Family Medicine

Answer 4 Osteopenia; alendronate for treatment. The T score compares the patient’s BMD in standard deviations with the average BMD of a young adult of the same gender. T score between 0 and –1 indicates normal bone density. A T score between –1 and –2.5 are consistent with osteopenia. T score 2.5 standard deviations or more below the standard (T score < –2.5) indicates osteoporosis. This patient is at increased risk for osteoporosis owing to her race and family history. Her BMD testing results indicate that she has osteopenia. Alendronate, a bisphosphonate compound proven to increase bone mass and decrease fractures, would be an appropriate choice for this patient.4 The NOF recommends pharmacologic therapy to reduce fracture risk in all postmenopausal women with T scores of –2 or below. Drug therapy also is recommended for postmenopausal women with risk factors for osteoporosis and T scores below –1.5 or those with hip or vertebral fractures. Because the case patient’s T score is below –1.5, her treatment should include reduction of modifiable risk factors, calcium and vitamin D supplementation, and additional drug therapy. REFERENCE 4. Brunader R, Shelton DK. Radiologic bone assessment in the evaluation of osteoporosis. Am Fam Physician 2002;65:1357–64. Click here to return to the questions

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