Exam 3 practice test
7. Which of the following hormones acts on its target tissues by a steroid hormone mechanism of action? (A) Thyroid hormone (B) Parathyroid hormone (PTH) (C) Antidiuretic hormone (ADH) on the collecting duct (D) β1-adrenergic agonists (E) Glucagon
The answer is A [II E; Table 7.2]. Thyroid hormone, an amine, acts on its target tissues by a steroid hormone mechanism, inducing the synthesis of new proteins. The action of antidiuretic hormone (ADH) on the collecting duct (V2 receptors) is mediated by cyclic adenosine monophosphate (cAMP), although the other action of ADH (vascular smooth muscle, V1 receptors) is mediated by inositol 1,4,5-triphosphate (IP3). Parathyroid hormone (PTH), β1-agonists, and glucagon all act through cAMP mechanisms of action.
11. Which of the following substances is derived from proopiomelanocortin (POMC)? (A) Adrenocorticotropic hormone (ACTH) (B) Follicle-stimulating hormone (FSH) (C) Melatonin (D) Cortisol (E) Dehydroepiandrosterone
The answer is A [III B 1, 2; Figure 7.5]. Proopiomelanocortin (POMC) is the parent molecule in the anterior pituitary for adrenocorticotropic hormone (ACTH), β-endorphin, α-lipotropin, and β-lipotropin (and in the intermediary lobe for melanocyte-stimulating hormone [MSH]). Follicle-stimulating hormone (FSH) is not a member of this "family"; rather, it is a member of the thyroidstimulating hormone (TSH) and luteinizing hormone (LH) "family." MSH, a component of POMC and ACTH, may stimulate melatonin production. Cortisol and dehydroepiandrosterone are produced by the adrenal cortex.
32. A 62-year-old man at sea level breaths a gas mixture containing 21% O2. He has the following arterial blood gas values: PaO2 = 60 mm Hg PaCO2 = 70 mm Hg Which of the following is the cause of his hypoxemia? (A) Hypoventilation (B) Fibrosis (C) V/Q defect (D) Right-to-left shunt (E) Anemia
The answer is A [IV D]. Anemia (or decreased hemoglobin concentration) is eliminated because it causes decreased O2 content of blood but does not cause hypoxemia. The remaining choices all cause hypoxemia. Calculating the A-a gradient distinguishes between these causes as follows: PIO2 = (760 mm Hg − 47 mm Hg) × 0.21 = 150 mm Hg. PaO2 = 150 mm Hg − 70 mm Hg/0.8 = 63 mm Hg. A-a gradient = 63 mm Hg − 60 mm Hg = 3 mm Hg, which is normal. Among the choices, the only cause of hypoxemia with a normal A-a gradient is hypoventilation, whereby PaO2 is lowered by hypoventilation and PaO2 equilibrates with that lowered value; since PaO2 and PaO2 are equilibrated (but lower than normal), they are essentially equal and A-a is close to zero, or normal. Fibrosis, V/Q defect, and right-to-left shunt all cause decreased PaO2 that is not equilibrated with PaO2 and thus cause increased A-a gradient.
29. Which person would be expected to have the largest A-a gradient? (A) Person with pulmonary fibrosis (B) Person who is hypoventilating due to morphine overdose (C) Person at 12,000 feet above sea level (D) Person with normal lungs breathing 50% O2 (E) Person with normal lungs breathing 100% O2
The answer is A [IV D]. Increased A-a gradient signifies lack of O2 equilibration between alveolar gas (A) and systemic arterial blood (a). In pulmonary fibrosis, there is thickening of the alveolar/pulmonary capillary barrier and increased diffusion distance for O2, which results in lack of equilibration of O2, hypoxemia, and increased A-a gradient. Hypoventilation and ascent to 12,000 feet also cause hypoxemia, because systemic arterial blood is equilibrated with a lower alveolar PO2 (normal A-a gradient). Persons breathing 50% or 100% O2 will have elevated alveolar PO2, and their arterial PO2 will equilibrate with this higher value (normal A-a gradient).
20. Which of the following changes occurs during strenuous exercise? (A) Ventilation rate and O2 consumption increase to the same extent (B) Systemic arterial PO2 decreases to about 70 mm Hg (C) Systemic arterial PCO2 increases to about 60 mm Hg (D) Systemic venous PCO2 decreases to about 20 mm Hg (E) Pulmonary blood flow decreases at the expense of systemic blood flow
The answer is A [IX A]. During exercise, the ventilation rate increases to match the increased O2 consumption and CO2 production. This matching is accomplished without a change in mean arterial PO2 or PCO2. Venous PCO2 increases because extra CO2 is being produced by the exercising muscle. Because this CO2 will be blown off by the hyperventilating lungs, it does not increase the arterial PCO2. Pulmonary blood flow (cardiac output) increases manifold during strenuous exercise.
10. Which of the following functions of the Sertoli cells mediates negative feedback control of follicle-stimulating hormone (FSH) secretion? (A) Synthesis of inhibin (B) Synthesis of testosterone (C) Aromatization of testosterone (D) Maintenance of the blood-testes barrier
The answer is A [IX B 2, 3]. Inhibin is produced by the Sertoli cells of the testes when they are stimulated by follicle-stimulating hormone (FSH). Inhibin then inhibits further secretion of FSH by negative feedback on the anterior pituitary. The Leydig cells synthesize testosterone. Testosterone is aromatized in the ovaries.
13. Selective destruction of the zona glomerulosa of the adrenal cortex would produce a deficiency of which hormone? (A) Aldosterone (B) Androstenedione (C) Cortisol (D) Dehydroepiandrosterone (E) Testosterone
The answer is A [V A 1; Figure 7.10]. Aldosterone is produced in the zona glomerulosa of the adrenal cortex because that layer contains the enzyme for conversion of corticosterone to aldosterone (aldosterone synthase). Cortisol is produced in the zona fasciculata. Androstenedione and dehydroepiandrosterone are produced in the zona reticularis. Testosterone is produced in the testes, not in the adrenal cortex.
