Clerkship Qstream Flashcards
An outbreak of bloody diarrhea accompanied by fever occurred at a day care center. Gram-negative bacilli that do not ferment lactose were isolated from 7 of 10 children with the illness. No common food had been shared among the children. However, an iguana was brought to the infected children’s classroom three days previously as part of a show-and-tell assignment and was confirmed by culture as the source of the infection.
Which ONE of the following is the MOST LIKELY cause of the outbreak?
Campylobacter jejuni
Salmonella enterica serovar Marina
Escherichia coli O157:H7
Shigella flexneri
The correct answer is Salmonella enterica serovar Marina
Explanation: Salmonella is a zoonotic pathogen that is a frequent cause of blood diarrhea in humans. Contact with reptiles such as turtles and iguanas has been associated with infection.
Other answers: All of the other agents listed can cause bloody diarrhea. Diarrhea associated with E. coli O157:H7 is not as likely as the others to be accompanied by fever, and is usually associated with undercooked beef or leafy vegetables. Campylobacter is generally contracted from food that has been contaminated with raw poultry. Shigella is a strictly human pathogen.
A 53-year-old Army colonel reported passing bright red blood per rectum. On colonoscopy, a sessile 3.1 cm diameter mass was observed protruding into the distal sigmoid colon.
This lesion is most likely which of the following?
Answer: Adenomatous polyp.
Explanation: The patient’s clinical history is most consistent with a left-sided intestinal polyp. An adenomatous polyp arises from the epithelial surface of the gastrointestinal tract. Adenomatous polyps are not a frequent cause of lower GI bleed, but they can present in this fashion.
Other answers: Meckel’s diverticulum is a congenital diverticulum in the ileum resulting from incomplete closure of the omphalomesenteric duct. Approximately half of Meckel’s diverticula have gastric mucosa. GI bleeding can occur in association with Meckel’s diverticula because the acid produced from the gastric mucosa in the diverticulum can produce an ulcer that bleeds. Meckel’s is not the correct answer because 1) it is not located in the colon, 2) it does not typically cause bright red blood per rectum (because the bleeding source is more proximal, the blood is either black or maroon in color in the stool), and 3) it typically presents in childhood.
A teratoma is a neoplasm originating in the testis, ovary or, rarely, the mediastinum. The neoplasm contains recognizable mature or immature cells or tissues representative of one or more germ cell layers. Usually all three germ cell layers are represented in the neoplasm.
A lipoma is a benign neoplasm of adipocytes. It can occur almost anywhere, and can occur in the colon. However, it is very unusual for intestinal lipomas to bleed. When intestinal lipomas become very bulky, they can cause intususseption and obstruction.
Robbins PBOD 8ed. pp 260-262
A 70 kg man who has been exhibiting symptoms of a stroke for 30 minutes is brought to the emergency room. You determine that administration of the thrombolytic agent alteplase is the appropriate course of treatment. The volume of distribution of alteplase is 0.1 L/kg, and its elimination half-life is 5 minutes. The product instructions state that an intravenous bolus dose of 70 μg/kg should be administered, followed by an intravenous infusion of 700 μg/kg over the next 60 minutes. What is the serum concentration immediately after the bolus dose?
4.9 mg/L
70 ug/L
700 ug/L
460 ug/L
The correct answer is 700 µg/L.
Explanation: The question asks for the serum concentration after administering a loading dose of 70 µg/kg.
To determine the serum concentration immediately after the bolus infusion of the drug, use the equation: Loading dose = Vd x TC, where Vd is the volume of distribution (the volume in which the drug will be diluted) and TC is the target concentration or the diluted concentration of the drug.
A simple way to think of this equation is by using the following units:
Loading dose in µg = Vd in L x TC in µg/L.
However, people of different sizes are going to have different volumes of distribution. So the units of Vd are often given as L/kg. Similarly, as the loading dose of a drug will also depend on the size of a person, loading dose is usually referred to as µg/kg.
Thus, we end up with the equation Loading dose in µg/kg = Vd in L/kg x Tc in µg/L.
In the question above, the Vd is given as 0.1 L/kg and the loading dose (the initial IV bolus) is given as 70 µg/kg. Thus, Tc = (70 µg/kg) / (0.1L/kg) = 700 µg/L.
In a study of trivalent inactivated influenza vaccine in adults aged 18–64 years, 24 of 2,000 vaccinated participants developed influenza, compared with 72 of 2,000 who received placebo. What is the relative risk of influenza in the vaccinated group compared to controls?
- 012
- 67
- 33
- 0
0.33
A young woman is brought to the emergency department by her friends, who say she participated in a witchcraft ritual in which she ingested belladonna potion (which contains atropine). She has a beet red face and dilated pupils. She is psychotic and difficult to handle and repeatedly pulls out her intravenous line containing diazepam. To reduce the belladonna-induced psychosis, you administer:
physostigmine
pralidoxime (2-PAM)
clozapine
pyridostigmine
The correct answer is physostigmine.
Explanation: To counter the blockade of muscarinic receptors by atropine, you should administer a cholinesterase inhibitor that can enter the brain. The cholinesterase inhibitor will bring about an elevation of endogenous acetylcholine levels that can competitively overcome the blockade of cholinergic muscarinic receptors by atropine. Since the patient has psychotic symptoms, an antidote that can reach the brain is needed. Although both physostigmine and pyridostigmine are anticholinesterases, only physostigmine can rapidly cross the blood-brain barrier.
Incorrect answers:
Clozapine is an atypical antipsychotic drug that binds to serotonergic and dopamine receptors. Additionally, clozapine can antagonize some cholinergic receptors, and thus could potentially worsen the effects of the belladonna potion.
Pralidoxime reverses acetylcholinesterase inhibition caused by organophosphate poisoning. By rescuing acetylcholinesterase, pralidoxime increases acetylcholine breakdown, and thus works as an anticholinergic agent. Since atropine is a competitive inhibitor of muscarinic acetylcholine receptors, pralidoxime would augment the toxic effects of belladonna (which contains atropine).
Pyridostigmine does not cross the BBB and therefore would not elevate acetylcholine levels in the brain. Thus, pyridostigmine would reverse the peripheral effects of atropine but would not reverse the central effects of atropine.
A high incidence of skin infections occurred among members of a high school wrestling team. Several of the teenagers were treated with intravenous nafcillin or oral dicloxacillin (semisynthetic penicillins related to methicillin), but the infections did not resolve. Gram-positive bacteria were isolated from pus within the inflamed skin lesions and the anterior nares of two team members, and bacteria of similar morphology were isolated from the wrestling mats and a tube of taping gel in the locker room.
All of the isolates were resistant to methicillin. Which ONE of the following is responsible for methicillin resistance in this organism?
An rRNA gene that underwent spontaneous mutation
Acquistion of a plasmid that encodes β-lactamase
Integration of an altered penicillin-binding protein PBP2a gene
An altered RNA polymerase
The correct answer is Integration of an altered penicillin-binding protein PBP2a gene.
Explanation: Most Staphylococcus aureus strains are resistant to penicillin by virtue of beta-lactamase production. Methicillin was developed as an antibiotic to overcome beta-lactamase. Like other antibiotics in the class, methicillin blocks the transpeptidation that cross-links the bacterial cell wall because it is an analogue to D-ala D-ala. Over time, resistance to methicillin has emerged because S. aureus acquired the mecA gene via a transposon (passed among S. aureus strains via conjugation). The mecA gene encodes for an alternative transpeptidase, penicillin-binding protein PBP2a, that has decreased affinity for methicillin. S. aureus strains with the mecA gene (PBP2a) are resistant to methicillin.
A 55-year-old female is admitted to an outpatient clinic with a 39°C fever and flank pain. She was recently discharged from a 3-day hospital visit during which she was catheterized. Many leukocytes and several white blood cell casts are seen in a microscopic examination of her urine. A Gram-negative organism that demonstrates a spreading morphology on blood agar was isolated in pure culture from her urine.
Which ONE of the following is the MOST LIKELY diagnosis?
Uncomplicated cystitis
Pelvic inflammatory disease
Pyelonephritis
Urethritis
The correct answer is Pyelonephritis.
Explanation: This patient was at risk of acquiring a urinary tract infection when hospitalized because she was catheterized. The symptoms of fever and flank pain are consistent with pyelonephritis and the appearance of renal casts, along with bacteria and white blood cells in the urine provides evidence that the urinary tract infection has ascended to the kidneys. Furthermore, the presence of a Gram-negative bacillus that exhibits spreading motility on blood agar suggests that Proteus mirabilis is the causative agent. P. mirabilis is notorious for its ability to persist in the urinary tract through pili and to ascend the ureters from the bladder due to its hyper-motility provided by this swarming mechanism.
A 45 year old man with long standing gastric acid reflux presents to his physician with a complaint of dysphagia. A biopsy of the lower third of the esophagus is performed. Examination of the biopsy by a pathologist shows the presence of columnar epithelium replacing the stratified squamous epithelium which typically lines the distal esophagus.
Which of the following choices describes the process present?
metaplasia
atrophy
dysplasia
hyperplasia
Correct answer: Metaplasia.
The question describes the replacement of the normal squamous epithelium of the distal esophagus with columnar epithelium. This phenomenon is known as metaplasia, defined as a change in a tissue from one cell type to another. Metaplasia of the distal esophagus from normal squamous epithelium to columnar epithelium occurs as a result of chronic gastroesophageal reflux.
Other answers: Atrophy refers to a decrease in size due to decrease in cell number or cell size. Dysplasia refers to changes in the epithelium where cells are pleomorphic and lose their normal orientation. Hyperplasia refers to an increase in cell number. Hypertrophy refers to an increase in cell size. Robbins PBOD 8ed. pp 10-11
During cold weather, many families use space heaters to stay warm. Space heaters, improperly used, lead to an increased risk of carbon monoxide poisoning.
You obtain a CBC, a pulse oximetry reading at the bedside, and an arterial blood gas (ABG) run on a co-oximeter. Which of the following best characterizes the findings one would expect in the blood of a patient suffering from acute carbon monoxide poisoning?
Normal hemoglobin levels, normal SaO2 by pulse oximetry, normal SaO2 and low PaO2 by ABG
Low hemoglobin levels, low SaO2 by pulse oximetry, low SaO2 and normal PaO2 by ABG
Low hemoglobin levels, normal SaO2 by pulse oximetry, normal SaO2 and low PaO2 by ABG
Normal hemoglobin levels, normal SaO2 by pulse oximetry, low SaO2 and normal PaO2 by ABG
Answer: Normal hemoglobin levels, normal SaO2 by pulse oximetry, low SaO2 and normal PaO2 by ABG
Hemoglobin has a higher affinity for CO than for oxygen. Therefore, CO binds to hemoglobin in erythrocytes and blocks oxygen from binding. Because CO blocks oxygen binding to hemoglobin, oxygen saturation of hemoglobin is low. CO poisoning does not affect hemoglobin levels. The amount of dissolved oxygen in the blood, measured as the partial pressure of oxygen on an arterial blood gas, is also unaffected by CO.
A pulse oximeter measures the percentage of hemoglobin which is saturated with oxygen (SaO2). However, a standard bedside pulse oximeter is unable to differentiate between oxyhemoglobin and carboxyhemoglobin. Therefore, standard bedside pulse oximeters typically do not read low in cases of carbon monoxide poisoning even though the true oxygen saturation of hemoglobin is low in this setting. In contrast to standard pulse oximetry, certain blood gas analyzers have co-oximeters that can directly measure concentrations of oxygenated hemoglobin, deoxygenated hemoglobin, carboxyhemoglobin, and methemoglobin as percentages of total hemoglobin. These units will report low oxygen saturation of hemoglobin in patients with CO poisoning. Please note, though, that not all ABG machines have co-oximeter functionality. Standard blood gas analyzers that lack co-oximetry calculate oxygen saturation of hemoglobin by inputting measured values for PaO2 and PH into standard oxygen dissociation curves. Like pulse oximeters, standard ABG analyzers may thus report erroneously normal saturation values in patients with CO poisoning.
For these reasons, when considering CO poisoning in a patient it is important to ask the respiratory technicians who run the ABG machines whether ABG values for oxygenated hemoglobin are being calculated or directly measured. At WRNMMC, the ABG machines in the ICU have co-oximetry capability to directly measure oxygen saturation. Thus, the oxygen saturation values reported from ABGs at WRNMMC are often directly measured values. However, for a week in February of 2013 the co-oximetry function on the ABG machines in the ICU at WRNMMC wasn’t working, and so the oxygen saturation values reported from the machines were calculated. So, in cases of suspected CO poisoning, it’s prudent to ask how the O2 saturation is being determined (calculated or measured) on the ABG machine that day.
The beginning of implantation occurs when the conceptus is at which of the following stages?
2-cell zygote
Morula
Blastocyst
4-cell zygote
Explanation: The blastocyst implants into the uterine lining at ~6 days. The human blastocyst contains from 70-100 cells. A blastocyst has an inner cell mass (the embryoblast), which will develop into the embryo, and an outer cell mass (the trophoblast), which will develop into a large part of the placenta.
The other options are earlier stages in embryonic development. During these stages of development the conceptus is located in the Fallopian tube. Implantation in the Fallopian tube is abnormal and may lead to an ectopic pregnancy. The morula has 32 cells, and is the point at which a zygote resembles a mulberry (from the Latin “morus” for mulberry). The morula is the stage immediately before the blastocyst.
An infant, born at 26 weeks gestation, is rushed into the intensive care unit at the hospital with significant respiratory difficulties. The consequences of the premature delivery on the respiratory system include all of the following except
severe arterial hypoxemia.
areas of atelectasis.
difficulty with inflation of the lungs.
atrophy of the respiratory muscles.
