The graph below shows the yearly incidence of viral hepatitis per 100,000 population from 1982-2012 in the United States (US).
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During this period, the total population in the US continued to increase, as shown in the table below:
Year | 1982 | 1987 | 1992 | 1997 | 2002 | 2007 | 2012 |
US population (in millions) | ~230 | ~243 | ~258 | ~275 | ~290 | ~303 | ~315 |
At any given time point, the number of individuals at risk for hepatitis infection is assumed to be equal to the total US population. Based on these data, which of the following is most likely to be true?
Incidence corresponds to the number of new cases of a disease diagnosed in a population at risk over a given time period. In 1997, incidence was >10 new cases/100,000 population for hepatitis A virus (HAV), <5/100,000 for hepatitis B virus (HBV), and <3/100,000 for hepatitis C virus (HCV). Assuming, as instructed, that the entire United States (US) population is at risk (reasonable theory given low number of hepatitis cases among the total population), then, in 1997 (US population ~275 million):
New HAV cases in 1997 = >10 new cases/100,000 population × 275,000,000 population = >27,500
New HBV cases in 1997 = <5/100,000 × 275,000,000 = <13,750
New HCV cases in 1997 = <3/100,000 × 275,000,000 = <8,250
There were thousands more new HAV (>27,500) than HBV and HCV cases combined (<13,750 + <8,250 = <22,000). The calculations incorporating total population (included for completion) are not needed to answer the question.
After 2006, HBV incidence surpassed HAV and HCV incidence. However, no information can be deduced regarding prevalence, which is the total number of cases in the population (Choice A). Compared to new HBV cases, there were approximately just as many new HAV cases in 1987 and more new HAV cases in 1989. However, the total number of HAV or HBV cases in the population cannot be calculated without additional information. In fact, there were likely more total cases (ie, higher prevalence) of HBV as it can progress to a chronic form (Choices B and C).
A simplified diagram compares incidence and prevalence to adding new drops (incident cases) into a water-containing sink (prevalent cases).
(Choice E) HAV incidence per 100,000 population was similar in 1982 and 1998. However, the overall US population, and consequently the number of new HAV cases, was different. Assuming HAV incidence was ~10 new cases/100,000 population for both years, then, in 1982 (US population ~230 million):
New HAV cases in 1982 = 10/100,000 × 230,000,000 = 23,000
Also, in 1998 (US population ~275 million at least):
New HAV cases in 1998 = 10/100,000 × 275,000,000 = 27,500
Even with different incidences, it is unlikely that exactly the same number of new HAV diagnoses were made.
Educational objective:
Incidence corresponds to the number of new cases of a disease in a certain population at risk over a given time period. Prevalence is the total number of cases in the population over a given period.