Zusammenfassung
Summary Jawetz Melnick&Adelbergs Medical Microbiology 26/E(MCB); Mycobacterium
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Microbiology and Parasitology Exam (elaborations) USMLE - United States Medical Licensing Examination Jawetz Melnick&Adelbergs Medical Microbiology 26/E, ISBN: 9780071815789
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Summary Jawetz Melnick&Adelbergs Medical Microbiology 26/E (MCB); Clostridium
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Summary Jawetz Melnick&Adelbergs Medical Microbiology 26/E(MCB); Brucella, Francisella
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Inhaltsvorschau
CHAPTER 23: MYCOBACTERIUM(E) Mutations in the genes encoding the S12 protein
and 16S rRNA is a characteristic of streptomycin
1. A 60-year-old-man has a 5-onth history of progressive resistant M. tuberculosis
weakness and a weight loss of 13kg along with
intermittent fever, chills, and a chronic cough 3. A 47-year-old woman presents with a 3-month history
productive of yellow sputum, occasionally streaked of progressive cough, weight loss, and fever. Chest
with blood. A sputum specimen is obtained, and radiography shows bilateral cavitary disease
numerous acid-fast bacteria are seen on the smear. suggestive of tuberculosis. Sputum culture grows an
Culture of the sputum is positive for Mycobacterium acid-fast bacillus that is a photochromogen (makes an
tuberculosis. Which treatment regimen is most orange pigment when exposed to light). The organism
appropriate for initial therapy? most likely is
(A) Isoniazid and rifampin (A) Mycobacterium tuberculosis
(B) Sulfamethoxazole-trimethoprim and streptomycin (B) Mycobacterium kansasii
(C) Isoniazid, rifampin, pyrazinamide, and (C) Mycobacterium gordonae
ethambutol (D) Mycobacterium avium complex
(D) Isoniazid, cycloserine, and ciprofloxacin (E) Mycobacterium fortuitum
(E) Rifampin and streptomycin
ANSWER: (B) Mycobacterium kansasii because it
ANSWER: (C) Isoniazid, rifampin, pyrazinamide is a photochromogen (produce pigment when grown
and ethambutol because these are the 4-drug with light)
regimen recommended for initial treatment.
WHY NOT THE REST:
WHY NOT THE REST: (A) M. tuberculosis does not belong in the
(A) Isoniazid (INH) and Rifampin (RMP) can’t be photochromogen group, it belongs in the
used because there are multi-drug resistant M. Tuberculosis complex
tuberculosis which are both resistant to INH and (C) M. gordonae does not produce pigment with light.
RMP. It only produces pigment in the dark which is why
(B) Sulfamethoxazole-trimethoprim and streptomycin it belongs to the scotochromogens
because there are streptomycin resistant tubercle (D) M. avium complex are nonpigmented or have
bacilli. SXT also isn’t stated as a treatment for M. light tan or buff-colored colonies
tuberculosis (E) M. fortuitum are rapid growers
(D) INH, cycloserine and ciprofloxacin are not
recommended because cycloserine and 4. A 31-year-old Asian woman is admitted to the hospital
ciprofloxacin are only used as a second-line with a 7-week history of increasing malaise, myalgia,
drugs which is used only when the initial non-productive cough, and shortness of breath. She
treatment failed or if it is drug resistant. has daily fevers of 38-39°C and a recent 5-kg weight
(E) Rifampin and streptomycin are not recommended loss. She had a negative chest radiograph when she
because there multi-drug resistant tubercle bacilli entered the United States 7 years ago. The patient’s
that are resistant to both due to its mutations in grandmother died of tuberculosis when the patient
the genes of the ribosomal S12 protein and 16S was an infant. A current chest radiograph is normal;
rRNA, rpsL and rrs (streptomycin resistant results of other tests show a decreased hematocrit
tubercle bacilli). Alterations in the b subunit of and liver function test abnormalities. Liver and one
RNA polymerase, the rpoB (rifampin resistant marrow biopsies show granulomas with giant cells
tubercle bacilli). and acid-fast bacilli. She is probably infected with
2. If the patient’s Mycobacterium tuberculosis isolate (A) Mycobacterium leprae
(question 1) proves to be resistant to isoniazid, the (B) Mycobacterium fortuitum
likely mechanism for resistant is (C) Mycobacterium ulcerans
(D) Mycobacterium gordonae
(A) β-lactamase (E) Mycobacterium tuberculosis
(B) Mutations in the catalase-peroxidase gene
(C) Alterations in the β subunit of RNA polymerase ANSWER: (E) Mycobacterium tuberculosis is the
(D) Mutations in the DNA gyrase gene diagnosis because the signs and symptoms shown
(E) Mutations in the genes encoding the S12 protein are characteristics of Tuberculosis which is caused by
and 16S rRNA the organism.
ANSWER: (B) Mutations in the catalase- WHY NOT THE REST:
peroxidase gene because these isolates become (A) M. leprae is not the diagnosis because it does
catalase negative or have decreased catalase activity. involve the lungs. S/S could be found on the skin.
(B) M. fortuitum can only cause superficial and
WHY NOT THE REST: systemic disease on rare occasions.
(A) β-lactamase is not a mechanism for this (C) M. ulcerans can infect fish and can only cause
resistance because M. tuberculosis is not a β- superficial skin lesions (ulcers, “swimming pool
lactamase producer granulomas”) in humans.
(C) Alterations in the β subunit of RNA polymerase is (D) M. gordonae is frequently found on plants, in soil,
a characteristic of RMP resistant M. tuberculosis or in water (similar to M. phlei)
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