22. Which step in steroid hormone biosynthesis is stimulated by adrenocorticotropic hormone (ACTH)? (A) Cholesterol → pregnenolone (B) Progesterone → 11-deoxycorticosterone (C) 17-Hydroxypregnenolone → dehydroepiandrosterone (D) Testosterone → estradiol (E) Testosterone → dihydrotestosterone
The answer is A [V A 2 a (2)]. The conversion of cholesterol to pregnenolone is catalyzed by cholesterol desmolase. This step in the biosynthetic pathway for steroid hormones is stimulated by adrenocorticotropic hormone (ACTH).
18. Increased adrenocorticotropic hormone (ACTH) secretion would be expected in patients (A) with chronic adrenocortical insufficiency (Addison disease) (B) with primary adrenocortical hyperplasia (C) who are receiving glucocorticoid for immunosuppression after a renal transplant (D) with elevated levels of angiotensin II
The answer is A [V A 2 a (3); Table 7.6; Figure 7.12]. Addison disease is caused by primary adrenocortical insufficiency. The resulting decrease in cortisol production causes a decrease in negative feedback inhibition on the hypothalamus and the anterior pituitary. Both of these conditions will result in increased adrenocorticotropic hormone (ACTH) secretion. Patients who have adrenocortical hyperplasia or who are receiving exogenous glucocorticoid will have an increase in the negative feedback inhibition of ACTH secretion.
24. Which of the following causes increased aldosterone secretion? (A) Decreased blood volume (B) Administration of an inhibitor of angiotensin-converting enzyme (ACE) (C) Hyperosmolarity (D) Hypokalemia
The answer is A [V A 2 b]. Decreased blood volume stimulates the secretion of renin (because of decreased renal perfusion pressure) and initiates the renin-angiotensin-aldosterone cascade. Angiotensin-converting enzyme (ACE) inhibitors block the cascade by decreasing the production of angiotensin II. Hyperosmolarity stimulates antidiuretic hormone (ADH) (not aldosterone) secretion. Hyperkalemia, not hypokalemia, directly stimulates aldosterone secretion by the adrenal cortex.
22. In the transport of CO2 from the tissues to the lungs, which of the following occurs in venous blood? (A) Conversion of CO2 and H2O to H+ and HCO3 − in the red blood cells (RBCs) (B) Buffering of H+ by oxyhemoglobin (C) Shifting of HCO3 − into the RBCs from plasma in exchange for Cl− (D) Binding of HCO3 − to hemoglobin (E) Alkalinization of the RBCs
The answer is A [V B; Figure 4.9]. CO2 generated in the tissues is hydrated to form H+ and HCO3 − in red blood cells (RBCs). H+ is buffered inside the RBCs by deoxyhemoglobin, which acidifies the RBCs. HCO3 − leaves the RBCs in exchange for Cl− and is carried to the lungs in the plasma. A small amount of CO2 (not HCO3 −) binds directly to hemoglobin (carbaminohemoglobin
30. Which of the following pancreatic secretions has a receptor with four subunits, two of which have tyrosine kinase activity? (A) Insulin (B) Glucagon (C) Somatostatin (D) Pancreatic lipase
The answer is A [VI C 2]. The insulin receptor in target tissues is a tetramer. The two β subunits have tyrosine kinase activity and autophosphorylate the receptor when stimulated by insulin.
27. In a maximal expiration, the total volume expired is (A) tidal volume (VT) (B) vital capacity (VC) (C) expiratory reserve volume (ERV) (D) residual volume (RV) (E) functional residual capacity (FRC) (F) inspiratory capacity G total lung capacity
The answer is B [I B 3]. The volume expired in a forced maximal expiration is forced vital capacity, or vital capacity (VC).
An infant born prematurely in gestational week 25 has neonatal respiratory distress syndrome. Which of the following would be expected in this infant? (A) Arterial PO2 of 100 mm Hg (B) Collapse of the small alveoli (C) Increased lung compliance (D) Normal breathing rate (E) Lecithin:sphingomyelin ratio of greater than 2:1 in amniotic fluid
The answer is B [II D 2]. Neonatal respiratory distress syndrome is caused by lack of adequate surfactant in the immature lung. Surfactant appears between the 24th and the 35th gestational week. In the absence of surfactant, the surface tension of the small alveoli is too high. When the pressure on the small alveoli is too high (P = 2T/r), the small alveoli collapse into larger alveoli. There is decreased gas exchange with the larger, collapsed alveoli; and ventilation/perfusion (V/Q) mismatch, hypoxemia, and cyanosis occur. The lack of surfactant also decreases lung compliance, making it harder to inflate the lungs, increasing the work of breathing, and producing dyspnea (shortness of breath). Generally, lecithin:sphingomyelin ratios greater than 2:1 signify mature levels of surfactant.
25. Secretion of oxytocin is increased by (A) milk ejection (B) dilation of the cervix (C) increased prolactin levels (D) increased extracellular fluid (ECF) volume (E) increased serum osmolarity
The answer is B [III C 2]. Suckling and dilation of the cervix are the physiologic stimuli for oxytocin secretion. Milk ejection is the result of oxytocin action, not the cause of its secretion. Prolactin secretion is also stimulated by suckling, but prolactin does not directly cause oxytocin secretion. Increased extracellular fluid (ECF) volume and hyperosmolarity are the stimuli for the secretion of the other posterior pituitary hormone, antidiuretic hormone (ADH).
