Saddleback College Human Physiology Biology 12 Lab Unit Three (updated Sp2010) Table of Contents
Title
Page
An exercise in the use of the clinical spectrophotometer- clinical assay laboratory
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DNA, RNA, and protein electrophoresis, and ELISA
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PhysioEx: Exercise 10: Acid Base balance- complete all activities and 13-Intro worksheet, but be sure to read the introduction presented in this packet only An exercise in the determination of urine specific gravity and Long’s coefficient
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An exercise in the methods of urine analysis- chemical analysis
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An exercise in the methods of urine analysis- Microscopic sediments
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Renal problems
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PhysioEx: Exercise 9: Renal system physiology- complete all activities and See worksheet PhysioEx book PhysioEx-Exercise 7: Respiratory system mechanics. Perform all activities and See worksheet Physioex book Physio Ex 8: Chemical and Physical processes of digestion- Perform all See activities and worksheet Physioex book An exercise in the use of the spirometer-Lung capacities and volumes 33 Measurement of lung function: forced expiratory volume (FEV1.0)
38 Total points possible (48)
Grading rubric for lab binders: Each lab, within the three lab units, is worth 4 possible points. • 4 pts: A lab is thoroughly completed and highlighted/underlined and no data are missing. All questions/exercises are complete. Answers to questions were written thoughtfully and with insight. Lab notebook is neat and organized. • 3.75 pts: A lab question or two were not answered very well, but everything else looked great. • 3.5 pts: A lab is not highlighted, or maybe a question or two wasn’t answered, but the rest of the lab was completed with thought and insight. • 3 pts: lab is complete but one or two of the following were found: Answers to some questions display little care and thought, some data may be missing; highlighting may be missing; answers to some questions are missing Lab notebook is overall neat and organized, however. • 2 pts: Lab notebook is not neat and organized and is somewhat completed but some data are missing; answers to questions are not complete thoughts. Little to no evidence of pre-write highlighting/underlining; • 1 pt: Significant amounts of the lab are not completed or data missing. Little thought put toward answering lab questions. • 0 pts: lab notebook turned in late or lab not completed at all
AN EXERCISE IN THE USE OF THE CLINICAL SPECTROPHOTOMETER - CLINCIAL ASSAY LABORATORY INTRODUCTION The purpose of this exercise is to introduce to you how physiological substances such as glucose, protein, cholesterol, etc. are assayed (quantitatively analyzed) in the clinical laboratory. The principle involved is really quite simple if you use the analogy of the "ice tea concentration" example. Suppose you were given two glasses of ice tea and told that one glass contained more tea than the other glass. How would you determine which glass had the higher concentration of tea? Quite simple isn't it? You would look to see which glass contained the darker mixture of tea and water. One glass would contain the stronger tea because the more tea "molecules" in the glass, the more light is absorbed by these "molecules" and less light transmitted. Since more light is absorbed by the substance in solution, the more concentrated solution is the darker solution. This is easily within our everyday experience of selecting various solutions in our environment for drinking, cleaning, etc. The principle involved here is the basis of how the spectrophotometer in the laboratory works. This instrument measures how concentrated a solution is by how much light is absorbed by the solution. There is then, a direct relationship between the ABSORBANCE OF LIGHT and the CONCENTRATION of a solution. When blood or urine samples are sent to the clinical laboratory to assay for substances in them, a type of spectrophotometer is used for the assay. The Spectrophotometer used in our Physiology Laboratory will demonstrate the principles involved. MATERIALS NEEDED
(1) bottle of 0.1 M cobalt chloride (CoCl2) (1) bottle of 1% hydrochloric acid (HCl) (1) bottle of "Unknown A" (1) bottle of "Unknown B" (2) glass cuvettes (small glass tubes) (5) glass test tubes (large glass tubes)
(1) graduated 10 ml cylinder or 10 ml pipette with ProPipette (1) medicine dropper (1) spectrophotometer (spectronic 20+) (1) packet of lens paper
Because there are so many glass items, be sure none of them are broken to begin with. When you return the materials, be sure none of the glass items are broken.
PROCEDURES 1. The following procedures should be used to operate the spectrophotometer (spectronic 20+) AFTER YOU HAVE READ THE ENTIRE EXERCISE:
a. Turn on the unit and allow a 15 minute warm-up 2
b. Find the 0%TI knob and turn it all the way to the right. c. Adjust WAVELENGTH to 520 nanometers - This particular wavelength of light, 520 nanometers, is what CoCl2 absorbs best. Each substance has a particular "Absorbance Maximum" (Amax) which is the wavelength the substance shows the highest absorbance for maximum test accuracy. d. Place glass cuvette with 1% HCl (blank) into CUVETTE HOLDER e. Adjust the absorbance so that the readout gives .000 absorbance f. Replace blank cuvette with sample cuvette and read Absorbance from DISPLAY
2. Examine the general spectrophotometer diagram at the end of the Exercise. The "light source" is the white light you will turn on when you turn on the instrument. The "lens" will focus the light through the "diffraction grating" which spreads the white light into all of the different wavelengths or colors of the spectrum (ROYGBIV). The movement of the "plate with slit" is controlled by the wavelength knob. When the wavelength knob is at the "520 nm" setting, the plate is moved in position so that only the wavelength "520 nm" is allowed to go through the slit. This selected wavelength is then allowed to strike the test solution in the "cuvette". The "phototube" reads how much light is absorbed by the solution and the results are displayed on the digital ABSORBANCE scale on the front of the instrument.
3. Take one of the clean cuvettes from the kit and fill it nearly full with 1% HCl. This will serve as the "blank". A "blank" is always used to subtract out the absorbance of the diluting solution so that changes in absorbance can only be due only to the changes in concentration of the test solution (ex. cobalt chloride). It is similar to “taring” a weighing paper on a scale. Since the CoCl2 is dissolved in the 1% HCl, the HCl serves as the "blank". Insert the "blank" in the cuvette holder in the instrument. Place the light cover (plastic cap) over the holder to prevent any background light from entering the holder. Consider your “blank”, which has no cobalt chloride. Be sure to use this zero absorbance value for
the 0.00 M CoCl2 solution when you plot a graph later. 4.
You now need to prepare some standard known solutions of cobalt chloride that will serve to give you a standard curve to be used later when you look for the concentration of CoCl2 in two Unknown solutions. From your stock solution of 0.1 M CoCl2, you will prepare standard solutions of 0.02 M, 0.04 M, 0.06 M, and 0.08 M of CoCl2. Use the five test tubes, the graduated cylinder, and the medicine dropper (to be exact) to prepare the solutions as follows: 0.02 M = use 1 ml of 0.1 M CoCl2 and 4 ml of 1% HCl. 0.04 M = use 2 ml of 0.1 M CoCl2 and 3 ml of 1% HCl. 0.06 M = use 3 ml of 0.1 M CoCl2 and 2 ml of 1% HCl. 0.08 M = use 4 ml of 0.1 M CoCl2 and 1 ml of 1% HCl. 0.10 M = use 5 ml of 0.1 M CoCl2 stock solution, with no HCl.
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5. Be sure that you have a total of 5 ml of each of the standard solutions. Use the other cuvette in the kit for your standard solutions. Remove the "blank" cuvette from the cuvette well. Fill your other cuvette nearly full with the 0.02 M solution. Place this cuvette into the cuvette well and read the ABSORBANCE and record the value. Clean out the cuvette and repeat using the 0.04 M solution. Repeat this process for all of your standard solutions AND also include a reading for the 0.1 M stock solution from your kit. This should give you a total of five ABSORBANCE readings to be used to draw the standard curve graph. 6. Now take the "Unknown A" and "Unknown B solutions from your kit. These two solutions contain an unknown concentration of cobalt chloride. It will be your job to determine what the concentration of CoCl2 is in each of these unknown solutions. The task is similar to finding, for example, the amount of glucose in a blood serum sample. The principle is the same. Fill the cuvette with unknown A solution and read the ABSORBANCE. Repeat for Unknown B. Why are you supposed to use the same cuvette each time?
7. Now you are ready to draw the standard curve for the experiment. Use the following graph paper to plot your known absorbance readings and concentrations. Draw a graph with the Molarity concentration for the "X" axis and the ABSORBANCE for the "Y" axis. Plot each of your known standard solutions against your ABSORBANCE readings. Do not include your "unknown solutions" yet.
8. After you have plotted all of the five points on the graph, draw a STRAIGHT line through the points (see diagram). If your points fall in such a way that you cannot draw a straight line through them, draw a straight "best line" through as many of them as you can. The line should pass through the 0 ABSORBANCE and ) Molarity apex of the curve (see diagram). When you have drawn the standard curve you now have a curve that relates the ABSORBANCE to the CONCENTRATION (molarity) for CoCl2.
9. Now take your two unknown solution ABSORBANCE figures. Put a point for unknown A on the ABSORBANCE axis where it belongs. Now draw a straight line to the right of that point until it hits the standard curve line (see diagram). Make a small point on the standard curve where the two lines meet. Draw a straight line down from that point until it crosses the molarity axis. Draw a small point where the lines cross. This is your concentration for Unknown A CoCl2. Now read the molarity for unknown A and record. Repeat this for Unknown B. If needed refer to the diagram.
