







Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Material Type: Notes; Class: Introductionto Biochemistry; Subject: Chemistry; University: University of Wisconsin - Milwaukee; Term: Spring 2008;
Typology: Study notes
1 / 13
This page cannot be seen from the preview
Don't miss anything!
S-
1. Equation for the Preparatory Phase of Glycolysis Write balanced biochemical equations for all the reactions in the catabolism of glucose to two molecules of glyceraldehyde 3-phosphate (the prepara- tory phase of glycolysis), including the standard free-energy change for each reaction. Then write the overall or net equation for the preparatory phase of glycolysis, with the net standard free-energy change.
Answer The initial phase of glycolysis requires ATP; it is endergonic. There are five reactions in this phase:
1. Glucose ATP 88n^ glucose 6-phosphate ADP G 16.7 kJ/mol 2. Glucose 6-phosphate 88n^ fructose 6-phosphate G 1.7 kJ/mol 3. Fructose 6-phosphate ATP 88n^ fructose 1,6-bisphosphate G 14.2 kJ/mol 4. Fructose 1,6-bisphosphate 88n dihydroxyacetone phosphate glyceraldehyde 3-phosphate G 23.8 kJ/mol 5. Dihydroxyacetone phosphate 88n^ glyceraldehyde 3-phosphate G 7.5 kJ/mol The net equation for this phase is Glucose 2ATP 88n^ 2 glyceraldehyde 3-phosphate 2ADP 2H The overall standard free-energy change can be calculated by summing the individual reac- tions: G 2.1 kJ/mol (endergonic). 2. The Payoff Phase of Glycolysis in Skeletal Muscle In working skeletal muscle under anaerobic conditions, glyceraldehyde 3-phosphate is converted to pyruvate (the payoff phase of glycolysis), and the pyruvate is reduced to lactate. Write balanced biochemical equations for all the reactions in this process, with the standard free-energy change for each reaction. Then write the overall or net equa- tion for the payoff phase of glycolysis (with lactate as the end product), including the net standard free-energy change.
Answer The payoff phase of glycolysis produces ATP, and thus is exergonic. This phase con- sists of five reactions, designated 6 to 10 in the text:
6. Glyceraldehyde 3-phosphate Pi NAD^ 88n^ 1,3-bisphosphoglycerate NADH H G 6.3 kJ/mol 7. 1,3-Bisphosphoglycerate ADP 88n^ 3-phosphoglycerate ATP G 185 kJ/mol 8. 3-Phosphoglycerate 88n^ 2-phosphoglycerate G 4.4 kJ/mol 9. 2-Phosphoglycerate 88n^ phosphoenolpyruvate G 7.5 kJ/mol 10. Phosphoenolpyruvate ADP 88n^ pyruvate ATP G 31.4 kJ/mol
The pyruvate is then converted to lactate: Pyruvate NADH H^ 88n^ lactate NAD^ G 25.1 kJ/mol
Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway S-
The net equation is
Glyceraldehyde 3-phosphate 2ADP Pi 88n^ lactate NAD^ G 57 kJ/mol Because the payoff phase uses two glyceraldehyde 3-phosphate molecules from each glucose entering glycolysis, the net equation is
2 Glyceraldehyde 3-phosphate 4ADP 2Pi 88n^ 2 lactate 2NAD and the energetic payoff for the net reaction is G 114 kJ/mol.
3. GLUT Transporters 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 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 and enters 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 glucose only when in- sulin levels are elevated in response to high blood glucose.
4. Ethanol Production in Yeast When grown anaerobically on glucose, yeast ( S. cerevisiae ) converts pyruvate to acetaldehyde, then reduces acetaldehyde to ethanol using electrons from NADH. Write the equation for the second reaction, and calculate its equilibrium constant at 25 C, given the standard reduction potentials in Table 13–7.
