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Glycogen metabolism qa

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1. Carbohydrateconversion. Write a balanced equation for the formation of
glycogen from galactose.
See answer
2. If a little is good, a lot is better. -Amylose is an unbranched glucose polymer.
Why would this polymer not be as effective a storage form for glucose as
glycogen?
See answer
3. Telltale products. A sample of glycogen from a patient with liver disease is
incubated with orthophosphate, phosphorylase, the transferase, and the
-1,6-glucosidase). The ratio of glucose 1-phosphate to
glucose formed in this mixture is 100. What is the most likely enzymatic
deficiency in this patient?
See answer
4. Excessive storage. Suggest an explanation for the fact that the amount of
glycogen in type I glycogen-storage disease (von Gierke disease) is increased.
See answer
5. A shattering experience. Crystals of phosphorylase a grown in the presence of
glucose shatter when a substrate such as glucose 1-phosphate is added. Why?
6. Recouping an essential phosphoryl. The phosphorylgroup on
phosphoglucomutaseis slowly lost by hydrolysis. Proposea mechanism that
utilizes a known catalytic intermediate for restoring this essential phosphoryl
group. How might this phosphoryldonorbe formed?
See answer
7. Hydrophobia. Why is water excluded from the active site of phosphorylase?
Predict the effect of a mutation that allows water molecules to enter.
See answer
8. Removing all traces. In human liver extracts, the catalytic activity of glycogenin
-amylase (S -
amylase necessary to reveal the glycogenin activity?
See answer
9. Two in one. A single polypeptide chain houses the transferase and debranching
enzyme. Cite a potential advantage of this arrangement.
10. How did they do that? A strain of mice has been developed that lack the
enzyme phosphorylasekinase. Yet, after strenuous exercise, the glycogen stores of
a mouse of this strain are depleted. Explain how this is possible.
See answer
11. Metabolic mutants. Predict the major consequenceof each of the following
mutations:
(a) Loss of the AMP-binding site in muscle phosphorylase.
(b) Mutation of Ser 14 to Ala 14 in liver phosphorylase.
(c) Overexpression of phosphorylase kinase in the liver.
(d) Loss of the gene that encodes inhibitor 1 of protein phosphatase1.
(e) Loss of the gene that encodes the glycogen-targeting subunit of protein
phosphatase1.
(f) Loss of the gene that encodes glycogenin.
See answer
12. More metabolic mutants. Briefly, predict the major consequences of each of the
following mutations affecting glycogen utilization.
(b) Loss of the gene that encodes inhibitor 1 of protein phosphatase1.
(c) Loss of phosphodiesteraseactivity.
See answer
13. Multiple phosphorylation. Protein kinase A activates muscle phosphorylase
kinase by rapidly phosphorylating its
subunits susceptible to the action of protein phosphatase1. What is the
functional significance of the slow
See answer
14. The wrong switch. What would be the consequences to glycogen mobilization
of a mutation in phosphorylase kinase that leads to the phosphorylation -
See answer
MechanismProblem
15. Family resemblance. Proposemechanisms for the two enzymes catalyzing
steps in glycogen debranching based on -
amylase family.
Chapter Integration and Data Interpretation Problems
16. Glycogen isolation 1. The liver is a major storage site for glycogen. Purified
from two samples of human liver, glycogen was either treated or not treated with
-amylase and subsequently analyzed by SDS-PAGE and Western blotting with
the use of antibodies to glycogenin. The results are presented in the adjoining
illustration.
(a) Why are no proteins visible in the lanes without amylase treatment?
(b) What is the effect of treating the sampl -amylase? Explain the results.
(c) List other proteins that you might expect to be associated with glycogen. Why
are other proteins not visible?
See answer
17. Glycogen isolation 2. The gene for glycogenin was transfected into a cell line
that normally stores only small amounts of glycogen. The cells were then
manipulated according to the following protocol, and glycogen was isolated and
analyzed by SDS-PAGE and Western blotting using an antibody to glycogenin
- amylase treatment. The results are presented in the adjoining
illustration.
The protocol:Cells cultured in growth medium and 25 mM glucose (lane 1) were
switched to medium containing no glucose for 24 hours (lane 2). Glucose-starved
cells were re-fed with medium containing 25 mM glucose for 1 hour (lane 3) or 3
-amylase, as indicated, before being loaded on the gel.
(a) Why did the Western analysis producea "smear" that is, the high-molecular-
weight staining in lane 1(-)?
(b) What is the significance of the decrease in high-molecular-weight staining in
lane 2(-)?
(c) What is the significance of the difference between lanes 2(-) and 3(-)?
(d) Suggest a plausible reason why there is essentially no difference between lanes
3(-) and 4(-)?
