INFECTION/ EXCLUSIVE/PGQUESTIONS

INFECTION/EXCLUSIVE:

Q. The duration of treatment for infections caused by rapidly growing mycobacteria is usually?
A. 7 to 10 days
B. 10 to 14 days
C. 14 to 28 days
D. 1 to 3 months
E. 4 to 6 months

Ans: E.

As with TB therapy, treatment of rapidly growing mycobacterial infections usually is very prolonged. Most skin and soft tissue infections require a combination of debridement and long courses of therapy; pulmonary infections may require even more than 6 months of appropriate chemotherapy for cure.

Q. What biochemical feature of Salmonella distinguishes it from most of the enteric flora (other enterobacteraciae)?
A. Lactose nonfermenter
B. Glucose fermenter
C. Oxidase negative
D. Nitrate reducer
E. Failure to produce hydrogen sulfide

Ans: A.

 Enteric pathogens, such as Salmonella and Shigella, are often distinguished from normal fecal flora by using plates such as MacConkey agar or Hektoen agar, which allow easy selection of lactose nonfermenters. The enterobacteraciae and Salmonella are glucose fermenters, oxidase negative, and nitrate reducing (B–D). Salmonella does produce hydrogen sulfide, which distinguishes it from Shigella and most enterobacteraciae.

Q. Which of the following combination would be most appropriate for a serious P. aeruginosa pneumonia with bacteremia in a bone marrow transplant recipient?
A. Penicillin and gentamicin
B. Cefepime and tobramycin
C. Cefepime and piperacillin
D. Ciprofloxacin and cefuroxime
E. Imipenem and ticarcillin

Ans: B.

The combination of an appropriate β-lactam antibiotic and an aminoglycoside is the most effective combination against P. aeruginosa. Penicillin does not have activity against P. aeruginosa. Both (C) and (E) are incorrect answers because combinations of two β-lactam antibiotics, even if they both have activity against P. aeruginosa, are antagonistic. Cefuroxime in (D) is a secondgeneration cephalosporin without activity against the organism

Q. Which of the following is a risk factor for P. aeruginosa associated necrotizing enterocolitis?
A. Use of prior antibiotics
B. Mesenteric ischemia
C. Neutropenia
D. Ulcerative colitis
E. Pseudomembranous colitis

Ans: C.

Neutropenia in cancer patients is a risk factor for necrotizing enterocolitis. The syndrome can also occur in young infants and is often fatal.

Q. Which of the following organisms, if found on a sputum culture, definitely indicates infection?

A. Chlamydia pneumoniae

B. Pneumococcus

C. Haemophilus influenzae

D. Legionella pneumophila

E. Moraxella

Ans: D.

Legionella is not known to be a colonizer; if it is found on culture, it should be treated. While Mycoplasma and Chlamydia cultures are rarely sent, these organisms can be found in  asymptomatic  subjects, so their mere presence does not require treatment. Moraxella, Haemophilus, and Pneumococcus can also be colonizers in asymptomatic patients. However, they should be treated in a patient with clinical signs of pneumonia and one of these organisms

predominating in a sputum Gram stain and culture.

Q. Which of the following host factors is most strongly linked to increased morbidity and mortality of WNV encephalitis?

A. HIV

B. Diabetes mellitus

C. Iatrogenic immunosuppression

D. Older age

E. Young age

Ans: D.

Convincing epidemiologic evidence, especially from the 1999 New York outbreak, shows that older age is a risk factor for death from WNV infection, as is an increased incidence of chronic neurologic sequelae.

Q. When should a patient with syphilis be treated with a drug other than penicillin?
A. Pregnancy
B. Penicillin-resistant syphilis is suspected
C. Neurosyphilis
D. Patient has a penicillin allergy
E. Topical treatment of syphilitic chancre in primary syphilis

Ans: D.

Syphilis is one of the very few pathogens in which drug resistance has not become a clinical problem. It is still exquisitely sensitive to penicillin. The only reason not to use penicillin is if

the patient has a severe allergy to the drug. Primary syphilis is a systemic infection and topical treatment is not possible. Neurosyphilis is still treated with penicillin, although the antibiotic

is given via continuous IV to achieve higher CNS levels. Penicillin is the only drug approved for use in pregnant patients.

Q. Which HIV drug is also active against HBV?

A. ddI

B. Indinavir

C. AZT

D. 3TC (lamivudine)

E. Nevirapine

Ans:  D.

Lamivudine is active against both HBV and HIV, as is tenofovir. Adefovir is also active against both viruses, but nephrotoxicity prevents adefovir from being used at the higher HIV dose.These

drugs have activity against HBV because HBV replication includes a reverse transcriptase step. Other HIV drugs, including the rest of the nucleoside analogs like AZT and ddI, are not active against HBV.

