -
17. Why is gene regulation more complex in a...Ch. 15 -
18. List and define the levels of eukaryotic gene...Ch. 15 -
19. Distinguish between the cis-acting regulatory...Ch. 15 - Prob. 20PDQCh. 15 - Compare the control of gene regulation in...Ch. 15 - Many eukaryotic promoter regions contain CAAT...Ch. 15 -
23. What is RNA-induced gene silencing in...Ch. 15 - Although it is customary to consider...Ch. 15 - DNA methylation is commonly associated with a...Ch. 15 - The interphase nucleus appears to be a highly...Ch. 15 - It has been estimated that at least two-thirds of...
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- "Upstream" "Downstream" Exons Start of transcription Termination codon 5 3' Promoter initiator codon Introns Polyadenylation signal (intervening sequences) 5' untranslated region 3' untranslated region Direction of transcription Please study the diagram above on eukaryotic gene expression. In order to provide instructions for gene expression, a eukaryotic gene should have the following sequences except for O A. Promoter B. Start codon also known as initiator codon C. Splicing signals (dinucleotide sequence in the intron) O D. 5' CAP sequencearrow_forwardThe following logo plot represents the preferred cis-regulatory sequences (i.e. transcription factor binding site) of bHLH transcription factor FOSL1. C 1 2 3 4 5 6 7 8 9 10 11 position Would you expect this sequence to be recognized by a monomer, a homodimer, or a heterodimer of the protein? Explain your answer. (short phrases are sufficient; please write your answer into the template below) A- В I A -l expect FOSL1 to bind as a: (monomer, homodimer, heterodimer; please choose) B - short explanation: information content (bit) !!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
- The 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_forwardIdentify the statements that are features of a promoter. In prokaryotes, the promoter contains a −35 and −10 region upstream of the transcription start site. In prokaryotes, the promoter is recognized by general transcription factors (GTF), which recruit the RNA polymerase holoenzyme. In both prokaryotes and eukaryotes, the promoter is located in the 5' direction, upstream from the transcription start site. In eukaryotes, the promoter recruits the preinitiation complex, which includes the TATA-binding protein. In eukaryotes, the promoter attracts the small and large ribosomal subunits with the help of initiation factors.arrow_forward
- What are the essential components for the transcription control in prokaryotes? Please explain brieflyarrow_forward4.1 Name and discuss two transcription regulatory elements that can be found in the figure. (6)4.2. During the activation of eukaryotic transcription the promoter region needs to be accessible for the binding of transcription factors. Describe in detail one of the mechanisms involved in this process.arrow_forwardThe consensus sequence for the –35 sequence of a bacterial promoter is 5′–TTGACA–3′. The –35 sequence of a particular bacterial gene is 5′– TTAACA–3′. A mutation changes the fifth base from a C to a G. Would you expect this mutation to increase or decrease the rate of transcription?arrow_forward
- The interphase nucleus is a highly structured organelle with chromosome territories, interchromatin compartments, and transcription factories. In cultured human cells, researchers have identified approximately 8000 transcription factories per cell, each containing an average of eight tightly associated RNAP II molecules actively transcribing RNA. If each RNAP II molecule is transcribing a different gene, how might such a transcription factory appear? Provide a simple diagram that shows eight different genes being transcribed in a transcription factory and include the promoters, structural genes, and nascent transcripts in your presentation.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 1.Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds. 2.Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. 3.Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. 4.Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination.arrow_forwardThe following DNA nucleotides are found near the end of a bacterial transcription unit. 3′–AGCATACAGCAGACCGTTGGTCTGAAAAAAGCATACA–5′ Q. Is this terminator rho independent or rho dependent?arrow_forward
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