7.05 S20 Solutions to Problem Set 8

7.05 Spring 2010 Problem Set 8 KEY Problem 1: Sugar Isomerization a) Convert α-glucose to its furanose form. b) Is it more likely for it to be found in the cyclized or linearized form? Cyclized c) Draw the linear depiction of D-glucose. See picture above Problem 2: Synthesis I would like to make D-galactose (shown) from D-fructose (also shown). Show which steps and enzymes are necessary for this. Enzymes you can use: Ketose-aldose Isomerase, Epimerase 1 Problem 3: Sugar Characteristics and Nomenclature (from previous exam) A) Answer the following by circling the number(s) corresponding to the sugar(s) above which best answer each question (there may be 0, 1, or >1 answer for each question). Provide BRIEF explanation only for the last question where requested. Which sugar(s) is/(are) a pyranose? Iii Which sugar(s) is/(are) drawn as an α-pyranose OR as an α-furanose? I, iii, v Identify all the aldose sugar(s)? ii, iii, v Identify all the reducing sugars(s)? I, ii, iii, iv, v Which sugar(s) is/are not found in nature? Briefly explain your answer. Ii – nature uses D sugars not L sugars 2 B) Name the following sugar: α-D-ribulofuranosyl-(2-4)-β-D-xyulofuranose C) Draw α-D-glucopyranosyl-(1-3)-D-glyceraldehyde D) Specify which of the sugars named/drawn in parts B and C above are reducing sugars. If neither is a reducing sugar, make this clear. Briefly explain your answer. Both sugars are reducing since both have an accessible aldehyde or ketone Problem 4: Chemistry of glycolysis A) The reactants in the net reaction of glycolysis are given below. Complete the reaction by writing the products and balancing the equation (including charges). Assume that all inorganic phosphate (Pi) is in the form HPO42-. glucose (C6H12O6) + 2 ADP + 2 Pi + 2 NAD+ → 2 pyruvate (C3H3O3-) + 2 ATP + 2 NADH + 2 H2O + 2 H+ 3 B) One component of the net reaction of glycolysis is the synthesis of ATP from ADP and Pi. For each of the following reactions, write the balanced chemical equation. Assume that all inorganic phosphate (Pi) is in the form HPO42-. Which step of glycolysis consumes Pi? Which step uses that same Pi to generate ATP? For each step, name the enzyme and draw the balanced reaction using names, formulas, or structures of the reactants and products. 1) The synthesis of ATP from ADP and Pi (a.k.a. reverse of ATP hydrolysis): ADP + Pi + H+ → ATP + H2O 2) The step of glycolysis that consumes Pi: GA3P + Pi + NAD+ → 1,3-BPG + NADH + H+ 3) The step of glycolysis that makes ATP from a product in question 4B2: 1,3-BPG + ADP → 3PG + ATP C) Red blood cells express a special enzyme, 2,3-bisphosphoglycerate mutase, that turns 1,3-bisphosphoglycerate (1,3-BPG) into 2,3-bisphosphoglycerate (2,3-BPG). (You may recall that 2,3-BPG stabilizes the T state of hemoglobin and decreases its oxygen affinity.) 2,3-BPG re-enters glycolysis when it is dephosphorylated by 2,3-BPG phosphatase to 3-phosphoglycerate (3PG). These two reactions are named the Luebering-Rapoport pathway: 1) If a glucose molecule enters glycolysis, and both of the 1,3-BPG molecules it yields go through the Leubering-Rapoport pathway instead of through phosphoglycerate kinase, what is the net number of ATP produced by glycolysis of this molecule of glucose? 0 4 2) Genetic defects in glycolysis can disrupt levels of 2,3-PBG and diminish the ability of red blood cells to transport oxygen. How would 2,3-BPG level and hemoglobin oxygen affinity be affected by defects in the following enzymes? Enzyme 2,3-BPG ( + / - / 0 ) Hb O2 affinity ( + / - / 0 ) hexokinase + pyruvate kinase + 3) In 2006, scientists discovered that cells in the outermost layer of the placenta also express 2,3-bisphosphoglycerate mutase. 