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POLYCYSTIC RENAL DISEASE. 1 in 500 autopsies 1 in 3000 hospital admissions Accounts for ≈10% of end-stage renal failure Autosomal dominant inheritance. CYSTIC FIBROSIS. 1/2000 births in white Americans Median age for survival late 30s Autosomal recessive inheritance.
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POLYCYSTIC RENAL DISEASE 1 in 500 autopsies 1 in 3000 hospital admissions Accounts for ≈10% of end-stage renal failure Autosomal dominant inheritance
CYSTIC FIBROSIS 1/2000 births in white Americans Median age for survival late 30s Autosomal recessive inheritance
COMPARISON OF ION CONCENTRATIONS INSIDE AND OUTSIDE A TYPICAL MAMMALIAN CELL Intracellular Extracellular Concentration Concentration Component (mM) (mM) Cations Na 5-15 145 K 140 5 Mg 0.5 1-2 Ca 10-4 1-2 H 8 x 10-5 (pH 7.1) 4 x 10-5 (pH 7.4) Anions Cl 5-15 110 Because the cell is electrically neutral the large deficit in intracellular anions reflects the fact that most cellular constituents are negatively charged. The concentrations for Mg and Ca are given for free ions.
A Cross between Human Beings and Plants . . .SCIENTISTS ON VERGE OF CREATING PLANT PEOPLE . . .Bizarre Creatures Could do Anything You Want Tuesday, July 1, 1980
Simple Diffusion • Flux is proportional to external concentration • Flux never saturates Flux [S]o
PROTEIN MEDIATED MEMBRANE TRANSPORT • PRIMARY ACTIVE • SECONDARY ACTIVE TRANSPORT • FACILITATED DIFFUSION • ENDOCYTOSIS/TRANSCYTOSIS
Membrane Flux (moles of solute/sec) • Simple Diffusion • Carrier Mediated Transport • Facilitated Diffusion • Primary Active Transport • Secondary Active Transport • Ion Channels
Membrane Potential Review • The lipid bilayer is impermeable to ions and acts like an electrical capacitor. • Cells express ion channels, as well as pumps and exchangers, to equalize internal and external osmolarity. • Cells are permeable to K and Cl but nearly impermeable to Na. • Ions that are permeable will flow toward electrochemical equilibrium as given by the Nernst Equation. Eion = (60 mV / z) * log ([ion]out / [ion]in) @ 30°C • The Goldman-Hodgkin-Katz equation is used to calculate the steady-state resting potential in cells with significant relative permeability to sodium.
Carrier-Mediated Transport • Higher flux than predicted by solute permeability • Flux saturates • Binding is selective (D- versus L-forms) • Competition • Kinetics: [S]o << KmM a [S] [S]o = KmM = Mmax / 2 [S]o >> KmM = Mmax Mmax Flux 0.5 Km [S]o
k + k - So + Co SCo Si S = Solute C = Carrier Transport Kinetics dSCo/dt = k+ [S]o [C]o – k- [SC]o = 0 at equilibrium Þ k+ [S]o [C]o = k- [SC]o k-/k+ = ([S]o [C]o)/[SC]o = KmÞ [SC]o = ([S]o [C]o)/Km Fractional Rate = M / Mmax = [SC]o / ([C]o + [SC]o) M = Mmax / (1 + [C]o/[SC]o) = Mmax / (1 + Km/[S]o)
Co Ci So Si SCo SCi Mnet = Min – Mout = Mmax ) ( 1 1 1 + Km / [S]o 1 + Km / [S]i - Reversible Transport
Facilitated Diffusion • Uses bidirectional, symmetric carrier proteins • Flux is always in the directions you expect for simple diffusion • Binding is equivalent on each side of the membrane
Primary Active Transport: Driven by ATP • Class P – all have a phosphorylated intermediate • Na,K-ATPase • Ca-ATPase • H,K-ATPase • Cu-ATPase • Class V • H+ transport for intracellular organelles • Class F • Synthesize ATP in mitochondria
3 Na ATP ADP + Pi 2 K Primary Active Transport:Na,K-ATPase • 3 Na outward / 2 K inward / 1 ATP • Km values: Nain = 20 mM Kout = 2 mM • Inhibited by digitalis and ouabain • Palytoxin “opens” ion channel • 2 subunits, beta and alpha (the pump) • Two major conformations E1 & E2 • Turnover = 300 Na+ / sec / pump site @ 37 °C