Ans: atrophy of the respiratory muscles
The premature delivery occurs before the full development of the lung alveolar structure, including importantly the Type II alveolar cells and the secretion of surfactant. This occurs at about 26 weeks (the terminal sac stage). Because surfactant is absent, surface tension is high and the normal inflation of the lungs is difficult. This increases areas of atelectasis in the lungs, reduces the exchange of gases, and reduce the levels of oxygen in the blood. Muscle development would be expected to be normal for this developmental stage and not be atrophied.
A 22 year old soldier presents with an acute abdomen with rebound tenderness . At exploratory laparotomy infarcted bowel is found secondary to a volvulus. The type of infarct that would result from this type of obstruction is due to which of the following mechanisms?
Proliferation of the endothelium at the site of the volvulus
Venous compression
Activation of the clotting cascade distal to the obstruction
Liquefactive necrosis in the infarcted area
Correct answer: Venous compression.
Explanation: The volvulus will preferentially compress the thinner-walled veins, occluding venous return, while arterial perfusion continues (even if to a lesser degree).
Other answers: In epithelial injury, the clotting cascade is activated at the site of injury, not distally. Liquefactive necrosis is seen in abscess formation and in ischemic injury to the central nervous system. A volvulus constricts vascular flow due to pressure of the twisting of the intestines, not a hyperplastic process. Robbins PBOD 8ed. pp 127-129.
A 45-year-old female received a nicotinic acetylcholine receptor antagonist to produce neuromuscular blockade during a major surgical procedure. At the end of the surgery administration of neostigmine and atropine to this patient will MOST LIKELY
enhance neuromuscular blockade and reduce the bradycardia caused by neostigmine.
enhance neuromuscular blockade and reduce the tachycardia caused by neostigmine.
reverse neuromuscular blockade and reduce the bradycardia caused by neostigmine.
reverse neuromuscular blockade and reduce the tachycardia caused by neostigmine.
The correct answer is , reverse neuromuscular blockade and reduce the bradycardia caused by neostigmine.
Explanation: To overcome the nicotinic blockade following the surgery, neostigmine is given to inhibit acetylcholinesterase, elevate acetylcholine at the neuromuscular junction, and overcome the nicotinic blockade. However, acetylcholine will also be elevated at muscarinic receptors where there is no block. Therefore, a muscarinic antagonist, like atropine is given to prevent the elevated level of acetylcholine from acting at M2 muscarinic receptors on the heart.
A 30 year old G1P0 woman presents to her physician at 30 weeks gestation with chief complaint of swollen legs and increased weight gain. On physical examination, her blood pressure is elevated. Urinalysis reveals the presence of protein. Labor is induced and a small for gestational age infant is delivered. On examination of the placenta, there is extensive infarction. Which of the following types of necrosis would be seen in the spiral arterioles that supply the placenta?
coagulative
fibrinoid
caseous
fat
The correct answer is Fibrinoid necrosis.
Explanation: Development of hypertension and proteinuria after 20 weeks of gestation is most consistent with a diagnosis of pre-eclampsia. While swollen legs can occur for a number of reasons during pregnancy, peripheral edema is also often a component of pre-eclampsia (though not necessary for the diagnosis). The exact mechanisms underlying pre-eclampsia remain unknown, but it is believed that both placental hypoxia and inadequate immune tolerance towards the placenta (resulting in increased maternal immune response to paternal antigens in the placenta) are involved. Pre-eclampsia can progress to eclampsia, a life-threatening condition which includes the development of generalized tonic-clonic seizures in the mother. The only cure for pre-eclampsia is delivery by either Caeserean section or induction of labor. Examination of the spiral arteries in pre-eclampsia typically reveals fibrinoid necrosis. Fibrinoid necrosis is characteristic of immune–mediated injuries.
Other answers: Caseous necrosis is associated with granulomatous inflammation associated with tuberculosis and fungal infections. Coagulative necrosis is seen in organs, except for the central nervous system, when there is ischemic injury. Fat necrosis is seen in acute pancreatitis. Robbins PBOD 8ed page 16
A 50 yo male presents for a routine clinic visit. He is a nonsmoker, drinks 2-3 glasses of wine a week, and is mildly overweight with a BMI of 28. His systolic blood pressure has been running in the high 130’s to low 140’s with diastolic pressures in the mid 80’s for the past few years. He has been trying to improve his diet and has only been exercising sporadically. Which of the following would include only evidence-based US Preventive Services Task Force recommended (A or B) clinical preventive services for this patient?
Screen for Type 2 Diabetes
Screen for abdominal aortic aneurysm
Screen for prostate cancer with a PSA
Screen for heart disease with an ECG
Correct answer: A, screen for diabetes.
Explanation: The USPSTF recommends screening for type 2 diabetes in asymptomatic adults with sustained blood pressure (either treated or untreated) greater than 135/80 mm Hg, a “B” recommendation.
The USPSTF recommends one-time screening for abdominal aortic aneurysm (AAA) with ultrasonography in men ages 65 to 75 years who have ever smoked, a “B” recommendation, but no recommendation exists for this patient’s age.
The U.S. Preventive Services Task Force (USPSTF) recommends against prostate-specific antigen (PSA)-based screening for prostate cancer, a “D” recommendation, which means that the harms outweigh the benefits.
The USPSTF recommends against screening with resting or exercise electrocardiography (ECG) for the prediction of coronary heart disease (CHD) events in asymptomatic adults at low risk for CHD events, a “D” recommendation. Even if you think he is at high risk for CHD, The USPSTF concludes that the current evidence is insufficient to assess the balance of benefits and harms of screening with resting or exercise ECG for the prediction of CHD events in asymptomatic adults at intermediate or high risk for CHD events, an “I” recommendation.
A dehydrated 25 year old serviceman comes into the clinic with acute diarrhea and leg cramps having returned three days prior from deployment to Haiti. You suspect Cholera may be the cause and immediately start intravenous fluids. The toxin released by Vibrio cholerae causes the symptoms of Cholera by covalently modifying which signaling molecule?
The monomeric G-protein Ras
The second messenger cAMP
The guanine nucleotide exchange factor SOS
A subunit of a heterotrimeric G protein
Answer: a subunit of a heterotrimeric G protein
Cholera toxin acts by covalently modifying the Gαs subunit of a heterotrimeric G-protein. This modification prevents the Gαs subunit from hydrolyzing GTP to GDP, and thus the Gαs subunit is always “ON” and the signal cannot be terminated. This causes a rise in cAMP level, activation of Protein Kinase A, and subsequent dephosphorylation of the cystic fibrosis transmembrane conductance regulator (CFTR) chloride channel, causing efflux of chloride ions out of the cells followed by secretion of water, sodium, potassium, and bicarbonate into the intestinal lumen. This results in massive diarrhea, with a loss of up to two liters of fluid per hour.
A 30-year-old male presents in the emergency room complaining of chest pain and shortness of breath. The triage nurse orders a chest radiograph prior to your examination of the patient. You look at his portable chest radiograph on the way to the exam room and see significant leftward shift of the mediastinum and no lung markings on the right side. He is hypoxic with distended vessels along the right side of his neck and there are diminished breath sounds over his right hemithorax. Your next best step would be:
Place a left anterior axillary line thoracostomy tube
Perform a right anterior chest wall needle decompression
Obtain a chest CT to evaluate for vascular compression
Intubate the patient for ventilator support
CORRECT ANSWER: Perform a right anterior chest wall needle decompression of a pneumothorax is the correct answer for this question.
EXPLANATION: The complaint of chest pain and shortness of breath combined with hypoxia, jugular venous distention and diminished breath sounds over the right hemithorax suggest a pneumothorax. The chest radiograph demonstrating a right hemithorax lucency (ie. no lung markings) with leftward shift of the mediastinal structures further implies a tension pneumothorax. Although a thoracostomy tube will likely eventually become necessary, the best next step would be an emergent right-sided needle decompression along the anterior chest wall at the midclavicular line over the 3rd rib using a large bore needle.
OTHER ANSWERS: The clinical presentation, physical exam, and radiographic findings all suggest a right-sided pneumothorax. Intubating the patient for ventilator support will not decompress the pneumothorax that exists within the right hemithorax. This option would also be premature in a patient that is still conscious not requiring ventilator support.
Obtaining a chest CT will end up wasting valuable time further delaying decompression of the patient’s pneumothorax. A chest CT might be considered if there is diagnostic confusion in a clinically stable patient.
All of the findings listed in the question stem indicate a right-sided pneumothorax. Placing a thoracostomy tube on the left would be incorrect and will only result in bilateral pneumothoraces. Although this patient will eventually need a right-sided thoracostomy tube.
A 76-year old male with a history of hypertension was seen in an emergency room with a complaint of sudden-onset weakness of the right arm. Upon arrival he had difficulty speaking and was only able to utter single syllables with great effort. Examination revealed weakness, hypertonia, hyperreflexia and loss of sensation on the right upper limb. The corneal (blink) reflex was present bilaterally, but there was weakness of the musculature on the lower half of the face on the right side.
The observed neurological deficits are MOST LIKELY due to a stroke involving the
right anterior cerebral artery
left middle cerebral artery
left anterior cerebral artery
left posterior cerebral artery
The clinical signs are consistent with a lesion involving the lateral aspect of the frontal lobe on the left side, which is in the territory of the MCA. Damage to Broca’s area causes motor aphasia and damage to the primary motor cortex on the lateral aspect of the hemisphere causes weakness of the right upper limb and lower half of the face.
Other answers: Stroke of the anterior cerebral artery often results in opposite leg weakness. Among the many other possible presentations of anterior cerebral artery stroke are anosmia, apraxia (inability to execute purposeful movements), and sensory deficits of the opposite lower extremity. While the clinical symptoms associated with stroke of a posterior cerebral artery can also be varied and depend on the exact location of the occlusion, they often include visual field defects opposite the side of the lesion.
You are caring for a patient who requires mechanical ventilation. The ventilator is currently delivering 10 breaths per minute and a Tidal Volume (VT) of 500 mL. On morning work rounds the attending physician says “the PCO2 is too high, increase the Minute Alveolar Ventilation (VA) by 25%.” Assuming the patient’s Anatomic Dead Space (VD) to be 100 mL, which of the following ventilator settings will achieve a 25% increase in Minute Alveolar Ventilation (VA)?
Ventilator Rate - 10 breaths per minute, Tidal Volume – 625 mL
Ventilator Rate - 15 breaths per minute, Tidal Volume – 600 mL
Ventilator Rate - 10 breaths per minute, Tidal Volume – 600 mL
Ventilator Rate - 15 breaths per minute, Tidal Volume – 500 mL
Answer: Ventilator Rate - 10 breaths per minute, Tidal Volume – 600 mL
Minute alveolar ventilation is the product of respiratory rate x air exchange volume. To calculate this volume, we must substract out the dead space volume which does not come in contact with the lung alveoli. In this example, the initial MAV would be (500-100 mL) x 10 breaths per minute = 4000 mL. A 25% increase would be (600-100 mL) x 10 bpm = 5000 mL.
A term male infant of birth weight 2450 grams is born to a 27-year old woman who is G 3, P 2. The infant is noted to have muscular hypotonia. He receives his first feeding at 2 hours of age and again at 4 hours. Shortly afterwards he develops an enlarging abdomen and vomits bile-stained material. On further examination, he is noted to have a significant cardiac murmur. Which of the following chromosomal abnormalities is most likely to be present?
45,X
47,XY,+21
47,XX,+18
47,XXY
Correct answer: 47, XY, +21.
Explanation: The infant has Down Syndrome (DS), Trisomy 21 with a heart defect (most frequently an endocardial cushion defect), muscular hypotonia, and intestinal atresia. Infants with Trisomy 21 have an increased frequency of a number of gastrointestinal malformations including duodenal atresia, Hirschsprung’s disease and congenital megacolon, esophageal atresia, and anorectal malformations. The areas of intestinal atresia, seen in almost 10% of infants with Down syndrome, include the duodenum, the rest of the small bowel, and rarely the large bowel. Trisomy 21 has an incidence of 1 in 700 newborns. Trisomy 21 is due to nondisjunction in 95% of cases, with an additional 4% of cases due to translocation with 50% having a balanced translocation carrier parent, most frequently 21:14 (rarely 21q;21q translocation with risk of 100%). 1% of infants with DS are mosaics (mixture of 46 and 47 chromosomes). Risk of having an infant with Downsyndrome increases with increasing maternal age from 1:1000 infants at age 20 to 1:30 at age 45 years. Still, 75% of Down syndrome babies are born to women under 35 since that is the age at which the vast majority of mothers have children. Risk of recurrence in subsequent pregnancies is 1% but goes to 16% if the mother is a carrier of translocation between 21 and other chromosomes (5% if father is carrier). DS males are infertile, but DS females may reproduce with a 50% chance of DS occurring in the infant.
Other answers:
45,X is Turner syndrome. Characteristics of patients with Turner syndrome include short stature, webbed neck, and low-set ears.
47,XX,+18 is trisomy 18. This is due to presence of an extra whole or partial chromosome 18 and typically results in major congenital abnormalities in a number of organs, including the heart (especially ventrical and septal defects), kidneys, intestines (including omphalocele and esophageal atresia), central nervous sytem, and liver. While infants with trisomy 18 can present with the findings present in the infant described in the question stem, trisomy 18 is a less likely diagnosis because it is far less common at birth than Down syndrome. Whereas Down syndrome occurs in 1 per 700 newborns, trisomy 18 occurs in only about 1 per 6000 newborns. The majority (about 80%) of fetuses with trisomy 18 die in utero. Once born, the lifespan of a an infant with trisomy 18 is typically only 1-2 weeks, though a small percentage survive longer than one year
47,XXY is Klinefelter syndrome, in which males have an extra X chromosome. While patients with this karyotype are often asymptomatic, they can have a variety of clinical manifestations including hypogonadism and decreased fertility.
