appendix

Seeley−Stephens−Tate: Anatomy and Physiology, Sixth Edition

Back Matter Appendix © The McGraw−Hill Companies, 2004

A-1

Appendix A

Very large numbers with many zeros such as1,000,000,000,000,000 or very small numbers such as0.0000000000000001 are very cumbersome to workwith. Consequently, the numbers are expressed in akind of mathematical shorthand known as scientificnotation. Scientific notation has the following form:

M � 10n

Scientific Notation

where n specifies how many times the number M israised to the power of 10. The exponent n has twomeanings, depending on its sign. If n is positive, M ismultiplied by 10 n times. For example, if n � 2 and M� 1.2, then

1.2 � 102 � 1.2 � 10 � 10 � 120

In other words, if n is positive, the decimal point of M

Appendix B

Table of Measurements

Table A.1

Unit Metric Equivalent Symbol U.S. Equivalent

Measures of Length

1 kilometer � 1000 meters km 0.62137 mile

1 meter � 10 decimeters or 100 centimeters m 39.37 inches

1 decimeter � 10 centimeters dm 3.937 inches

1 centimeter � 10 millimeters cm 0.3937 inch

1 millimeter � 1000 micrometers mm

1 micrometer � 1/1000 millimeter or 1000 nanometers �m

1 nanometer � 10 angstroms or 1000 picometers nm

1 angstrom � 1/10,000,000 millimeter Å

1 picometer � 1/1,000,000,000 millimeter pm

Measures of Volume

1 cubic meter � 1000 cubic decimeters m3 1.308 cubic yards

1 cubic decimeter � 1000 cubic centimeters dm3 0.03531 cubic foot

1 cubic centimeter � 1000 cubic millimeters or 1 milliliter cm3 (cc) 0.06102 cubic inch

Measures of Capacity

1 kiloliter � 1000 liters kL 264.18 gallons

1 liter � 10 deciliters L 1.0567 quarts

1 deciliter � 100 milliliters dL 0.4227 cup

1 milliliter � volume of 1 gram of water at standard mL 0.3381 ouncetemperature and pressure

Measures of Mass

1 kilogram � 1000 grams kg 2.2046 pounds

1 gram � 100 centigrams or 1000 milligrams g 0.0353 ounce

1 centigram � 10 milligrams cg 0.1543 grain

1 milligram � 1/1000 gram mg

Table of Measurements

Note that a micrometer was formerly called a micron (�), and a nanometer was formerly called a millimicron (m�).

is moved to the right n times. In this case the decimalpoint of 1.2 is moved two places to the right.

If n is negative, M is divided by 10 n times.

1.2 1.21.2 � 10�2 � (10 � 10) � 100 � 0.012

1.20.



drogen ion concentration [H�] times the hydroxideion concentration [OH�] is a constant that is equalto 10�14.

[H�] � [OH�] � 10�14

Consequently, as the hydrogen ion concentration

Pure water weakly dissociates to form small numbersof hydrogen and hydroxide ions:

H2O ←→ H� � OH�

At 25°C the concentration of both hydrogen ionsand hydroxide ions is 10�7 mol/L. Any solution that

pH

has equal concentrations of hydrogen and hydroxideions is considered neutral. A solution is an acid if ithas a higher concentration of hydrogen ions thanhydroxide ions, and a solution is a base if it has alower concentration of hydrogen ions than hydrox-ide ions. In any aqueous solution (at 25°C) the hy-

Appendix D

Solution Concentrations

Physiologists often express solution concentration interms of percent, molarity, molality, and equivalents.

Percent

The weight-volume method of expressing percentconcentrations states the weight of a solute in a givenvolume of solvent. For example, to prepare a 10% so-lution of sodium chloride, 10 g of sodium chloride isdissolved in a small amount of water (solvent) toform a salt solution. Then additional water is addedto the salt solution to form 100 mL of salt solution.Note that the sodium chloride was dissolved in waterand then diluted to the required volume. Thesodium chloride was not dissolved directly in 100mL of water.

Molarity

Molarity determines the number of moles of solutedissolved in a given volume of solvent. A 1 molar (1 M)solution is made by dissolving 1 mole (mol) of a sub-stance in enough water to make 1 L of solution. For ex-ample, 1 mol of sodium chloride solution is made bydissolving 58.44 g of sodium chloride in enough waterto make 1 L of solution. One mol of glucose solution ismade by dissolving 180.2 g of glucose in enough waterto make 1 L of solution. Both solutions have the samenumber (Avogadro’s number) of formula units (NaCl)and molecules (glucose) in solution.

Molality

Although 1 M solutions have the same number ofsolute molecules, they don’t have the same number ofsolvent (water) molecules. Because 58.5 g of sodiumchloride occupies less volume than 180 g of glucose,the sodium chloride solution has more water mole-cules. Molality is a method of calculating concentra-tions that takes into account the number of soluteand solvent molecules. A 1 molal solution (1 m) is1 mol of a substance dissolved in 1 kg of water. Thusall 1-molal solutions have the same number of sol-vent molecules.

When sodium chloride, which is an ionic com-pound, is dissolved in water it dissociates to form twoions, a sodium cation (Na�) and a chloride anion(Cl�). Glucose does not dissociate when dissolved inwater, however, because it’s a molecule. Thus, thesodium chloride solution contains twice as many par-ticles as the glucose solution (one Na� and one Cl� foreach glucose molecule). To report the concentration ofthese substances in a way that reflects the number ofparticles in a given mass of solvent the concept of os-molality is used. The osmolality of a solution is themolality of the solution times the number of particlesinto which the solute dissociates in 1 kg of solvent.Thus 1 mol of sodium chloride in 1 kg of water is a2 osmolal (osm) solution because sodium chloridedissociates to form two ions.

Appendix C

The osmolality of a solution is a reflection of thenumber, not the type, of particles in a solution. Thus a1 osm solution contains 1 osm of particles per kilo-gram of solvent, but the particles may be all one typeor a complex mixture of different types.

The concentration of particles in body fluids isso low that the measurement milliosmole (mOsm),1/1000 of an osmole, is used. Most body fluids have anosmotic concentration of approximately 300 mOsmand consist of many different ions and molecules. Theosmotic concentration of body fluids is important be-cause it influences the movement of water into or outof cells (see chapter 3).

Equivalents

Equivalents are a measure of the concentrations ofionized substances. One equivalent (Eq) is 1 mol of anionized substance multiplied by the absolute value ofits charge. For example, 1 mol of NaCl dissociates into1 mol of Na� and 1 mol of Cl�. Thus there is 1 Eq ofNa� (1 mol � 1) and 1 Eq of Cl� (1 mol � 1). Onemole of CaCl2 dissociates into 1 mol of Ca2� and2 mol of Cl�. Thus there are 2 Eq of Ca2� (1 mol � 2)and 2 Eq of Cl� (2 mol � 1). In an electrically neutralsolution the equivalent concentration of positivelycharged ions is equal to the equivalent concentrationof the negatively charged ions. One milliequivalent(mEq) is 1/1000 of an equivalent.

In other words, if n is negative, the decimal point of Mis moved to the left n times. In this case the decimalpoint of 1.2 is moved two places to the left.

If M is the number 1.0, it often is not expressed in sci-entific notation. For example, 1.0 � 102 is the samething as 102, and 1.0 � 10�2 is the same thing as 10�2.

0.01.2

Two common examples of the use of scientificnotation in chemistry are Avogadro’s number and pH.Avogadro’s number, 6.023 � 1023, is the number ofatoms in 1 molar mass of an element. Thus

6.023 � 1023 � 602,300,000,000,000,000,000,000

which is a very large number of atoms.

The pH scale is a measure of the concentration ofhydrogen ions in a solution. A neutral solution has10�7 moles of hydrogen ions per liter. In other words

10�7 � 0.0000001

which is a very small amount (1 ten-millionth of agram) of hydrogen ions.

A-2 Appendix C



A-3Appendix E

Reference Laboratory Values

Appendix E

Table E.1

Test Normal Values Clinical Significance

Acetoacetate plus 0.32–2 mg/100 mL Values increase in diabetic acidosis, fasting, high-fat diet, and toxemia of acetone pregnancy

Ammonia 80–110 �g/100 mL Values decrease with proteinuria and as a result of severe burns and increase in multiple myeloma

Amylase 4–25 U/mL* Values increase in acute pancreatitis, intestinal obstruction, and mumps; values decrease in cirrhosis of the liver, toxemia of pregnancy, and chronic pancreatitis

Barbiturate 0 Coma level: phenobarbital, approximately 10 mg/100 mL; most other drugs, 1–3 mg/100 mL

Bilirubin 0.4 mg/100 mL Values increase in conditions causing red blood cell destruction of biliary obstruction or liver inflammation

Blood volume 8.5%–9% of body weight in kilograms

Calcium 8.5–10.5 mg/dL Values increase in hyperparathyroidism, vitamin D hypervitaminosis; values decrease in hypoparathyroidism, malnutrition, and severe diarrhea

Carbon dioxide content 24–30 mEq/L Values increase in respiratory diseases, vomiting, and intestinal obstruction; 20–26 mEq/L in infants they decrease in acidosis, nephritis, and diarrhea(as HCO3

�)

Carbon monoxide 0 Symptoms with over 20% saturation

Chloride 100–106 mEq/L Values increase in Cushing’s syndrome, nephritis, and hyperventilation; they decrease in diabetic acidosis, Addison’s disease, and diarrhea and after severe burns

Creatine phosphokinase Female 5–35 mU/mL Values increase in myocardial infarction and skeletal muscle diseases such as (CPK) Male 5–55 mU/mL muscular dystrophy

Creatinine 0.6–1.5 mg/100 mL Values increase in certain kidney diseases

Ethanol 0 0.3%–0.4%, marked intoxication0.4%–0.5%, alcoholic stupor0.5% or over, alcoholic coma

Glucose Fasting 70–110 mg/100 mL Values increase in diabetes mellitus, liver diseases, nephritis, hyperthyroidism, and pregnancy; they decrease in hyperinsulinism, hypothyroidism, and Addison’s disease

Iron 50–150 �g/100 mL Values increase in various anemias and liver disease; they decrease in iron deficiency anemia

Blood, Plasma, or Serum Values

*A unit (U) is the quantity of a substance that has a physiologic effect. continued

decreases, the hydroxide ion concentration increases,and vice versa. For example:

[H�] [OH�]Acidic solution 10�3 10�11

Neutral solution 10�7 10�7

Basic solution 10�12 10�2

Although the acidity or basicity of a solution could beexpressed in terms of either hydrogen or hydroxide ion

concentration, it’s customary to use hydrogen ion con-centration. The pH of a solution is defined as

pH � �log10(H�)

Thus a neutral solution with 10�7 mol of hydrogenions per liter has a pH of 7

pH � �log10(H�)� �log10(10�7)

� �(�7)� 7

In simple terms, to convert the hydrogen ion concen-tration to the pH scale, the exponent of the concen-tration (e.g., �7) is used, and it’s changed from anegative to a positive number. Thus an acidic solutionwith 10�3 mol of hydrogen ions/L has a pH of 3,whereas a basic solution with 10�12 hydrogen ions/Lhas a pH of 12.



Table E.1


Lactic acid 0.6–1.8 mEq/L Values increase with muscular activity and in congestive heart failure, severe hemorrhage, shock, and anaerobic exercise

Lactic dehydrogenase 60–120 U/mL Values increase in pernicious anemia, myocardial infarction, liver diseases, acute leukemia, and widespread carcinoma

Lipids Cholesterol 120–220 mg/100 mL Increased values for cholesterol and triglycerides are connected with Cholesterol esters 60%–75% of increased risk of cardiovascular disease, such as heart attack and stroke

cholesterolPhospholipids 9–16 mg/100 mL as

lipid phosphorusTotal fatty acids 190–420 mg/100 mLTotal lipids 450–1000 mg/100 mLTriglycerides 40–150 mg/100 mL

Lithium Toxic levels 2 mEq/L

Osmolality 285–295 mOsm/kg water

Oxygen saturation (arterial) 96%–100%see Po2

Pco2 35–43 mm Hg Values decrease in acidosis, nephritis, and diarrhea; they increase in respiratory diseases, intestinal obstruction, and vomiting

pH 7.35–7.45 Values decrease as a result of hypoventilation, severe diarrhea, Addison’s disease, and diabetic acidosis; values increase due to hyperventilation, Cushing’s syndrome, and vomiting

Po2 75–100 mm Hg Values increase in polycythemia and decrease in anemia and obstructive (breathing room air) pulmonary diseases

Phosphatase (acid) Male: total 0.13–0.63 U/mL Values increase in cancer of the prostate gland, hyperparathyroidism, someFemale: total 0.01–0.56 U/mL liver diseases, myocardial infarction, and pulmonary embolism

Phosphatase (alkaline) 13–39 IU/L* (infants and Values increase in hyperparathyroidism, some liver diseases, and pregnancyadolescents up to 104 IU/L)

Phosphorus (inorganic) 3–4.5 mg/100 mL (infants in first Values increase in hypoparathyroidism, acromegaly, vitamin D year up to 6 mg/100 mL) hypervitaminosis, and kidney diseases; they decrease in hyperparathyroidism

Potassium 3.5–5 mEq/100 mL

Protein Total 6–8.4 g/100 mL Total protein values increase in severe dehydration and shock; they decreaseAlbumin 3.5–5 g/100 mL in severe malnutrition and hemorrhageGlobulin 2.3–3.5 g/100 mL

Salicylate 0Therapeutic 20–25 mg/100 mLToxic Over 30 mg/100 mL

Over 20 mg/100 mL after age 60

Sodium 135–145 mEq/L Values increase in nephritis and severe dehydration; they decrease in Addison’s disease, myxedema, kidney disease, and diarrhea

Sulfonamide 0Therapeutic 5–15 mg/100 mL

Urea nitrogen 8–25 mg/100 mL Values increase in response to increased dietary protein intake; values decrease in impaired renal function

Uric acid 3–7 mg/100 mL Values increase in gout and toxemia of pregnancy and as a result of tissue damage

continued

A-4 Appendix E



A-5Appendix E

Table E.2


Clotting (coagulation) 5–10 minutes Values increase in afibrinogenemia and hyperheparinemia, severe liver damagetime

Fetal hemoglobin Newborns: 60%–90% Values increase in thalassemia, sickle-cell disease, and leakage of fetal blood into Before age 2: 0%–4% maternal bloodstream during pregnancyAdults: 0%–2%

Hemoglobin Male: 14–16.5 g/100 mL Values decrease in anemia, hyperthyroidism, cirrhosis of the liver, and severe Female: 12–15 g/100 mL hemorrhage; values increase in polycythemia, congestive heart failure, obstructive Newborn: 14–20 g/100 mL pulmonary disease, high altitudes

Hematocrit Male: 40%–54% Values increase in polycythemia, severe dehydration, and shock; values decrease in Female: 38%–47% anemia, leukemia, cirrhosis, and hyperthyroidism

Ketone bodies 0.3–2 mg/100 mL Values increase in ketoacidosis, fever, anorexia, fasting, starvation, high-fat dietToxic level: 20 mg/100 mL

Platelet count 250,000–400,000/mm3 Values decrease in anemias and allergic conditions and during cancer chemotherapy; values increase in cancer, trauma, heart disease, and cirrhosis

Prothrombin time 11–15 seconds Values increase in prothrombin and vitamin deficiency, liver disease, and hypervitaminosis A

Red blood cell count Males: 4.6–6.2 million/mm3 Values decrease in systemic lupus erythematosus, anemias, and Addison’s disease; Females: 4.2–5.4 million/mm3 values increase in polycythemia and dehydration and following hemorrhage

Reticulocyte count 1%–3% Values decrease in iron-deficiency and pernicious anemia and radiation therapy; values increase in hemolytic anemia, leukemia, and metastatic carcinoma

White blood cell Neutrophils 60%–70% Neutrophils increase in acute infections; eosinophils and basophils increase in allergic count, differential Eosinophils 2%–4% reactions; monocytes increase in chronic infections; lymphocytes increase during

Basophils 0.5%–1% antigen–antibody reactionsLymphocytes 20%–25%Monocytes 3%–8%

White blood cell 5000–9000/mm3 Values decrease in diabetes mellitus, anemias, and following cancer chemotherapy; count, total values increase in acute infections, trauma, some malignant diseases, and some

cardiovascular diseases

Blood Count Values



A-6 Appendix E

Table E.3


Urine Values

Acetone and 0 Values increase in diabetic acidosis and during fastingacetoacetate

