Downloadable Solution Manual for PhysioEx 80 for Human Physiology Lab Simulations in Physiology 1st EditiIM Sample11

54899_Ex_01-88.qxd 1/11/08 10:14 AM Page 1 E X E R C I S E version 8 1 Cell Transport Mechanisms and Permeability...

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E X E R C I S E

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Cell Transport Mechanisms and Permeability Time Allotment: 2 hours minimum, 3 hours preferred.

The following minimum computer hardware is recommended for PhysioEx 8.0: WINDOWS OS: Windows® XP, Vista™ Resolution: 1024 × 768 Latest version of Adobe® Flash® Player Latest version of Adobe Reader® Browsers: Internet Explorer 6.0 (XP only); Internet Explorer 7.0; Firefox 2.0 Internet Connection: 56K modem minimum for website Printer

MACINTOSH OS: 10.3.x, 10.4.x Resolution: 1024 × 768 Latest version of Adobe Flash Player Latest version of Adobe Reader Browsers: Safari 1.3 (10.3.x only); Safari 2.0 (10.4.x only); Firefox 2.0 Internet Connection: 56K modem minimum for website Printer

Advance Preparation, Comments, and Pitfalls 1. If you are using PhysioEx in a computer lab: Because PhysioEx requires a browser (such as Firefox or Internet Explorer) and the Flash Player to run, it is recommended that you make sure these items are already installed on student computers before beginning this lab. 2. If you are using the web version of PhysioEx: Have your students go to http://www.physioex.com and follow the registration instructions found in the very front of their lab manual. 3. It is helpful to instruct students in the proper operation of the mouse and menu system as part of the lab introduction. 4. When considering an upgrade for computer systems, memory (RAM) offers the largest performance increase for the least cost. 5. Having students work in pairs results in the most successful lab experience. If students must work in larger groups, have them each get keyboard and mouse experience with the program.

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6. Occasionally, data in the Cell Transport Mechanisms and Permeability module will appear with “#” symbols next to numbers. In the Simple Diffusion, Facilitated Diffusion, and Osmosis experiments, the “#” symbol after “rate” data indicates that equilibrium was not reached for that solute. In the Active Transport experiment, the symbol after “rate” data indicates 1) for glucose data, that equilibrium was not reached for glucose; 2) for NaCl and KCl, that transport was interrupted for that solute. In the Osmosis experiment, the symbol after “pressure” data means that osmotic equilibrium was not reached.

Answers to Questions Activity 1: Simulating Simple Diffusion (pp. 3–5) The molecular weight of Na+ is 23. The molecular weight of Cl– is 35. The following membranes allowed both of these ions to pass through them: membranes with a MWCO of 50, MWCO 100, and MWCO 200. The following materials diffused from the left beaker to the right beaker with one or more of the membranes: Na+, Cl–, urea, and glucose. Albumin did not diffuse. The molecular weight of albumin is larger than any of the MWCO membranes will allow to pass through them. Activity 2: Simulating Dialysis (p. 5) The urea concentration decreases to 50% of the original amount. The urea has diffused from the left beaker to the right beaker. The urea moved from an area of higher concentration (the left beaker) to an area of lower concentration (the right beaker, which originally contained no urea). Activity 3: Facilitated Diffusion (pp. 6–7) As the number of carriers increases, the time it takes to reach equilibrium decreases. No, the diffusion rate of Na+/Cl– does not change with the number of receptors. Simple diffusion. No, you would not observe any diffusion. Yes, diffusion is still taking place. Activity 4: Osmosis (pp. 7–9) Yes, pressure changes were observed with the MWCO 20 membrane. Pressure was only found in the left beaker. In the left beaker, the concentration of solute was higher and the membrane allowed no solute diffusion, so water from the right vessel was “drawn” in, creating pressure. Yes, with all membranes above MWCO 20.

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PhysioEx™ Exercise 1

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With the larger pore membranes, Na+/Cl– was allowed through via simple diffusion. As solute concentration increases, pressure increases. No, the solute particles are moving so that concentrations are equal on both sides of the membrane and there is no net water movement. No. The equal solute concentrations would have equal amounts of water moving in both directions, so there would be no pressure generated. No—the size of the membrane allows both solutes to diffuse. Yes—because the albumin cannot diffuse, pressure is generated from the NaCl side to the albumin side. The solution that cannot diffuse through the membrane generates the pressure. Activity 5: Filtration (pp. 9–10) NaCl was found on the membrane, because only the water passed through it. Yes, the membrane MWCO affects filtration rate. Yes, the pressure applied affects filtration rate. No, not all the solutes passed through all the membranes. Charcoal did not pass through any of the membranes. Glucose did not pass through any of the membranes with an MWCO below 200. The molecular weight of the solutes was too large to pass through the membrane pores. By increasing blood vessel radius to increase fluid flow in the vessel going to a given organ. Activity 6: Active Transport (pp. 11–12) No. Sodium transport does not occur because potassium is not available. The Na+/K+ pump requires both a sodium and potassium presence on opposite sides of the membrane. The transport stops because of a lack of ATP. Yes, more solute is transported. Yes—the amount of solute transported increases with an increase in carriers or pumps. No. Yes. You can test this by adding different amounts of glucose to both beakers and observing simple diffusion occur. No. Because there is no potassium present. The Na+/K+ pump requires both a sodium and potassium presence on opposite sides of the membrane. Yes.