Robbins PBOD 8ed. pp 161-162
Sgt Rivera presents to you in clinic with persistent post-concussive symptoms including fatigue, irritability, memory problems, and sensitivity to noise following exposure to an IED blast approximately 6 months ago in theater. At the time of the injury, Sgt Rivera was reported to have experienced 15 minutes of loss of consciousness (LOC) and did not remember the hour following the blast until reaching the hospital for evaluation. 8 hours after the injury Glasgow Coma Scale score was 14 and a head CT was read as within normal limits (WNL). What would be your preliminary characterization of this injury event and recommendations?
Severe TBI. Begin physical therapy and suggest Medical Board Evaluation
Moderate TBI. Refer patient for neuropsychological assessment to characterize deficits and evaluate additional possible comorbid factors (e.g., PTSD, major depression).
Mild TBI. Refer patient for neuropsychological assessment to characterize deficits and evaluate additional possible comorbid factors (e.g., PTSD, major depression).
Mild TBI. Assess for psychological disorder and malingering
Correct answer: Mild TBI. Refer for further testing.
Rationale for answer: Mild TBI is defined as LOC less than or equal to 30 minutes and/or posttraumatic amnesia (PTA) less than or equal to 24 hours. Brain imaging is typically normal in mild TBI. GCS score within 48 hours of injury is generally 13 or greater with mild TBI, 9-12 with moderate TBI, and less than 9 with severe TBI.
While in most cases of mild TBI symptoms typically resolve within a few months of injury, the trajectory of healing can vary. It is important to also evaluate for other contributing factors, such as PTSD and depression, which can have significant impacts on cognition, sleep, and other aspects of functioning.
A rapid expansion of the intra-vascular volume by an infusion of 2 liters of isotonic saline will normally be compensated for by:
an increase in thirst
a decrease in angiotesin II production
a decrease in glomerular filtration rate
an increase in ADH secretion
Answer: a decrease in angiotensin II production.
Explanation: increased intravascular volume = increased GFR = decreased renin secretion by JG apparatus = decreased angiotensin I production = decreased substrate for the formation of angiotensin II = decreased angiotensin II production.
Incorrect answers:
The volume receptor mechanism would inhibit ADH secretion
A volume sensing mechanism would inhibit thirst
Increased vascular volume would increase GFR
Chromosome studies were conducted on a couple that has lost a baby soon after birth because of multiple abnormalities. The father was found to have a balanced reciprocal translocation involving chromosomes 4 and 11.
Given this, which one of the following statements is true?
All of their future babies will be abnormal.
The abnormalities in their baby were probably caused by polyploidy.
Balanced reciprocal translocations are usually harmless so it is very unlikely that this finding is relevant to their baby’s abnormalities.
It is very unlikely that this translocation will cause ill health in the father but there is a risk that he and his partner will have another abnormal baby.
The father has a balanced reciprocal translocation. Translocations occur through exchange of DNA between non-homologous chromosomes. The karytoype figure demonstrates the possible outcomes for future infants. Possibilities include normal karyotype, balanced karyotype, unbalanced karyotype with possibility for survival, and unbalanced karyotype not compatible with survival.
Chromosomes with translocations will pair through their homologous parts with each other and with the pair of normal homologous chromosomes leading to the formation of a quadrivalent (instead of two bivalents as in normal meiosis). During the first meiotic division they will segregate randomly to one of two cells. Note that chromosome combinations in the sperm that are shown on the figure only indicate spermatozoa that receive two chromosomes, whereas it is possible that some of spermatozoa may receive three chromosomes from the quadrivalent and some may receive only one.
Incorrect answers:
The answer that all of their future babies will be abnormal is incorrect. As shown on the diagram below at least half of the couple’s newborns will be normal (will have either a normal or balanced karyotype). Keep in mind that spermatozoa with some unbalanced karyotypes either do not survive or are not proficient in fertilization, and most embryos with trisomies and monosomies (e.g., the last two on the figure) are spontaneously aborted, often before the pregnancy is even recognized.
The abnormalities in their baby were probably not caused by polyploidy. Polyploidy is the increase in the number of complete chromosomal sets (i.e., chromosomes #1 through #22 and a sex chromosome). Balanced rearrangements may result in aneuploidies (e.g., one extra and one missing chromosome as shown on the figure) in the offspring, but they do not induce polyploidy.
Although balanced translocations are usually harmless to their carriers the random nature of segregation of the rearranged chromosomes and their corresponding homologous chromosomes in meiosis may lead to unbalanced karyotypes in the progeny.
A 5 year old girl has a history of recurrent bacterial infections. She is found to have an inherited condition in which a gene important in promoting the production of reactive oxygen radicals (O2.) is defective leading to poor bacterial killing. The diagnosis of her condition is:
Chediak-Higashi syndrome
Chronic granulomatous disease
Leukocyte adhesion deficiency type 1
Leukocyte adhesion deficiency type 2
Answer: Chronic granulomatous disease.
Chronic granulomatous disease involves a defect in genes encoding a component of phagocyte oxidase, important for generation of O2 radical (O2.). Since initial neutrophil defense is inadequate, a macrophage-rich chronic inflammatory reaction is recruited that tries to control the infection (granuloma). Chediak-Higashi syndrome involves decreased leukocyte function because of mutations affecting a protein involved in lysosomal membrane traffic. Leukocyte adhesion deficiency type 1 involves defective leukocyte adhesion because of mutations in the beta chain of CD11/CD18 integrins. Leukocyte adhesion deficiency type 2 involves defective leukocyte adhesion because of mutations in fucosyl transferase required for synthesis of sialylated oligosaccharide (ligand for selectins).
Robbins PBOD 8ed. pp 55-56.
A 17-year-old Airman who is scheduled to be deployed to Iraq presents to your office because of shortness of breath and wheezing on exertion, particularly during basketball season. He does not notice any shortness of breath with routine activity. There is no family history of asthma. On physical examination, he is in no respiratory distress. His lungs are clear, with no wheezing during either tidal breathing or forced expiration. His cardiac examination is normal. Baseline spirometry is normal. What is the BEST next diagnostic step?
Allergy testing
Exercise testing with postexercise spirometry
Overnight oximetry
Reassurance
The most likely diagnosis is exercise-induced asthma, and exercise testing with postexercise spirometry is the most appropriate diagnostic test. Exercise testing for exercise-induced asthma should attempt to mimic the activity that induces the shortness of breath. In addition to establishing the diagnosis of exercise-induced asthma, exercise testing can reveal exercise-induced laryngeal dysfunction that sometimes mimics exercise-induced asthma. All patients with significant disease should also be instructed to use prophylactic treatment 5 to 10 minutes before exercise, usually two puffs of a medium-acting inhaled beta2-agonist (albuterol). A better term for this condition is “exercise-induced bronchospasm,” because not all persons with the condition have asthma. The estimated prevalence of exercise-induced bronchospasm ranges from 7% to more than 20% in the general population.
Exercise-induced bronchospasm probably results from changes in airway physiology triggered by the large volume of relatively cool, dry air inhaled during vigorous activity. Bronchodilation is the more common first event during exercise and lasts for 1 to 3 minutes after exercise. In patients with exercise-induced bronchospasm, the initial bronchodilation is followed by bronchoconstriction, which begins within 3 minutes, generally peaks within 10 to 15 minutes, and resolves by 60 minutes. Acute bronchoconstriction was previously followed by late-phase bronchoconstriction in some patients; however, the risk and severity of late-phase bronchoconstriction due to exercise-induced bronchospasm is decreased when compared with allergen-induced asthma.
Other answers: Although some patients with exercise-induced asthma may also have an allergic component, there is no clear correlation between the development of exercise-induced asthma and allergies. Patients with exercise-induced asthma may have equivocal results on methacholine challenge testing, making it a less helpful test than postexercise spirometry. Overnight oximetry would provide no information regarding the onset of bronchospasm during exercise.
A 5 year old boy develops an acute bacterial pneumonia. His CBC shows a marked elevation of neutrophils with a “shift to the left” (i.e. increase in immature band forms). Gram stain shows Gram positive cocci in pairs and short chains. The mechanism for this change involves:
Direct IL-1 and TNF-alpha stimulation of the bone marrow post-mitotic reserve pool.
Induction of G-CSF from macrophages through bacterial stimulation of Toll-like receptors
Acute release of C-reactive protein from the liver
LPS-induced release of IL-1 and TNF-alpha resulting in stimulation of PGE2 production
Answer: Direct IL-1 and TNF-alpha stimulation of the bone marrow post-mitotic reserve pool.
IL-1 and TNF-a are cytokines with largely overlapping biologic properties that play a major role in mediating acute inflammation. One effect is to act directly on the bone marrow post-mitotic reserve pool resulting in a rise in the number of immature neutrophils (bands) in addition to rapid release of mature neutrophils.
Other answers: LPS (lipopolysaccharide) is found in the membranes of Gram negative bacteria. PGE2 plays a role in induction of fever and stimulation of vasodilation and increase vascular permeability. It has no role in causing a left-shift.
C-reactive protein (CRP) is an acute phase reactant released by the liver in response to IL-6 production. CRP binds to phosphocholine on certain bacterial polysaccharide capsules and on dead and dying cells. Once bound to a cell or bacterial surface, CRP activates the complement system and functions as an opsonin to enhance clearance of dead cells and bacteria by macrophages.
G-CSF stimulates bone marrow synthesis of mature neutrophils.
Robbins PBOD 8ed. p75.
A 46 year old man complains of gradually developing numbness on his left hand over a two-week period. You have him remove his watch, which is relatively tight. Your examination shows that the motor examination is normal and that the numbness involves the skin on the dorsum of his hand between the thumb and the index metacarpal bones. This area is innervated by the:
lateral antebrachial cutaneous nerve
superficial branch of the radial nerve
dorsal branch of the ulnar nerve
median nerve
The radial nerve has a specific sensory distribution that is consistent and that covers the dorsoradial aspect of the hand. These sensory fibers are transmitted via the superficial branch of the radial nerve (the deep branch transmits primarily motor impulses), with little if any “cross-over” from the other nerves distal to the wrist.
Because of a high prevalence of hepatitis C virus (HCV) infection (10–20%) among veterans seeking care in Department of Veterans Affairs hospitals, current US military forces were evaluated for HCV infection. Banked serum samples were randomly selected from 10,000 active-duty military personnel and were tested for antibody to HCV (anti-HCV). 42 samples were initially positive. Over 5 years of follow-up, 6 subjects who were initially negative for anti-HCV seroconverted. What is the 5-year incidence of a positive test result for HCV infection in active-duty military personnel?
Choices 48/9,958 6/9,958 6/10,000 42/10,000
Incidence is number of new cases divided by population at risk of disease. There were 6 new cases during the follow-up period. Although the entire population consists of 10,000 personnel, 42 were initially positive and thus could not become a new case, so the population at risk is 10,000 – 42 = 9,958. Incidence is 6/9,958 over the 5 year period.
An 18-year-old college freshman is brought to a local emergency room with a fever of 103.4°, a few petechiae on her trunk, and, on examination, positive Kernig and Brudzinski’s signs. A sample of her cerebrospinal fluid is obtained and Gram stain reveals a few Gram-negative cocci. The major component that allows this organism to resist complement-mediated phagocytosis is MOST LIKELY
lipooligosaccharide
a serotype of M protein
a polysaccharide capsule
an IgA protease
The correct answer is a polysaccharide capsule.
The disease described is most likely meningococcal meningitis. A primary virulence factor of Neisseria meningitidis is its anti-phagocytic polysaccharide capsule. Protection against common types of N. meningitidis is provided through vaccination of young people with a polyvalent vaccine that consists of several of these capsular antigens.
Other answers: The M protein is a helical protein component of the cell wall of Group A streptococci that has an antiphagocytic property as well. Several organisms produce IgA proteases that have activity against that class of secretory immunoglobulins, however IgA is a poor activator of complement and opsonizes very weakly. While N. meningitidis expresses lipooligosaccharide (LOS), this is not a mechanism by which the bacteria resists complement-mediated phagocytosis. Rather, LOS contributes to clinical disease by binding to toll-like receptor 4 on innate immune cells, resulting in the release of massive amounts of pro-inflammatory cytokines.
A Finnish woman is the mother of a child with Meckel syndrome, a rare autosomal recessive disorder occurring in 1/10000 births in her part of Finland. She is widowed and is now going to remarry. What is the increased risk of the disease in a fetus if she marries the brother of her deceased husband rather than an unrelated Finnish man from her geographic area?
¼ versus 1/200
1/8 versus 1/10000
1/8 versus 1/200
100% versus 50%
Answer: 1/8 versus 1/200
If the child is affected by autosomal recessive disease he must be a homozygote. Therefore, his parents are obligate heterozygotes. Either the deceased husband’s mother or father is also a heterozygote, because their son must have inherited the mutant gene from one of them. Then the chance that the deceased husband’s brother is a carrier is 1/2 (he can inherit either a normal or a mutant allele from the heterozygous parent). If the mother and the deceased husband’s brother are heterozygotes the chance of the fetus inheriting both mutant alleles is 1/4 so the total risk is 1/2 X 1/4 = 1/8.
The mother is an obligate heterozygote. The disease allele frequency in this population is √q = √ 1/10000 = 1/100. Normal allele frequency is 1–1/100 ~ 1. Carrier (heterozygote) frequency is 2pq = 2 x 1 x 1/100 = 1/50. This is the risk of a random person being a carrier. If the mother and a random Finnish male are both carriers for autosomal recessive disease the chance of their child being affected is 1/4. Total risk for the fetus is 1/4 x 1/50 = 1/200.