Albumin 0 to trace Values increase in glomerular nephritis and hypertension

Ammonia 20–70 mEq/L Values increase in diabetes mellitus and liver disease

Bacterial count Under 10,000/mL Values increase in urinary tract infection

Bile and bilirubin 0 Values increase in biliary tract obstruction

Calcium Under 250 mg/24 h Values increase in hyperparathyroidism and decrease in hypoparathyroidism

Chloride 110–254 mEq/24 h Values decrease in pyloric obstruction and diarrhea; values increase in Addison’s disease and dehydration

Potassium 25–100 mEq/L Values decrease in diarrhea, malabsorption syndrome, and adrenal cortical insufficiency; values increase in chronic renal failure, dehydration, and Cushing’s syndrome

Sodium 75–200 mg/24 h Values decrease in diarrhea, acute renal failure, and Cushing’s syndrome; values increase in dehydration, starvation, and diabetic acidosis

Creatinine clearance 100–140 mL/min Values increase in renal diseases

Creatinine 1–2 g/24 h Values increase in infections and decrease in muscular atrophy, anemia, and certain kidney diseases

Glucose 0 Values increase in diabetes mellitus and certain pituitary gland disorders

Urea clearance Over 40 mL of blood Values increase in certain kidney diseasescleared of urea per minute

Urea 25–35 g/24 h Values decrease in complete biliary obstruction and severe diarrhea; values increasein liver diseases and hemolytic anemia

Uric acid 0.6– 1 g/24 h Values increase in gout and decrease in certain kidney diseases

CastsEpithelial Occasional Increase in nephrosis and heavy-metal poisoningGranular Occasional Increase in nephritis and pyelonephritisHyaline Occasional Increase in glomerular membrane damage and feverRed blood cell Occasional Values increase in pyelonephritis; blood cells appear in urine in response to

kidney stones and cystitisWhite blood cell Occasional Values increase in kidney infections

Color Amber, straw, transparent Varies with hydration, diet, and disease statesyellow

Odor Aromatic Becomes acetonelike in diabetic ketosis

Osmolality 500–800 mOsm/kg water Values decrease in aldosteronism and diabetes insipidus; values increase in high-protein diets, heart failure, and dehydration

pH 4.6–8 Values decrease in acidosis, emphysema, starvation, and dehydration; values increase in urinary tract infections and severe alkalosis



A-7Appendix E

Table E.4

Test Normal Values

Steroid hormones

Aldosterone Excretion: 5–19 mg/24 h*

Fasting at rest, 210 mEq sodium diet Supine: 48 � 29 pg/mL†

Upright: 65 � 23 pg/mL

Fasting at rest, 10 mEq sodium diet Supine: 175 � 75 pg/m

Upright: 532 � 228 pg/mL

Cortisol

Fasting 8 A.M.: 5–25 �g/100 mL

At rest 8 P.M.: Below 10 �g/100 mL

Testosterone Adult male: 300–1100 ng/100 mL‡

Adolescent male: over 100 ng/100 mL

Female: 25–90 ng/100 mL

Peptide hormones

Adrenocorticotropin (ACTH) 15–170 pg/mL

Calcitonin Undetectable in normals

Growth hormone (GH)

Fasting, at rest Below 5 ng/mL

After exercise Children: over 10 ng/mL

Male: below 5 ng/mL

Female: up to 30 ng/mL

Insulin

Fasting 6–26 �U/mL

During hypoglycemia Below 20 �U/mL

After glucose Up to 150 �U/mL

Luteinizing hormone (LH) Male: 6–18 mU/mL

Preovulatory or postovulatory female: 5–22 mU/mL

Midcycle peak 30–250 mU/mL

Parathyroid hormone Less than 10 microl equiv/L

Prolactin 2–15 ng/mL

Renin activity

Normal diet

Supine 1.1 � 0.8 ng/mL/h

Upright 1.9 � 1.7 ng/mL/h

Low-sodium diet

Supine 2.7 � 1.8 ng/mL/h

Upright 6.6 � 2.5 ng/mL/h

Thyroid-stimulating hormone (TSH) 0.5–3.5 �U/mL

Thyroxine-binding globulin 15.25 �g T4/100 mL

Total thyroxine 4–12 �g/100 mL

Hormone Levels

*1 microgram (1 �g) is equal to 10�6 g.†1 picogram (1 pg) is equal to 10�12 g.‡1 nanogram (1 ng) is equal to 10�9 g.



Answers To Review and Comprehension Questions

Chapter One

1. a; 2. b; 3. a; 4. c; 5. d; 6. e; 7. a; 8. b; 9. c; 10. d; 11. d;12. c; 13. d; 14. d; 15. a; 16. b; 17. b; 18. a; 19. e; 20. c;21. a; 22. b; 23. a; 24. b; 25. e

Chapter Two

1. e; 2. a; 3. b; 4. b; 5. a; 6. d; 7. b; 8. e; 9. c; 10. e; 11. c;12. d; 13. a; 14. b; 15. c; 16. c; 17. d; 18. c; 19. d; 20. e;21. b; 22. a; 23. b; 24. d; 25. b; 26. a; 27. c; 28. d; 29. a;30. e

Chapter Three

1. a; 2. e; 3. c; 4. d; 5. e; 6. c; 7. e; 8. d; 9. b; 10. b; 11. a;12. b; 13. b; 14. a; 15. d; 16. d; 17. b; 18. b; 19. c; 20. e;21. c; 22. c; 23. b; 24. c; 25. d; 26. c; 27. a; 28. d; 29. e;30. e

Chapter Four

1. e; 2. c; 3. a; 4. a; 5. b; 6. d; 7. c; 8. d; 9. d; 10. b; 11. a;12. b; 13. b; 14. e; 15. a; 16. d; 17. b; 18. b; 19. d; 20. d;21. a; 22. b; 23. e; 24. b; 25. c; 26. c; 27. b; 28. c; 29. b;30. d

Chapter Five

1. e; 2. a; 3. b; 4. a; 5. b; 6. b; 7. d; 8. e; 9. b; 10. a; 11. c;12. d; 13. e; 14. c; 15. d; 16. b; 17. c; 18. c; 19. d; 20. b;21. c; 22. b; 23. a; 24. b.; 25. a; 26. d; 27. c; 28. d; 29. c;30. c

Chapter Six

1. e; 2. e; 3. d; 4. a; 5, c; 6. e; 7. d; 8. b; 9. d; 10. c; 11. e;12. a; 13. a; 14. b; 15. e; 16. c; 17. a; 18. c; 19. e; 20. d;21. a; 22. b; 23. c; 24. e; 25. a; 26. e; 27. c; 28. e; 29. d;30. a

Chapter Seven

1. c; 2. c; 3. e; 4. d; 5. c; 6. d; 7. c; 8. a; 9. d; 10. a; 11. b;12. a; 13. a; 14. c; 15. a; 16. b; 17. c; 18. a; 19. d; 20. e;21. c; 22. b; 23. c; 24. a; 25. b; 26. c; 27. b; 28. c; 29. a;30. a

Chapter Eight

1. e; 2. b; 3. d; 4. d; 5. a; 6. e; 7. d; 8. e; 9. c; 10. d; 11. b;12. a; 13. b; 14. e; 15. a; 16. d; 17. c; 18. d; 19. e; 20. a;21. c; 22. b; 23. c; 24. d; 25. b; 26. b; 27. a; 28. c; 29. c

Chapter Nine

1. c; 2. e; 3. a; 4. d; 5. c; 6. e; 7. b; 8. a; 9. b; 10. b; 11. d;12. b; 13. d; 14. c; 15. e; 16. e; 17. d; 18. c; 19. c; 20. a;21. d; 22. b; 23. d; 24. a; 25. d; 26. c; 27. c; 28. c; 29. c;30. b

Chapter Ten

1. d; 2. b; 3. c; 4. c; 5. c; 6. c; 7. d; 8. d; 9. b; 10. c; 11. a;12. a; 13. c; 14. d; 15. a; 16. b; 17. d; 18. b; 19. a; 20. d;21. a; 22. c; 23. a; 24. a; 25. d; 26. c; 27. b; 28. b; 29. d

Chapter Eleven

1. a; 2. b; 3. b; 4. a; 5. c; 6. c; 7. b; 8. a; 9. c; 10. a; 11. b;12. d; 13. a; 14. c; 15. b; 16. a; 17. a; 18. b; 19. d; 20. c;21. e; 22. b; 23. d; 24. d; 25. b; 26. b; 27. e; 28. e; 29. b;30. a

Chapter Twelve

1. d; 2. c; 3. c; 4. d; 5. b; 6. b; 7. d; 8. c; 9. c; 10. b; 11. e;12. c; 13. d; 14. d; 15. c; 16. e; 17. c; 18. d; 19. c; 20. d

Chapter Thirteen

1. c; 2. b; 3. e; 4. c; 5. d; 6. b; 7. b; 8. b; 9. c; 10. a; 11. e;12. d; 13. d; 14. a; 15. c; 16. d; 17. b; 18. b; 19. a; 20. a;21. d; 22. a; 23. b; 24. b; 25. e; 26. b; 27. c; 28. e; 29. c;30. b; 31. b

Chapter Fourteen

1. c; 2. d; 3. d; 4. c; 5. a; 6. d; 7. b; 8. e; 9. a; 10. b; 11. b;12. e; 13. c; 14. c; 15. b; 16. d; 17. a; 18. b; 19. b; 20. d;21. b; 22. b; 23. d; 24. e; 25. d; 26. b; 27. b; 28. d; 29. e;30. e

Chapter Fifteen

1. e; 2. e; 3. c; 4. a; 5. b; 6. e; 7. b; 8. a; 9. c; 10. d; 11. b;12. b; 13. c; 14. d; 15. c; 16. b; 17. a; 18. d; 19. c; 20. d;21. e; 22. a; 23. d; 24. c; 25. a; 26. a; 27. b; 28. d; 29. a;30. c

Chapter Sixteen

1. e; 2. d; 3. d; 4. a; 5. e; 6. b; 7. d; 8. e; 9. d; 10. d; 11. d;12. a; 13. a; 14. c; 15. e; 16. a; 17. d; 18. a; 19. e; 20. c

Chapter Seventeen

1. c; 2. c; 3. b; 4. e; 5. d; 6. b; 7. a; 8. e; 9. b; 10. a; 11. c;12. e; 13. c; 14. e; 15. a; 16. d; 17. c; 18. d; 19. a; 20. c

Chapter Eighteen

1. e; 2. d; 3. e; 4. b; 5. a; 6. c; 7. b; 8. b; 9. e; 10. c; 11. d;12. e; 13. a; 14. d; 15. b; 16. c; 17. a; 18. a; 19. d; 20. c;21. d; 22. e; 23. e; 24. d; 25. b; 26. d; 27. a; 28. d; 29. a;30. b; 31. e; 32. c; 33. c; 34. e; 35. d

Chapter Nineteen

1. e; 2. c; 3. a; 4. e; 5. d; 6. a; 7. e; 8. d; 9. b; 10. c; 11. d;12. b; 13. b; 14. a; 15. c; 16. b; 17. b; 18. e; 19. e; 20. b;21. a; 22. a; 23. d; 24. a; 25. d; 26. c; 27. c; 28. c; 29. d;30. b

Chapter Twenty

1. d; 2. e; 3. c; 4. a; 5. b; 6. d; 7. c; 8. c; 9. a; 10. a; 11. c;12. d; 13. a; 14. b; 15. e; 16. a;17. a; 18. b; 19. e; 20. b;21. c; 22. d; 23. a; 24. a; 25. d; 26. c; 27. c; 28. c; 29. e;30. c

Chapter Twenty-One

1. b; 2. a; 3. a; 4. d; 5. d; 6. b; 7. c; 8. d; 9. a; 10. a; 11. d;12. e; 13. c; 14. a; 15. c; 16. b; 17. b; 18. d; 19. a; 20. e;21. d; 22. b; 23. b; 24. b; 25. d; 26. a; 27. d; 28. c; 29. b;30. e

Chapter Twenty-Two

1. d; 2. c; 3. a; 4. b; 5. a; 6. e; 7. d; 8. d; 9. e; 10. b; 11. a;12. d; 13. d; 14. e; 15. d; 16. b; 17. e; 18. e; 19. a; 20. d;21. c; 22. a; 23. b; 24. c; 25. a; 26. d; 27. d; 28. d; 29. d;30. c

Chapter Twenty-Three

1. a; 2. b; 3. e; 4. d; 5. e; 6. c; 7. d; 8. b; 9. c; 10. b; 11. d;12. a; 13. d; 14. c; 15. c; 16. d; 17. c; 18. d; 19. b; 20. d;21. b; 22. c; 23. b; 24. b; 25. c; 26. a; 27. c; 28. a; 29. d;30. d

Chapter Twenty-Four

1. a; 2. d; 3. e; 4. d; 5. a; 6. b; 7. a; 8. b; 9. a; 10. c; 11. b;12. d; 13. c; 14. d; 15. e; 16. b; 17. d; 18. d; 19. e; 20. e;21. b; 22. b; 23. b; 24. a; 25. e; 26. e; 27. e; 28. a; 29. c;30. a

Chapter Twenty-Five

1. d; 2. c; 3. a; 4. e; 5. b; 6. d; 7. e; 8. e; 9. b; 10. a; 11. d;12. b; 13. d; 14. a; 15. e; 16. c; 17. a; 18. b; 19. a; 20. b

Chapter Twenty-Six

1. d; 2. b; 3. d; 4. c; 5. e; 6. e; 7. b; 8. b; 9. d; 10. e; 11. a;12. b; 13. b; 14. b; 15. a; 16. a; 17. c; 18. d; 19. c; 20. b;21. e; 22. d; 23. e; 24. d; 25. d; 26. c; 27. e; 28. a; 29. b;30. c

Chapter Twenty-Seven

1. b; 2. a; 3. b; 4. c; 5. d; 6. a; 7. a; 8. a; 9. b; 10. a; 11. b; 12. d;13. a; 14. a; 15. b; 16. d; 17. a; 18. c; 19. c; 20. d

Chapter Twenty-Eight

1. e; 2. a; 3. a; 4. b; 5. d; 6. e; 7. b; 8. b; 9. d; 10. c; 11. b;12. c; 13. a; 14. e; 15. a; 16. b;17. d; 18. c; 19. b; 20. a;21. a; 22. c; 23. a; 24. c; 25. e; 26. e; 27. c; 28. d; 29. c;30. a

Chapter Twenty-Nine

1. d; 2. e; 3. a; 4. a; 5. d; 6. a; 7. c; 8. c; 9. b; 10. c; 11. a;12. a; 13. e; 14. d; 15. c; 16. c; 17. e; 18. a; 19. b; 20. d;21. a; 22. c; 23. d; 24. c; 25. c; 26. b; 27. c; 28. c; 29. d

Appendix F

A-8



Chapter 1

1. Student B is correct. Body temperature begins torise as a result of exposure to the hotenvironment. Sweating eliminates heat from thebody and lowers body temperature. Bodytemperature returning to its ideal normal valueis an example of negative feedback. Student Aprobably thought that it was positive feedbackbecause sweating continued to increase.Sweating, however, is the response. The variablebeing regulated by sweating is body temperature.

2. Answer e is correct. Positive-feedbackmechanisms result in movement away fromhomeostasis and are usually harmful. Thecontinually decreasing blood pressure is anexample. Negative-feedback mechanisms resultin a return to homeostasis. The elevated heartrate is a negative-feedback mechanism thatattempts to return blood pressure back to anormal value. In this case, the negative-feedbackmechanism was inadequate to restorehomeostasis, and medical intervention (atransfusion) was necessary.

3. When a boy is standing on his head, his nose issuperior to his mouth. Remember that directionalterms refer to a person in the anatomic position,not to the body’s current position.

4. a. Inferior or caudalb. Anterior, ventral, or superficialc. Proximal, superior, or cephalicd. Medial

5. The esophagus is located in the left-upperquadrant and the epigastric region. The urinarybladder is located in the left-lower and right-lower quadrants and is in the hypogastric region.

6. Answer a is correct. The best way to reach theanterior surface of the heart begins with thepatient lying on his or her back so that theanterior surfaces of the thorax and heart arefacing the surgeon. The heart is located in theanterior portion of the thoracic cavity within themediastinum and is surrounded by thepericardial cavity. The pericardial cavity is linedwith the pericardial serous membranes, throughwhich a cut can be made to reach the heart.

7. The uterus is located in the pelvic cavity. Thepelvic cavity, however, is surrounded by thebones of the pelvis and doesn’t increase in sizeduring pregnancy. Instead, as the fetus grows theexpanding uterus must move into theabdominal cavity, thereby crowding abdominalorgans and dramatically increasing the size ofthe abdominal cavity.

