The concentration of glucose in your circulatory system is maintainednear 5.0 mM by the actions of the pancreatic hormones glucagon andinsulin. Glucose is imported into cells by protein transporters that arehighly specific for binding glucose. Inside the liver cells the importedglucose is rapidly phosphorylated to give glucose-6-phosphate (G-6-P).This is an ATP-dependent process that consumes 1 mol ATP per molof glucose. Given the steady-state intracellular concentrations below,calculate the theoretical maximum concentration of G-6-P inside aliver cell at 37 °C, pH = 7.2 when the glucose concentration outsidethe cell (i.e., [glucoseloutside) is 5.0 mM:ATP =4.7 mM; ADP = 0.15 mM; P, = 6.1 mMFor: ATP + H,O ADP + P + H* AG" = -30.5 kJ/mol andG-6-P + H,0 -→ Glucose + PAG" = -13.8 kJ/molThe glucose phosphorylation reaction isATP + glucosenside» ADP + glucose-6-phosphate + H+

Phosphorylation of glucose is the first step of Glycolysis which is a catabolic pathway of cellular…

Answered: The concentration of glucose in your…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Kt of glucose for GLUT1=3 mM, GLUT2=17 mM, GLUT 3=1.3 mM, and GLUT11=0.3 mM. Sketch a graph with [glucose] on the x-axis and rate of transport on the y-axis. Sketch a line for each GLUT transporter above. Which of these GLUT transporters will be working at its maximum rate at [glucose]=5 mM? Which of these GLUT transporters will be working closest to ½ Vmaxwhen [glucose]=5 mM? What will be the approximate rate of glucose transport for each of these transporters if [glucose]=17 mM?
Glucose is phosphorylated to glucose-6-phosphate by hexokinase in the first step of the glycolytic pathway to trap it in the cell, as G6P cannot diffuse across the lipid bilayer. This reaction also decreases the concentration of free glucose, favoring additional import of the molecule. However, this has a postive standard free energy of 3 kcal per mole. To favor this reaction, hydrolysis of ATP is coupled, which has a standard free energy of -7 kcal per mole. Determine the actual free energy change in kcal/mole for the following conditions: Glucose concentration: 4.5 mM ATP conc: 3.79 mM G6P conc. 0.052 mM ADP conc: 0.12 mM inorganic phosphate conc: 1.3 mM temperature: 313 K Ans. in 3 SFs.
D-glucose is added to cells and the rate of glucose transport is plotted against D- glucose concentration. In the presence of the D-galactose derivative 6-0-benzyl-D- galactose, the curve is shifted to the right, indicating that D-glucose and the D- galactose derivative both compete for a place in the binding site. External factors such as the addition of ATP or ion gradients doesn't change the rate of absorption into cells. Which of the following is true, select all that apply: O the enzyme is a uniport O mediated transport is occurring O non-mediated transport occurring O passive transport is occurring O active transport is occurring
Some anaerobic bacteria use alternative pathways for glucose catabolism that convert glucose to acetate rather than to pyruvate. Shown below is one possible metabolic pathway. The first part of this pathway (glucose to fructose-1,6-bisphosphate) is identical to the glycolytic pathway. In the second part of the alternative pathway, Enzymes 1–6 all have mechanisms/ activities analogous t enzymes in glycolysis. Note that there are two C¬C bond cleavage reactions in this new pathway: A → B + C (Enzyme 1) and C → B + D. All the steps where ATP is consumed or generated have been shown; however, the addition or loss of NAD+/NADH, Pi , H2O, or H+ has not been shown explicitly. Draw the structures for the intermediates B, F, G, H, and I, and include other reaction participants as needed.
Anaerobic glycolysis of glucose potentially produces a fewer number of ATPs per glucose moleculein red blood cells than in muscle cells, because -Red blood cells contain bisphosphoglycerate mutase -Red blood cells lack phosphoglycerate kinase -Red blood cells do not have mitochondria -Red blood cells do not have a nucleus Which of the following is likely to be correct if all mutases are completely inhibited? -Removal of glucosyl residues from nonreducing ends of glycogen in the liver is compromised;therefore, release of free glucose into blood from the liver would decrease -Glycerol can be efficiently converted into glucose but cannot be efficiently converted into glucosylresidues in glycogen in the liver -Both A and B -Neither A nor B
