question 6 st your set of assigned enzymes andthe expected fragment sizes here.Base Pairs1.Using the gel picture to the left, show where your digestion of the paper1,000 -900 -cutting model showing cut fragments would appear. The linear uncut DNA is 640base pairs in length.800 -700 -600 -500 -400 -2.Explain why the restriction endonucleases BglII (A^GATCT) and Xbal(T^CTAGA) will not recognize and cut the restriction site of the other.300 -200 -100 -3.What are the last three of six nucleotides (bases) of the palindromic site for50 -the enzyme Ndel if the first three are C AT?4.Why would you NOT expect a restriction endonuclease to exist that would recognize the site AAGGAA?5.If the hypothetical restriction endonuclease CutVI were to recognize and cleave a circular piece of DNA at 6different sites, how many fragments of DNA would result?6.Answer the above question for CutVI if the starting DNA were linear instead of circular.7.If the electrical leads were reversed by mistake (red connected to black), what would be the result?The gel would run backwards, with DNA moving the other direction.The DNA would separate as usual, only now the larger pieces would travel farther and smallerpieces shorter distances.The DNA would not move at all.A.В.С.8.Explain your answer choice for the previous question in rational terms.DNA ladderPaper DNA digest

Restriction enzymes are enzymes that identify specific sequences in DNA and cut DNA at or near these…

Answered: Answer the above question for CutVI if…

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Biochemistry

Answer the above question for CutVI if the starting DNA were linear instead of circular.

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 1P: Semiconservative or Conservative DNA Replication If 15N-Iabeled E. coli DNA has a density of 1.724...

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Homologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesnâ€™t the mismatch repair system of E. coli eliminate these mismatches?
Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.
Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.
Chemical Mutagenesis of DNA Bases Show the nucleotide sequence changes that might arise in a dsDNA (coding strand segment GCTA) upon mutagenesis with (a) HNO2, (b) bromouracil, and (C) 2-aminopurine.
The Enzymatic Activities of DNA Polymerase I (a) What are the respective roles of the 5 -exonudease and 3 -exonuclease activities of DNA polymerase I? (b) What might be a feature of an E. coli strain that lacked DNA polymerase I 3 -exonuclease activity?
Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
During proofreading, which of the following enzymes reads the DNA? primase topoisomerase DNA pol helicase
DNA Profiles as Tools for Identification A PCR-based paternity test is conducted using STRs that consistently produce a unique DNA fragment pattern from a single chromosome. Examining the results of the following Southern blot, which male(s) can be excluded as the father of the child? Which male(s) could be the father of the child?
Multiple Replication Forks in E. coli I Assuming DNA replication proceeds at a rate of 750 base pairs per second, calculate how long it will take to replicate the entire E. coli genome. Under optimal conditions, E. coli cells divide every 20 minutes. What is the minimal number of replication forks per E. coli chromosome in order to sustain such a rate of cell division?
Analyzing Cloned Sequences A base change (A to T) is the mutational event that created the mutant sickle cell anemia allele of beta globin. This mutation destroys an MstII restriction site normally present in the beta globin gene. This difference between the normal allele and the mutant allele can be detected with Southern blotting. Using a labeled beta globin gene as a probe, what differences would you expect to see for a Southern blot of the normal beta globin gene and the mutant sickle cell gene?
DNA contains many hydrogen bonds. Are hydrogen bonds stronger or weaker than covalent bonds? What are the consequences of this difference in strength?