Your patient is a 46-year-old male who tells you that he is having difficulty with his vision. He also tells you that his breasts have recently enlarged and started to lactate. What would be the most appropriate drug to treat this patient?
pegvisomant
testosterone
bromocriptine
sermorelin
The correct answer is bromocriptine.
The patient’s symptoms, breast enlargement with lactation and loss of vision, are compatible with hyperprolactinemia resulting from an anterior pituitary prolactin-secreting adenoma. A blood test for serum prolactin should be performed to confirm your diagnosis. A long-acting dopamine agonist, like bromocriptine, will inhibit prolactin secretion and cause the adenoma to shrink. None of the other drugs [testosterone, pegvisomant (a GH antagonist), or sermorelin (GHRH1-29)] would affect prolactin levels or the adenoma. The patient may eventually want to have the adenoma surgically removed, but it would be appropriate to shrink the adenoma before proceeding with the surgery.
A cheese-linked outbreak of bacteremia and meningitis in the elderly and stillbirths in pregnant women was revealed through the diligent work of a nurse epidemiologist. The cheese was made from raw milk and stored for weeks in the cold. The MOST LIKELY cause of this outbreak was
Brucella abortus
Listeria monocytogenes
Mycobacterium bovis
Escherichia coli O157:H7
The correct answer is Listeria monocytogenes.
Although all of these agents may be found in unpasteurized milk, Listeria monocytogenes is capable of actually replicating at refrigerator temperatures and hence is often linked to refrigerated meats and dairy products. Generally speaking, healthy people develop only mild self-limiting flu-like or gastrointestinal symptoms from foods contaminated with Listeria, but individuals that lack good cell mediated immunity are at risk from this facultatively intracellular pathogen. Unlike the other agents listed, Listeria has the capacity to cross the placenta in pregnant patients and cause stillbirth or meningitis in the newborn. Meningitis is also seen in AIDS patients or the elderly infected with listeria. Brucella causes a recurrent febrile syndrome, E. coli causes hemorrhagic colitis and hemolytic uremic syndrome, and M. bovis is associated with miliary TB, especially where unpasteurized milk is a major food source.
Willamena Whiner is a 40-year-old female who sprained her wrist playing golf. She tells her doctor that she has been taking aspirin for the pain, but her wrist still hurts. Her doctor prescribes a mild opiate that she is to use in combination with the aspirin. Now, she returns to her doctor claiming that the prescribed opiate is completely ineffective. In a follow-up test, it is discovered that she is deficient in active CYP2D6. Which opiate had her doctor MOST LIKELY prescribed that was ineffective in this patient?
fentanyl
codeine
loperamide
morphine
The correct answer is codeine.
Explanation: In order to be an effective analgesic, codeine must be converted to morphine by CYP2D6. In patients deficient in CYP2D6, codeine is ineffective as an analgesic.
Incorrect answers: Morphine and fentanyl are full mu opiate agonists that need no modification for activity. Loperamide lacks any analgesic property because it is rapidly pumped out of the CNS by the MDR transporter. Loperamide is an opiate that is used as an anti-diarrheal agent.
A 50 year old male with chronic diverticulitis eventually developed a fibrotic stricture in his colon resulting in gastrointestinal obstruction. Which of the following was the most likely sequence of events leading to this outcome?
Activation of macrophages and lymphocytes, production of TGF-b and PDGF, and decrease in metalloproteinase activity.
Activation of mast cells, secretion of histamine, increased vascular permeability
Secretion of IL-5 by Th2 cells, recruitment of eosinophils, release of major basic protein
Secretion of IFN-γ by Th1 cells, macrophage activation, focal collections of epithelioid cells
Answer: Activation of macrophages and lymphocytes, production of TGF-b and PDGF, and decrease in metalloproteinase activity.
Explanation: Chronic inflammation as observed in chronic diverticulitis results in tissue fibrosis. Activated macrophages and lymphocytes release a variety of mediators, including PDGF and TGF-β, which play a role in promoting fibroblast proliferation and collage synthesis as well as inhibiting metalloproteinase synthesis (with resultant decreased collagen degradation).
Other answers: Activation of mast cells, secretion of histamine, and increased vascular permeability are observed in type 1 hypersensitivity reactions. Secretion of IFN-γ by Th1 cells, macrophage activation, and focal collections of epithelioid cells are involved in granulomatous inflammation. Secretion of IL-5 by Th2 cells, recruitment of eosinophils, and release of major basic protein are observed in allergic reactions and in response to helminth infections.
Robbins PBOD 8ed. pp 107-108
A patient with severe Chronic Obstructive Pulmonary Disease (emphysema) presents to the Emergency Department because of increasing shortness of breath due to an acute respiratory infection. You know the patient and are aware that her arterial PaCO2 is usually about 60 mmHg. Arterial blood gases obtained in the Emergency Department show a PaO2 of 30mm Hg, a PaCO2 of 70 mm Hg, and an arterial pH of 7.15. You administer a high concentration of inspired oxygen by face mask. Repeat arterial blood gases, obtained with the patient on the inspired oxygen, reveal a PaO2 of 40mm Hg, a PaCO2 of 90 mm Hg, and an arterial pH of 6.90. Which of the following choices would be the most appropriate at this time?
Have the patient inhale CO2 to stimulate breathing.
Have the patient breathe 100% oxygen.
Place the patient on a ventilator and adjust the settings to achieve adequate arterial PO2 and PCO2.
Return the patient to breathing room air.
Ans: Place the patient on a ventilator and adjust the settings to achieve adequate arterial PO2 and PCO2.
The patient is suffering from an inability to provide sufficient exchange of CO2 and oxygen. Administering increased oxygen signals to the respiratory centers to slow down the ventilatory rate. However, because of her COPD, she needs to breathe more rapidly to exchange CO2. Increasing CO2 does not stimulate increased breathing because the patient has been at elevated CO2 levels for so long that her receptors have accommodated to the increased CO2. Therefore, mechanical intervention is required to get her balance of ventilation rate and volume to the point that her PaCO2 will be reduce to acceptable levels.
The development of protective antibodies specific for which ONE of the following bacterial surface components is often protective and is also the basis of several successful vaccines against bacterial pathogens?
Capsule
Multidrug resistance efflux pumps
Lipopolysaccharide
Colonization pili
The correct answer Capsule.
Explanation: Capsular polysaccharides are the basis for a number of vaccines including Stretococcus pneumoniae, Haemophilus influenzae and Neisseria meningitidis. To develop an anamnestic immune response, these polysaccharides are often conjugated to proteins to stimulate a T-cell response. Protection against colonization with pili is a potential strategy but thus far there are no such vaccines available. The toxic nature of lipopolysaccharide rules it out as a potential immunogen, although it is easy to envision that anti-LPS antibody would be useful in blocking shock associated with Gram-negative infections. Antibodies against drug resistance pumps have not been used as a vaccine strategy.
A Marine normally stationed at Camp Pendleton, CA is deployed to Afghanistan and helicoptered to a forward operating base situated at 2500m. He is uncomfortable with headache and nausea but the next day is feeling well enough to perform his duties. Which of the following changes accounts for his accommodation to the altitude of his new base?
Increased red cell production
Central chemoreceptor adaptation
Hemoglobin isoform alteration
Angiogenesis
Ans: Central chemoreceptor adaptation.
Hypoxemia is the cause of the observed symptoms. Lowered levels of oxygen will cause the peripheral chemoreceptors to increase the rate of breathing in an attempt get more oxygen into the system. This will cause a decrease in PaCO2 which will cause the central chemoreceptors to slow the breathing rate, again decreasing PaO2 . Breathing rate must increase to maintain adequate PaO2. This back and forth feedback continues until the central chemoreceptor adapt to lower PaCO2 levels at the higher respiratory rates. One would expect that red cell production will be stimulated but the differentiation of erythrocytes will take about a week before the increased numbers will show up in the bloodstream.
An 11-month old male infant is brought to his pediatrician because he seems to be much “slower” in doing things than his siblings. He has just learned to roll over but is unable to sit up. He is noted to have a large head and mildly low set ears that are posteriorly rotated. His sternum is somewhat concave and his fingers show hyperextensibility. He has an older brother with a prominent mandible who is doing poorly in school. Which of the following types of gene mutations is most likely to produce these findings?
Frameshift
Nonsense (stop codon)
Trinucleotide repeat
Point
Correct answer: Trinucleotide repeat.
Explanation: These findings are typical of fragile X syndrome, a condition in which there are 250-4000 tandem repeats of the trinucleotide sequence CGG. In normal individuals there are between 10-50 CGG repeats but in families with this disorder the number of repeats tends to increase with succeeding generations (“genetic anticipation”). The gene is on chromosome X at q27.3. Other diseases with unstable trinucleotide repeats of C_G include Myotonic Dystrophy – CTG, Huntington Disease – CAG, and Kennedy’s Disease – CAG. In Fragile X Syndrome, the number of CGG repeats affects the phenotype with 6-60 repeats (normal allele) producing a normal phenotype, 60-200 repeats (premutation allele) also producing a normal phenotype but one who is an obligate carrier, and >200 repeats (full mutation allele) leading to various expression of the Fragile X phenotype.
Robbins PBOD 8ed. pp 169-171
A 55 year old woman complains of blurred vision and burning and itching of her eyes. She also has difficulty swallowing solid foods and reports a decrease in the ability to taste. On physical exam she has conjunctival erythema and cracks and fissures in the mouth. Serum anti-SS-A antibody is elevated. She has no other associated autoimmune diseases. Which of the following histological changes is most likely to be seen in a salivary gland biopsy?
granulomatous inflammation
periductal lymphocytes
eosinophilic infiltrates and interstitial edema
metaplastic changes in acinar epithelium
Correct answer: Periductal and perivascular lymphocytes and germinal centers.
Explanation: This is a case of Sjogren’s syndrome, an autoimmune exocrinopathy. In this condition CD4+ T cells reacting to exocrine gland self-antigens are believed to stimulate ductal epithelial cell hyperplasia, with resulting duct obstruction and acinar cell atrophy and fibrosis. This leads to reduced production of exocrine gland secretions, accounting for many of the signs and symptoms of this disease, including dry eyes and dry mouth. Most of the lymphocytes infiltrating the salivary gland are CD4+ T-cells, and 10% are B cells.
Other answers: Eosinophils are typically seen in allergic conditions and infections with nematodes. Some autoimmune diseases exhibit granulomatous inflammation (e.g. Wegener granulomatosis, temporal arteritis, etc.), but not Sjogren’s syndrome. Although acinar epithelial cells are injured in Sjogren’s syndrome, they do not undergo metaplasia.
Robbins PBOD 8ed. pp 221-223.
A 44-year-old patient develops fevers while recovering from surgery to repair a bowel perforation that developed secondary to a complicated appendicitis. The microbiology lab calls and informs you that the blood cultures are growing a germ-tube positive yeast. The most appropriate antibiotic to treat this organism initially is:
ceftazidime
fluconazole
ampicillin/sulbactam
amphotericin B
The correct answer Fluconazole.
Explanation: The patient has candidemia (Candida infection of the blood). The only two antifungal agents listed are amphotericin B and fluconazole. Clinical studies in immunocompetent patients have demonstrated that fluconazole and amphotericin B are equally effective at treating bloodstream infections with Candida albicans. Of the two agents, amphotericin B has a substantially worse side effect profile, with significant risk of renal toxicity. Having ascertained by the germ-tube test that the yeast is most likely Candida albicans (which, in addition to the rare Candida dublinensis is the only Candida species which is germ tube positive), fluconazole would be a more appropriate choice of drug than amphotericin B in this situation. An echinocandin would also be a reasonable choice. Ampicillin/sulbactam and ceftazidime target bacterial cell wall synthesis and therefore would not be useful against this yeast infection.
If the yeast was germ tube negative, one would be concerned that the patient could possibly be infected with a Candida species that is resistant to fluconazole, such as Candida glabrata or Candida krusei. Thus, in the setting of a germ-tube negative yeast, one would not start with fluconazole. Rather, one would use an echinocandin or amphotericin B. Additionally, because of the widespread use of azoles in the prophylaxis of cancer patients that become neutropenic, patients with neutropenia who develop a Candidemia are frequently started on an echinocandin or amphotericin for initial coverage.
An 82 year old retired Army Colonel with a history of diabetes, hypertension and obesity presents to the Emergency Department at 4 am with severe pain in his right ankle which woke him from sleep. He states that even having the bedsheet rest on his ankle elicits pain. He has an oral temperature of 101F. The affected ankle appears erythematous. Even slight movement produces excruciating pain. An arthrocentesis yields a small amount of amber fluid which microscopically reveals numerous PMN’s and positively birefringent crystals. Which of the following diagnoses is BEST supported by the findings?
Gout
Intraarticular fracture of the MTP joint
Septic arthritis
Pseudogout
Answer: pseudogout
The presentation is suggestive of either a septic joint or crystal-induced synovitis, with severe pain of rapid onset and low-grade fever. This presentation is not really compatible with a fracture (no history of trauma). The appropriate diagnostic test was the joint fluid aspiration, which demonstrated crystals. Uric acid crystals are needle-shaped on polarized microscopy and are negatively birefringent. Calcium pyrophosphate crystals, in contrast, are blue-yellow birefringent. Because the aspiration revealed positively birefringent crystals, the correct answer is pseudogout.
Which one of the following can infect the joints via hematogenous spread from the reproductive tract?
Chlamydia trachomatis, serovars L1, L2, L2a and L3
Neisseria gonorrhoeae
Haemophilus ducreyi
Chlamydia trachomatis, serovars D-K
The correct answer is Neisseria gonorrhoeae
Although all these agents are pathogenic in the reproductive tract, the one associated with hematogenous spread, joint infection, and disseminated gonorrheal infection (DGI) is Neisseria gonorrhoeae. In fact, the leading cause of septic arthritis in people of reproductive age is Neisseria gonorrhoeae. It may be difficult to confirm the diagnosis by culturing synovial fluid or the skin lesions that accompany such an infection. Culture of the urethra, cervix, rectum or pharynx is often positive and would confirm the diagnosis.