Answer CH 3 CHO NADH H^ CH 3 CH 2 OH NAD Acetaldehyde Ethanol
Solve for K eq using the E values in Table 13–7 and Equations 13–3 and 13–7. G RT ln K eq G n E RT ln K eq n E
ln K eq
In this reaction, n 2, and E 0.123 V (calculated from values in Table 13–7 as shown in Worked Example 13–3). Substitute the standard values for the faraday and R , and 298 K for the temperature:
ln K eq 9.
K eq e 9.58^ = 1.45 10 4
5. Energetics of the Aldolase Reaction Aldolase catalyzes the glycolytic reaction
Fructose 1,6-bisphosphate 88n^ glyceraldehyde 3-phosphate dihydroxyacetone phosphate The standard free-energy change for this reaction in the direction written is 23.8 kJ/mol. The con- centrations of the three intermediates in the hepatocyte of a mammal are: fructose 1,6-bisphosphate, 1.4 10 ^5 M; glyceraldehyde 3-phosphate, 3 10 ^6 M; and dihydroxyacetone phosphate, 1.6 10 ^5 M. At body temperature (37 C), what is the actual free-energy change for the reaction?
2(96,480 J/V ^ mol)(0.123 V) (8.315 J/mol ^ K)(298 K)
n E RT
y^ z
9. Equivalence of Triose Phosphates^14 C-Labeled glyceraldehyde 3-phosphate was added to a yeast extract. After a short time, fructose 1,6-bisphosphate labeled with 14 C at C-3 and C-4 was isolated. What was the location of the 14 C label in the starting glyceraldehyde 3-phosphate? Where did the sec- ond 14 C label in fructose 1,6-bisphosphate come from? Explain.
Answer Problem 1 outlines the steps in glycolysis involving fructose 1,6-bisphosphate, glyc- eraldehyde 3-phosphate, and dihydroxyacetone phosphate. Keep in mind that the aldolase re- action is readily reversible and the triose phosphate isomerase reaction catalyzes extremely rapid interconversion of its substrates. Thus, the label at C-1 of glyceraldehyde 3-phosphate would equilibrate with C-1 of dihydroxyacetone phosphate ( G 7.5 kJ/mol). Because the aldolase reaction has G 23.8 kJ/mol in the direction of hexose formation, fructose 1,6-bisphosphate would be readily formed, and labeled in C-3 and C-4 (see Fig. 14–6).
10. Glycolysis Shortcut Suppose you discovered a mutant yeast whose glycolytic pathway was shorter because of the presence of a new enzyme catalyzing the reaction
Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway S-
Would shortening the glycolytic pathway in this way benefit the cell? Explain.
Answer Under anaerobic conditions, the phosphoglycerate kinase and pyruvate kinase reac- tions 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-phosphoglyc- erate kinase, no ATP would be formed. Under anaerobic conditions, the net reaction for gly- colysis normally produces 2 ATP per glucose. In the mutant yeast, net production 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.
11. Role of Lactate Dehydrogenase During strenuous activity, the demand for ATP in muscle tissue is vastly increased. In rabbit leg muscle or turkey flight muscle, the ATP is produced almost exclusively by lactic acid fermentation. ATP is formed in the payoff phase of glycolysis by two reactions, promoted by phosphoglycerate kinase and pyruvate kinase. Suppose skeletal muscle were devoid of lactate dehy- drogenase. Could it carry out strenuous physical activity; that is, could it generate ATP at a high rate by glycolysis? Explain.
Answer The key point here is that NAD^ must be regenerated from NADH in order for gly- colysis to continue. Some tissues, such as skeletal muscle, obtain almost all their ATP through the glycolytic pathway and are capable of short-term exercise only (see Box 14–2). In order to generate ATP at a high rate, the NADH formed during glycolysis must be oxidized. In the ab- sence of significant amounts of O 2 in the tissues, lactate dehydrogenase converts pyruvate and NADH to lactate and NAD. In the absence of this enzyme, NAD^ could not be regenerated and glycolytic production of ATP would stop—and as a consequence, muscle activity could not be maintained.
