Assignment : a. In a particular negative supercoiled plasmid, pTOP, there is 11.

Question

Assignment :

a. In a particular negative supercoiled plasmid, pTOP, there is 11.7 bp per gain in the DNA helix. After relaxation of this plasmid with topoisomerase I there is 10.4 bp per recovery. Are helix windings removed or added to the plasmid during this reaction?

b. A particular topoisomer of the plasmid pTOP has Twist (Tw) = 50 and Writhe (Wr) = 4. What is the linker number (Lk) of the topoisomer?

c. The "relayed" pTOP plasmid is incubated with gyrase and ATP in a suitable buffer. What will happen to the plasmid's Lk and Wr if we assume that Tw remains unchanged?

d. Describe briefly the role of the gyras during the replication in bacteria. A laboratory works with four different E. coli strains that have the following characteristics (genotypes) relative to the DNA gyras:

• A strain that is wild type (wt), that is, normal.

• A strain resistant to novobiocin, which is an inhibitor of the gyrogenic (gyrogenic) enzymatic activity. That is, novobiocin does not affect the activity of gyras in this strain.

• A strain that has a temperature-sensitive mutation in the gyraset (gyrasets), which causes the gyrous to function normally at 37 ° C but has reduced activity at 42 ° C.

• A strain that is both resistant to novobiocin and has the temperature-sensitive gyrase mutation (a double mutant gyraser, ts).

e. The following experiment is added to the laboratory: The four strains A, B, C and D grow at 37 ° C or at 42 ° C in the presence of varying concentrations of the gyrozic inhibitor novobiocin according to the table below. The concentration of novobiocin is measured in g / ml. The results in the table indicate the amount of DNA synthesis measured in percent after 25 minutes. Based on these results, determine the genotype of the four tribes labeled A, B, C and D in the table. Use the black table below. Use the notation wt, gyraseres, gyrasets or gyraseres, ts.

Temperatur Novobiocin (g/ml)0 Stamme A Stamme B Stamme C Stamme D 37 °C 10 100 60 100 42 °C 10 45 50 100 50 40 100 100 100 100 0 45 100 100 40 100 10 100 25 100 Svartabel E. coli stamme Genotvpe

Explanation / Answer

a) Topoisomearse I removes supercoiling, so the helix windings are removed

b)Linking number (Lk) = twist (Tw) + writhe (Wr)

Lk = 50 + 4 = 54

c) DNA is incubated with gyrase then the DNA will be relaxed. In this case linking number will be egual to twist number and writhe will be 0

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