DNA gyrase is a type of Type Il topoisomerase found in bacteria that induces negative supercoiling. A bacterial chromosome that has a twist of 200 and a writhe of -2 is acted upon by DNA gyrase. What is the resulting twist and writhe after a single DNA gyrase reaction? OA) twist: 200; writhe: -4 OB) twist: 201; writhe: -2 OC) twist: 199; writhe: -2 OD) twist: 200; writhe: 0 Q3 Lk Number what is the Lk number of the resulting DNA molecule? Enter your answer here

DNA gyrase is a type of Type Il topoisomerase found in bacteria that induces negative supercoiling. A bacterial chromosome that has a twist of 200 and a writhe of -2 is acted upon by DNA gyrase. What is the resulting twist and writhe after a single DNA gyrase reaction? OA) twist: 200; writhe: -4 OB) twist: 201; writhe: -2 OC) twist: 199; writhe: -2 OD) twist: 200; writhe: 0 Q3 Lk Number what is the Lk number of the resulting DNA molecule? Enter your answer here

Biology Today and Tomorrow without Physiology (MindTap Course List)

5th Edition

ISBN:9781305117396

Author:Cecie Starr, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Christine Evers, Lisa Starr

Chapter10: Biotechnology

Section: Chapter Questions

Problem 1CT

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All are correct about DNA gyrase in E. coli EXCEPT: It works to remove positive supercoiling introduced by the DnaB protein (helicase). It is a topoisomerase that hydrolyzes ATP during its reaction mechanism. Its mechanism involves the breaking of a single phosphoester bond in one strand of dsDNA. It works to relieve supercoiling in DNA to overcome the torsion stress imposed upon unwinding.
Supercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?
What distinguishes topoisomerase type I from topoisomerase type II (“gyrase”) in E. coli bacteria? Both types of topoisomerases participate in the formation of nucleosomes. Topoisomerase type I cleaves one strand of a DNA duplex to remove DNA supercoils, and topoisomerase type II cleaves both strands of the DNA duplex when introducing negative supercoils into the molecule. Topoisomerase type I works to replicate DNA and topoisomerase type II works to transcribe RNA. Topoisomerase type I works on double-stranded DNA, and topoisomerase type II ("gyrase") works on single-stranded DNA. Topoisomerase type I adds negative DNA supercoils, and topoisomerase type II removes them.
Hundreds of DNA double strand breaks are created by SPO11, a topoisomerase type II like protcin. (i) Draw and explain how SPO11 can generate double strand breaks.
If deoxyribonucleotides that lack the 3’-OH groups are added during the replication process, what do you expect will occur? Describe what happens when a nonsense mutation is introduced into the gene encoding transposase within a transposon. A pure culture of an unknown bacterium was streaked onto plates of a variety of media. You notice that the colony morphology is strikingly different on plates of minimal media with glucose compared to that seen on trypticase soy agar plates. How can you explain these differences in colony morphology?
#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
a) A plasmid DNA in bacteria has a length of 14,000 bp and an Lk of 1300. Calculate the superhelical density o for this plasmid. Show your work for partial credit, round to one digit after the decimal point. b) You use a Type II topoisomerase to change the linking number of this plasmid to 1310. How many turnovers must the topoisomerase perform? Is this resulting plasmid underwound or overwound?
A mixture of four a-[*p]-labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular- weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribo- nucleoside triphosphates was added, with the results shown in the graph.
A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. Incorporation of radioactivity, cpm Excess (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?
A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?
For the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’
It is possible to "reconstitute" nucleosomes by mixing DNA and his- tone octamers in 2M NaCl and then dialyzing to low salt. When such experiments were carried out using a specific 208-bp fragment of sea urchin DNA, the following results were obtained: Digestion of the product with micrococcal nuclease gave quantitative production of 146-bp DNA. Upon cleavage of this DNA with a restriction nucle- ase having a single site in the fragment, several sharp bands were obtained, with sizes as follows: 29 bp, 39 bp, 107 bp, 117 bp. How would you interpret these data in terms of nucleosome positioning on this DNA?