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(a)
Interpretation:
It should be determined that whether the given base sequences are sticky or not sticky.
Concept Introduction:
A base is nitrogen containing heterocyclic compound which is found in DNA and RNA.
There are mainly four nitrogen bases found in DNA and they are,
- (1) Adenine
- (2) Guanine
- (3) Cytosine
- (4) Thymine
Short stretches of single stranded DNA are sticky (complementary) to each other. If both ends are cut with the same enzyme, the sticky ends will stick together by complementary base pairing, forming hydrogen bonds.
In DNA, Adenine always makes a double bond with thymine (
(b)
Interpretation:
It should be determined that whether the given base sequences are sticky or not sticky.
Concept introduction:
A base is nitrogen containing heterocyclic compound which is found in DNA and RNA.
There are mainly four nitrogen bases found in DNA and they are,
- (1) Adenine
- (2) Guanine
- (3) Cytosine
- (4) Thymine
Short stretches of single stranded DNA are sticky (complementary) to each other. If both ends are cut with the same enzyme, the sticky ends will stick together by complementary base pairing, forming hydrogen bonds.
In DNA, Adenine always makes a double bond with thymine (
(c)
Interpretation:
It should be determined that whether the given base sequences are sticky or not sticky.
Concept introduction:
A base is nitrogen containing heterocyclic compound which is found in DNA and RNA.
There are mainly four nitrogen bases found in DNA and they are,
- (1) Adenine
- (2) Guanine
- (3) Cytosine
- (4) Thymine
Short stretches of single stranded DNA are sticky (complementary) to each other. If both ends are cut with the same enzyme, the sticky ends will stick together by complementary base pairing, forming hydrogen bonds.
In DNA, Adenine always makes a double bond with thymine (
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Chapter 27 Solutions
Fundamentals of General, Organic, and Biological Chemistry, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (8th Edition)
- 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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