12. Efficiency of ATP Production in Muscle The transformation of glucose to lactate in myocytes re- leases only about 7% of the free energy released when glucose is completely oxidized to CO 2 and H 2 O. Does this mean that anaerobic glycolysis in muscle is a wasteful use of glucose? Explain.
Glyceraldehyde 3-phosphate H 2 3-phosphoglycerate
NAD^ NADH H
S-164 Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
Answer The transformation of glucose to lactate occurs when myocytes are low in oxygen, and it provides a means of generating ATP under oxygen-deficient conditions. Because lactate can be transformed to pyruvate, glucose is not wasted: the pyruvate can be oxidized by aero- bic reactions when oxygen becomes plentiful. This metabolic flexibility gives the organism a greater capacity to adapt to its environment.
13. Free-Energy Change for Triose Phosphate Oxidation The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate, catalyzed by glyceraldehyde 3-phosphate dehydrogenase, proceeds with an unfavorable equilibrium constant ( K eq 0.08; G 6.3 kJ/mol), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfa- vorable equilibrium?
Answer In organisms, where directional flow in a pathway is required, exergonic reactions are coupled to endergonic reactions to overcome unfavorable free-energy changes. The ender- gonic glyceraldehyde 3-phosphate dehydrogenase reaction is followed by the phosphoglycer- ate kinase reaction, which rapidly removes the product of the former reaction. Consequently, the dehydrogenase reaction does not reach equilibrium and its unfavorable free-energy change is thus circumvented. The net G of the two reactions, when coupled, is 18.5 kJ/mol 6.3 kJ/mol 12.2 kJ/mol.
14. Arsenate Poisoning Arsenate is structurally and chemically similar to inorganic phosphate (Pi), and many enzymes that require phosphate will also use arsenate. Organic compounds of arsenate are less stable than analogous phosphate compounds, however. For example, acyl arsenates decompose rapidly by hydrolysis:
O
O O
O
R C O As^ H2O O
O O
O
R C O^ HO As^ ^ H
On the other hand, acyl phosphates, such as 1,3-bisphosphoglycerate, are more stable and undergo fur- ther enzyme-catalyzed transformation in cells. (a) Predict the effect on the net reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase if phosphate were replaced by arsenate. (b) What would be the consequence to an organism if arsenate were substituted for phosphate? Ar- senate is very toxic to most organisms. Explain why.
Answer (a) In the presence of arsenate, the product of the glyceraldehyde 3-phosphate dehydroge- nase reaction is 1-arseno-3-phosphoglycerate, which nonenzymatically decomposes to 3- phosphoglycerate and arsenate; the substrate for the phosphoglycerate kinase is there- fore bypassed. (b) No ATP can be formed in the presence of arsenate because 1,3-bisphosphoglycerate is not formed. Under anaerobic conditions, this would result in no net glycolytic synthesis of ATP. Arsenate poisoning can be used as a test for the presence of an acyl phosphate intermediate in a reaction pathway.
15. Requirement for Phosphate in Ethanol Fermentation In 1906 Harden and Young, in a series of classic studies on the fermentation of glucose to ethanol and CO 2 by extracts of brewer’s yeast, made
S-166 Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
In fact, the enzyme glycerol 3-phosphate dehydrogenase catalyzes this reaction (see Fig. 21–17).
18. Severity of Clinical Symptoms Due to Enzyme Deficiency The clinical symptoms of two forms of galactosemia—deficiency of galactokinase or of UDP-glucose:galactose 1-phosphate uridylyltransferase— show radically different severity. Although both types produce gastric discomfort after milk ingestion, deficiency of the transferase also leads to liver, kidney, spleen, and brain dysfunction and eventual death. What products accumulate in the blood and tissues with each type of enzyme deficiency? Estimate the relative toxicities of these products from the above information.
