Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 17, Problem 9RQ
Summary Introduction
To review:
Definition of the following terminologies:
1. Histones
2. Heterochromatin
3. Euchromatin
4. Intergenic sequences
5. Tandem repeats
Introduction:
The DNA (deoxyribonucleic acid) needs to be packaged within the cellular regions. This packaging requires that the DNA should be condensed into minute structural forms that can be contained in the cell. This compaction is essentially brought by histone proteins. The extent of DNA condensation determines the euchromatin or heterochromatin nature of the DNA. Moreover, the genome of many organisms is predominantly composed of intergenic sequences and tandem repeats.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Question 2:
Part a: Complete the table describing different components of intron removal from mRNA. Nu:, X and Y refer to B-type chemistry shown on the previous page. (YELLOW table shown)
Part b: Complete the table describing different components of group I self-splicing intron removal from 26S rRNA in Tetrahymena. (BLUE table shown)
Part c: Draw the intron with an all atom structure for Branchpoint A after intron removal from mRNA
Part d: Draw the Group I self-splicing intron with an all atom structure for the Guanosine cofactor after intron removal from 26S rRNA in Tetrahymena.
QUESTION NO. 1
Patients with the rare genetic disease xeroderma pigmentosum (XP) are very sensitive to light and are highly susceptible to skin cancers. The study of such patients has enhanced our knowledge of DNA repair because XP is caused by defective DNA repair nucleotide excision repair. (A variant, XP-V, is deficient in postreplication repair.) In nucleotide excision repair
A. removal of the damaged bases occurs on only one strand of the DNA.
B. only thymine dimers generated by UV light can be removed .
C. the excision nuclease is an exonuclease.
D. a single multifunctional enzyme carries out the repair process.
E. only the damaged nucleotides are removed.
QUESTION NO.2
Homologous recombination:
A. occurs only between two segments from the same DNA molecule.
B. requires that a specific DNA sequence be present.
C. requires one of the duplexes undergoing recombination be nicked in both strands.
D. involves a…
The genome’s functional gene products are either _____________ or _____________.
Chapter 17 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 17 - Prob. 1QCh. 17 - Prob. 2QCh. 17 - Prob. 3QCh. 17 - Prob. 4QCh. 17 - Prob. 5QCh. 17 - Prob. 6QCh. 17 - Prob. 7QCh. 17 - Prob. 8QCh. 17 - Prob. 1RQCh. 17 - Prob. 2RQ
Ch. 17 - Prob. 3RQCh. 17 - Prob. 4RQCh. 17 - Prob. 5RQCh. 17 - Prob. 6RQCh. 17 - Prob. 7RQCh. 17 - Prob. 8RQCh. 17 - Prob. 9RQCh. 17 - Prob. 10RQCh. 17 - Prob. 11RQCh. 17 - Prob. 12RQCh. 17 - Prob. 13RQCh. 17 - Prob. 14RQCh. 17 - Prob. 15RQCh. 17 - Prob. 16RQCh. 17 - Prob. 17RQCh. 17 - Prob. 18RQCh. 17 - Prob. 19RQCh. 17 - Prob. 20RQCh. 17 - Prob. 21RQCh. 17 - Prob. 22RQCh. 17 - Prob. 23RQCh. 17 - Prob. 24RQCh. 17 - Prob. 25RQCh. 17 - Prob. 26RQCh. 17 - Prob. 27RQCh. 17 - Prob. 28RQCh. 17 - Prob. 29RQCh. 17 - Prob. 30RQCh. 17 - Prob. 31RQCh. 17 - Prob. 32RQCh. 17 - Prob. 33RQCh. 17 - Prob. 34RQCh. 17 - Prob. 35RQCh. 17 - Prob. 36RQCh. 17 - Prob. 37RQCh. 17 - Prob. 38RQCh. 17 - Prob. 39RQCh. 17 - Prob. 40RQCh. 17 - Prob. 41FBCh. 17 - Prob. 42FBCh. 17 - Prob. 43FBCh. 17 - Prob. 44FBCh. 17 - Prob. 45FBCh. 17 - Prob. 46FBCh. 17 - Prob. 47FBCh. 17 - Prob. 48FBCh. 17 - Prob. 49FBCh. 17 - Prob. 50FBCh. 17 - Prob. 51SACh. 17 - Prob. 52SACh. 17 - Prob. 53SACh. 17 - Prob. 54SACh. 17 - Prob. 55SACh. 17 - Prob. 56TQCh. 17 - Prob. 57TQCh. 17 - Prob. 58TQCh. 17 - Prob. 59TQCh. 17 - Prob. 60TQCh. 17 - Prob. 61TQCh. 17 - Prob. 62TQCh. 17 - Prob. 63TQCh. 17 - Prob. 64TQCh. 17 - Prob. 65TQCh. 17 - Prob. 66TQCh. 17 - Prob. 