10. You now have the answer to the Exercise. You set out to find the concentration of CoCl2 in these unknown solutions. By comparing known absorbance readings and concentrations with the unknown values, you can determine what the CoCl2 concentrations in your Unknown solutions are—as long as your standard solutions obey the Lambert-Beer law. (A= єcl; A1/C1= A2/C2)
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Discussion and Study Questions: What is the Lambert-Beer Law and how can you use it to mathematically calculate an unknown’s concentration?
What is the design of a spectrophotometer and how does this design allow it to detect a solution’s concentration?
What is the importance of a standard curve? (make sure you REALLY understand this question).
How did you generate your standard curve?
How did you use your standard curve to determine the concentration of your unknown?
What was the importance of using a blank?
What was the importance of using the proper wavelength?
Why is using the linear range of your standard curve to determine your “unknown” concentration so important?
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Electrophoresis and Serological Testing I.
INTRODUCTION
One of the most important facts you will learn in lecture is that blood serum or plasma contains about 7 grams of protein per 100 ml (7 gm% or 7%). This protein will be referred to many times during the course as a source of different physiologically important proteins and as the basis for the plasma colloid osmotic pressure which will be discussed later in the course. The clinical laboratory is very interested in being able to separate the serum protein into distinct observable bands that can be used to diagnose various disease states. In 1937, a Swedish biochemist, Tiselius, introduced the process of electrophoresis. Electrophoresis is a fast, accurate way to separate these proteins. The basic principle of electrophoresis is that serum protein molecules are charged with a negative charge and that these molecules will migrate towards the positive pole (anode) in an electric field. Proteins are negatively charged because they contain an abundance of amino acids with R groups containing carboxyl units (COOH, which are acidic groups because they can give up their H+ ion). Amino acids such as aspartic and glutamic acids have such 'R" groups. When these amino acids are in a basic pH medium, such as blood serum which has a pH of 7.4 the "R" group dissociates the hydrogen ion (H+) and leaves the electron behind on the oxygen of the 'R" group (COO-). If enough of the amino acids on a protein have this property, then the net charge on the protein is negative. When electrophoresis is used for the separation of serum proteins, a buffer of pH 8.8 is used to maintain the basic pH and therefore, the negative charge on the proteins. II.OBJECTIVES 1. To become familiar with the fact that serum/plasma proteins are negatively charged.
2. To understand principles of electrophoresis as they pertain to the separation of DNA, RNA, proteins such as serum/plasma proteins. 3. To understand the principles of enzyme-linked immunosorbant assays and their advantages and limitations compared to western blotting. III. CONCEPTS DNA, RNA and proteins all have molecular groups that can cause the overall molecule to have a net charge. For example, proteins have charges associated with the R groups of each amino acid that make up the protein strand. Depending on the particular R group of each amino acid, it may carry a negative charge, a positive charge or no charge at all. (Consult your textbook for the structure of the 20 amino acids, and determine which are acidic, basic or neutral amino acids.) The net charge on any protein is the sum of all the charges displayed on that protein. Placing a protein in a buffer solution of a specific pH may help to expose charges on R groups by dissociating or binding H+ ions. Similarly, placing DNA or RNA into a basic buffer will cause the phosphate backbone to dissociate H+ ions and the DNA/RNA molecule will have a net negative charge. In electrophoresis, we make use of the net charge on DNA, RNA, or proteins to help us separate or sort the DNA and RNA into various sized molecules or various proteins found in human serum. In the case of serum protein electrophoresis (SPE), a small volume of a patient’s serum is placed on an absorbent paper strip (nitrocellulose.) The strip is then placed in a chamber containing buffer of a particular pH (8.8 in this case.) A voltage is applied across the chamber using a power supply hooked to electrodes (wires) in the chamber. A voltage is applied, such that the anode (+ electrode) and the cathode (- electrode) are at opposite ends of the absorbent paper strip. Negatively charged proteins move across the strip toward the anode (by charge attraction.) Protein mobility is favored by the voltage applied to the strip, but is hindered by the size of the protein (larger proteins move slower than smaller proteins of the same charge.) Below are the major bands or peaks produced in serum protein electrophoresis of the normal person.
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The general principle that governs electrophoresis is given by the following formula: Mobility of a molecule= applied voltage x net charge on the molecule Friction of the molecule The rate of movement of a molecule is increased by increasing the applied voltage or increasing the net electrical charge of the molecule. The rate of movement decreases with increasing friction caused by molecular size and shape. The net electrical charge of a molecule also determines the direction it will migrate. A molecule with a net positive charge will migrate toward the negative side of the chamber. The net negatively charged molecule will move toward the positive side of the chamber. DNA, RNA, and amino acids, have chemical characteristics that make them easy to separate by electrophoresis. Their molecular weight is low and their physical size offers little to no friction to decrease mobility. The amphoteric nature of amino acids may be exploited to great advantage in electrophoresis. Amphoteric molecules may possess a net positive charge, a net negative charge, or be electrically neutral depending on their environment.
The pH at which an amino acid exists in solution as a neutral molecule, a zwitterion, is called its isoelectric point (pI). As zwitterions, the amino acid is electrically neutral and will not migrate to either side of the electrophoresis chamber. As depicted below the pH at which an amino acid exists determines its net electrical charge and the type of charge (+ or -).
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If you increase the pH of a solution of an amino acid by adding hydroxyl ions, the hydrogen ion is removed from the -NH3+ group. During electrophoresis, this amino acid would move toward the anode (the positive electrode)
If you decrease the pH by adding an acid to a solution of an amino acid, the -COO- part of the zwitterion picks up a hydrogen ion.
This time, during electrophoresis, the amino acid would move towards the cathode (the negative electrode).
Each amino acid has its own isoelectric point so its net charge will vary with the pH of the solution it is in. It should be noted that amino acids in their crystalline form are electrically neutral (zwitterions).
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INTRODUCTION: ELISA (Enzyme Linked Immunosorbant Assay) The purpose of ELISAs is to detect specific protein antigens using antibodies. In laboratory medicine, ELISAs are commonly used to screen serum samples from animals to detect antibodies to a specific pathogen. In the research environment, ELISAs can be modified to detect a protein of interest. Method: 1. 96 well plates are coated with an antigen. Sites unoccupied by the antigen are blocked with a blocking buffer to prevent non-specific binding of sample. Draw this below:
2. Serum samples are added and incubated to allow specific antibodies to bind to specific antigen. Antibodies that do not recognize and bind specifically to the antigen remain unbound. Draw this below:
Positive sample
Negative sample
3. Non-specific antibodies that do not bind to the antigen are washed away. Draw this below.
Positive sample
Negative sample
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4. An enzyme labeled (e.g. alkaline phosphatase or horseradish peroxidase) secondary antibody is incubated with the samples and binds tightly to sample antibody. Draw this below.
Positive sample
negative sample
5. Unbound labeled antibody is washed away and a colorimetric substrate is added. Draw this below.
Positive sample
negative sample
6. The enzyme cleaves the substrate causing a color change of the substrate solution. The intensity of the color is quantified using a spectrophotometer and is proportional to the amount of antibody present in the sample. (Obviously, a standard curve would have to be generated, could you think of how to do that?).
Positive sample
negative sample
Follow the instructions on the PhysioEx handout, provided on the next few pages. The following link has a very nice animation of ELISA: http://www.biology.arizona.edu/IMMUNOLOGY/activities/elisa/technique.html?
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Compare and contrast the ELISA and the Western Blot Technique.
Which test do you think would be more specific, the indirect ELISA or the Western blot technique? Why?
Why are proteins generally considered to be negatively charged? What are “R” groups? How can proteins be manipulated so that the entire protein is as negatively charged as it can be? Once a protein is negatively charged, how can a sample of proteins be separated electrophoretically? What happens to the proteins as they “run” out on the gel?
Did you know that pregnancy tests are a modified ELISA (capture ELISAs). Can you think of how the antibodies are arranged to detect human chorionic gonadotropin, the “pregnancy hormone?”
If a woman used a standard pregnancy test, very early in her pregnancy, why could she get a “false negative?”
When might these tests be used for clinical purposes?
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PhysioEx: Acid/Base Balance Introduction: (Although lengthy, this introduction was written to help you with the next 2 labs). pH is the measure of the hydrogen concentration of the solution: -log[H+] where [H+] is usually expressed in moles/L. Don’t let the word log scare you- it simply means the number over the base ten (i.e., the exponent). Also remember, that each increase of pH is a jump of ten H+ concentration (moles/L) 1 x 10 -2 1 x 10 -3 1 x 10 -4 1 x 10 -5 1 x 10 -6 1 x 10 -7 1 x 10 -8 1 x 10 -9 1 x 10 -10 1 x 10 -11 1 x 10 -12 1 x 10 -13 1 x 10 -14
pH 2 3 4 5 6 7 8 9 10 11 12 13 14
Decimal points .01 .001 .0001 .00001 .000001 .0000001 .00000001 .000000001 .0000000001 .00000000001 .000000000001 .0000000000001 .00000000000001
An acid is any substance that will release “free hydrogen ions (free in the solution)”. Hydrogen is what changes the pH- it will cause it to go down in number. For strong acids- the dissociation of the hydrogen ion is very easy and quick and essentially all the hydrogen ions will dissociate (only a few will re-bind). (See “A” below). A weak acid (Acetic acid) is a weak acid- maybe only a small percentage of the molecules will dissociate. And the likelihood of the hydrogen ions re-associating is about equal to dissociation. (See “B” below). A.
