Ch 16 Final Review Essays

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Ch 16 Final Review Essays
2014-12-06 11:44:27
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  1. 1)      Why is glucose used?
    a.       It is one of several monosaccharides formed from formaldehyde under prebiotic conditions, and so it may have been available as a fuel source for primitive biochemical systems. Glucose also has a low tendency, relative to other monosaccharides, to nonenzymatically glycosylate proteins. 
  2. 1)      Briefly explain stage one.
    a.       Stage one is the trapping and prepararation phase. No ATP is generated in this stage. Stage 1 begins with the conversion of glucose into fructose 1,6-bisphosphate, which undergoes a phosphorylation, an isomerization, and a second phosphorylation. It is completed with the cleavage of fructose1,6-bisphosphate into two three-carbon fragments. 
  3. 1)      Briefly explain stage two. 
    a.       ATP is harvested when the three-carbon fragments are oxidized to pyruvate. 
  4. 1)      What is step one of glycolysis; why is it notable?              
    a.       Conversion of glucose to glucose 6-phosphate by hexokinase. It is notable because glucose 6-phosphate cannot pass through the membrane because it is not a substrate for the glucose transporters and the addition of the phosphoryl group acts to destabilize glucose, facilitating its further metabolism. ATP is used in this step. Hexokinase transfers a phosphoryl group from ATP to glucose
  5. 1)      What is the structure of hexokinase? What is special about it? 
    a.       It consists of two lobes, which move toward each other when glucose is bound. When glucose binds, one lobe rotates 12 degrees with respect to the other, resulting in movements of the polypeptide backbone of as much as eight angstroms. The cleft between the lobes closes, and the bound glucose becomes surrounded by protein, except for the hydroxyl group of carbon 6, which will accept the phosphoryl group from ATP. 
  6. 1)      What does the structural change of hexokinase do? 
    • a.       The environment aroud glucose becomes more nonpolar, which favors reactin between the hydrophilic hydroxyl group of glucose and the terminal phosphoryl group of ATP.
    • b.      The conformational changes enable the kinase to discriminate against H2O as a substrate. The closing of the cleft keeps water out of the active site.
  7. 1)      What is step two of glycolysis? 
    a.       The second step is isomerization of glucose 6-phosophate to fructose 6-phosphate,  which occurs because glucose 6-phosophate does not have a free carbon at the end. So, a keto group is formed. This is catalyzed by phosphoglucose isomerase, which opens up glucose 6-phosphate, catalyzes the isomerization, and then promotes the formation of the five-embered ring of fructose 6-phosphate.
  8. 1)      What is step three of glycolysis?
    a.       Frucose 6-phosphate is phosphorylated at the expense of ATP to fructose 1,6-bisphosphate. This is catalyzed by phosphofructokinase (PFK). 
  9. 1)      What is step four of glycolysis? 
    a.       Fructose 1,6-bisphosphate is cleaved into glyceraldehyde 3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP), competing stage 1 of glycolysis. This reaction is catalyzed by aldolase. Glyceraldehyde 3-phosphate can enter Stage 2 but DHAP must be isomerized to DHAP by triose phosphate isomerase. 
  10. 1)      What is the mechanism for TPI? 
    a.       Glutamate plays the role of a general acid-base catalyst by abstracting a proton from carbon 1 and donating it to carbon2. Because it is not strong enough to pull a proton away from carbon adjacent to the carbonyl, histidine donates a proton to stabilize the negative charge that develops on the C2 carbonyl group. Glutamic acid, ow acting as the general acid, donates a proton to C2 while histidine removes a proton from OH group of C1. The product is formed. 
  11. 1)      What makes TPI great? 
    a.       First, it is kinetically perfect and can catalyze the substrate upon contact. Diffusion is the only limiting factor. Second, it suppresses an undesired reaction, the decomposition of the enediol intermediate into methyl glyoxal and orthophosphate. TPI, to prevent the enediol from leaving the active site, traps it int eh active site by the movement of a loop of ten residues, which acts as a lid for the active site, shutting it when the enediol is present and reopening it when isomerization is completed. 
