. A closed circular supercoiled DNA is relaxed by treatment withtopoisomerase. No matter how much enzyme is used, or how longthe experiment is run, the experimenter always finds a gel elec-trophoresis pattern indicating some DNA with one, two, and threesuperhelical turns in addition to the relaxed (nicked) circle (see fig-ure). Suggest an explanation for this observation.Nickedcircle+

DNA topoisomerases are enzymes that relieve the topological strains in the DNA by creating single or…

Answered: . A closed circular supercoiled DNA is…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Please explain in detail....
Can you please help with 1c please picture with 1 graph is for question 1a) picture with 4 graphs is for question 1b) 1a) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? 1b) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC…
Homologous recombination in E. coli forms heteroduplex regions of DNA containing mismatched bases. Why are these mismatches not eliminated by the mismatch repair system?
A. Diagram a short single strand of DNA 5’ -AA-GG- 3’. Show the chemical structure of the phosphoribosyl backbone and the attachment point for nucleotides added as “A” or “G”. B. Diagram the product of digestion was a restriction enzyme to cut this sequence between the A and G.
This shows where EcoR1 cuts, where HindIII cuts, and where BstEII cuts. Given that the entire chromosome is 48,502 base pairs (as it says), figure out the fragment sizes you will expect if you cut it with EcoR1, or with HindIII, or with BseEII. Draw a map like the example attached " Show the sizes of each piece. Color code it according to which enzyme cut it.
5'-GCGGTACGTTACGGCTTTACTGACCTGCAGGC-3' A. Convert this to a double strand DNA molecule by writing the complementary sequence. Be sure to correctly label the ends of the double stranded molecule B. Pick one of the two strands and write the sequence of an RNA molecule that could be transcribed from it. Again, you must correctly label the ends of the molecule.
Explain why a DSB-repair homologous recombination can occur between any two DNA molecules that share homology rather than only between two DNA molecules that carry a specific sequence.
None
. Propose a mechanism by which a type II topoisomerase could use the energy of ATP hydrolysis to scan a large DNA molecule and, thereby, to direct that the enzyme will catalyze largely “disentan- gling" reactions (decatenation and unknotting).
Explain why the active site of poly(A) polymerase is much narrower than that of DNA and RNA polymerases.
Can you help solve this problem?

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