When odds ratio approximates relative risk — MCQs

10 questions

Recently, clarithromycin was found to have an increased risk of cardiac death in a Danish study. This study analyzed patients who were previously treated with clarithromycin or another antibiotic, and then they were followed over time to ascertain if cardiac death resulted. What type of study design does this represent?

A 25-year-old man with a genetic disorder presents for genetic counseling because he is concerned about the risk that any children he has will have the same disease as himself. Specifically, since childhood he has had difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy. He has also had diarrhea and malabsorption requiring enzyme replacement therapy. If his wife comes from a population where 1 in 10,000 people are affected by this same disorder, which of the following best represents the likelihood a child would be affected as well?

A 6-month-old male presents for a routine visit to his pediatrician. Two months ago, the patient was seen for tachypnea and wheezing, and diagnosed with severe respiratory syncytial virus (RSV) bronchiolitis. After admission to the hospital and supportive care, the patient recovered and currently is not experiencing any trouble breathing. Regarding the possibility of future reactive airway disease, which of the following statements is most accurate?

You are reading through a recent article that reports significant decreases in all-cause mortality for patients with malignant melanoma following treatment with a novel biological infusion. Which of the following choices refers to the probability that a study will find a statistically significant difference when one truly does exist?

The incidence of a relatively benign autosomal recessive disease, X, is 1 in 25 in the population. Assuming that the conditions for Hardy Weinberg Equilibrium are met, what is the probability that a male and female, who are carriers, will have a child expressing the disease?

A study is conducted to find an association between serum cholesterol and ischemic heart disease. Data is collected, and patients are classified into either the "high cholesterol" or "normal cholesterol" group and also into groups whether or not the patient experiences stable angina. Which type of data analysis is most appropriate for this study?

A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk (RR) of 3.0 for people with elevated LDL levels compared to individuals with normal LDL levels. The p-value was 0.04 with a 95% confidence interval of 2.0-4.0. According to the study results, what percent of heart disease in these patients can be attributed to elevated LDL?

A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk of 2.3 for people with elevated LDL levels compared to individuals with normal LDL levels. The 95% confidence interval was 1.05-3.50. This study is most likely to have which of the following p values?

A randomized controlled trial was initiated to evaluate a novel DPP-4 inhibitor for blood glucose management in diabetic patients. The study used a commonly prescribed sulfonylurea as the standard of care treatment. 2,000 patients were enrolled in the study with 1,000 patients in each arm. One of the primary outcomes was the development of diabetic nephropathy during treatment. This outcome occurred in 68 patients on the DPP-4 inhibitor and 134 patients on the sulfonylurea. What is the relative risk reduction (RRR) for patients using the DPP-4 inhibitor compared with the sulfonylurea?

Q10

A medical research study is evaluating an investigational novel drug (medication 1) as compared with standard therapy (medication 2) in patients presenting to the emergency department with myocardial infarction (MI). The study enrolled a total of 3,000 subjects, 1,500 in each study arm. Follow-up was conducted at 45 days post-MI. The following are the results of the trial: Endpoints Medication 1 Medication 2 P-Value Primary: death from cardiac causes 134 210 0.03 Secondary: hyperkalemia 57 70 0.4 What is the relative risk of death from a cardiac cause, expressed as a percentage? (Round to the nearest whole number.)

When odds ratio approximates relative risk US Medical PG Practice Questions and MCQs

Question 1: Recently, clarithromycin was found to have an increased risk of cardiac death in a Danish study. This study analyzed patients who were previously treated with clarithromycin or another antibiotic, and then they were followed over time to ascertain if cardiac death resulted. What type of study design does this represent?

