Physiologic serum 1,25 dihydroxyvitamin D is inversely associated with prostatic Ki67 staining in a diverse sample of radical prostatectomy patients

生理血清 1,25 二羟基维生素 D 与不同根治性前列腺切除术患者样本中的前列腺 Ki67 染色呈负相关

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作者:Adrian Rosenberg, Oluwarotimi S Nettey, Pooja Gogana, Ujalla Sheikh, Virgilia Macias, Andre Kajdacsy-Balla, Roohollah Sharifi, Rick A Kittles, Adam B Murphy

Conclusion

The correlation between physiologic serum levels of 25 hydroxyvitamin D with both prostatic 25 hydroxyvitamin D and serum 1,25 dihydroxyvitamin D suggests that serum levels are reasonable biomarkers of vitamin D status. Furthermore, serum 1,25 dihydroxyvitamin D has an inverse association with Ki67 staining in tumor cells at physiologic levels and may protect against tumor progression.

Methods

In this cross-sectional analysis of 119 men undergoing radical prostatectomy, serum from whole blood and expressed prostatic fluid was collected on the day of surgery. Tumor proliferation was measured in the dominant tumor on formalin-fixed prostatectomy tissues by immunohistochemical staining for Ki67 and quantified by Aperio imaging analysis.

Purpose

To investigate the correlation between serum 25 hydroxyvitamin D, prostatic 25 hydroxyvitamin D, and serum 1,25 dihydroxyvitamin D, and their respective associations with prostatic tumor proliferation at the time of radical prostatectomy.

Results

The sample included 88 African Americans (74%) and 31 (26%) European Americans. Serum and prostatic levels of 25 hydroxyvitamin D were correlated with each other (Spearman's rho (ρ) = 0.27, p = 0.004), and there was also a correlation between serum 25 hydroxyvitamin D and 1,25 dihydroxyvitamin D (ρ = 0.34, p < 0.001). Serum and prostatic 25 hydroxyvitamin D levels were not correlated with Ki67 staining in tumor cells. Serum 1,25 dihydroxyvitamin D was inversely correlated with Ki67 staining in tumor cells (ρ = - 0.30, p = 0.002). On linear regression, serum 1,25 dihydroxyvitamin D was negatively associated with Ki67 staining in tumor cells (β - 0.46, 95% CI - 0.75, - 0.04, p = 0.04).

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