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vazorg [7]
3 years ago
11

Identify the true statements regarding alpha-1,6 linkages in glycogen.

Chemistry
1 answer:
frozen [14]3 years ago
3 0

Answer:

a. New alpha- 1,6 linkages can only form if the branch has a free reducing end

b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages

c. At least four glucose residues separate alpha-1,6 linkages

e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.

Explanation:

Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.

The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.

The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.

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A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
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3 0
3 years ago
A mixture that has two or more substances that are spread out evenly is called _____
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D

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What is the average kinetic energy of 1 mole of a gas that is at 100 degrees Celsius?
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3 0
3 years ago
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