23. Which of the following causes of hypoxia is characterized by a decreased arterial PO2 and an increased A-a gradient? (A) Hypoventilation (B) Right-to-left cardiac shunt (C) Anemia (D) Carbon monoxide poisoning (E) Ascent to high altitude
The answer is B [IV A 4; IV D; Table 4.4; Table 4.5]. Hypoxia is defined as decreased O2 delivery to the tissues. It occurs as a result of decreased blood flow or decreased O2 content of the blood. Decreased O2 content of the blood is caused by decreased hemoglobin concentration (anemia), decreased O2-binding capacity of hemoglobin (carbon monoxide poisoning), or decreased arterial PO2 (hypoxemia). Hypoventilation, right-to-left cardiac shunt, and ascent to high altitude all cause hypoxia by decreasing arterial PO2. Of these, only right-to-left cardiac shunt is associated with an increased A-a gradient, reflecting a lack of O2 equilibration between alveolar gas and systemic arterial blood. In right-to-left shunt, a portion of the right heart output, or pulmonary blood flow, is not oxygenated in the lungs and thereby "dilutes" the PO2 of the normally oxygenated blood. With hypoventilation and ascent to high altitude, both alveolar and arterial PO2 are decreased, but the A-a gradient is normal.
16. Compared with the systemic circulation, the pulmonary circulation has a (A) higher blood flow (B) lower resistance (C) higher arterial pressure (D) higher capillary pressure (E) higher cardiac output
The answer is B [VI A]. Blood flow (or cardiac output) in the systemic and pulmonary circulations is nearly equal; pulmonary flow is slightly less than systemic flow because about 2% of the systemic cardiac output bypasses the lungs. The pulmonary circulation is characterized by both lower pressure and lower resistance than the systemic circulation, so flows through the two circulations are approximately equal (flow = pressure/resistance).
In which vascular bed does hypoxia cause vasoconstriction? (A) Coronary (B) Pulmonary (C) Cerebral (D) Muscle (E) Skin
The answer is B [VI C]. Pulmonary blood flow is controlled locally by the PO2 of alveolar air. Hypoxia causes pulmonary vasoconstriction and thereby shunts blood away from unventilated areas of the lung, where it would be wasted. In the coronary circulation, hypoxemia causes vasodilation. The cerebral, muscle, and skin circulations are not controlled directly by PO2
21. If an area of the lung is not ventilated because of bronchial obstruction, the pulmonary capillary blood serving that area will have a PO2 that is (A) equal to atmospheric PO2 (B) equal to mixed venous PO2 (C) equal to normal systemic arterial PO2 (D) higher than inspired PO2 (E) lower than mixed venous PO2
The answer is B [VII B 1]. If an area of lung is not ventilated, there can be no gas exchange in that region. The pulmonary capillary blood serving that region will not equilibrate with alveolar PO2 but will have a PO2 equal to that of mixed venous blood.
28. Which of the following results from the action of parathyroid hormone (PTH) on the renal tubule? (A) Inhibition of 1α-hydroxylase (B) Stimulation of Ca2+ reabsorption in the distal tubule (C) Stimulation of phosphate reabsorption in the proximal tubule (D) Interaction with receptors on the luminal membrane of the proximal tubular cells (E) Decreased urinary excretion of cyclic adenosine monophosphate (cAMP)
The answer is B [VII B 2]. Parathyroid hormone (PTH) stimulates both renal Ca2+ reabsorption in the renal distal tubule and the 1α-hydroxylase enzyme. PTH inhibits (not stimulates) phosphate reabsorption in the proximal tubule, which is associated with an increase in urinary cyclic adenosine monophosphate (cAMP). The receptors for PTH are located on the basolateral membranes, not the luminal membranes.
18. Compared with the apex of the lung, the base of the lung has (A) a higher pulmonary capillary PO2 (B) a higher pulmonary capillary PCO2 (C) a higher ventilation/perfusion (V/Q) ratio (D) the same V/Q ratio
The answer is B [VII C; Figure 4.10; Table 4.5]. Ventilation and perfusion of the lung are not distributed uniformly. Both are lowest at the apex and highest at the base. However, the differences for ventilation are not as great as for perfusion, making the ventilation/perfusion (V/Q) ratios higher at the apex and lower at the base. As a result, gas exchange is more efficient at the apex and less efficient at the base. Therefore, blood leaving the apex will have a higher PO2 and a lower PCO2.
15. Which step in steroid hormone biosynthesis, if inhibited, blocks the production of all androgenic compounds but does not block the production of glucocorticoids? (A) Cholesterol → pregnenolone (B) Progesterone → 11-deoxycorticosterone (C) 17-Hydroxypregnenolone → dehydroepiandrosterone (D) Testosterone → estradiol (E) Testosterone → dihydrotestosterone
The answer is C [Figure 7.11]. The conversion of 17-hydroxypregnenolone to dehydroepiandrosterone (as well as the conversion of 17-hydroxyprogesterone to androstenedione) is catalyzed by 17,20-lyase. If this process is inhibited, synthesis of androgens is stopped.
A 12-year-old boy has a severe asthmatic attack with wheezing. He experiences rapid breathing and becomes cyanotic. His arterial PO2 is 60 mm Hg and his PCO2 is 30 mm Hg. To treat this patient, the physician should administer (A) an α1-adrenergic antagonist (B) a β1-adrenergic antagonist (C) a β2-adrenergic agonist (D) a muscarinic agonist (E) a nicotinic agonist
The answer is C [II E 3 a (2)]. A cause of airway obstruction in asthma is bronchiolar constriction. β2-adrenergic stimulation (β2-adrenergic agonists) produces relaxation of the bronchioles
11. Which of the following is the site of highest airway resistance? (A) Trachea (B) Largest bronchi (C) Medium-sized bronchi (D) Smallest bronchi (E) Alveoli
The answer is C [II E 4]. The medium-sized bronchi actually constitute the site of highest resistance along the bronchial tree. Although the small radii of the alveoli might predict that they would have the highest resistance, they do not because of their parallel arrangement. In fact, early changes in resistance in the small airways may be "silent" and go undetected because of their small overall contribution to resistance.
