Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 19, Problem 5TYK
Summary Introduction
Introduction:
Escherichia coli (E. coli) is an example of prokaryotic bacteria present in human guts. All prokaryotes have a simple cellular structure containing chromosomes, plasmid, and the nucleus is not surrounded by a nuclear membrane. The protein synthesis takes place in the cytoplasm of the cells.
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Define the following terms: a. promoter b. consensus sequence c. operon d. chromatin-remodeling complex e. general transcription factors
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.
Which of the following statements best describes the promoter of a protein-coding gene?
A.
The promoter is a protein that recruits RNA polymerase.
B.
The promoter is part of the RNA molecule itself.
C.
The promoter is a site found on RNA polymerase.
D.
The promoter contains the AUG start codon.
E.
The promoter is a nontranscribed region of a gene.
Chapter 19 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 19.1 - What additional biological questions can be...Ch. 19.2 - What is the principle behind whole-genome shotgun...Ch. 19.2 - Prob. 2SBCh. 19.2 - Prob. 3SBCh. 19.2 - Prob. 4SBCh. 19.3 - Prob. 1SBCh. 19.3 - Prob. 2SBCh. 19.3 - Prob. 3SBCh. 19.4 - Prob. 1SBCh. 19.4 - Prob. 2SB
Ch. 19 - Prob. 1TYKCh. 19 - How do pseudogenes differ from genes? a. They are...Ch. 19 - Prob. 3TYKCh. 19 - Prob. 4TYKCh. 19 - Prob. 5TYKCh. 19 - Prob. 6TYKCh. 19 - About 95% of the average human transcription unit...Ch. 19 - Prob. 8TYKCh. 19 - Prob. 9TYKCh. 19 - When two protein-coding genes have very similar...Ch. 19 - Prob. 11TYKCh. 19 - Prob. 12TYKCh. 19 - Prob. 13TYKCh. 19 - Discuss Concepts The genome of the yeast...Ch. 19 - Prob. 15TYKCh. 19 - Prob. 16TYKCh. 19 - Prob. 17TYKCh. 19 - Below is a sequence of 540 bases from a genome....
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- Which of the following statements correctly describes promoters in E. coli a.A promoter may be present on either side of a gene or in the middle of it. b.All promoters have the same sequence, which is that recognized by RNA polymerase holoenzyme. c.Many promoters are similar and resemble a consensus sequence. d.Every promoter has a different sequence, with little or no resemblance to other promoters.arrow_forwardWhich of the following is not true regarding gene regulation that involves DNA bending? a. The precise distance between the regulatory sequence and the promoter is important. b. Effect can be to repress transcription c. Effect can be to activation transcription d. Regulated genes can be thousands of base pairs away from the regulatory sitesarrow_forwardImagine 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_forward
- 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_forward1a) Why is it possible for you to study the eye colour gene by extracting cheek cells? a. Because the nucleus of every cell in the human body contains the same genetic information. b. Because the cheek cells are located near the cells of the eye and so they are able to exchange DNA. c. Because all genes in the human body are expressed at all times so it is easy to study them. d. All of the above are possible explanations. 1b) What is the purpose of heating the sample to 75°C following addition of the 0.2M NaOH solution? a. To denature the histone proteins that are keeping the DNA tightly coiled. b. To ensure that all the DNA is removed from the swab in preparation for PCR. c. To breakdown the cheek cell membrane to release the DNA from the cell. d. It breaks down the circular DNA down into linear fragments so that they will be easier to visualize.iarrow_forwardWhich of the following statements about RNA processing in eukaryotes is INCORRECT? A. The excision of introns from pre-mRNA is the only modification required to produce a mature mRNA. B. A protein/RNA complex is used to remove introns from the pre-mRNA. C. A poly A tail is added on to the 3′ end of the mRNA. D. A 7-methylguanosine cap is added on to the 5′end of the mRNA. E. Modification occurs in the nucleus.arrow_forward
- Which of the following mutations would have the greatest negative impact on the protein product of a gene? A. a single base deletion close to the end of the coding region of a gene. B. a single base insertion near the start of the coding region of the gene C. a base-pair substitution D. a deletion of three bases near the middle of the genearrow_forwardWhich of the following is NOT a DIFFERENCE between prokaryotic and eukaryotic gene regulation? A. prokaryotic mRNA is NOT capped by a 5’mG after transcription, but eukaryotic mRNA is so capped B. eukaryotic mRNAs are monocistronic (encode single proteins), whereas prokaryotic mRNAs are polycistronic (encode multiple proteins) C. prokaryotic mRNA is not modified by polyadenylation after transcription, but eukaryotic mRNA is so modified. D. translation of eukaryotic mRNA into protein is not coupled to transcription of the mRNA from DNA, but in prokaryotes it is so coupled E. prokaryotic translation and eukaryotic translation use different genetic codes to translate mRNA codons into amino acid sequences of proteinsarrow_forwardThere is Hyaluronic acid synthesis occuring in Group X Strep and it is controlled by an operon with 3 genes, called hasXYZ. Based on the 3-line diagram model, a. How many ribosome binding sites are there for the protein? b. How many promoters are there for the genes? c. How many start codons are there for the protein? d. How many RNA Polymerase binding locations are there for the genes? e. How many proteins will be fully functional? f. How many mRNA strands are made?arrow_forward
- Which 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_forwardGeneticists isolated six trains of E. coli (strains 1-6) whose lac operons had one of the following mutations (a-f): a. deletion of lacY b. oº mutation c. missense mutation in lacz d. inversion of the lac operon (which does not include the lacl gene) e. superrepressor mutation f. inversion of lacZ, Y, and Z, but not lacl, P, o Which of these mutations above would prevent cells from utilizing lactose?arrow_forwardWhich of the following is an RNA processing event that increases the number of potential polypeptides that can be encoded by a single gene. a. transcription b. alternative splicing c. alternative transcription d. polyadenylationarrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY