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- 21. Based upon the following reactions, what would be the AG" for the formation of ATP from phosphoenolpyruvate and ADP? ATP → ADP + Pi phosphoenolpyruvate pyruvate + Pi AG" = -31.5 kJ/mol AG"=-62.2 kJ/mol1. Assuming that everything that’s needed to make tripalmitin comes from glucose, how many glucose would be used by adipose tissue in the synthesis of 1 mol tripalmitin? 2. How many ATP would be used (net of produced and used)? 3. If an animal absorbs 35 g glucose (MW = 180 g/mol) from drinking a can of pop, how many grams of tripalmitin (MW = 807 g/mol) can be produced from it in adipose tissue? Please provide only typed answer solution no handwritten solution needed allowedfructose-6-phosphate + ATP fructose-1,6- biphosphate + ADP AG° = 30.5 and 16.3 respectively Standard Free Energy change: -14.2 kj/mol ATP: 3mM, ADP: 0.4mM, fructose-1,6-biphosphate: 7mM %3D At what concentration of fructose-1,6-biphosphate would the reaction proceed backwards at 35°C?
- 5. Given the following information tables, determine in which cell type cleavage of the terminal phosphate from ADP yields the greatest actual change in free energy. (Please work in kJ/mol.) TABLE 13-6 Standard Free Energies of Hydrolysis of Some Phosphorylated Compounds and Acetyl-CoA (a Thioester) TABLE 13-5 Total Concentrations of Adenine Nucleotides, Inorganic Phosphate, and Phosphocreatine in Some Cells AG (kJimol) (kcalimol) Concentration (MM) Phosphoenolpyruvate 1,3-Bisphosphoglycerate (+3-phosphoglycerate + P) Phosphocreatine -61.9 -14.8 ATP ADP AMP P, PCr Rat hepatocyte Rat myocyte 49.3 -11.8 3.38 1.32 0.29 4.8 -43.0 -103 8.05 0.93 0.04 8.05 28 ADP (+ AMP + P) -32.8 -7.8 -7.3 Rat neuron 2.59 0.73 0.06 2.72 4.7 ATP (+ ADP + P) Location of -30.5 Human erythrocyte 225 0.25 0.02 1.65 ATP (+ AMP • PP) AMP (+ adenosine P) cleavage is important! 45.6 -10.9 E. coli cell 7.90 1.04 0.82 7.9 -14.2 -34 For erythrocytes the concentrations are those of the cytosol (human erythrocytes lack a…5) A certain aerobic organism is able to metabolize the following glycolipid: "CH,OH OH HO OH A. Draw the 2 resulting structures that would occur upon initial hydrolysis of the O-glycosidic bond. B. Calculate how much ATP is formed upon complete aerobic oxidation of one mole of the compound. Assume that no ATP is produced when one mole of the glycosidic bond in the above compound is hydrolyzed. Show calculation below.1. Acid phosphatases are an important group of enzymes that can be detected in human bloodserum. Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyzebiological phosphate esters as follows:R-O-P-O3-2 + H 2O R-OH + HO-P-O3-2.Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver,and prostrate gland. The enzyme from the prostrate gland is clinically important because anincreased activity in the blood is frequently an indication of cancer of the prostrate gland.Tartrate ion can strongly inhibit the phosphatase from the prostrate gland, but not acidphosphatases from other tissues. How can you use the information above to develop a specificprocedure for measuring the activity of the acid phosphatase of the prostrate gland in humanblood serum?
- 1. Outline the first round of lipid catabolism using a C18 saturated fatty acid. Indicate cofactors and type of chemistry that takes place. a. How much NADH, FADH2 and ACCOA are you getting from complete catabolism of this fatty acid? b. How many moles of ATP are you getting from the breakdown of this fatty acid? Keep in mind that in the mitochondria 1 mole FADH2 gives about 1.5 moles of ATP while 1 mole NADH yields about 2.5 moles of ATP.3. (a) In Bakers yeast there are two variants of cytochrome c, a heme containing protein of 108 amino acid residues that catalyzes electron transfer reactions in mitochondria. The two forms of cyto- chrome c known as iso-1 and iso-2 cytochrome c differ in 17 of the 108 amino acids. The content of proton dissociable side chains for the two cytochromes c is given in the table below: Side chain pka iso-1 cytochrome c Ziso-1 iso-2 cytochrome c 12.5 3 4.0 4 8.0 4.0 6.0 10.0 9.0 7.0 4.0 Arginine Aspartate Cysteine Glutamate Histidine Lysine Tyrosine -NH3Ⓡ -COOH 1 4 4 16 5 1 1 What is the overall electrostatic charge on each cytochrome at pH 8? (b) To the right is a diagram of the chemical groups in carboxymethyl-cellulose and DEAE-cellulose that are used for ion exchange chromatography to purify proteins. At pH 8 which chromatographic resin has to be used to separate iso-1-cytochrome c from iso-2-cytochrome c from a lysate of yeast cells? If the proteins are separated by application of a…1. Acid phosphatases are an important group of enzymes that can be detected in human blood serum.Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyze biological phosphate esters as follows: R-O-P-O3-2 + H 2O R-OH + HO-P-O3-2. Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver, and prostate gland. The enzyme from the prostate gland is clinically important because an increased activity in the blood is frequently an indication of cancer of the prostate gland. Tartrate ion can strongly inhibit the phosphatase from the prostate gland, but not acid phosphatases from other tissues. How canyou use the information above to develop a specific procedure for measuring the activity of the acid phosphatase of the prostate gland in human blood serum?
- 3. In muscle cells, the AG for glucose+ ATP 2 glucose – 6-P+ADP is –33.5 kJ mol-1. In contrast, the AG for glucose – 6 – P fructose – 6 – P is -2.5 kJ · mol-1. (a) In a physiological context, which reaction is faster? How do you know? (b) The enzyme that catalyzes glucose + ATP 2 glucose - 6 – P + ADP (hexokinase) is an important point of regulation in glycolysis, while the enzyme that catalyzes glucose – 6 – P2 fructose – 6 – P (phosphoglucose isomerase) is not. Why is hexokinase a good step at which to regulate glycolysis relative to phosphoglucose isomerase?(d) of glucose oxidation in diabetic human patients treated with Metformin (●) and in (nondiabetic) control human patients (0). At –150 min both groups of subjects were started on an intravenous feed of 3-(®H)-glucose, and at t = 0 min they were started on an oral glucose tolerance test whereby a measured amount of glucose in water (a syrupy mixture) was swallowed followed by measure- ment of blood glucose levels at 30 min intervals. The flux of glucose oxidation was measured by the appear- ance of 3H2O in the blood stream. While the information The diagram to the right compares the rate 8000 ORAL GLUCOSE 6000- 4000- 2000- -120 -60 60 120 180 240 300 Minutes cannot be directly extracted from the reaction mecha- nism diagrams in the textbook, the glycolytic step in which the tritium is first released into water is that catalyzed by TPI, as illustrated at the beginning of Question #3. Explain why this step is suitable for measuring the flux of glycolysis through the release of °H…2. If glyceraldehyde-3-phosphate dehydrogenase in red blood cells is completely inhibited, which intermediates in glycolysis accumulate most? Please use the following equation and the provided information to explain your reasoning. AG=AG° + RTln [C][D]d [A] [B]¹ Fructose 1,6-bisphosphate → Glyceraldehyde 3-phosphate + Dihydroxyacetone phosphate The standard free energy for the above reaction is: AG" = +23.9 KJ/mol T° = 298 R = 8.314 J mol-¹ K-¹