8. A 38-year-old man who has galactorrhea is found to have a prolactinoma. His physician treats him with bromocriptine, which eliminates the galactorrhea. The basis for the therapeutic action of bromocriptine is that it (A) antagonizes the action of prolactin on the breast (B) enhances the action of prolactin on the breast (C) inhibits prolactin release from the anterior pituitary (D) inhibits prolactin release from the hypothalamus (E) enhances the action of dopamine on the anterior pituitary
The answer is C [III B 4 a (1), c (2)]. Bromocriptine is a dopamine agonist. The secretion of prolactin by the anterior pituitary is tonically inhibited by the secretion of dopamine from the hypothalamus. Thus, a dopamine agonist acts just like dopamine—it inhibits prolactin secretion from the anterior pituitary.
30. Which of the following sets of data would have the highest rate of O2 transfer in the lungs?
The answer is C [III D]. The diffusion of O2 from alveolar gas to pulmonary capillary blood is proportional to the partial pressure difference for O2 between inspired air and mixed venous blood entering the pulmonary capillaries, proportional to the surface area for diffusion and inversely proportional to diffusion distance or thickness of the barrier.
31. A 16-year-old, seemingly normal female is diagnosed with androgen insensitivity disorder. She has never had a menstrual cycle and is found to have a blind-ending vagina; no uterus, cervix, or ovaries; a 46 XY genotype; and intra-abdominal testes. Her serum testosterone is elevated. Which of the following characteristics is caused by lack of androgen receptors? (A) 46 XY genotype (B) Testes (C) Elevated serum testosterone (D) Lack of uterus and cervix (E) Lack of menstrual cycles
The answer is C [IX C]. The elevated serum testosterone is due to lack of androgen receptors on the anterior pituitary (which normally would mediate negative feedback by testosterone). The presence of testes is due to the male genotype. The lack of uterus and cervix is due to anti-müllerian hormone (secreted by the fetal testes), which suppressed differentiation of the müllerian ducts into the internal female genital tract. The lack of menstrual cycles is due to the absence of a female reproductive tract.
16. A 46-year-old woman has hirsutism, hyperglycemia, obesity, muscle wasting, and increased circulating levels of adrenocorticotropic hormone (ACTH). The most likely cause of her symptoms is (A) primary adrenocortical insufficiency (Addison disease) (B) pheochromocytoma (C) primary overproduction of ACTH (Cushing disease) (D) treatment with exogenous glucocorticoids (E) hypophysectomy
The answer is C [V A 5 b]. This woman has the classic symptoms of a primary elevation of adrenocorticotropic hormone (ACTH) (Cushing disease). Elevation of ACTH stimulates overproduction of glucocorticoids and androgens. Treatment with pharmacologic doses of glucocorticoids would produce similar symptoms, except that circulating levels of ACTH would be low because of negative feedback suppression at both the hypothalamic (corticotropin-releasing hormone [CRH]) and anterior pituitary (ACTH) levels. Addison disease is caused by primary adrenocortical insufficiency. Although a patient with Addison disease would have increased levels of ACTH (because of the loss of negative feedback inhibition), the symptoms would be of glucocorticoid deficit, not excess. Hypophysectomy would remove the source of ACTH. A pheochromocytoma is a tumor of the adrenal medulla that secretes catecholamines.
When a person is standing, blood flow in the lungs is (A) equal at the apex and the base (B) highest at the apex owing to the effects of gravity on arterial pressure (C) highest at the base because that is where the difference between arterial and venous pressure is greatest (D) lowest at the base because that is where alveolar pressure is greater than arterial pressure
The answer is C [VI B]. The distribution of blood flow in the lungs is affected by gravitational effects on arterial hydrostatic pressure. Thus, blood flow is highest at the base, where arterial hydrostatic pressure is greatest and the difference between arterial and venous pressure is also greatest. This pressure difference drives the blood flow
15. Which volume remains in the lungs after a maximal expiration? (A) Tidal volume (VT) (B) Vital capacity (VC) (C) Expiratory reserve volume (ERV) (D) Residual volume (RV) (E) Functional residual capacity (FRC) (F) Inspiratory capacity (G) Total lung capacity
The answer is D [I A 3]. During a forced maximal expiration, the volume expired is a tidal volume (VT) plus the expiratory reserve volume (ERV). The volume remaining in the lungs is the residual volume (RV).
17. A healthy 65-year-old man with a tidal volume (VT) of 0.45 L has a breathing frequency of 16 breaths/min. His arterial PCO2 is 41 mm Hg, and the PCO2 of his expired air is 35 mm Hg. What is his alveolar ventilation? (A) 0.066 L/min (B) 0.38 L/min (C) 5.0 L/min (D) 6.14 L/min (E) 8.25 L/min
The answer is D [I A 5 b, 6 b]. Alveolar ventilation is the difference between tidal volume (VT) and dead space multiplied by breathing frequency. VT and breathing frequency are given, but dead space must be calculated. Dead space is VT multiplied by the difference between arterial PCO2 and expired PCO2 divided by arterial PCO2. Thus, dead space = 0.45 × (41 − 35/41) = 0.066 L. Alveolar ventilation is then calculated as (0.45 L − 0.066 L) × 16 breaths/min = 6.14 L/min.
12. Which of the following inhibits the secretion of growth hormone by the anterior pituitary? (A) Sleep (B) Stress (C) Puberty (D) Somatomedins (E) Starvation (F) Hypoglycemia
The answer is D [III B 3 a]. Growth hormone is secreted in pulsatile fashion, with a large burst occurring during deep sleep (sleep stage 3 or 4). Growth hormone secretion is increased by sleep, stress, puberty, starvation, and hypoglycemia. Somatomedins are generated when growth hormone acts on its target tissues; they inhibit growth hormone secretion by the anterior pituitary, both directly and indirectly (by stimulating somatostatin release).
