Concept explainers
(a)
To determine: The situation in which a protein has multiple subunits, each with a single ligand-binding site. Binding to one site decreases the binding affinity of other sites for the ligand.
Introduction:
The cooperative binding of oxygen to the hemoglobin was first described by Archibald Hill in 1910. The cooperative binding of oxygen to Hb is a form of allosteric binding in which the binding of one ligand (oxygen) affects the binding affinities of other remaining binding sites of the receptor molecule. The measure of the degree of cooperativity is denoted by nH, the Hill coefficient. If nH is equal to 1, the binding of ligand is not cooperative, if nH is greater than 1, the ligand binding indicates positive cooperativity, and if nH is less than 1 the ligand binding indicates the negative cooperativity. The plotting of hill equations on a plot of log Y versus log L is called a Hill plot.
(b)
To determine: The situation in which a protein is a single polypeptide with two ligand-binding site, each having a different affinity for the ligand.
Introduction:
The cooperative binding of oxygen to the hemoglobin was first described by Archibald Hill in 1910. The cooperative binding of oxygen to Hb is a form of allosteric binding in which the binding of one ligand (oxygen) affects the binding affinities of other remaining binding sites of the receptor molecule. The measure of the degree of cooperativity is denoted by nH, the Hill coefficient. If nH is equal to 1, the binding of ligand is not cooperative, if nH is greater than 1, the ligand binding indicates positive cooperativity, and if nH is less than 1 the ligand binding indicates the negative cooperativity. The plotting of hill equations on a plot of log Y versus log L is called a Hill plot.
(c)
To determine: The situation in which a protein is a single polypeptide with a single ligand-binding site. As purified, the protein preparation is heterogenous, containing some protein molecules that are partially denatured and thus have a lower binding affinity for the ligand.
Introduction:
The cooperative binding of oxygen to the hemoglobin was first described by Archibald Hill in 1910. The cooperative binding of oxygen to Hb is a form of allosteric binding in which the binding of one ligand (oxygen) affects the binding affinities of other remaining binding sites of the receptor molecule. The measure of the degree of cooperativity is denoted by nH, the Hill coefficient. If nH is equal to 1, the binding of ligand is not cooperative, if nH is greater than 1, the ligand binding indicates positive cooperativity, and if nH is less than 1 the ligand binding indicates the negative cooperativity. The plotting of hill equations on a plot of log Y versus log L is called a Hill plot.
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Chapter 5 Solutions
SAPLINGPLUS FOR PRINCIPLES OF BIOCHEMIS
- 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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