Membrane Transport and Cellular Functions that Depend on the Na,K-ATPase
Secondary Active Transport • Energy stored in the Na+ gradient is used to power the transport of a variety of solutes glucose, amino acids and other molecules are pumped in (cotransport) Ca2+ or H+ are pumped out 2 or 3 Na+ / 1 Ca2+ ; 1 Na+ / 1 H+ (countertransport) • These transport proteins do not hydrolyze ATP directly; but they work at the expense of the Na+ gradient which must be maintained by the Na,K-ATPase
Energy available from ATP H2O ATP ADP + Pi DG = Gproducts – G reactants Chemical Energy (G) = RT ln [C] DG = DG° + 2.3 RT (log ([ADP] [Pi]) – log [ATP]) 2.3 RT = 5.6 kiloJoules / mole @ 20° C DG° = -30 kiloJoules /mole @ 20°C, pH 7.0 and 1M [reactants] and [products] “Standard Conditions”
DG = -30 – 5.6 log [ATP] kJ / mole [ADP] [Pi] Energy Depends on Substrate Concentrations • The energy available per molecule of ATP depends on: • [ATP] @ 4mM, [ADP] @ 400 µM, [Pi] @ 2 mM • Þ per mole of ATP hydrolyzed: • DG = -30 kJ – 5.6 kJ * log 4 x 10-3 • 2 x 10-3 * 4 x 10-4 • = -30 kJ - 21 kJ = -51 kiloJoules per mole of ATP • Converting to approximately -530 meV/molecule of ATP
Energy in the Sodium Gradient Consider Na+ movement from outside to inside: G = Gproducts – Greactants = Ginside – Goutside Gtotal = Gelectrical + Gchemical Conditions for our sample calculation: Vm = -60 mV [Na+]out = 140 mM [Na+]in = 14 mM and 2.3 RT = 60 meV / molecule
Energy in the Na Gradient: Electrical Term • Gelectrical = e mVin – e mVout • = +1e -60 mV – (+1e) 0 mV • = -60 meV • negative sign means energy is released moving from outside to inside • 60 meV is the energy required to move a charged ion (z=1) up a voltage gradient of 60 mV (assuming zero concentration gradient) * * * *
Energy in the Na Gradient: Chemical Term • DGchemical = 2.3 RT (log [Na+]in – log [Na+]out) • = 60 meV * (-1) • = -60 meV • negative sign means energy is released moving from outside to inside • 60 meV is the energy required to move a molecule up a 10 fold concentration gradient (true for an uncharged molecule or for a charged molecule when there is no voltage gradient)
Energy in the Na Gradient: Total • DGtotal = DGelectrical + DGchemical = -120 meV • 120 milli-electron-Volts of energy would be required to pump a single Na+ ion out of the cell up a 10 fold concentration gradient and a 60 mV voltage gradient. • Hydrolysis of a single ATP molecule can provide at least 500 meV of energy – enough to pump 4 Na+ ions. • A single Na+ ion moving from outside to inside would be able to provide 120 meV of energy, which could be used to pump some other molecule, such as glucose, an amino acid, Ca2+ or H+ up a concentration gradient
Example: Na+/Ca2+ exchange Compare the internal [Ca2+] for exchange ratios of 2 Na+ : 1 Ca2+ vs. 3 Na+ : 1 Ca2+ Vm = -60 mV, [Ca2+]out = 1.5 mM [Ca2+]in = ? Ca2+ moves from inside to outside DG = Gproducts – Greactants = Goutside – Ginside DGelectrical = (+2e) * (0 mV) – (+2e) * (-60 mV) = +120 meV DGchemical = 60 meV (log 1.5 – log ?)
Na+/Ca2+ exchange DGtotal = DGE + DGC = 120 meV + 60 meV log (1.5 / ?) 2 Na+Þ 240 meV 240 = 120 + 60 log (1.5 / ?) 120 / 60 = log (1.5 / ?) 102 = 1.5 / ? ? = 15 µM 3 Na+Þ 360 meV 360 = 120 + 60 log (1.5 / ?) 240 / 60 = log (1.5 / ?) 104 = 1.5 / ? ? = 0.15 µM Internal [Ca2+]can be reduced 100 fold lowerfor 3 Na : 1 Cavs 2 Na : 1 Ca
Summary: Energetics Transport Energetics • A molecule of ATP donates about 500 meV • It takes 60 meV to transport up a 60 mV electrical gradient • It takes 60 meV to transport up a 10 fold concentration gradient • A single sodium ion donates approximately 120 meV