Estimate the flux (mg/cm2/s) by diffusion of estrogen (a steroid) through a lipid bilayer cell membrane when assuming the diffusion coefficient for estrogen across the lipid bilayer is 10^–6 cm2/s, and that the initial concentration of estrogen in the extracellular fluid is 1 ng/mL and 0 in the cytoplasm.
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Estimate the flux (mg/cm2/s) by diffusion of estrogen (a steroid) through a lipid bilayer cell membrane when assuming the diffusion coefficient for estrogen across the lipid bilayer is 10^–6 cm2/s, and that the initial concentration of estrogen in the extracellular fluid is 1 ng/mL and 0 in the cytoplasm.
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- The average time it takes for a molecule to diffuse adistance of x cm is given byt = x2/2D where t is the time in seconds and D is the diffusioncoefficient. Given that the diffusion coefficient ofglucose is 5.7 × 10−7cm2/s, calculate the time it wouldtake for a glucose molecule to diffuse 10 μm, which isroughly the size of a cell.In the Nernst equation [V = 62 log10 (Co/ Ci)], the term Co represents: the intracellular concentration of calcium the extracellular concentration of potassium the extracellular concentration of sodium the intracellular concentration of potassium the membrane potential (in millivolts)Below find the structures for ibogaine and cocaine. Ibogaine and cocaine inhibit the dopamine active transporter (DAT). This transporter is a secondary active transporter, and depends on the primary active transporter Na+/K+ ATPase. Ibogaine had a Kι = 2 μM, and cocaine a Kι = 0.64 μM respectively. (a) Define secondary active transport. (b) Is ibogaine an effective treatment for cocaine based on DAT binding?
- Calculate the change in Gibbs free energy for transport of Ca2+ from outside to inside the cell. The extracellular Ca2+ concentration is 135 uM, and the intracellular Ca2+ concentration is 98 uM. The membrane potential is -22 mV and the temperature is 37°C. O. -5.1 kJ/mol O 1.2 kJ/mol -410 kJ/mol 3.4 kJ/molIn the Nernst equation [V = 62 log10 (Co / Ci)], the term Co represents: cell bio the intracellular concentration of calcium the extracellular concentration of potassium the extracellular concentration of sodium the intracellular concentration of potassium the membrane potential (in millivolts)The following table shows experimental results of the glucose transport rate, mM/sec, following facilitated diffusion by glucose carrier proteins. (Recall: the starting conc. L represents glucose added to one side of the membrane; distilled water, omM of glucose was added to the other side of the membrane). The rate of glucose transport was 0.0031 mm/sec with 8mM of glucose (run number 4, highlighted); the rate decreased to 0.0017 mM/sec with 10mM of glucose (run 5, highlighted). Why was the rate of glucose transport slower when the concentration gradient was increased? Experiment Results Run Number Solute 1 1 2 2 3 33 4 4 5 6 6 Na Ch Glucose Na Ch Glucose Na Ch Glucose Nat Ch Glucose Na Ch Glucose Nat Cl Glucose Start Conc. L Start Conc. R (MM) (mM) 0.00 0.00 2.00 0.00 0.00 0.00 8.00 0.00 0.00 0.00 2.00 0.00 0.00 0.00 8.00 0.00 0.00 0.00 10.00 0.00 2.00 0.00 2.00 0.00 Carriers 500 500 500 500 700 700 700 700 100 100 700 700 Rate (mm/sec) 0.0000 0.0008 0.0000 0.0023 0.0000 0.0010…
- Calculate the equilibrium membrane potentials to be expected across a membrane at 37 ∘C, with a NaCl concentration of 0.50M on the "right side" and 0.08 M on the "left side", given the following conditions. In each case, state which side is (+) and which is (−). Membrane permeable only to Cl−.yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding media. suppose that a sample of yeast cells were placed in a 4% sodium chloride solution by weight. The density of solution is at 25 C = 1.02 g/ml, Molecular weight of solute = 58.44 g/mol, i of glycerol = 1 and R=0.08205 L-atm/mol-K What is the weight of solute in grams What is the moles of solute What is the volume of the solution in liters What is the molarity of the solution What is the value of the temperature to be used to solved for the osmotic pressure of the solution What is the osmotic pressure of solutionIntestinal epithelial cells pump glucose into the cell against its concentration gradient using the Na+– glucose symporter. Recall that the Na+ concentration is significantly higher outside the cell than inside the cell. The symporter couples the "downhill" transport of two Na+ ions into the cell to the "uphill" transport of glucose into the cell. If the Na+ concentration outside the cell ([Na+]out) is 163 mM and that inside the cell ([Na+]in) is 21.0 mM, and the cell potential is −54.0 mV (inside negative), calculate the maximum energy available for pumping a mole of glucose into the cell. Assume the temperature is 37 °C.
- Calculate the energy required for, or released in, a transport of 20 Na+ ions and of 100 molecules of glucose into a biological cell at 37 oC if the membrane potential is –50 mV (negative inside the cell), the concentrations of Na+ and glucose inside the cell are 0.001mol L-1 and 0.01mol L-1 consequently and the concentrations of Na+ and glucose outside of the cell are 0.1mol L-1 and 0.001mol L-1 consequently.For a typical vertebrate cell with a membrane potential of −0.070 V (inside negative), what is the free-energy change for transporting 1 mol of Na+ from the cell into the blood at 37 °C? Assume the concentration of Na+ insidethe cell is 12 mM and in blood plasma it is 145 mM.In the following situations, what is the free energy change if 1 mole of Na+is transported across a membrane from a region where the concentration is1 μM to a region where it is 100 mM?(Assume T = 37 °C.) (a) In the absence of a membrane potential. (b) When the transport is opposed by a membrane potential of 70 mV. (c) In each case, will hydrolysis of 1 mole of ATP suffice to drive the transport of 1 mole of ion, assuming pH 7.4 and the following cytoplasmic concentrations: ATP = 4.60 mM, Pi = 5.10 mM, ADP = 310 μM?