PhysioEx™ Exercise 1

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R E V I E W

S H E E T

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EXERCISE

NAME ____________________________________ LAB TIME/DATE _______________________

1.

Cell Transport Mechanisms and Permeability

Match each of the definitions in Column A with the appropriate term in Column B. Column A e

term used to describe a solution that has a lower concentration of solutes compared to another solution

g

term used to describe a solution that has a higher concentration of solutes compared to another solution

a

the movement of molecules from an area of higher concentration to an area of lower concentration as a result of random thermal motion

d

the movement of molecules across a membrane that requires the expenditure of cellular energy (ATP)

c

2.

1

the transport of water across a semipermeable membrane

f

term used to describe two solutions that have the same concentration of solutes relative to one another

b

the movement of molecules across a selectively permeable membrane with the aid of specialized transport proteins

Column B a.

diffusion

b.

facilitated diffusion

c.

osmosis

d.

active transport

e.

hypotonic

f.

isotonic

g.

hypertonic

What is the main difference between simple diffusion and facilitated diffusion? In simple diffusion, solutes move unaided from an area of higher concentration to an area of lower concentration across a biological membrane. In facilitated diffusion, solutes require the aid of specialized transport proteins in order to move from an area of higher concentration to an area of lower concentration across a biological membrane.

3.

What is the main difference between facilitated diffusion and active transport? Facilitated diffusion does not require the use of cellular energy (ATP) expenditure. Active transport requires the expenditure of cellular energy.

4.

In the “Simple Diffusion” experiment, which solute(s) passed through the MWCO 20 membrane? none of them

Why? The molecular weight of each of the solutes was higher than the molecular weight cutoff of 20.

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5. List three examples of passive transport mechanisms. simple diffusion, facilitated diffusion, osmosis

6. Describe the relationship of solute concentration to solvent concentration in osmosis. The solute concentration is the inverse of the solvent concentration; if the solute concentration is high, the solvent concentration is low.

7. What is the equation for Fick’s First Law of Diffusion? J = –DA Δc /Δx

Explain Fick’s First Law of Diffusion. The rate of diffusion is proportional to the area of the membrane and the difference in concentration of the solute on both sides of the membrane.

8. In the mock dialysis activity, what was the only solute removed from the beaker representing the patient’s blood? urea

Why is it important that this solute be removed from diabetic patients? In a diabetic patient, kidney function is compromised, and thus the body cannot effectively remove urea (a waste product) from the bloodstream. Too much urea in the blood will lead to nitrogen narcosis and eventually death.

9. How can the concentration of water in a solution be decreased? Increasing the solute concentration will reduce the water concentration.

10. Suppose that a membrane separates a solution of higher osmolarity and a solution of lower osmolarity. To prevent osmotic flow of water across the membrane, pressure should be applied to which of the two solutions? Pressure should be applied to the solution of higher osmolarity. The solution with higher osmolarity has a higher concentration of solute and lower concentration of water; therefore, this solution would need to have pressure applied in order to block water from coming in from the solution of lower osmolarity.

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Review Sheet 1

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11. What change in cell volume will occur when a cell is placed in a hypotonic solution? Water will move from the solution into the cell, causing the cell to expand. If enough water enters the cell, the cell will eventually burst.

12. What change in cell volume will occur when a cell is placed in a hypertonic solution? Water will move out of the cell, causing the cell to become crenated.

13. By what mechanism does the active transport of sodium lead to osmotic flow of water across a membrane? When sodium is actively transported across the membrane, a higher solute concentration will be set up on one side of the membrane which will draw water toward that side of the membrane.

14. If two solutions having different osmolarities are separated by a water-permeable membrane, will there be a change in the volume of the two compartments if the membrane is impermeable to solutes? yes

Will there be a change in the volume of the two compartments if the membrane is permeable to solutes? no

Explain your answers. When solutions are separated by a water-permeable membrane that is impermeable to solutes, water moves from an area of higher concentration to an area of lower concentration (osmosis). When the membrane is permeable to solutes, the solutes may move from an area of higher concentration to an area of lower concentration (diffusion) to establish equilibrium.

Review Sheet 1

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E X E R C I S E

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Skeletal Muscle Physiology

Time Allotment: 2–2.5 hours minimum if students are well-prepared, 3 hours preferred.