(e) Why are the bands at 66 kd the same in the lanes treated with amylase, despite
the fact that the cells were treated differently?
Answers
1. Galactose + ATP + UTP + H2O + glycogenn →glycogen n +1 + ADP + UDP +
2 Pi + H+
See question
2. -amylose has only one nonreducing end.
-
amylose molecule. Because glycogen is highly branched, there are many
nonreducing ends per molecule. Consequently, many phosphorylase molecules can
release many glucose molecules per glycogen molecule.
See question
3. The patient has a deficiency of the branching enzyme.
See question
4. The high level of glucose 6-phosphate in von Gierke disease, resulting from the
absence of glucose 6-phosphataseor the transporter, shifts the allosteric
equilibrium of phosphorylated glycogen synthase toward the active form.
5. Glucose is an allosteric inhibitor of phosphorylase a. Hence, crystals grown in
its presence are in the T state. The addition of glucose 1-phosphate, a substrate,
shifts the R T equilibrium toward the R state. The conformational differences
between these states are sufficiently large that the crystal shatters unless it is
stabilized by chemical crosslinks.
See question
6. The phosphoryldonoris glucose 1,6-bisphosphate, which is formed from
glucose 1-phosphate and ATP in a reaction catalyzed by phosphoglucokinase.
See question
7. Water is excluded from the active site to prevent hydrolysis. The entry of water
could lead to the formation of glucose rather than glucose 1-phosphate. A site-
specific mutagenesis experiment is revealing in this regard. In phosphorylase, Tyr
573 is hydrogen bonded to the 2 -OH of a glucose residue. The ratio of glucose 1-
phosphateto glucose productis 9000:1 for the wild-type enzyme, and 500:1 for the
Phe 573 mutant. Model building suggests that a water molecule occupies the site
normally filled by the phenolic OH of tyrosine and occasionally attacks the
oxocarbonium ion intermediate to form glucose.
See question
8. The amylase activity was necessary to remove all of the glycogen from the
glycogenin. Recall that glycogenin synthesizes oligosaccharides of about eight
glucose units, and then activity stops. Consequently, if the glucose residues are not
removed by extensive amylase treatment, glycogenin would not function.
9. The substrate can be handed directly from the transferase site to the debranching
site.
See question
10. During exercise, [ATP] falls and [AMP] rises. Recall that AMP is an allosteric
activator of glycogen phosphorylase b. Thus, even in the absence of covalent
modification by phosphorylase kinase, glycogen is degraded.
See question
11. (a) Muscle phosphorylase b will be inactive even when the AMP level is high.
Hence, glycogen will not be degraded unless phosphorylase is converted into the a
form by hormone-induced or Ca2+-induced phosphorylation.
(b) Phosphorylase b cannot be converted into the much more active a form. Hence,
the mobilization of liver glycogen will be markedly impaired.
(c) The elevated level of the kinase will lead to the phosphorylation and activation
of glycogen phosphorylase.
Because glycogen will be persistently degraded, little glycogen will be present in
the liver.
(d) Protein phosphatase1 will be continually active. Hence, the level of
phosphorylase b will be higher than normal, and glycogen will be less readily
degraded.
(e) Protein phosphatase1 will be much less effective in dephosphorylating
glycogen synthase and glycogen phosphorylase. Consequently, the synthase will
stay in the less active b form, and the phosphorylase will stay in the more active a
form. Both changes will lead to increased degradation of glycogen.
(f) The absence of glycogenin will block the initiation of glycogen synthesis. Very
little glycogen will be
synthesized in its absence.
See question
12. is always produced. Glycogen
degradation always occurs, and glycogen synthesis is always inhibited.
(b) Glycogen phosphorylasecannot be covalently activated. Glycogen degradation
is always inhibited; nothing can remain phosphorylated. Glycogen synthesis is
always active; nothing can remain phosphorylated.
(c) Phosphodiesterasedestroys cAMP. Therefore, glycogen degradation would
always be active and glycogen synthesis would always be inhibited.
See question
13. of phosphor-ylasekinase serves
to prolong the degradation of glycogen. The kinase cannot be deactivated until its
that the kinase and, in turn, phosphorylase stay active for an extended interval.
See question
14.
degradation. Subsequent
substrates for protein phosphatase. Thus, if the
and little glycogen degradation would take place.
15.