12. A 49-year-old man has a pulmonary embolism that completely blocks blood flow to his left lung. As a result, which of the following will occur? (A) Ventilation/perfusion (V/Q) ratio in the left lung will be zero (B) Systemic arterial PO2 will be elevated (C) V/Q ratio in the left lung will be lower than in the right lung (D) Alveolar PO2 in the left lung will be approximately equal to the PO2 in inspired air (E) Alveolar PO2 in the right lung will be approximately equal to the PO2 in venous blood
The answer is D [VII B 2]. Alveolar PO2 in the left lung will equal the PO2 in inspired air. Because there is no blood flow to the left lung, there can be no gas exchange between the alveolar air and the pulmonary capillary blood. Consequently, O2 is not added to the capillary blood. The ventilation/perfusion (V/Q) ratio in the left lung will be infinite (not zero or lower than that in the normal right lung) because Q (the denominator) is zero. Systemic arterial PO2 will, of course, be decreased because the left lung has no gas exchange. Alveolar PO2 in the right lung is unaffected.
76-year-old man with lung cancer is lethargic and excreting large volumes of urine. He is thirsty and drinks water almost constantly. Laboratory values reveal an elevated serum Ca2+ concentration of 18 mg/dL, elevated serum osmolarity of 310 mOsm/L, and urine osmolarity of 90 mOsm/L. Administration of an ADH analogue does not change his serum or urine osmolarity. 33. The cause of the patient's excess urine volume is (A) dehydration (B) syndrome of inappropriate ADH (C) central diabetes insipidus (D) nephrogenic diabetes insipidus
The answer is D [III C 1; VII, B 3; Chapter 7, VI]. The man's urine osmolarity is very dilute, while his serum osmolarity is increased. In the face of increased serum osmolarity, there should be increased ADH secretion, which should then act on the collecting duct principal cells to increase water reabsorption and concentrate the urine. The fact that the urine is dilute, not concentrated, suggests that ADH either is absent (central diabetes insipidus) or is ineffective (nephrogenic diabetes insipidus). Administration of an exogenous ADH analogue separates these two possibilities—it was ineffective in changing serum or urine osmolarity; thus, it can be concluded that ADH unable to act on the collecting ducts, that is, nephrogenic diabetes insipidus. One cause of nephrogenic diabetes insipidus is hypercalcemia, which is present in this patent secondary to his lung cancer; he likely has humoral hypercalcemia of malignancy, due to secretion of PTH-rp by the tumor. Dehydration would cause increased ADH secretion and increased urine osmolarity. Syndrome of inappropriate ADH would cause increased urine osmolarity and subsequently decreased serum osmolarity, due to excess water reabsorption.
76-year-old man with lung cancer is lethargic and excreting large volumes of urine. He is thirsty and drinks water almost constantly. Laboratory values reveal an elevated serum Ca2+ concentration of 18 mg/dL, elevated serum osmolarity of 310 mOsm/L, and urine osmolarity of 90 mOsm/L. Administration of an ADH analogue does not change his serum or urine osmolarity. 32. The man's serum ADH level is (A) decreased because excess water-drinking has suppressed ADH secretion (B) decreased because his posterior pituitary is not secreting ADH (C) normal (D) increased because the elevated serum osmolarity has stimulated ADH secretion (E) increased because his extreme thirst has directly stimulated ADH secretion
The answer is D [III C 1; also Chapter 7, VII]. The man is excreting large volumes of dilute urine, which has raised his serum osmolarity and made him very thirsty. The increase in serum osmolarity would then cause an increase in serum ADH levels. The fact that exogenous ADH administration did not change his serum or urine osmolarity suggests that the collecting duct of the nephron is unresponsive to ADH. Thirst does not directly increase ADH secretion.
20. Blood levels of which of the following substances is decreased in Graves disease? (A) Triiodothyronine (T3) (B) Thyroxine (T4) (C) Diiodotyrosine (DIT) (D) Thyroid-stimulating hormone (TSH) (E) Iodide (I−)
The answer is D [IV B 2; Table 7.5]. In Graves disease (hyperthyroidism), the thyroid is stimulated to produce and secrete vast quantities of thyroid hormones as a result of stimulation by thyroid-stimulating immunoglobulins (antibodies to the thyroid-stimulating hormone [TSH] receptors on the thyroid gland). Because of the high circulating levels of thyroid hormones, anterior pituitary secretion of TSH will be turned off (negative feedback).
31. A 48-year-old woman at sea level breaths a gas mixture containing 21% O2. She has the following arterial blood gas values: PaO2 = 60 mm Hg PaCO2 = 45 mm Hg Her measured DLCO is normal. Which of the following is the cause of her hypoxemia? (A) The values demonstrate normal lung function (B) Hypoventilation (C) Fibrosis (D) Carbon monoxide poisoning (E) Right-to-left shunt
The answer is D [IV D]. Since the woman is hypoxemic at sea level and breathing a mixture containing a normal % of O2, she cannot have normal lung function. Also, because she is hypoxemic, she does not have carbon monoxide poisoning (which would decrease O2 content of blood but would not decrease PaO2. Right-to-left shunt as the cause of the woman's hypoxemia is further supported by calculating the A-a gradient as follows. PIO2 = (760 mm Hg − 47 mm Hg) × 0.21 = 150 mm Hg. PaO2 = 150 mm Hg − 45 mm Hg/0.8 = 94 mm Hg. A-a gradient = 94 mm Hg − 60 mm Hg = 34 mm Hg, which is increased and consistent with right-to-left shunt.
