BICH 441 SPRING 2006

Exam #1

02/23/06
Questions 1-21 are 3 points each, questions 21-27 vary for a total of 110pts.

pXX refers to the page number in G&G


1. The standard free energies of hydrolysis for creatine phosphate, ATP and glucose-6-phosphate are -43, -31 and -14 kJ/mol, respectively. In which direction will phosphoryl group transfer occur spontaneously between CP, ATP and G6P?

CP --> ATP --> G6P
p63


2. Three reasons for the large negative free energy change for ATP hydrolysis are:

competing resonance in reactant
resonance stabilization of products
greater solvation of products
charge density in reactant relieved - reverse reaction requires combination of negatively charged molecules
formation of stronger acids that deprotaonate
p66


3. Why does phosphoenolpyruvate have a large negative free energy of hydrolysis?

resonance stabilized enolate anion
protonation of enolate anion to form enolpyruvate
tautomerization to ketopyruvate
p69


4. During muscular exercise the intracellular pH may decrease by as much as 1pH unit. As a result the phosphoryl group transfer potential of ATP decreases because:

lower pH protonates ATP and reduces charge denity. For pH 6-7 this calculates to be about 6 kJ/mol
p72


5. The conflicting demands of catabolism and anabolism are managed in two ways which are:

compartmentation
unique enzymes - differential regulation
p543


6. Why are the metabolic pathways for anabolism and catabolism different?

pathways must have a net negative free energy change
glycolysis has a net 2 ATP so gluconeogenesis must have an input of 2 ATP for stoichiometry PLUS more ATP to make the pathway spontaneous
p545


7. The conversion of glucose to lactate involves no net oxidation. So where does chemical the energy for net synthesis of 2 ATP come from?

rearrangements and C-C cleavage
secondary alcohols at C3,4 oxidized to acids
secondary alcohols at C2,5 oxidized to carbonyls
C1 and C6 reduced to methyls
p579


8. What are the advantages of phosphorylating glucose?

transport of glucose in
negative charge reduces diffusion of Glc6P out
negative charge provides a good handle for binding to enzymes
p582


9. What are two characteristics of hexokinase not shared by glucokinase?

Km 0.1mM vs 10mM
inhibited by Glc6P vs inhibited by GK Binding protein
p582


10. Name both substrate-level phosphorylation steps in glycolysis.

phosphoglycerate kinase
pyruvate kinase


11. Name the substrate-level phosphorylation step in the TCA cycle.

succinyl-CoA synthetase
p619


12. When the rate of production of NADH2 in glycolysis outstrips the ability of mitochondria o re-oxidize it, the excess NADH2 is re-oxidized by which enzyme?
Which other dehydrogenase, operating at the triose level, also carries out this function?

lactate dehydrogenase, alcohol dehydrogenase
glycerol-3P dehydrogenase
p597


13. The three enzymes responsible for the metabolism of fructose to triose phosphates are:

fructokinase, aldolase B or fructose-1P aldolase, triose kinase.
p600


14. Explain why a bolus of fructose produces acute lactic acidosis but a bolus of galactose or glucose does not.

Frc metabolism is NOT regulated
bypasses the regulatory steps of glycolysis at HK/GK and PFK-1 and rapidly metabolized to pyruvate
excess pyruvate is reduced to lactate producing the lactic acidosis
Glc and Gal metabolism both share regulation at PFK-1


15. In the pyruvate dehydrogenase complex the free energy of the oxidative decarboxylation is first conserved as:

formation of HETPP followed by formation of the thioester of lipoamide
p614


16. In the mechanism of succinyl-CoA synthetase, the free energy of the thioester is first conserved as:

succinyl-phosphate
p620


17. Why is the succinate dehydrogenase reaction completely stereospecific?

proR and proS H removed to form a trans product


18. Aldolase will catalyze an aldol cleavage of sedoheptulose-7-phosphate. Draw the structures of the products.

dihydroxyacetone (NOT phosphate) and erythrose 4-P, a 4 C aldose


19. The std.dG for malate dehydrogenase is +30 kJ/mol. Explain what pulls or pushes the reaction to completion.

product OAA rapidly removed by citrate synthase
high malate/OAA and NAD/NADH2 ratios
p621


20. The plant glycoside "atractyloside" inhibits the antiporter for transport of ATP out of, and ADP into, the mitochondrial matrix. Explain why this glycoside is often fatal.

ATP is synthesized in mito matrix and must be transported out. Atractyloside prevents this.
If matrix ATP/ADP ratio is high, ETC is inhibited (because of thermodynamic considerations) and NADH2 is not re-oxidized (NAD/NADH2 is low). This inhibits the TCA cycle and PDC.
A lack of ATP in cytosol results in low a ATP/AMP ratio which activates glycolysis but this often is not ewnough to compensate and death results.


