Pulmonary Diffusion Flashcards
Describe Henry’s Law for equilibrium gas solubility*
NOTE: the starred* flashcards mean that this is a determinant of gas exchange in the lung (next two)
In order to understand the relationship between alveolar gas composition and the concentration Of 02 and C02 in pulmonary capillary blood, first consider how much gas would be in the blood if alveolar gas is in equilibrium with the blood.
Henry’s law states that when a liquid and gas phase are in equilibrium at a given temperature, the concentration of gas in solution is directly proportional to the partial pressure of the gas. At equilibrium, the partial pressure of the gas is identical in the gas and liquid phases. For example, for carbon dioxide,
[CO2] = 0.03• PCO2
[gas]solution = (solubility coefficient) x P[gas]
where [CO2] is the gas concentration in solution (mlMol/L solution)
0.03 is the solubility coefficient of CO2 (every gas has a different solubility coefficient)
PCO2 is the partial pressure of CO2 (mm Hg)
The values of the solubility coefficient depend on the specific gas, the liquid in which the gas is dissolved, and also on temperature*.
The concentration of a gas dissolved in blood is expressed in vol% (or mmolar), or as the partial pressure in mm Hg of gas that would be in equilibrium with the solution in accordance with Henry’s law if a gas phase were present.
Partial pressure is thus used for both the gaseous and liquid phases. @ equillibrium, the partial pressure of the gas in the gaseous & liquid phases are identical (you can have partial pressures of gasses in solutions). However, partial pressures of gasses in water, does not contribute to the pressure of the water (e.g., blood pressure). Note that partial pressure of a gas only refers to the dissolved portion (i.e., dissolved O2, not O2 bound to hemoglobin).
In contrast, the dry gas fraction, or F, is used only for the gas phase; the concentration, or C, is used only for the solution phase.
The partial pressure is related to the chemical potential of the gas. If the partial pressures are unequal in the alveolar gas and blood, there will be a net movement of gas from the phase with high partial pressure to the phase with low partial pressure until diffusional equilibrium is reached, as defined by equality of the partial pressure in the two phases. Since the partial pressure of oxygen in alveolar gas (PA02) is greater than in pulmonary artery blood (PVO2), and since PAC02 is less than PVCO2, these gases are transferred by diffusion across the alveolar-capillary membrane.
Henry’s law describes the equilibrium between gaseous and liquid phases, while Fick’s law describes the rate of movement of gas between two compartments containing gases of differing partial pressures.
See figure, pg. 198
Describe Fick’s Law for diffusion of gases*
This explains gas diffusion & flow (O2 & CO2).
Fick’s law states that the flow of gas across the membrane is directly proportional to the area of the membrane and the difference in partial pressure of the gas in the alveoli and capillary blood, and inversely proportional to the thickness of the membrane.
V=[A x D x (PA -PC)]/ T
where V = the flow (ml/min) of a gas
A = the alveolar surface area available for diffusion (about 50 - 100 m2)
T = the thickness of the alveolar membrane (< 0.5 micron)
(PA - PC) = the difference in partial pressure (mm Hg) of the gas between the alveoli and the pulmonary capillary blood. THIS IS THE MOST IMPORTANT FACTOR. If PA - PC = 0, there is DIFFUSION–BUT NO NET DIFFUSION.
D = the diffusion coefficient (cm2/min/mm Hg) of the gas = CHEMICAL PROPERTY
The diffusion coefficient of a gas is in turn proportional to its solubility and inversely proportional to the square root of its molecular weight.
D is proportional to solubility/(MW)^.5
BY CONVENTION HOWEVER WE MEASURE:
V = Dlung x (PA -PC)
where D = (A x D)/T
When the lung gets edema, diffusion time increases to reach equillibrium since the membrane size increases.
