What is the T score cut-off value for diagnosing osteoporosis?

The T score cut-off value for diagnosing osteoporosis is set at -2.5. This standard is defined by the World Health Organization (WHO) and signifies that a T score of -2.5 or lower indicates a significantly reduced bone mineral density compared to the peak bone mass of a healthy young adult. Osteoporosis is characterized by weakened bones and an increased risk of fractures, making accurate diagnosis important for the management and prevention of this condition.

A T score of -2.5 suggests that the bone density is 2.5 standard deviations below the average peak bone density, which is clinically relevant for determining the risk of fracture along with other clinical factors. In practice, a healthcare provider will use this cut-off as one of the criteria to initiate treatment or preventive measures to enhance bone health and reduce the likelihood of osteoporotic fractures. Understanding this threshold is crucial for both diagnosis and guiding patient care in addressing osteoporosis.