67TQCh. 17 - Prob. 68TQCh. 17 - Prob. 69TQCh. 17 - Prob. 70TQCh. 17 - Prob. 71TQCh. 17 - Prob. 72TQCh. 17 - Prob. 73TQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- The purpose of mitosis in human cells is for the production of two diploid, _________________ cells.arrow_forwardQuestion #3: CRISPR has been used to cure an individual from sickle cell. Below is a Sanger electropherogram of a sequence from a patient without sickle cell and one with sickle cell. Sequence from a normal individual mmmm Sequence from the diseased individual G T GIIC A GC A Se SCIENCEphe A G A SCIENCE SCIENCEphoto G a) Where is the change in the sequence and what is the consequence to the protein sequence of this mutation? b) Below is an image of the normal and diseased quaternary hemoglobin protein. What is different about the protein shape and why does that structure have a huge impact on its function (please name the function!)? Adult haemogBRAR G G G G A G Sickle Cell haemoglobin S Structure a s RARY COLIBRARY c) If you were to use CRISPR to modify the genome of a diseased individual, to which nucleotides might you design your guide RNA? Why? d) RNA Seq is used to determine off-target effects of Cas9 cleavage. Why is this an appropriate tool to determine these effects? e) Data on…arrow_forward. ___________________________ factors are proteins thatregulate or initiate RNA synthesis by binding directly orindirectly to specific DNA sequences called responseelements.arrow_forward
- QUESTION 8 Consider the pathway for the synthesis of the amino acid arginine in Neurospora: ARG-E → citrulline ARG-F ARG-H ornithine argininosuccinate arginine Mutant strains of Neurospora may carry one or more mutations. Neurospora mutant strain b is grown on minimal media plus supplements as shown. Growth is shown by (+) and no growth is shown by (o). Supplements ornithine mutant nothing citrulline arginino- succinate arginine strain What can you conclude from these data? O Strain a has only one mutation and it is in ARG-E. O Strain b has only one mutation and it is in ARG-H. O Strain a has a mutation in ARG-F and strain b has a mutation in ARG-E. O Strain b has mutations in ARG-E, ARG-F, and ARG-H. + +arrow_forwardQuestion 49 The Sanger method of DNA sequencing follows the principle of complementarity just like in the replication process. A) True B) Falsearrow_forwardQuestion 1A Explain why DNA replication requires RNA primers. What do they do and why do they have to be removed?arrow_forward
- Question 20 : The restriction map of the plasmid pSC48 is presented below. The numbers in parentheses correspond to the position of the respective restriction site with respect to position (0). (0) BamHI ЕcoRI (4640) (620) PSC48 5200 pb BamHI (1130) Smal ЕcoRI (3420) (1350) The plasmid has been digested by one or several restriction enzyme(s). The digestion products have been analysed by agarose gel electrophorese. (see figure below). 1 2 10,000 8,000 6,000 5,000 4,000 3,000 2,500 –2,000 - 1,500 -1,000 -750 - 500 -250 1. Digestion products 2. Size marker base pairs Which restrcition enzyme(s) have been used for the digestion? (Indicate the right answer) А- ЕсoRI. B- EcoRI/BamHI. C- EcoRI/Smal. D- BamHI. E- Smal. | |arrow_forward_____________ are the linear sequences of nucleotides in an organism’s genetic information.arrow_forwardQuestion 47 (a) Use the following figure to determine the changes to the amino acids that correspond to the normal and mutated DNA sequences. Normal DNA sequence: 3 CAT TCA AAC ATT 5 Mutated DNA sequence: 3 CAT AGT GAG GTC 5 (Hint: Write the mRNA first, then identify the amino acids.) (b) What type of mutation is shown? First base of codon U C A G U UUU UUC UUA UUG CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC GUA GUG UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA Met ACG GCU GCC GCA GCG Phe -Leu Leu lle Second base of codon A C Val -Ser Pro Thr Ala UAUTYT туг UAC UAA UAG CAU CAC CAA CAG. His Gin AAU AAC Asn AAA GAUT GAC GAA GAG Asp G AAGLYS AGG. GGU GGC GGA GGG Glu UGU UGC UGA UGG CGU CGC CGA CGG Bys A Trp G U C -Arg AGU AGC AGA аса за Ser Arg U C Gly AG A U C A G U C A G Third base of codonarrow_forward