HCl
B. CH3COOH
H+ + Cl CH3COO- + H+
A base is any substance that accepts free hydrogen ions (i.e. binds them and takes them out of solution) and will make the solution less acidic. Hydroxyl radicals are very strong bases (NaOH) NaOH + HCl = NaCl + H2O pKa- refers to the “dissociation” constant for a particular weak acid or base. A weak acid and its conjugate base can be used together to make a buffer; similarly a weak base and its conjugate acid. If a 13
buffered solution suddenly has a lot of acid dumped into it, the weak base in the buffer could quickly bind the excess hydrogen ions and prevent drastic changes in the pH of the solution. A similar effect would be seen if a strong base were added: the weak acid in the solution would donate hydrogen ions and these would bind to the excessive hydroxyl radicals to form water. The buffer works best if its pKa is closely matched to the desired pH. In other words, we pick the buffer whose pKa is closest to the pH we want. Chemists use pKas to know where the buffer works the best. If a cell has a pH 7.40, we would want a buffer with a pKa closest to 7.40. To make a buffer, take a beaker of water and add the weak acid/base and conjugate base/acid. The pKa is the pH that is produced by adding equal amounts of both partners.
Strong acid
Let’s review: To the left is a picture of a beaker of water and pH is 7.0. If we drop in HClthe acidity of the solution would climb due to the increase in hydrogen ions in solution (we have introduced a strong acid that dissociates quickly and adds H+ to the solution). The pH would drop. Reverse this for a strong base. 2
7 pH
water
So how do buffers work? In our solution we have a weak acid and a conjugate (weak base). We need both partners for our solution to be ‘buffered.” One partner donates hydrogen ions and the other accepts them. Now if we add a strong acid- the weak base (conjugate base) will quickly sponge up the hydrogen ions from the strong acid and the pH will not change drastically. If we add a strong base, releasing hydroxyl ions, the weak acid will donate hydrogen ions to neutralize the strong base (to make water). (see picture to the right).
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Conjugate base Weak acid
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We have three major buffer systems in our body that maintain the pH of the blood. 3 Major physiologic buffer systems: 1. PROTEINS- can act as buffers because proteins are long chains of amino acids. Each amino acid has a variable R group- we have neutral, acidic and basic R groups. An acidic amino acid can dissociate and release H+ ions. Basic amino acids have an R group that can pick up an H+. So proteins, to a certain degree, can act as a buffer. Too much of an acid/base change will end up denaturing the proteins (you will unravel them). The best buffering proteins have many histidines in them because histidines have an optimal pH that is closest to that of blood. Hemoglobin has many histidines and works as a good buffer. Albumin doesn’t have many histidines, but because we have so much, it still can act as a weak buffer. This protein system isn’t strong enough alone- if we relied on it completely, it wouldn’t be enough for our body. pKa of histidine is 6.5 2. PHOSPHATE –Our biggest extracellular source is calcium phosphate from bone (calcium hydroxyapatite). pKa of 6.8 This system is particularly important in buffering acids within the renal tubules.
Na2HPO4 + H+
NaH2PO4 + Na+
3. BICARBONATE/CARBONIC ACID BUFFER SYSTEM- Remember that the blood’s set point is pH is 7.40. All our waste products will lower the pH- so our body needs buffers to maintain the pH. Cells are constantly disturbing the pH system- and our lungs and kidney are in charge of maintaining the acid base system. Kidneys will remove, reabsorb, or create bicarbonate and the lungs (breathing rate) can control the amount of carbonic acid in the body. Both systems will help buffer the pH. This is the system we can regulate with our kidney and respiratory system. Carbonic acid is regulated by the respiratory system and the bicarbonate is regulated by the renal system. The pKa for the bicarbonate buffer system is 6.1. Notice that this pKa is the worst for our pH! So why does our body choose this one? It’s because we can control it physiologically. This means we have a better than predicted (chemically) buffer system. A chemist can’t change constantly what is in the beaker. Our body acts as a chemist that can constantly change the dynamics of the solutions, and therefore, this is a huge advantage. Carbonic anhydrase
CO2 + H2O Carbon Dioxide
H2CO3 Carbonic acid (weak acid)
Respiratory System
H+ + HCO3-
Bicarbonate ion (conjugate base)
Renal System
Carbonic anhydrase is an enzyme that can speed up this reaction. It is found in several places throughout the body: kidney tubules, RBC membrane, and epithelial cells of the stomach. Once carbonic acid forms it spontaneously dissociates to bicarbonate ion and H+. From now on, we have license to know if we have 15
carbon dioxide, carbonic anhydrase will automatically transform it into carbonic acid. We can think of carbon dioxide as a weak acid (because it will be converted so quickly). Bicarbonate is a weak base.
How does the kidney regulate the plasma levels of bicarbonate? -
it can maintain plasma [HCO3-]: freely filter and then reabsorb all of it. No bicarbonate in the urine and the plasma levels will stay the same, hence maintained.
-
it can lower the plasma [HCO3-]: freely filter the bicarbonate and don’t reabsorb any. Now the urine will have bicarbonate and plasma levels will drop. Now urine pH will be more basic and plasma will be more acidic.
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it can raise the plasma [HCO3-]: Freely filter and then reabsorb all of it, but the distal convoluted tubule can manufacture new molecules (so we generate more). Just because it has carbonic anhydrase, it can synthesize bicarbonate, that is reabsorbed by the peritubular capillary, and the blood gets more basic, and the hydrogen ion gets pumped into the tubule lumen, and the urine becomes more acidic. HCO3- concentration is measured in mEq/L
Peritubular capillary This reaction is occurring in the PCT or DCT Carbonic anhydrase
CO2 + H2O
H2CO3
H+ + HCO3-
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How does the respiratory system regulate the plasma levels of CO2? Carbon dioxide is a gas, and it is measured in Torr or mm of Hg (the partial pressure it exerts: pCO2).
expiration
pCO2 Glucose
CO2 + H2O pCO2
cell
capillary
alveolus
Normal ventilation is when the rate of CO2 production exactly matches what is blown off. Hyperventilation occurs when the rate of removal of CO2 from the blood stream is higher than the rate at which the cells are dumping CO2 into the blood stream. So, plasma levels decrease. Hence, blood will become more basic and the pH will rise. Hypoventilation occurs when the rate of removal is lower than the rate of production. Hence plasma levels of CO2 will increase and blood will become more acidic and the pH will drop. There are chemoreceptors (they sense levels of chemicals in the blood and CSF) in the same places as the baroreceptors (aortic arch and carotid sinus). They report to the medulla and pons- both pressure and chemical makeup (pCO2). The medulla and pons then alter your ventilation to maintain proper levels of pCO2. Ventilation involves the rate and depth of inspiration and expiration. So blood gases are controlled by the respiratory system (lung and brain wiring). Every time we exhale, we blow off acid. If we hold our breath (or slow our breathing) we retain more acid. Together the kidney and lung help to keep us in acid/base balance. Acidosis (pH<7.40) Too much acid (High CO2) Too little base (Low HCO3-
Alkalosis (pH >7.40) Problem Too little acid (low CO2) Respiratory
Compensation Kidneys
Too much base (High metabolic HCO3-)
Kidneys and lungs
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Although buffers in body fluids help resist changes in the pH of those body fluids, the respiratory system and the kidneys are the main regulators of pH in the body fluids. Malfunctions of either the respiratory system or the kidneys can result in acidosis or alkalosis. The major effect of acidosis is depression of the central nervous system (CNS). When the pH of the blood falls below 7.35, the CNS malfunctions, and the individual becomes disoriented and possibly comatose as the condition worsens. The major effect of alkalosis is hyperexcitability of the nervous system. Peripheral nerves are affected first, resulting in spontaneous nervous stimulation of muscles. Spasms and tetanic contractions and possibly extreme nervousness or convulsions result. Severe alkalosis can cause death as a result of tetany of the respiratory muscles. Acidosis and alkalosis are categorized by the CAUSE of the condition. Respiratory acidosis or respiratory alkalosis results from abnormalities of the respiratory system which, in turn, allow too much CO2 to be retained or too much CO2 to be eliminated. Metabolic acidosis or metabolic alkalosis results from all causes other than abnormal respiratory functions. Inadequate ventilation of the lungs causes respiratory acidosis (see Table below), because the rate at which CO2 is eliminated from the body fluids through the lungs decreases. The result of this is an increase in the concentration of carbon dioxide in body fluids. As carbon dioxide levels increase, excess carbon dioxide reacts with water to form carbonic acid. The carbonic acid then dissociates to form hydrogen ions (H+) and bicarbonate ions (HCO3-) as per the following (reversible) reaction: H2CO3 → H+ + HCO3-. The increase in H+ concentration causes the pH of body fluids to decrease. If the pH of body fluids falls below 7.35, symptoms of respiratory acidosis become apparent. TABLE: Clinical Observations of Acidosis and Alkalosis Causes Respiratory Acidosis (elevated CO2) • Hypoventilation due to depression of respiratory centers via narcotic, drugs, anesthetics • Hypoventilation due to CNS disease and depression, trauma • Interference with respiratory muscles by disease, drugs, toxins • Restrictive, obstructive lung disease: severe emphysema or advanced asthma Respiratory Alkalosis (lowered CO2) • Hyperventilation syndrome (psychological) • Bacteremias, Fevers • Overventilation on mechanical respirators • Ascent to high altitudes Metabolic Acidosis (lowered HCO3-) • Ingestion, infusion or production of fixed acids. Direct reduction of body fluid pH as acid is absorbed (ie, ingestion of acidic drugs such as aspirin) • Decreased excretion of acid by kidney • Diahrrea (loss of intestinal HCO3-) • Accumulation of lactic acid in severe hypoxemia • Diabetic ketoacidosis, production of large amounts of fatty acids/other acidic metabolites such as ketone bodies (ie, untreated diabetes mellitus) • Inadequate oxygen delivery to tissues resulting in anaerobic respiration and lactic acid build-up (ie, exercise, heart failure, or shock) Metabolic Alkalosis (increased HCO3-) • Excessive loss of fixed acids due to ingestion, infusion, or renal reabsorption of bases • Loss of gastric juice during vomiting • Intake of stomach antacids • Diuretic therapy (loss of H+ ions)
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AN EXERCISE IN THE DETERMINATION OF URINE SPECIFIC GRAVITY AND LONG'S COEFFICIENT INTRODUCTION The specific gravity of a liquid is the ratio of the density of the liquid to the density of pure water (1.0 gram/ml). The closer the specific gravity is to 1.000, the closer it is to pure water. When solutes are added to a liquid, the specific gravity increases. Therefore, the specific gravity of a urine sample is indicative of the amount of solutes present in the urine. The higher the value, the more solutes are dissolved in the urine. The use of Long's coefficient is a relative mathematical means of comparing the specific gravity of a urine sample to the actual amount of solute present in the sample. The normal specific gravity for various urine samples are given below: Adults: Random Specimen = 1.010-1.025 24 hour Specimen* = 1.015-1.018 * total urine collected during a 24 hour period Newborn babies Random Specimen = 1.002-1.004 There is a progressive decrease in the urine specific gravity as one reaches middle age and over. When using the urinometer to measure the specific gravity of a sample, it is important to note that the scale on the urinometer is calibrated only for a temperature of 15.5 degrees C. In order to obtain a correct reading of any given urine sample the temperature of the sample must be known and corrected to the calibrated temperature. MATERIALS NEEDED The kit in the metal cabinet contains the following: (1) urinometer (special hydrometer for urine specific gravity) (1) urinometer cylinder (1) bottle of sodium chloride crystals Thermometers are available on the front table. PROCEDURES
1. Before actually measuring the specific gravity of a urine sample, you should practice reading the urinometer
2.
