Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 10, Problem 18P
Following restriction digestion, DNA fragments produced by digestion with certain enzymes have “sticky ends,” while fragments produced by digestion using other enzymes have “blunt ends.” Distinguish the meaning of these two terms.
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Restriction endonucleases are bacterial enzymes that cleave duplex (double-stranded) DNA at specific nucleotide sequences. The mode of replication of the animal virus SV40 has been investigated by using restriction endonucleases that cleave SV40 DNA into a number of unique segments. Like most viruses, SV40 DNA is circular.
The map positions of the 11 fragments produced by a pair of restriction endonucleases are shown on the next page. Immediately following a 5 or 10 minute pulse of radioactively labeled thymidine, labeled SV40 molecules that have completed replication during the pulse are isolated. These newly replicated DNA molecules are digested by the restriction endonucleases and the resulting fragments are analyzed for the relative amounts of pulse label they contain. The results are in the table below. Assume that at the time the label was added there was a random population of replicating SV40 DNA molecules in all possible stages of synthesis.
From the information given below,…
When circular DNA is sequenced, the nucleotide base pairs are numbered starting from a fixed position on the DNA, all the way around, usually in a clockwise manner. a DNA molecule that is 3133 base pairs long is digested with RsaI restriction enzyme recognition sites at base numbers 366, 1534, and 2207. What are the sizes of the DNA fragments that will be produced after the DNA is digested with RsaI?
Consider the following image of a stained agarose gel representing the separation of a plece of DNA that has been digested by a series of restriction
enzymes. Assume no digest fragment is of identical size to another digest fragment.
nucleotide pairs (kb)
8
S
45
3.5
AA
size
Hindill +
markers EcoRI Hindill EcoRI
-
1st attempt
Based on all data presented in the stained agarose gel, which of the following statements is correct?
Choose one:
OA. The original piece of DNA was circular and was cut by EcoRI twice and Hindill only once.
OB. The original piece of DNA was linear and was cut by EcoRI twice and Hindill only once.
OC. The original piece of DNA was circular and was cut by EcoRI once and not at all by Hind!!!.
Chapter 10 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 10 - Define the following terms as described in this...Ch. 10 - 2. Using sickle cell disease as an example,...Ch. 10 -
3. Compare and contrast the contributions of...Ch. 10 - Why do differences in protein electrophoretic...Ch. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - 8. Wildtype βglobin protein is composed of amino...Ch. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - 11. How is an autoradiograph produced from a...Ch. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - The family represented in the pedigree and...Ch. 10 - Suppose the mating couple (I-1 and I-2) shown in...Ch. 10 - What are restriction endonucleases, and why are...Ch. 10 - 18. Following restriction digestion, DNA fragments...Ch. 10 - 19. The doublestranded DNA sequence below is part...Ch. 10 - 20. Restriction enzymes recognize specific...Ch. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - 25. A second strain of dwarf plants has a...Ch. 10 - During gel electrophoresis of linear DNA...Ch. 10 - Prob. 27PCh. 10 - 28. In molecular biology, restriction...Ch. 10 - A complete plant gene containing four introns and...Ch. 10 - Prob. 30PCh. 10 - The map below illustrates three alleles in a...Ch. 10 - Prob. 32PCh. 10 - 33. Northern blot analysis is performed on mRNA...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- A new restriction enzyme is discovered. It recognizes a 6 base pair pallindromic sequence. The first three bases of the target sequence are are given below with the cut identified by ^. Answer the following questions: Target sequence: 5' G^TA _ _ _ 3'3' _ _ _ _ _ _ 5' 1) Complete the target sequence with all missing bases. Also include the position of the cut in the lower sequence(^) 2) What kind of clevage does the enzyme use?arrow_forwardNucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?arrow_forwardA 10 kb DNA fragment digested with the restriction endonuclease EcoRI yields fragments of 4 kb and 6 kb. When the 10 kb fragment is digested with BamHI, three fragments of 1, 3.5 and 5.5 kb are generated. Digestion with both enzymes yields four fragments of 0.5, 1, 3 and 5.5 kb. Draw the restriction map for the 10 kb fragment based on the data. Label the cut sites for the two enzymes, and the lengths between the cut sites.arrow_forward
- Using the “target DNA” sequence provided in the attached image, what sizes of inserts can be prepared using HindIII digestion? Which one contains the coding sequence?arrow_forwardA DNA strand was sequenced using the Sanger method (https://www.youtube.com/watch?v=KTstRrDTmWI). The reaction tube contained the DNA strand, fluorescently labelled dideoxynucleotide triphosphates (ddATP – yellow, ddGTP – green, ddCTP – blue, ddTTP - red), deoxynucleotide triphosphates, DNA polymerase, or its Klenow fragment. Synthesis of DNA is allowed to proceed, and the results are shown on the right: 15 14 13 12 11 10 (a) What is the sequence of the copy and the template strands? (b) If the template strand were in the 5'-3' direction, what will be the sequence of the DNA copy? Nucleotide Lengtharrow_forwardA 12 kb linear DNA fragment is subject to single or double RE digest and agarose gelelectrophoresis, to yield the gel profile shown below. The first lane contains the size marker(M).a) Explain how the name of the enzyme EcoRI is derived.b) How many sites are there for EcoRI and PvuII respectively on this DNA fragment?c) Use the sizes of the DNA bands on the gel to compile a restriction enzyme map of the DNAfragment. Indicate the positions of the restriction enzymes sites for EcoRI and PvuII on themap.arrow_forward
- A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HindIII (recognition sequence is AAGCTT)arrow_forwardDescribe the possible outcome of a PCR experiment in which (a) there is a single-stranded break in the target DNA sequence, which is present in only one copy in the starting sample, and (b) there is a doublestranded break in the target DNA sequence, which is present in only one copy in the starting sample.arrow_forwardAfter restriction enzymes cut, they contain unpaired bases. Type II restriction enzymes leave ends that may be 5' overhanging, 3' overhanging, or blunt. In all cases each end is left with a 3' OH and a 5' phosphate. All blunt ends, and any complementary overhanging ends may be re-ligated with T4 DNA ligase, as long as at least one 5'- phosphate is present. In the tables below G^AATTC means that the end after cutting with enzyme will be: -----G 3' -----CTTAA 5' GTGCA^C means that the end will be: -----GTGCA 3' -----C 5' Which RE’s from table below have a 5’ overhang? Which ones have a 3’ Overhang? AccI GT^CGAC BamHI G^GATCC ClaI AT^CGAT NsiI ATGCA^T PstI CTGCA^G BglII A^GATCT TaqI T^CGAarrow_forward
- A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HpaII (recognition sequence is CCGG)arrow_forwardIf restriction endonucleases are produced by bacteria within a host, why don’t these enzymes chew up the genomic DNA of their host? What is the role of DNA methyltransferase in this? Indicate the answerarrow_forwardA geneticist is interested in determining the locations of methylated cytosines within a fragment of DNA. She treats some copies of the fragment with sodium bisulfite and leaves some copies untreated. She then sequences the treated and untreated copies of the fragment and obtains the following results. Give the original sequence of the DNA fragment and indicate the locations of methylated cytosines. Sequence without treatment: — AATTGCCCGATCGATTAAGCCA — Sequence with treatment: — AATTGTTTGATCGATTAAGCTA —arrow_forward
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