Nernst: Ex =RT In [X]。 [X] Goldman: Em =RT In PNa [Na]o+ Pk+[K+]o + Pai-[CI-]i PNa [Na]+Pk+ [K] + Pai-[CI]o 1. Calculate the reversal potential for sodium, potassium, and cloride at human body temperature if: [Na]o = 464 mM, [Na] 11 mM, [K] 36 mM, [K]=499 mM, [Cl] = 424 mM, [CI]=34 mM ENa+ = Ex+ = Ecl-
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- Table Q1(a) shows typical values for the intracellular and extracellular concentrations of the major ion species (in millimoles per litre) for frog skeletal muscle. Table Q1(a) Permeability (cm/s) 2 x 10-8 2х 10 Ions Intracellular Extracellular Na* 12 145 K+ 155 4 4 120 4 x 106 By referring to Table Q1(a), compute the equilibrium resting potential for this membrane by assuming the room temperature is 20° C. Given the Boltzman's constant, k = 1.38 x 1023 J/K and an electronic charge, q = 1.602 x 10-19 C.The simple form of |Hoff equation is: II = [B]RT In this equation the [B] is the molar concentration of solute. So: n m [B] = v MV = cg /MA Where c, the mass concentration of the solute is in the total volume of solution and M, is the molar mass of the solute. This equation can be replaced in the previous one to get: RT II = MA In this equation molar mass of given solute can be detemined from the slope of the II vs Cz plot. This equation applies only to solutions that are sufficiently dilute to behave as ideal-dilute solutions. In the case of non-ideal solutions, however, the extended formula is: II = [B]RT{1+ k. [B] + n. [B]² + ...} Biological macromolecules dissolve to produce solutions that are far from ideal, but we can still calculate the osmotic pressure by assuming that the van't Hoff equation is only the first term of a lengthier expression: II [B]RT(1+ b. [B]) II = RT + bRT. [B] [B] II = RT + bRT./M. */Ma п RT ÞRT Ca MA MA In this equation molar mass of given biomolecule can…Calculate the effective quantity (g) of sodium chloride related to tonicity in 100 mL of an intravenous fluid labeled "5% dextrose in 0.45% sodium chloride," and indicate whether the solution is isotonic, hypotonic, or hypertonic.
- A one-to-one protein (P)-ligand (L) complexation (P + L PL) has a dissociation equilibrium constant (Kd) value of 100 nM at 25°C, and the Kd remains the same at 37°C. 1) What is AS of binding at 25°C? Assume ACp of the binding is 0 over the temperature range. AS = 1.34E2 kJ/(mol*K) (note the unit!!) (sig. fig =3) 2) What is the concentration of the PL complex formed at equilibrium when you mix 0.20 uM (microM) of Protein and 1.0 uM of Ligand together at 37°C? PL at equilibrium = 8.1E-1 uM (note the unit!!) (sig. fig =2)In the Nernst equation [V = 62 log10 (Co/ Ci)], the term Ci represents: the extracellular concentration of potassium the extracellular concentration of sodium the membrane potential (in millivolts) the intracellular concentration of calcium the intracellular concentration of potassium Which of the following ions must be kept to very low concentrations within the cell cytoplasm in order to allow for enough substrate molecules to synthesize nucleotides and nucleic acids? HCO3- (bicarbonate) Ca2+ (calcium) PO43- (phosphate) Na+ (sodium) K+ (potassium)A 100 ml of a 1% sol'n of boric acid (freezing point = -0.29) isotonic w/ blood by NaCl (freezing point = -0.58). what grams is needed to make it an isotonic? what weight of NaCl is needed to a 2% sol'n of atropine sulfate (E=0.13) made isotonic with blood plasma?
- When the Cu2+ concentration is 1.01 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.138V. What is the Cr3+ concentration? 3Cu2(aq) + 2Cr(s) 3Cu(s) + 2Cr3+(aq) Answer: MCalculate the equilibrium membrane potentials to be expected across a membrane at 37 ∘C, with a NaCl concentration of 0.50 M 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 equally permeable to both ions.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−.
- 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 0.10 M on the “right side” and 0.01 M on the “left side”, given the following conditions. In each case, state which side is (+) and which is (-). (a) Membrane permeable only to Na+ (b) Membrane permeable only to Cl– (c) Membrane equally permeable to both ionsEstimate 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.