Metabolism
To produce the maximum number of ATPs, post-glycolytic pathways in aerobes must receive from glycolysis: NADH and pyruvate.
The post-glycolytic pathways are the Krebs cycle and electron transport, which require oxygen, NADH, and pyruvate. Pyruvate is converted into acetyl-CoA to enter the Krebs cycle (choice D is correct).
In another experiment, a mutase was used to transform the glycolytic intermediate 3-phosphoglycerate into 1-phosphoglycerate, which was immediately removed from the pathway and not able to become a substrate in subsequent steps. One would expect to find levels of ATP produced under these conditions to be: identical to levels produced by the arsenate-poisoned system
If 3-phosphoglycerate is converted into 1-phosphoglycerate, then no energy will be extracted as ATP later in the pathway by pyruvate kinase. Thus, the net ATP production per glucose will be 0 ATP. Per glucose, the ATP produced at each step is: -1 Hexokinase -1 Phosphofructokinase +2 Phosphoglycerate kinase 0 Pyruvate kinase (normally +2) This is the same amount of ATP as in the presence of arsenate (choice C is correct and choice B is wrong).
n glycolysis, how many moles of inorganic phosphate are consumed in the oxidation of 3 moles of glucose to pyruvate, as pictured in Figure 1? 6 For every glucose molecule, 2 molecules of glyceraldehyde-3-phosphate are produced. Every molecule of glyceraldehyde-3-phosphate requires one molecule of Pi to be converted into 1,3-bisphosphoglycerate. For 3 moles of glucose, 6 moles of glyceraldehyde-3-phosphate are produced, requiring 6 moles of Pi.
1 glucose = 2 X glyceraldehyde-3-phosphate 1 glyceraldehyde-3-phosphate = requires one molecule of Pi
Glucose-6-phosophate dehydrogenase (G6PDH) deficiency is an inheritable metabolic disorder that can result in the destruction of red blood cells, and in severe cases, can lead to kidney and liver failure. Which of the following best describes the reason why G6PDH deficiency can result in premature red blood cell destruction? G6PDH deficiency results in a deficiency of NADPH, which results in increased damage from reactive oxygen species and causes premature cell death. Glucose-6-phosphate dehydrogenase (G6DPH) catalyzes the first step in the pentose phosphate pathway, where NADPH is produced. NADPH plays a major role in preventing damage due to oxidative stress/reactive oxygen species. A deficiency of G6PDH prevents the production of NADPH, thus individuals with this deficiency would experience increased damage from reactive oxygen species (choice C is correct). G6PDH deficiency would not be expected to alter the intracellular osmolarity (choice A is incorrect) or to increase the intracellular pH (choice D is also incorrect). Choice B is incorrect: macrophages are not involved in the death of red blood cells in G6PDH cells.
> pentose phosphate pathway > Glucose-6-phosphate dehydrogenase > NADPH (damage decreases due to oxidative stress/reactive oxygen species)
Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP
>kinase (transfers an inorganic phosphate from somewhere to ADP form am ATP) > pyruvate kinase (phosphate from phosphoenolpyruvate) > phosphoenolpyruvate (converted to pyruvate) > Pyruvate (catalyzes the last step of glycolysis)
During peak cardiac activity, heart muscle utilizes the anaerobic pathway for only 10% of its energy as opposed to up to 85% for maximally contracting skeletal muscle. With respect to skeletal muscle, cardiac muscle would: consume less glucose per ATP produced.