3. 4.
5.
scale in different solutions. Do this by filling the cylinder about 3/4 full of deionized water (white DI-on laboratory sink). Add a small pinch of sodium chloride to give some solute to the pure water. Read the temperature of the water sample using the centigrade thermometer. Grasp the urinometer float at its top and slowly insert into the cylinder (insert weighted end first - see diagram). Avoid wetting the float stem above the liquid line. Excessive wetting of the stem will cause the float to sink below the true test reading. Be very careful of the urinometer float because they can break easily. Using your thumb and index finger, impart a slight spin to the float as it is released into the sample. Remember that the urinometer must be able to float in the liquid, add more liquid if necessary to insure a free floating urinometer. This spinning helps keep the float in the middle of the sample and not against the cylinder sides. Read the scale of the urinometer at the lowest portion of the meniscus (see diagram). To correct the reading for the temperature variation, use the following method: Take the recorded temperature of the sample in either C or F degree value. If you work with F (Fahrenheit) degrees, for every 5.4°F above 60°F add 0.001 to the float reading. For sample temperatures below 60°F, subtract 0.001 for every 5.4°F. If you are working with C (Centigrade or Celsius) degrees, for every 3.0°C above 15.5°C, add 0.001 to the float reading and for every 3.0°C below 15.5°C subtract 0.001 from the reading.
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For example: The urinometer float shows a specific gravity of 1.015 and the measured temperature is 82°F. Subtract: 82°F 60°F = 22°F. Divide 22 degrees by 5.4 degrees = 4.07 = 4.00 (rounded). Multiply 4.00 x 0.001 = 0.004. Corrected specific gravity =1.015 + 0.004 = 1.019
6. After the above procedure, clean out the cylinder. If you like, add a little more or less sodium chloride than in the previous procedure and determine the temperature corrected specific gravity. Practice this until you understand the procedure well. 7. In the bin marked "Urine Containers" in the front of the laboratory, there are some clean urine cups. Use them to collect some real urine and determine the corrected specific gravity of the urine sample. 8. Long's coefficient may be used to show the relationship between the specific gravity of a sample and the amount of solute dissolved in the sample. Take the last two figures of the corrected specific gravity sample and multiply by 2.6 (Long's coefficient). This will give you the approximate grams of dissolved solute in a liter of the sample. 9. For example: 1.019: 19 x 2.6 = 49.4 grams/liter 10. Calculate the gram/liter values of your various samples you tested in the laboratory. Record all of the values here:
11. Be sure to clean all equipment well and return to the metal cabinets. Rinse out the urine cup in the laboratory sinks and discard in the wastebasket.
Discussion and Study Questions: How did you perform this experiment?
What is specific gravity of a liquid?
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What does a specific gravity of 1.0 gram/ml mean?
If a solution has a higher specific gravity what does this mean?
Why would a urine sample be collected over 24 hour period for a urine specific gravity (USG)? Why not just one sample?
How do we have to adjust for temperature? In other words, why would a change in temperature cause the specific gravity to change?
How do we correct for temperature fluctuations above 15.5 degrees Celsius or 60 degrees F?
What does the Long’s coefficient (2.6) tell you?
How would the following conditions alter USG? Sweating?
Glucosuria?
Dehydration?
Excessive fluid intake?
Diabetes insipidus?
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AN EXERCISE IN THE METHODS OF URINE ANALYSISCHEMICAL ANALYSIS INTRODUCTION The chemical analysis of a urine sample is usually done using "dip-stixs." These include CLINISTIXS, KETOSTIXS, MULTISTIXS and COMBISTIXS. Also, a tablet called CLINITEST is used. These are chemically treated paper sticks and tablets that are placed into a urine sample. Each "stix" has a particular chemical test to detect the presence or absence of a substance in the urine. For example: CLINSTIX tests for glucose, KETOSTIX tests for ketone bodies such as acetoacetic acid, MULTISTIX tests for pH, protein, glucose, ketones, bilirubin, blood and urobilinogen, and COMBISTIX tests for pH, glucose and protein in the urine. The CLINITEST tablets are used to test for the presence of glucose in a urine sample. The Hay's test is for bile salts in the urine and uses the detergent action of the bile salts in the urine to lower the surface tension of the water in the urine. If a small amount of sulfur is added to the urine sample, the presence of bile salts will cause the sulfur to sink. MATERIALS NEEDED (It is not necessary to try every dip stick version, as long as you obtain results for most urine products. They are redundant.) The kit in the metal cabinets will contain the following: (1) Multistix bottle (1) Ketostix bottle (1) Combistix bottle (1) Clinistix bottle (1) Clinitest bottle (this may not be available check with instructor)
(1) small bottle of sulfur powder (1) metal scoop for sulfur powder (1) medicine dropper (1) test tube (1) forceps
PROCEDURES 1. Obtain one of the urine cups from the bin marked "Urine Containers" and get a fresh real urine sample. 2. Take the bottles of Multistix, Combistix, Ketostix and Clinistix and read the instructions on the side of the bottles. Each instruction sheet is complete with a color chart that explains the chemical test. Be sure you can read each of the "stix" tests. Be sure that you replace the cap on the bottle after you have removed the strip for the test. The cap should be replaced "promptly" and "tightly". Use the urine in the urine container for your tests and record the results in your laboratory book. You might note here that a high level of ascorbic acid (vitamin C) in the diet may alter all of the tests for the presence of glucose. Complete this chart to document your test results. Stix used Substance tested results
3. To perform the Clinitest for the presence of glucose follow the following procedure: a. Place 10 drops of water and 5 drops of fresh urine in the test tube. b. Using the forceps in the kit, add one Clinitest tablet to the test tube. Be careful-the tablet contains sodium hydroxide and may cause burns. The sodium hydroxide in the tablet generates heat so that the liquid will boil in the test tube.
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c. A high concentration of phosphates may produce a white precipitate. If glucose is present in the sample, a reddish-colored precipitate (cuprous oxide) is formed. The solution may be turned green, yellow, orange, or red depending on the relative concentration of glucose in the urine sample. The following code is usually used for recording the results: SOLUTION COLOR:
Blue Green Yellow Orange Red
Test is Negative Record "+" Record "++" Record "+++" Record "++++"
The relative number of "+" indicates the degree of glucose concentration, green is lowest and red being the greatest amount of glucose in the urine. Be able to relate these findings with the concept of the Tm for glucose as discussed in the lecture. Be sure to clean the test tube with soap and water. Record the results here:
4. Use the following procedure for the Hay's test for the presence of bile salts in the urine: i. fill the clean test tube 1/2 full of urine. Sprinkle a small amount of sulfur powder (use the metal scoop) on the surface of the cool urine. ii. if the sulfur sinks at once, bile salts are present in the amount of 0.01% or more. If the sulfur sinks only after gentle agitation, bile salts are present in the amount of 0.0025% or more. If the sulfur remains floating even after gentle shaking, bile salts are absent. 5. Be sure to wash and dry the test tube before returning it to the kit. Discard urine down the sink. Clean the urine cup in the sink and discard in the trash. Record your findings for bile salts here:
Discussion and Study Questions: How did you perform this experiment?