8. After passing through the left thoracic wall, thefirst membrane encountered is the parietal pleura.Continuing through the pleural cavity the visceralpleura of the left lung and then the left lung arepierced. Leaving the lung the bullet penetrates thevisceral pleura, the pleural cavity, and the parietal

A-9

Answers to Critical Thinking Questions

pleura (remember that the lung is surrounded bya double-membrane sac). Next the parietalpericardium, the pericardial cavity, the visceralpericardium, and the heart are penetrated.

9. After passing through the abdominal wall, theparietal peritoneum is pierced. In passingthrough the stomach, the visceral peritoneum,the stomach itself, and the visceral peritoneumon the other side of the stomach are penetrated.Because the diaphragm is lined inferiorly byparietal peritoneum and superiorly by parietalpleura, these are the next two membranespierced. The pole then passes through the pleuralspace and visceral pleura to enter the lung.

Chapter 2

1. An atom of iron has 26 protons (the atomicnumber), 30 neutrons (the mass number minusthe atomic number), and 26 electrons (becausethe number of electrons is equal to the numberof protons). If an atom of iron loses threeelectrons, it has three more protons (positivecharges) than electrons (negative charges).Therefore the iron ion has an overall charge of�3, which is represented symbolically as Fe3�.

2. The formation of free fatty acids and glycerolfrom a triglyceride is a decomposition reactionbecause a larger molecule breaks down intosmaller molecules. All of the decompositionreactions in the body are collectively referred toas catabolism. This reaction can also be classifiedas a hydrolysis reaction because as part of thereaction a water molecule is split into hydrogen,which becomes part of the glycerol molecule,and hydroxide, which becomes part of a fattyacid molecule.

3. The slight amount of heat functions as activationenergy and starts a chemical reaction. Thereaction releases a large amount of heat, thuscausing the solution to become hot.

4. Heating (boiling) has destroyed the ability of themolecules in one or both of the solutions tofunction in the chemical reaction. This is calleddenaturation. There are two possibilities as to whatis denatured. It could be the reactants themselvesor an enzyme that catalyzes the reaction.

5. Muscle contains proteins. To increase muscle mass,proteins must be synthesized from amino acids.The synthesis of molecules in living organismsrequires the input of energy. That energy comesfrom the potential energy stored in the chemicalbonds of food molecules, which is released duringthe decomposition of food molecules.

6. Remember that pH is a measure of hydrogen ionconcentration. If equal amounts of solutions Aand B are mixed, the resulting hydrogen ionconcentration is the average value of the twosolutions, that is, the pH is (8 � 2)/2 = 5. A pHof 5 is acidic. This question illustrates an

important point: The pH of a solution can bechanged by adding a more acidic or basicsolution to it.

7. The sodium bicarbonate dissociates, therebyincreasing the amount of bicarbonate ions in thesolution. The bicarbonate ions combine withhydrogen ions to form carbonic acid, whichbecomes carbon dioxide and water. The decreasein hydrogen ions causes the pH of the solution toincrease (become more alkaline).

8. As A and B are added to the solution, the enzymeE catalyzes the formation of C. However, when Cbinds to the active site of E, the ability of E tocatalyze the formation of C is blocked. As moreand more C is produced, the rate of formation ofC is slowed. Because the reaction of C with E isreversible, there will always be some E that has afunctional (not blocked) active site, and some Awill therefore always combine with B.

9. One might try heating the substances becauseproteins can be denatured and can coagulate (likefrying an egg). Another possibility is to trydissolving the substances in water. Most lipids areinsoluble in water, while many proteins are eithersoluble in water or form colloids with water.

10. Most proteins (i.e., a typical protein) containsulfur, which is not found in phospholipids.Typical phospholipids and proteins containcarbon, hydrogen, oxygen, nitrogen, andphosphate.

Chapter 3

1. The cells within the wound swell with water andlyse by the introduction of a hypotonic solution.This kills potentially metastatic cells that maystill be present in the wound.

2. Water moves by osmosis from solution B intosolution A. Because solution A is hyperosmoticto solution B, solution A has more solutes andless water than does solution B. Water thereforemoves from solution B (with more water) tosolution A (with less water).

3. Answer b is correct. Because the solution isisotonic, there is no exchange of water. Becausethe solution contains the same concentration ofall substances except that it has no urea, only anet movement of urea occurs across themembrane.

4. Answer b is correct. At point A on the graph, theextracellular concentration is equal to theintracellular concentration. If movement were bysimple diffusion or by facilitated diffusion, atthis point the rate of movement would be zero.Because it is not zero, it’s reasonable to concludethat the mechanism involved is active transport.

5. A reduced intracellular K� concentration reducesthe concentration difference for K� across theplasma membrane. Thus, the rate at which K�

diffuses out of the cell is reduced, and a smaller

Appendix G



charge difference is required across the plasmamembrane to oppose the diffusion of the K� outof the cell. The potential difference across theplasma membrane is therefore reduced.

6. Because the drug inhibits mRNA synthesis,protein synthesis is stopped. If the cell releasesproteins as they were synthesized, the rate ofprotein secretion dramatically decreasesfollowing the administration of the drug. On theother hand, if the cell releases stored proteins,the rate of secretion at first is normal and thengradually declines.

7. The well-developed rough endoplasmicreticulum is indicative of protein synthesis, and awell-developed Golgi apparatus is indicative ofsecretion. It’s likely that this cell synthesizes andsecretes proteins.

Chapter 4

1. The tissue is epithelial tissue, as it is lining a freesurface, and the epithelium is stratified becauseit consists of more than one layer. The types ofstratified epithelium are stratified squamous,stratified cuboidal, stratified columnar, ortransitional epithelium. The structure of the cellsin the surface layers enables the determination ofa specific tissue type. Flat cells in the surfacelayer indicate stratified squamous epithelium.Cuboidal cells in the surface layer indicatestratified cuboidal epithelium, and columnarcells in the surface layer point to stratifiedcolumnar epithelium. The surface cells oftransitional epithelium are roughly cuboidalwith cubelike or columnar cells beneath them.When transitional epithelium is stretched, thesurface cells are still roughly cuboidal, butunderlying layers can be somewhat flattened.

2. In general, epithelial cells undergo cell division(mitosis) in response to injury, and the newlyproduced cells replace the damaged cells. If thebasement membrane is destroyed, however,nothing is present to provide scaffolding for thenewly formed epithelial cells. Without thebasement membrane, there’s not an effective wayfor the newly formed epithelial cells to repair astructure such as a kidney tubule. Since thebasement membranes appear to be mostlypresent, the person is likely to survive and thekidney will regain most of its ability to function.

3. Epithelium that functions to resist abrasion isstratified squamous epithelium. The moiststratified squamous epithelium lining the mouthand the keratinized stratified squamousepithelium of the skin are examples. The cells atthe surface are flattened, and when scraped awaydue to abrasion they are replaced by the cellsbeneath them. In contrast epithelial cells thatcarry out absorption are either simple cuboidalor simple columnar. Because they are one layerthick, they are more susceptible to damage andare not resistant to abrasion. In addition, thesecells are large in volume, which allows them tocontain the organelles involved in transport,such as mitochondria to produce ATP in the caseof active transport. The surface of the cells thatabsorb are likely to contain microvilli, whichincreases the surface area for absorption. The flatcells that resist abrasion have no microvilli.

4. Glands producing merocrine secretions do sowith no loss of actual cellular material, whereasglands producing holocrine secretions shedentire cells. The cells rupture and die, and theentire cell becomes part of the secretion. Youcould chemically analyze the secretions for thetypes of molecules found in cellular organelles.For example, if phospholipids and proteinsnormally found in membranes are in thesecretion, then the secretion is a holocrinesecretion. If the secretion is watery or containsproducts that are not found in membranes ororganelles, it’s a merocrine secretion.

5. The statement is not appropriate. A tissuecapable of contracting is muscle. Both cardiacmuscle and smooth muscle cells aremononucleated, although some cardiac musclecells can have two nuclei, and they are bothunder involuntary control. Cardiac muscle isstriated, and smooth muscle is not, however.

6. Histamine is one of the mediators ofinflammation released in response to tissuedamage. Several other mediators ofinflammation, however, are released duringinflammation in addition to histamine.Antihistamines might reduce the inflammatoryresponse somewhat, but it’s not likely to have amajor effect because of the other mediators ofinflammation released at the same time. Incertain types of inflammatory responses, such asallergic responses, histamines are released in largeamounts. Under these conditions, antihistaminesdo reduce the inflammatory response.

Chapter 5

1. Yes, the skin (dermis) can be overstretched dueto obesity.

2. The stratum corneum, the outermost layer ofthe skin, consists of many rows of flat, deadepithelial cells. The many rows of cells, whichare continuously shed and replaced, areresponsible for the protective function of theintegument. In infants, there are fewer rows ofcells, resulting in skin that’s more easilydamaged than that in adults.

3. Melanocytes produce melanin, which protectsunderlying tissue from ultraviolet radiation.Therefore, we expect melanocytes to be assuperficial as possible. Also recall that melaninproduction varies depending on exposure to thesun. Response to stimulation is a characteristic ofliving cells. Thus, melanocytes should be foundin the most superficial living layer of theepidermis, the stratum basale.

4. When first exposed to the cold temperature justbefore starting the run, the blood vessels in theskin constrict to conserve heat. This produces apale skin color. Dilation of the skin blood vesselsdoesn’t occur at this time because the skin hasnot been exposed to the cold long enough tocause the skin temperature to fall below 15°C.After running for awhile, as a result of the excessheat generated by the exercise, the blood vesselsin the skin dilate. This results in heat loss andhelps to prevent overheating. Increased bloodflow through the skin causes it to turn red. Afterthe run, the body still has excess heat toeliminate, so the skin remains red for some time.

5. Eyelashes have a short growth stage (30 days)and are therefore short. Fingernails growcontinuously but are short because they are cut,broken off, or worn down.

6. Several methods have some degree of success intreating acne: (a) Kill the bacteria. One effectiveagent is benzoyl peroxide, found in some acnemedications. (b) Prevent blockage of the hairfollicle. A vitamin A derivative (tretinoin;Retin-A) has proven effective in keeping thefollicular epithelial cells and sebum frombuilding up and closing off the hair follicle.(c) Unplug the follicle. Some sulfur compounds(Acnederm) speed up peeling of the skin andthus unplug the follicle.

7. Probably not, because following removal of thenail from the nail fold, it may grow back into thenail fold and the ingrown toenail would reoccur.One solution is to remove the small part of thenail responsible for the ingrown toenail. Prior tothis drastic approach, sterile gauze can be placedbetween the nail and the nail fold to force thenail away from the nail fold. After the nail fold ishealed, the gauze can be removed.

Chapter 6

1. Normally bone matrix and bone trabeculae areorganized to be strongest along lines of stress.Random organization of the collagen fibers ofbone matrix results in weaker bones. In addition,the reduced amount of trabecular bone makesthe bone weaker. Fractures of the bone can occurwhen the weakened bone is subjected to stress.

2. Replacement of cartilage of the epiphyseal plateby bone normally occurs on the diaphyseal sideof the plate. As growth ceases, the cartilage cellsstop dividing and producing new cartilage.Replacement of cartilage with bone continuesfrom the diaphyseal side, and eventually all ofthe cartilage of the plate is bone.

3. Mechanical stress applied to bone stimulatesosteoblast activity, so the patient with a walkingcast should heal faster.

4. Osteoporosis is depletion of bone matrix thatresults when more bone is destroyed than isformed. Because mechanical stress stimulatesbone formation (osteoblast activity), runninghelps to prevent osteoporosis in the bones beingstressed. This includes the bones of the lowerlimbs and the spine.

5. The loss of bone density results because thebones are not bearing weight in the weightlessenvironment. Therefore osteoblasts are notsufficiently stimulated and bone resorptionexceeds bone building. Bone loss can be slowedby stressing the bones using exercises againstresistance such as cycling.

6. The kidneys are the site of production of activevitamin D (see chapter 5), which is needed forcalcium absorption in the small intestine. Kidneyfailure can result in inadequate vitamin Dproduction, too little uptake of calcium, andtherefore osteomalacia.

7. Testosterone normally causes a spurt of growthat puberty followed by slower growth andclosure of the epiphyseal plate. Withouttestosterone, growth is slower, but proceedslonger, resulting in a taller-than-normal person.

A-10 Appendix G



A-11Appendix G

8. Blood vessels in central canals run parallel to thelong axis of the bone, and perforating canals runat approximately a right angle to the centralcanals. Thus, perforating canals connect tocentral canals, which allows blood vessels in theperforating canals to connect with blood vesselsin the central canals. After a fracture, blood flowthrough the central canals stops back to thepoint where the blood vessels in the centralcanals connect to the blood vessels in theperforating canals. The regions of bone on eitherside of the fracture associated with this lack ofblood delivery die.

9. Hyperthyroidism stimulates increased bonebreakdown and could cause osteitis fibrosacystica, a condition in which the bone is eatenaway as calcium is released from the bone. Theresult can be a deformed bone that is likely tofracture. Vitamin D therapy might help becausevitamin D promotes an increase in bloodcalcium levels (see chapter 5) and thereforeincreased deposition of calcium in bone.

Chapter 7

1. An infection in the nasal cavity could spread toadjacent cavities and fossae, including theparanasal sinuses: (1) frontal, (2) maxillary, (3)ethmoidal, and (4) sphenoidal; (5) the orbit(through the nasolacrimal duct); (6) the cranialcavity (through the cribriform plate); and (7) thethroat (through the posterior opening of thenasal cavity).

2. Falling on the top of the head could drive theoccipital condyles into the superior articulatingprocesses of the atlas, causing a fracture. Anuppercut to the jaw would slightly lift theoccipital condyles away from the superiorarticulating processes of the atlas and usuallydoesn’t result in a fractured atlas. Such a blow tothe jaw can, however, fracture the temporal bonewhere it articulates with the mandible.

3. Forceful rotation of the vertebral column is mostlikely to damage the articular processes, especiallyin the lumbar region, where the articularprocesses tend to prevent excessive rotation (thesuperior articular processes face medially and theinferior articular processes face laterally).

4. Weaker back muscles on one side could cause thevertebral column to bend laterally (scoliosis)toward the opposite side. Lordosis can result frompregnancy. As the fetus causes the abdomen tomove anteriorly, the thorax and head tend to pullposteriorly, to restore the center of gravity. Thisposture increases the lumbar curvature. The sameeffect can be seen in people who are “pot-bellied.”

5. If the ulna and radius become fused, the radiuscan no longer rotate relative to the ulna, and, as aresult, most of the rotation of the forearm andhand is lost.

6. Measure from the anterior superior iliac spine (a“stationary” point relative to the limb, which canbe easily found as a surface landmark) to thelateral malleolus. The inferior side of the footcould also be used, if the person is standing on aflat surface. A defect of the foot or ankle mayoccur, however, in which the ankle on one side iselevated. If the length of the thigh is the only partto be measured, measure to the lateral epicondyle.

7. The ischial tuberosity is the bony protuberance.8. Women’s hips are wider than men’s. As the knees

are positioned toward the midline the slope ofthe femur from its proximal end toward its distalend is greater in women, and as a result, womentend to be knock-kneed more often than men.

9. The lateral malleolus extends further distallythan does the medial malleolus, thus making itmore difficult to turn the foot laterally thanmedially. The styloid process of the radiusextends further distally than the styloid processof the ulna, thus making it more difficult to cockthe wrist toward the thumb (laterally) thantoward the little finger (medially).

10. Landing on the heels could fracture thecalcaneus. Heavy objects, such as Hefty Stomper,landing on the dorsal surface of the foot couldfracture the metatarsals or even the tarsals.

Chapter 8

1. If the sternocostal synchondrosis were to ossify,becoming a synostosis, there would no longer beany stretch through the costal cartilage, thethorax could not expand, and, as a result,respiration would be severely hampered.

2. a. Suture, little or no movement.b. Syndesmosis, some movement.c. Complex synovial joints: the

humeroulnoradial joint is a hinge joint, theradioulnar joint is a pivot joint. All haveconsiderable movement.

3. a. Flexion and supinationb. Flexion of the thigh and extension of the legc. Abduction of the armd. Flexion of the leg and plantar flexion of the

foot4. The anterior drawer test determines the integrity

of the anterior cruciate ligament, and the posteriordrawer test determines the integrity of theposterior cruciate ligament. Unusual movementduring the posterior drawer test indicates damageto the posterior cruciate ligament.

