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2011 BIOC221 Problem Set (3) Due date: Feb 8 (Tue, 11:30 am)

1.Glucose Transporters Read Lehninger p. 391393, p. 539, and p. 542.

Glucose enters cells by facilitated diffusion via a specific glucose transporter, at a rate ~5000

times greater than uncatalyzed transmembrane diffusion. The transporter is, in principle,equally able to move glucose into or out of the cell. Since the glucose conc. in blood plasma is

~5 mM and the intracellular glucose conc. is kept low (owing to fast metabolism of glucose),

glucose always moves down its conc. gradient (thus, passive transport). Human genomecontains 12 glucose transporter genes. Each transporter shows unique specificity toward

different hexose, different pattern of tissue distribution, and function (see Table 11-3, p. 391).

Compare the localization of GLUT4 with that of GLUT2 and GLUT3, and explain why these

localizations are important in the response of muscle, adipose tissue, brain, and liver to insulin.

Answer GLUT2 (and GLUT1) is found in liver and is always present in the plasma membrane

of liver cells (hepatocytes). GLUT3 is always present in the plasma membrane of certain brain

cells. GLUT4 is normally sequestered in vesicles in cells of muscle and adipose tissue andenters the plasma membrane only in response to insulin. Thus, liver and brain can take up

glucose from blood regardless of insulin level, but muscle and adipose tissue take up glucoseonly when insulin levels are elevated in response to high blood glucose. This makes sense

because muscle and adipose tissues, being not vital, are the tissues that should get glucose

only when it is abundant.

2. Pathway of Atoms in Fermentation (1) A pulse-chase experiment using 14C-labeled

carbon sources is carried out on a yeast extract maintained under strictly anaerobic conditions

to produce ethanol. The experiment consists of incubating a small amount of14C-labeledsubstrate (the pulse) with the yeast extract just long enough for each intermediate in the

fermentation pathway to become labeled. The label is then chased through the pathway bythe addition of excess unlabeled glucose. The chase effectively prevents any further entry oflabeled glucose into the pathway.

(a) If [1-14C]glucose (glucose labeled at C-1 with 14C) is used as a substrate, what is the location

of14C in the product ethanol? Explain. (Hint: Handout(3), p. 11 and p. 24, or Fig. 14-6)

(b) Where would 14C have to be located in the starting glucose to ensure that all the 14C

activity is liberated as 14CO2 during fermentation to ethanol? Explain.

Answer

Anaerobiosis requires the regeneration of NAD+ from NADH in order to allow glycolysis

to continue.

(a) Figure 146 illustrates the fate of the carbon atoms of glucose. C-1 (or C-6) becomes C-3of glyceraldehyde 3-phosphate and subsequently pyruvate. When pyruvate is decarboxylated

and reduced to ethanol, C-3 of pyruvate becomes the C-2 of ethanol (14CH3CH2OH).

(b) If all the labeled carbon from glucose is converted to 14CO2 during ethanol fermentation,the original label must have been on C-3 and/or C-4 of glucose, because these are converted

to the carboxyl group of pyruvate.

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3. Pathway of Atoms in Fermentation (2) Through the magic of time travel, you find

yourself working as a scientist in Arthur Hardens group. (For who Harden is, see Lehninger

p.531.) You found yourself trying to convince Harden that the product of glycolysis, pyruvate,can be converted to lactic acid when no oxygen is available. Hardens group has the expertise

to detect lactic acid, and so to prove your point you radiolabel carbon four of glucose and feed

this to a microorganism under anaerobic conditions. See p. 24 of Handout (3), or Lehninger p.547 for help, if needed.

(a) When you isolate lactic acid, which carbon do you find has been radiolabelled?

(b) To Hardens surprise, only half the lactic acid produced contains radiolabel, but this is nota mystery to you. Explain.

(c) Harden is now really excited about this new technique, and therefore conducts the same

experiment under aerobic conditions later that evening. To his complete surprise, the

radiolabel completely disappears from the system. When you return in the morning, you canexplain his results. What is your explanation?