Q. Which enzyme is inhibited by zanamivir?

A. Reverse transcriptase

B. Hemagglutinin

C. Protease

D. Neuraminidase

E. Dihydrofolate reductase

Ans:   D.

 Zanamivir and oseltamivir are both neuraminidase inhibitors. Hemagglutinin (B) is the other major surface protein of influenza virus, used for binding to target cells. No drugs currently target it. Protease and reverse transcriptase (C and A, respectively) are drug targets in HIV. Dihydrofolate reductase (E) is inhibited by sulfa drugs and is a target in bacteria and parasites.

Q. Which of the following is a contraindication to influenza vaccination?

A. AIDS

B. Receipt of the vaccine within the previous 5 years

C. Pregnancy

D. Allergy to eggs

E. Severe pulmonary or cardiac disease

Ans: D.

Allergy to eggs is a contraindication to the vaccine because eggs are used to grow the virus for vaccine production. Becauseinfluenza virus is a killed virus, it is safe for AIDS patients (A)

and pregnant women (C) to receive. Patients with severe pulmonary or cardiac disease (E) benefit the most from vaccination because they are the most likely to die from influenza, and vaccination programs should target them. Whereas the Pneumovax should not be repeated more than every 5 to 7 years (D), the influenza vaccination needs to be given every year to maximize

protection against currently circulating virus.

Q. Which of the following drugs is active against influenza B?

A. Amantadine

B. Stavudine

C. Oseltamivir

D. Nelfinavir

E. Rimantadine

Ans:  C.

Oseltamivir (and zanamivir) are neuraminidase inhibitors active against both influenza A and B. Amantadine (A) and rimantadine (E) are both only active against influenza A. Stavudine (B)

is a nucleoside reverse transcriptase inhibitor active against HIV. Nelfinavir (D) is a protease inhibitor active against HIV.

Q. Mucormycosis can be distinguished from aspergillosis in that Mucor branches at angles of:

A. 15°

B. 30°

C. 60°

D. 90°

E. 150°

Ans: D.

Mucor and Rhizopus form nonseptate, true hyphae with broad irregular walls and branches that form at right angles (90º).

Q.A physician is obligated to treat yeast grown from which culture site?

A. Stool

B. Sputum

C. Skin

D. Urine

E. Blood

Ans: E. Blood cultures growing yeast could theoretically be contaminants, but the physician is obligated to treat them.Yeast can colonize the other listed sites without causing disease, although yeast skin and UTIs are often diagnosed and treated. True yeast pneumonias are exceedingly rare; usually yeast in the sputum is a colonizer of the oropharynx.

Q. Which of the following drugs acts by blocking yeast cell wall synthesis?

A. Amphotericin B

B. Penicillin

C. Fluconazole

D. Nystatin

E. Caspofungin

Ans: E.

Caspofungin, an echinocandin, blocks beta-glycan synthesis required to make yeast cell walls. This is analogous to the antibacterial antibiotic penicillin (B). Polyenes, such as amphotericin and nystatin (A and D, respectively), and azoles, such as fluconazole (C), target ergosterol, a component of the yeast cell membrane

ALCOHOLISM EXCLUSIVES/MCQ/PGQUESTIONS/NEET-PG

ALCOHOLISM EXCLUSIVES:

Q. 37-year-old man with a long history of alcoholism who is brought by ambulance to the ED severely agitated but oriented and cooperative. He is diaphoretic with vital signs stable at T 38.0°C; HR 98; BP 139/85; RR 24; SaO2 100%. He develops severe stuttering speech, a tongue wag, and generalized tremors, pronounced by intentional movement. He soon undergoes a generalized tonic-clonic seizure. Once he stops seizing, he is given longer-acting diazepam and is admitted to the hospital. He reports having had his last drink about 24 hours ago. What is his most likely diagnosis?

A. Withdrawal tremors and seizure

B. Delirium tremens

C. Alcoholic hallucinosis

D. Wernicke encephalopathy

E. Korsakoff syndrome

Ans: A. This presentation is consistent with withdrawal tremors and seizures, which occur within 48 hours after the last drink. Delirium tremens is unlikely, as this condition usually occurs after 48 hours and usually is accompanied by severe autonomic instability and vital sign fluctuations. He did not report visual hallucinations and therefore is not experiencing hallucinosis.Wernicke encephalopathy is an alcohol-induced organic brain symptom characterized by the classic triad of ataxia, ophthalmoplegia, and altered mental status. Korsakoff syndrome is a persistent amnesia with confabulation (mnemonic: K is for konfabulation). Both Wernicke and Korsakoff syndromes are caused by thiamine deficiency. These syndromes are usually present in the stable alcoholic patient. Given TN’s apparently normal mental status, these conditions are unlikely etiologies for his presentation.

