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2 years ago

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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.

1 answer:

Otrada [13]2 years 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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$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

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Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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