A. Cross-sectional study
B. Randomized controlled trial
C. Case control study
D. Cohort study (Correct Answer)

Explanation: ***Cohort study*** - This study design involves following a group of individuals (a **cohort**) over time to observe the incidence of specific outcomes, in this case, **cardiac death**. - The study identifies groups based on exposure (clarithromycin treatment vs. another antibiotic) and then tracks them for future events, which is characteristic of a **prospective cohort study**. *Cross-sectional study* - A **cross-sectional study** assesses exposure and outcome at a **single point in time**. - It does not involve following individuals over time, making it unsuitable for studying the temporal relationship between antibiotic use and subsequent cardiac death. *Randomized controlled trial* - A **randomized controlled trial (RCT)** involves randomly assigning participants to an intervention or control group to determine the effect of the intervention. - This study did not involve random assignment of clarithromycin but rather observed groups based on **prior treatment**, ruling out an RCT. *Case control study* - A **case-control study** starts with individuals who have the outcome (cases) and individuals who do not (controls) and then retrospectively looks back at their exposures. - This study started with exposed individuals (treated with clarithromycin) and then followed them forward, which is the opposite direction of a case-control study.

Question 2: A 25-year-old man with a genetic disorder presents for genetic counseling because he is concerned about the risk that any children he has will have the same disease as himself. Specifically, since childhood he has had difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy. He has also had diarrhea and malabsorption requiring enzyme replacement therapy. If his wife comes from a population where 1 in 10,000 people are affected by this same disorder, which of the following best represents the likelihood a child would be affected as well?

A. 0.01%
B. 2%
C. 0.5%
D. 1% (Correct Answer)
E. 50%

Explanation: ***Correct Option: 1%*** - The patient's symptoms (difficulty breathing requiring bronchodilators, inhaled corticosteroids, and chest physiotherapy; diarrhea and malabsorption requiring enzyme replacement therapy) are classic for **cystic fibrosis (CF)**, an **autosomal recessive disorder**. - For an autosomal recessive disorder with a prevalence of 1 in 10,000 in the general population, **q² = 1/10,000**, so **q = 1/100 = 0.01**. The carrier frequency **(2pq)** is approximately **2q = 2 × (1/100) = 1/50 = 0.02**. - The affected man is **homozygous recessive (aa)** and will always pass on the recessive allele. His wife has a **1/50 chance of being a carrier (Aa)**. If she is a carrier, she has a **1/2 chance of passing on the recessive allele**. - Therefore, the probability of an affected child = **(Probability wife is a carrier) × (Probability wife passes recessive allele) = 1/50 × 1/2 = 1/100 = 1%**. *Incorrect Option: 0.01%* - This percentage is too low and does not correctly account for the carrier frequency in the population and the probability of transmission from a carrier mother. *Incorrect Option: 2%* - This represents approximately the carrier frequency (1/50 ≈ 2%), but does not account for the additional 1/2 probability that a carrier mother would pass on the recessive allele. *Incorrect Option: 0.5%* - This value would be correct if the carrier frequency were 1/100 instead of 1/50, which does not match the given population prevalence. *Incorrect Option: 50%* - **50%** would be the risk if both parents were carriers of an autosomal recessive disorder (1/4 chance = 25% for affected, but if we know one parent passes the allele, conditional probability changes). More accurately, 50% would apply if the disorder were **autosomal dominant** with one affected parent, which is not the case here.

Question 3: A 6-month-old male presents for a routine visit to his pediatrician. Two months ago, the patient was seen for tachypnea and wheezing, and diagnosed with severe respiratory syncytial virus (RSV) bronchiolitis. After admission to the hospital and supportive care, the patient recovered and currently is not experiencing any trouble breathing. Regarding the possibility of future reactive airway disease, which of the following statements is most accurate?

A. “There is no clear relationship between RSV and the development of asthma.”
B. “Your child has a greater than 20% chance of developing asthma” (Correct Answer)
C. “Your child’s risk of asthma is less than the general population.”
D. “Your child has a less than 5% chance of developing asthma”
E. “Your child’s risk of asthma is the same as the general population.”

Explanation: ***“Your child has a greater than 20% chance of developing asthma”*** - Severe **RSV bronchiolitis** in infancy is a significant risk factor for the development of **recurrent wheezing** and **childhood asthma**. - Studies estimate that a substantial proportion, often greater than 20%, of infants with severe RSV bronchiolitis will go on to develop **asthma** later in childhood. *“There is no clear relationship between RSV and the development of asthma.”* - This statement is incorrect as there is a **well-established link** between severe RSV infection in early life and an increased risk of developing **asthma**. - Numerous epidemiological and longitudinal studies have documented this association. *“Your child’s risk of asthma is less than the general population.”* - This is incorrect, as severe RSV infection **increases** the risk of asthma, not decreases it. - Children with a history of severe RSV have a **higher incidence** of asthma compared to the general pediatric population. *“Your child has a less than 5% chance of developing asthma”* - This percentage is **too low** given the known association between severe RSV bronchiolitis and subsequent asthma. - The actual risk is considerably higher, typically falling into the range of 20-50% for those with severe RSV. *“Your child’s risk of asthma is the same as the general population.”* - This statement is inaccurate because severe RSV infection in infancy is a recognized independent **risk factor** for **asthma development**. - Therefore, the child's risk is elevated above that of the general population.

Question 4: You are reading through a recent article that reports significant decreases in all-cause mortality for patients with malignant melanoma following treatment with a novel biological infusion. Which of the following choices refers to the probability that a study will find a statistically significant difference when one truly does exist?

A. Type II error
B. Type I error
C. Confidence interval
D. p-value
E. Power (Correct Answer)

Explanation: ***Power*** - **Power** is the probability that a study will correctly reject the null hypothesis when it is, in fact, false (i.e., will find a statistically significant difference when one truly exists). - A study with high power minimizes the risk of a **Type II error** (failing to detect a real effect). *Type II error* - A **Type II error** (or **beta error**) occurs when a study fails to reject a false null hypothesis, meaning it concludes there is no significant difference when one actually exists. - This is the **opposite** of what the question describes, which asks for the probability of *finding* a difference. *Type I error* - A **Type I error** (or **alpha error**) occurs when a study incorrectly rejects a true null hypothesis, concluding there is a significant difference when one does not actually exist. - This relates to the **p-value** and the level of statistical significance (e.g., p < 0.05). *Confidence interval* - A **confidence interval** provides a range of values within which the true population parameter is likely to lie with a certain degree of confidence (e.g., 95%). - It does not directly represent the probability of finding a statistically significant difference when one truly exists. *p-value* - The **p-value** is the probability of observing data as extreme as, or more extreme than, that obtained in the study, assuming the null hypothesis is true. - It is used to determine statistical significance, but it is not the probability of detecting a true effect.

Question 5: The incidence of a relatively benign autosomal recessive disease, X, is 1 in 25 in the population. Assuming that the conditions for Hardy Weinberg Equilibrium are met, what is the probability that a male and female, who are carriers, will have a child expressing the disease?

A. 1/5
B. 8/25
C. 1/4 (Correct Answer)
D. 1/25
E. 4/5

Explanation: ***1/4*** - If both parents are **carriers** for an autosomal recessive disease, each parent has one copy of the normal allele (A) and one copy of the recessive allele (a). - When two heterozygous (Aa) individuals mate, the probability of their child inheriting two recessive alleles (aa) and expressing the disease is 1 in 4 (25%), according to Mendelian genetics. *1/5* - This value represents the **allele frequency (q)** in the population for the recessive allele, given an incidence of 1 in 25 (q^2 = 1/25, so q = 1/5). - However, this is not the probability of a child being affected if both parents are already known to be carriers. *8/25* - This option is incorrect and does not directly relate to the probability of an affected child from two known carriers. - It might represent a miscalculation involving carrier frequencies or a different genetic scenario. *1/25* - This is the **incidence of the disease (q^2)** in the general population, which means 1 out of 25 individuals express the disease. - It is not the probability of a child inheriting the disease from two parents already identified as carriers. *4/5* - This value represents the **allele frequency (p)** of the dominant allele (p = 1 - q = 1 - 1/5 = 4/5). - It is not the probability of a child expressing the disease from two carrier parents.

Question 6: A study is conducted to find an association between serum cholesterol and ischemic heart disease. Data is collected, and patients are classified into either the "high cholesterol" or "normal cholesterol" group and also into groups whether or not the patient experiences stable angina. Which type of data analysis is most appropriate for this study?

