NMS Q&A Family Medicine

252 NMS Q&A Family MedicineExamination Answers1. The answer is B. A treatment group of 50 would preventone colorectal cancer case employing the new preventiveregimen in 10 years. As a simple matter ofmathematics, one method of arriving at this conclusion isas follows: the difference of incidence between the treatmentgroup (3%) and the control group (5%) is 2%. Thus,the treatment is assumed to prevent colorectal cancer in2% of the population as defined in the body of the question.Two percent of the population in the treatmentgroup is eight subjects. Thus, to prevent colorectal cancerin one patient, the group size needed is 1/8 of 400 or 50.2. The answer is B. Papanicoulau smears are examplesof secondary prevention. The malignant disease processof carcinoma of the cervix will have been underway if orwhen the pap smear has become positive, leading to anintervention such as conization of the cervix. Salt restrictionto prevent hypertension is primary prevention, if it iseffective; if the hypertension is established, salt restrictionmay alleviate the process, effectively preventing progressionof hypertension, thus being a form of secondaryhypertension if there be no symptoms, such as headaches,precipitation of angina, etc. In the latter events, preventionis tertiary in type because not only is the disease processunderway but it is symptomatic. Aspirin to preventplatelet aggregation is a form of primary prevention,assuming that no vascular events have occurred. Angioplastyto alleviate angina and prevent future events is aform of tertiary prevention since the disease process ispresent and symptomatic. Simvastatin to correct dyslipidemiais a form of tertiary prevention in the patient whohas experienced a vascular event.3. The answer is A, primary. Cigarette smoking cessationprevents the vast majority of lung cancers from developingif embarked upon early enough. Thus, a disease processnot underway and prevented from developing is anexample of primary prevention.4. The answer is A. Resection of a carcinoma of the colon,found to be in Duke’s stage A in a patient who presentedwith hematochezia is definitely tertiary prevention. Thecancerous disease process is underway and symptomaticwith bright red blood in the stool. Finding an elevatedTSH in a euthyroid patient without goiter and correctingthe metabolic status by prescribing exogenous thyroidhormone is an example of secondary prevention, interruptionof an asymptomatic disease process. Removal ofan adenomatous polyp during screening colonoscopy isprimary prevention of colon cancer. Voluntary weight lossin a centrally obese man with family history of type IIdiabetes is primary prevention of diabetes, assuming thepatient would have been destined to contract the disease.Medically indicated carotid endarterectomy in an asymptomaticperson is an example of secondary prevention.5. The answer is C, 90%. 63 cancers were discovered and70 cancer occurred in the study group (Table 41–1).TABLE 41–1 Presentation of Data from Which theShown Functions Can Be Calculatedfrom Questions 5 and 8Subjectswith TestsSubjectswith TestsTotals ofCases(Columns2 and 3)Cancer present 63 7 70Cancer absent 110 820 930Totals 173 827 1,000Sensitivity 63/70 0.9 100 90%Specificity 820/930 .88 100 88%PPV 63/173 .36 100 36%NPV 820/827 0.991 100 99.1%6. The answer is B. The specificity of the test is 88%, thatis, the chances of a patient who is negative for the test nothaving the disease (Table 41–1).7. The answer is D. The PPV is 36%, that is, the chancesof someone with a positive test actually having the disease.This illustrates the fact that the value of a screening test isheavily dependent on the proportion of the populationwho will be positive. Such tests must be evaluated on thebasis of cost and discomfort attendant to the test and thecost, both emotional and financial, of the false-positivetests that will occur, and of the subsequent testing thatmust be done to rule out the false positives and avoidchronic and terminal care avoided (Table 41–1).8. The answer is D. 99% is the chance of actually nothaving the disease if the test is negative (Table 41–1). Severalobservations should be apparent from the foregoingproblems. First might be that even with an impressivesensitivity (e.g., 90%) when testing for a disease with lowincidence or prevalence in the population, the predictivepower of a positive test may be less than 50%. Conversely,in that situation even a modest specificity may convey a