21. In 1935 Szent-Gyorgyi, using pigeon breast muscle mitochondria, observed that succinate addition caused more oxygen uptake than could be accounted for by its complete oxidation. In 1938 Krebs explained this observation as due to:

a cycle!
Succinate will produce OAA which combines with AcCoA produced from endogenous pyruvate to produce more succinate.
Note that succinate to OAA is an oxidation but there is no loss of C as CO2. This only occurs after condensation.
p608


22.(12 pts) Asn + H2O --> Asp + ammonia, dG -14 kJ/mol
ATP + H2O --> AMP + PPi , dG -32 kJ/mol
Where Asn = asparagine and Asp = aspartate
A. Calculate the dG for the coupled reaction synthesizing Asn.

Asp + ammonia --> Asn + H2O, dG +14 kJ/mol
ATP + H2O --> AMP + PPi, dG = -32 kJ/mol
SUM : Asp + ATP + ammonia --> Asn + AMP + PPi, dG -18 kJ/mol

B. Write equations for the two partial reactions involved in the coupled synthesis of Asn.

the formation of AMP and PPi is diagnostic of nucleophilic attack on P-alpha, a common reaction of ATP
Asp side chain carboxyl is activated by formation of an adenylate
Asp + ATP --> Asp-AMP + PPi, dG = x kJ/mol
Asp-AMP + ammonia --> Asn + AMP, dG = y kJ/mol
SUM: Asp + ATP + ammonia --> Asn + AMP + PPi, dG = x+y = -18 kJ/mol

C. If the DG°' for the hydrolysis of the intermediate formed in the first partial reaction is -42 kJ/mol, what is the DG°' for each step of the reaction?

Asp-AMP + H2O --> Asp + AMP, dG -42 kJ/mol
Asp + ammonia --> Asn + H2O, dG +14 kJ/mol
SUM : Asp-AMP + ammonia --> Asn + AMP, dG = -28 kJ/mol = y
Therefor: Asp + ATP --> Asp-AMP + PPi, dG = +10 kJ/mol = x as x+y = -18 kJ/mol

D. Draw a plausible mechanism for the formation of the intermediate.

Asp carboxylate attacks P alpha
trigonal bipyramidal pentacovalent P intermediate
attacking Asp and leaving PPi (Pgamma and P beta) are apical for Sn2 in-line addition-elimination


23.(10pts) A variety of clinical disorders have been described resulting from a deficiency of the pyruvate dehydrogenase complex (PDC).

A. There are many different biochemical reasons for a deficiency of PDC. List six (6).

mutations leading to:
binding of coenzymes TPP, lopoic acid and FAD to E1, E2 and E3 respectively
active site mutations removing key groups (eg bases for H+ extraction)
binding of substrates - pyruvate, CoASH and NAD
protein folding mutations
protein-protein interactions in complex
defective kinase and phosphatase

B. Explain why they all result in an elevation of pyruvate, lactate and alanine, and a chronic (long term) lactic acidosis.

If Pyruvate cannot enter the TCA cycle then it is diverted to other products - lactate and alanine. The production of lactate will lead to the chronic lactic acidosis.

C. The PDC kinase is inhibited by dichloroacetic acid (DCA) Explain why DCA has been used to treat the chronic lactic acidosis of PDC deficient patients?

PDC kinase phosphorylates and inhibits E1.
This makes any mutation reducing PDC activity even worse.
Inhibition of the kinase by DCA will eventually result in dephosphorylation of E1 by the PDC phosphatease with activation of PDC and relief of the acidosis.


24.(10pts) Draw the mechanism for the decarboxylation of pyruvate catalyzed by the pyruvate dehydrogenase complex or pyruvate decarboxylase.

see p614 and PS#3.


25.(5pts) Describe the regulation of phosphofructokinase-1 using these kinetic plots as templates.

see Figs. 18-8, -9 and -10.


26.(5pts) Malic enzyme catalyzes the reaction:
Malate + NADP ý pyruvate + totCO2 + NADPH2
Draw a plausible reaction mechanism for this oxidative decarboxylation.

see mechanism for isocitrate DH


27.(5pts) Substrate analogs used to measure rates of transport should be poorly metabolized so that they accumulate largely unchanged inside cells. Explain why 2-deoxyglucose is used for measuring rates of glucose transport?

2DOG is transported in and phosphorylated by HK to 2DOG-6P
But, because it lacks the 2-OH it cannot be isomerized to 2-deoxyfructose-6P.
2DOG and 2DOG-6P accumulate inside cells.