Answers to end-of-chapter questions



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Saladin/Nolan


Clinical Applications Manual

ANSWERS TO END-OF-CHAPTER QUESTIONS

Chapter 1
1. The mental and physiological signs of the disease result from ingestion and inhalation of lead by the children. These children were exposed to lead (and other heavy metals) at the manufacturing plant where they scavenged materials.
2. The children’s risk factors for lead poisoning include the presence of lead-based paint, dust containing crushed and pulverized paint chips, and metal-contaminated soil around the factory. The risk factor for the elderly woman is her habit of chewing on the lead foil wrapping from wine bottles.
3. Lead is a poison but not a toxin, because toxins are of animal or plant origin.
4. Idiopathic diseases are those with no known cause. Since there is an identifiable cause in all the cases in this scenario, none of these people has idiopathic lead poisoning.
5. Yes, lead poisoning could be considered a syndrome. The signs, symptoms and changes in physiology are characteristic of lead poisoning.
6. Since blood is a tissue, the diagnostic findings on the blood (traces of lead and reduced RBC count) are histopathological.
7. High morbidity? Yes, among the children but not among adults, because a large percentage of children in the housing project exhibit signs of the disease. High mortality? No, because no deaths were reported. High prevalence? Yes, high morbidity in this case is equivalent to the high prevalence of the disease among the children. High incidence? There is insufficient information to evaluate incidence in this case, because no data were available to document the number of new cases in a given time period.

8. Yes, lead poisoning among the children of this community can be considered an epidemic because it occurs in a far higher percentage of individuals (more than 70 of the 112 children examined) than it does in the general population. Lead poisoning is not an infectious disease, however, because it is not caused by an infectious agent (such as a virus or bacterium) and it is not transmitted from person to person.

9. (a) joint pain = symptom; (b) difficulty walking = sign; (c) excessive salivation = sign; (d) personality changes = sign; (e) low RBC count = sign; (f) subnormal intelligence = sign; (g) dimness of vision = symptom; (h) lead in the urine = sign. (Remember that the criterion for a symptom is something that can only be known by the patient, while a sign is capable of being objectively verified by another observer.)
10. You could advise the daughter to buy wine without lead foil or to dispose of the foil wrappers carefully so that her mother cannot get at them. You might also do a blood test on the mother to check for iron-deficiency anemia and prescribe a multivitamin and mineral supplement if necessary. If the mother insists on chewing on something, you could recommend substituting something nontoxic such as ice or gum.

Chapter 2

1. Ellen’s symptoms include sore throat, pain while swallowing, headache, exhaustion, and the feeling of being cold. Her hoarseness, flushed face, watery eyes, and fever (elevated body temperature) are specific signs that she is ill.


2. Ellen’s chief complaint is a sore throat.
3. Questions that could be asked of Ellen include: When did you first develop a sore throat? Have you recently been around anyone with similar symptoms? Have you had trouble eating or drinking? How long have you had a headache? Is the headache worse in the presence of light?
4. Ellen’s risk factors for disease include lack of sleep, inadequate diet, contact with a roommate who may have transmitted the illness, stress (due to final exams and other factors), exertion, and exposure to inclement weather.
5. Yes, a sphygmomanometer will be needed because it is used to determine blood pressure, which is one of the vital signs and therefore a routine step in the initial examination of all patients.

6. When feeling the sides of Ellen’s neck, the physician is using palpation.

7. Using the stethoscope to listen to Ellen’s heart and lungs is an example of auscultation.
8. An otoscope is used to examine the ears.
9. The physician is obtaining cells for culture to test for the presence of microorganisms that may be causing the disease. In this case, he is most likely going to test for Streptococcus, a bacterium commonly associated with throat infections (strep throat).
10. The physician would most likely complete a physical examination and order additional tests, such as urinalysis and various blood tests, including a CBC.

Chapter 3
1. Phase I is not designed to show whether the drug has an effect on cancer pain. It is designed only to establish that there are no unacceptable side effects of the drug and to establish the maximum safe dosage. The volunteers in phase I are not suffering from cancer or pain. Therefore, there is no harm in the subjects knowing that they are receiving the drug. Phase II, however, is designed to determine whether the drug really has a measurable effect on the pain felt by cancer patients. Pain relief is highly subject to psychosomatic effects, so a placebo group is needed to control for this effect. Otherwise, if patients reported that NoPain did not relieve their discomfort, the researchers would have no way of knowing whether this relief was due to a pharmacological action of NoPain or merely to a placebo (psychosomatic) effect.