Estimate the transit time* for gases in the pulmonary capillaries and to describe the time course of oxygen delivery and carbon dioxide uptake along the length of the pulmonary capillaries
Transit time is how much time it takes for the blood move through the capillaries in the lung.
x = vt also, t = V/Q
Under resting conditions, the blood stays in the pulmonary capillary for about 0.75 second and acquires the alveolar PO2 during the first 0.3 second. In exercise, the cardiac output increases, and the blood passes through he pulmonary capillaries more rapidly. When the cardiac output is increased 3-fold or less, there still is an equilibrium reached between the blood PO2 and the alveolar PO2 before the blood leaves the lungs. In other words, in the lung of the healthy subject, oxygen transfer is seldom if ever diffusion-limited.
Distinguish between perfusion – and diffusion – limited transfer of gases
Uptake of oxygen from the lung by the blood is perfusion limited, rising from 40 mm Hg in mixed venous blood to 100 mm Hg during equilibration with alveolar gas. Uptake of nitrous oxide is also perfusion limited. This occurs because they reach equilibrium far before their transit time is over. Carbon dioxide is also perfusion limited.
The transfer of carbon monoxide is diffusion limited since hemoglobin takes it up before equillibrium can occur.
CO2, O2, and nitrous oxide transfer in the healthy subject is not diffusion-limited but is perfusion-limited.
If there is a diffusion problem in a patient, transit time may increase for oxygen, & it may become diffusion limited.
**See pg. 202 - 203
Define pulmonary diffusion capacity; how it is determined
Diffusing capacity (DLCO) is the part of a comprehensive series of tests (the pulmonary function tests) that is done to determine the overall ability of the lung to transport gas into and out of the blood.
We use carbon monoxide to measure diffusion capacity since it does not reach equillibrium (cannot use O2 or N2O) since hemoglobin serves as a sink.
We use equation: V = Dlung x (PA -PC) & solve for D
All you do is ask the patient to breathe some air w CO in it for a few minutes (we know V-ventilation & PA & PC) & then measure CO expired. The difference of inspired from expired CO is VCO (ventilation CO).
PC = 0 since unbound CO concentration in blood = 0
(ViCO - VeCO) = Dlung x (PA - 0)
(ViCO - VeCO)/PA = Dlung
Describe:
1) how exercise can result in hypoxemia in patients with impaired diffusion capacity
2) the main features of inspiratory hypoxia of altitude
3) how breathing gas enriched in oxygen can aid the oxygenation of blood in patients with impaired diffusion capacity
1) In exercise, the transit time in the pulmonary capillaries is reduced, say by 2/3 to 0.25 sec since cardiac output increases. In this case, a normal individual would still equilibrate the pulmonary capillary blood with alveolar gas. But an individual with a diffusion problem would have exercise-induced hypoxemia, i.e. normoxemia at rest but hypoxemia during exercise. See pg. 204
2) At high altitude, the partial pressure of inspired oxygen is reduced, thereby decreasing the partial pressure of oxygen in the alveolar gas. A normal individual will equilibrate their end-capillary blood with alveolar gas, resulting in hypoxic hypoxemia. An individual with a diffusion problem will be even more hypoxemic since the end-capillary partial pressure of oxygen will be even lower than the alveolar partial pressure of oxygen. See pg. 205.
In all regions of the body except the lung, the body responds to hypoxemia by vasodialating to try & extract as much O2 as possible. But in the lung, the opposite occurs–vasoconstriction occurs in hypoxia because it needs to make sure that it can trap blood in the lung.
3) The hypoxemia that is caused by hypoventilation can be remedied by administering air that is only slightly enriched with O2 which will increase partial pressure of oxygen in the alveoli & increase partial pressure of oxygen in the capillaries to a better value to saturate hemoglobin. This case is similar to patients with abnormal righ to left shunts that have hypoxia without hypercapnea
Since the partial pressure of oxygen in the thin capillaries of the lung is higher, it diffuses into the ______ easily. Since pCO2 is higher in the blood, it diffuses easily into the _______ compartment. PO2 = ____mmHg, PCO2 = ___mmHg.
