Can someone please anwer this questions BOTH parts and SHOW work on how to get t
ID: 891028 • Letter: C
Question
Can someone please anwer this questions BOTH parts and SHOW work on how to get the answer
2. a. At what rate does glucose enter a cell by passive diffusion alone? Estimate a typical cell as an ideal sphere with a diameter of 20 m. Remember that the surface area of a cell is 4r. The permeability coefficient(P-KD/x) for glucose is 3x10 8 cm/sec. For normal, steady-state values, use [glcextr-5 mM, and [glcJyto 0.1 mM. Use Fick's Law and give your answer in molecules per second. b. If the same cell is expressing 10' GLUT3 proteins (KM-1.5 mM) in its plasma membrane, each of which can maximally cycle 10f glucose molecules into the cell per second, at what rate will glucose enter the cell by facilitated transport? For easy comparison, give your answer in molecules per second.Explanation / Answer
Passive transport across membrane requires no energy input from the cell..for glucose,
facilitated diffusion, Vmax=500 micromoles per hour and K=1.5 mmol/cm3. For simple diffusion, A x P is 3 cm3/hour.
Fick’s Law-
Diffusion surface area x concentration difference/ distance
dn/dt = PxA(dc/dx)
where A is the membrane area and P is the permeability constant
dn/dt= 3x10^-10m/s x 4 x22/7 x 100x10^-12m^2(5-0.1mM/20x10^-6m)= 37.71428x10^-17 x2.45
= 9.4x10^-20Mcm^2/s = 9.4x10^-20 mol/s= 6.023x10^23x9.4x10^-20 =56.616x10^3molecules/s
b-dn/dt = Vmax /1+K(dc/dx) = 10^4/ 1+1.5mMx 0.245x10^6mM/m= 10^4/1.3675x10^6mM^2/m = 0.73126