Session 1: Transport across cellular membranes

Human physiology is the integrated

study of the normal function

of the human body

ING338: Human Physiology for Engineers. 2017-1

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Themes across the topics of the course 1. Living organisms are causal mechanisms whose functions are

to be understood by applications of the laws of physics and chemistry.

2. Living organisms must obtain matter and energy from the external world. This matter and energy must be transformed and transferred in varied ways to build the organism and to perform work.

3. The cell is the basic unit of life. 4. Life requires information flow within and between cells and

between the environment and the organism. 5. Homeostasis (and “stability” in a more general sense)

maintains the internal environment in a more or less constant state compatible with life.

Mechanisms of cellular transport Fanny Casado, Ph.D. [email protected]


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Nature Cell Biology 15,1019–1027 (2013)

Cells are the organizational unit of life


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https://genographic.nationalgeographic.com/science-behind/genetics-overview/

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Notas de la presentación

Common shared features. Form follows function.

Learning objectives Explain the transport of biological material

as fluxes of heat, solute concentration and volume.

Contrast passive and active transport, and osmosis as responsible for the transport of small molecules across cellular membranes.


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Pressure driven flow; : ELECTRICAL FORCE, POTENTIAL, CAPACITANCE AND CURRENT

Activity: Intensive or extensive? A. Temperature B. Heat C. Volume D. Density E. Mass F. Concentration G. Moles H. Pressure I. Area J. Flow K. Flux L. Viscosity


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Extensive properties depend on the amount of

matter

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Intensive Extensive Extensive Intensive Extensive Intensive (as long as the solution is well-mixed) Extensive Intensive Extensive Extensive Intensive Intensive

Body Fluid Compartments The human body stripped of fat is ~60%

water Total body water (TBW) = 0.6 x 70kg = 42 L TBW= Intracellular fluid (ICF) + Extracellular

fluid (ECF) ECF = Interstitial space water (IS) +

Intravascular space water (IVS) ¾ ECF = IS


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The fluid volumes of the body are measured by the isotope dilution method. In this procedure, a known quantity of a radioactive substance (marker) is administered, allowed to equilibrate within the body compartments and then its concentration is measured in a known volume of plasma. As the total amount administered is known, the volume of the diluted marker can be calculated from its final concentration in the plasma. This quantity is corrected for any of the substance excreted during equilibration and for the half-life (decay) of the radioactive isotope over time. To measure the intravascular space (plasma), radiolabeled albumin is infused; for extracellular markers, inulin is used. The whole body water volume is determined by infusing tritiated water.

How much of the total body water is blood plasma?


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1/4x1/3=1/12

Activity http://www.wiley.com/legacy/college/boyer/0470003790/animations/membrane_transport/membrane_transport.htm


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Transport across cellular

membranes


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Coulomb’s Law describes electrical forces between charged particles

κ

Vacuum: κ=1 Other materials: k>1

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Electrical charge is a fundamental property of some subatomic particles such electrons (-) and protons (+). Signs are assigned arbitrarily. Coulomb’s law assumes electrical charges are separated in a vacuum F: force q: electrical point charges in coulombs r: distance in meters Epsilon 0: electrical permittivity of space =8.85x10^-12C^2N^-1m^-2

Electrical forces between charged particles


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Fig. 1.3.1. Electrical forces between separated point charges. Q indicates charge. Like charges repel, so that the force of q1 on q2 is directed away from q1 on a line connecting their centers. The force of q1 on q2 is exactly opposite to the force of q2 on q1. Unlike charges attract, with forces opposite but in line with the vector connecting their centers.

The electrical potential at A is the work per unit + charge from infinity to A


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A separation of charge produces an electric potential

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Potential energy U is the work to bring a positive unit charge from infinite separation to point A. The unit of potential is joules coulomb-1=volts Fint is the interacting electrostatic force. Coulomb’s law can be used because the force is directed along dS

Potentials are limited to conservative forces

Potentials are path-independent

Potential difference depends only on the initial and final state


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Fig. 1.3.2 Potential is independent of path. Any path from start to finish can be successively approximated by a series of paths oriented either parallel to the vector connecting the charges or perpendicular to it. Those components of the path perpendicular to the vector require no force and therefore contribute nothing to the potential at point A, which is defined as the work necessary to a bring a unit positive charge (here shown as q2) from infinite separation to point A. Components of the path oriented parallel to the vector connecting the point charges contribute Fdr to the force. Therefore, the total work (potential) moving the charge dependsonly on the radial separation and not the path taken

The electric field (force per unit charge) is the negative gradient of the potential