Why are the “dip-stixs” clinically important?
What did the CLINITEST tablets test for and what did the MULTISTIX test for? What did the Hay’s test tell you?
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Is glucose normally found in the urine? Protein? Ketones? Blood? Bilirubin? Urobilinogen? Bile salts?
If these items (above) are found in the urine and SHOULD NOT be, what does this tell you about the patient (specifically)?
What is the pH of urine?
What is the relevance of the colors for the Clinitest? If the color is red, what would this suggest about the patients Tm for glucose?
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AN EXERCISE IN THE METHODS OF URINE ANALYSISMICROSCOPIC SEDIMENTS INTRODUCTION The normal sediment in urine varies considerably with the sex of the subject, the time of the day when the specimen was taken, the specimen pH and how the specimen was collected and stored. The urine sediment usually will contain some squamous epithelial cells. These originate in the mucosa of the bladder and urethra, but may also originate in the vaginal mucosa of females. Various sediment crystals may also be present. Calcium oxalate and yellow uric acid crystals are common in acidic urine. Often, a few leukocytes are seen; however, amounts greater than 6 to 10 per high power field IN THE MALE are regarded with suspicion and may be diagnostic of a Urinary Tract Infection (UTI). Slightly higher numbers of leukocytes are considered normal IN FEMALES because of them originating from the vaginal mucosa. Erythrocytes are usually not present; however they may be present after exercising or perhaps following menses. Bacteria are not normally present in freshly voided, properly handled urine. However, because of contamination, bacteria may multiply rapidly over a relatively short period of time. Large numbers of bacteria in freshly voided urine may be indicative of UTI. Casts are results of solidification of albumin that has entered the uriniferous tubule. Casts are usually hollow and they retain the shape of the tubule. The casts may contain any material which was lying free in the tubule or loosely attached to its wall. Types of casts include: granular, RBC, WBC, epithelial and hyaline (clear). Casts may be recognized by their parallel sides and rounded ends. Large numbers of casts are usually associated with albuminuria and may indicate the presence of renal disease. Ova of certain parasites, sperm, trichimonads, yeast, hair and tissue paper fibers may also be found in a urine specimen. Slides are normally prepared from fresh urine samples. A 10 ml urine sample is centrifuged for 5 minutes to collect sediment. The urine supernatant is poured off, and the sediment is stained with 2 drops of Sedi-stain, a dye which allows one to visualize the cells and other elements. A drop of stained sediment is placed on a glass slide, covered with a coverslip and examined directly under a compound microscope. You will be examining a prepared slide of a fixed, stained urine sample. MATERIALS NEEDED Obtain a prepared urine slide and a compound microscope. PROCEDURES 1. Examine the specimen under both low and high power. Be sure to adjust the light, the condenser and the iris diaphragm to maximize your view. 2. Refer to the diagram and explanation sheets below and on the following pages to identify the various sediments. 3. Sketch and identify each of the observed sediments below in the space provided. 4. Return the slide to the slide tray. Replace the compound microscope in its proper location.
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Urinary Sediments
Refer to http://www.ec.upstate.edu/path /urinalysis/frame.htm
Diagram Key Number
Description and Pathology
1,2
Epithelial cells are the covering cells from the urethra and prepuce in males and the external genitalia in females. If leukocytes occur in large numbers this may indicate a renal infection Squamous cells are considered the same as epithelial cells Mucus threads are present in small numbers normally. Increased numbers are found in chronic inflammation of the urethra and bladder Casts are cylindrical structures with parallel sides and blunt rounded ends. They are formed in the renal tubules. Casts consist of mucoprotein matrix with formed elements imbedded. If the formed elements are RBC’s, this would indicate glomerular lesions. If the formed elements are leukocytes, this would indicate renal infection. Epithelial casts have epithelial cells in the mucoprotein matrix.
3 4 5 6,8
7
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9,10 11 12 13 14 15
16 17
18 19
Granular casts in the urine usually indicate serious renal damage. There is a high concentration of protein in the matrix. Hyaline casts usually indicate mild renal damage. There are clear casts with little imbedded materials in the matrix. Uric acid crystals are rhombic, six-sided plates, often arranged in rosette-like clusters. A presence of cholesterol crystals would indicate the level of saturated fats in the body. The presence of these crystals is rare. The presence of calcium phosphate crystals would indicate an alkaline urine. Cystine crystals are flat, hexagonal plates with unequal sides. The presence of these crystals may indicate cystinuria. Cystinuria is a congenital defect involving the reabsorption of cystine by the renal tubules. The presence of tyrosine crystals indicates serious liver damage. The crystals appear as fine needles arranged in radiating sheaves. Calcium oxalate crystals are colorless, octahedral crystals with a highly refractile cross connecting the corners. They are present in acidic urine but occasionally in alkaline urine. Hipuuric acid crystals will occur seldom in urine samples. Ammonium –magnesium phosphate crystals are prism shaped crystals presenting three, four or six sides. These crystals are found in alkaline urine.
Study Questions: What objects would you expect to see in urine? Sketch them here.
Which objects would be a sign of pathology?
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AN EXERCISE IN SOLVING RENAL CLEARANCE PROBLEMS INTRODUCTION This laboratory exercise is a series of renal clearance problems. The method of calculating the answers will be discussed in lecture. Solve each of the problems as presented on the worksheets here on these pages.
Renal Physiology Problems 1. A person is suspected of renal failure and it is decided to run an inulin renal clearance test to determine the Glomerular Filtration Rate (GFR.) The person is asked to empty their bladder, then an IV is started with inulin. A catheter is used to collect the urine sample during the test and when completed both the urine sample and a blood sample are sent to the laboratory. Using an ACCU-STAT instrument the following laboratory results were obtained: Plasma inulin = 0.14 mg/ml urine
Urine inulin = 20 mg/ ml
urine volume = 9 ml / 10 minutes
Calculate the GFR for this person. Is the GFR normal? Does the patient’s GFR indicate any possible Goldblatt Renal Hypertension? (Ans. 128.5 ml/min; yes; no)
2. Given that the Tm (reabsorption) for glucose is 320 mg/min, calculate how much glucose will spill over in the urine of the same person described in question 1, given that an ACCU-STAT reading of the plasma glucose is: Plasma glucose = 450 mg % (Ans. 258.3 mg/min)
3. A person with a BP of 163/105 mm Hg was tested for suspected renal (Goldblatt) hypertension. PAH and inulin tests were run on the person with the following laboratory results: Plasma inulin = 1.0 mg/ml plasma
PAH plasma = 0.1 mg/ml
Urine inulin = 41 mg /ml urine
PAH urine = 20 mg /ml
Urine flow rate = 2.0 ml/min
hematocrit (HCT) = 40%
Tm (secretion) for PAH = 80 mg/min
MCV = 93 cubic microns
Use 90% bladder clearance for PAH a) Calculate the renal clearance value for PAH. (Ans a) 400 ml/ min;
b) Calculate the secretion rate for PAH. (ans: b) 31.8 mg /min)
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c) Calculate the total whole blood flow through the renal system per minute. ( ans: c) 741 ml/min)
What role did the Tm for PAH play in the calculation? Explain.
Does the data indicate a possible renal hypertension?
If this person does have renal hypertension, how would you explain the 160/105 torr BP?
Do these additional renal problems One of the roles of the kidney is to remove substances from the blood. Clearance is defined as the volume of plasma per minute (ml plasma / min) from which a particular substance is totally removed or “cleared.” For any substance, A, renal clearance value (RCV) is determined as follows: RCV A = UA x V PA Where, U = urine concentration of substance A V = urine flow rate (ml/min) P = plasma concentration of substance A Marker substances: Inulin is an exogenous substance that is freely filtered but is neither reabsorbed nor secreted by renal tubules. Therefore, the clearance of inulin represents the volume of plasma filtered each minute (ml plasma/min), or GFR. Any substance that is freely filtered, but that also undergoes net reabsorption will have a clearance value less than inulin. Any substance that is freely filtered but undergoes net secretion will have a clearance value higher than inulin. Creatinine is a byproduct of muscle metabolism. It is endogenous to the body. It is freely filtered and is only slightly secreted. Concentrations of this substance are fairly level in people with normal renal function. Creatinine clearance can be used to estimate GFR, but since it undergoes some net secretion, its value will be a slight overestimate of GFR. Ex. = CCR = 140 ml/min when GFR is 125 ml/min PAH (para amino hippurate) is an exogenous substance and is a weak acid. It is freely filtered and is secreted by a Tm limited process along the renal tubules. Since it enters the nephron by two routes (filtration and secretion), it is a marker for any blood reaching the nephron (at the glomerulus and peritubular capillaries) but not for blood that reaches other non-nephron regions of the kidney. The clearance value of PAH represents effective renal plasma flow (ERPF.)