Intracellular concentrations in resting muscle are as follows: fructose- 6-phosphate, 1.0 mM; fructose-1,6-bisphosphate, 10 mM; AMP, 0.1 mM; ADP, 0.5 mM; ATP, 5 mM; and P, 10 mM. Is the phosphofructokinase reac- tion in muscle more or less exergonic than under standard conditions? By how much?
Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]
After the training of marathon, lactic acid is accumulated in the athlete's thigh muscle. Compare the physiological functions and Vmax of glycogen phosphorylase in actively working skeletal muscle and that in liver After vigorous exercise, the athlete faints due to low blood glucose level. A liver biopsy test shows excess amount of glycogen is present in the liver. Suggest with explanation, which type of enzyme is defective in this case.
It has been shown recently that the African naked GLUT5-O Fructose mole-rat can withstand up to 18 min of complete anoxia (ab- sence of O2) in the presence of high levels of CO2 without ap- parent brain damage. The basis of its capacity to tolerate such unusual conditions is that the brain, heart, and liver in the mole- rat switch to fructose-driven glycolysis. In contrast to normal mice, these tissues in the naked mole-rat have high levels of GLUT5 specific for cellular uptake of fructose. The available pathways for breakdown of fruct ose for metabolic energy production are shown on the right. KHK stands for ketohexosekinase, an enzyme that is markedly upregulated in the mole-rat compared to normal mice. Conse- quently mice must metabolize fructose largely via hexokinase (HK). In the diagram ALDOB, ALDOA, etc. represent aldolase isozymes. KHK HK Fructose-6-P PFK Fructose-1-P Fructose-1,6-BP ALDOB ALDOC ALDOA ALDOC Glyceraldehyde DHAP GAЗP Glyc-3-P 3-PGA Lactate Pyruvate mouse naked…
It has been shown recently that the African naked GLUT5-.O Fructose (b) mole-rat can withstand up to 18 min of complete anoxia (ab- sence of O2) in the presence of high levels of CO2 without ap- parent brain damage. The basis of its capacity to tolerate such unusual conditions is that the brain, heart, and liver in the mole- rat switch to fructose-driven glycolysis. In contrast to normal mice, these tissues in the naked mole-rat have high levels of GLUT5 specific for cellular uptake of fructose. The available pathways for breakdown of fruct ose for metabolic energy production are shown on the right. KHK stands for ketohexosekinase, an enzyme that is markedly upregulated in the mole-rat compared to normal mice. Conse- quently mice must metabolize fructose largely via hexokinase (HK). In the diagram ALDOB, ALDOA, etc. represent aldolase isozymes. KHK HK Fructose-6-P PFK Fructose-1-P Fructose-1,6-BP ALDOA ALDOC ALDOB ALDOC Glyceraldehyde DHAP GAЗP Glyc-3-P 3-PGA Lactate Pyruvate mouse…
During glycogen synthesis, glucose-1P is converted into a molecule called UDPG. This reaction also cleaves uridine triphosphate (UTP) forming uridine monophosphate and pyrophosphate (PPi). Provide four reasons why UTP can be used to power this reaction (no diagrams necessary).
The glutamate dehydrogenase (GDH) catalyses the following reaction: H +H₂N- -C -COO CH₂ CH₂ COO™ acide glutamique + NAD + H₂O GDH Time (min) A340 COO™ c=o CH₂ 5 1 2 3 4 1.760 1.718 1.675 1.635 1.595 CH₂ COO + NH4+ NADH + H+ acide a-cétoglutarique The activity of GDH is monitored in the sense of the formation of glutamate using the following conditions: 0.2 mL of 5 M ammonium sulphate 2.4 mL of buffer at pH 8 0.1 mL of NADH at 6.15 mg.mL-¹ (M = 709 g.mol-¹) 0.2 mL of 1 M a-ketoglutarate solution Warm mixture at 25 °C for 5 min Add 0.1 mL of GDH solution containing 1.6 mg.mL-¹protein to start the reaction. The change in absorbance at 340 nm is monitored, in a 1-cm cuvette, every minute for 10 min. Results are given in the table below: Data ENADH at 340 nm = 6220 M¹.cm7 6 7 8 9 10 1.550 1.510 1.489 1.476 1.451 Calculate ammonium sulphate, NADH, concentrations in the reaction medium at t = 0. proteins - Draw the graph A = f(t). Calculate A340 at t = 0 and place this point on the curve. -…

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