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E: electric field intensity is a vector -dU/dS is a scalar so for the equation to be equivalent we take a particular kind of derivative: gradient to convert scalar potential into vector force to obtain the three dimensional vector equation I,j,k are unit vectors in x,y,z The gradient is a vector that does not align with any axis, but it point in the direction of the steepest slope of the potential surface in three dimensions. The force points down this slope being the negative of the gradient of the potential

Gauss’s Law is another way of stating Coulomb’s Law


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Fig. 1.3.3 Gauss’s law for a sphere. Electric field surrounding a spherically symmetrical distribution of positive charge. ds is a vector having a magnitude of the area increment and directed normal to the closed surface. In this case, we take the Gaussian surface, indicated here by a dashed line, to be a sphere centered on the symmetrically distributed charge. The electric field vector and the surface normal vector are pointing in the same direction, so that the angle between them, theta is zero and the dot product of E and ds is Eds, because costheta=1

Capacitance of parallel plates depends on the area and plate separation


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Capacitance: ability to store electric charge Fig. 1.3.4 A parallel plate capacitor. Two plates, each of are A are separated by a distance delta. They are charged by connecting them to a battery that produces a capacitance current until the potential difference between the two plates is equal to that across the two poles of the battery, so that the net potential difference across the entire circuit loop is zero. At this point there is no more current flow. The separation of charges produces a uniform electric field between the two plates


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Fig. 1.3.5. Gaussian surface on a parallel plate capacitor. The Gaussian surface is the box indicated by the dashed lines. The electric field is constant within the capacitor and oriented as shown. The integral of E ds in the plane is zero because E is zero there. The integral of Eds in the dielectric between the plates is EA. The integral of Eds on the sides of the enclosed surface is zero because E and ds are orthogonal in this reqion.

Fig. 1.3.6. Biological membranes are capacitors


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Biological membranes are electrical capacitors

https://www.youtube.com/watch?v=moPJkCbKjBs

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Lipid bilayer membrane consisting of various lipid molecules arranged with their hydrocarbon tails toward the interior of the bilayer and their water-soluble parts facing the water phase

Activity A muscle cell approximates a right circular cylinder 10 cm long and 70 micron in diameter. The specific capacitance of the membrane is 1 micron f cm-2. A. What is the capacitance of the muscle membrane? B. How much charge is separated by this membrane to give a potential of -85 mV?


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A. Here C = Cm A = 1 micron f cm-2 x 2 phi r l. The surface area, 2 phi r l = 70 x 10-4 cm x phi x 10 cm = 0.2199 cm2. So the capacitance is 1 micron f cm-2 x 0.2199 cm2 = 0.2199 x 10-6 f B. The charge is given as q = V C = 0.085 volts x 0.2199 x 10-6 f = 1.87 x 10-8 coulomb

Electrical charges move in response to electrical forces


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Fig. 1.3.8. Forces on a charged particle in solution subjected to a constant electric field. The electrical force Fe is the product of the charge q on the particle and the electric field E. This electrical force accelerates the charged particle and it moves through the solution. This movement produces a drag force Fd which is proportional to the velocity v. The particle reaches a terminal velocity when the net force on the particle is zero Fe+Fd =0 Beta is a drag coefficient or frictional coefficient

Movement of ions in response to electrical forces makes a current and a solute flux


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The solute flux Js is related to the velocity and the concentration of the solute C Each solute particle carries the charge ze. Z is the valence +/- integral number of charges per particle and e is the unit charge of the electron The current density i is also related to the solute flux J Ending up with an analogue for the one-dimensional Ohm’s law


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Fig. 1.3.9 Relation between J, v and C. If solutes have an average velocity v they sweep out a distance vdt in time dt and this corresponds to an entire volume of solute equal Avdt moving to the right. The number of solute particles in this volume is Cavdt. The flux is this number per unit area per unit time J=Cadt/Adt=Cv

DIFFUSION


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A bolus of soluble material will gradually spread out in its solvent until a uniform solution results. This diffusion process must have been familiar to humans in antiquity. However,a mathematical description of these macroscopic changes in concentration was not available until the 1850s (Fick,1855),and a microscopic or particle-level model,not until the turn of this century (Einstein,1906). Diffusion plays an important role in such a wide range of disciplines,that it is important for students of science and engineering to develop an understanding of the macroscopic laws of diffusion and their microscopic basis. We will review some important characteristics of the macroscopic laws of diffusion and their relation to random-walk models. A fuller treatment is available elsewhere (Weiss, 1996a).