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Total RBF is the value of interest. To arrive at RBF, ERPF must be corrected by taking into account: a) the ratio of whole blood to plasma (use the HCT) b) 90% of the total blood flow to the kidney reaches nephrons. 10% does not. Use this information to solve the following problem set: 4. What is the tubular load of glucose (TL glu) when plasma glucose concentration is 450 mg% and GFR is normal (125 ml/min)?
5. Knowing the Tm for glucose (320 mg/min) what is the rate of glucose excretion in this person?
6. Substance “A” is freely filtered and undergoes net secretion, but no reabsorption. The following data were obtained from the patient: Plasma concentration A = 200 mg % Urine concentration A 190 mg/ ml Clearance inulin (RCV inulin) = 115 ml/min Urine flow rate = 20 ml urine collected over 10 min Determine a) This person’s GFR
b) The tubular load of substance A
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c) The rate of excretion of substance A
d) The rate of secretion of substance A
7. Determine the clearance of the following substance: Plasma concentration = 0.4 mg/ml
urine concentration = 40 mg /ml
Urine flow rate = 1ml/min
Does this substance undergo any other processes besides filtration? Explain.
8. A person presents with the following lab results. Urine flow rate = 15 ml / 10 min Plasma PAH = 0.2 mg /ml
Urine PAH = 110 mg / ml HCT = 42%
Determine a) RPF
b) RBF
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c) Is this RBF normal?
9. A patient received a piggyback IV infusion of inulin and PAH over a 20 min interval. 15 ml of urine was collected. Lab values returned as follows: Urine inulin = 160 mg/ml Plasma inulin = 2 mg / ml HCT = 43%
Urine PAH = 78 mg/ml Plasma PAH = 0.2 mg /ml
Determine a) GFR
b) RBF
c) Discuss these values relative to normal values.
Study Questions: Make sure you know how to define and perform all calculations for: CO, RBF, ERBF, RPF, ERPF, GFR, RCV, and excretion.
What is the importance of PAH? Inulin? Creatinine?
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AN EXERCISE IN THE USE OF THE SPIROMETER-LUNG CAPACITIES & VOLUMES INTRODUCTION The spirometer is an instrument that is normally used to measure the flow of air in and out of the lungs under varying conditions. The wet spirometer is based on the simple mechanical principle that air, exhaled from the lungs, will cause displacement of a closed chamber which is partially submerged in water. Basically, it consists of two vessels: a larger vessel containing water and having a breathing hose attached to it; and a smaller vessel inverted and suspended in the water. A counterweight and indicator are attached to the inverted chamber. Air blown into the inverted chamber will cause it to rise, thus moving an indicator arrow which is calibrated in liters, to give lung volume measurements. In this exercise, you will use the wet spirometer to measure only exhalations. Therefore all of the inhalation data will be calculated using mathematical relationships.
Lung Volumes: There are four pulmonary lung volumes that, when added together, equal the maximum volume to which the lungs can be expanded. These are: 1. Tidal Volume (TV) is the volume of air inspired or expired with each normal breath; it amounts to about 500 milliliters in the adult male. 2. Inspiratory Reserve Volume (IRV) is the extra volume of air that can be inspired over and above the normal tidal volume when the person inspires with full force; it is usually equal to about 3100 milliliters. 3. Expiratory Reserve Volume (ERV)is the maximum extra volume of air that can be expired by forceful expiration after the end of a normal tidal expiration; this normally amounts to about 1200 milliliters. 4. Residual Volume (RV) is the volume of air remaining in the lungs after the most forceful expiration; this volume cannot be voluntarily exhaled, so it must be measured by other methods than those available to us; this volume averages about 1200 milliliters Pulmonary Capacities are the sum of two or more of the volumes. Capacities are measured in volume units. There are four, and these are: 1. Inspiratory Capacity (IC) equals the tidal volume plus the inspiratory reserve volume. This is the amount of air (about 3600 milliliters) a person can breathe in, beginning at the normal expiratory level and distending the lungs to the maximum amount. (IC = TV + IRC)
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2. Functional Residual Capacity (FRC) equals the expiratory reserve volume plus the residual volume. This is the amount of air that remains in the lungs at the end of normal expiration (about 2400 milliliters). (FRC = ERV + RV) 3. Vital Capacity (VC) equals the inspiratory reserve volume plus the tidal volume plus the expiratory reserve volume. This is the maximum amount of air a person can expel from the lungs after first filling the lungs to their maximum extent and then expiring to the maximum extent (about 4800 milliliters). (VC = IRV +TV + ERV) 4. Total Lung Capacity (TLC) is the maximum volume to which the lungs can be expanded with the greatest possible effort (about 6000 milliliters); it is equal to the vital capacity plus the residual volume. (TLC = IRV +TV + ERV +RV) MATERIALS NEEDED A bell (wet) spirometer provided on the countertop. PROCEDURES Work as a small group and collect the data from one subject. Insert a clean disposable mouthpiece into the oneway valve in the hose. Once a student has exhaled into the spirometer, the bell will rise with the air in it. To start with an empty bell, one must gently remove the one-way valve at the end of the hose before the bell will sink and reset to the bottom of the housing. Reattach the one-way valve and mouthpiece after the bell has been released. 1. Measure TIDAL VOLUME (TV) The tidal volume represents the amount of air that is inspired or expired during a normal breathing cycle. To directly measure this volume, start with the dial of the spirometer set to zero. Rotate the outer, larger circular scale dial at the top of the spirometer until the scale is horizontal and the pointer reads at the zero mark. Hold or clip your nose shut while exhaling to prevent any air from leaking through the nostrils. The subject should be standing and facing forward, but not looking at the dial. Relax and breathe normally, inhaling through the nose. When ready, attach to the mouthpiece sealing the lips around it, and exhale THREE normal breaths into the spirometer, inhaling room air through the nose between each exhalation. Do not forcibly exhale. The dial on the spirometer should move slightly with each exhalation. Record the dial reading after the three exhalations. Take the reading and divide this value by three to get the average tidal volume. The unit will be in cubic centimeters (cc) or milliliters (ml.) Why is it good practice to take three breaths and then divide by three to get the average TV? For reference, a normal TV is about 400 to 500cc.
Calculate your average TV here: Bell (Wet)-spirometer
2. Measure MINUTE RESPIRATORY VOLUME (MRV) The minute respiratory volume (MRV) is the volume of air that passes in or out of the lungs per minute. To calculate this value, count the number of normal breaths the subject takes when breathing normally over one minute and multiply this by the average TV: MRV = Ventilation Rate x TV
Calculate your MRV here: 34
3. Measure EXPIRATORY RESERVE VOLUME (ERV) Measure the amount of air that can be forcibly breathed out after normal expiration (ERV) After taking three normal breaths and exhaling normally into the room after each, on the last breath, attach to the mouthpiece and continue to forcibly exhale as much air as possible. Record the number of cc (or ml). The normal values for males, aged 20-30 years is about 1200 cc and from 30-60 years is about 1000 cc. Females aged from 20-30 years, the normal value is 1000 cc and for females over age 30, the normal ERV is 900 cc.
Write your ERV here:
4. Measure VITAL CAPACITY (VC) Measure the maximum amount of air which can be forcibly exhaled immediately following a maximal inhalation (VC) (Max in to Max out.) Set the spirometer dial to "zero". Determine the VC by taking three deep breaths and exhaling completely after each inspiration. On the fourth deepest full breath, exhale all of the air into the spirometer. Do not "puff" hard through the spirometer but instead, use a slow forced exhalation and "give it all you got". For norms, see charts in the back of this exercise, or use the VC formulas below: MALES: VC (in liters) = (0.052 x H) - (0.022 x A) - 3.60 FEMALES: VC (in liters) = (0.041 x H) - (0.018 x A) - 2.69 Where "H" = height in centimeters (2.54 cm = 1inch) "A" = age in years to nearest birthday Example: calculate the VC of a 28 year-old female who is 5'7" (170.18 cm) in height. Answer: 3.786 liters. Calculate your VC using the equation above:
What was your VC from the spirometer?
What is your VC from the chart at the end of this exercise?
Quite often, the VC is expressed as a percentage of the spirometery-obtained VC compared to the calculated VC. Therefore, VC Percentage = VC (in liters) from Spirometer Reading x 100 VC (in liters) from formula or chart Your Vital Capacity Percent should be very close to 100% because this would be the normal percentage for a healthy person or non Chronic Obstructive Pulmonary Disease (COPD) person. Calculate your VC% here:
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5. Determine INSPIRATORY CAPACITY (IC) The inspiratory capacity (IC) is the total amount of air one can draw in after a normal exhalation. You will obtain this value indirectly, as we do not wish to inhale air contained in the spirometer. The formula is: IC = VC - ERV
Calculate your IC here:
6. Measure INSPIRATORY RESERVE VOLUME (IRV) The inspiratory reserve volume (IRV) is the amount of air that can be inhaled after normal inspiration. You will obtain this value indirectly, as we do not wish to inhale air contained in the spirometer. The formula is: IRV = IC - TV Calculate your IRV here:
7. Determine RESIDUAL VOLUME (RV) The residual volume is the amount of air left in the lungs after the most forceful expiration. This cannot be voluntarily exhaled, but can only be squeezed out of a nonliving lung. It is measured by indirect methods, and has been experimentally determined for normal persons of different age/sex. We can assume you fit the normal. See chart below and highlight your RV range:
FOR MEN Age RV 14-19: 830 cc 20-29: 1410 cc 30-39: 1510 cc 40-49: 1690 cc 50-59: 1740 cc 60-75: 1760 cc
FOR WOMEN Age RV 14-19: 820 cc 20-29: 1290 cc 30-39: 1310 cc 40-49: 1390 cc 50-59: 1300 cc 60-64: 1200 cc
8. Determine FUNCTIONAL RESIDUAL CAPACITY (FRC) The functional residual capacity (FRC) is defined as the amount of air remaining in the lungs after normal expiration. This is a very important volume, as it represents the volume of air in the lung at the equilibrium position in the tug of war between opposing forces of the lung and chest wall. It too, cannot be completely exhaled, as it is made up partly of the residual volume from the previous step. The formula is: FRC = ERV + RV
Calculate your FRC here:
9. TOTAL LUNG CAPACITY (TLC) The total lung capacity (TLC) is the total amount of air in the lungs after maximal inspiration. This value may be determined by either formula: TLC = FRC + IC OR TLC = RV + VC. If you calculated correctly in previous steps, you should get the same answer for both methods. The normal TLC for males is about 6000 cc and for females, about 4200 cc.