2,3-BPG binds more tightly to adult hemoglobin than to fetal hemoglobin. Why would expressing this enzyme in the placenta help the fetus get oxygen? Production of 2,3-BPG in the placenta would cause the mother’s blood to release oxygen to the fetus in the vicinity of the placenta, and the fetus’s blood would absorb the oxygen because 2,3-BPG does not reduce the oxygen affinity of fetal hemoglobin as much as adult hemoglobin. D) Organisms growing anaerobically cannot perform glycolysis for long without reducing the pyruvate from glycolysis into another compound, most commonly to lactate or to ethanol plus CO2. Both of these reactions are given below in their unbalanced forms. Explain in one sentence why one of these reducing steps is needed to sustain anaerobic glycolysis. 1) pyruvate + NADH → lactate + NAD+ 2) pyruvate + NADH → ethanol + CO2 + NAD+ This step is required to replenish the NAD+ that glycolysis consumes. E) In mammals, muscles produce lactate during intense exercise because glucose availability for glycolysis is in excess of oxygen delivery to support complete glucose oxidation. Lactate is transported via the bloodstream to the liver, where it is converted back to pyruvate and then to glucose via gluconeogenesis, which returns to the muscles (the Cori Cycle). Write the balanced reaction that converts lactate back into pyruvate in the liver. lactate (C3H5O3-) + NAD+ → pyruvate (C3H3O3-) + NADH + H+ 5 Problem 5: Energetics of glycolysis A) The complete combustion of glucose (C6H12O6) in oxygen (O2) yields water and CO2. Write the balanced reaction for the combustion of glucose. glucose (C6H12O6) + 6 O2 → 6 H2O + 6 CO2 B) The free energy (∆G°’) of this reaction under standard conditions (25°C, each species at 1M) is -2930.4 kJ/mol [equilibrator.weizmann.ac.il]. Calculate the free energy under physiological conditions, assuming that T = 37°C, [glucose] = 1.0 mM, [O2] = 2.0 mM, [CO2] = 2.5 mM. [��/ ]1 ∆� = ∆�°% + �� ln + 9 [�������][�/ ]1 0.00251 = −2930.4 + 0.008314 × 310 × ln + 9 0.001 × 0.00201 = −2909 ��/��� C) Assume that under physiological conditions, hydrolysis of ATP has a ∆G of -46 kJ/mol. What percentage of the chemical energy in glucose (your answer to 5B) gets stored in the molecules of ATP produced from glycolysis? Does glycolysis by itself seem like an efficient way to get energy from glucose? 2 * -46 / -2909 = 3.2% Glycolysis does not efficiently obtain energy from glucose because only about 3% of the energy in glucose winds up in the two product ATP molecules. D) The free energy change of each step of glycolysis is given in the table below. ∆G°’ is the free energy under standard conditions (25°C, 1M each reactant, pH 7), while ∆G is the free energy change at presumed physiological conditions. Step Enzyme ∆G°’ (kJ/mol) ∆G (kJ/mol) 1 HK - 20.9 - 27.2 2 PGI 2.2 - 1.4 3 PFK - 17.2 - 25.9 4 aldolase 22.8 - 5.9 5 TIM 7.9 ~ 0.0 6 + 7 GAPDH + PGK - 16.7 -1.1 8 PGM 4.7 - 0.6 9 enolase - 3.2 - 2.4 10 PK - 23.0 - 13.9 6 a. Why must no step have a positive ∆G under physiological conditions? If a step had a positive ∆G under physiological conditions, it would go in the reverse direction, so glycolysis would not happen. b. Conversion of F1,6BP to GA3P and DHAP by aldolase is striking in that it is strongly unfavorable when all species are at 1 M (∆G°’ = +22.8 kJ/mol) but favorable when the species are at their physiological conditions (∆G = -5.9 kJ/mol). Based on ∆G°’ and ∆G of aldolase and TIM, and using a reasonable physiological value of 15 mM for the concentration of F1,6BP, calculate the concentrations of GA3P and DHAP under physiological conditions (use T = 37°C). Why is the reaction so much more favorable under these conditions than when all species are at 1 M? [��3�][����] ∆�JKL = ∆�°% JKL + �� ln + 9 [�1,6��] ∆UVWX Y∆U°Z VWX Y^._Y//.