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Damm [24]
1 year ago
11

Arrange the steps of glycogen degradation in their proper order. Hormonal signals trigger glycogen breakdown. Glucose 6‑phosphat

e undergoes further metabolic processing. Answer Bank Glucose 1‑phosphate is cleaved from the nonreducing ends of glycogen and converted to glucose 6‑phosphate.Blocks consisting of three glucosyl residues are moved by remodeling of α‑1,4‑glycosidic linkages.Glycogen is branched by hydrolysis of α‑1,6‑glycosidic linkages.
Chemistry
1 answer:
Otrada [13]1 year ago
4 0

Answer: Please see answer below

Explanation:

The steps of glycogen degradation is as follows from this order.

--->Hormonal signals trigger glycogen breakdown.

1. Glycogen is (de)branched by hydrolysis of α‑1,6‑glycosidic linkages.

2. Blocks consisting of three glucosyl residues are moved by remodeling of α‑1,4‑glycosidic linkages.

3.[Glucose 1‑phosphate is cleaved from the non reducing ends of glycogen and converted to glucose 6‑phosphate.

--->Glucose 6‑phosphate undergoes further metabolic processing

The  degradation  of Glycogen follows  three steps:

(1) the release of glucose 1-phosphate from glycogen,

(2) the remodeling of the glycogen substrate to permit further degradation, and

(3) the conversion of glucose 1-phosphate into glucose 6-phosphate for further metabolism.

(https://www.ncbi.nlm.nih.gov/books/NBK21190)

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Watch the video to determine which of the following relationships are correct according to Boyle’s law.
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Answer:

Part A

Boyle's Law is given mathematically as

P ∝(1/V) or V ∝(1/P)

Options 4 and 5, if they are properly written.

Part B

At constant temperature, and according to the Boyle's law for an ideal gas,

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

The pressure when the gas occupies a volume of 5.0 L = 40 atm

Part D

The pressure when the gas occupies a volume of 4.5 L = 36 atm

Explanation:

Part A

Boyle's Law states that at constant temperature, the pressure of an ideal gas is inversely proportional to the volume occupied by the gas.

So, mathematically, Boyle's Law is given as

P ∝(1/V) or V ∝(1/P)

Part B

Inverse relationship between two quantities means that the higher the value of one of the quantities go, the lower the value of the other quantity goes and vice versa.

So, at constant temperature, and according to the Boyle's law for an ideal gas.

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

A certain gas occupies a volume of 20 L when the applied pressure is 10 atm, find the pressure when the gas occupies a volume of 5.0 L.

According to Boyle's Law for an ideal gas,

P ∝(1/V)

P = (k/V)

where k is the constant of proportionality

PV = k

Therefore,

P₁V₁ = P₂V₂ = k

P₁ = 10 atm

V₁ = 20 L

P₂ = ?

V₂ = 5.0 L

10 × 20 = P₂ × 5

P₂ = 40 atm

Part D

If a certain gas occupies a volume of 18 L when the applied pressure is 9.0 atm , find the pressure when the gas occupies a volume of 4.5 L

P₁V₁ = P₂V₂ = k

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V₁ = 18 L

P₂ = ?

V₂ = 4.5 L

9 × 18 = P₂ × 4.5

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Functional group and bond hybridization of vanillin
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Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

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Answer:

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Explanation:

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