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Calculate your TLC here:
Study Questions:
What is a spirometer?
Be sure you know the definitions for ALL volumes and capacities and how to calculate them: Tidal volume, MRV, ERV, Vital capacity, VC percentages, IC, IRV, FRC, RV, and TLC. PRACTICE SPIROMETRY PROBLEMS: 1. Given: VC = 4300 ml IC = 3000 ml TV = 400 ml RR = 16/min Calculate: a) IRV b) ERV c) MRV
2. Given: TV = 400 ml IRV = 2600 ml ERV = 1000 ml RR = 12/min Calculate: a) VC b) IC c) MRV
3. Given: IRV IC RV VC Calculate:
= 2800 ml = 3250 ml = 1200 ml = 5375 ml a) TV b) ERV c) TLC
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MEASUREMENT OF LUNG FUNCTION: FORCED EXPIRATORY VOLUME (FEV1.0) INTRODUCTION This is a rapid but valuable test of lung function. This is a dynamic lung test, one that takes into account how rapidly you can move air out of the lungs across airways. Rate of airflow is volume/ time. You will be using a research grade recording spirometer for this exercise. This consists of a wet spirometer with an attached rotating drum with pen for recording volume changes.
Pulmonary function tests are often used in diagnosis of respiratory disease. The particular test you are doing today is a more valuable test in diagnostics than are simple lung volume and capacity measurements (static lung tests) because it examines the rate or speed of exhalation (dynamic testing.) The rate of exhalation is often affected in obstructive lung disease. This test is also known as a "forced vital capacity" or "FVC" because the subject first maximally inhales and then maximally exhales into the spirometer. You are asked to work in pairs or small groups, one of you being the subject and the other(s) acting as equipment operator and coach. The operator / coach will set the recording paper speed at its fastest setting just before the subject performs the vital capacity maneuver (inhales maximally and then exhales maximally.) At this rapid paper speed, the exhalation is expanded out so that you can analyze the data by looking at the slope of the curve. By referring to the graph, you can calculate the percent of the total vital capacity that is exhaled in the first (FEV 1.0.) Pulmonary Disorders Chronic pulmonary disorders can be divided into two major classes: obstructive disorders and restrictive disorders. These two classes can be differentiated by the use of the spirometry tests performed in this exercise. According to Poiseuille's Law, the resistance to airflow through a tube is inversely proportional to the fourth power of the radius of the tube. Even a small obstruction in the airways can result in a greatly increased resistance to airflow. Diseases like emphysema, bronchitis and asthma are classified as obstructive disorders. Bronchiolar obstruction can result from inflammation, edema, smooth muscle constriction, and secretions, making it very difficult to pass air through the airways. These patients usually have a problem with exhaling or gas exchange across the respiratory membrane and their FEV1/FVC is below 80% (80% is normal). In restrictive disorders, lung damage such as alveolar destruction and scarring can result in changes in lung elasticity and thus the total amount of air one can exhale. If a disease is purely restrictive, as in pulmonary fibrosis, the lung loses compliance, becomes difficult to inflate, and the vital capacity decreases. In this disease, the airways may remain unobstructed, resulting in a normal FEV test (rate of airflow), however the vital capacity is reduced. These
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patients have an increased problem with inhaling sufficient air, therefore their VC will be overall reduced, as will the FEV1. But, their ratio of FEV1/FVC will likely remain as 80%. In a disease like emphysema, which is both obstructive and restrictive, the FEV test will be abnormal. The obstruction stems from weakened bronchioles and alveoli that have lost much of their elastic fibers. Lack of elastic recoil results in a need for the person to actively exhale, using the muscles of exhalation. The resulting elevated intrathoracic pressure during exhalation actually acts to close off the smaller bronchioles, increasing the resistance of the airways and blocking the very air one was trying so hard to exhale. The outcome is called "airtrapping." MATERIALS NEEDED Recording spirometer with pen attached Recording graph paper Washable rubber mouthpieces Nose-clips PROCEDURE
1. This exercise is best done with two persons, one to operate the spirometer and to act as the “coach” and the other to be the subject.
2. Turn on the main power switch found at the back of the unit. 3. Remove the one-way valve at the end of the hose where the mouthpiece will be fitted. Gently press the metal
4. 5. 6. 7.
8.
9.
release valve inside the hose and slowly move the tank up or down, allowing room air to exit or fill the tank. Note that as you move the tank up or down, the pen's position on the recording-drum changes. Fill the tank partially with room air, and position the tank so that the pen sits near the top of the recording drum. Wrap a sheet of graph paper around the recording drum, sticking the prepared gummy edge to the free edge of paper, not to the drum. Several measurements can be made on each sheet. Attach a clean, plastic mouthpiece to the hose. The mouthpiece can be cleaned by washing with mild soapy water and additionally by wiping with alcohol. At the completion of this exercise, wash the mouthpiece and return it to its storage container. Remove the pen cap, and place it in its storage receptacle on the unit. Use the metal lever at the base of the unit to touch the pen to the paper. (You may need to “fix” it in place using a wedge between it and the drum.) During the exercise, the student will pinch or clip their nose shut using a nose-clip during the entire exercise. When mouthpiece is inserted into the mouth, the teeth are positioned on the bite plate and lips extend over the front edge (just like a scuba mouthpiece.) Both of these last steps prevent air from leaking through the mouth and nose. The operator turns the drum to its slowest speed (Note the speed: mm/min.) Have the subject breathe several breaths normally (tidal volumes) into the room. Whenever the subject is ready, he inhales room air maximally and quickly inserts the mouthpiece into his mouth. (It may help if the subject visually cues the operator at this point.) Just as the subject begins this deep inhalation, the operator turns the drum to its highest speed (note paper’s speed: mm/sec) and the subject exhales as forcefully and as rapidly as possible into the spirometer until there is no air left to exhale. It may help flex at the waist or squat toward the end of the exhale. Coaching the subject to be sure he's exhaled every bit of air is helpful! The test is complete. Stop the drum. Recap the pen, and remove your graph paper. (Note that several students’ tracings can be recorded on the same paper in the event you need to repeat the exercise, or wish to obtain recordings from other group members.) The last group will turn of the main power switch located on the lower, rear of the unit.
Calculations Goal: To measure the vital capacity. Do this by subtracting the maximal exhalation volume from the peak inhalation volume. You do not have to correct for temperature changes from body to spirometer. On the graph, note the point of maximal inhalation (the highest point on the tracing.) This is known as peak inhalation volume. Now find the lowest point on the tracing (when the subject maximally exhaled.)
39
Determine the vertical distance between these points in mm. Note on the graph paper: in the vertical direction, 1 mm of pen deflection corresponds to 30 ml of air. Compute the vital capacity. Your subject's forced vital capacity (FVC) equals ________________ml Goal: Measure the amount of air exhaled in the first second. Find the point on the curve that corresponds to one second after peak inhalation, and determine how much air was exhaled over that interval. For this step, you must know the paper speed of your recording, and you must use the x-axis to determine where the “one second” mark lies from the start of the exhalation. a. b.
Find the point on the tracing one second after peak inhalation. At this rapid paper speed, how many mm in the horizontal direction corresponds to 1 second? ________________ c. Find the point of peak inhalation, and drop a vertical line to meet the x-axis. Now measure in the horizontal direction from peak inhalation the mm representing one second and mark the point on the tracing. d. Determine the volume of air exhaled over this interval by measuring the vertical drop in mm between this point and peak inhalation. Compute the number of ml air this represents. Your FEV 1.0, which is the volume of air exhaled in the first second, equals _______________ml.
Goal: Determine the percent of vital capacity that is exhaled in the 1st second by dividing your FEV1.0 by FVC. This ratio is usually expressed as a percentage, so multiply the ratio by 100. Ex.