Chlamydia trachomatis, serovars D-K, cause urethritis, cervicitis, pelvic inflammatory disease, and neonatal pneumonia and conjunctivitis. Chlamydia trachomatis, serovars L1, L2, L2a and L3 cause lymphogranuloma venereum (LGV), which can cause a painless genital ulceration, inguinal lymphadentis/lymphangitis, and proctitis. Haemophilus ducreyi causes painful genital ulcerations.
Which of the following structures disappears just before the start of implantation?
Zona pellucida
Cilia on the endometrial epithelium
Allantois
Sperm acrosome
Explanation: The zona pellucida is a glycoprotein shell that surrounds the ovum, acts as receptors for sperm and induces the acrosome reaction. At the end of the fourth day the zona pellucida disappears and the conceptus implants into the uterine lining at ~day 6. The conceptus would not be able to implant if it was still surrounded by the zona pellucida.
Incorrect answers:
The allantois forms after implantation.
Cilia on the endometrial epithelium do not disappear.
The sperm acrosome disappears during the process of fertilization, an event that occurs ~6 days before implantation.
A three year old girl presents with a lesion at the lower palpebral conjunctiva that has been present for two weeks . Periauricular nodes are enlarged. A large mass in the left axilla shows reddening of the skin over the mass. Her six year old brother is noted to have enlarged right inguinal lymph nodes and a lesion on his ankle. The regional lymphadenopathy occurred along with fever, malaise, headache and bone and joint pain. The axillary and inguinal lymph nodes are enlarged, matted, nodular and adherent to surrounding tissue. Histology of an axillary lymph node displays confluent necrotizing granulomas. The children live with their family on a farm and are responsible for care of the chickens, three dogs and two kittens. To identify the organism causing this disease, a biopsy of the tissue must be prepared using which procedure:
Culture the tissue on blood agar in 100% oxygen
Stain the sections with a connective tissue stain (e.g. trichrome stain)
A touch preparation of the tissue stained with Wright’s stain
Culture the tissue in heart-brain infusion broth at 32⁰C
ANSWER: Culture the tissue in heart-brain infusion broth at 32⁰C .
Explanation: The lymphadenopathy with the necrotizing granulomas is typical of cat scratch disease (CSD), typically seen in children, but also in adults, who have been in contact with cats, usually kittens. The diagnosis is usually made by positive serology in combination with certain clinical criteria*. To identify the organism, however, the organism typically needs to be grown in culture.
The organism causing CSD (Bartonella henselae) is a fastidious bacillus that grows best in heart-brain infusion broth at 32⁰C - 35⁰C. The typical culture medium (blood agar) used to culture most bacterial and fungal organisms has a low yield for identifying Bartonella henselae. Stains such as H& E, trichrome, gram and Wright will not stain the organism. Silver stains such as Warthin-Starry will show clusters of the bacilli in area of necrosis within the necrotizing granulomas. PCR identification of B. henselae is also useful, but not that sensitive (thus negative PCR does not rule out the diagnosis). It is important to recognize that special stains and culture approaches are needed when considering the diagnosis of CSD, so that the appropriate tests can be ordered on biopsy samples.
- Clinical criteria supportive of a diagnosis of cat scratch disease include:
- -> cat or flea contact of any kind
- -> negative serology for other causes of adenopathy OR + Bartonella PCR OR + liver and spleen lesions on CT OR sterile pus from lymph node aspiration
- -> biopsy with granulomatous inflammation OR a positive Warthin-Starry silver stain
A 60-year-old female has recently been diagnosed with Parkinson’s disease. Levodopa has been effective in relieving her rigidity and akinesia, but she is now experiencing severe emesis and orthostatic hypotension. What drug, acting only in peripheral tissues, would be the MOST appropriate addition to her therapy to decrease these side effects of levodopa?
rasagiline
carbidopa
benztropine
pramipexole
The correct answer is carbidopa.
Emesis and orthostatic hypotension result from the peripheral actions of dopamine. Carbidopa inhibits the decarboxylation of l-dopa to dopamine, but since carbidopa does not enter the brain, it prevents the conversion of l-dopa to dopamine only in peripheral tissue.
None of the other drugs listed would prevent the peripheral side effects associated with l-dopa administration.
Benztropine is an anticholinergic agent that can be used as a 2nd line agent for Parkinson’s disease to reduce tremor. However, it would not reverse the peripheral effects of l-dopa administration.
Pramiprexole is a dopamine agonist that is sometimes used in early treatment of Parkinson’s disease. Like levodopa, its side effects include nausea, vomiting, and orthostatic hypotension.
Rasagiline is an irreversible monoamine oxidase inhibitor that can be used in early Parkinson’s or as an additional agent in late Parkinson’s disease. It would not reverse the peripheral effects of l-dopa administration.
A 4 year old child presents to her physician for eye discomfort. Physical examination reveals dry eyes with corneal softening. Which of the following vitamin deficiencies is the most likely cause for this presentation?
Folic acid
Vitamin A
Vitamin B12
Vitamin C
The correct answer is Vitamin A.
Explanation: Vitamin A is needed for epithelial maturation. Absence of vitamin A leads to squamous metaplasia of the lacrimal ducts leading to dry eyes and corneal softening (keratomalacia). Other symptoms of vitamin A deficiency include night blindness, corneal ulceration, pruritis, and growth retardation.
Other answers: B12 and folic acid deficiencies are associated with megaloblastic anemia. Additionally, B12 deficiency can cause decreased vibratory sensation. Vitamin C deficiency leads to scurvy, which is characterized by vascular fragility and bone changes. Some other notable vitamin deficiencies include Vitamin D deficiency (which causes rickets), Vitamin K deficiency (which causes hemorrhagic diathesis), and Niacin deficiency (which causes pellagra – diarrhea, dementia and dermatitis).
Robbins PBOD 8ed pp 430-431
A 70-year-old male was hospitalized for surgical repair of a femur fracture. On hospital day six he developed fever to 40°C, productive cough and hypoxia. Chest X-ray confirmed a right upper lobe pneumonia, and sputum Gram stain showed numerous neutrophils and gram-negative rods. Intravenous ceftazadime (a third generation cephalosporin) therapy was initiated, but the patient failed to improve and developed respiratory failure. Sputum culture, taken at the onset of his pneumonia, yielded Enterobacter cloacae resistant to all of the extended spectrum penicillins and cephalosporins. What is the MOST LIKELY mechanism of resistance manifested by this organism to these antibiotics?
An altered penicillin binding protein
A DNA-gyrase (topoisomorase) mutation
An extended spectrum beta-lactamase
Altered permeability of the bacterial membrane
The correct answer An extended spectrum beta-lactamase.
Explanation: It is not unusual, especially in the hospital setting, to encounter members of the family Enterobacteriaceae that have acquired the genes to encode an extended spectrum beta-lactamase (ESBL). This enzyme is capable of inactivating numerous beta-lactam antibiotics of the penicillin and Cephalosporin families. In addition Enterobacter cloacae, other Enterobacteriaceae that often express ESBLs include Escherichia coli, Klebsiella pneumonia, Klebsiella oxytoca, Salmonella species, Proteus mirabilis.
Pseudomonas aeruginosa, which is not a member of the Enterobacteriaceae family, is another common nosocomial gram negative organism that can also express an ESBL.
A 7 year old boy is brought to his physician for difficulty concentrating at school. A complete blood count performed as part of his workup demonstrates a microcytic anemia, and a peripheral smear of his blood shows basophilic stippling of the red blood cells. An assay for zinc protoporphyrin shows marked elevation. Exposure to which of the following substances is the most likely cause for his presentation?
Mercury
Cadmium
Arsenic
Lead
The correct answer is Lead toxicity.
Lead toxicity is associated with CNS injury leading to intellectual and psychologic impairment in children. Lead interferes with enzymes important in heme synthesis, including aminolevulinic dehydratase, aminolevulinic acid synthetase, and ferrochelatasae. Consequently, lead intoxication can result in decreased iron incorporation into heme, causing a microcytic anemia and increased levels of zinc protoporphyrin in the blood. Basophilic stippling is often seen in anemia due to lead poisoning. This phenotype is believed to be due to lead inhibition of a pyrimidine 5’ nucleotidase, which results in increased degradation of ribosomal RNA. Other features of lead toxicity include abdominal colic, growth retardation, neuropathy and nephrotoxicity. Lead lines may be observed in the gums on physical examination. These typically appear as a blue line along the gums, with a blue/black coloring of the gum tissue adjacent to teeth. Deposition of lead in the bones of children can often be observed as a transverse hyperdense line in the metaphysis of long bones.
Other answers: Cadmium toxicity is associated with obstructive lung disease and renal tubular injury.
Arsenic toxicity is associated with gastrointestinal, cardiovascular, and CNS dysfunction. Transverse white lines in the nails (Mees lines) are seen with arsenic exposure. There is also an increase in skin and lung cancer from chronic exposure.
Mercury toxicity is associated with diarrhea, visual field defects and nephrotoxicity.
Robbins PBOD 8ed pp406-407
The mother of a 2 month old female infant noticed a palpable mass in the upper portion of the abdomen along with numerous blue/black slightly raised spots on the skin of her trunk and face. The infant displayed mild respiratory distress. A CT scan revealed a markedly enlarged liver and a mass in the left upper portion of the abdomen. Laboratory studies showed that the 24-hour urine levels of homovanillic acid (HVA) and vanillylmandelic acid (VMA) were increased. A bone marrow aspirate displayed numerous clusters of “small round blue cells.” Bone scans were unremarkable for any bony destruction. Which of the following should be evaluated to assist in therapeutic decision making?
Alpha Fetoprotein (AFP) staining of the cells in the liver
Ultrasound to check for malformations of the kidney
Assessing for blastemal cells in the biopsy of the mass in the left upper abdomen
Presence of N-MYC amplification
ANSWER: Presence of N-MYC amplification.
Explanation: The patient has a neuroblastoma, a tumor that arises from primitive sympathetic ganglion cells. These tumors can arise anywhere within the sympathetic nervous system, including the adrenal glands, abdomen, and thoracic cavity. About 2/3 of neuroblastomas originate in the abdomen (and 2/3 of those are from an adrenal gland). Neuroblastoma cells typically have defective catecholamine synthesis, and thus release increased amounts of catecholamine intermediates such as HVA, VMA, and dopamine. There are many types of neuroblastomas, and they can cause a wide variety of clinical diseases (from benign, mature tumors, to aggressive and metastatic ones). Neuroblastomas can metastasize by both lymphatic and hematogenous routes. Tumors can spread to the skin, where they typically appear as firm, non-tender, bluish nodules.
The infant in the clinical stem most likely has a neuroblastoma with involvement of the left adrenal gland, the liver, the skin and the bone marrow (but not cortical bone). N-MYC (or MYCN) is an oncogene that is an important prognostic feature of neuroblastomas. Presence of the N-MYC oncogene portends a poor prognosis, while absence of N-MYC amplification is associated with a more favorable prognosis. Of note, the child in this question stem most likely had a 4S tumor. While metastatic neuroblastomas typically have a very poor prognosis, children under 1 year of age who have metastatic neuroblastoma without spread to cortical bone typically do quite well with frequent spontaneous regression of the tumors. In these special 4S cases, when the skin lesions regress a biopsy will show only mature ganglion cells. The liver and bone marrow involvement will also disappear.
Other answers: Malformations of the kidney are not associated with neuroblastoma. Blastemal cells are a component of nephroblastoma (Wilms tumor). AFP staining of hepatocytes is associated with hepatoblastoma.
Robbins PBOD 8ed. pp 475-479
20-year-old U.S. solider is brought to a clinic in El Paso with severe, non-bloody diarrhea and dehydration. No leukocytes are seen in stained fecal smears. Incubation of stool from the patient results in a pure culture of Gram-negative bacilli that ferment lactose. Bacteria are sent to a health department for further identification. The bacteria responsible for his illness are MOST LIKELY to produce which one of the following virulence factors?
Shiga toxin (ST)
Cholera toxin (CT)
Heat labile toxin (LT)
Cytolethal distending toxin
The correct answer is Heat labile toxin.
By the description of the agent we know that it is a Gram-negative organism that ferments lactose. That distinguishes E. coli (which is lactose fermenting) from non-lactose fermenters such as Vibrio (which makes cholera toxin) and Campylobacter (which produces cytolethal distending toxin). Heat labile toxin and Shiga toxin are both associated with E. coli, which is lactose fermenting. However Shiga toxin is associated with bloody diarrhea and the Hemolytic Uremic Syndrome. The description in this case is of a secretory diarrhea which is the classic ETEC, or traveler’s diarrhea picture associated with heat labile toxin, an analogue of Cholera toxin.
A 23-year old woman gave birth to an 1100 gram infant at 31 weeks gestation. The newborn male had APGAR scores of 4 and 6 at one and five minutes and shortly afterwards developed increasing respiratory distress for which therapy with oxygen and high frequency ventilation was initiated. He was weaned from the oxygen and mechanical ventilation over 8 days and began gaining weight. He was sent home at one month of age. Which of the following factors was most likely responsible for this infant’s respiratory disease?
Congenital lobar emphysema
Congenital pulmonary airway malformation
Surfactant deficiency
Tetralogy of Fallot
ANSWER: Surfactant deficiency.
Explanation: Prematurity (birth weight under 2500 gms) is frequently complicated by respiratory distress (also called hyaline membrane disease) if the infant is less than 35-36 weeks gestation. Prior to that gestational age the lung has not developed a sufficient number of Type 2 alveolar cells to produce an adequate amount of surfactant to prevent the infant’s lungs from collapsing when exhaling. In fact, any event that may lead to the destruction of Type 2 alveolar lining cells (e.g. anoxia during delivery, congenital infection) can lead to surfactant deficiency even in term infants.