  12. 1)      What is step one of stage two? What is up with this reaction? 
    • a.       Glyceraldehyde 3-phosphate is converted into 1,3-bisphosphoglycerate by glyceraldehyde 3-phosphate dehydrogenase
    • It can be viewed as the sum of two processes that do not occur in succession: oxidation of the aldehyde to a carboxylic acid by NAD+ and the joining of the carboxylic acid and orthophosphate to form the acyl-phosphate product. While the first reaction is thermodynamically favorable,t he second is not. So, the processes must be coupled so that the favorable aldehyde oxidation can be used to drive the formation of the acyl phosphate. The key is an intermediate, formed as a result of the aldehyde oxidation, that is linked to the enzyme by a thioester bond
  13. 1)      What is the mechanism of glyceraldehyde 3-phosphate dehydrogenase? 
    • a.       It has a reactive cysteine resiude, as well as NAD+ and a crucial histidine.
    • b.      In step one, the aldehyde substrates reacts with the sulfhydryl group of cysteine to form a hemithioacetal.
    • c.       In step two, the hydriode ion is transferred to NAD_ that is tightly bound to the enzyme and adjacent to the cysteine residue.
    •                                                               i.      This reaction is favored by the deprotonation fot he hemithioacetal by histidine.
    •                                                             ii.      The products are NADH and a thioester intermediate, which has free energy close to that of the reactants
    • d.      In step three, the NADH leaves the enzyme and is replaced by another NAD+, which polarizes the thioester intermediate to facilitate attack by orthophosphate.
    • e.      In step four, orthophosphate attacks the thioester to form 1,3-BPG and free the cysteine residue. 
  14. 1)      What is step two of stage two? 
    a.       1,3-bisphosphoglycerate is converted to 3-phosphoglycerate by phosphoglycerate kinase, which catalyzesthe transfer of a phosphoryl group from 1,3-BPG to ADP, forming ATP and 3-phosphoglycerate. 
  15. 1)      What is step three of stage two? 
    a.       3-phosphoglycerate is converted to 2-phosphoglycerate by phosphoglycerate mutase. This is a rearrangement. The position of the phosphoryl group shifts in the conversion of 3-phosphoglycerate into 2-phosphoglycerate. Catalytic aounts of 2,3-bisphosphoglycerate to maintain an active site histidine residue in a phosphorylated form. Once 2,3-bisphosphoglycerate is formed, the mutase then functions as a phosphatase, converting 2,3-BPG to 2-phosphoglyceate. 
  16. 1)      What is step four of stage two? 
    a.       2-phosphoglycerate is dehydrated to introduce a double bond, creating an enol called phosphoenolpyruvate (PEP). This is catalyzed by enolase. It has high phosphoryl-transfer potential due to its unstable enol form. 
  17. 1)      What is step five of stage two? 
    a.       Phosphoenolpyruvate is converted to a more stable ketone—pyruvate. This is done by pyruvate kinase. 
  18. 1)      What are three reactions of pyruvate that are of primary importance? 
    • a.       Alcohol fermentation: ethanol is formed from pyruvate.
    •                                                               i.      Step one is decarboxylation of pyruvate catalyzed pyruvate decarboxylase, which requires TPP
    •                                                             ii.      Step two is the reduction of acetaldehyde to ethanol by NADH in a reaction catalyzed by alcohol dehydrogenase.
    •                                                           iii.      This process regenerates NAD+; a zinc ion was used to polarize the carbonyl group of the substrate to favor the transfer of a hydride from NADH
    • b.      Lactic acid fermentation: takes place when lactate is formed from pyruvate; takes place when oxygen is limiting; reduction of pyruvate done by lactate dehydrogenase. ; no net oxidation-reduction. The NADH formed in the oxidation of glyceraldehyde 3-phosphate is consumed in the reduction of pyruvate. This regenerates NAD+.
    • c.       Conversion of carbon dioxide: extraction of the most energy occurs through oxidative phosphorylation
  19. 1)      If fermentation yields a small amount of energy, why is it useful? 
    a.       It does not require oxygen, which allows organisms that do not use oxygen to obtain energy. 