% FVC = FEV1.0 FVC
= 2400 ml exhaled in the first second ___X 100 = 50% 4800 ml exhaled over several seconds
Show your calculation here:
Your subject's %FVC
exhaled in the first second
equals ______________%
Goal: Compare your values to the predicted values for the subject's age/sex. a) The ratio (percent) FEV1.0/FVC is an indicator of obstructive disease. Obstruction may be present if the subject’s ratio is lower than normal. b) The FVC value is an indicator of restrictive disease. Restriction may be present if the subject’s values are lower than normal. Examine the subject’s FVC value, and compare it with predicted values for the subject’s sex, age and height in cm found in the following tables. Your subject's predicted Forced Vital Capacity is ____________. What percent is this of normal?_________________ Your subject's predicted Percent Forced Vital Capacity is ____________% What percent is this of normal?___________ How does your subject's vital capacity compare to the predicted value for his or her age? If your subject did not reach the predicted range for their age, what factors could influence your subject’s VC? (e.g., do they smoke? Have asthma?A cold or stuffiness?)
40
Discussion Study Questions: How did you perform this experiment? How is FVC important in determining obstructive lung diseases?
What are obstructive disorders?
What are restrictive disorders?
How does Poiseuille’s Law affect FEV1.0 with respect to bronchial radius?
What is normal FVC? How does the FVC change in someone with obstructive vs. restrictive lung disease?
41
Taken from http://www.nationalasthma.org.au/HTML/management/spiro_book/sp_bk010.asp Appendix B - Predicted Normal Values There is a vast literature of normal population studies, many of which have deficiencies in sample size, ethnicity, definition of normality, inclusion of smokers and choice of equipment. In the absence of population-specific reference values, measurements for patients of mainly European origin are better predicted by European values while those for patients of mainly British and Irish origin are better predicted by North American values. The following tables provide mean predicted values for Caucasians of both sexes for FEV1, FVC, FEV1/FVC, FEF25–75% and PEF. The tables are based on a large and well-conducted study in asymptomatic, lifelong non-smokers aged 8 to 80 years who participated in the 2,3 National Health and Nutrition Examination Survey (NHANES III). It is not recommended to extrapolate beyond the age and height range of the population used to obtain the reference equations. Age is shown in years, height is shown in centimetres.
Respiratory function tables From the National Asthma Council Australia. Asthma Management Handbook 2006. Melbourne: National Asthma Council Australia, 2006. Used with permission.
FEV1 (L) Male Age
20
25
30
35
40
45
50
55
60
65
70
75
80
145 cm
3.19
3.08
2.97
2.85
2.72
2.58
2.44
2.28
2.12
1.94
1.76
1.57
1.37
150 cm
3.40
3.29
3.18
3.06
2.93
2.79
2.64
2.49
2.32
2.15
1.97
1.78
1.58
155 cm
3.61
3.51
3.39
3.27
3.14
3.01
2.86
2.70
2.54
2.37
2.19
2.00
1.80
160 cm
3.83
3.73
3.62
3.50
3.37
3.23
3.08
2.93
2.76
2.59
2.41
2.22
2.02
165 cm
4.06
3.96
3.85
3.73
3.60
3.46
3.31
3.15
2.99
2.82
2.64
2.45
2.25
170 cm
4.30
4.19
4.08
3.96
3.83
3.69
3.55
3.39
3.23
3.05
2.87
2.68
2.48
175 cm
4.54
4.44
4.33
4.20
4.07
3.94
3.79
3.63
3.47
3.30
3.12
2.93
2.73
180 cm
4.79
4.69
4.58
4.45
4.32
4.19
4.04
3.88
3.72
3.55
3.37
3.18
2.98
185 cm
5.05
4.95
4.83
4.71
4.58
4.44
4.30
4.14
3.98
3.80
3.62
3.43
3.24
190 cm
5.31
5.21
5.10
4.98
4.85
4.71
4.56
4.41
4.24
4.07
3.89
3.70
3.50
195 cm
5.58
5.48
5.37
5.25
5.12
4.98
4.83
4.68
4.51
4.34
4.16
3.97
3.77
20
25
30
35
40
45
50
55
60
65
70
75
80
145 cm
3.63
3.57
3.50
3.42
3.32
3.21
3.09
2.95
2.80
2.63
2.45
2.26
2.06
150 cm
3.91
3.85
3.78
3.69
3.60
3.49
3.36
3.22
3.07
2.91
2.73
2.54
2.33
155 cm
4.19
4.13
4.06
3.98
3.88
3.77
3.64
3.51
3.36
3.19
3.01
2.82
2.62
160 cm
4.48
4.43
4.36
4.27
4.17
4.06
3.94
3.80
3.65
3.48
3.31
3.11
2.91
165 cm
4.79
4.73
4.66
4.57
4.48
4.37
4.24
4.10
3.95
3.79
3.61
3.42
3.21
170 cm
5.10
5.04
4.97
4.89
4.79
4.68
4.55
4.42
4.26
4.10
3.92
3.73
3.52
175 cm
5.42
5.36
5.29
5.21
5.11
5.00
4.88
4.74
4.59
4.42
4.24
4.05
3.85
180 cm
5.75
5.69
5.62
5.54
5.44
5.33
5.21
5.07
4.92
4.75
4.57
4.38
4.18
185 cm
6.09
6.03
5.96
5.88
5.78
5.67
5.55
5.41
5.26
5.09
4.91
4.72
4.52
190 cm
6.44
6.38
6.31
6.23
6.13
6.02
5.90
5.76
5.61
5.44
5.26
5.07
4.87
195 cm
6.80
6.74
6.67
6.59
6.49
6.38
6.25
6.12
5.97
5.80
5.62
5.43
5.22
FVC (L) Male Age
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FEV1/FVC (%) Male Age All Heights
20
25
30
35
40
45
50
55
60
65
70
75
80
83.9
82.9
81.9
80.8
79.8
78.8
77.7
76.7
75.7
74.6
73.6
72.6
71.5
FEV1 (L) Female Age
18
20
25
30
35
40
45
50
55
60
65
70
75
80
145 cm
2.72
2.70
2.64
2.57
2.49
2.40
2.30
2.18
2.06
1.94
1.80
1.65
1.49
1.32
150 cm
2.89
2.87
2.81
2.74
2.66
2.57
2.46
2.35
2.23
2.10
1.97
1.82
1.66
1.49
155 cm
3.07
3.05
2.98
2.91
2.83
2.74
2.64
2.53
2.41
2.28
2.14
1.99
1.83
1.66
160 cm
3.25
3.23
3.16
3.09
3.01
2.92
2.82
2.71
2.59
2.46
2.32
2.17
2.01
1.85
165 cm
3.44
3.41
3.35
3.28
3.20
3.11
3.01
2.90
2.78
2.65
2.51
2.36
2.20
2.03
170 cm
3.63
3.61
3.54
3.47
3.39
3.30
3.20
3.09
2.97
2.84
2.70
2.55
2.39
2.23
175 cm
3.83
3.80
3.74
3.67
3.59
3.50
3.40
3.29
3.17
3.04
2.90
2.75
2.59
2.42
180 cm
4.03
4.01
3.95
3.88
3.79
3.70
3.60
3.49
3.37
3.24
3.10
2.95
2.80
2.63
185 cm
4.24
4.22
4.16
4.08
4.00
3.91
3.81
3.70
3.58
3.45
3.31
3.16
3.01
2.84
190 cm
4.46
4.43
4.37
4.30
4.22
4.13
4.03
3.92
3.80
3.67
3.53
3.38
3.22
3.05
195 cm
4.68
4.65
4.59
4.52
4.44
4.35
4.25
4.14
4.02
3.89
3.75
3.60
3.44
3.27
18
20
25
30
35
40
45
50
55
60
65
70
75
80
145 cm
2.97
2.98
2.99
2.98
2.95
2.90
2.83
2.74
2.63
2.51
2.36
2.20
2.01
1.81
150 cm
3.19
3.20
3.21
3.19
3.16
3.11
3.05
2.96
2.85
2.72
2.58
2.41
2.23
2.03
155 cm
3.42
3.42
3.43
3.42
3.39
3.34
3.27
3.18
3.08
2.95
2.80
2.64
2.46
2.25
160 cm
3.65
3.66
3.67
3.65
3.62
3.57
3.50
3.42
3.31
3.18
3.04
2.87
2.69
2.49
165 cm
3.89
3.90
3.91
3.89
3.86
3.81
3.75
3.66
3.55
3.42
3.28
3.11
2.93
2.73
170 cm
4.14
4.15
4.15
4.14
4.11
4.06
3.99
3.91
3.80
3.67
3.53
3.36
3.18
2.98
175 cm
4.39
4.40
4.41
4.40
4.37
4.32
4.25
4.16
4.05
3.93
3.78
3.62
3.43
3.23
180 cm
4.66
4.67
4.67
4.66
4.63
4.58
4.51
4.42
4.32
4.19
4.05
3.88
3.70
3.50
185 cm
4.93
4.94
4.94
4.93
4.90
4.85
4.78
4.69
4.59
4.46
4.32
4.15
3.97
3.77
190 cm
5.21
5.21
5.22
5.21
5.18
5.13
5.06
4.97
4.87
4.74
4.59
4.43
4.25
4.04
195 cm
5.49
5.50
5.51
5.49
5.46
5.41
5.35
5.26
5.15
5.02
4.88
4.71
4.53
4.33
18
20
25
30
35
40
45
50
55
60
65
70
75
80
87.0
86.6
85.5
84.4
83.4
82.3
81.2
80.2
79.1
78.1
77.0
75.9
74.9
73.8
FVC (L) Female Age
FEV1/FVC (%) Female Age All Heights
43