Q.  A 52-year-old woman with a long history of alcoholism who presents complaining of fatigue and palpitations. She notes having passed foul-smelling, black stool in the last 2 days. Her recent history is significant for three episodes of vomiting and dry retching after an alcoholic binge. Generally, she appears well, with an HR of 96 and BP of 125/87. Her exam is significant for general pallor, diffuse epigastric tenderness, and heme-positive stool. Laboratory findings are significant for Hct of 32%, Plt 150,000/μL, and MCV of 85 fL. What is the most likely cause of her anemia?

1.  Folate and vitamin B12 deficiency causing a megaloblastic anemia
2.  Anemia of chronic disease from chronic alcoholism 
3.   Iron deficiency anemia 
4.  Occult bleeding from Mallory-Weiss esophageal    tear  
5. Ruptured gastroesophageal varices

Ans: D. Given the recent appearance of black stool and her normal MCV, it is likely that her anemia is caused by an acute event such as a Mallory-Weiss tear in her mucosa at the gastroesophageal junction caused by severe retching from excessive alcohol intake. Such a mucosal tear would produce moderate upper GI hemorrhage that appears as melena (black stool).

Chronic etiologies include iron, folate, and vitamin B12 deficiencies as well as anemia of chronic disease. Since her MCV is normal, it is unlikely that she has an iron deficiency anemia or anemia of chronic disease, which produce microcytic anemias. Folate or vitamin B12 deficiency would produce a macrocytic anemia. Lastly, although retching can rupture gastric varices, this would present with more profuse bleeding with frank hematochezia (bright red blood from the rectum), as well as a more fulminant clinical course.

Q. A 19-year-old boy is brought by ambulance after being found collapsed in his college dorm room by his roommate, “blue and not breathing on his own.” On the field, he is found to have an O2 saturation of 75% and is quickly intubated, which resolves his cyanosis, bringing his saturation to 100%. He hasa heart rate of 55 and a blood pressure of 95/55, for which he is given fluids. He has a GCS of 3, glucose of 85. He smells of alcohol and shows no signs of trauma. His pupils are equaland reactive, and the rest of his exam is normal. What is his most likely diagnosis?

A. Acute heroin overdose

B. Acute alcohol toxicity

C. Acute cocaine overdose

D. Insulin overdose

E. Acute BZD overdose

Ans: B. Ingestion of large amounts of alcohol  has been associated with CNS depression severe enough  to completely depress respiratory drive and result in death. Such severe intoxication must be treated with respiratory support until the patient can metabolize the alcohol. Heroin intoxication is less likely in this case, given the lack of stigmata like pinpoint pupils and track marks, although he can be given naloxone empirically. Benzodiazepine overdose is also possible but less likely. His vital signs and physical exam are inconsistent with a cocaine overdose. Lastly, his glucose level does not suggest an insulin overdose.

Q. 72-year-old man with a long history of alcoholism who is brought in by ambulance after having collapsed in his chair at home. He is eventually diagnosed with a hemorrhagic stroke. What is the most likely cause of this stroke?

A. Thiamine deficiency

B. Hypoglycemia

C. Korsakoff syndrome

D. Hepatic encephalopathy

E. Vitamin K deficiency

Ans: E.Vitamin K deficiency is associated with coagulopathy due to inability to produce sufficient coagulation factors. Such coagulopathy can predispose an individual to hemorrhagic strokes. Thiamine deficiency can produce altered mental status along with paresthesias. Long-term organic brain syndromes such as Korsakoff syndrome can cause amnestic symptoms with confabulation. Chronic liver disease can predispose someone to hypoglycemic events, which can manifest as syncope. Lastly, chronic liver insufficiency can produce high levels of serum ammonia, which can produce a hepatic encephalopathy and decreased sensorium.

STATISTICS EXCLUSIVES/MCQS/PGQUESTIONS

STATISTICS EXCLUSIVES/MCQS/PGQUESTIONS

Q. A screening test is performed and there are 10 positive results. Of these, six are false positives. Of note, there are exactly 10 people with disease in the population screened. Which of the following is true?

A. This test has a 40% sensitivity.

B. This test has a 60% sensitivity.

C. This test has a 100% sensitivity.

D. This test has a 60% specificity.

E. This test has a 40% specificity.

Ans:. A.