Answer

(a) The C-4 radiolabel of glucose ends up as C-1 of lactate (the carboxyl group).(b) Half the glucose is split into glyceraldehyde-3-P and half into dihydroxyacetone

phosphate, which is isomerized to glyceraldehyde-3-P. C-4 will end up as part of theglyceraldehyde-3-P, while the other glyceraldehyde-3-P will originate from dihydroxyacetone

phosphate that would have contained no label. Since glyceraldehyde-3-P, is eventually

converted to pyruvate and finally lactate, only half the lactate will be radiolabelled (1mark).

(c) The carbon originating from C-4 will be immediately lost as CO2 in the pyruvate

decarboxylase reaction, and therefore has bubbled out of the reaction mixture. A gas trap

would be required to isolate the radiolabel.

4. Glycolysis There are two reactions in glycolysis in which an aldose is isomerized to a ketose. For

one of these reactions draw the structures of the aldose and the ketose. For both reactions the G' is

positive. Briefly explain how the reactions are able to proceed without the input of additional energy.

Ans: The two reactions are those catalyzed by phosphohexose isomerase and triose phosphate

isomerase:

glucose 6-phosphate fructose 6-phosphate

(aldose) (ketose)

dihydroxyacetone phosphate glyceraldehyde 3-phosphate

(ketose) (aldose)

Although both of these reactions have standard free-energy changes (G') that are positive, they can

occur within cells because the products are immediately removed by the next step in the pathway. The

result is a very low steady-state concentration of the products, making the actual free-energy changes

(G) negative:

G = G' + RT ln ([products]/[substrates])

In other words, the rxn is driven forward by mass action.

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C

C

CH2OP

O

H OH

H

C

HC

CH2OP

OH

H OC

C

CH2OP

OH

OH

H

enediol

intermediate

B

HA

BH

A

5. Glycolysis Define fermentation and explain, by describing relevant reactions, how it

differs from glycolysis. Your explanation should include a discussion of the role of NADH inthe reaction(s).

Ans: Fermentation is the operation of the glycolytic pathway under anaerobic conditions.Under aerobic conditions, the pyruvate produced by glycolysis is oxidized to acetyl-CoA,

which passes through the citric acid cycle. NADH produced in the oxidations passes electrons

to O2, and is thus recycled to NAD+ allowing the continuation of the glycolytic reactions.

When no O2 is available to reoxidize the NADH produced by the glyceraldehyde 3-phosphate

dehydrogenase reaction, electrons from NADH must be passed to one of the products of

glycolysis, such as pyruvate or acetaldehyde, forming lactate or ethanol.

6. Glycolysis Shortcut Suppose you discovered a mutant yeast whose glycolytic pathway was

shorter because of the presence of a new enzyme catalyzing the reaction

Glyceraldehyde 3-P + H2 + NAD+ 3-phosphoglycerate + NADH + H+

Would shortening the glycolytic pathway in this way benefit the cell? Explain.

Consider metabolic outcomes only in anaerobic conditions.

Answer Under anaerobic conditions, the phosphoglycerate kinase and pyruvate kinase

reactions are essential. The shortcut in the mutant yeast would bypass the formation of an acyl

phosphate by glyceraldehyde 3-phosphate dehydrogenase and therefore would not allow the

formation of 1,3-bisphosphoglycerate. Without the formation of a substrate for 3-phosphoglycerate kinase, no ATP would be formed. Under anaerobic conditions, the net

reaction for glycolysis normally produces 2 ATP per glucose. In the mutant yeast, netproduction of ATP would be zero and growth could not occur. Under aerobic conditions,however, because the majority of ATP formation occurs via oxidative phosphorylation, the

mutation would have no observable effect.

7. Mechanism Propose a mechanism for the conversion of dihydroxyacetone phosphate into

glyceraldehyde 3-P by triose phosphate isomerase based on the mechanism of phosphoglucose

isomerase that catalyzes the conversion of glucose 6-P into fructose 6-P. Indicate any generalacid/base group at the enzyme active site that is involved in the catalysis.

Answer The mechanism is analogous to that for phosphoglucose isomerase. It proceeds

through an enediol intermediate. The active site would be expected to have a general base anda general acid.