25. A 38-year-old woman moves with her family from New York City (sea level) to Leadville, Colorado (10,200 feet above sea level). Which of the following will occur as a result of residing at high altitude? (A) Hypoventilation (B) Arterial PO2 greater than 100 mm Hg (C) Decreased 2,3-diphosphoglycerate (DPG) concentration (D) Shift to the right of the hemoglobin-O2 dissociation curve (E) Pulmonary vasodilation (F) Hypertrophy of the left ventricle (G) Respiratory acidosis
The answer is D [IX B; Table 4.9]. At high altitudes, the PO2 of alveolar air is decreased because barometric pressure is decreased. As a result, arterial PO2 is decreased (<100 mm Hg), and hypoxemia occurs and causes hyperventilation by an effect on peripheral chemoreceptors. Hyperventilation leads to respiratory alkalosis. 2,3-Diphosphoglycerate (DPG) levels increase adaptively; 2,3-DPG binds to hemoglobin and causes the hemoglobin-O2 dissociation curve to shift to the right to improve unloading of O2 in the tissues. The pulmonary vasculature vasoconstricts in response to alveolar hypoxia, resulting in increased pulmonary arterial pressure and hypertrophy of the right ventricle (not the left ventricle).
26. The pH of venous blood is only slightly more acidic than the pH of arterial blood because (A) CO2 is a weak base (B) there is no carbonic anhydrase in venous blood (C) the H+ generated from CO2 and H2O is buffered by HCO3 - in venous blood (D) the H+ generated from CO2 and H2O is buffered by deoxyhemoglobin in venous blood (E) oxyhemoglobin is a better buffer for H+ than is deoxyhemoglobin
The answer is D [V B]. In venous blood, CO2 combines with H2O and produces the weak acid H2CO3, catalyzed by carbonic anhydrase. The resulting H+ is buffered by deoxyhemoglobin, which is such an effective buffer for H+ (meaning that the pK is within 1.0 unit of the pH of blood) that the pH of venous blood is only slightly more acid than the pH of arterial blood. Oxyhemoglobin is a less effective buffer than is deoxyhemoglobin.
27. A 39-year-old man with untreated diabetes mellitus type I is brought to the emergency room. An injection of insulin would be expected to cause an increase in his (A) urine glucose concentration (B) blood glucose concentration (C) blood K+ concentration (D) blood pH (E) breathing rate
The answer is D [VI C 3; Table 7.7]. Before the injection of insulin, the woman would have had hyperglycemia, glycosuria, hyperkalemia, and metabolic acidosis with compensatory hyperventilation. The injection of insulin would be expected to decrease her blood glucose (by increasing the uptake of glucose into the cells), decrease her urinary glucose (secondary to decreasing her blood glucose), decrease her blood K+ (by shifting K+ into the cells), and correct her metabolic acidosis (by decreasing the production of ketoacids). The correction of the metabolic acidosis will lead to an increase in her blood pH and will reduce her compensatory hyperventilation.
6. A 41-year-old woman has hypocalcemia, hyperphosphatemia, and decreased urinary phosphate excretion. Injection of parathyroid hormone (PTH) causes an increase in urinary cyclic adenosine monophosphate (cAMP). The most likely diagnosis is (A) primary hyperparathyroidism (B) vitamin D intoxication (C) vitamin D deficiency (D) hypoparathyroidism after thyroid surgery (E) pseudohypoparathyroidism
The answer is D [VII B 3 b]. Low blood [Ca2+] and high blood [phosphate] are consistent with hypoparathyroidism. Lack of parathyroid hormone (PTH) decreases bone resorption, decreases renal reabsorption of Ca2+, and increases renal reabsorption of phosphate (causing low urinary phosphate). Because the patient responded to exogenous PTH with an increase in urinary cyclic adenosine monophosphate (cAMP), the G protein coupling the PTH receptor to adenylate cyclase is apparently normal. Consequently, pseudohypoparathyroidism is excluded. Vitamin D intoxication would cause hypercalcemia, not hypocalcemia. Vitamin D deficiency would cause hypocalcemia and hypophosphatemia.
A 12-year-old boy has a severe asthmatic attack with wheezing. He experiences rapid breathing and becomes cyanotic. His arterial PO2 is 60 mm Hg and his PCO2 is 30 mm Hg. 4. Which of the following statements about this patient is most likely to be true? (A) Forced expiratory volume1/forced vital capacity (FEV1/FVC) is increased (B) Ventilation/perfusion (V/Q) ratio is increased in the affected areas of his lungs (C) His arterial PCO2 is higher than normal because of inadequate gas exchange (D) His arterial PCO2 is lower than normal because hypoxemia is causing him to hyperventilate (E) His residual volume (RV) is decreased
The answer is D [VIII B 2 a]. The patient's arterial PCO2 is lower than the normal value of 40 mm Hg because hypoxemia has stimulated peripheral chemoreceptors to increase his breathing rate; hyperventilation causes the patient to blow off extra CO2 and results in respiratory alkalosis. In an obstructive disease, such as asthma, both forced expiratory volume (FEV1) and forced vital capacity (FVC) are decreased, with the larger decrease occurring in FEV1. Therefore, the FEV1/FVC ratio is decreased. Poor ventilation of the affected areas decreases the ventilation/perfusion (V/Q) ratio and causes hypoxemia. The patient's residual volume (RV) is increased because he is breathing at a higher lung volume to offset the increased resistance of his airways.
24. A 42-year-old woman with severe pulmonary fibrosis is evaluated by her physician and has the following arterial blood gases: pH = 7.48, PaO2 = 55 mm Hg, and PaCO2 = 32 mm Hg. Which statement best explains the observed value of PaCO2? (A) The increased pH stimulates breathing via peripheral chemoreceptors (B) The increased pH stimulates breathing via central chemoreceptors (C) The decreased PaO2 inhibits breathing via peripheral chemoreceptors (D) The decreased PaO2 stimulates breathing via peripheral chemoreceptors (E) The decreased PaO2 stimulates breathing via central chemoreceptors
The answer is D [VIII B; Table 4.7]. The patient's arterial blood gases show increased pH, decreased PaO2, and decreased PaCO2. The decreased PaO2 causes hyperventilation (stimulates breathing) via the peripheral chemoreceptors, but not via the central chemoreceptors. The decreased PaCO2 results from hyperventilation (increased breathing) and causes increased pH, which inhibits breathing via the peripheral and central chemoreceptors.