Chapter 9

1. Botulism poisoning results from theconsumption of botulism toxin produced by thebacterium Clostridium botulinum. The toxinbinds to presynaptic nerve terminals andprevents the release of acetylcholine. Thus,action potentials in nerves cannot produceaction potentials in skeletal muscles, and theresult is paralysis of skeletal muscles, whichexplains the difficulty in breathing andswallowing. Other reasonable explanations arethat the toxin binds to and blocks the receptorsfor acetylcholine, that the toxin blocks the entryof Ca2� into the presynaptic terminal and thusprevents acetylcholine release, or that the toxinspecifically prevents entry of ions through Na�

channels of skeletal muscle cells.2. Muscular dystrophy results from gradual atrophy

of skeletal muscle fibers and their replacementwith connective tissue. Myasthenia gravis resultsfrom the degeneration of the receptors foracetylcholine on the postsynaptic membranes ofskeletal muscle cells. If an inhibitor ofacetylcholinesterase is administered, the resultshould be an increase in the concentration of

acetylcholine in the nerve muscle synapse. Thus,more acetylcholine is available to bind toacetylcholine receptors. In people suffering frommyasthenia gravis, the increased concentration ofacetylcholine in the synapse allows acetylcholine tobind a greater percentage of the acetylcholinereceptors present and causes the musclecontractions to increase in strength. In people whohave muscular dystrophy, the muscle contractionsdon’t increase in strength because muscle atrophyis the cause of the weakness. The additionalacetylcholine in the neuromuscular synapse has noeffect on the weakened muscle fibers.

3. Placing sarcoplasmic reticulum from skeletalmuscle cells into the beaker would removecalcium from the solution because sarcoplasmicreticulum transports Ca2� from the solutioninto the sarcoplasmic reticulum. In addition,ATP would have to be added for two reasons: (1)the sarcoplasmic reticulum actively transportscalcium and, therefore, requires ATP; and (2)ATP must bind to the heads of the myosinmolecules before the myosin heads can releasefrom the active sites on the actin molecules.

4. A lower-than-normal temperature decreases therate of all of the processes that occur in the lagphase of muscular contraction because a lowertemperature decreases the rate of all chemicalreactions and the rate of ion diffusion. As aconsequence, the lag phase requires a longer time.

5. Start with a subthreshold stimulus and increasethe stimulus strength by very small increments.Apply the stimulus to the nerve of muscle A andmuscle B. If the number of motor units is thesame for both preparations, each time thestimulus strength is increased the degree oftension produced by the muscles would alsoincrease to the same degree in each muscle. Ifone muscle has more motor units than the other,the muscle with the greater number of motorunits would exhibit a greater number of separateincreases in tension, and the magnitude of theincreases in tension would be smaller than thoseseen in the muscle with fewer motor units.

6. When a muscle slowly lifts an object, thecontraction starts with a small number of motorunits being stimulated. Each motor unit isstimulated tetanically. As the contractioncontinues, more and more motor units arerecruited to lift the object slowly. To lower theobject, the number of motor units stimulatedtetanically is reduced slowly and the tensionproduced by the muscle decreases. In a muscletwitch, a stimulus causes a single action potentialin all of the muscle fibers responding to thestimulus. The stimulated muscle fibers contractin an all-or-none fashion and then relax. Bothcontraction and relaxation occur quickly.

7. The shape of an active tension curve for skeletalmuscle can be seen in figure 9.20. In contrast, anactive tension curve is much flatter for smoothmuscle. That is, for each increase in the length ofa muscle fiber there is little change in the activetension produced by the smooth muscle fiber.Smooth muscle has, as one of its majorcharacteristics, the ability to increase in lengthwithout much increase in the tension producedby the smooth muscle cells.



8. Both the 100 m run and weight lifting involverapid and intense contractions of skeletalmuscles that are completed quickly. Thesecontractions depend on anaerobic metabolismfor a significant amount of the ATP produced. Incontrast, the 10,000 m run involves sustainedmuscular contractions that are not as rapid, butthe slower contractions are repeated many timesduring the run. Aerobic metabolism producesthe majority of the ATP for the 10,000 m run.Anaerobic metabolism is associated with adecrease in creatine phosphate, an increase increatine, an increase in lactic acid, and a decreasein glycogen, and the enzymes responsible foranaerobic metabolism function more rapidly.Aerobic metabolism is associated with increasedenzyme activity in the mitochondria and anincrease in carbon dioxide production. Oxygen isused more rapidly during aerobic metabolism.

9. During intense exercise it’s possible to experiencephysiologic contracture. Being unable to eithercontract or relax the muscles for a short timewhile exercising suggests the existence ofphysiologic contracture.

10. Smooth muscle depends almost entirely onaerobic metabolism to produce the ATP requiredfor muscle contraction. If the blood supply tosmooth muscle fibers is decreased the smoothmuscle, therefore, cannot maintain contractions.

11. During the 100 m race Shorty depended on ATPproduced by anaerobic metabolism. Thatproduced an oxygen debt at the end of the run,which resulted in an elevated rate of respirationfor a time. During the longer and slower runmost of the ATP for muscle contractions wasproduced by aerobic respiration, and very littleoxygen debt developed. Prolonged aerobicrespiration is required to “pay back” the oxygendebt. Shorty’s rate of respiration was, therefore,prolonged after the 100 m race but not after thelonger but slower run.

12. High blood K� concentration also results indepolarization of smooth muscle plasmamembranes. Depolarization of the smoothmuscle plasma membrane results in increasedmuscle contractions and increased permeabilityof the plasma membrane to both Na� and Ca2�,which results in further depolarization and anincrease in the intracellular concentration ofCa2�. These changes result in the production ofaction potentials and muscle contractions.

13. The muscles would contract. ATP would beavailable to bind to the myosin heads, thusallowing myosin molecules to be released fromactin molecules. The cross-bridges wouldimmediately re-form, and complete cross-bridgecycling would result in contraction of the musclefibers. As long as Ca2� were present at highconcentrations in the sarcoplasm, contraction ofthe muscles would occur. If the sarcoplasmicreticulum were intact, ATP would be available todrive the active transport of Ca2� into thesarcoplasmic reticulum. As the Ca2� decreased inthe sarcoplasm, relaxation would result. If thesarcoplasmic reticulum were not intact, however,and could not transport Ca2� into thesarcoplasmic reticulum as fast as they leak out, themuscle would remain contracted until it fatigued.

14. Hormones can bind to ligand-gated Ca2�

channels, and the channels, in response, open.Ca2� diffuse into the cell and cause contractionto occur. Only a small amount of depolarizationresults as Ca2� diffuse into the cell, and sinceNa� channels don’t open, a large change in theresting membrane potential doesn’t occur.

15. In experiment A, the students used anaerobicrespiration as they started to run in place, butaerobic respiration also increased to meet mostof their energy needs. When they stopped, theirrespiration rate was increased over resting levelsbecause of repayment of the oxygen debt due toanaerobic respiration. In experiment B almost allof the student’s respiration came from anaerobicrespiration because the students held theirbreath while running place. Consequently, thestudents had a much larger oxygen debt. Thestudent’s respiratory rate and depth was greaterthan inexperiment A, or that their respirationrates were elevated for a longer period of timethan in experiment A.

Chapter 10

1. Muscle Action Synergist AntagonistLongus Flexes Rectus Most of thecapitis neck capitis posterior

anterior neckLongis musclescoli

Erector Extends Interspinales Mostspinae vertebral Multifidus anterior

column Semispinalis abdominalthoracis muscles

Coraco- Adducts Latissimus Deltoidbrachialis arm dorsi Supra-

Pectoralis spinatusmajorTeres majorTeres minor

Flexes arm Deltoid Deltoid(anterior) (posterior)Pectoralis Latissimusmajor dorsiBiceps Teres majorbrachii Teres minor

Infra-spinatusSub-scapularisTricepsbrachii

2. Biceps brachii: Pull-ups with hands supinatedTriceps brachii: Push-upsDeltoid: Abduction of the arms to shoulderheight, with weights in the hands (abductionpast shoulder height involves mostly scapularrotation by the trapezius)Rectus abdominis: Sit-ups to 45 degrees (sit-upspast 45 degrees involve mostly the psoas major)Quadriceps femoris: Extending the legs against aforceGastrocnemius: Plantar flexion of the feetagainst a force, such as toe raises with a weighton the shoulders

3. The brachioradialis originates on the humerusand inserts onto the distal end of the radius. Thefulcrum of this lever system is the elbow joint.With a weight held in the hand, the force,applied between the weight and the fulcrum, is aclass III lever system. With the weight on theforearm, the weight is between the force and thefulcrum and is a class II lever system. A greaterweight can be lifted if placed on the forearmrather than in the hand, but weights placed onthe forearm cannot be lifted as far.

4. The muscles that flex the head also opposeextension of the neck. In an accident causinghyperextension of the neck, these muscles couldbe stretched and torn. The muscles involvedcould include the sternocleidomastoid, longuscapitis, rectus capitis anterior, and longus coli.Automobile headrests are designed so that, ifadjusted correctly, the back of the head hits theheadrest during a rear-end accident, therebypreventing hyperextension of the neck.

5. The only muscle that elevates the lower eyelid isthe orbicularis oculi, which “closes the eye.” Withthis muscle not functioning, the lower eyelidwould droop. The levator anguli oris, whichelevates the angle of the mouth, was alsoapparently affected allowing the corner of themouth to droop. The zygomaticus major mayalso have been affected, as it inserts onto thecorner of the mouth (see figure 10.7).

6. The genioglossus muscle protrudes the tongue.If it becomes relaxed, or paralyzed, the tonguemay fall back and obstruct the airway. This canbe prevented or reversed by pulling forward anddown on the mandible, thus opening themouth. The genioglossus originates on the genuof the mandible. As the mandible is pulled downand forward, the genioglossus is pulled forwardwith the mandible, thus pulling the tongueforward also.

7. The rotator cuff muscles are the primary musclesholding the head of the humerus in the glenoidfossa, especially the supraspinatus. In fact, a tornrotator cuff, which usually involves a tear of thesupraspinatus muscle, often results in dislocationof the shoulder.

8. With the quadriceps femoris paralyzed, the legcould not be extended, and the lower limb couldnot bear weight unless the knee were passivelyextended, such as by pushing back on the distalend of the thigh with the hand. Walking wouldbe almost impossible, except by taking very smallsteps and by pushing back on the knee with eachstep, or by bracing the knee in an extendedposition.

9. Speedy has ruptured the calcaneal tendon, andthe gastrocnemius and soleus muscles haveretracted, thereby causing the abnormal bulgingof the calf muscles. Because the major plantarflexors are no longer connected to the calcaneus,the runner cannot plantar flex the foot, and thefoot is abnormally dorsiflexed because theantagonists have been disconnected.

Chapter 11

1. A reduced intracellular concentration of K�

causes depolarization of the resting membranepotential. Because the intracellular concentration

A-12 Appendix G



of K� is reduced, the concentration gradient forpotassium from the inside to the outside of theplasma membrane is also reduced. Thus the rateat which K� diffuse out of the cell is reduced,and a smaller charge difference across the plasmamembrane is required to oppose the diffusion ofthe K� out of the cell. Therefore, the potentialdifference across the plasma membrane isreduced, and the cell is depolarized.

2. Because the plasma membrane is much lesspermeable to Na� than to K�, changes in theextracellular concentration of Na� effect theresting membrane potential less than do changesin the extracellular concentration of K�.Therefore, increases in extracellular Na� have aminimal effect on the resting membranepotential. Because the membrane is much morepermeable to Na� during the action potential,the elevated concentration of Na� in theextracellular fluid results in Na� diffusing intothe cell at a more rapid rate during the actionpotential, resulting in a greater degree ofdepolarization during the depolarization phaseof the action potential.

3. Because lithium ions reduce the permeability ofplasma membranes to Na�, the Na� channels inthe plasma membrane tend to remain closed. Anormal stimulus causes Na� channels to open,allowing Na� to diffuse into the cell, therebyresulting in depolarization. The cell is lesssensitive to stimuli because the membrane is lesspermeable to Na�.

4. Smooth muscle cells contract spontaneously inresponse to spontaneous depolarizations thatproduce action potentials. One way actionpotentials can be produced spontaneously is ifmembrane permeability to Na� spontaneouslyincreases. As a result, a few Na� enter thesmooth muscle cells and cause a slightdepolarization of the plasma membrane. Thesmall depolarization can cause voltage-gatedNa� channels to open, which results in furtherdepolarization, thereby stimulating additionalvoltage-gated ion channels to open. Thispositive-feedback cycle can continue until theplasma membrane is depolarized to its thresholdlevel and an action potential is produced.

5. Action potential conduction along a myelinatednerve fiber is more energy efficient because theaction potential is propagated by saltatoryconduction, which produces action potentials atthe nodes of Ranvier. Compared to anunmyelinated nerve fiber, only a small portion ofthe myelinated neuron’s membrane has actionpotentials. Thus there is less flow of sodium intothe neuron (depolarization) and less flow ofpotassium out of the neuron (repolarization).Consequently, the sodium–potassium exchangepump has to move fewer ions in order to restoreion concentrations. Because thesodium–potassium exchange pump requiresATP, myelinated axons use less ATP thanunmyelinated axons.

6. The inhibitory neuromodulator causes thepostsynaptic neuron to become less sensitive toexcitatory stimuli, probably by causinghyperpolarization of the postsynaptic neuron. Asa result, the excitatory neurotransmitter released

from the excitatory neuron is less likely toproduce postsynaptic action potentials.

7. With aging, there’s a decrease in the amount ofmyelin surrounding axons, which decreases thespeed of action potential propagation. Atsynapses there’s also an increase in the time ittakes for action potentials in the presynapticterminal to cause the production of actionpotentials in the postsynaptic membrane. It’sbelieved this results from a reduced release ofneurotransmitter by the presynaptic terminaland a reduced number of receptors in thepostsynaptic membrane.

8. Organophosphates inhibit acetylcholinesterase,thereby causing an increase in acetylholine in thesynaptic cleft leading to overproduction ofaction potentials, tetany of muscles, and possibledeath resulting from respiratory failure (seechapter 11). Curare is the best antidote because itblocks the effect of acetylcholine and acts tocounteract the organophosphate. Too muchcurare, however, could cause flaccid paralysis ofthe respiratory muscles. Injecting acetylcholinewould make the effect of the organophosphateworse. Potassium chloride causes depolarizationof muscle cell membranes, thereby making themmore sensitive to acetylcholine.

9. If the motor neurons supplying skeletal muscleare innervated by both excitatory and inhibitoryneurons, then blocking the activity of theinhibitory neurons with strychnine results inoverstimulation of the motor neurons by theexcitatory neurons.

Chapter 12

1. If the neuron with its cell body in the cerebrumis an inhibitory neuron and if it also synapseswith the motor neuron of a reflex arc, thenstimulation of the cerebral neuron could inhibitthe reflex.

2. The phrenic nerve is cut in the thorax, and thesurgery is performed while the lung is beingremoved.

3. The ulnar nerve supplies the medial third of thehand, little finger, and medial half of the ringfinger. The median nerve supplies the lateraltwo-thirds of the palm and thumb, and thesurface of the index, middle, and lateral half ofthe ring finger. The radial nerve supplies thelateral two-thirds of the dorsum of the hand.

4. Pulling on the upper limb when it is raised overthe head can damage the lower brachial plexus,in this case, the origin of the ulnar nerve. Theulnar nerve innervates muscles thatabduct/adduct the fingers and flex the wrist.

5. The ischiadic nerve has rootlets from L4–S3.Depending on the rootlet compressed, pain canbe felt in different locations.

6. a. Obturator nerveb. Femoral nervec. Ischiadic (tibial) nerved. Obturator nervee. Obturator nerve, some from femoral nerve

Chapter 13

1. A condition in which a patient looses all sense offeeling in the left side of the back, below theupper limb, and extending in a band around tothe chest, also below the upper limb, but whereall sensation on the right is normal, suggests thatthe patient’s dorsal roots have been damaged onthe left side adjacent to the part of the spinalcord supplying that part of the body. (The basisof this condition is explained more fully inchapter 14.)

2. The skull restricts the growth of the brain. Thesurface area of the cerebral cortex increases asmore neurons migrate into the cortex and asmore synapses are formed. As the cerebral cortexincreases in area, it becomes folded. This foldingallows a greater surface area to be housed in amuch smaller volume.

3. If CSF does not drain properly, the fluidaccumulates and exerts pressure on the brain(hydrocephalus). In the developing fetus, theventricles enlarge because of the excess fluidpressure. The head also enlarges because theskull bones have not fused. The expansion of thehead is not sufficient, however, to relieve all thepressure exerted on the developing brain by theexpanding ventricles. As a result, the cerebralcortex becomes proportionately thinner as it’scompressed between the ventricles and skull. Inmany cases, less gyri form in the cerebral cortex.Brain damage may or may not result, dependingon the amount of excess CSF, the ventricularpressure generated, and the areas of the braindamaged by the pressure.