2. Variation from patient to patient and atypical responses would have a strong effect on averages based on a small number of patients. Testing a small sample of people also may fail to reveal side effects and unusual responses that could affect a significant number of people if a drug is approved for the market and used by millions of people. Therefore, clinical trials must use a large enough sample that such responses have a high probability of being detected and individual variations do not greatly skew the data and lead to false conclusions.

3. NoPain is a palliative treatment because it is able to relieve patients’ pain and thereby make them more comfortable, but not able to cure or prevent their cancer.
4. Explain to Mae that U.S. regulations governing the safe development of drugs require years of clinical trials and the expenditure of millions of dollars before the drug company can market the product. Also, at any point during the process, the drug may fail to qualify for human use, in which case the expenses incurred during clinical trials may never be recovered. Even after a drug is approved, unexpected side effects can arise and force a company to withdraw it from the market. Therefore, drug companies must price a drug so as to recover the costs of its development, cover the risks of failure, and still make a sufficient profit to stay in business.
5. Available sterilization techniques include pressurized steam, dry heat, irradiation, and chemicals. The technique used depends on what equipment is available and how soon the sterile swabs are needed. Pressurized steam and irradiation are both relatively easy to accomplish with the appropriate equipment, and the sterilization process can be completed in less than 2 hours in most cases. Chemical sterilization, especially with alcohol and phenol, would not be suitable because it most often requires placing the item to be sterilized in direct contact with the chemical. In the case of cotton swabs, the chemical would then come in contact with patients’ skin.
6. Immunotherapy, chemotherapy, and radiotherapy are all used to treat cancer.

7. Although the hospital tries to minimize the presence and spread of infectious agents in many different ways, it is impossible to maintain a sterile environment in a hospital room. This infection probably occurred when the patient was exposed to a person carrying the infectious agent, such as another patient or a visitor, or the infectious agent may have been airborne.

8. The physician would most likely use an injection to rapidly produce a high circulating antibiotic concentration. For a more prolonged dose, the physician would probably follow the injection with oral medication to be taken at regular intervals.
9. International units provide a way of determining the response expected from a given amount of a compound. The biological activity of the same compound may differ based on the means of preparation. As a result, using the mass (weight) may not produce reproducible effects. Therefore, international units provide a means of producing a uniform response, regardless of the means of preparation.
10. Gastrotomy refers to an incision in the stomach. Gastrectomy means removal of part or all of the stomach. Gastroscopy means viewing the stomach, normally with an endoscope. Gastrocentesis means puncturing the stomach to remove fluid. Gastroclysis means irrigating the stomach with fluid.

Chapter 4
1. bradycardia = slow heart rate, usually less than 60 beats per minute; hypotension = decreased blood pressure; lymphadenopathy = chronic or excessive enlargement of the lymph nodes; cervical lymph nodes = lymph nodes located in the cervical region of the body; inguinal lymph nodes = lymph nodes located in the inguinal region; albuminuria = the presence of albumin in the urine; petechiae = small, purplish spots on the skin resulting from hemorrhages.
2. Jason and Mary’s WBC counts are abnormally low. They are exhibiting leukopenia. The change in white blood cell count is relevant to the diagnosis because it suggests a pathology affecting the immune system.

3. Attempts to control dengue fever usually focus on controlling the reproduction of the Aedes mosquito, the vector for the causative agent, Flavivirus. Methods of limiting the mosquito population include minimizing the presence of standing water and using insecticides. Tourists might be advised to avoid areas of known mosquito infestation, to use insect repellent, and if possible to sleep under bed nets at night as a protective barrier against insects.