DIFFUSION


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Macroscopic laws of diffusion and their microscopic basis


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A bolus of soluble material gradually spreads out in its solvent until a uniform solution results. This diffusion process is quite intuitive to humans and people must have been familiar with it since antiquity. However, a mathematical description of these macroscopic changes in concentration was not available until the 1850s (Fick,1855),and a microscopic or particle-level model, not until the turn of this century (Einstein,1906).

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The macroscopic laws of diffusion for the simple case when the particles are not subject to a body force,the medium does not convect the particles,the diffusion coefficient is a constant,and the particles are conserved are summarized in one and three dimensions in Table 2.1. These equations relate the flux of particles (φ),which is the number of moles of particles transported through a unit area in a unit time,to the concentration of particles (c),which is the number of moles of particles per unit volume. Fick’s first law implies that a solute concentration gradient causes a solute flux in a direction to reduce the concentration gradient. The continuity equation follows from conservation of particles,and the diffusion equation is obtained by combining Fick’s first law with the continuity equation.

Definitions


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• Passive movement of materials down their concentration gradient.

• It takes place in an open system or across a permeable partition.

• There are two types of diffusion: simple and facilitated.

Difussion SIMPLE

Membrane permeable(lipid soluble) molecules cross membranes moving from a high to low concentration

Requires no energy expenditure.

Continues until equilibrium is reached

Occurs rapidly over short distances and slowly over long distances

FACILITATED

Membrane impermeable molecules can either enter or leave cells using transport proteins.

Transport proteins are classified as either transporters or channels


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Fick’s First Law of Diffusion


With the continuity equation applies to steady states The molar flux due to diffusion is proportional to the concentration gradient.

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D: diffusion coefficient

Fick’s Second Law can be derived from the random walk (concentration changes with time)


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Consider one dimension Particles travel in a straight line between collisions Step length 8 to the left or to the right Probability of going to the right or to the left = 1/2

Probability of finding a particle

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Fig. 1.6.1. Binomial probability density for the probability of finding a particle that initially was placed at m=0 as a function of the number of steps, N=10, 20, 50 and 100


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Fig. 1.6.2. Binomial probability distribution for N=50 with the envelope of the Gaussian distribution with N=50


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Fig. 1.6.3 Blow up of Fig. 1.6.2.


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The time for one-dimensional diffusion increases as the square of the distance


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Diffusion in cells is less rapid than in water


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Activity The free diffusion coefficient of oxygen in aqueous solutions is about 1.5 x 10-5 cm2 s-1. If the diffusion distance between air and blood is 0.5 µm, about how long is the diffusion time?


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The time is estimated from )t = x2 / 2 D = ( 0.5 x 10-4 cm)2 / 2 x 1.5 x 10-5 cm2s-1 = 0.25 x 10-8 cm2 / 3 x 10-5 cm2 s-1 = 0.083 x 10-3 s = 0.083 ms

Transporter proteins Membrane proteins that bind specific molecules

(substrates) and then carry them across the membrane by changing conformation (shape)

Small organic molecules such as glucose and amino acids cross membranes using transporters.

A transporter is open to one side of the membrane at any time.


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Transporter proteins


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Transporters never form a direct connection between the ECF and ICF compartments.

Channel proteins Membrane proteins that create a water filled

passageway that connects the ECF and ICF compartments

An open channel forms an aqueous pore across the membrane

Water and ions move through these channels. The open and closed state of the channels is determined by a part of the channel that acts as a “gate”.

The gating of the channel is controlled by ligands (chemically gated), electrical state of the cell (voltage gated) and by tension (mechanically gated) ING338: Human Physiology for Engineers. 2017-

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Activity The intestinal enterocytes form a covering over the intestinal lining which, to the first approximation, can be considered to be a plane. Assuming no binding or sequestration within the cell, what is the estimated time of diffusion of Ca2+ across the intestinal enterocyte? Consider that the effective diffusion coefficient of Ca2+ is about 0.4 x 10-5 cm2s-1. And the length of the enterocyte is 20 µm.


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The length of the enterocyte is 20 :m and assume that the effective diffusion coefficient of Ca2+ is about 0.4 x 10-5 cm2s-1. The time for one-dimensional diffusion is related to the distance by where x is the distance, D is the diffusion coefficient and t is the time of diffusion. In the problem, D= 0.4 x 10-5 cm2s-1 and x = 20 x 10-4 cm. Inserting these values, we calculate t = (20 x 10-4 cm)2 / ( 2 x 0.4 x 10-5 cm2s-1) = 0.5s

OSMOSIS Movement of water across membranes.

Osmosis only refers to the movement of water and is facilitated diffusion.

The water channel is called aquaporin. Water is never actively transported. Water

flows from compartments with “dilute” solutions (where concentration of water is high) to “concentrated” solutions (where concentration of water is low) until the concentrations of water and solute are equal in both compartments.