Aerobic respiration includes glycolysis, the PDC, the Krebs cycle, and electron transport, to produce a total of 32 ATP per glucose (assuming the malate-aspartate shuttle is used to transfer the electrons from glycolytic NADH into the mitochondria). By contrast, anaerobic respiration involves only glycolysis and fermentation of pyruvate to lactate or alcohol, to produce 2 ATP per glucose. Anaerobic respiration is therefore much less efficient and must consume much more glucose to produce the same amount of ATP. Since cardiac muscle uses mostly aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct). Conversion of pyruvate to lactate occurs during fermentation, which cardiac muscle carries out less of, not more (choice A is wrong). The amount of contractile force generated per ATP is independent of the source of ATP used (choice B is wrong). Cardiac muscle uses more aerobic respiration, so it will use more oxygen per ATP produced, not less (choice D is wrong). ========================= Since cardiac muscle = aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct) skeletal muscle = fermentation = less efficient = use more glucose
The loss of net production of ATP during glycolysis that occurred in the presence of arsenate and Pi was due to: decreased formation of 1,3-bisphosphogylcerate.
Arsenate competes with Pi as a substrate for glyceraldehyde-3-phosphate dehydrogenase. In the presence of arsenate, 1-arseno-3-phosphoglycerate is produced instead of 1,3-bisphosphoglycerate. =================== glyceraldehyde-3-phosphate dehydrogenase = formation of 1,3-bisphosphogylcerate.
Which of the following would be true about cis-oleic acid, a monounsaturated fatty acid with the formula CH3(CH2)7(CH)2(CH2)7COOH? It will generate approximately 119 ATP after 8 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain. Recognize the ability to treat this question as a 2x2 elimination. cis-Oleic acid has 18 carbons and thus will undergo 8 rounds of β-oxidation (eliminate choices A and C). This will generate 9 molecules of acetyl-CoA, 8 molecules of NADH and 7 molecules of FADH2 since it is a monounsaturated fatty acid. Each of the 9 acetyl-CoAs will go through the Krebs cycle and this will generate 27 NADH (which will give 67.5 ATP), 9 FADH2 (which will give 13.5 ATP) and 9 GTPs (9 ATP equivalents). This means the acetyl-CoAs alone generate 90 ATP equivalents. Since the NADH and FADH2 made in β-oxidation will generate even more ATP, the total will exceed 90 (eliminate choice D and choice B is correct).
Beta oxidation > # of carbons > #/2 = # acetyl-CoA > # acetyl-CoA - 1 = # of rounds & NADH > # acetyl-CoA - 2 = # of FADH2 ========================= Kreb cycle 1 acetyl-CoA = 3 NADH, 1 FADH2, 1GTP =================== ETC 1 NADH = 2.5 ATP 1 FADH2 = 1.5 ATP 1 GTP = 1 ATP
Which of the following is the LEAST characteristic of the Krebs cycle? Most of the energy is produced through substrate-level phosphorylation.
Choice B is the only false statement (which is the correct answer for this question). Only two of the 20 ATP made from the Krebs cycle are from substrate-level phosphorylation.
Isolate genomic DNA from a cancer cell line and sequence the gene for pyruvate carboxylase, to confirm which amino acid is altered in the mutant form.
F The question stem states that a frameshift mutation in pyruvate carboxylase has already been found. There would therefore be little point in sequencing the gene again. In addition, "confirm[ing] which amino acid is altered in the mutant form" is a better match to confirming a point mutation rather than a frameshift mutation (choice A is wrong). ====================== pyruvate carboxylase = pyruvate oxaloacetate = gluconeogenesis = glucose increases oxaloacetate = either glucose ↓ or ↑
Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis.
F Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong) ======================= (reactant) Glucose-6-phosphate = pentose phosphate pathway
High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism.
F The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). ====================== (product) ribose-5-phosphate = pentose phosphate pathway
If the researchers had used a luminescent probe to visualize the location of FAS itself in Experiment 3, where would it predominantly be found? Cytosol
Fatty acid synthesis takes place in the cytosol and requires the presence of FAS (choice A is correct). No part of the fat metabolism takes place in the nucleus of the cell (eliminate choice B). The mitochondrial matrix is the location for beta-oxidation (fatty acid break down; eliminate choice C). Elongation and desaturation of fatty acids takes place inside the smooth ER using separate enzymes. FAS is not used during extensions of fatty acids beyond carbon-16 nor is it involved in the introduction of any double bonds (eliminate choice D). ============= Fatty acid synthesis = takes place in the cytosol
Based on Figure 1, what is the net reaction for the transformation of glucose to pyruvate?