A. Attributable risk
B. Analysis of variance
C. Chi-squared (Correct Answer)
D. T-test
E. Pearson correlation

Explanation: ***Chi-squared*** - The **chi-squared test** is ideal for analyzing two **categorical variables**, such as cholesterol levels (high/normal) and the presence of stable angina (yes/no), to see if there's an association between them. - It assesses whether the observed frequencies in each category differ significantly from the expected frequencies, under the assumption of no association. *Attributable risk* - **Attributable risk** quantifies the proportion of disease in an exposed group that is directly due to the exposure. - While it might be calculated *after* establishing an association (e.g., using a chi-squared test), it's a measure of actual impact rather than a method for *finding the association* between two categorical variables. *Analysis of variance* - **Analysis of variance (ANOVA)** is used to compare the means of **three or more groups** for a continuous outcome variable. - It works when you have a categorical independent variable with multiple levels and a continuous dependent variable, which is not the case here as both variables are categorical. *T-test* - A **t-test** is used to compare the means of **two groups** for a continuous outcome variable. - It is not appropriate for analyzing the association between two categorical variables like cholesterol categories and angina presence. *Pearson correlation* - **Pearson correlation** measures the linear relationship between **two continuous variables**. - It is unsuitable for this study as both cholesterol status and angina presence are categorical variables, not continuous.

Question 7: A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk (RR) of 3.0 for people with elevated LDL levels compared to individuals with normal LDL levels. The p-value was 0.04 with a 95% confidence interval of 2.0-4.0. According to the study results, what percent of heart disease in these patients can be attributed to elevated LDL?

A. 67% (Correct Answer)
B. 50%
C. 100%
D. 33%
E. 25%

Explanation: ***67%*** - The **attributable risk percent (AR%)** in the exposed group represents the proportion of disease in the exposed population that can be attributed to the exposure. - It is calculated using the formula: **AR% = [(RR - 1) / RR] × 100** - With an RR of 3.0: [(3.0 - 1) / 3.0] × 100 = (2.0 / 3.0) × 100 = 66.67%, which rounds to **67%** - This means that 67% of heart disease cases among those with elevated LDL can be attributed to the elevated LDL levels. *50%* - This value would result from an RR of 2.0: [(2.0 - 1) / 2.0] × 100 = 50% - The study reports an RR of 3.0, not 2.0, making this incorrect. *100%* - This would imply that all cases of heart disease in the exposed group are due solely to elevated LDL (RR approaching infinity). - An RR of 3.0 indicates elevated LDL increases risk threefold, but does not account for all cases. *33%* - This might result from incorrectly calculating 1/RR = 1/3.0 = 33.3% - The correct formula is (RR - 1)/RR, not 1/RR. *25%* - This would correspond to an RR of 1.33: [(1.33 - 1) / 1.33] × 100 ≈ 25% - The given RR of 3.0 yields a much higher attributable risk percent.

Question 8: A prospective cohort study was conducted to assess the relationship between LDL and the incidence of heart disease. The patients were selected at random. Results showed a 10-year relative risk of 2.3 for people with elevated LDL levels compared to individuals with normal LDL levels. The 95% confidence interval was 1.05-3.50. This study is most likely to have which of the following p values?

A. 0.20
B. 0.06
C. 0.08
D. 0.04 (Correct Answer)
E. 0.10

Explanation: ***0.04*** - A 95% confidence interval that **does not include 1 (one)** suggests a **statistically significant** association, meaning the p-value is likely to be **less than 0.05**. - The given CI of 1.05-3.50 for the relative risk (RR) is entirely above 1, indicating a significant positive association, and therefore, a p-value less than 0.05. *0.20* - A p-value of 0.20 is **greater than 0.05**, which would imply the finding is **not statistically significant**. - If the p-value were 0.20, the 95% confidence interval would likely **include 1**, suggesting no significant difference in risk. *0.06* - A p-value of 0.06 is **greater than 0.05**, indicating that the association is **not statistically significant at the conventional alpha level**. - If the p-value were 0.06, the 95% confidence interval would likely **include 1**, or be very close to including it, contradicting the given CI of 1.05-3.50. *0.08* - A p-value of 0.08 is **greater than 0.05**, indicating that the finding is **not statistically significant**. - If the p-value were 0.08, the 95% confidence interval would almost certainly **include 1**, which is inconsistent with the provided interval. *0.10* - A p-value of 0.10 is **greater than 0.05**, which signifies that the finding is **not statistically significant**. - If the p-value were 0.10, the 95% confidence interval for the relative risk would typically **include 1**, contradicting the given confidence interval.

