(a)
Interpretation:
The structure of D-Allose: Epimeric at C-3 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. Anomers are the glycosides or cyclic monosaccharides that are epimers. These are different from each other in the configuration of C-1 (aldoses) or at C-2 if (ketoses).
Answer to Problem 9P
The structure of D-Allose:
Explanation of Solution
When two sugars vary in their configuration around one carbon atom then those sugars are known as epimers of each other.
The structure of D-Allose: Epimeric at C-3 is given below:
(b)
Interpretation:
The structure of D-Altrose: Isomeric at C-2 and C-3 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D-Altrose:
Explanation of Solution
Glucose is an aldose as it contains one
The structure of D-Altrose: Isomeric at C-2 and C-3 is given below:
(c)
Interpretation:
The structure of D-Mannose: Eplmeric at C-2 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D-Mannose:
Explanation of Solution
Anomers are the glycosides or cyclic monosaccharides that are epimers. These are different from each other in the configuration of C-1 (aldoses) or at C-2 if (ketoses). In a pair of anomers, the anomer containing the alkoxy group or hydroxy group is pointing up on the anomeric carbon is known as the ß-anomer.
The structure of D-Mannose: Eplmeric at C-2 is given below:
(d)
Interpretation:
The structure of D- Gulose: Isomeric at C-3 and C-4
should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D- Gulose:
Explanation of Solution
Glucose is an aldose as it contains one aldehyde group per molecule in its acyclic form. Mannose is a sugar of the hexose group. It occurs as a constituent of several natural polysaccharides.
The structure of D- Gulose: Isomeric at C-3 and C-4 is given below:
(e)
Interpretation:
The structure of D-Idose: Isomeric at C-2, C-3, and C-4 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D-Idose:
Explanation of Solution
Glucose is an aldose as it contains one aldehyde group per molecule in its acyclic form. Mannose is a sugar of the hexose group. It occurs as a constituent of several natural polysaccharides.
The structure of D-Idose: Isomeric at C-2, C-3, and C-4 is given below:
(f)
Interpretation:
The structure of D-Galactose: Epimeric at C-4 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D-Galactose:
Explanation of Solution
Glucose is an aldose as it contains one aldehyde group per molecule in its acyclic form. When two sugars vary in their configuration around one carbon atom then those sugars are known as epimers of each other.
The structure of D-Galactose: Epimeric at C-4 is given below:
(g)
Interpretation:
The structure of D-Talose: Isomeric at C-2 and C-4 should be determined.
Concept introduction:
Carbohydrates are a large group of organic compounds that are present in foods, living tissues. These include sugars, cellulose, and starch. These are comprised of hydrogen and oxygen in a similar ratio as in water (2:1). They are usually broken down to release the energy in the animal body. These are of three types: monosaccharides, disaccharides, and polysaccharides.
Answer to Problem 9P
The structure of D-Talose:
Explanation of Solution
Glucose is an aldose as it contains one aldehyde group per molecule in its acyclic form. When two sugars vary in their configuration around one carbon atom then those sugars are known as epimers of each other.
The structure of D-Talose: Isomeric at C-2 and C-4 is given below:
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Chapter 11 Solutions
BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
- 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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