` [��3�][����] [\ =� = � ]a.aa`bcd×bcae = 1.46 × 10Y^ � [�1,6��] [��3�][����] = 1.46 × 10Y^ × [�1,6��] = 1.46 × 10Y^ � × 0.015 � = 2.2 × 10Yg �/ ∆�\hi = ∆�°% \hi + �� ln j [��3�] k [����] a.aYg._ ∆Ulmn Y∆U°Z lmn [��3�] [\ =� = � ]a.aa`bcd×bcae = 4.66 × 10Y/ [����] [��3�] [����] = = 21.5 × [��3�] 4.66 × 10Y/ 2.2 × 10Yg �/ = [��3�][����] = [��3�] × (21.5 × [��3�]) = 21.5 × [��3�]/ 2.2 × 10Yg �/ [��3�] = q = 1.0 × 10Yd � = 0.1 �� 21.5 [����] = 21.5 × [��3�] = 21.5 × 10Yd � = 2.2 × 10Yb � = 2.2 �� Because [GA3P] and [DHAP] are much lower than [F1,6BP], the reaction becomes favorable, even though it would be very unfavorable if all concentrations were 1 M. This is conceptually similar to Le Chatelier’s principle, which says that if you start at equilibrium and then add more reactant, the net reaction will go forward until reaching equilibrium again. 7 E) The step of glycolysis catalyzed by PFK is called the committed step because it is the first reaction in glycolysis that is practically irreversible and doesn’t produce a product that is used for anything else. As such, PFK activity is regulated to control the rate of glycolysis. If the products of glycolysis accumulate, PFK is inhibited; if the products begin to deplete, PFK is activated. This type of regulation is negative feedback. Based on this, propose whether each of the following chemicals activates (+), inhibits (-), or has no effect (0) on PFK activity: Chemical Effect ( + / - / 0) ATP AMP + + H (i.e. low pH) hexokinase 0 phosphoenolpyruvate (PEP) fructose 2,6-bisphosphate (made from + F6P by phosphofructokinase 2) F) Humans actually use different versions of hexokinase in different tissues. For example, muscle cells express a form known simply as hexokinase, while liver cells express a form known as glucokinase. There are many notable differences between hexokinase and glucokinase: a. The phosphorylation of glucose by hexokinase or glucokinase traps it inside cells. Muscle cells may need to absorb glucose no matter its concentration in the blood. Liver cells store glucose but do not use it as a main fuel source, so they only take up glucose when concentrations in the blood are high. Based on this, which enzyme (hexokinase or glucokinase) do you predict has a higher Michaelis constant (KM), and why? Glucokinase has a higher KM because it only needs to be active at high glucose concentrations. b. Like PFK, some hexokinases can be inhibited by their product, G6P. One role of the liver is to store large quantities of glucose (in the form of glycogen), while muscle cells metabolize glucose only when they have a large need for ATP. Do you predict that hexokinase is inhibited by G6P? What about glucokinase? Hexokinase is inhibited by G6P because muscle cells only need to make more G6P when G6P is depleted. Glucokinase is not inhibited by G6P because liver cells need to regulate the level of sugar in the blood. Even if they have already generated a lot of G6P, if blood glucose levels are high they will need to absorb and phosphorylate the excess glucose using glucokinase. 8