28. A person with a ventilation/perfusion (V/Q) defect has hypoxemia and is treated with supplemental O2. The supplemental O2 will be most helpful if the person's predominant V/Q defect is (A) dead space (B) shunt (C) high V/Q (D) low V/Q (E) V/Q = 0 (F) V/Q = ∞
The answer is D [VII]. Supplemental O2 (breathing inspired air with a high PO2) is most helpful in treating hypoxemia associated with a ventilation/perfusion (V/Q) defect if the predominant defect is low V/Q. Regions of low V/Q have the highest blood flow. Thus, breathing high PO2 air will raise the PO2 of a large volume of blood and have the greatest influence on the total blood flow leaving the lungs (which becomes systemic arterial blood). Dead space (i.e., V/Q = ∞) has no blood flow, so supplemental O2 has no effect on these regions. Shunt (i.e., V/Q = 0) has no ventilation, so supplemental O2 has no effect. Regions of high V/Q have little blood flow, thus raising the PO2 of a small volume of blood will have little overall effect on systemic arterial blood.
14. Which of the following explains the suppression of lactation during pregnancy? (A) Blood prolactin levels are too low for milk production to occur (B) Human placental lactogen levels are too low for milk production to occur (C) The fetal adrenal gland does not produce sufficient estriol (D) Blood levels of estrogen and progesterone are high (E) The maternal anterior pituitary is suppressed
The answer is D [X F 5]. Although the high circulating levels of estrogen stimulate prolactin secretion during pregnancy, the action of prolactin on the breast is inhibited by progesterone and estrogen. After parturition, progesterone and estrogen levels decrease dramatically. Prolactin can then interact with its receptors in the breast, and lactation proceeds if initiated by suckling
Which of the following lung volumes or capacities can be measured by spirometry? (A) Functional residual capacity (FRC) (B) Physiologic dead space (C) Residual volume (RV) (D) Total lung capacity (TLC) (E) Vital capacity (VC)
The answer is E [I A 4, 5, B 2, 3, 5]. Residual volume (RV) cannot be measured by spirometry. Therefore, any lung volume or capacity that includes the RV cannot be measured by spirometry. Measurements that include RV are functional residual capacity (FRC) and total lung capacity (TLC). Vital capacity (VC) does not include RV and is, therefore, measurable by spirometry. Physiologic dead space is not measurable by spirometry and requires sampling of arterial PCO2 and expired CO2.
Which volume remains in the lungs after a tidal volume (VT) is expired? (A) Tidal volume (VT) (B) Vital capacity (VC) (C) Expiratory reserve volume (ERV) (D) Residual volume (RV) (E) Functional residual capacity (FRC) (F) Inspiratory capacity (G) Total lung capacity
The answer is E [I B 2]. During normal breathing, the volume inspired and then expired is a tidal volume (VT). The volume remaining in the lungs after expiration of a VT is the functional residual capacity (FRC).
Which of the following is true during inspiration? (A) Intrapleural pressure is positive (B) The volume in the lungs is less than the functional residual capacity (FRC) (C) Alveolar pressure equals atmospheric pressure (D) Alveolar pressure is higher than atmospheric pressure (E) Intrapleural pressure is more negative than it is during expiration
The answer is E [II F 2]. During inspiration, intrapleural pressure becomes more negative than it is at rest or during expiration (when it returns to its less negative resting value). During inspiration, air flows into the lungs when alveolar pressure becomes lower (due to contraction of the diaphragm) than atmospheric pressure; if alveolar pressure were not lower than atmospheric pressure, air would not flow inward. The volume in the lungs during inspiration is the functional residual capacity (FRC) plus one tidal volume (VT).
9. Which of the following hormones originates in the anterior pituitary? (A) Dopamine (B) Growth hormone-releasing hormone (GHRH) (C) Somatostatin (D) Gonadotropin-releasing hormone (GnRH) (E) Thyroid-stimulating hormone (TSH) (F) Oxytocin (G) Testosterone
The answer is E [III B; Table 7.1]. Thyroid-stimulating hormone (TSH) is secreted by the anterior pituitary. Dopamine, growth hormone-releasing hormone (GHRH), somatostatin, and gonadotropin-releasing hormone (GnRH) all are secreted by the hypothalamus. Oxytocin is secreted by the posterior pituitary. Testosterone is secreted by the testes.
26. A 61-year-old woman with hyperthyroidism is treated with propylthiouracil. The drug reduces the synthesis of thyroid hormones because it inhibits oxidation of (A) triiodothyronine (T3) (B) thyroxine (T4) (C) diiodotyrosine (DIT) (D) thyroid-stimulating hormone (TSH) (E) iodide (I−)
The answer is E [IV A 2]. For iodide (I−) to be "organified" (incorporated into thyroid hormone), it must be oxidized to I2, which is accomplished by a peroxidase enzyme in the thyroid follicular cell membrane. Propylthiouracil inhibits peroxidase and, therefore, halts the synthesis of thyroid hormones
29. Which step in steroid hormone biosynthesis occurs in the accessory sex target tissues of the male and is catalyzed by 5α-reductase? (A) Cholesterol → pregnenolone (B) Progesterone → 11-deoxycorticosterone (C) 17-Hydroxypregnenolone → dehydroepiandrosterone (D) Testosterone → estradiol (E) Testosterone → dihydrotestosterone
The answer is E [IX A]. Some target tissues for androgens contain 5α- reductase, which converts testosterone to dihydrotestosterone, the active form in those tissues.