4. The invasion of a cell by viruses increases the cell’s osmolarity. This occurs because the viruses act like any other particle. As the number of viruses increases, the number of particles in the cells increases as well. This increased particle number gives the cells an increased osmolarity that could eventually lead to cellular lysis.
5. Cells with high metabolic rates are highly dependent on ATP and therefore on functionally normal mitochondria. Thus, they are the most vulnerable to cytopathies that disturb mitochondrial function.
6. Prokaryotic organisms are all bacteria and are unicellular. All multicellular organisms are eukaryotic. Parasitic worms and arthropods are multicellular; therefore, they are eukaryotic.
7. Children do not inherit Kearns-Sayre syndrome from their fathers because only the mitochondria in the egg cell are passed on to the next generation. The mitochondria of the sperm cell that fertilizes the egg do not survive.
8. Since giardiasis is transmitted through water contaminated with sewage, a knowledgeable epidemiologist would hypothesize that the school’s water supply is contaminated with sewage and the students are getting Giardia from sources such as the drinking fountains or cafeteria food. Since the school is old and run-down, it is possible that its water and sewage lines are not in good repair, and these should be inspected for leaks.
9. The main risk factor for Gaucher disease is family history because it is a hereditary disease; this risk factor is unavoidable. A main risk factor for malaria is exposure to the vector, which is avoidable. Risk factors for toxoplasmosis include diet (undercooked meat, unpasteurized milk) and exposure to infected cats or their feces; these risk factors are avoidable.

10. If this disorder were expressed in the zygote (fertilized egg), the first cell divisions after fertilization could not occur. All such zygotes would perish before a woman was even aware of her pregnancy. Some genes, however, are expressed later in life or only under certain conditions. In such a case, the hereditary disorder might affect growth at any stage from embryo to adult, or it might affect the growth of selected organs. It could cause sterility due to the inability to produce eggs and sperm through meiosis, or it could produce premature aging because organs would be unable to replace cells at a sufficient rate to keep pace with their normal death.


Chapter 5
1. The following individuals demonstrate principle 1: Bill, Terry, Greg, and Jenny.
2. The following individuals represent principle 2: Carl, Terry, and Greg. (The fact that Terry and Greg have the disease whereas their father was not even a carrier is just as relevant as the fact that Bill had the disease but did not pass it on to his son Carl.)
3. The following individuals represent principle 3: Florence, Jessica, Terry, and Greg.
4. Males cannot carry hemophilia because it is an X-linked disease and males only have a single X chromosome. If the male has an X with the recessive allele for hemophilia, he has the disease.
5. The probability of Marcy and Tom having a child with hemophilia is 12.5%. This is because Marcy has a 50% chance of being a carrier (receiving a recessive X allele from her mother). Marcy’s children in turn have a 50% chance of receiving a copy of the X chromosome with the recessive allele from her. Thus, there is a 25% chance that any of the children would receive a copy of the recessive X allele from Marcy—(0.5)(0.5) = 0.25. Tom does not have hemophilia, so we know that the gene on his X chromosome is normal. Thus, there is no chance that his daughters could receive an X chromosome with the allele for hemophilia from him. Therefore, only Marcy and Tom’s sons could have hemophilia. Since there is a 50% chance that any child will be male, the total probability of Marcy and Tom having a child with hemophilia is 12.5%—(0.5)(0.5)(0.5) = .125.

6. Marcy should not be worried that the baby girl will have hemophilia. Even if Marcy is a carrier (a 50% probability), Tom does not have hemophilia. Therefore, a baby girl would receive a copy of Tom’s X chromosome that does not have the allele for hemophilia. If Marcy is indeed a carrier of hemophilia and passes the recessive allele to her daughter, any future grandsons could have hemophilia. But only if Marcy’s daughter were to marry a man with hemophilia, could any future granddaughters have the disease.

7. No, the evidence does not rule out the man as the father of the child. A person with blood type A may have both A (Ia) and O (i) alleles, while a person with type B blood may have B (I B) and i alleles. If both the woman and the man each have one O allele, the child could inherit an A allele from the mother and an O allele from the father. The child would then have blood type A, even though its father’s blood type was B.
8. Hair loss is often a side effect of chemotherapy because hair growth occurs by mitosis in the hair bulb. Since the drugs used for chemotherapy target rapidly dividing cells (cells with a high mitotic rate), these drugs would also kill the mitotic cells of the hair bulb, and lost hairs would not be replaced with new growth from below.
9. According to table 5.1, oral cancer has a better prognosis because 27.3% (or 8,400 out of 30,750) of people with oral cancer die compared to almost 50% (46,600 out of 94,100) of people with colon cancer.
10. The cancerous cells exhibit relatively enlarged nuclei that stain darkly because of their dense chromatin.