We know that there are six false positives. Thus, the remaining four positive tests must be true positives. Since there are exactly 10 people with disease, there must be six false negatives. Using this information we can calculate the sensitivity, but not the specificity. Sensitivity is a/(a + c)= 4/10 =0.4, or 40%.

Q. A screening test for diabetes is being evaluated. You know that 900 nondiabetics and 100 diabetics are being screened. There are 100 positive tests and of these, 90 have diabetes.

Which of the following is true in this population?

A. The NPV of the test is 0.9.

B. The NPV of the test is 0.5.

C. The PNV of the test is 0.1.

D. The PPV of the test is 0.1.

E. The PPV of the test is 0.9.

Ans: E.

Ninety of the 100 positive screens have diabetes, so they must be true positives. The remaining

10 must be false positives. Since there are 100 people with diabetes, there must be 10 false negatives as well. So we can calculate PPV but not NPV. The PPV is equal to 90/100 = 0.9.

Q. Given the following collection of data {1,2,4,6,7,8,9}, which is correct?
A. The mean is 5.0.
B. The median is 5.0.
C. The median is 5.5.
D. The median is 6.0.
E. The mean is 6.0.

Ans:. D.

 Given the set of numbers {1,2,4,6,7,8,9}, the sum of these numbers is 37 and there are seven values, so the mean is 37/7 =5and2/7, or 5.28. The median, which is the center of an odd number of values, is the number 6, with 1, 2, and 4 below and 7, 8, and 9 above.

Q. Given the following collection of data {1,1,1,1,2,3,5,5, 6,6,7,7,9,9}, which of the following is correct?
A. The mode is 1.
B. The mean is 4.5.
C. The median is 6.
D. A and B are correct.
E. A and C are correct

Ans. D.

Given the set of numbers {1,1,1,1,2,3,5,5,6,6,7,7,9,9}, there are 14 numbers, and their sum is 63. Thus the mean is 4.5; clearly, the median is 5; and the mode is 1. Thus, the answer is (D), which is a mode of 1 and a mean of 4.5.

Q. You are told that your patient has a cholesterol result that is the mean of that in the population. Which of the following distributions would also mean that his result is necessarily
greater than half of the population?
A. Skewed to the left
B. Skewed to the right
C. Uniform
D. Normal
E. Bimodal

Ans:  B.

The question asks, “In what type of distribution is the mean greater than the median?”The mean will equal the median in normal and uniform distributions. In a symmetric bimodal distribution this will also be true. In a skewed-to-the-left distribution, the median will be greater than the mean, whereas in a skewed-tothe- right distribution, the mean will be greater than the median

Q. Which of the following statements is true about a 95% confidence interval?
A. There is a 95% chance that the true value is in the interval.
B. Ninety-five percent of the time, the true value will fall in the interval.
C. If you repeated the study 20 times, the true value would be in the interval at least once.
D. If the interval does not contain the null hypothesis, you can be 95% certain that your result is the true value.
E. If you repeat the study an infinite number of times, the results would fall in the interval 95% of the time

Ans;E.

What a 95% CI tells you is that if you repeated the same experiment that you did to get your summary statistic (e.g., the mean or proportion), an infinite number of times, 95% of the time you would get a value that falls within the 95% CI. It does not tell you that the true value you are seeking is going to fall within your CI, particularly because your experiment may be faulty or biased; thus, the true value may be missed by the CI. Even with a perfectly performed experiment, the CI is not used to tell you about the true value; it gives just the results of that particular experiment.

 Q. You are designing a study to investigate whether ethnicity is associated with type II DM. You have a cohort of 1000 patients of Asian, African American, Caucasian, and Latino ethnicity. Which of the following tests would be the best to compare the results of the study?

A. Student t test
B. Power analysis
C. z test
D. Chi-square test
E. Bonferroni correction

Ans. D.

In this study design, you are attempting to determine whether there is a difference in any of these groups of patients with respect to the risk for type 2 diabetes. The test that is needed must be able to compare proportions between multiple groups. The only one that can do that is the chi-square test. The student t test is to compare two means; the z test is to compare the proportions of two groups. A power analysis is performed to measure the power of a particular study to reject the null hypothesis. A Bonferroni correction is performed when making more than one comparison, using, for example, a student t test. Using a cutoff of the test that allows a 5% chance of type I error goes awry when you make many comparisons of outcomes in the same study. For example, if you compared 100 different means, on average you would expect five of them to meet the p _ .05 requirement. Thus, a Bonferroni or, more commonly, the student- Newman-Keuls test can be used to adjust the cutoff when making multiple comparisons.