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 14, Problem 16TYK
Summary Introduction
To review:
The fact that the evolutionary relationship of the amino acid sequences of the DNA polymerases found in Archaea show great similarity to Eukaryote but show little similarity to those of the DNA polymerases of bacteria.
Introduction:
DNA polymerases are the enzymes that carry out the process of
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Which of the following best explains the production of Okazaki fragments in replicating DNA
(a) DNA is stressed when it unwinds (b) DNA is anti-parallel and can only be synthesized 5’ to 3’ (c) DNA contains once less oxygen in its sugar while RNA has an OH attached to its 2’ carbon (d) Template strands are complementary and have a tendency to reform hydrogen bonds (e) both a and d
Which of the following statements most accurately describes the action of the enzyme RNA polymerase?Select one
1.) RNA polymerase will transcribe only the exons by skipping over the introns within a eukaryotic gene sequence
2.) RNA polymerase will transcribe both DNA strands, moving in the 3' to 5' direction for one strand and 5' to 3' on the other
3.) RNA polymerase will transcribe both DNA strands, but only one RNA molecule will be used during translation
4.)None of the statements accurately describe the function of RNA polymerase
Explain the statement "In a comparison between the DNAS of related organisms such as humans and mice, identifying the conserved DNA sequences facilitates the search for functionally important regions" is true or false.
Chapter 14 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 14.1 - Prob. 1SBCh. 14.2 - Prob. 1SBCh. 14.2 - Prob. 2SBCh. 14.2 - Prob. 3SBCh. 14.2 - Prob. 4SBCh. 14.3 - What is the importance of complementary base...Ch. 14.3 - Why is a primer needed for DNA replication? How is...Ch. 14.3 - DNA polymerase III and DNA polymerase I are used...Ch. 14.3 - Prob. 4SBCh. 14.4 - Why is a proofreading mechanism important for DNA...
Ch. 14 - Working on the Amazon River, a biologist isolated...Ch. 14 - Prob. 2TYKCh. 14 - Pyrimidines built from a single carbon ring are:...Ch. 14 - Which of the following statements about DNA...Ch. 14 - Which of the following statements about DNA is...Ch. 14 - Prob. 6TYKCh. 14 - Prob. 7TYKCh. 14 - Prob. 8TYKCh. 14 - Prob. 9TYKCh. 14 - Prob. 10TYKCh. 14 - Discuss Concepts Eukaryotic chromosomes can be...Ch. 14 - Prob. 12TYKCh. 14 - Prob. 13TYKCh. 14 - Discuss Concepts During replication, an error...Ch. 14 - Design an Experiment Design an experiment using...Ch. 14 - Prob. 16TYKCh. 14 - Prob. 1ITDCh. 14 - Prob. 2ITD
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- The technique of fluorescence in situ hybridization (FISH) is described. This is another method for examining sequence complexity within a genome. In this method, a DNA sequence, such as a particular gene sequence, can be detected within an intact chromosome by using a DNA probe that is complementary to the sequence.For example, let’s consider the β-globin gene, which isfound on human chromosome 11. A probe complementary to theβ-globin gene binds to that gene and shows up as a brightly colored spot on human chromosome 11. In this way, researchers can detectwhere the β-globin gene is located within a set of chromosomes. Becausethe β-globin gene is unique and because human cells are diploid(i.e., have two copies of each chromosome), a FISH experimentshows two bright spots per cell; the probe binds to each copy ofchromosome 11. What would you expect to see if you used thefollowing types of probes?A. A probe complementary to the Alu sequenceB. A probe complementary to a tandem array near…arrow_forwardplease do (i)arrow_forwardA researcher sequences the genome of a variety of bacterial and eukaryotic cells. She finds that the bacterial genome is smaller, but that there are more genes for a given number of base pairs in the eukaryotic cells. In other words, there are fewer genes per unit of length of DNA in the eukaryotic cells. What do you predict she will find if she examines the DNA more closely? A. All of the bacterial DNA consists of coding sequences, but this is not true of the eukaryotic DNA. B. There are more repetitive sequences in the eukaryotic DNA than in the bacterial DNA. C. There are densely packed genes in the eukaryotic DNA that were not immediately distinguishable during the first analysis. D. The bacteria have larger quantities of noncoding DNA than the eukaryotic cells.arrow_forward
- the one above: Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTT 2. Using this mutated DNA strand, express it as a polypeptide by using the correct reading frame. When you get to the stop codon – you may write an “*” to denote the stop codon. 3. How many amino acids were changed in the mutated polypeptide?arrow_forwardPolymerase Chain Reaction (PCR) was invented by Kary Mullis in 1983. This technique had indeed facilitated research in various areas of molecular biology and genetics. Why is the annealing temperature vital in this technique? Explain how annealing temperature will affect the efficiency of this reaction.arrow_forwardplease do (ii)arrow_forward
- Suppose you have been directed to find new enzymes to use in the breakdown of wood in order to process biofuel (switchgrass, for example). Suppose you wanted to use fungal or bacterial DNA from the environment in order to do so. DNA can be unwound from the double stranded double helix into single strands, amplified, separated on gels by size, stained with dyes. It can be mutated by a variety of means. It can be sequenced. Describe one or more of the ways that you might manipulate DNA towards the stated goal. Relate the technology you plan to utilize to the structure of DNA. (You can break this into multiple posts, as multiple procedures might be used).arrow_forwardWhich of the comparisons between DNA polymerase and RNA polymerase is false? -Both DNA polymerase and RNA polymerase have proofreading activity. -DNA polymerase requires an additional enzyme to unwind double-stranded DNA molecules, but RNA polymerase does not. -Both DNA and RNA polymerase catalyze the addition of nucleotides to the 3’ end. -DNA polymerase needs a free 3’-OH to begin the polymerization reaction, but RNA polymerase does not.arrow_forwardThe image below shows the base cytosine and a methylated form of cytosine that occurs frequently in the human genome. Use your knowledge of DNA structure to answer the following question: a) Does methylation of cytosine affect its ability to base-pair with guanine? Explain b) Could methylation of cytosine affect the binding of a protein that interacts with a C-G base-pair in the major groove? Explain your answer.arrow_forward
- DNA polymerases cannot act as primers for replication, yet primase and other RNA polymerases can. Some geneticists have speculated that the inability of DNA polymerase to prime replication is a result of its proofreading function. This hypothesis argues that proofreading is essential for the faithful transmission of genetic information and that because DNA polymerases have evolved the ability to proofread, they cannot prime DNA synthesis. Explain why proofreading and priming functions in the same enzyme might be incompatible.arrow_forwardAssume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…arrow_forwardLocate as accurately as possible the listed items that are shown on the following figure. Some items are not shown. (a) 5′ end of DNA template strand; (b) 3′ end of mRNA; (c) ribosome; (d) promoter; (e) codon; (f) an amino acid; (g) DNA polymerase; (h) 5′ UTR; (i) centromere; (j) intron; (k) anticodon; (l) N terminus; (m) 5′ end of charged tRNA; (n) RNA polymerase; (o) 3′ end of uncharged tRNA; (p) a nucleotide; (q) mRNA cap; (r) peptide bond; (s) P site; (t) aminoacyl-tRNA synthetase; (u) hydrogen bond; (v) exon; (w) 5′ AUG 3′; (x) potential wobble interaction.arrow_forward
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