Chapter 6
1. The tumor is encapsulated (surrounded by a layer of connective tissue), and its cells have a “normal” appearance for the organ. These are indications that Ms. Bennett’s tumor is benign.
2. Ms. Malcolm’s physician recommends a biopsy of the axillary lymph nodes because the tumor is malignant and therefore capable of metastasizing. Since the axillary lymph nodes are the closest lymph nodes to the mammary glands, these would be the first lymph nodes biopsied for possible metastasis. Wandering cancer cells from the breast can easily lodge there and “seed” the lymph node, resulting in the development of a new tumor.

3. The mitotic index is a measure of cell division. An abnormally high mitotic index for a specific tissue or organ suggests cancer.

4. Some tumors increase in size more rapidly in the presence of certain chemicals that stimulate cell growth and division. In the case of breast cancer, one of these chemicals is estrogen. By using tamoxifen to block estrogen’s ability to bind to its receptor, the rate of tumor growth would be decreased.
5. Even though X rays can induce mutations and thereby increase a person’s risk of developing cancer, X rays are also able to detect very small masses that may not be palpable (either by the woman during a BSE or by a clinician during a physical examination). The sooner the mass is detected, the sooner a diagnosis can be made—and if the woman has cancer, the sooner treatment can be started. These factors combine to increase the patient’s odds of survival. Thus, X rays carry both risks and benefits, but this is a case in which the benefits considerably outweigh the risks.
6. Anaerobic fermentation generates lactic acid. As lactic acid accumulates, the intracellular pH decreases. This drop in pH can change the shape of cellular proteins. Since enzymes are proteins, the shape of the enzyme is altered. This conformational change can lead to decreased enzymatic activity. As more and more enzymes are affected by the pH, there is increased inhibition of catalytic reactions, eventually leading to cessation of the metabolic pathways in the cell.
7. In both hypothermic injury and dystrophic calcification, crystals (ice and calcium crystals, respectively) are formed in the cell. These crystals can damage the plasma membrane and disrupt the osmotic balance of the cell.

8. Gouty arthritis more commonly affects the great toe than other joints because every time a person takes a step, the toe moves, thus being subjected to a great deal of mechanical stress. Activities such as running, jumping, and standing on tiptoe increase the mechanical stress on the great toe still further.

9. Drinking more water induces the body to produce more urine. This increases the rate of urate excretion from the body, thus helping to prevent urate from accumulating to the level that it accumulates in the tissues and causes joint inflammation.
10. The rehydration step would be unnecessary.

Chapter 7
1. The risk factor is Norma’s frequent occupational exposure to latex. Her symptom is pruritis (itching). The signs of her latex allergy include erythema, edema, bullae, vesicles, and a positive response to latex in the patch test.
2. Skin scrapings are taken to rule out other possible pathologies. The signs and symptoms may point to a latex allergy, but those same signs and symptoms could also be caused by bacterial or parasitic infestations. Examining the skin scrapings allows the physician to rule out these alternatives.
3. Since Norma often works with patients with infectious diseases, skin rashes, and wounds, she has an increased risk of infection if the skin becomes broken through removal of the tops of the blisters.
4. Norma may be experiencing a cross-reaction between her latex allergy and an allergy to avocados. (Avocados, bananas, and kiwi fruits are known to trigger allergic reactions in some people with latex allergies.) Norma should consult her physician immediately, since her next reaction could be more serious, and she may have to eliminate some fruits from her diet.
5. In a patch test, different allergens are introduced to discrete regions of the skin. Inflammation occurring at one or more sites indicates which substance(s) the patient is allergic to. In Norma’s case, the site where the latex was applied produced a positive result.

6. Anabolic steroids (testosterone and its agonists) often cause acne for the same reason that elevated testosterone secretion predisposes adolescents to acne. High concentrations of testosterone increase the activity of the sebaceous glands, leading to an increased production of sebum, one of the three factors that contribute to acne.

7. (c) A fissure would mostly likely result from tinea.
8. Controlling pruritis is important in curing skin diseases because itching leads to scratching. Scratching damages the skin and increases the chance of infection. Also, scratching an infected or infested area allows the causative agent to be transferred to other regions of the body (or even to other people), especially if the patient does not wash his or her hands after scratching.
9. (b) The elderly white man with a broken hip would be most likely to develop a pressure ulcer.
10. (d) The pruritis of scabies leads to persistent scratching, which is a common cause of lichenification.