Glucose + 2 Pi + 2 ADP + 2 NAD+Equilibrium2 pyruvate + 2 ATP + 2 NADH + 2 H+ + 2 H2O
A physician treating a patient with Von Gierke's disease notes elevated lipid, lactic acid and uric acid levels in the patient's blood. Which of the following additional findings regarding the patient is the physician most likely to also note? Decreased blood glucose concentrations
Glucose-6-phosphatase, the enzyme that the passage states is deficient in Von Gierke's disease patients, hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. When this enzyme is absent, glucose-6-phosphate, a product of the breakdown of glycogen in the liver, will remain trapped within the liver and will not pass into circulation. As a result, Von Gierke's patients are at an increased risk of hypoglycemia and require regular administration of exogenous glucose to meet metabolic demands. This is consistent with choice D which is the correct answer. ===================================== Glucose-6-phosphatase = hydrolyzes glucose-6-phosphate = glucose increases
Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? Glycogen phosphorylase
Glycogen phosphorylase catalyzes the rate limiting step of glycogenolysis—the release of glucose-1-phosphate from glycogen via its action on terminal α(1→4) linkages. This makes choice A true and the correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. Choice B is false and an incorrect answer. Phosphoglucomutase is involved in both glycogenolysis and glycogenesis by interconverting glucose 1-phosphate and glucose 6-phosphate, thus choice C is false and the incorrect answer. Pyrophosphatases are members of the acid hydrolase family of enzymes that cleave diphosphate bonds. Choice D is false and an incorrect answer. ============================= Glycogen phosphorylase = glycogenolysis = release of glucose-1-phosphate from glycogen =
2,4-dinitrophenol (2,4-DNP) is a highly toxic substance which was sold to the public as a weight loss drug in the 1930s. It acts by permeabilizing the inner mitochondrial membrane (IMM) to ions. Which of the following is true of 2,4-DNP's effects on oxidative phosphorylation? It causes a decrease in the electromotive potential built up by the electron transport chain.
If the inner mitochondrial membrane became permeable to ions, then hydrogen ions would not need to go through the ATP synthase in order to re-enter the matrix. This would dissipate the proton gradient established by the electron transport chain, and decrease its potential to generate ATP (choice D is correct). FADH2 and NADH would still be consumed by the transport chain proteins, since the electron transport chain is not shut down; it's just that the proton gradient would be more easily dissipated (choice C is wrong). The dissipation of the gradient would result in heat production; note that this is similar to what happens in brown fat, used by hibernating animals to stay warm (choice B is wrong). Electrons do not flow through the ATP synthase (choice A is wrong). =================== permeable to ions = dissipate the proton gradient
Thirteen amino acids, including methionine, valine and proline, are glucogenic in humans. This means their α-keto acid carbon skeleton is converted to pyruvate during amino acid catabolism. After deamination, valine can therefore: Be converted into CO2 and H2O to generate ATP. Generate at least three NADH and two FADH2. Enter gluconeogenesis to generate glucose. Be converted into CO2 and H2O to generate ATP. Generate at least three NADH and two FADH2. Enter gluconeogenesis to generate glucose.