The following minimum computer hardware is recommended for PhysioEx 8.0: WINDOWS OS: Windows XP, Vista Resolution: 1024 × 768 Latest version of Adobe Flash Player Latest version of Adobe Reader Browsers: Internet Explorer 6.0 (XP only); Internet Explorer 7.0; Firefox 2.0 Internet Connection: 56K modem minimum for website Printer

MACINTOSH OS: 10.3.x, 10.4.x Resolution: 1024 × 768 Latest version of Adobe Flash Player Latest version of Adobe Reader Browsers: Safari 1.3 (10.3.x only); Safari 2.0 (10.4.x only); Firefox 2.0 Internet Connection: 56K modem minimum for website Printer

Advance Preparation, Comments, and Pitfalls 1. If you are using PhysioEx in a computer lab: Because PhysioEx requires a browser (such as Firefox or Internet Explorer) and the Flash Player to run, it is recommended that you make sure these items are already installed on student computers before beginning this lab. 2. If you are using the web version of PhysioEx: Have your students go to http://www.physioex.com and follow the registration instructions found in the very front of their lab manual. 3. It is helpful to instruct students in the proper operation of the mouse and menu system as part of the lab introduction. 4. When considering an upgrade for computer systems, memory (RAM) offers the largest performance increase for the least cost. 5. Having students work in pairs results in the most successful lab experience. If students must work in larger groups, have them each get keyboard and mouse experience with the program.

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6. Prior to the lab, suggest to the students that they become familiar with the exercise before coming to class. If students have a home computer, or access to a computer on campus, they can become familiar with the general operation of the simulations before coming to class. 7. You might do a short introductory presentation with the following elements: a. Describe the basics of muscle contraction at the cellular level, focusing on the sarcomere. This explanation is especially important for the isometric part of the simulation. b. Students often have problems distinguishing between in vivo stimulation via the nervous system versus the electrical stimulation we apply to whole skeletal muscle in an experiment. Mention that increasing the intensity of an electrical stimulus to the surface of whole muscle is not the same as stimulation via the nervous system, but that the outcome of increased force production is similar in both methods. c. Encourage students to try to apply the concepts from the simulation to human skeletal muscles as they work through the program. d. If a demonstration computer screen is available, briefly show students the basic equipment parts. 8. Keep in mind that many students in an introductory science course are deficient in their graphing skills. Review the principles of plotting before the class begins may prove helpful. 9. Be prepared to help the students answer the more difficult “What if...” questions.

Answers to Questions Activity 1: Identifying the Latent Period (p. 18) The latent period is 2.78 msec long. The latent period does not change with changes in stimulus voltage. Activity 2: Identifying the Threshold Voltage (pp. 18–20) You see 0.00 V in the Active Force display. The threshold voltage is 0.8 V. A small rise in the graph is observed, as opposed to the flat line generated by voltages under the threshold. The Active Force display also shows a value higher than 0.00 V at threshold voltage. Activity 3: Effect of Increases in Stimulus Intensity (p. 20) The peaks increased in height as voltage increased. The active force generated by the muscle increased with increases in voltage. The maximal voltage is 8.2 V. At the maximal voltage, all muscle fibers are responding (contracting) to the stimulus applied. Below this voltage, some fibers remain that will not be stimulated by the applied stimulus.

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No, because the muscle we are working with consists of many, many fibers—not one individual fiber. Activity 4: Treppe (p. 20) You see an increase in the peak of the graph with each successive application of stimulus. Activity 5: Summation (pp. 21–22) The active force is 1.82 gms. The active force will be higher than before. (Actual values will vary with each student.) The active force is slightly lower when the second stimulus is applied after the graph begins to fall versus when the second stimulus is applied before the graph begins to fall. Yes, the active force was higher. The fibers that responded to the first stimulus were in their refractory period, so new fibers were stimulated by the second stimulus; with more fibers responding, there was greater force generated. Yes, you see the same pattern of changes at lower voltages. Yes, the force increases with each additional stimulus. Activity 6: Tetanus (p. 22) At about 80 msec, the peaks begin to plateau—that is, no further increases in the height of the peaks are observed. Tetanus. There are almost no peaks and valleys visible at 130 stimuli/sec. Complete (fused) tetanus. There is no further increase in force past 146 stimuli/sec. Maximal tetanic tension. Activity 7: Fatigue (p. 22) In fatigue, force production decreases over time. Activity 8: Isometric Contractions (p. 24) The most active force was generated at muscle length = 70 to 80 mm. At muscle length = 90 mm. At muscle length = 80 mm. At muscle length = 90 mm, the muscle was being stretched too far for the actin and myosin in the sarcomere to fully interact, so active force decreased, thus decreasing the total force. The muscle length.

PhysioEx™ Exercise 2

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