  20. 1)      What is special about the structure of dehydrogenases? 
    a.       Their NAD+ binding domains are very similar. It is made up of four alpha helices and a sheet of six parallel Beta strands. The bound NAD+ displays nearly the same conformation; and, it is called the Rossmann fold
  21. 1)      How does fructose enter the cycle? 
    • a.       It can take one of two pathways.
    •                                                               i.      Fructose 1-phosphate pathway
    • 1.       Phosphorylation of fructose to fructose 1-phosphate by fructokinaseà split into GAP and DHAP, catalyzed by fructose 1-phosphate aldolaseà GAP is then phosphorylated to G3P by triose inase.
    •                                                             ii.      Other pathway: fructose is phosphorylated to fructose 6-phosphate by hexokinase
  22. 1)      How does galactose enter the pathway? 
    • a.       It does so in four steps.
    •                                                               i.      The first reaction is a galactose-glucose interconversion pathway: phosphorylation of galactose to galactose 1-phosphate by galactokinaseà acquires uridyl from UDP-glucoseà a series of other reactions that yield glucose 1-phosphate, which is isomerized to glucose 6-phosphate by phosphoglucomutase
  23. 1)      What are the forms of control in glycolysis in muscle tissue? 
    • a.       The enzymes that are controlled are phosphofructokinase (the on-off switch), hexokinase, and pyruvate kinase.
    • b.      Phosphofructokinase is the most important control site. It is controlled by:
    •                                                               i.      High levels of ATP, which bind to a specific regulatory site that is distinct form the catalytic site, thus lowering its  affinity for fructose 6-phosphate.
    •                                                             ii.      AMP increases enzyme activity
    • 1.       The reason it is AMP and not ADP is because, under stressful energetic conditions, adenylate kinase can convert two ADP into ATP and AMP
    •                                                           iii.      Decrease in pH (such as due to lactic acid in muscles) inhibits it by augmenting the inhibitory effect of ATP
    •                                                            iv.      This enzyme catalyzes the committed step in glycolysis
    • c.       Hexokinase is inhibitedby its product., glucose 6-phosphate.
    •                                                               i.      High concentrations of glucse 6-phosphate signal that the cell no longer nees glucose for energy or for storage as glycogen.
    •                                                             ii.      When phosphofructokinase is inhibited, the concentration fo fructose 6-phosphate rises. Fructose 6-phosphate is in equilibrium with glucose 6-phosphate, so its concentration rises, thus inhibiting hexokinase.
    • d.      Pyruvate kinase controls the outflow from glycolysis, with a final step yielding ATP and pyruvate.
    •                                                               i.      ATP inhibits it to slow glycolysis when the energy charge is high
    •                                                             ii.      Alanine also allosterically inhibits it to signal that building blocks are abundant. 
  24. 1)      What are the forms of control in glycolysis int eh liver? 
    • a.       Phosphofructokinase:
    •                                                               i.      ATP inhibits it, but low pH has no effect
    •                                                             ii.      Citrate inhibits it because it means that biosynthetic precursors are abundant, and so there is no need to degrade additional glucose; it enhances the inhibitory effect of ATP
    •                                                           iii.      Fructose 2,6-bisphosphate activates phosphfructokinase; its production is accelerated by the abundance of fructose 6-phosphate
    • b.      Hexokinase:
    •                                                               i.      Glucokinase is an isozyme of hexokinase in the liver; glucokinase provides glucose 6-phosphate for the synthesis of glycogen and for the formation of fatty acids.
    • c.       Pyruvate kinase: there are L and M forms of it that have properties in common
    •                                                               i.      Only the L form is controlled by reversible phosphorylation.
    •                                                             ii.      When blood glucose levels are low, the cAMP cascade leads to phosphorylation of pyruvate kinase, which diminishes its activity
    •                                                           iii.      This prevents the liver from consuming glucose when it is more urgently needed by the brain and muscle
  25. 1)      What are the similarities between cancer and exercise? 
    a.       Tumors metabolize glucose to lactate even in the presence of oxygen, called aerobic glycolysis (Warburg effect)