Question 9: A randomized controlled trial was initiated to evaluate a novel DPP-4 inhibitor for blood glucose management in diabetic patients. The study used a commonly prescribed sulfonylurea as the standard of care treatment. 2,000 patients were enrolled in the study with 1,000 patients in each arm. One of the primary outcomes was the development of diabetic nephropathy during treatment. This outcome occurred in 68 patients on the DPP-4 inhibitor and 134 patients on the sulfonylurea. What is the relative risk reduction (RRR) for patients using the DPP-4 inhibitor compared with the sulfonylurea?

A. 23%
B. 49% (Correct Answer)
C. 33%
D. 59%
E. 43%

Explanation: ***49%*** - To calculate **relative risk reduction (RRR)**, first determine the **event rate (ER)** for each group. - ER (DPP-4 inhibitor) = 68/1000 = 0.068. ER (Sulfonylurea) = 134/1000 = 0.134. - Next, calculate the **absolute risk reduction (ARR)**: ARR = ER (Sulfonylurea) - ER (DPP-4 inhibitor) = 0.134 - 0.068 = 0.066. - Finally, calculate RRR: RRR = ARR / ER (Sulfonylurea) = 0.066 / 0.134 ≈ 0.4925 or **49%**. *23%* - This value is incorrect and does not result from the proper application of the **relative risk reduction (RRR)** formula. - A common mistake is to reverse the subtrahend and minuend in the numerator or denominator. *33%* - This value is incorrect and does not result from the proper application of the **relative risk reduction (RRR)** formula. - Incorrect calculations in either the numerator or denominator of the **RRR formula** would lead to this incorrect result. *59%* - This value is incorrect and is likely the result of an error in calculating either the **absolute risk reduction (ARR)** or dividing it by the wrong **event rate**. - Always ensure the correct event rates are used for the control group and the intervention group. *43%* - This value is incorrect and does not align with the correct calculation of **relative risk reduction (RRR)**. - Errors in setting up the formula or executing the division could lead to this result.

Question 10: A medical research study is evaluating an investigational novel drug (medication 1) as compared with standard therapy (medication 2) in patients presenting to the emergency department with myocardial infarction (MI). The study enrolled a total of 3,000 subjects, 1,500 in each study arm. Follow-up was conducted at 45 days post-MI. The following are the results of the trial: Endpoints Medication 1 Medication 2 P-Value Primary: death from cardiac causes 134 210 0.03 Secondary: hyperkalemia 57 70 0.4 What is the relative risk of death from a cardiac cause, expressed as a percentage? (Round to the nearest whole number.)

A. 64% (Correct Answer)
B. 42%
C. 72%
D. 36%
E. 57%

Explanation: ***64%*** - The **relative risk (RR)** is calculated as the event rate in the exposed group divided by the event rate in the unexposed (control) group. - For cardiac death, the event rate for Medication 1 is 134/1500 = 0.0893, and for Medication 2 is 210/1500 = 0.14. Therefore, RR = 0.0893 / 0.14 = 0.6378. - Expressing as a percentage: 0.6378 × 100 = 63.78%, which rounds to **64%**. - This indicates that Medication 1 has 64% of the risk of cardiac death compared to Medication 2, representing a **36% relative risk reduction**. *42%* - This option is incorrect as it does not reflect the accurate calculation of **relative risk** using the provided event rates. - A calculation error or conceptual misunderstanding of the relative risk formula would lead to this value. *72%* - This percentage is higher than the calculated relative risk, suggesting an incorrect application of the formula or a misinterpretation of the event rates. - It does not represent the ratio of risk between the two medication groups for cardiac death. *36%* - This value represents the **relative risk reduction** (100% - 64% = 36%), not the relative risk itself. - This is a common error where students confuse relative risk with relative risk reduction. *57%* - While closer to the correct answer, this value is not the precise result when rounding to the nearest whole number. - Small calculation discrepancies or rounding at intermediate steps could lead to this slightly different percentage.

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