Answer In galactokinase deficiency, galactose accumulates; in UDP-glucose:galactose 1-phosphate uridylyltransferase deficiency, galactose 1-phosphate accumulates (see Fig. 14–12). The latter metabolite is clearly more toxic.
19. Muscle-Wasting in Starvation One consequence of starvation is a reduction in muscle mass. What happens to the muscle proteins?
Answer Muscle proteins are selectively degraded by proteases in myocytes, and the resulting amino acids move, in the bloodstream, from muscle to liver. In the liver, glucogenic amino acids are the starting materials for gluconeogenesis, to provide glucose for export to the brain (which cannot use fatty acids as fuel).
20. Pathway of Atoms in Gluconeogenesis A liver extract capable of carrying out all the normal meta- bolic reactions of the liver is briefly incubated in separate experiments with the following 14 C-labeled precursors:
(a) [^14 C]Bicarbonate,
(b) [1-^14 C]Pyruvate,
HO 14 C O C O
CH^14 CO 3
O
O
Trace the pathway of each precursor through gluconeogenesis. Indicate the location of 14 C in all inter- mediates and in the product, glucose.
Answer (a) In the pyruvate carboxylase reaction, 14 CO 2 is added to pyruvate to form [4-^14 C]oxaloac- etate, but the phosphoenolpyruvate carboxykinase reaction removes the same CO 2 in the next step. Thus, 14 C is not (initially) incorporated into glucose.
O C P
HO
CH 2 OH
Dihydroxyacetone phosphate (^) Glycerol 3-phosphate
CH 2 O
O
CH 2 OH C CH 2 O P
O H
NADH H^ NAD
O^ O^
O^ O^
Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway S-
21. Energy Cost of a Cycle of Glycolysis and Gluconeogenesis What is the cost (in ATP equiva- lents) of transforming glucose to pyruvate via glycolysis and back again to glucose via gluconeogenesis?
Answer The overall reaction of glycolysis is Glucose 2ADP 2P (^) i 2NAD^ 88n^ 2 pyruvate 2ATP 2NADH 2H^ 2H 2 O The overall reaction of gluconeogenesis is 2 Pyruvate 4ATP 2GTP 2NADH 2H 4H 2 O 88n glucose 2NAD^ 4ADP 2GDP 6P (^) i
The cost of transforming glucose to pyruvate and back to glucose is given by the difference between these two equations: 2ATP 2GTP 2H 2 O 88n^ 2ADP 2GDP 4P (^) i The energy cost is four ATP equivalents per glucose molecule.
22. Relationship between Gluconeogenesis and Glycolysis Why is it important that gluconeogenesis is not the exact reversal of glycolysis?
Answer If gluconeogenesis were simply the reactions of glycolysis in reverse, the process would be energetically unfeasible (highly endergonic), because of the three reactions with large, nega- tive standard free-energy changes in the catabolic (glycolytic) direction. Furthermore, if the same
OPO 32
H
3,4- 14 C-glucose
C
CH 2
C
POH 2 C O
COO
CH 2 OPO 32 14
COO
14
COO
14
C COO
OPO 32 14
CH 2
CH 2 OH
O
O
C C
CH 2 OPO 32
OH
H
14
OPO 32 H C
CH (^2) OH
14
O
C
H
2 O (^3)
OPO 32 H C
CH (^2) OH
(^14) COO 2-Phospho- glycerate
Phosphoenolpyruvate
Dihydroxyacetone phosphate
Oxaloacetate
3-Phospho- glycerate
Fructose 1,6-bisphosphate
Glyceraldehyde 3-phosphate
CH 2 OPO 32
HO
C OH 14
C
H
(^14) C OH
C O
CH 2 OPO 32
1,3-Bisphospho- glycerate
H
CH (^3) C O (^14) COO 1-^14 C-pyruvate
HOH C
(b)
Explain how this reaction inhibits the transformation of lactate to pyruvate. Why does this lead to hypoglycemia?
