Answer :USMLE step 2 Mcq 315:A 23 year old female has had type I diabetes since she was 8 years old
on Saturday, December 25, 2010
Correct Answer: B Explanation: This patient is presenting with symptoms of diabetic ketoacidosis. The fruity smell to her breath is seen with diabetic ketoacidosis. The rapid respirations (Kussmaul respiration) are also seen with diabetic acidosis. She has the production of increased metabolic acids, so this is considered a metabolic acidosis. The correct answer is therefore either (a) or (b).The anion gap is calculated as follows: NA - (CL + HCO3). (Potassium is usually ignored clinically in the calculation of the anion gap because the value is low.) A normal anion gap is 8-16 mEq/L . Her anion gap would be [170-(115 + 14)] = 41. Her anion gap is elevated. Therefore, choice (b) is incorrect because her anion gap is elevated, not normal. She has a metabolic acidosis with an elevated anion gap.Choice (c) is incorrect because she does not have an alkalosis, she has an acidosis.Her initial problem is not due to a pathology in CO2 and respiratory problems, so choices (d) and choice (e) are incorrect.
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