4. Enlargement of the lateral and third ventricles,without enlargement of the fourth ventricle,suggests a blockage between the third and fourthventricles in the cerebral aqueduct. This defect iscalled aqueductal stenosis and is a commoncongenital problem.

5. Blood in the CSF taken through a spinal tapindicates the presence of blood in thesubarachnoid space and suggests that thepatient has a damaged blood vessel in thesubarachnoid space.

Chapter 14

1. The first sensations that occur when a womanpicks up an apple and bites into it are visual(special), tactile (general), and proprioceptive(general). The woman holds the apple in herhand and looks at it. The tactile sensations frommechanoreceptors in the hand tell her that theapple is firm and smooth. The proprioceptivesensations originating in the joints of the handtell the woman the size and shape of the apple.Visual input also tells her the size of the apple,and that it has a smooth surface, as well as itscolor. As the woman bites into the apple andbegins to chew, proprioceptive sensations fromthe teeth and jaw provide information as to howwidely the jaws must be opened to accommodatethe bite and how hard to bite down. Tactilesensations originating in the tongue and cheekstell the person the location of the bite of appleand its texture as it is moved about in the mouth.Taste sensations (special, chemoreceptor) from

A-13Appendix G



the tongue provide information that the applehas characteristics of being both sweet and sour.Olfactory sensations (special) provide morespecific information that the “fruity taste” is thatof an apple.

2. a. The most likely explanation is that theolfactory neurons accommodate and nolonger respond to odor stimulus.

b. The fact that one can hear the sound whenone tries indicates that the hair cells in thespiral organ have not accommodated andare still able to detect the sound stimulus.Many action potentials arriving in the brainare prevented from causing consciousperception, until we consciously “payattention” to the stimulus. For example, youmay not be paying attention to generalconversations in a crowded room or hall,until someone says your name. The sound ofyour name leaps out of the surroundingbabble, and you are suddenly interested inwhat was being said by the person whospoke your name.

3. The fibers of the dorsal-column/medial-lemniscalsystem carry two-point discrimination andproprioceptive information. Primary neuronsfrom the right side of the body ascend the spinalcord in the dorsal column and synapse withsecondary neurons in the medulla oblongata. Thesecondary neurons cross over in the uppermedulla and ascend through the left side of thepons to the thalamus. A patient suffering from aloss of two-point discrimination andproprioception on the right side of the body as aresult of a lesion in the medial lemniscal system inthe pons has a lesion in the left side of the pons.

4. The fibers of the lateral spinothalamic tract carryimpulses for pain and temperature. A lesion inthe area where these fibers decussate results inthe bilateral loss of pain and temperaturesensations only at the level of the lesion, andthere is no loss of sensation below the lesion.This occurs because fibers decussating above orbelow the lesion, as well as tracts that pass lateralto the lesion, are unaffected.

5. The damaged tracts are the lateral corticospinaltract, controlling motor functions on the rightside of the body, and the lateral spinothalamictract for pain and temperature sensations fromthe left side of the body. Damage to these tracts inthe right side of the spinal cord produces theobserved symptoms, because, in the cord, thelateral spinothalamic tract crosses over at thelevel of entry, and is, therefore, located on theopposite side of the cord from its peripheralnerve endings, whereas the corticospinal tract lieson the same side of the cord as its target muscles.

6. Complete unilateral transection of the right sideof the spinal cord results in loss of motorfunction (lateral corticospinal tract),proprioception, and two-point discrimination(dorsal-column/medial-lemniscal system) on thesame side of the body as the lesion, below thelevel of the lesion. Pain and temperaturesensations (lateral spinothalamic tract) are loston the opposite side of the body from below thelevel of the lesion. These symptoms describe theBrown-Séquard syndrome. Light touch is not

greatly affected on either side because of thelarge number of collateral branches in theanterior spinothalamic tract.

7. The right cerebral cortex controls the left sideof the body. The motor cortex has atopographic representation of the opposite sideof the body, with the hand, forearm, arm, andshoulder located approximately in the center ofthe precentral gyrus. The lesion is therefore inthe center of the right precentral gyrus of thecerebrum. Some grosser control of the left-upper limb may still exist because of theindirect pathways, but there would be spasticparalysis.

8. Damage to the cerebellum can result indecreased muscle tone, balance impairment, atendency to overshoot when reaching for ortouching something, and an intention tremor.These symptoms are opposite to those seen withbasal ganglia dysfunction. Cerebellar dysfunctionexhibits very similar symptoms to those seen inan inebriated person, and the same tests could beapplied, such as having the person touch theirnose or walk a straight line.

9. Memory storage for the 10 minutes prior to theaccident was in short-term memory and wasdisrupted before it could be transferred to long-term memory.Additional information: Anytime a person suffersa concussion there’s a possibility that he or shemay later develop postconcussion syndrome.Symptoms include muscle tension or migraineheadaches, reduced alcohol tolerance, difficultylearning new things, reduction in creativity andmotivation, fatigue, and personality changes, andthe syndrome may last a month to a year.Postconcussion syndrome may be the result of aslowly occurring subdural hematoma, which maybe missed by an early examination.

Chapter 15

1. The first sensations that occur when a personpicks up an apple and bites into it are visual. Theperson holds the apple in his hand and looks atit. Visual input (which stimulates light receptors)tells him the size of the apple, and that it has asmooth surface, as well as its color. As the personbites into the apple and begins to chew, tastesensations (chemoreceptors) from the tongueprovide information that the apple hascharacteristics of being both sweet and sour.Olfactory sensations (hemoreceptors) providemore specific information that the “fruity taste”is that of an apple.

2. The lens of the eye is biconvex and causes lightrays to converge. If the lens is removed, then thereplacement lens should also cause light rays toconverge. A biconvex lens or a lens with a singleconvex surface would work. Bifocals or trifocalscould also be recommended because of the lossof accommodation.

3. The light reflected by the tapetum lucidum couldstimulate photoreceptors and increase thesensitivity of the eye to light, which could be anadvantage when light levels are low. Because thesame light image could stimulate differentphotoreceptors, however, there is a loss of visualacuity and a blurring of vision.

4. Carrots contain vitamin A (retinoic acid), whichcan be used to form retinal. Retinal and opsincombine to form rhodopsin, which is found inrods. Rhodopsin is necessary for rods to respondto low levels of light. Lack of vitamin A canresult in lack of rhodopsin and night blindness.

5. By looking a few inches to the side, the image ofthe needle and thread is projected to theperiphery of the retina, where there is the highestconcentration of rods. The rods function betterthan cones in low-light intensities. If Jean looksdirectly at the needle and thread, their image fallson the macula, which has few rods and mostlycones, which don’t function well in dim light. Bylooking to the side, however, she is using a part ofthe retina where the photoreceptor cells are not asdensely packed as in the macula, and the image isfuzzy rather than sharp.

6. This phenomenon is called a negativeafterimage. While staring at the clock, thedarkest portion of the image (the black clock)causes dark adaptation in part of the retina.That is, part of the retina becomes moresensitive to light. At the same time the lightestpart of the image (the white wall) causes lightadaptation in the rest of the retina, and that partof the retina becomes less sensitive to light.When the man looks at a black wall, the darkadapted portion of the retina, which is moresensitive to light, produces more actionpotentials than does the light adapted part ofthe retina. Consequently, he perceives a lightclock against a darker background.

7. A lesion of the optic chiasma results in visualloss in both the right and left temporal fields, acondition called bitemporal hemianopsia, ortunnel vision. Tunnel vision can cause problemsfor normal functions, such as when driving a car,because the peripheral vision is severely limited.The occurrence of this condition can also suggesta much more serious problem, such as apituitary tumor, which sits just posterior to theoptic chiasma.

8. The most likely area damaged is the spiral organ,where waves result in the production of actionpotentials. The action is much like ocean wavesbreaking on the shore during a violent storm ascompared to those breaking in from a calmocean. Specifically, damage likely occurs in thepart of the spiral organ near the oval window,because it is this part of the basilar membranethat vibrates the most in response to high-frequency sounds.

9. Normally, as pressure changes, the auditorytubes open to allow an equalization of pressurebetween the middle ear and the externalenvironment. If this doesn’t occur, then thebuildup of pressure in the middle ear canrupture the tympanic membrane, or the pressurecan be transmitted to the inner ear and causesensoneural damage.

10. Normally, airborne sounds cause the tympanicmembrane to vibrate, resulting in movement ofthe middle ear ossicles and the production ofwaves in the perilymph of the scala vestibuli.Vibration of the skull bones can also causevibration of the perilymph in the scalavestibuli.

A-14 Appendix G



Chapter 16

1. The sympathetic division of the ANS isresponsible for dilation of the pupil.Preganglionic fibers from the upper thoracicregion of the spinal cord pass through spinalnerves (T1 and T2), into the white ramicommunicantes, and into the sympathetic chainganglia. The preganglionic fibers ascend thesympathetic chain and synapse withpostganglionic neurons in the superior cervicalsympathetic chain ganglia. The axons of thepostganglionic neurons leave the sympatheticchain ganglia as small nerves that project to thepupil of the eye.

2. Reduced salivary and lacrimal gland secretionscould indicate damage to the facial nerves,which innervate the submandibular, sublingual,and lacrimal glands. The glossopharyngealnerves innervate the parotid glands but not thelacrimal glands.

3. Cutting the preganglionic fibers in the white ramiof T2–T3 is the best way to eliminate innervationof the blood vessels in the skin. Cutting the grayrami at levels T2–T3 is inappropriate because thepostganglionic fibers that innervate the handblood vessels exit from the first thoracic andinferior cervical sympathetic chain ganglia.Cutting the spinal nerves is inappropriate becauseit eliminates all sensory and motor functions tothe area supplied.

4. a. Pelvic nervesb. Gray ramic. Vagus nervesd. Cranial nervese. Pelvic nerves

5. The parasympathetic division innervates theheart through the vagus nerves. Thepostganglionic nerve fibers of the vagus nervesrelease acetylcholine, which reduces heart rate.Methacholine can bind to the same receptors asacetylcholine and reduce heart rate. Side effectsresult from stimulating other parasympatheticeffector organs. For example, stimulating thesalivary glands results in increased salivation.Dilation of the pupils and sweating are effectsexpected from sympathetic stimulation. Themuscles of respiration are not regulated by theANS, but they do respond to acetylcholinethrough somatic neurons. Methacholine wouldbe expected to make contractions of respiratorymuscles more likely.

6. One would expect mostly parasympathetic effects,because the effects of acetylcholine are enhanced:blurring of vision as a result of contraction ofciliary muscles, excess tear formation because ofoverstimulation of the lacrimal glands, frequentor involuntary urination because ofoverstimulation of the urinary bladder. Pallorresulting from vasoconstriction in the skin is asympathetic effect that would not be expectedbecause skin blood vessels respond tonorepinephrine. Muscle twitching or cramps ofskeletal muscles might occur because theynormally respond to acetylcholine.

7. Epinephrine causes vasoconstriction andconfines the drug to the site of administration.This increases the duration of action of the drug

locally and decreases systemic effects.Vasoconstriction also reduces bleeding if a dryfield (an area clear of blood on its surface) isrequired.

8. Because normal action potentials are produced,the drug doesn’t act at the synapse between thepreganglionic and postganglionic neurons.Because injected norepinephrine works,sympathetic receptors in the heart arefunctioning and are not affected by the drug.Therefore, the drug must somehow affect thepostganglionic neurons. Possibly it inhibitsneurotransmitter production or release from thepostganglionic neurons.

9. Because cutting the white rami of T1–T4 doesn’taffect the action of the drug, sympatheticpreganglionic neurons in the spinal cord andsympathetic centers in the brain can be ruled outas a site of action. Because cutting the vagusnerves eliminates the effect of the drug, the drugcannot be acting at the synapse between thepreganglionic neurons and the postganglionicneurons, or between the synapse of thepostganglionic neuron and the effector organ ofeither division of the ANS. The drug must,therefore, excite parasympathetic centers in thebrainstem, resulting in a decrease in heart rate.

10. a. Responses in a person who is extremelyangry are primarily controlled by thesympathetic division of the ANS. Theseresponses include increased heart rate andblood pressure, decreased blood flow to theinternal organs, increased blood flow toskeletal muscles, decreased contractions ofthe intestinal smooth muscle, flushed skin inthe face and neck region, and dilation of thepupils of the eyes.

b. For a person who has just finished eatingand is now relaxing, the parasympatheticreflexes are more important thansympathetic reflexes. The blood pressure andheart rate are at normal resting levels, theblood flow to the internal organs is greater,contractions of smooth muscle in theintestines are greater, and secretions thatachieve digestion are more active. If theurinary bladder or the colon becomesdistended, autonomic reflexes that result inurination or defecation can result. Bloodflow to the skeletal muscles is reduced.

Chapter 17

1. Liver and kidney disease increases theconcentration of this hormone in the blood, andthe concentration would remain high for alonger time. The liver modifies the hormone tocause it to be excreted by the kidney morerapidly. In the case of liver disease, the hormoneis not modified and excreted rapidly. Therefore,the concentration becomes higher than normaland the concentration of the hormone remainshigh for longer than normal. A similar result isseen if the kidney is diseased and the hormonecannot be excreted rapidly.

2. Secretion of hormones is usually controlled by anegative-feedback mechanism. If a hormonecontrols the concentration of a substance in thecirculatory system, the hormone is secreted in

smaller amounts if the substance increases in thecirculatory system. If a tumor begins to secretethe substance in large amounts, the presence ofthe substance has a negative-feedback effect onthe secretion of the hormone and theconcentration of the hormone in the circulatorysystem is very low.

3. Usually intracellular mediator mechanismsrespond quickly, and the effect of the hormone isbrief. Intracellular receptor mechanisms usuallytake a long time (several hours) to respond, andtheir effects last much longer. If the hormone islarge and water-soluble, it’s probably functioningthrough an intracellular mediator mechanism, orif the hormone is lipid-soluble, it’s probably anintracellular receptor mechanism. If you have theability to monitor the concentration of asuspected intracellular mediator and it increasesin response to the hormone, or if you can inhibitthe synthesis of an intracellular mediator and itprevents the target cells’ response to the hormone,it’s an intracellular mediator mechanism. If youcan inhibit the synthesis of mRNA and thisinhibits the action of the hormone, or if you canmeasure an increase in mRNA synthesis inresponse to the hormone, then the mechanism isan intracellular receptor mechanism.

4. When the hormone binds to its receptor, the �subunit of the G protein is released. GTP mustbind to the � subunit, however, before it canhave its normal effect. If the � subunit cannotbind GTP, the hormone. therefore, has no effecton the target tissue.

5. Inhibitors of prostaglandin synthesis reduceprostaglandin synthesis in all tissues, not just inthose tissues in which prostaglandins produceundesirable effects. Symptoms such asinflammation, vomiting, and fever are reduced.Because prostaglandins also play a role inproducing beneficial effects in some tissues,however, these benefits would not occur normally.Inhibitors of prostaglandin synthesis may causelabor to be delayed or produce other undesirableresponses due to their inhibitory effects on thesynthesis of prostaglandins.

6. Phosphodiesterase causes the conversion ofcAMP to AMP, thus reducing the concentrationof cAMP. A drug that inhibits phosphodiesterase,therefore, increases the amount of cAMP in cellswhere cAMP is produced. Therefore, an inhibitorof phosphodiesterase increases the response of atissue to a hormone that has cAMP as anintracellular mediator.

7. A short half-life for epinephrine allowsepinephrine to produce a short-lived response.The response to a potentially harmful ordangerous situation is terminated shortly afterthe harmful or dangerous situation passes. Ifepinephrine had a long half-life, the heart rateand blood glucose would be elevated for a longtime, even if the harmful or dangerous situationwas very brief.

8. Because thyroid hormones are important inregulating the basal metabolic rate, a long half-life is an advantage. Thyroid hormones aresecreted and have a prolonged effect withoutlarge fluctuations in the basal metabolic rate. Ifthyroid hormones had a short half-life, the basal

A-15Appendix G



metabolic rate might fluctuate with changes inthe rate of secretion of thyroid hormones.Certainly the rate of secretion of thyroidhormones would have to be controlled withinnarrow limits if it did have a short half-life.