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Glycogen metabolism qa

  • 1. 1. Carbohydrateconversion. Write a balanced equation for the formation of glycogen from galactose. See answer 2. If a little is good, a lot is better. -Amylose is an unbranched glucose polymer. Why would this polymer not be as effective a storage form for glucose as glycogen? See answer 3. Telltale products. A sample of glycogen from a patient with liver disease is incubated with orthophosphate, phosphorylase, the transferase, and the -1,6-glucosidase). The ratio of glucose 1-phosphate to glucose formed in this mixture is 100. What is the most likely enzymatic deficiency in this patient? See answer 4. Excessive storage. Suggest an explanation for the fact that the amount of glycogen in type I glycogen-storage disease (von Gierke disease) is increased. See answer 5. A shattering experience. Crystals of phosphorylase a grown in the presence of glucose shatter when a substrate such as glucose 1-phosphate is added. Why? 6. Recouping an essential phosphoryl. The phosphorylgroup on phosphoglucomutaseis slowly lost by hydrolysis. Proposea mechanism that utilizes a known catalytic intermediate for restoring this essential phosphoryl group. How might this phosphoryldonorbe formed? See answer 7. Hydrophobia. Why is water excluded from the active site of phosphorylase? Predict the effect of a mutation that allows water molecules to enter. See answer 8. Removing all traces. In human liver extracts, the catalytic activity of glycogenin -amylase (S - amylase necessary to reveal the glycogenin activity? See answer 9. Two in one. A single polypeptide chain houses the transferase and debranching enzyme. Cite a potential advantage of this arrangement.
  • 2. 10. How did they do that? A strain of mice has been developed that lack the enzyme phosphorylasekinase. Yet, after strenuous exercise, the glycogen stores of a mouse of this strain are depleted. Explain how this is possible. See answer 11. Metabolic mutants. Predict the major consequenceof each of the following mutations: (a) Loss of the AMP-binding site in muscle phosphorylase. (b) Mutation of Ser 14 to Ala 14 in liver phosphorylase. (c) Overexpression of phosphorylase kinase in the liver. (d) Loss of the gene that encodes inhibitor 1 of protein phosphatase1. (e) Loss of the gene that encodes the glycogen-targeting subunit of protein phosphatase1. (f) Loss of the gene that encodes glycogenin. See answer 12. More metabolic mutants. Briefly, predict the major consequences of each of the following mutations affecting glycogen utilization. (b) Loss of the gene that encodes inhibitor 1 of protein phosphatase1. (c) Loss of phosphodiesteraseactivity. See answer 13. Multiple phosphorylation. Protein kinase A activates muscle phosphorylase kinase by rapidly phosphorylating its subunits susceptible to the action of protein phosphatase1. What is the functional significance of the slow See answer 14. The wrong switch. What would be the consequences to glycogen mobilization of a mutation in phosphorylase kinase that leads to the phosphorylation - See answer MechanismProblem
  • 3. 15. Family resemblance. Proposemechanisms for the two enzymes catalyzing steps in glycogen debranching based on - amylase family. Chapter Integration and Data Interpretation Problems 16. Glycogen isolation 1. The liver is a major storage site for glycogen. Purified from two samples of human liver, glycogen was either treated or not treated with -amylase and subsequently analyzed by SDS-PAGE and Western blotting with the use of antibodies to glycogenin. The results are presented in the adjoining illustration. (a) Why are no proteins visible in the lanes without amylase treatment? (b) What is the effect of treating the sampl -amylase? Explain the results. (c) List other proteins that you might expect to be associated with glycogen. Why are other proteins not visible? See answer 17. Glycogen isolation 2. The gene for glycogenin was transfected into a cell line that normally stores only small amounts of glycogen. The cells were then manipulated according to the following protocol, and glycogen was isolated and analyzed by SDS-PAGE and Western blotting using an antibody to glycogenin - amylase treatment. The results are presented in the adjoining illustration. The protocol:Cells cultured in growth medium and 25 mM glucose (lane 1) were switched to medium containing no glucose for 24 hours (lane 2). Glucose-starved cells were re-fed with medium containing 25 mM glucose for 1 hour (lane 3) or 3 -amylase, as indicated, before being loaded on the gel. (a) Why did the Western analysis producea "smear" that is, the high-molecular- weight staining in lane 1(-)? (b) What is the significance of the decrease in high-molecular-weight staining in lane 2(-)? (c) What is the significance of the difference between lanes 2(-) and 3(-)? (d) Suggest a plausible reason why there is essentially no difference between lanes 3(-) and 4(-)? (e) Why are the bands at 66 kd the same in the lanes treated with amylase, despite the fact that the cells were treated differently? Answers