Answer The first step in the synthesis of glucose from lactate in the liver is oxidation of the lactate to pyruvate; like the oxidation of ethanol to acetaldehyde, this requires NAD. Con- sumption of alcohol forces a competition for NAD^ between ethanol metabolism and gluco- neogenesis, reducing the conversion of lactate to glucose and resulting in hypoglycemia. The problem is compounded by strenuous exercise and lack of food because at these times the level of blood glucose is already low.
26. Blood Lactate Levels during Vigorous Exercise The concentrations of lactate in blood plasma before, during, and after a 400 m sprint are shown in the graph.
Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway S-
(a) What causes the rapid rise in lactate concentration? (b) What causes the decline in lactate concentration after completion of the sprint? Why does the decline occur more slowly than the increase? (c) Why is the concentration of lactate not zero during the resting state?
Answer (a) Rapid depletion of ATP during strenuous muscular exertion causes the rate of glycolysis to increase dramatically, producing higher cytosolic concentrations of pyruvate and NADH; lactate dehydrogenase converts these to lactate and NAD^ (lactic acid fermentation). (b) When energy demands are reduced, the oxidative capacity of the mitochondria is again adequate, and lactate is transformed to pyruvate by lactate dehydrogenase, and the pyruvate is converted to glucose. The rate of the dehydrogenase reaction is slower in this direction because of the limited availability of NAD^ and because the equilibrium of the reaction is strongly in favor of lactate (conversion of lactate to pyruvate is energy- requiring). (c) The equilibrium of the lactate dehydrogenase reaction
Pyruvate NADH H^ 88n^ lactate NAD is strongly in favor of lactate. Thus, even at very low concentrations of NADH and pyru- vate, there is a significant concentration of lactate.
Blood [lactate] (
M)
0
150
Time (min)
100
50
40 60
Before
0
200
Run After
20
S-170 Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
27. Relationship between Fructose 1,6-Bisphosphatase and Blood Lactate Levels A congenital defect in the liver enzyme fructose 1,6-bisphosphatase results in abnormally high levels of lactate in the blood plasma. Explain.
Answer In the liver, lactate is converted to pyruvate and then to glucose by gluconeogenesis (see Figs 14–15, 14–16). This pathway includes the glycolytic bypass step catalyzed by fruc- tose 1,6-bisphosphatase (FBPase-1). A defect in this enzyme would prevent the entry of lac- tate into the gluconeogenic pathway in hepatocytes, causing lactate to accumulate in the blood.
28. Effect of Phloridzin on Carbohydrate Metabolism Phloridzin, a toxic glycoside from the bark of the pear tree, blocks the normal reabsorption of glucose from the kidney tubule, thus causing blood glucose to be almost completely excreted in the urine. In an experiment, rats fed phloridzin and sodium succinate excreted about 0.5 mol of glucose (made by gluconeogenesis) for every 1 mol of sodium succinate ingested. How is the succinate transformed to glucose? Explain the stoichiometry.
O (^) C O
OH
OH
OH
HO
OH H
H (^) H
H OH
HOCH 2
CH 2
H
O
Phloridzin
CH 2
Answer Excretion of glucose promoted by phloridzin causes a drop in blood glucose, which stimulates gluconeogenesis. The ingested succinate enters the mitochondrion via the dicar- boxylate transport system and is transformed to oxaloacetate by enzymes of the citric acid cycle. The oxaloacetate passes into the cytosol and is transformed to phosphoenolpyruvate by PEP carboxykinase. Two moles of PEP are then required to produce a mole of glucose by the route outlined in Figure 14–16, consistent with the observed stoichiometry. Note that the rate of glucose production must be much higher than the rate of utilization by tissues because almost 100% of the glucose is excreted.