9. If liver disease results in a decrease of plasmaproteins to which thyroid hormones bind,higher-than-normal concentrations of free(unbound) thyroid hormones occur in thecirculatory system. Because of the higher-than-normal concentration of thyroid hormones thatare unbound, the responses to thyroid hormonesincrease. In addition, the half-life of the thyroidhormones is shortened. Thus, as thyroidhormone secretion increases, the concentrationof thyroid hormone also increases. As thethyroid hormone secretion decreases, theconcentration of thyroid hormone alsodecreases. Thyroid hormones fluctuate inconcentration in the circulatory system morethan normal.

10. Elevated GnRH levels in the blood as a result ofthe GnRH-secreting tumor causes down-regulation of GnRH receptors in the anteriorpituitary. This decreases the ability of GnRH tostimulate the anterior pituitary, and the rate ofluteinizing hormone and follicle-stimulatinghormone secretion by the anterior pituitarydecreases and remains decreased as long as theGnRH levels are chronically elevated. Therefore,the functions of the reproductive systemcontrolled by luteinizing hormone and follicle-stimulating hormone decrease.

11. Insulin levels normally change in order tomaintain normal blood sugar levels, despiteperiodic fluctuations in sugar intake. A constantsupply of insulin from a skin patch might result ininsulin levels that are too low when blood sugarlevels are high (after a meal) and might be toohigh when blood sugar levels are low (betweenmeals). In addition, insulin is a protein hormonethat would not readily diffuse through the lipidbarrier of the skin (see Chapter 5). Estrogen is alipid soluble steroid hormone.

Chapter 18

1. The hypothalamohypophyseal portal systemallows neurohormones that function as releasingand inhibiting hormones, which are secreted byneurons in the hypothalamus, to be carrieddirectly from the hypothalamus to the anteriorpituitary gland. Consequently, the releasing andinhibiting hormones are not diluted nor are theydestroyed by the enzymes, which are abundant inthe kidneys, liver, lungs, and general circulation,before they reach the anterior pituitary. Also thetime it takes for releasing and inhibiting hormonesto reach the anterior pituitary is less than if theywere secreted into the general circulation.

2. A hot environment increases ADH secretion.Because the amount of water lost in the form ofsweat can be quite large, and because sweat ismore dilute than the body fluids, sweatinggradually increases the osmolality of the bodyfluids. The increasing osmolality of body fluidsstimulates an increase in ADH secretion. Thus, ahot environment can result in increased ADHsecretion because of an increasing osmolality of

the body fluids. Increased ADH secretion in a hotenvironment reduces the amount of water lost inthe form of urine. Therefore, water is conserved.

3. Polydipsia and polyuria are consistent with eitherdiabetes mellitus or diabetes insipidus. Diabetesmellitus, however, is consistent with an increasedurine osmolality because of the large amount ofglucose lost in the urine. Diabetes insipidus isconsistent with urine with a low specific gravitybecause little water is reabsorbed by the kidney.Thus urine has an osmolality close to that of thebody fluids, and the rapid loss of dilute urineresults in a decrease in blood pressure. Thuspolyuria with a low specific gravity is notconsistent with diabetes mellitus but is consistentwith diabetes insipidus. Administration of ADHreverses the symptoms of diabetes insipidus.Neither polydipsia nor polyuria results from alack of glucagon or aldosterone.

4. The symptoms are consistent with acromegaly,which is a consequence of elevated GH secretionafter the epiphyses have closed. Increased GHcauses enlarged finger bones, growth of bonyridges over the eyes, and increased growth of thejaw. The anterior pituitary tumor increasespressure at the base of the brain near the opticnerves as it enlarges. The pituitary rests in thesells turcica of the sphenoid bone; as it enlargespressure increases because the pituitary is nearlysurrounded by rigid bone and the brain is locatedjust superior to the pituitary. As the anteriorpituitary enlarges because of a tumor, it pushessuperiorly and pressure is applied to the ventralportion of the brain. The GH also causes bonedeposition on the inner surface of skull bones,which also increases the pressure inside the skull.

5. If hyperthyroidism results from a pituitaryabnormality, laboratory tests should showelevated TSH levels in the circulatory system inaddition to elevated T3 and T4 levels. Ifhyperthyroidism results from the production ofa nonpituitary thyroid-stimulating substance,laboratory tests should also show elevated T3 andT4 levels, but TSH levels would be low because ofthe negative-feedback effects of T3 and T4 on thehypothalamus and pituitary gland.

6. The second student is correct. Low levels ofvitamin D reduce calcium uptake in thegastrointestinal tract, which results in adecreased blood level of calcium ions. As bloodcalcium levels decrease, the rate of PTHsecretion would increase. Parathyroid hormoneincreases bone breakdown, which maintainsblood calcium levels, even if vitamin Ddeficiency exists for a prolonged time.Osteomalacia results because of the increasedbone reabsorption necessary to maintain normalblood calcium levels.

7. A glucose tolerance test can distinguish betweenthese conditions. The person would consumeglucose after a period of fasting. Over the nextfew hours the blood glucose levels in a healthyperson increase and then return to fasting levels.The blood glucose levels always remain withinthe normal range, however. In a person withdiabetes, the blood glucose levels increase toabove-normal levels and remain elevated forseveral hours. In a person who secretes large

amounts of insulin, blood glucose levels wouldincrease, and then they would decrease to below-normal levels within a relatively short time.

8. Because the person is a diabetic and probably istaking insulin, the condition is more likely to beinsulin shock than a diabetic coma. To confirmthe condition, however, a blood sample should betaken. If the condition is due to a diabetic coma,then the blood glucose levels will be elevated. Ifthe condition is due to insulin shock, the bloodglucose levels will be below normal. In the case ofinsulin shock, glucose can be administeredintravenously. In the case of diabetic coma,insulin should be administered. An isotonicsolution containing insulin can be administeredto reduce the osmolality of the extracellular fluid.

9. Adrenal diabetes results from elevated anduncontrolled secretion of glucocorticoidhormones, such as cortisol, from the adrenalgland. Because glucocorticoid hormones increaseblood glucose levels, elevated secretion of thesehormones results in elevated blood glucose levelsand symptoms similar to diabetes mellitus.Pituitary diabetes results from elevated secretionof GH from the anterior pituitary. Elevated GHcauses an increase in blood glucose levels and,therefore, produces symptoms similar to diabetesmellitus. Prolonged elevation of bothglucocorticoids and growth hormone secretioncan lead to the development of diabetes mellitusif the insulin-secreting cells of the pancreaticislets degenerate because of the prolonged needto secrete insulin in response to the elevatedblood glucose levels.

10. Elevated epinephrine from the adrenal medullapromotes elevated blood pressure and increasesthe work load on the heart, increases the rate ofmetabolism, and results in increased sweatingand nervousness. The risk of heart attack andstroke are increased. Elevated cortisol causeshyperglycemia and can lead to diabetes mellitus,a depressed immune system with increasedsusceptibility to infections, and destruction ofproteins leading to tissue wasting.

Chapter 19

1. Because of the rapid destruction of the red bloodcells we would expect erythropoiesis to increasein an attempt to replace the lost red blood cells.The reticulocyte count would therefore be abovenormal. Jaundice is a symptom of hereditaryhemolytic anemia because the destroyed redblood cells release hemoglobin, which isconverted into bilirubin. Removal of the spleencures the disease because the spleen is the majorsite of red blood cell destruction.

2. Blood doping increases the number of red bloodcells in the blood, thereby increasing its oxygen-carrying capacity. The increased number of redblood cells also makes it more difficult for theblood to flow through the blood vessels,increasing the heart’s workload.

3. Symptoms resulting from decreased red bloodcells are associated with a decreased ability of theblood to carry oxygen: shortness of breath,weakness, fatigue, and pallor. Symptomsresulting from decreased platelets are associatedwith a decreased ability to form platelet plugs

A-16 Appendix G



and clots: small areas of hemorrhage in the skin(petechiae), bruises, and decreased ability to stopbleeding. Symptoms resulting from decreasedwhite blood cells could include an increasedsusceptibility to infections.

4. Hypoventilation results in decreased bloodoxygen levels, which stimulates erythropoiesis.Therefore, the number of red blood cellsincreases and produces secondary polycythemia.

5. Removal of the stomach removes intrinsic factor,which is necessary for vitamin B12 absorption.Therefore, the patient develops perniciousanemia. Lack of stomach acid can decrease ironabsorption in the small intestine and result iniron-deficiency anemia.

6. Vitamin B12 and folic acid are necessary forblood cell division. Lack of these vitamins resultsin pernicious anemia. Iron is necessary for theproduction of hemoglobin. Lack of iron resultsin iron-deficiency anemia. Vitamin K isnecessary for the production of many bloodclotting factors. Lack of vitamin K can greatlyincrease blood clotting time, resulting inexcessive bleeding.

Chapter 20

1. The walls of the ventricles are thicker than thewalls of the atria because the ventricles mustproduce a greater pressure to pump blood into thearteries. Only a small pressure is required to pumpblood from the atria into the ventricles duringdiastole. The wall of the left ventricle is thickerthan the wall of the right ventricle because the leftventricle produces a much greater pressure toforce blood through the aorta than the rightventricle produces to move blood through thepulmonary trunk and pulmonary arteries.

2. During systole, the cardiac muscle in the rightand left ventricles contracts, which compressesthe coronary arteries. During diastole, thecardiac muscle of the ventricles relaxes andblood flow through the coronary arteriesincreases. The diastolic pressure is sufficient tocause blood to flow through coronary arteriesduring diastole.

3. A drug that prolongs the plateau of cardiacmuscle cell action potentials prolongs the timeeach action potential exists and increases therefractory period. Therefore, the drug wouldslow the heart. A drug that shortens the plateaushortens the length of time each action potentialexists and shortens the refractory period.Therefore, the drug could allow the heart rate toincrease further.

4. Endurance-trained athletes have decreased heartrates because their cardiac muscle undergoeshypertrophy in response to exercise. Thehypertrophied cardiac muscle causes the strokevolume to increase substantially. The increasedstroke volume is sufficient to maintain anadequate cardiac output and blood pressure eventhough the heart rate is slower.

5. The two heartbeats occurring close together canbe heard through the stethoscope, because theheart valves open and close normally duringeach of the heartbeats even if they are closetogether. The second heartbeat, however,produces a greatly reduced stroke volume

because there’s not enough time for theventricles to fill with blood between the first andsecond contraction of the heart. Thus, thepreload is reduced. Because the preload isreduced, the second heartbeat has a greatlyreduced stroke volume. The reduced strokevolume fails to produce a normal pulse. Thepulse deficit, therefore, results from the reducedstroke volume of the second of the two beats thatare very close together.

6. Aerobic training causes hypertrophy of thecardiac muscle in the heart and causes the heartto produce a greater stroke volume as aconsequence. The heart rate can decrease whilethe cardiac output remains the same becausecardiac output is equal to the stroke volumetimes the heart rate. If the stroke volumeincreases, the heart rate can decrease and thecardiac output can remain the same.

7. Atrial contractions complete ventricular filling,but atrial contractions are not primarilyresponsible for ventricular filling. Therefore, if theatria are fibrillating, blood can still flow into theventricles and ventricular contractions can occur.As long as the ventricles contract rhythmically theheart can pump an adequate amount of bloodeven though the atria are fibrillating. If theventricles undergo fibrillation, however, theycannot fill with blood and cannot function aspumps. Thus the stroke volume will become toolow to maintain adequate blood flow to tissues.

8. The results depend on Cee Saw's response to theconditions of the laboratory. First, as Cee Saw’shead is lowered, gravity causes blood pressure inthe carotid sinuses and aortic arch to increase.The increased blood pressure stimulatesbaroreceptors, which detect the increased bloodpressure and send action potentials indicatingthat blood pressure increased to thecardioregulatory center in the medulla oblongataalong sensory nerve fibers. The cardioregulatorycenter increases parasympathetic stimulationand reduces sympathetic stimulation of theheart. Thus the heart rate decreases. Second, if, asher head is lowered, Cee Saw becomes excited,the sympathetic division of the ANS becomesmore active. The resulting increase insympathetic stimulation of the heart causes theheart rate to increase.

9. After Cee Saw is tilted so that her head is higherthan her feet for a few minutes, the regulatorymechanisms that control blood pressure adjust sothat the heart pumps sufficient blood to supplythe needs of her tissues. If she is then tilted sothat her head is higher than her feet, gravitywould cause blood to flow toward her feet, andthe blood pressure in the carotid sinus and aorticarch would decrease. The decrease in bloodpressure would be detected by the baroreceptorsin these vessels and would activate baroreceptorreflexes. The result is increased sympathetic anddecreased parasympathetic stimulation of theheart and an increase in the heart rate. Theincreased heart rate would function to increasethe blood pressure to its normal value.

10. An ECG measures the electrical activity of theheart and does not indicate a slight heartmurmur. Heart murmurs are detected by

listening to the heart sounds. The boy may havea heart murmur, but the mother does notunderstand the basis for making such adiagnosis.

11. When both common carotid arteries areclamped, the blood presure within the internalcarotid arteries drops dramatically. Thedecreased blood pressure is detected, and thebaroreceptor reflex increases heart rate andstroke volume. The resulting increase incardiac output causes the increase in bloodpressure.

12. Venous return declines markedly inhemorrhagic shock because of the loss of bloodvolume. With decreased venous return, strokevolume decreases (Starling’s law of the heart).The decreased stroke volume results in adecreased cardiac output, which produces adecreased blood pressure. In response to thedecreased blood pressure, the baroreceptorreflex causes an increase in heart rate in anattempt to restore normal blood pressure.However, with inadequate venous return theincreased heart rate is not able to restore normalblood pressure.

Chapter 21

1. a. Aorta, left coronary artery, circumflex artery,posterior interventricular artery; or aorta,right coronary artery, posteriorinterventricular artery

b. Aorta, brachiocephalic artery, right commoncarotid artery, right internal carotid artery;or aorta, left common carotid artery, leftinternal carotid artery

c. Aorta, brachiocephalic artery, rightsubclavian artery, right vertebral artery,basilar artery; or aorta, left subclavian artery,left vertebral artery, basilar artery

d. Aorta, left or right common carotid artery,left or right external carotid artery

e. Aorta, left subclavian artery, axillary artery,brachial artery, radial or ulnar artery, deepor superficial palmar arch, digital artery (onthe right: the brachiocephalic artery wouldbe included)

f. Aorta, common iliac artery, external iliacartery, femoral artery, popliteal artery,anterior tibial artery

g. Aorta, celiac artery, common hepatic arteryh. Aorta, superior mesenteric artery, intestinal

branchesi. Aorta, left or right internal iliac artery

2. a. Great cardiac vein, coronary sinus; oranterior cardiac vein

b. Transverse sinus, sigmoid sinus, internaljugular vein, brachiocephalic vein, superiorvena cava

c. Retromandibular vein, external jugular vein,subclavian vein, brachiocephalic vein,superior vena cava

d. Deep: vein of hand, radial or ulnar vein,brachial vein, axillary vein, subclavian vein,brachiocephalic vein, superior vena cavaSuperficial: vein of hand, radial or ulnarvein, cephalic or basilic vein, axillary vein,subclavian vein, brachiocephalic vein,superior vena cava

A-17Appendix G



e. Deep: vein of foot, dorsalis veins of foot,anterior tibial vein, popliteal vein, femoralvein, external iliac vein, common iliac vein,inferior vena cavaSuperficial: vein of foot, great saphenousvein, external iliac vein, common iliac vein,inferior vena cava; or vein of foot, smallsaphenous vein, popliteal vein, femoral vein,external iliac vein, common iliac vein,inferior vena cava

f. Gastric vein or gastroepiploic fein, hepaticportal vein, hepatic sinusoids, hepatic vein,inferior vena cava

g. Renal vein, inferior vena cavah. Hemiazygous vein or accessary hemiazygous

vein, azygous vein, superior vena cava3. A superficial vessel would be easiest, such as the

right cephalic or basilic vein. The catheter ispassed through the cephalic (or brachial) veinand the superior vena cava to the right atrium.Because the pulmonary veins are not readilyaccessible, dye would not normally be placeddirectly into them. Instead, the dye would beplaced in the right atrium using the procedurejust described. The dye passes from the rightatrium into the right ventricle, the pulmonaryarteries, the lungs, the pulmonary veins, and intothe left atrium. If the catheter has to be placed inthe left atrium, it could be inserted through anartery, such as the femoral artery, and passed viathe aorta to the left ventricle and then into theleft atrium.

4. The viscosity of the blood is affected primarily bythe hematocrit. As hematocrit increases, theviscosity of the blood increases logarithmically, sothat even a small increase in hematocrit results ina large increase in viscosity. Greater force istherefore not needed to cause blood to flowthrough the blood vessels. With the increasedblood volume, blood flow through vessels isadequate without an increase in viscosity.