  • 4. 1. Galactose + ATP + UTP + H2O + glycogenn →glycogen n +1 + ADP + UDP + 2 Pi + H+ See question 2. -amylose has only one nonreducing end. - amylose molecule. Because glycogen is highly branched, there are many nonreducing ends per molecule. Consequently, many phosphorylase molecules can release many glucose molecules per glycogen molecule. See question 3. The patient has a deficiency of the branching enzyme. See question 4. The high level of glucose 6-phosphate in von Gierke disease, resulting from the absence of glucose 6-phosphataseor the transporter, shifts the allosteric equilibrium of phosphorylated glycogen synthase toward the active form. 5. Glucose is an allosteric inhibitor of phosphorylase a. Hence, crystals grown in its presence are in the T state. The addition of glucose 1-phosphate, a substrate, shifts the R T equilibrium toward the R state. The conformational differences between these states are sufficiently large that the crystal shatters unless it is stabilized by chemical crosslinks. See question 6. The phosphoryldonoris glucose 1,6-bisphosphate, which is formed from glucose 1-phosphate and ATP in a reaction catalyzed by phosphoglucokinase. See question 7. Water is excluded from the active site to prevent hydrolysis. The entry of water could lead to the formation of glucose rather than glucose 1-phosphate. A site- specific mutagenesis experiment is revealing in this regard. In phosphorylase, Tyr 573 is hydrogen bonded to the 2 -OH of a glucose residue. The ratio of glucose 1- phosphateto glucose productis 9000:1 for the wild-type enzyme, and 500:1 for the Phe 573 mutant. Model building suggests that a water molecule occupies the site normally filled by the phenolic OH of tyrosine and occasionally attacks the oxocarbonium ion intermediate to form glucose. See question 8. The amylase activity was necessary to remove all of the glycogen from the glycogenin. Recall that glycogenin synthesizes oligosaccharides of about eight
  • 5. glucose units, and then activity stops. Consequently, if the glucose residues are not removed by extensive amylase treatment, glycogenin would not function. 9. The substrate can be handed directly from the transferase site to the debranching site. See question 10. During exercise, [ATP] falls and [AMP] rises. Recall that AMP is an allosteric activator of glycogen phosphorylase b. Thus, even in the absence of covalent modification by phosphorylase kinase, glycogen is degraded. See question 11. (a) Muscle phosphorylase b will be inactive even when the AMP level is high. Hence, glycogen will not be degraded unless phosphorylase is converted into the a form by hormone-induced or Ca2+-induced phosphorylation. (b) Phosphorylase b cannot be converted into the much more active a form. Hence, the mobilization of liver glycogen will be markedly impaired. (c) The elevated level of the kinase will lead to the phosphorylation and activation of glycogen phosphorylase. Because glycogen will be persistently degraded, little glycogen will be present in the liver. (d) Protein phosphatase1 will be continually active. Hence, the level of phosphorylase b will be higher than normal, and glycogen will be less readily degraded. (e) Protein phosphatase1 will be much less effective in dephosphorylating glycogen synthase and glycogen phosphorylase. Consequently, the synthase will stay in the less active b form, and the phosphorylase will stay in the more active a form. Both changes will lead to increased degradation of glycogen. (f) The absence of glycogenin will block the initiation of glycogen synthesis. Very little glycogen will be synthesized in its absence. See question 12. is always produced. Glycogen degradation always occurs, and glycogen synthesis is always inhibited. (b) Glycogen phosphorylasecannot be covalently activated. Glycogen degradation is always inhibited; nothing can remain phosphorylated. Glycogen synthesis is always active; nothing can remain phosphorylated. (c) Phosphodiesterasedestroys cAMP. Therefore, glycogen degradation would always be active and glycogen synthesis would always be inhibited.
  • 6. See question 13. of phosphor-ylasekinase serves to prolong the degradation of glycogen. The kinase cannot be deactivated until its that the kinase and, in turn, phosphorylase stay active for an extended interval. See question 14. degradation. Subsequent substrates for protein phosphatase. Thus, if the and little glycogen degradation would take place. 15.
  • 7. 16. (a) The glycogen was too large to enter the gel and, because analysis was by Western blot with the use of an antibody specific to glycogenin, one would not expect to see background proteins. -Amylase degrades glycogen, releasing the protein glycogenin so that it can be visualized by the Western blot. (c) Glycogen phosphorylase, glycogen synthase, and protein phosphatase1. These proteins might be visible if the gel were stained for protein, but a Western analysis reveals the presence of glycogenin only. See question 17. (a) The smear was due to molecules of glycogenin with increasingly large amounts of glycogen attached to them. (b) In the absence of glucose in the medium, glycogen is metabolized, resulting in a loss of the high-molecularweight material. (c) Glycogen could be resynthesized and added to the glycogenin when the cells were fed glucose again. (d) No difference between lanes 3 and 4 suggests that, by 1 hour, the glycogen molecules had attained maximum size in this cell line. Prolonged incubation does not apparently increase the amount of glycogen. -Amylase removes essentially all of the glycogen, and so only the glycogenin remains.