Item I is true: Amino acids are catabolized via deamination into α-keto acids and ammonia. Based on the information in the question stem, the α-keto acid formed from valine will be converted to pyruvate. Pyruvate can keep going through cellular respiration to generate CO2, H2O and ATP. Eliminate choice B. Item II is false: Pyruvate is converted into one acetyl-CoA (during which 1 NADH is made), and the acetyl-CoA would then generate three NADH and only one FADH2 as it cycles through the Krebs cycle. Eliminate choice C. Item III is true: Pyruvate can also enter gluconeogenesis to generate glucose. Eliminate choice A and choice D is correct. ====================== catabolized Amino acids = deamination into α-keto acids and ammonia Pyruvate = gluconeogenesis = generate glucose Pyruvate = glycolysis/ Kreb cycle = generate ATP/CO2/H2O
Increased levels of which of the following foods would likely be recommended for individuals on a Standard diet, but NOT for individuals on a Paleo diet? Bananas and squashes Whole grain breads Cheese and yogurt Whole grain breads
Item III is false. Cheese, being high in fat, would unlikely be part of a Standard diet (choice D can be eliminated and choice A is the correct answer choice).
After recovering from a severe illness caused by O157:H7, a person who could previously eat dairy products without difficulty has become lactose intolerant. Which of the following is the most likely cause of this? The disruption of the membrane during infection has also impacted the brush border enzymes.
Lactose is a found in dairy products. It is a disaccharide consisting of glucose and galactose. Lactose is hydrolyzed by lactase, which is a brush border enzyme. The passage states that the infection by O157:H7 damages the epithelium; by doing so the brush border and the brush border enzymes are also damaged (choice A is correct). ====================== Lactose = sugar destroy = border enzyme
Most biological unsaturated fatty acids are cis and contain non-conjugated double bonds. Because of this, additional steps are required in β-oxidation. These most likely include: Combining two double bonds via a reductase enzyme (which uses NADPH as a reducing agent), then changing the resultant cis double bond to trans via an isomerase enzyme.
Polyunsaturated fatty acids require both an isomerase and a reductase enzyme to complete β -oxidation; this also requires the reducing agent NADPH (choice D is correct). ===================== >beta-oxidation >unsaturated fatty acids = 1st reduced = then isomerase to trans (from cis) = become conjugated
PDC deficiency is a rare genetic disease. Which of the following conditions would be expected in patients with the disease?Lactic acidosis In PDC deficiency, the pyruvate cannot be converted to acetyl-CoA and therefore, the cells can only undergo glycolysis and anaerobic respiration. The end product of anaerobic respiration in humans is lactic acid. Thus, patients with PDC deficiency have lactic acidosis (choice B is correct). Glucose is still being utilized for energy so there is no reason to assume its circulating levels would increase (eliminate choice C) and PDC does not play a role in processing lipids (eliminate choice D). A lack of PDC activity would not increase the body's pH (eliminate choice A).
Pyruvate dehydrogenase complex deficiency = glycolysis and anaerobic respiration = lactic acid (increases) ===================== Pyruvate dehydrogenase complex = pyruvate converted to acetyl-CoA
A series of cell lines were generated, each missing expression of one or more IDH isoform. When grown in the presence of 14C-D-glucose: large amounts of 14C-lactate is detectable in the growth media of IDH3-null lines but not IDH1 or IDH2-null lines.
Since IDH3 is important in the TCA cycle, cells that lack IDH3 will resort to anaerobic metabolism to survive. This will generate large amounts of lactate (choice C is correct).
Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides.
T Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis
If the measurements taken in Step 4 of Experiments 1 and 2 had shown no change in bacterial lactose concentration, then it is likely that within the medium:hydrogen ion concentration had remained unchanged in Experiments 1 and 2.
T These experiments show that a gradient of either potassium or protons can drive the transport of lactose into the cell against its gradient. If there is no lactose transport, it is safe to assume that the concentration in the media of the driving ion will remain the same (choice A is correct). ================ no transport = concentration driving ion will remain the same
fructose-2,6-bisphosphate, It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase
T ======================= > gluconeogenesis > reactant ( fructose-1,6-bisphosphATE ) > product ( fructose-6-phosphate ) > enzyme ( fructose-1,6-bisphosphatase) ======================= > glycolysis > reactant (fructose-6-phosphate) > product (fructose-1,6-bisphosphATE) > enzyme (PFK = phosphofructokinase-1) > secondary enzyme (fructose-2,6-bisphosphate)