5. The resistance to blood flow is less in the venacavae for two reasons: first, the diameter of onevena cava is greater than the diameter of theaorta and second, an increased diameter of ablood vessel reduces resistance to flow (seePoiseuille’s law). In addition, there are two venaecavae, the superior vena cava and the inferiorvena cava, but only one aorta. The blood flowthrough the aorta and the venae cavae is aboutequal, but the velocity of blood flow is muchhigher in the aorta than it is in the venae cavae.

6. According to Laplace’s law, as the diameter of ablood vessel increases, the force applied to thevessel wall increases, even if the pressure remainsconstant. The increased connective tissue foundin the walls of the large blood vessels thereforemakes the wall of those vessels stronger and morecapable of resisting the force applied to the wall.

7. Veins and lymphatic vessels have one-way valvesin them. Massage creates a cycle of increasingand decreasing pressure to the veins, whichrhythmically compresses them. The compressionof the veins forces fluid to move out of the limbthrough both veins and lymphatic vessels. Themovement of fluid through the veins lowers thepressure within the venous end of the capillary.Thus the forces that move fluid into the

capillaries at their venous ends are greater andthey move more interstitial fluid into thecapillaries. Compression of the lymphaticcapillaries also causes more lymphatic fluid tomove into the lymphatic vessels. Because there isless fluid in the limb, the edema decreases.

8. The nursing student’s diagnosis was incorrect.Blood pressure measurements are normallymade in either the right or left arm, both ofwhich are close to the level of the heart. Bloodpressure taken in the leg is influenced bypressure created by the pumping action of theheart, but the effect of gravity on the blood, as itflows into the leg, also influences the bloodpressure in a substantial way. In this case gravityincreases blood pressure from about 120 mm Hgfor the systolic pressure to 200 mm Hg.

9. Decreased liver function includes a decrease inthe synthesis of plasma proteins. Consequentlythe concentration of plasma proteins decreases,and the colloid osmotic pressure of the blooddecreases. Therefore, less water moves byosmosis into the capillaries at the venous endsand the result is edema.

10. Chemoreceptors in the medulla oblongata detectcarbon dioxide and the pH of the blood. Thenormal blood levels of CO2 and pH stimulatethese chemoreceptors, which in turn stimulatethe vasomotor center. The vasomotor centerkeeps blood vessels partially constricted underresting conditions. This basal level of activity iscalled the vasomotor tone. Blowing off CO2

reduces the blood levels of carbon dioxide andincreases the pH of the body fluids. Thesechanges reduce vasomotor tone and result invasodilation. If a person hyperventilates andblows off CO2, the stimulus to the vasomotorcenter decreases, which results in a decrease invasomotor tone. The decrease in vasomotor toneresults in a decrease in systemic blood pressure.If the blood pressure decreases enough, theblood flow to the brain decreases and can cause asensation of dizziness or can even cause a personto lose consciousness.

11. Epinephrine is secreted from the adrenalmedulla in response to stressful stimuli and theepinephrine stimulates responses that areconsistent with increased physical activity.Vasoconstriction of the blood vessels in the skinshunts blood away from the skin to skeletalmuscles. Vasodilation occurs in blood vessels ofexercising skeletal muscles. Blood flow throughthe exercising skeletal muscles thereforeincreases. Because epinephrine causesvasodilation of the blood vessels of cardiacmuscle, blood flow through the cardiac muscleincreases. This response is consistent with theincreased work performed by the heart underconditions of increased physical activity.

12. The hot Jacuzzi increases Skinny’s skin and bodytemperature. As a result, the blood vessels of theskin dilate. Because the blood vessels dilate,peripheral resistance decreases, causing theblood pressure to decrease. The baroreceptors ofthe carotid sinus and aortic arch detect thedecrease in blood pressure and send actionpotentials to the cardioregulatory center in themedulla oblongata. As a result, the sympathetic

stimulation to the heart increases and the heartrate, in response, increases also. The increasedheart rate elevates the blood pressure back towithin its normal range of values.

Chapter 22

1. Elevation of the limb reduces blood pressure inthe limb, resulting in less fluid movement fromthe blood into the tissues (see chapter 21). Thus,the edema is reduced as the lymphatic systemremoves fluid from the tissues faster that it entersthem. Massage moves lymph through thelymphatic vessels in the same fashion ascontraction of skeletal muscle. The applicationof pressure periodically to lymphatic vesselsforces lymph to flow toward the trunk of thebody, but valves prevent the flow of lymph in thereverse direction. The removal of lymph fromthe tissue helps to relieve edema.

2. Normally T cells are processed in the thymus andthen migrate to other lymphatic tissues. Withoutthe thymus this processing is prevented. Becausethere are normally five T cells for every one B cell,the number of lymphocytes is greatly reduced.The loss of T cells results in an increasedsusceptibility to infection and an inability toreject grafts because of the loss of cell-mediatedimmunity. In addition, since helper T cells areinvolved with activation of B cells, antibody-mediated immunity is also depressed.

3. That there is no immediate effect indicates thereis a reservoir of T cells in the lymphatic tissue. Asthe reservoir is depleted through time, thenumber of lymphocytes decreases and cell-mediated immunity is depressed, the animal ismore susceptible to infections, and the ability toreject grafts decreases. The ability to produceantibodies decreases because of the loss of helperT cells that are normally involved with theactivation of B cells.

4. Injection B results in the greatest amount ofantibody production. At first, the antigen causesa primary response. A few weeks later, the slowlyreleased antigen causes a secondary response,resulting in a greatly increased production ofantibodies. Injection A doesn’t cause a secondaryresponse because all of the antigen is eliminatedby the primary response.

5. If the patient has already been vaccinated, thebooster shot stimulates a memory (secondary)response and rapid production of antibodiesagainst the toxin. If the patient has never beenvaccinated, vaccinating now is not effectivebecause there’s not enough time for the patientto develop his or her own primary response.Therefore, antiserum is given to provideimmediate, but temporary, protection.Sometimes both are given: The antiserumprovides short-term protection and the tetanusvaccine stimulates the patient's immune systemto provide long-term protection. If the shots aregiven at the same location in the body, theantiserum (antibodies against the tetanus toxin)could cancel the effects of the tetanus vaccine(tetanus toxin altered to be nonharmful).

6. The infant’s antibody-mediated immunity is notfunctioning properly, whereas his cell-mediatedimmunity is working properly. This explains the

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susceptibility to extracellular bacterial infectionsand the resistance to intracellular viral infections.It took so long to become apparent because IgGfrom the mother crossed the placenta andprovided the infant with protection. The infantbegan to get sick after these antibodies degraded.

7. Bone marrow is the source of the lymphocytesresponsible for adaptive immunity. If successful,the transplanted bone marrow starts producinglymphocytes and the baby has a functioningimmune response. In this case, there’s a graftversus host rejection in which the lymphocytes inthe transplanted red marrow mount an immuneattack against the baby’s tissues, resulting in death.

8. At the first location an antibody-mediatedresponse results in an immediate hypersensitivityreaction, which produces inflammation. Mostlikely the response resulted from IgE antibodies.At the second location a cell-mediated responseresults in a delayed hypersensitivity reaction,which produces inflammation. This probablyinvolves the release of cytokines and the lysis ofcells. At the other locations there is neither anantibody-mediated nor a cell-mediated response.

9. The ointment is a good idea for the poison ivy,which causes a delayed hypersensitivity reaction,for example, too much inflammation. For thescrape it’s a bad idea, because a normal amountof inflammation is beneficial and helps to fightinfection in the scrape.

10. Because antibodies and cytokines both produceinflammation, the fact that the metal in thejewelry results in inflammation is not enoughinformation to answer the question. However, thefact that it took most of the day (many hours) todevelop the reaction indicates a delayedhypersensitivity reaction and therefore cytokines.

Chapter 23

1. Minute respiratory volume is equal to therespiration rate times the tidal volume. With arespiration rate of 12 breaths per minute and atidal volume of 500 mL per breath, normalminute ventilation is 6000 mL/min (12 � 500).Rapid (24 breaths per minute), shallow (250 mLper breath) breathing results in the same minuteventilation, that is, 6000 mL/min (24 � 250).Alveolar ventilation rate (VA) is the respiratoryrate (frequency; f) times the difference betweenthe tidal volume (VT) and dead space (VD).

VA � f(VT � VD)

Normal resting VA � 12 � (500 � 150) �

4200mL/minIn this case of rapid shallow breathing,

VA � 24 � (250 � 150) � 2400mL/min

Thus, even though the minute ventilation is thesame in both cases, the alveolar ventilation rate isless during rapid, shallow breathing becausethere’s less effective exchange of gases betweenthe atmosphere and the dead space. Becausethere’s less exchange of gases, the partialpressures of alveolar gases become closer to thepartial pressure of blood gases. Consequently,the alveolar partial pressure of O2 decreases andthe alveolar partial pressure of CO2 increases.This decreases the concentration gradients for

gases, resulting in less gas exchange betweenalveolar air and blood.

2. We expect vital capacity to be greatest whenstanding because the abdominal organs moveinferiorly, thereby allowing greater depressionof the diaphragm and a greater inspiratoryreserve volume.

3. The hose increases dead space and thereforedecreases alveolar ventilation. Ima Diver has tocompensate by increasing respiratory rate ortidal volume. If the hose is too long, she won’t beable to compensate. Furthermore, with a longhose, air is simply moved back and forth in thehose with little exchange of air between theatmosphere and the lungs taking place. Anotherconsideration is the effect of water pressure onthe thorax, which decreases compliance andincreases the work of ventilation. In fact, a fewfeet underwater there’s enough pressure on thethorax to prevent the intake of air through evena short hose connected to the atmosphere.

4. The increase in atmospheric pressure increases thepartial pressure of oxygen. According to Henry’slaw, as the partial pressure of oxygen increases, theamount of oxygen dissolved in the body fluidsincreases. The increase in dissolved oxygen isdetrimental to the gangrene bacteria. Becausehemoglobin is already saturated with oxygen, theHBO treatment doesn’t increase the ability ofhemoglobin to pick up oxygen in the lungs.

5. Compression causes a decrease in thoracicvolume and therefore lung volume.Consequently, pressure in the lungs increasesover atmospheric pressure and air moves out ofthe lungs. Raising the arms expands the thoraxand lungs. This results in a lower-than-atmospheric pressure in the lungs, and air movesinto the lungs.

6. The victim’s lungs expand because of thepressure generated by the rescuer’s muscles ofexpiration. This fills the lungs with air that has agreater pressure than atmospheric pressure. Airflows out of the victim’s lungs as a result of thispressure difference and because of the recoil ofthe thorax and lungs. Although the partialpressure of oxygen of the rescuer’s expired air isless than atmospheric, enough oxygen can beprovided to sustain the victim. The lower partialpressure of oxygen could also activate thechemoreceptor reflex and stimulate the victim tobreathe. In addition, the rescuer’s partialpressure of carbon dioxide is higher thanatmospheric and this could activate thechemosensitive area in the medulla.

7. All else being equal (i.e., the thickness of therespiratory membrane, the diffusion coefficientof the gas, and the surface area of the respiratorymembrane), diffusion is a function of the partialpressure difference of the gas across therespiratory membrane. The greater thedifference in partial pressure, the greater the rateof diffusion. The greatest rate of oxygendiffusion should therefore occur at the end ofinspiration when the partial pressure of oxygenin the alveoli is at its highest. The greatest rate ofcarbon dioxide diffusion should occur at the endof inspiration when the partial pressure ofcarbon dioxide in the alveoli is at its lowest.

8. Because the partial pressure of oxygen at highaltitudes decreases, a shift to the left isadvantageous. Such a shift enables hemoglobinto pick up more oxygen at a lower partialpressure of oxygen.

9. Cutting the phrenic nerves eliminates contractionof the diaphragm. Tidal volume decreasesdrastically, and death probably results. Cuttingthe intercostal nerves eliminates raising of theribs and sternum and decreases tidal volume,unless the diaphragm compensates. Cutting thevagus nerves eliminates the Hering-Breuer reflexand results in a greater-than-normal inspiration.This increases tidal volume.

10. While hyperventilating and making ready toleave your instructor behind, you might makethe following arguments:• Hyperventilation increases the oxygen

content of the air in the lungs; therefore, youwould have more oxygen to use when holdingyour breath.

• It’s hemoglobin that is saturated.Hyperventilation increases the amount ofoxygen dissolved in the blood plasma.

• Hyperventilation decreases the amount ofcarbon dioxide in the blood. This makes itpossible to hold one’s breath for a longer timebecause of a decreased urge to take a breath.

• Hyperventilation activates alveoli not in usebecause increasing alveolar oxygen anddecreasing alveolar carbon dioxide causeslung arterioles to relax, thereby increasingblood flow through the lungs.

Chapter 24

1. With the loss of the swallowing reflex, the vocalfolds no longer occlude the glottis.Consequently, vomit can enter the larynx andblock the respiratory tract.

2. Without adequate amounts of hydrochloric acid,the pH in the stomach is not low enough for theactivation of pepsin. This loss of pepsin functionresults in inadequate protein digestion. If the foodis well chewed, however, proteolytic enzymes inthe small intestine (e.g., trypsin, chymotrypsin)can still digest the protein. If the stomachsecretion of intrinsic factor decreases, theabsorption of vitamin B12 is hindered. Inadequateamounts of vitamin B12 can result in decreasedred blood cell production (pernicious anemia).

3. Even though ulcers are apparently ultimatelycaused by bacteria, overproduction ofhydrochloric acid due to stress is a possiblecontributing factor. Reducing hydrochloric acidproduction is recommended. In addition toantibiotic therapy, commonly recommendedsolutions include relaxation, drugs that reducestomach acid secretion, and antacids toneutralize the hydrochloric acid. Smaller mealsare also advised because distension of thestomach stimulates acid production. Proper dietis also important. The patient is also advised toavoid alcohol, caffeine, and large amounts ofprotein because they stimulate acid production.Ingestion of fatty acids is recommended becausethey inhibit acid production by causing releaseof gastric inhibitory polypeptide andcholecystokinin. Stress also stimulates the

A-19Appendix G



sympathetic nervous system, which inhibitsduodenal gland secretion. As a result, theduodenum has less of a mucous coating and ismore susceptible to gastric acid and enzymes.Relaxing after a meal helps decrease sympatheticactivities and increase parasympathetic activities.

4. Lack of bile due to blockage of the common bileduct can result in jaundice (due to anaccumulation of bile pigments in the blood) andclay-colored stools (due to lack of bile pigments inthe feces). Blockage of the bile duct causesabdominal pain, nausea, and vomiting. Fatabsorption is impaired because of the absence ofbile salts in the duodenum and a loose, bulky stoolwould result. Lack of fat absorption reduces theabsorption of fat-soluble vitamins such a vitaminK, resulting in lack of normal clotting function.

5. The patient would still be able to defecate.Following a meal the gastrocolic andduodenocolic reflexes could initiate massmovement of the feces into the rectum. In therectum, local reflexes and the defecation reflex(integrated in the sacral level of the cord and notrequiring connections to high brain centers)would cause defecation. Awareness of the need todefecate would be lost (due to loss of sensoryinput to the brain) and the ability to voluntarilyprevent defecation via the external anal sphincterwould also be lost.

6. The accumulation of materials above the site ofimpaction and the action of bacteria on thematerial would result in an increase in osmoticpressure in the area. Water would move byosmosis into the colon above the site ofimpaction. Bowel impaction is very dangerousand must be treated quickly. The increasedvolume and distention of the digestive tractabove the site of impaction causes compressionof the mucosa. This compression can occludeblood vessels in the mucosa and lead to necrosis.Necrosis of the mucosa results in increasedpermeability of the mucosa, thus allowing toxicorganisms and substances in the digestive tractto enter the circulation, resulting in septic shock.

Chapter 25

1. In figure 25.2, the Daily Value for saturated fat islisted as less than 20 g for a 2000 kcal/day diet.The % Daily Values appearing on food labels arebased on a 2000 kcal/day diet. Therefore, the %Daily Value for saturated fat for one serving ofthis food is 10% (2/20 � .10, or10%).

2. According to the Daily Value guidelines, total fatsshould be no more than 30% of total kilocaloricintake. For someone consuming 3000 kcal/daythis is 900 kcal (3000 kcal � 0.30). There are9 kcal in a gram of fat. Therefore, the maximumamount (weight) of fats the active teenage boyshould consume is 100 g (900/9).

3. The % Daily Value is the amount of the nutrient inone serving divided by its Daily Value. Thereforethe % Daily Value is 10% (10/100 � .10, or 10%).