Surfactant is a mixture of lipids (about 90%, mostly phosphatidylcholine), proteins, and glycoproteins. Surfactant decreases the surface tension within alveoli, which decreases the pressure required to expand them. Decreased surfactant production increases alveolar collapse. In alveolar cells, surfactant is stored within specialized granules called lamellar bodies. As fetal lungs mature, the ratio of lecithin to sphingomyelin (both glyoproteins in surfactant) increases. Tests on amniotic fluid can be helpful in predicting degree of pulmonary immaturity before birth. These include measuring the lecithin-sphingomyelin ratio, fluorescence polarization, and lamellar body counts. The recent development of synthetic surfactant has allowed treatment of pulmonary immaturity immediately after birth.
Other answers:
Congenital pulmonary airway malformation, also known as congenital cystic adenomatoid malformation, is the most common primary congenital abnormality of the lung. CPAMs occur as focal growths in the developing lung that can have cystic and adenomatous components. Large lesions can inhibit development and growth of normal alveolar tissue in the lung by compression. In contrast to surfactant deficiency, these do not typically improve as a baby gets older. Thus, symptomatic CPAMS at birth are frequently treated by surgical resection.
Congenital lobar emphysema is a rare congenital abnormality of the lungs that presents with hyperinflation and air trapping. Chest x-ray shows distention of one lung and compression of the contralateral one. As with CPAMs, congenital lobar emphysema does not typically resolve on its own. Asymptomatic cases can be followed. Treatment for symptomatic congenital lobar emphysema usually involves surgical resection of the affected lobe.
Tetralogy of Fallot (TOF) includes RV outflow tract obstruction, ventricular septal defect, overriding aorta to the right (such that the origin of the aorta overrides the VSD), and RV hypertrophy. Infants presenting with respiratory distress due to TOF would not be expected to improve substantially without surgical correction.
Robbins PBOD 8ed. pp Robbins 456-458
A 57 year old male presents with crushing chest pain, dyspnea, and diaphoresis of one hour duration. He is brought to the emergency department by his family where EKG reveals marked ST elevations in the lateral leads. If the patient dies 48 hours after the onset of the above event, which of the following would be the most classic histologic finding in the involved area of the heart?
Early granulation tissue formation and edema
Coagulative necrosis and neutrophilic infiltrate
Edema and hemorrhage
Wavy fibers throughout the infarcted area
Late granulation tissue and collagen formation
Phagocytosis and macrophages at the border
The correct answer is Coagulative necrosis and neutrophilic infiltrate.
Explanation: The patient is presenting with classic symptoms and EKG findings of a myocardial infarction. Histopathological changes seen during a myocardial infarction are as follows.
Wavy fibers are seen at the border of infarcts in the first 4 hours.
Edema and hemorrhage is maximal in the first 12 hours.
Neutrophilic infiltrate and coagulative necrosis are maximal in days 1-3.
Phagocytosis and macrophages are seen most in days 3-7.
Later findings include development of granulation tissue (days 7-10), and collagen formation(weeks 2-8).
Robbins PBOD 8ed. pp page 550, table 12-5.
A 63 year old female is found dead in her home by her husband. No signs of external trauma are visible. Her past medical history was significant for angina, severe pulmonary hypertension, and a recent viral illness. In addition to aortic valve stenosis, dilated cardiomyopathy, and pulmonary hypertension, which of the following conditions is most likely to cause sudden cardiac death?
Myocarditis
Cardiac myxoma
Atrial septal defect
Papillary Fibroelastoma
The correct answer is Myocarditis.
Explanation: Myocarditis can be seen after viral illnesses and is a cause of sudden cardiac death. Myocarditis (inflammation of the muscular tissue of the heart) has both infectious and non-infectious causes. The most common cause is viral infection, especially enteroviruses including Coxsackie B. Other viral infectious causes include adenovirus, HIV, parvovirus B19, herpes virus 6, and hepatitis C. While myocarditis often presents with heart failure, it can also be a cause of sudden death, most likely due to ventricular tachycardia or ventricular fibrillation).
Other answers: Cardiac myxoma is a primary cardiac tumor. These often cause symptoms such as dyspnea and orthopnea by obstructing flow of blood, but are not a common cause of sudden cardiac death. Atrial septal defect is a congenital cardiac defect which is generally well tolerated and often not recognized until the patient is in their 30s. Papillary fibroeleastoma is a benign neoplasm usually found as an incidental finding at autopsy.
Robbins PBOD 8ed. p 558.
Two and a half weeks after a 9-year-old boy misses school because of a sore throat and fever he develops hematuria (blood in his urine). On physical exam, his physician discovers that he also has elevated blood pressure and edematous fingers. Throat culture does not reveal any pathogens and the urine culture is negative. The physician’s presumptive diagnosis is acute glomerulonephritis. Which ONE of the following tests would be most helpful in establishing the diagnosis for the cause of this boy’s disease?
Assaying the boy’s serum for an elevated titer of streptolysin O antibodies
Analyzing the boy’s serum for antibodies to Shiga toxin
Examining the child’s urine for the presence of a specific capsular antigen by ELISA
Culturing the child’s stool for Escherichia coli O157:H7
The correct answer Assaying the boy’s serum for an elevated titer of streptolysin O antibodies.
Explanation: Glomerulonephritis is a post-suppurative complication of streptococcal pharyngitis. As such, the bacteria responsible for the infection are likely to have been cleared by the time of clinical presentation with glomerulonephritis. However, antibody to the bacterial enzyme streptolysin O will remain elevated, and thus provides a useful tool for diagnosis of this syndrome.
Other answers: Capsular antigens are not useful for diagnosis of Group A streptococci, which produce a hyaluronic acid capsule that is not highly immunogenic. The other options listed focus on Shiga toxin-producing E. coli, which would have been more likely had the patient presented with prior or ongoing blood diarrhea rather than a pharyngitis. Infections with Shiga-toxin producing E. coli typically occur after consumption of contaminated food or water can result in bloody diarrhea. A dangerous sequela that may occur following infection with Shiga toxin-producing E. coli is the hemolytic uremic syndrome (HUS). Symptoms of HUS include kidney failure, thrombocytopenia, and hemolytic anemia.
A 60 y/o retired captain reports to clinic with a chronic productive cough. Physical exam reveals non-tender cervical and axillary lymph node enlargement. Examination of a deep cough sputum shows gram positive diplococci. The patient is placed on a 10 day treatment with oral amoxicillin. A subsequent hemogram shows hemoglobin 9.0 g/dL; hematocrit 24%; MCV 90 µm3; platelet count 140,000/mm3; and WBC count 68,000/mm3. The peripheral blood smear shows a predominance of mature-looking lymphocytes. On flow cytometry the predominant cells are TdT-, CD19+, and CD5+. Additional laboratory tests are ordered: which of the following findings could be expected.
Beta globulin spike consistent with free L-chains
t (8,14) associated with c-Myc overexpression
X-linked mutation
Partial deletion of Chromosome 13 (or chromosome 13q deletion)
The correct answer is Partial chromosomal deletion.
Explanation: The patient most likely has chronic lymphocytic leukemia, a chronic lymphoproliferative disorder (neoplasm) of mature B cells. CLL is the most common leukemia in the U.S. It is more common in men than women, but is not X-linked. It typically occurs in older individuals, with the median age at diagnosis being around 70. Patients can present with fevers, sweats, weight loss, and fatigue. Physical signs include lymphadenopathy and splenomegaly. Invovlement of skin (and other organs) can also occur. Blood counts demonstrate an absolute lymphocytosis, and frequently the disease is diagnosed in asymptomatic patients after detection of lymphocytosis on a routine blood count. In some cases it can present with an unexpected infection such as streptococcal pneumonia. Flow cytometry of CLL cells demonstrates expression of B-cell antigens such as CD19 (or CD20), CD5 (a T-cell antigen that is also present on B1 B cells), and low levels of surface immunoglobulin. Cells in CLL are also typically TdT negative (TdT is terminal deoxynucleotidyl transferase, a DNA polymerase typically expressed in immature pre-T and pre-B cells).
In this case, the lymph node enlargement is consistent with the high WBC count and the finding of CD5+ lymphocytes. In ~20% of cases there is a mild hemolytic anemia which would cause a reticulocytosis in early stages of the disease. While the lymphocytes are mature and may produce immunoglobulin they do not produce free light chains as in multiple myeloma. A partial deletion of chromosome 13 (13q14 = deletion of region 14 on the long “q” arm of chromosome 13) occurs in over half of CLL cases and is a positive prognostic indicator (mechanism unknown). Unless immunosuppressed, the patient is unusually old for Burktt’s lymphoma and associated t(8;14).
Robbins PBOD 8ed. pp 603-605
During a routine physical examination, you determine that Colonel Needapee, a 55-year-old male, has an enlarged prostate. The colonel mentions that he has difficulty initiating urination. Also, you determine that his blood pressure is 100/70. Which drug would be most appropriate for treating the colonel’s benign prostatic hyperplasia while having only a minimal effect on his blood pressure?
albuterol.
tamsulosin.
prazosin.
phenylephrine.
The correct answer is D, tamsulosin.
Explanation: One way to increase urine flow in a male with benign prostatic hyperplasia is to administer a drug that will relax smooth muscle lining his urethra. Blockade of a1 adrenergic receptors on urethral smooth muscle will prevent norepinephrine from causing constriction of the urethra. Both prazosin and tamsulosin block a1 receptors and both would relieve the symptoms of benign prostatic hyperplasia. However, tamsulosin is selective for a 1A receptors while prazosin blocks both a 1A and a1B receptors. Since urethral smooth muscle expresses only a 1A receptors and vascular smooth muscle expresses both a1A and a 1B receptors, tamsulosin can effectively relax urethral smooth muscle without preventing vasoconstriction since a1B receptors remain unblocked. On the other hand, prazosin would block both a1A and a 1B receptors on the vasculature and may result in an unsafe drop in blood pressure.
A 67-year-old man smoked 2 packs of cigarettes daily since the age of 17. He had a prior history of many “bad colds” that he could not shake off. He would cough every morning until he would have his first cigarette puff. He presented with recent significant weight loss, weakness, constipation, and a serum calcium of 12 mg/dL. A central lung shadow was found on chest X-ray that had not been present one year earlier. A diagnostic transbronchial biopsy was obtained. The patient’s most likely diagnosis was:
Bronchiectasis
Squamous cell carcinoma
Tuberculous granuloma
Silicotic nodule
Answer: Squamous cell carcinoma.
The patient’s age and heavy smoking history are important clues. His clinical history is also significant for chronic bronchitis, which also is consistent with his smoking history. The features that favor a diagnosis of squamous cell carcinoma are his prior smoking history, recent significant weight loss and the presence of a radiographic lung shadow. His history of constipation, weakness and an elevated serum calcium level are all consistent with a hypercalcemia paraneoplastic syndrome, also a feature of squamous cell carcinoma. None of the other possibilities have a relationship to cigarette smoking although they may coincidentally occur in smokers. Silicotic nodules tend to be multiple and bilateral. Bronchiectasis typically is associated with production of copious, foul-smelling sputum. Tuberculosis can be associated with significant weight loss. However, it frequently also presents with fevers and/or night sweats, and hypercalcemia is not a typical feature of tuberculosis.
Robbins PBOD 8ed. pp 721-729
A 66 year old woman presents with abdominal pain, increased thirst, constipation and fatigue. She has a history of depression for which she takes multiple over the counter medications, including vitamins. Her examination is notable for a blood pressure of 160/72. Laboratory studies included:
Total calcium 12.5 mg/dL (8.2-10.4)
Phosphorus 2.5 mg/dL (3.0-4.5)
Creatinine 0.8 mg/dL (0.7-1.2)
Intact PTH 80 pg/mL (10-60)
24 hour urine:
Calcium 350 mg/dL
Creatinine 1.2 gm/dL
What is the MOST likely cause of this patient’s hypercalcemia?
Hypercalcemia of malignancy
Sarcoidosis
Multiple myeloma
Primary hyperparathyroidism
Answer: primary hyperparathyroidism
Hypercalcemia should normally cause suppression of PTH; therefore this patient has hyperparathyroidism. While frequently asymptomatic, hyperparathyroidism can cause a range of symptoms in patients.
The high PTH levels lead to development of kidney stones, bone disease, proximal renal tubular acidosis, and hyperuricemia with associated gout. High PTH levels also cause electrolyte abnormalities such as low phosphate and magnesium levels.
High calcium levels from any cause can cause disease in multiple organ systems, including the kidneys (kidney stones, nephrogenic diabetes insipidus causing increased thirst and polyuria and polydipsia, distal RTA, and renal insufficiency), the gastrointestinal system (nausea, vomiting, anorexia, pancreatitis, constipation due to decreased bowel motility), the musculoskeletal system (osteopenia, bone pain, muscle aches and weakness), the neurologic system (confusion, decreased concentration), and the cardiovascular system (decreased QT interval, hypertension, and bradycardia).
Other answers: Hypercalcemia of malignancy is mediated through a PTH-related protein that mimics PTH effects on receptors, but is not measured by the PTH immunoassay. Sarcoidosis and other granulomatous diseases cause hypercalcemia through increased conversion of 25 hydroxyvitamin D to 1,25 dihydroxyvitamin D. All of these conditions should present with low PTH levels.