21. Which of the following hormones acts by an inositol 1,4,5-triphosphate (IP3)-Ca2+ mechanism of action? (A) 1,25-Dihydroxycholecalciferol (B) Progesterone (C) Insulin (D) Parathyroid hormone (PTH) (E) Gonadotropin-releasing hormone (GnRH)
The answer is E [Table 7.2]. Gonadotropin-releasing hormone (GnRH) is a peptide hormone that acts on the cells of the anterior pituitary by an inositol 1,4,5-triphosphate (IP3)-Ca2+ mechanism to cause the secretion of folliclestimulating hormone (FSH) and luteinizing hormone (LH). 1,25- Dihydroxycholecalciferol and progesterone are steroid hormone derivatives of cholesterol that act by inducing the synthesis of new proteins. Insulin acts on its target cells by a tyrosine kinase mechanism. Parathyroid hormone (PTH) acts on its target cells by an adenylate cyclase-cyclic adenosine monophosphate (cAMP) mechanism.
17. Which of the following decreases the conversion of 25- hydroxycholecalciferol to 1,25-dihydroxycholecalciferol? (A) A diet low in Ca2+ (B) Hypocalcemia (C) Hyperparathyroidism (D) Hypophosphatemia (E) Chronic renal failure
The answer is E [VII C 1]. Ca2+ deficiency (low Ca2+ diet or hypocalcemia) activates 1α-hydroxylase, which catalyzes the conversion of vitamin D to its active form, 1,25-dihydroxycholecalciferol. Increased parathyroid hormone (PTH) and hypophosphatemia also stimulate the enzyme. Chronic renal failure is associated with a constellation of bone diseases, including osteomalacia caused by failure of the diseased renal tissue to produce the active form of vitamin D.
76-year-old man with lung cancer is lethargic and excreting large volumes of urine. He is thirsty and drinks water almost constantly. Laboratory values reveal an elevated serum Ca2+ concentration of 18 mg/dL, elevated serum osmolarity of 310 mOsm/L, and urine osmolarity of 90 mOsm/L. Administration of an ADH analogue does not change his serum or urine osmolarity. 34. The most appropriate treatment is (A) ADH antagonist (B) ADH analogue (C) PTH analogue (D) half-normal saline (E) pamidronate plus furosemide
The answer is E [VII, B 3]. The man's nephrogenic diabetes insipidus is caused by hypercalcemia secondary to increased PTH-rp secreted by his lung tumor. PTH-rp has all of the actions of PTH, including increased bone resorption, increased renal Ca2+ reabsorption, and decreased renal phosphate reabsorption; all of these actions lead to increased serum Ca2+ concentration. The treatment should be directed at lowering serum Ca2+ concentration, which can be achieved by giving an inhibitor of bone resorption (e.g., pamidronate) and an inhibitor of renal Ca2+ reabsorption (furosemide). Giving an ADH antagonist would be ineffective because the man's nephrogenic diabetes insipidus has made his collecting ducts insensitive to ADH. Giving half-normal saline could lower his serum osmolarity temporarily, but would not address the underlying problem of hypercalcemia.
19. Hypoxemia produces hyperventilation by a direct effect on the (A) phrenic nerve (B) J receptors (C) lung stretch receptors (D) medullary chemoreceptors (E) carotid and aortic body chemoreceptors
The answer is E [VIII B 2]. Hypoxemia stimulates breathing by a direct effect on the peripheral chemoreceptors in the carotid and aortic bodies. Central (medullary) chemoreceptors are stimulated by CO2 (or H+). The J receptors and lung stretch receptors are not chemoreceptors. The phrenic nerve innervates the diaphragm, and its activity is determined by the output of the brainstem breathing center.
A 35-year-old man has a vital capacity (VC) of 5 L, a tidal volume (VT) of 0.5 L, an inspiratory capacity of 3.5 L, and a functional residual capacity (FRC) of 2.5 L. What is his expiratory reserve volume (ERV)? (A) 4.5 L (B) 3.9 L (C) 3.6 L (D) 3.0 L (E) 2.5 L (F) 2.0 L (G) 1.5 L
The answer is G [I A 3; Figure 4.1]. Expiratory reserve volume (ERV) equals vital capacity (VC) minus inspiratory capacity [inspiratory capacity includes tidal volume (VT) and inspiratory reserve volume (IRV)].
23. The source of estrogen during the second and third trimesters of pregnancy is the (A) corpus luteum (B) maternal ovaries (C) fetal ovaries (D) placenta (E) maternal ovaries and fetal adrenal gland (F) maternal adrenal gland and fetal liver (G) fetal adrenal gland, fetal liver, and placenta
The answer is G [X F 3]. During the second and third trimesters of pregnancy, the fetal adrenal gland synthesizes dehydroepiandrosterone sulfate (DHEA-S), which is hydroxylated in the fetal liver and then transferred to the placenta, where it is aromatized to estrogen. In the first trimester, the corpus luteum is the source of both estrogen and progesterone.
19. Which of the following would be expected in a patient with Graves disease? (A) Cold sensitivity (B) Weight gain (C) Decreased O2 consumption (D) Decreased cardiac output (E) Drooping eyelids (F) Atrophy of the thyroid gland (G) Increased thyroid-stimulating hormone (TSH) levels (H) Increased triiodothyronine (T3) levels
The answer is H [IV B 2; Table 7.5]. Graves disease (hyperthyroidism) is caused by overstimulation of the thyroid gland by circulating antibodies to the thyroid-stimulating hormone (TSH) receptor (which then increases the production and secretion of triiodothyronine [T3] and thyroxine [T4], just as TSH would). Therefore, the signs and symptoms of Graves disease are the same as those of hyperthyroidism, reflecting the actions of increased circulating levels of thyroid hormones: increased heat production, weight loss, increased O2 consumption and cardiac output, exophthalmos (bulging eyes, not drooping eyelids), and hypertrophy of the thyroid gland (goiter). TSH levels will be decreased (not increased) as a result of the negative feedback effect of increased T3 levels on the anterior pituitary.
