Calculate glucose concentration. Na (sodium) and glucose secondary active transport. Na transport (which drive glucose import) G=R*T*In(Na in/ Na out)+Z*F*Y(psi symbol) Na in=14mM Na out=145MM Z=+1 F=96.5 KJ/V*mol Y(psi)=-0.05V What is the glucose in and out concentration? Please be very through when explaining this calculation. (I am stuck at why the energy sign changes from negative to positive when using the calculated energy from sodium to glucose)
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Gibbs free energy:
The change in free energy for a process equals the maximum work that can be done by the system on the surroundings in a spontaneous process that occur at constant temperature and pressure.
The sign of will depend on the symbol of .
In some cases, the temperature will result in the spontaneity of a reaction.
For a system at equilibrium,
The standard free energy for the reaction is directly related to the equilibrium constant for the reaction.
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Solved in 2 steps
- 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?Intracellular concentrations in resting muscle are as follows: fructose6-phosphate, 1.0 mM; fructose-1,6-bisphosphate, 10 mM; AMP, 0.1 mM;ADP, 0.5 mM; ATP, 5 mM; and Pi, 10 mM. Is the phosphofructokinasereaction in muscle more or less exergonic than under standard conditions?By how much?these are KT values. Which transporter that is a GLUT will transport glucose from blood which is 5nm at the maximum or Vmax? Is it (GLUT 11,KT= 0.1M), (is it GLUT1, KT = 3nm), (is it GLUT 2, KT= 17nm), (or GLUT 4, KT = 5nm)
- Calculate delta G of glucose transport into the cell if the concentration of clugoce is 1mM inside the cell and 100uM outside. R=2cal/mol*K and T=300KCalculate all the glucose data into cmol. What is the number of cmol glucose at timepont 10.42 (19.65 g/L)?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?
- Computation: Ratio Strength, PPM, mg%Show your complete solution.1. Diabetes is diagnosed by any of the following: two consecutive fasting blood glucose tests that are equal to or greater than 125 mg/dL, any random blood glucose that is greater than 200 mg/dL, and a two hour oral glucose tolerance test with any value over 200 mg/dL.A.) what is yje equivalent value express in terms of milligrams percent?Equivalent value of 126 mg/dL in mg% = ?Equivalent value of 200 mg/dL in mg% = ?B.) how many milligrams of glucose would be present in 10-mL sample serum (use the FASTING BLOOD SUGAR LEVEL)?Define glucose-sparing effectDrink # Glucose Concentration in mM 1 2.455 mM 2 0.307 mM 3 3.750 mM 4 1.100 mM 5 1.500 mM From here, each group needs to convert the concentration of their drink from mM of the diluted drink into g/100mL of the undiluted drink. This is where you come in! The students are stuck at this point and really need your help! Main Goal: Convert the experimental glucose concentration (provided in the table below) from mM (millimolar) to g/100 mL. Hints & Tips Take the 1/100 dilution factor into consideration Molarity will open the doors for you to get to grams. Once at M assume 1L of solution. (Hint: this will help you when converting from M to moles) You will also need to calculate the molar mass of glucose which has the formula C6H12O6 There are 5 steps in total Task 1: Write a step-by-step guide with key words for the Chem 2 students on how to do these calculations. You can use diagrams, a table or just type the steps out. Ensure the steps…
- Drink # Glucose Concentration in mM 1 2.455 mM 2 0.307 mM 3 3.750 mM 4 1.100 mM 5 1.500 mM From here, each group needs to convert the concentration of their drink from mM of the diluted drink into g/100mL of the undiluted drink. This is where you come in! The students are stuck at this point and really need your help! Main Goal: Convert the experimental glucose concentration (provided in the table below) from mM (millimolar) to g/100 mL. Hints & Tips Take the 1/100 dilution factor into consideration Molarity will open the doors for you to get to grams. Once at M assume 1L of solution. (Hint: this will help you when converting from M to moles) You will also need to calculate the molar mass of glucose which has the formula C6H12O6 There are 5 steps in total Task 1: Write a step-by-step guide with key words for the Chem 2 students on how to do these calculations. You can use diagrams, a table or just type the steps out. Ensure the steps…Reaction centers PSI and PSII are also called as per the wavelengths at which they have a maximumabsorbance. What are these wavelengths?If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATP concentration = 5 mM and ADP concentration = 5 mM , what is the theoretical concentration of glucose-6-phosphate at equilibrium at pH=7.4 and 37 ∘C ? Express your answer to three significant figures and include the appropriate units. Please type answer note write by hend.