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Concept explainers
(a)
To determine: The Lk of a 5000 bp circular duplex DNA molecule with a nick in one strand.
Introduction:
Linking number (Lk) is a topological characteristic of a DNA molecule. It determines the numbers of helical turns within a closed-circular DNA molecule without any supercoiling. It does not vary whether the DNA is deformed or bent. It causes both DNA strands to remain intact.
(b)
To determine: The Lk of a
Introduction:
Linking number does not vary, as long as both strands of DNA molecule remain intact. It defines the number of times in which each strand is underwinds around each other. During the relaxed form, the linkage number (Lk) is equal to the number of turns found in the DNA double helix.
(c)
To determine: The way through which Lk of a 5’000 bp circular DNA duplex when it is in relaxed form, would be affected by the action of a single molecule of E. coli topoisomerase I.
Introduction:
Topoisomers are formed when two forms of a given circular DNA molecule differ only in a topological characteristic. Topoisomerase enzymes can increase or reduce the numbers of DNA underwinding. They can also influence the topological characteristics.
(d)
To determine: The Lk of a 5,000 bp circular DNA duplex molecule after eight enzymatic turnovers by a single molecule of DNA gyrase in the presence of ATP.
Introduction:
DNA topoisomerase causes the reduction in topological stress that is formed during the strand separation within the DNA by helicases.
(e)
To determine: Lk of a 5000 bp circular DNA duplex molecule be after four enzymatic turnovers by a single molecule of bacterial type I topoisomerase.
Introduction:
Type I topoisomerase break one strand of double helical DNA molecule. The unbroken strand is further transferred to the single stranded breaks, and resealed.
(f)
To determine: The Lk of a 5000 bp circular DNA duplex molecule be after binding of one nucleosome core.
Introduction:
When a protein complex is associated for forming a nucleosome, the linking number does not change because the strand of DNA does not break and rejoin.
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Chapter 24 Solutions
SaplingPlus for Lehninger Principles of Biochemistry (Six-Month Access)
- Problem 15 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 4 of 15 4G 54% Done On the following Lineweaver-Burk -1 plot, identify the by dragging the Km point to the appropriate value. 1/V 40 35- 30- 25 20 15 10- T Км -15 10 -5 0 5 ||| 10 15 №20 25 25 30 1/[S] Г powered by desmosarrow_forward1:30 5G 47% Problem 10 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a competitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' s mM¹ with 10 mg pe 20 V' 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 14 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:36 CO Problem 9 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of the catalytic efficiency (specificity constant) given that the enzyme concentration in this experiment is 5.0 μ.Μ. 1 [S] ¨‚ μM-1 1 V sμM-1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| O Гarrow_forwardProblem 11 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a noncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 13 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 8 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of kcat given that the enzyme concentration in this experiment is 5.0 μM. 1 [S] , мм -1 1 V₁ s μM 1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| Гarrow_forward1:33 5G. 46% Problem 12 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without an uncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
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