A 73-year-old retired insulator gave a history of progressive dyspnea over many years and subsequently presented with a non-productive cough and weakness. During WW II he did extensive repairs to the insulation in the boiler rooms aboard ships. Physical examination disclosed bibasilar inspiratory crackles. Chest radiographs demonstrated bilateral irregular lung opacities in the lower lung zones as well as calcification of both hemidiaphragms. An open lung biopsy of this patient is most likely to demonstrate:
Birefringent silica particles associated with concentric, whorled fibrosis
Interstitial fibrosis associated with ferruginous bodies
Caseous necrosis of the pleura
Hyaline membranes lining the alveoli
Answer: Interstitial fibrosis associated with ferruginous bodies.
The patient’s occupational history is consistent with prior heavy asbestos exposure that has been well described in insulators and Navy personnel working aboard ships during WW II. The clinical picture is highly suggestive of asbestosis, as reflected by weakness and a non-productive cough in association with bibasilar inspiratory rales. Asbestosis classically is associated with bilateral irregular opacities in the lower lung zones, findings that are characterized by interstitial pulmonary fibrosis associated with dumbbell-shaped, beaded, brown asbestos (ferruginous) bodies. Persons occupationally exposed to asbestos also may have bilateral parietal or diaphragmatic pleural plaques that frequently calcify. These asbestos-related pleural plaques/calcifications occur independently from asbestosis but both types of asbestos-related lesions may coexist in the same patient.
Other answers: None of the other diagnostic possibilities have any relationship to asbestos exposure. Exposure to silica typically occurs in miners, sandblasters (individuals working in quarries), and artists (especially those working with ceramics). Silicosis is characterized by birefringent silica particles within concentric fibrous nodules, mainly in the upper lobes. This is distinct from the lower lobe location of asbestosis. Caseous necrosis is the hallmark of tuberculous granulomas and alveolar hyaline membranes are pathognomonic of adult respiratory distress syndrome.
Robbins PBOD 8ed. pp 696-701
A 65-year-old retiree presented with a lesion on his tongue that could not be scraped off. He stated that he first noticed it several months previously but he did not feel the need to have it examined because it did not hurt. He had been a pipe smoker for approximately 20 years. He was referred to a surgeon for evaluation. This was a photograph of the lesion. The probable diagnosis was:
Pleomorphic adenoma
Squamous cell carcinoma
Oral thrush
Leukoplakia
Answer: Leukoplakia.
Leukoplakia refers to firm, white, warty plaques in the oral cavity, generally of long duration, which cannot be scraped off. They are typically seen in cigarette, pipe or cigar smokers. They must be regarded as premalignant lesions although only about 5-25% actually are premalignant. They occur most commonly in males between 40 and 70 years of age.
Other answers: Oral thrush, which can be seen in young infants, individuals with AIDS or other severe immunodeficiencies, and individuals on systemic or inhaled corticosteroids, is caused by the yeast Candida albicans and is characterized by painful, creamy white, slightly raised lesions on the tongue, inner cheeks, roof of the mouth, gums, tonsils, or back of the throat. Lesions of oral thrush typically bleed slightly when scraped. Oral squamous cell carcinoma manifests either as firm, exophytic or ulcerative growths on the tongue, inner surface of the cheeks or soft palate.
Pleomorphic adenoma is a generally benign, slow-growing tumor of the major salivary glands and usually is characterized by painless, progressive enlargement of the parotid gland. It is not seen on the dorsal surface of the tongue.
Robbins PBOD 8ed. pp 743-745
A USUHS student went on a summer experience to rural Brazil. During his stay there, he forgot what he’d learned in Medical Parasitology and bathed daily in a small stream. After several days he discontinued these noontime swims because of marked pruritis and the onset of a rash immediately following his exposure to the water. He later learned that the stream was the main swimming and washing area for a small village 200 yards upstream. Given his bathing habits, which of the following is MOST likely to be responsible for his skin symptoms:
Sarcoptic itch
Cercarial dermatitis
Pustular dermatitis secondary to Staphylococcus aureus
Contact dermatitis
Answer: Cercarial dermatitis
Cercariae are the infectious stage of parasites of the genus Schistosoma, the causative agent of schistosomiasis. The species of schistosome prevalent in Brazil is Schistosoma mansoni. In this case, the village upstream has inhabitants who are infected and who are passing eggs in their stools. The eggs can be washed into the stream by rainfall. Once in water, the eggs hatch and the small ciliated miracidium stage seeks out the appropriate species of snail which serves as the intermediate host. It is from this snail host that the cercariae emerge, then swim tadpole-like in the water seeking a human host. A cercaria has enzymes in its anterior end that allow it to penetrate intact human skin, enter the capillary bed, and eventually end up in the liver where maturation to an adult worm takes place. The process of penetration of skin can invoke a local inflammatory response in the human host, causing a punctate, pruritic rash that is referred to as cercarial dermatitis. A similar skin rash can occur in humans exposed to water infected with various types of bird schistosomes, including bodies of water such as the Chesapeake Bay and the Great Lakes. This dermatitis is referred to as swimmer’s itch. These non-human schistosomes cannot complete their life cycle, so persons who have swimmer’s itch in the U.S. are not in danger of developing the disease schistosomiasis.
Other answers:
The itch referred to as ‘sarcoptic’ refers to the arthropod that causes it, the scabies mite Sarcoptes scabei. This is transmitted primarily by direct contact with an infested person and their fomites (clothing, bedding, etc.).
Contact dermatitis is typically due to a delayed hypersensitivity response to an irritant (such as a detergent or chemical solvent) or allergen (such as poison ivy or adhesives or certain metals such as nickel) and requires that the inciting (noninfectious) agent be in direct contact with the skin.
Pustular dermatitis would not have the same chronology of water exposure and the lesions (pustules) are distinctly different in appearance from the punctate, nonpustular lesions seen in cercarial dermatitis.
A 57 y/o post-menopausal woman G4P4 has blood drawn for routine testing prior to elective repair of a uterine prolapse. Except for the prolapse, she has had no significant medical problems. Her diet is well balanced; however, the anesthesiologist elicits a history of recent weight loss and ‘digestive problems.’ Liver and renal functional tests are normal, but her hemoglobin is only 9.4 gm/dL. Surgery is postponed while iron supplements are prescribed. A month later the hemoglobin has failed to increase and she has not gained weight. Gastroscopy and colonoscopy reveal no malignancy or source of gastrointestinal bleeding, and there is no family history of cancer. The reticulocyte count remains low. The WBC is 3,800 cells/mm3. Which of the following tests would likely prove most useful in differential diagnosis of her condition?
Serum cobalamin
Serum gastrin
Serum erythropoietin
Serum TSH
The correct answer is Serum cobalamin.
‘Digestive problems’ raise the possibility of chronic gastritis or intestinal malabsorption associated with folate or cobalamin (B12) deficiencies. Microscopic examination of the blood smear for evidence of oval macrocytosis, enlarged polylobulated neutrophils, or giant bands forms would be useful but not sufficient to distinguish between folate and cobalamin deficiencies. The latter is most critical to exclude since treatment with folate alone can mask disease progression toward neuropathy (which occurs with vitamin B12 deficiency). The lack of evidence for blood loss, the failure of iron supplements to correct the anemia, and the low WBC count also raise the possibilities of bone marrow dysfunction due to chronic disease or a sideroblastic anemia associated with aging. Exclusion of these conditions might require bone marrow biopsy if the serum tests were not informative.
Other answers: Provided healthy kidney function, serum EPO can be expected to increase in many types of anemia. It can be useful in distinguishing between aplastic anemia (increase) and anemia of chronic renal disease (decrease). Gastrin levels are often elevated in pernicious anemia. However, high gastrin levels are not specific for pernicious anemia or B12 deficiency, and would not be obtained as a first line test for determining whether a pt is B12 deficient. Additionally, autoantibody damage to gastrin producing glands can occur in pernicious anemia, decreasing gastrin levels. Autoantibodies in pernicious anemia may cross-react with the thyroid gland epithelium and lead to a compensatory increase of TSH, but TSH levels alone would not illuminate the underlying cause.
Robbins PBOD 8ed. pp 656-658
A 38-year-old Caucasian non-smoker presented with a firm, discrete, painless mass in the right parotid region that had been present for approximately 8 years. Diagnosis was obtained by fine needle aspiration. The likely diagnosis was:
Mumps
Nasopharyngeal carcinoma
Pleomorphic adenoma
Warthin tumor
Answer: Pleomorphic adenoma.
The parotid gland is the favored site of pleomorphic adenomas but is also the exclusive location of Warthin tumor. However, the patient’s age favors pleomorphic adenoma since the highest incidence of Warthin tumor is in the 5th-7th decades of life. Warthin’s tumor also is closely linked to smoking. Mumps occurs in much younger patients, is acute in onset and is associated with painful enlargement of the parotid gland. Nasopharyngeal carcinoma presents with neck swelling and/or nasal obstruction but does not involve the parotid region. This tumor is rare in Caucasians and is especially common in Chinese nationals.
Robbins PBOD 8ed. pp 757-759
A 72-year-old man presents with tremor in the left hand that is lessened during voluntary movements but becomes more pronounced at rest. He complains that he has been having trouble with his balance, and that it has been taking him a longer time to complete daily activities. Walking has become more difficult, and he exhibits short “shuffling” strides. He reports that while walking he often hesitates or “freezes up,” which has caused him to fall twice in the last month. A decision is made to prescribe a combination drug therapy consisting of L‑dihydroxyphenylalanine (L‑dopa) and an inhibitor compound, carbidopa, that does not cross the blood-brain barrier but prevents the conversion of L-dopa to dopamine in peripheral tissues. Because the conversion is promoted by a specific cofactor derived vitamin, the patient was advised to avoid abnormally high intake of vitamin supplements. Large quantities of which vitamin would be most important to avoid?
Vitamine B12
Pyridoxine
Thiamine
Biotin
Answer: Pyridoxine
The clinical vignette is characteristic of Parkinson disease. The conversion of L-dopa to dopamine is a decarboxylation reaction that normally takes place in the brain as well as in peripheral tissues. Amino acid decarboxylases require pyridoxal phosphate (PLP) derived from the vitamin pyridoxine (vitamin B6)*. Carbidopa is an analog of L-dopa that acts as a dopa decarboxylase (DDC) inhibitor. Because it does not pass the blood-brain barrier, carbidopa acts peripherally, but does not block dopamine formation in the brain. The peripheral action of carbidopa prevents adverse systemic effects of dopamine and helps to maintain high levels of dopa sufficient for entry into brain tissue via aromatic amino acid transporters (minimizing the pharmacological first pass effect). Consumption of high levels of vitamin B6 has the potential to promote systemic decarboxylation of L-dopa by increasing activity of residual decarboxylase, counteractive to the effect of the DDC inhibitor.
Biotin is incorrect because biotin is used for carboxylation reactions, not decarboxylations. Thiamine is involved in decarboxylations, but with alpha-keto acids (like pyruvate and alpha-ketoglutarate) not amino acids. Vitamin B12 is used for only two reactions in humans; methyl transfer reactions (methionine synthase) and a radical rearrangement that converts methylmalonyl-CoA to succinyl-CoA (methylmalonyl-CoA mustase) involved in odd-chain fatty acid and branched-chain amino acid catabolism.
*Decarboxylation reactions are also involved in converting other amino acids to a number of other neurotransmitters (e.g., histamine, GABA, serotonin). PLP is also used for other types of reactions performed on amino acids such as racemization, transamination and beta/gamma eliminations. Major metabolic pathways that require PLP include: amino acid metabolism, heme biosynthesis, neurotransmitter biosynthesis and the formation of sphingolipids. Vitamin B6 is therefore essential for normal function of the nervous system and hematopoiesis.
A 15-year-old boy is brought by his parents to their family doctor with a complaint of weight loss and watery diarrhea. Physical examination is unremarkable. A stool sample is negative for blood, ova, and parasites. A small bowel biopsy obtained on endoscopy reveals blunting of villi with increased chronic inflammatory cells. His symptoms improve with elimination of gluten from his diet. Which of the following is associated with this disease?
T-cell lymphoma
Hemochromatosis
Invasive aspergillosis
Hepatitis B virus
The correct answer is T-cell lymphoma.
The child has Celiac disease, an immune mediated disease due to hypersensitivity to gluten-containing products in genetically predisposed individuals. Most individuals who develop celiac disease carry the Class II HLA allelles, HLA-DQ2 or HLA-DQ8. Clinically they present with anemia, chronic diarrhea and fatigue. Many patients are asymptomatic. The morphology of celiac disease is villous atrophy, crypt hyperplasia and intraepithelial lymphocytosis composed of CD8+ lymphocytes. T cell lymphoma and digestive tract cancers appear to occur with greater frequency in patients with Celiac disease than in the general population. Dermatitis herpetiformis (a skin condition characterized by papules and vesicles on the extremities and trunk), Type 1 diabetes, and IgA deficiency are non-malignant diseases associated with Celiac disease.
Other answers: Hemochromatosis is a multisystem disease due to mutation in the HFE gene leading to iron overload that leads to cirrhosis, heart failure and secondary diabetes. Aspergillosis is seen in immunodeficient individuals leading to blood vessel obstruction and infarction. Hepatitis B is an enveloped DNA virus that causes acute and chronic hepatitis. While chronic infection with hepatitis B virus is associated with some extrahepatic diseases, celiac disease is not one of them. Extrahepatic manifestations of hepatitis B are typically due to circulating immune complexes and include serum sickness, polyarteritis nodosa, membranous nephropathy, and membranoproliferative.
Robbins PBOD 8ed pp 795-796
Treatment with a single drug failed to control a patient’s complex partial seizures. You decide to treat this patient with two anti-seizure drugs simultaneously. Your patient’s current antiseizure medication is lamotrigine. What is the MOST APPROPRIATE choice of adjunct therapy if your goals are to MAXIMIZE control of your patient’s seizures and MINIMIZE the possibility of drug interactions and an overlap in mechanism of action?
valproate
phenytoin
gabapentin
carbamazepine
The correct answer is gabapentin.