Chapter 8
1. The emergency room physician suspects osteoporosis because Susan is postmenopausal, a risk factor for osteoporosis. On the other hand, the fact that Susan is physically active makes her less likely to have osteoporosis and leads to the possibility that bone cancer could have weakened her hip bone and led to the fracture.
2. If Susan had bone cancer, the scans would have revealed the presence of a tumor, such as an osteosarcoma, chondrosarcoma, fibrosarcoma, or myeloma. The bone density scan would have revealed local changes in bone density associated with a tumor rather than the overall decrease in bone density characteristic of osteoporosis.
3. Vitamin D is needed for the proper absorption of dietary calcium. Although the body produces vitamin D on its own, the goal of treatment in Susan’s case is to maximize her calcium absorption by having her take supplementary vitamin D.
4. Susan’s fracture is nondisplaced, meaning that the portions of the bone remain in the correct anatomical alignment with each other, and oblique, meaning that the fracture is diagonal.

5. Osteomyelitis developed after Susan’s surgery. Characteristic signs and symptoms observed in Susan include fever, swollen lymph nodes, elevated lymphocyte count, and redness, inflammation, edema, and increased pain at the surgical site. The diagnosis of osteomyelitis is confirmed by the observation of a subperiosteal abscess through a CT scan.

6. In all likelihood, Susan’s case is one of exogenous osteomyelitis, a bacterial bone infection introduced during the open reduction of her hip fracture. Bacteria might have been introduced by way of imperfectly sterilized screws inserted to internally fix the femur or by other contamination of the surgical field.
7. John has a better prognosis than Marvin because chondrosarcomas can only be successfully removed by amputating the affected bone. Amputation of the lower leg is possible, but amputation of the pelvic bones is not possible.
8. A bone tumor predisposes a bone to pathological fracture because the tumor weakens the bone, making it more likely to break. Bones are also weakened by other diseases, such as osteoporosis. A stress fracture is not caused by disease but by severe trauma to a bone such as a hard fall or a blow.
9. Unlike the skin and other superficial body structures, bone cannot be directly observed during a routine physical examination. Palpation can be used to determine gross bone abnormalities and some fractures, but it is not as useful for discerning bone cancers, osteoporosis, osteomyelitis, or the different types of fractures. Imaging techniques enable a clinician to view the bone and determine whether abnormalities are present.
10. Osteochondroses are most prevalent in children because their bones are still growing. Adult bones do not have ossification centers that could be affected by osteochondrosis.

Chapter 9
1. Jamie must rest his feet so that they can heal. If he feels some pain, he is more likely to cut back on his physical activities and stop putting undue stress on his feet. Masking the pain with analgesics might enable him to remain active and thereby permanently injure his feet.

2. The pain subsides after a period of inactivity because stress is no longer being placed on the epiphyseal region of Jamie’s heel and pressure on the calcaneal tendon is decreased. This allows tissue regeneration to outpace the new tissue injury that would occur if Jamie continued his normal activity.

3. Sever disease occurs more frequently in active children because running, jumping, and other physical exercise place stresses on the heel and associated structures. Children who are sedentary place less stress on these regions.
4. Sever disease occurs in children because the epiphyseal plates in the heel are still cartilaginous and therefore subject to tearing. In adults, the epiphyseal plates have ossified.
5. When Jamie uses the heel pads, the pain will be relieved because the calcaneal tendon will be stretched less and there will be less stress on the calcaneus. When he stops using the pads, the pain is likely to return because of renewed stress on the tendon and bone.
6. Jamie’s X rays are normal because Sever disease does not cause abnormalities in the bones themselves; rather, it affects the cartilage of the epiphyseal plates of the heel, and cartilage does not show on X ray. If bone spurs were present, X rays would reveal a bony exostosis on the inner tuberosity of the calcaneus extending toward the plantar fascia.
7. Normally, the skull widens as the parietal bones add new osseous tissue along the margins of the sagittal suture, where they face each other. Once they become rigidly joined along this suture, further lateral expansion is minimal.
8. An experienced pediatrician could determine the type of craniosynostosis on sight, but its classification and quantification would be helped by measuring the circumference of the skull, its height, and its dimensions from right to left and from anterior to posterior, and then comparing those measurements to normal standards.
9. Children with craniosynostosis may exhibit mental retardation because premature fusion of the sutures restricts the normal, symmetric growth of the brain.

10. Paget disease results in malformation of bone due to excessive bone remodeling. This disease most often affects the bones of the axial skeleton. The vertebrae become thicker, narrowing the vertebral canal and intervertebral foramina and thus compressing the spinal cord and spinal nerves.


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