4. The % Daily Value for one serving of the food is10% (see answer to question 3). Since there arefour servings in the package, if the teenager eatshalf of the food in the package, he consumes twoservings. Thus, he eats 20% (10% � 2) of therecommended maximum total fat.

5. The protein in meat contains all of the essentialamino acids and is a complete protein food.Although plants contain proteins, a variety ofdifferent plants must be consumed to ensure thatall the essential amino acids are included inadequate amounts. Also, plants contain lessprotein per unit weight than meat, so a largerquantity of plants must be consumed to get thesame amount of protein.

6. Copper is necessary for proper functioning ofthe electron-transport chain. Inadequate copperin the diet results in reduced ATP production,that is, not enough energy.

7. Fasting can be damaging because proteins areused to produce glucose. The glucose enters theblood and provides an energy source for thebrain. This breakdown of proteins can damagetissues such as muscle and disrupt chemicalreactions regulated by enzyme systems. A singleday without food, however, is unlikely to causepermanent harm.

8. Weight is lost when kilocalories used per dayexceeds kilocalories ingest per day. About 60% ofthe kilocalories used per day is due to basalmetabolic rate. A person with a high basalmetabolic rate loses weight faster than a personwith a low basal metabolic rate, all else beingequal. Another factor to consider is the amountof physical activity, which accounts for about30% of kilocalories used per day. An activeperson loses more weight than a sedentaryperson does.

9. Amino acids, derived from ingested proteins, arenecessary to build muscles. As Lotta and herfriend discovered, excess proteins don’taccelerate this process. Excess proteins can beused as an energy source in oxidativedeamination, for the formation of theintermediate molecules of carbohydratemetabolism, or in gluconeogenesis. Excessproteins are also converted into storagemolecules through glycognesis or lipogenesis.Lotta is in positive nitrogen balance because theamount of nitrogen she gains from her diet isgreater than the amount she loses by excretion.Some of the nitrogen in the amino acids sheingests is incorporated into the proteins of hermuscles as they enlarge.

10. No, this approach doesn’t work because he is notlosing stored energy from adipose tissue. In thesauna, he gains heat, primarily by convectionfrom the hot air and by radiation from the hotwalls. The evaporating sweat is removing heatgained form the sauna. The loss of water willmake him thirsty, and he will regain the lostweight from fluids he drinks and food he eats.

Chapter 26

1. The large volume of hypoosmotic fluid ingestedincreases blood volume and causes bloodosmolality to decrease. The increased bloodvolume is detected by baroreceptors, and thedecreased blood osmolality is detected byosmoreceptors in the hypothalamus. The responseto these stimuli is inhibition of ADH secretion.The alcohol in the beer also inhibits ADHsecretion. The increased volume inhibits therenin-angiotensin-aldosterone mechanism,

which, in turn, inhibits aldosterone secretion. Thechanges in aldosterone, however, take muchlonger to influence kidney function than changesin ADH. As a result of these changes a largevolume of dilute urine is produced until the bloodosmolality and blood volume return to normal.

2. Once the salt is absorbed, the osmolality of theblood increases. The increased osmolality ofblood is detected by osmoreceptor neurons inthe hypothalamus, thereby stimulating ADHsecretion and inhibiting aldosterone secretion. Asmall volume of concentrate urine is produced asa result, until the excess salt is eliminated and theblood osmolality returns to its normal value.

3. The hypoosmotic sweat loss results in more loss ofwater than electrolytes. This simultaneouslydecreases plasma volume and increases bloodosmolality, thereby stimulating increased ADHsecretion. In addition, the decreased plasmavolume stimulates the renin-angiotensin-aldosterone mechanism, resulting in a decreasedglomerular filtration rate and increasedaldosterone secretion. The effect of the changes isto produce a small amount of concentrated urine.

4. The loss of sweat results in a loss of water andelectrolytes. Replacing just the water restoresblood volume and also decreases bloodosmolality. At first, the decreased osmolalityinhibits ADH secretion, and dilute urine isproduced. As blood volume decreases as a resultof urine production, however, ADH secretionand the renin-angiotensin-aldosteronemechanisms are stimulated. Consequently, urineconcentration increases, and only a smallamount of urine is produced.

5. As aldosterone levels decrease, sodiumreabsorption in the nephron decreases and,consequently, plasma sodium levels decrease. Thesodium is lost in the urine, and water follows thesodium by osmosis. Thus, a large amount ofurine that has a high concentration of sodium isproduced. The loss of water reduces bloodvolume, which causes the low blood pressure. Asaldosterone levels decrease potassium secretioninto the nephron decreases, resulting in anincrease in plasma potassium levels. Theincreased extracellular potassium causesdepolarization of nerve and muscle membranes,leading to tremors of skeletal muscles and cardiacarrhythmias including fibrillation.

6. There are several ways to decrease glomerularfiltration rate:a. Decrease hydrostatic pressure in the

glomerulus.1. Decrease systemic arterial blood

pressure.a. Decrease extracellular fluid volume.b. Decrease peripheral resistance.c. Decrease cardiac output.

2. Constrict or occlude the afferent arteriole.3. Relax the efferent arteriole.

b. Increase glomerular capsule pressure.c. Increase the colloid osmotic pressure of the

plasma.d. Decrease the permeability of the filtration

barrier.e. Decrease the total area of the glomeruli

available for filtration.

A-20 Appendix G



7. Assume that the ascending limb of the loop ofHenle and the distal tubules are impermeable tosodium and other ions but actively pump outwater. Other characteristics of the kidney areassumed to be unchanged. As the urine movesup the ascending limb it becomes hyperosmotic,because sodium remains behind as water ispumped out. Assuming that the collecting ductsare impermeable to sodium, upon reaching thecollecting ducts the presence or absence of ADHdetermines the final concentration of the urine.If ADH is absent, there’s little or no exchange ofwater as the urine passes down the collectingducts and a hyperosmotic urine will beproduced. On the other hand, if ADH is present,water moves from the interstitial fluid into thecollecting ducts, thus diluting the urine andproducing a hypoosmotic urine.

8. Urea is partially responsible for the highosmolality of the interstitial fluid in the medullaof the kidney. Since a high osmolality of theinterstitial fluid must exist for the kidney toproduce a concentrated urine, a small amount ofurea in the kidney results in the production ofdilute urine by the kidney.

9. A low-salt diet tends to reduce the osmolality ofthe blood. Consequently, ADH secretion isinhibited, producing dilute urine and thuseliminating water. This in turn reduces bloodvolume and blood pressure.

10. As the loops of Henle become longer, themechanisms that increase concentration of theinterstitial fluid of the medulla become moreefficient, thus raising the concentration of theinterstitial fluid. The maximum concentrationfor urine is determined by the concentration ofthe interstitial fluid deep in the medulla of thekidneys. The higher the concentration ofinterstitial fluid in the medulla of the kidney, thegreater the concentration of the urine the kidneyis able to produce.

Chapter 27

1. When excess glucose is not reabsorbed itosmotically obligates water to remain in thenephron. This results in a large production ofurine, called polyuria, with a consequent loss ofwater, salts, and glucose. The loss of water can becompensated for by increasing fluid intake. Theintense thirst that stimulates increased fluid intakeis called polydipsia. The loss of salts can becompensated for by increasing the salt intake. Thehigh glucose levels in the blood would increasethe blood osmolality, thus stimulating thesecretion of ADH. This increases the permeabilityof the distal convoluted tubule and collecting ductto water. Normally, this would allow reabsorptionof water from the collecting ducts and thusconserve water. If glucose levels in the urine arehigh enough, however, water loss increases evenwith high levels of ADH being present.

2. When ADH levels first increase the reabsorption ofwater increases and urinary output is reduced.This also causes an increase in blood volume and,therefore, an increase in blood pressure. Theincreased blood pressure increases glomerularfiltration rate, which increases urinary output tonormal levels. In addition, the increased blood

volume inhibits the renin-angiotensin-aldosteronemechanism, inhibits aldosterone secretion, andstimulates natriuretic hormone secretion. Theseresponses also increase urinary output.

3. Elevated ammonia ions in the urine results froman increased secretion of H�. Increased secretionof H� occurs in response to either metabolic orrespiratory acidosis. Because an elevatedrespiratory rate increases blood pH, the mostlogical conclusion is that the condition ismetabolic acidosis, and the observed increase inrespiration rate compensates for the metabolicacidosis by lowering H� levels.

4. Diarrhea is one of the most common causes ofmetabolic acidosis, resulting from the loss ofbicarbonate ions. Increasing the respiration rateand producing an acidic urine both help toincrease the blood pH.

5. Blocking H� secretion produces acidosis.Because H� are exchanged for Na�, the Na�

remain in the urine as sodium bicarbonate. Thiseffectively prevents the reabsorption of HCO3

�

and produces an alkaline urine. The blood pH isreduced because H� are not being secreted asrapidly by the nephron. The respiration rateincreases because of the stimulatory effect ofdecreased blood pH on the respiratory center.

6. Breathing through the glass tube increases thedead air space and decreases the efficiency of gasexchange. Consequently, blood carbon dioxidelevels increase and produce a decrease in bloodpH. Compensatory responses include anincreased respiration rate and the production ofacidic urine.

7. A major effect of alkalosis is hyperexcitability ofthe nervous system. If the girl is prone to havingconvulsions, then inducing alkalosis might resultin a seizure. This could be accomplished byhaving the girl hyperventilate. The resulting lossof carbon dioxide from the blood causes anincrease in blood pH.

8. At high altitudes, we expect stimulation of thechemoreceptor reflex and an increase inrespiration rate. This could result inhyperventilation, a decrease in blood carbondioxide, and respiratory alkalosis. The increasedsecretion of hydrochloric acid into the stomachcould also increase blood pH and contribute tothe problem. The kidney produces a morealkaline urine.

Chapter 28

1. Removing the testes would eliminate the majorsource of testosterone. Blood levels oftestosterone would therefore decrease. Becausetestosterone has a negative-feedback effect on thehypothalamus and pituitary gland, GnRH, FSH,and LH secretion would increase and the bloodlevels of these hormones would increase.

2. Prior to puberty, the levels of GnRH are very lowbecause the hypothalamus is very sensitive to theinhibitory effects of testosterone. Since GnRHlevels are low, so are FSH and LH levels. Loss of thetestes and testosterone production would result inan increase in GnRH, FSH, and LH levels. Becauselittle testosterone is produced the boy would notdevelop sexually and would have no sex drive.Small amounts of androgens would be produced

because the adrenal cortex produces someandrogens. He would be taller than normal as anadult, with thin bones and weak musculature. Hisvoice would not deepen and the normal masculinedistribution of hair would not develop.

3. Ideally the pill would inhibit spermatogenesis.Using the same approach as in females, inhibitionof FSH and LH secretion should work. It’s knownthat chronic administration of GnRH suppressesFSH and LH levels enough to cause infertility,through down-regulation. Lack of LH can alsoresult in reduced testosterone levels and a loss ofsex drive, however. Some evidence indicates thatadministration of testosterone in the properamounts would reduce FSH and LH secretion,thus leading to a reduced sperm cell production.The testosterone, however, maintains normal sexdrive. The technique appears to work for a largepercentage of males, resulting in a spermconcentration in the semen that’s too low toresult in fertilization. The technique is notsufficiently precise, however, to be used as astandard birth-control technique.

4. In a postmenopausal woman the ovaries havestopped producing estrogen and progesterone.Without the negative-feedback effect of thesehormones the levels of GnRH, FSH, and LHincrease. Removal of the nonfunctioning ovariesin a postmenopausal woman doesn’t change thelevel of any of these hormones or produce anysymptoms not already occurring due to the lackof ovarian function.

5. Answer e is correct. The secretory phase of themenstrual cycle occurs after ovulation. It isfollowing ovulation that the corpus luteumforms and produces progesterone. In addition,the progesterone acts on the endometrium of theuterus to cause its maximum development.Progesterone secretion therefore reaches itsmaximum levels and the endometrium reachesits greatest degree of development during thesecretory phase of the menstrual cycle.

6. The removal of the ovaries from a 20-year-oldwoman eliminates the major site of estrogen andprogesterone production, thereby causing anincrease in GnRH, FSH, and LH levels due tolack of negative feedback. One expects to see thesymptoms of menopause such as cessation ofmenstruation and reduction in the size of theuterus, vagina, and breasts. There may also be atemporary reduction in sex drive.

7. It’s clear that estrogen and progesteroneadministration resulted in a large decrease in theamount of LH in the plasma the day ofovulation. The differences in plasma LH levelsbetween the groups at other times are very small.The incidence of pregnancies suggests that thereduced plasma LH levels may result in noovulation.

8. The progesterone inhibits GnRH in thehypothalamus. Consequently, the anteriorpituitary is not stimulated to produce LH andFSH. Lack of LH prevents ovulation and lack ofFSH prevents development of the follicles. LHalso is required for maturation of follicles priorto ovulation. Without follicle development,there’s inadequate estrogen production, whichcauses the hot flash symptoms.

A-21Appendix G



9. GnRH administered either before or after thenormal time of ovulation doesn’t result inovulation, because the anterior pituitary is lesssensitive to the effect of GnRH during thosetimes. Also, follicles in the ovary are notadequately developed. The concentration ofGnRH must be controlled carefully because toolittle results in inadequate FSH and LH beingreleased from the anterior pituitary. Too littleFSH and LH fails to cause ovulation. Too muchGnRH given at the proper time results in thematuration of more than one follicle and therelease of an oocyte from more than one of thefollicles. If the oocytes are fertilized, multiplepregnancies can result.

Chapter 29

1. Postovulatory age, the approximate length oftime the embryo has been developing, is 14 daysless than the time since the last menstrual period(LMP). In this case, the postovulatory age is 30days (44 � 14). By this time the neural tube hasclosed, the somites have formed, the digestivetract is developing, the limb buds have appeared,a tubular beating heart is present, and the lungsare developing. Based on reproductivestructures, which are just forming, male andfemale embryos are indistinguishable at this age.

2. The fever would have occurred on day 21–31 ofdevelopment, which is during part of the time ofneural tube closure (days 18–25). If the fever

prevented neural tube closure, the child could beborn with anencephalus or spina bifida.

3. The limb buds develop in a proximal-to-distalsequence. If the apical ectodermal ridge isdamaged during embryonic development whenthe limb bud is about one-half grown, theproximal structures, the arm and forearm,develop normally, but the distal structures, thewrist and hand, do not form normally. Dependingon the degree of damage, the wrist and handcould be completely absent or underdeveloped.

4. The mesonephric duct system develops, becauseof testosterone, to form portions of the malereproductive duct system. Without theproduction of Müllerian-inhibiting hormone,the paramesonephric duct system also developsto form the uterus and uterine tubes. Althoughovaries are present, the clitoris may be enlargedbecause of testosterone to produce somewhat theappearance of male external genitalia. Theamount of masculinization would depend on thelevels of testosterone and how long it wasadministered. High levels of testosterone over anextended period would completely masculinizethe external genitalia.

5. This total Apgar score of 5 indicates: appearance(A, 0) white or blue; pulse (P, 1) low; grimace(G, 1) slight; activity (A, 1) little movement andpoor muscle tone; and respiration (R, 2) normal.The white or blue appearance (A, 0) is consistentwith a poor circulation indicated by a reduced

pulse (P, 1). The reduced heart rate, resulting inthe low pulse, may indicate a circulatory systemproblem. The reduced reflexes and motoractivity (G, 1; A, 1) could result from the lack ofoxygen in the muscles resulting from poorcirculation. Because the infant has poorcirculation despite a normal respiration, clearingthe airway (if obstructed) and administeringoxygen are in order. This Apgar score could haveseveral causes, and additional information isnecessary to determine the actual cause.

6. Suckling the breast stimulates the release ofoxytocin from the neurohypophysis (posteriorpituitary). Once the oxytocin is in the blood, ittravels to both breasts and causes milk letdown.

7. If both parents are heterozygous for dimpledcheeks, then the child could receive a recessivegene for no dimples from each parent, resultingin the homozygous recessive condition with nodimples in the cheeks.

8. It’s not possible at present to determine byphenotype if a child is homozygous orheterozygous for tongue rolling. Even if it werepossible to determine that the child washeterozygous, that’s not very strong evidence thatthe recessive allele came from the proposed father.

9. Hemophilia is a sex-linked trait. Since the fatherhas hemophilia he must be XhY. If the motherwere XHXH, all their children would be normal.For half of the children to have hemophilia, shemust be XHXh.

A-22 Appendix G

appendix

Documents

posterior cruciate ligament

inferior vena cava

superior vena cava

external iliac vein

common iliac vein

essential amino acids

superior articulating processes

lateral spinothalamic tract