Lamotrigine, phenytoin, valproate, and carbamazepine all share inhibition of Na+ channels as a mechanism of action. Gabapentin, in contrast, does not inhibit Na+ channels. While the mechanism of action of gabapentin is not yet fully understood, it appears that interaction with voltage sensitive calcium channels is important for its function. Gabapentin is: 1) approved for adjunct therapy in patients with complex partial seizures; 2) does not significantly overlap mechanisms of action with lamotrigine; and 3) of all the listed anti-seizure drugs, has the lowest propensity for drug interactions because it is excreted unchanged in the urine.
A 66 year old physician with a family history of cystic fibrosis is approaching retirement. Ten years prior, he was successfully treated for prostatic carcinoma. He drinks one or two glasses of red wine daily. Now, he is found to have an HbA1c of 7.2%. There is no response to carbohydrate restrictions and weight loss of 10 lbs so he is placed on an oral hypoglycemic agent. At a 6 month examination he reports excessive flatus and is noted to have an additional 5 lb weight loss. A stool sample appears fatty, but serum lipase is not elevated. He elects to try Pancreaze (lipase) as a dietary supplement. Ultrasound reveals no evidence of lithiasis and the serum bilirubin is not elevated; however CT detects a 1.5 cm nodule in the head of the pancreas. It is explored and totally removed. The pathology report notes that the lesion is confined within the pancreas but abuts the pancreatic duct and is attached to an adjacent vein. Which conclusion is most accurate with respect to the complex events described:
There was a drug reaction leading to chronic pancreatitis and complications
The pancreatic insufficiency and cyst removed were delayed expressions of cystic fibrosis
This patient is at increased risk for a pulmonary embolism
This was a typical progression of chronic pancreatitis induced by excessive alcohol intake
Submit
The correct answer is The patient is at increased risk for a pulmonary embolism.
The patient most likely had a pancreatic carcinoma. Total removal of the pancreas is the treatment for pancreatic carcinoma and a needle biopsy or frozen section during surgery would have confirmed the diagnosis. Attachment to a vein suggest vascular invasion. Hypercoagulability is a recognized risk in these cases.
Other answers: Pancreatitis due to bile reflux can be associated with excessive ethanol intake, but in this case it presumably occurred as a consequence of secretory stasis due to a slowly growing carcinoma impinging upon the intraglandular portion of the duct of Wirsung or the duct of Santorini. While chronic pancreatitis associated with alcohol can lead to pseudocyst formation there is no evidence of a relation to carcinogenesis. Islets of Langerhans are concentrated in the pancreatic tail. Except in far advanced cases function is not usually compromised either in pancreatitis. Cystic fibrosis is frequently associated with exocrine pancreas insufficiency (and thus could account for the patient’s malabsorptive syndrome) and often with impairment of islet cell function as well (causing diabetes). However, CF does not lead to mass lesions within the pancreas. In this case, the patient’s high HgbA1c is likely due to type II diabetes unrelated to the pancreatic cancer.
Robbins PBOD 8ed. pp 900-903
A 55-yo man presents with sudden loss of movement and sensation on part of the left side of his body. He currently smokes a pack of cigarettes per day and has a history of smoking since his early 20s. His vital signs include: Temperature, 99.0 ºF; Pulse, 80/minute; Respirations, 16/minute; and Blood pressure, 160/100 mm Hg. A cerebral angiogram reveals an occlusion in his middle cerebral artery. Laboratory results include a hemoglobin A1C of 9%. Which one of the following components of blood lipids is most important in contributing to his disease?
VLDL
Lipoprotein lipase
Oxidized LDL
Chylomicrons
Correct answer: oxidized LDL
A stroke is often the consequence of cerebral atherosclerosis. LDL is the major vehicle for the delivery of cholesterol to the tissues, including the arterial walls. When LDL levels are increased or when other factors including hypertension, smoking, and diabetes are present, there is a higher degree of LDL oxidation. Scavenger receptors (SR-A1 and SR-A2) expressed in macrophages nonspecifically take up oxidized LDL into the arterial wall. This initiates the formation of atheromas or plaques that can restrict blood flow, and eventually leads to the production of emboli or a rupture in the arterial wall. Of note, in contrast to LDL, HDL is thought to act as a protective factor against atherosclerosis. HDL is involved in VLDL and chylomicron maturation, and in reverse cholesterol transport. The latter process involves the transport of cholesterol from cholesterol‑rich cells back to the liver.
Incorrect answers:
Lipoprotein lipase (LPL) is an enzyme expressed on endothelial cells in heart, muscle, and adipose tissue. LPL is a triglyceride hydrolase that requires ApoCII as a cofactor. Deficiencies in LPL lead to elevated levels of triglycerides in the blood. Individuals with LPL deficiency generally have lower levels of LDL cholesterol.
Chylomicrons are formed in intestinal epithelial cells and deliver dietary fats to the liver. Triglycerides are the most abundant component of chylomicrons. Chylomicrons are the most elevated lipoprotein when LPL or ApoCII is defective.
VLDL particles are synthesized in the liver and transport triglycerides to extrahepatic tissues. Although elevated levels of VLDL increase the risk for premature atherosclerosis, the risk is far greater when LDL levels are high.
A 30 year old man is being evaluated by his physician for blood in his urine. As part of the work up a kidney biopsy is done and shows the following linear IgG pattern on immunofluorescence microscopy. Which of the following features is also consistent with the disease process shown?
crescents
hyaline arteriolosclerosis
mesangial deposits
subendothelial deposits
The correct choice is Crescents.
The immunofluorescence image shows a linear pattern consistent with anti-GBM disease. Most patients with anti-GBM disease (Goodpasture disease, if pulmonary involvement is present) develop crescents (crescentic glomerulonephritis). Mesangial, subendothelial and subepithelial deposits are not features of anti-GBM disease. Arteriolosclerosis is seen in hypertension.
Robbins PBOD 8ed pp 913, 920-921
Structures that control the paracellular transport of molecules between epithelial cells but do not allow direct passage of molecules from one epithelial cell to another are
desmosomes
gap junctions
occluding (tight) junctions
centrosomes
Answer: occluding (tight) junctions
Explanation: Paracellular transport is the movement of substances through an epithelial layer by moving between cells; in this case between the lateral sides of epithelial cells. Epithelial cells are joined together by junctional complexes. Occluding (tight) junctions are the most apical junctions and allow epithelial cells to function as a barrier. The amount of material that is transported along the paracellular pathway is dependent on the permeability of the occluding junctions.
Incorrect answers:
Desmosomes (macula adherens) are localized spotlike, anchoring junctions between cells. They do not provide a continuous structure around the cell so are not able to function as a barrier.
Gap junctions allow direct passage of molecules from the cytoplasm of one cell to the cytoplasm of the adjacent cell.
Centrosomes are located within the cell and are the microtubule-organizing center of the cell. (Note: this is different from basal bodies, which are located near the plasma membrane under the apical surface of epithelial cells and are the microtubule organizing centers of cilia.)
A patient with a seizure disorder that has been difficult to control is prescribed phenobarbital. For several weeks, this patient has been taking ibuprofen, a drug that is significantly metabolized (> 70%) by CYP2C9, to reduce peripheral joint pain. A week after starting the phenobarbital therapy, the patient complains of a return of the pain in his joints. A likely reason for the return of the joint pain is that chronic phenobarbital treatment
activates constitutive androstane receptors (CAR) resulting in an increase in the expression of CYP2C9 protein, increasing the elimination of ibuprofen.
inhibits the metabolism of ibuprofen by CYP3A4.
impairs the absorption of ibuprofen from the gastrointestinal tract.
inhibits the metabolism of ibuprofen by CYP2C9.
The correct answer is – “activates constitutive androstane receptors (CAR) resulting in an increase in the expression of CYP2C9 protein, increasing the elimination of iburofen”
Explanation: Since ibuprofen is metabolized to a significant extent by CYP2C9, a marked increased in the amount of this enzyme is likely to increase the clearance of ibuprofen from the blood by increasing its metabolism. Phenobarbital is a known inducer of the expression of this enzyme in the liver, acting by binding to the CAR receptor. The CAR-phenobarbital complex forms part of the transcriptional protein complex that increases the rate of transcription of the CYP2C9 mRNA for this protein, and thus the amount of active protein generated. The other answers are incorrect statements.
A 71-year old woman was seen by her physician with a complaint of left leg weakness for three months. On examination the patellar and Achilles tendon reflexes were exaggerated on the left side and the plantar extension reflex could be elicited in the left foot. Reflexes were normal on the right side of the body and the left upper limb. Vibration and joint position sense were decreased on the left side below the level of the inguinal ligament and pinprick sensation was decreased on the right leg and foot. Which of the following conditions is most likely the cause of the patient’s neurological deficits?
Compression of the left side of the spinal cord
Occlusion of the anterior spinal artery
Multiple sclerosis
Expansion of the central canal
Answer: compression of the left side of the spinal cord
Upper motor neuron signs in the left lower limb (including exaggerated reflexes and plantar extension reflex) suggest a lesion to the lateral corticospinal tract on the left side of the spinal cord or the primary motor cortex on the medial aspect of the frontal lobe on the right. The sensory deficits pinpoint the lesion to the spinal cord. Specifically, loss of vibration and position sense on one side and pain on the opposite side is a hallmark of spinal hemisection. In this case, the posterior columns and anterolateral spinothalamic tracts were damaged on the left side near L1. Since the posterior column system does not cross until the lower medulla, vibration and joint position sense were lost on the left. Since the anterolateral spinothalamic tract fibers are already crossed in the cord, pain sensation was lost on the side opposite the lesion (right side).
A 40 year old male developed proteinuria. He had a history of recurrent abdominal pain and arthritis of the knee. A kidney biopsy revealed AA amyloid. Of the following which is most likely in this patient?
Familial Mediterranean fever
Familial amyloidotic neuropathy
Medullary carcinoma of the thyroid
Multiple myeloma
Correct answer: Familial Mediterranean Fever.
Explanation: Familial Mediterranean Fever (FMF) is due to mutations in the MEFV gene and is associated with spontaneous attacks of fever and pain due to serosal inflammation. The most common pain syndrome is abdominal pain due to sudden peritonitis. Pleuritis and synovitis are other common clinical manifestations. Attacks typically last for a few days and then resolve. While the pathophysiology underlying FMF is not completely understood, it is clear that inflammation plays a key role. During attacks, there is an abnormally high production of IL-1 leading to the autoinflammatory syndrome. Over time, patients with FMF can develop secondary (AA) amyloidosis. AA amyloid is produced in the setting of chronic inflammation, due to increased hepatic synthesis of the acute phase protein serum amyloid A. The AA amyloid can then deposit in the kidneys, spleen, liver, and intestines. Renal amyloidosis in FMF often progresses to nephritic syndrome and then, over a period of many years, to end-stage renal disease. Intestinal amyloidosis in FMF can result in malabsorption.
Other answers: Patients with multiple myeloma and production of a monoclonal antibody (M spike) may have AL amyloid resulting from free light chains (i.e. Bence-Jones protein). Amyloid in medullary thyroid carcinoma is composed of calcitonin and is localized to the tumor. Familial amyloidotic neuropathy leads to deposition of ATTR amyloid composed of transthyretin. Of note, Alzheimer disease (not listed as an option) demonstrates Aβ amyloid, especially in the brain, derived from the APP protein.
Ann goes to the gym every other day, where she cross-trains on a stationary bike. She also takes the opportunity to weigh herself on the medical scale in the women’s locker room. One day, after a week of heavy exercise and strict dieting, she was mortified to see that she had gained a pound. Unknown to her (and the other scale users), the service technician had mistakenly changed the scale’s calibration, so it was reading high. Ann responded by starving herself for 2 days. Depressed by her obsession with weight, she visited her friend Jean Ann Tonic, a bad choice, because Jean convinced Ann to have a couple of shots of vodka. Ann soon collapsed, unconscious, and was rushed to the hospital. The ER doctors found that she was severely hypoglycemic, and started IV glucose. Ann regained consciousness when her blood glucose level was brought back to normal, and she recovered without incident. At a biochemical level, how did alcohol cause the hypoglycemia?
Inhibition of gluconeogenesis in her liver
Inhibition of insulin release by her pancreas
Activation of glycogen synthesis in her liver and muscle
Inhibition of glycogen breakdown in the liver
In a fasted state, blood glucose levels are maintained by liver glycogenolysis and gluconeogenesis. After about 24 hours of fasting, glycogen supplies are exhausted so gluconeogenesis is the only pathway that can maintain blood glucose. Ethanol is metabolized in the liver to acetate via two dehydrogenase reactions that convert NAD+ to NADH. The effect is to greatly elevate the NADH/NAD+ ratio in the liver, thereby affecting all pathways that use this pair of cofactors. One of the pathways is gluconeogenesis, which requires pyruvate and/or oxaloacetate as starting materials. When the ratio of NADH/NAD+ is high, pyruvate is reduced to lactate and oxaloacetate is reduced to malate, making the starting materials unavailable for gluconeogenesis. Therefore, one of the mechanisms by which ethanol causes hypoglycemia is inhibition of gluconeogenesis in the liver. Additionally, there is recent data that ethanol may also cause hypoglycemia by increasing insulin release from the pancreas.
Incorrect answers:
Inhibition of glycogen breakdown in the liver is incorrect because Ann’s 2-day fast depleted her of liver glycogen. Also, ethanol does not affect glycogen breakdown, which is why most people don’t faint when they drink moderately. Inhibition of insulin release would result in increased glucose levels. Also, there is recent data suggesting ethanol actually increases insulin release. One cannot rule out activation of glycogen synthesis on first principles; ethanol could have some regulatory effect on the pathways of glycogen synthesis, but this potential mechanism has not been demonstrated.