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OSMOTIC PRESSURE Prevents the movement of water across a

membrane. There is no osmotic pressure across a semi-

permeable membrane when the water concentration on each side of the membrane is equal.

Non-penetrating solutes (e.g., Na+) determine the movement of water and consequently the size of the fluid compartment.

Penetrating solutes (e.g., steroid hormones, urea) diffuse freely across cell membranes so no net water movement will occur


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Activity The extracellular fluid (ECF) volume varies with the size of the person. Suppose that in an individual, we determine that the ECF is 14 L. The average Na+ concentration in the ECF is about 143 mM. 1. What is the total amount of Na+ in the ECF. In moles? In grams? 2. Suppose this person works out and sweats 1.5 L with an

average Na+ concentration of 50 mM. During this time the urine output is 30 mL with an average concentration of Na+ of 600 mM. How much Na+ is lost during the workout?

3. If the person does not drink fluids at all during the workout, what will be the Na+ concentration in the plasma at the end of the workout? Assume that all of the fluid in the sweat and urine originated from the ECF.


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A. What is the total amount of Na+ in the ECF, in moles? In grams? Amount = Volume x concentration; = 14 L x 143 x 10-3 mol L-1 = 2002 x 10-3 mol = ; 2.002 mol. We convert this to grams by multiplying by the gram atomic weight of Na = 22.99 g mol-1: 2.002 mol x 22.99 g mol-1 = 46.03 g B. Suppose this person works out and sweats 1.5 L with an average [Na+] of 50 mM. During this time the urine output is 30 mL with an average [Na+] of 600 mM. How much Na+ is lost during the workout? 1.5 L x 50 x 10-3 mol L-1 = 0.075 mol = 1.72 g Na+ lost in the sweat 0.03 L x 600 x 10-3 mol L-1 = 0.018 mol = 0.41 g Na+ lost in the urine Total loss = 2.13 g Na C. If the person does not drink fluids at all during the workout, what will be the [Na+] in the plasma at the end of the workout? Assume that all of the fluid in the sweat and urine originated from the ECF. Final ECF = 14 L - 1.53 = 12.47 L; Final Na+ = 2.002 mol - 0.075 - 0.018 = 1.909 mol Final [Na+] = 1.909 mol/ 12.47 L = 0.153 M

Activity You suspect that you are anemic and your physician orders some tests. The clinical laboratory reports that your hemoglobin is 13 g %. A. What is the concentration of hemoglobin in molar units in your blood? B. Each hemoglobin molecule binds to four oxygen molecules. If the hemoglobin is saturated with oxygen, what is the concentration of O2 bound to Hb, in molar? C. What is the volume of molecular oxygen in your blood assuming that O2 is an ideal gas under conditions of standard temperature and pressure, dry (STPD)? ING338: Human Physiology for Engineers. 2017-

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A. What is the concentration of hemoglobin in molar in your blood? The [Hb] is 13 g% which is 13 g dL-1 x 10 dL L-1 = 130 g L-1. The molecular weight of Hb is 66,500 g mol-1 so the [Hb] in molar is [Hb] = 130 g L-1 / 66,500 g mol-1 = 1.95 x 10-3 M B. Each hemoglobin binds 4 oxygen molecules. If the hemoglobin is saturated with oxygen, what is the concentration of O2 bound to Hb, in molar? The oxygen concentration ought to be 4 times the [Hb] since each Hb molecule binds 4 oxygens. Thus it would be 7.82 x 10-3 M C. Convert the answer in B to volume using the ideal gas equation, PV = nRT where T is the absolute temperature, R = 0.082 L atm mol-1 °K-1, V is the volume that we seek and P = 1 atm. The conditions for volume of gas are usually STPD - standard temperature and pressure, dry. The standard temperature is 0 °C and pressure is 1 atm. We can convert this to units of volume %. One dL of blood would contain 7.82 x 10-4 mol (part C x 1L /10 dL). So PV = nRT where P = 1 atm, V is the unknown, n = 7.82 x 10-4 mole dL-1, R = 0.082 L atm mol-1 °K-1 and T = 273.1 °K. The volume per cent of gas at STPD is thus 0.082 L atm mol-1 °K-1 x 273.1 °K x 7.82 x 10-4 mol dL-1 = 175.1 x 10-4 L dL-1 = 17.51 mL dL-1

ACTIVE TRANSPORT Moves a molecule against its

concentration gradient and requires the input of energy (ATP). Active transporters are called “pumps”.

They exhibit specificity, competition, and saturation


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Session 1: Transport across cellular membranes

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Transcript of Session 1: Transport across cellular membranes