Since the partial pressure of oxygen in the thin capillaries of the lung is higher, it diffuses into the blood easily. Since pCO2 is higher in the blood, it diffuses easily into the alveolar compartment (capillaries). PO2 = 100mmHg, PCO2 = 40mmHg.
See pg. 196
Gas exchange in the lung occurs only in the _______.
Gas exchange in the lung occurs only in the capillaries.
______ventilation produces hypoxemia & hypercapnia. Why do MILD diffusion problems produce hypoxemia without hypercapnea?
Hypoventilation produces hypoxemia & hypercapnia. Why do MILD diffusion problems produce hypoxemia without hypercapnea?
This occurs because it is more difficult for O2 to diffuse than CO2, so the problem amplifies this for O2, but since CO2 diffusion is 20x stronger, this difference is not felt as much. Therefore, no hypercapnea but there is hypoxemia.
Do diffusion capacity problem, pg. 208
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What factors increase pulmonary diffusion capacity?
Remember DLCO = (D x A)/T
so anything that increases A or decreases T, increases DLCO. D is an intrinsic property & cannot be changed.
Body position: DLCO is greater in the recumbent position than in the uptight position because distension of the pulmonary vasculature by an increased pulmonary blood volume increases the area* available for diffusion.
Exercise: If pulmonary blood flow is increased 3-fold during exercise, P02 would still be 100 mm Hg in end-capillary blood. If cardiac output increases 5-fold during strenuous exercise, recruitment of capillaries and increased distension would still permit equilibration of capillary blood. In either case, increased alveolar distension, recruitment of capillaries, and a better matching of ventilation to perfusion all act to increase DLCO during exercise. Increase area.
What factors decrease pulmonary diffusion capacity?
DLCO = (D x A)/T
Pulmonary diseases and dysfunctions: DLCO may be decreased due to a decreased pulmonary surface area caused by (a) loss of lung tissue during surgery, as for removal of a tumor, (b) an airway obstruction caused by a foreign object in the airways, (c) a capillary obstruction such as an embolism, and (d) a mismatch of ventilation to perfusion.
DLCO may also be decreased by an increased diffusion distance due to (a) a thickened
alveolar membrane as might occur in pulmonary fibrosis, or (b) accumulation of fluids in pulmonary edema, such as might occur in pulmonary hypertension or during pulmonary inflammatory or allergic responses. Thickness increases here.
Loss of tissue = loss of area
DLCO decreases in anemia due to decrease in blood volume. Initially, increased breathing & increased CO (cardiac output) increase DLCO; however, there is a decreased hemoglobin concentration so sink concentration decreases so DLCO decreases. This does not mean that the diffusion capacity of the patient is decreased, it decreases only as measured by CO. So we correct this measurement for anemic patients.
The structures (or barriers) through which O2 must diffuse in passing from the surfactant containing liquid in the alveolar lumen to hemoglobin are in the following order:
Surfactant containing liquid –> alveolar membrane –> basement membrane –> capillary endothelium –> plasma –> erythrocyte membrane
Which of the following best describes the effects of a hypoxic environment on the pulmonary and systemic vascular resistances?
It is important for the blood to be distributed to those segment of those lungs where the alveoli are best oxygenated. When the oxygen tension of the alveoli decreases below normal, the adjacent blood vessels constrict causing their resistance to increase as much as fivefold at extremely low oxygen levels. This is opposite to the effect observed in systemic vessels, which dilate in response to low oxygen (i.i., resistance decreases).
Discuss the factors that increase diffusion.
Fick’s law of diffusion states that the rate of diffusion (D) of a gas through a biological membrane is proportional to ΔP, A, and S, and inversely proportional to d and the square root of the MW of the gas (i.e., D α ( ΔP X A X S) / (d X MW-2)
When the distance of the diffusion pathway is shorter, it will take less time for the molecules to diffuse the entire distance. When the molecular weight of the gas molecule is decreased, the velocity of kinetic movement of the molecule will be higher, which also increase the rate of diffusion.