29. Excess O 2 Uptake during Gluconeogenesis Lactate absorbed by the liver is converted to glucose, with the input of 6 mol of ATP for every mole of glucose produced. The extent of this process in a rat liver preparation can be monitored by administering [^14 C]lactate and measuring the amount of [^14 C]glucose produced. Because the stoichiometry of O 2 consumption and ATP production is known (about 5 ATP per O 2 ), we can predict the extra O 2 consumption above the normal rate when a given amount of lactate is administered. However, when the extra O 2 used in the synthesis of glucose from lactate is actually measured, it is always higher than predicted by known stoichiometric relationships. Suggest a possible explanation for this observation.
Answer If the catabolic and biosynthetic pathways operate simultaneously, a certain amount of ATP is consumed in “futile cycles” (or “substrate cycles”) in which no useful work is done. Examples of such cycles are that between glucose and glucose 6-phosphate and that between fructose 6-phosphate and fructose 1,6-bisphosphate. The net hydrolysis of ATP to ADP and P (^) i increases the consumption of oxygen, the terminal electron acceptor in oxidative phosphorylation.
S-172 Chapter 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
D-Xylose
H O C C C C CH 2 OH
H
HO
H OH
OH
H
(g) What additional enzymes would you need to introduce into the modified Z. mobilis strain de- scribed above to enable it to use xylose as well as arabinose to produce ethanol? You don’t need to name the enzymes (they may not even exist in the real world!); just give the reactions they would need to catalyze.
Answer (a) Ethanol tolerance is likely to involve many more genes, and thus the engineering would be a much more involved project. (b) L-Arabinose isomerase (the araA enzyme) converts an aldose to a ketose by moving the carbonyl of a nonphosphorylated sugar from C-1 to C-2. No analogous enzyme is dis- cussed in this chapter; all the enzymes described here act on phosphorylated sugars. An enzyme that carries out a similar transformation with phosphorylated sugars is phospho- hexose isomerase. L-Ribulokinase ( araB ) phosphorylates a sugar at C-5 by transferring the phosphate from ATP. Many such reactions are described in this chapter, including the hexokinase reaction. L-Ribulose 5-phosphate epimerase ( araD ) switches the —H and —OH groups on a chiral carbon of a sugar. No analogous reaction is described in the chapter, but it is described in Chapter 20 (see Fig. 20–1, p. 774). (c) The three ara enzymes would convert arabinose to xylulose 5-phosphate by the follow- ing pathway:
Arabinose L-ribulose L-ribulose 5-phosphate
xylulose 5-phosphate. (d) The arabinose is converted to xylulose 5-phosphate as in (c), which enters the pathway in Figure 14–22; the glucose 6-phosphate product is then fermented to ethanol and CO 2. (e) 6 molecules of arabinose 6 molecules of ATP are converted to 6 molecules of xylulose 5-phosphate, which feed into the pathway in Figure 14–22 to yield 5 molecules of glu- cose 6-phosphate, each of which is fermented to yield 3 ATP (they enter as glucose 6-phosphate, not glucose)—15 ATP in all. Overall, you would expect a yield of 15 ATP – 6 ATP 9 ATP from the 6 arabinose molecules. The other products are 10 molecules of ethanol and 10 molecules of CO 2. (f) Given the lower ATP yield, for an amount of growth (i.e., of available ATP) equivalent to growth without the added genes the engineered Z. mobilis must ferment more arabi- nose, and thus it produces more ethanol. (g) One way to allow the use of xylose would be to add the genes for two enzymes: an ana- log of the araD enzyme that converts xylose to ribose by switching the —H and —OH on C-3, and an analog of the araB enzyme that phosphorylates ribose at C-5. The result- ing ribose 5-phosphate would feed into the existing pathway.
Reference Deanda, K., Zhang, M., Eddy, C., & Picataggio, S. (1996) Development of an arabinose-fermenting Zymomonas mobilis strain by metabolic pathway engineering. Appl. Environ. Microbiol. 62, 4465–4470.
888888 n
88888888888888n 888888888 n
Another sugar commonly found in plant matter is xylose.
L-arabinose isomerase L-ribulokinase
epimerase