- Question 1: The restriction enzyme Sau3A recognizes the sequence 5'-/GATC-3' and cleaves on the 5' side of the G as indicated by the slash. Since the top and bottom strands of this restriction site are identical in the 5'-3' direction, only one strand of the site needs to be shown. The single-stranded ends produced by Sau3A cleavage are identical to those produced by BamHI, which recognizes the sequence 5'-G/GATCC-3' and cuts at the location of the slash. This means that an end cut by one enzyme can be ligated onto an end cut by the other enzyme by DNA Ligase. Given this information, answer the following questions: A) Write out both strands of the recognition sequences for BamHI and Sau3A, show what they would look like after they are cut, and what the DNA would look like after you ligate a BamHI site to a Sau3A site. Make sure that it is easy to follow the individual pieces of DNA. You may want to use two different colors to keep them separate. B) What fraction of BamHI sites can be…arrow_forward(Thanks for the help if you can break down the steps and provide drawings(needed) during the answer since i am trying to see if i have solved this in the right way not copy :) Let's consider a guanine - cytosine base pair, as observed in B- DNA. How does the anticancer medication cisplatin binds to such a base pair, and how many base pairs generally attach to a single Pt? Make sure to draw the resulting coordinationarrow_forwardQuestion 47 The supercoiled DNA can either be positively or negatively supercoiled. A) True B) Falsearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
BiochemistryBiochemistryISBN:9781319114671Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.Publisher:W. H. Freeman
Lehninger Principles of BiochemistryBiochemistryISBN:9781464126116Author:David L. Nelson, Michael M. CoxPublisher:W. H. Freeman
Fundamentals of Biochemistry: Life at the Molecul...BiochemistryISBN:9781118918401Author:Donald Voet, Judith G. Voet, Charlotte W. PrattPublisher:WILEY
BiochemistryBiochemistryISBN:9781305961135Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougalPublisher:Cengage Learning
BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning
Fundamentals of General, Organic, and Biological ...BiochemistryISBN:9780134015187Author:John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. PetersonPublisher:PEARSON
Biochemistry
Biochemistry
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:W. H. Freeman
Lehninger Principles of Biochemistry
Biochemistry
ISBN:9781464126116
Author:David L. Nelson, Michael M. Cox
Publisher:W. H. Freeman
Fundamentals of Biochemistry: Life at the Molecul...
Biochemistry
ISBN:9781118918401
Author:Donald Voet, Judith G. Voet, Charlotte W. Pratt
Publisher:WILEY
Biochemistry
Biochemistry
ISBN:9781305961135
Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal
Publisher:Cengage Learning
Biochemistry
Biochemistry
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Cengage Learning
Fundamentals of General, Organic, and Biological ...
Biochemistry
ISBN:9780134015187
Author:John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Publisher:PEARSON
Bacterial Endospore Formation -Biology Pundit; Author: Biology Pundit;https://www.youtube.com/watch?v=6_sinRhE8zA;License: Standard YouTube License, CC-BY
Taxonomy of Bacteria: Identification and Classification; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8IJRzcPC9wg;License: Standard YouTube License, CC-BY