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Direct titration by fluorescent polarization

This study explores direct titration by fluorescent polarization to analyze DNA interactions. The anisotropy values for free and fully bound DNA are derived from the equation A = Af - (Ab - Af)((D + E + Kd) - √((D + E + Kd)² - 4DE))/2D. The dissociation constants (Kd) for FAM-AniI and FAM-SceI are calculated as 17.71 ± 2.33 nM and 19.78 ± 3.34 nM respectively. Additionally, competitive titration is utilized to determine unlabeled DNA concentrations, revealing Ki values of 51.56 ± 20.37 nM and 120.8 ± 23.18 nM.

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Direct titration by fluorescent polarization

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  1. Direct titration by fluorescent polarization A = Af - (Ab-Af)[(D+E+Kd)-{(D+E+Kd)2-4DE}1/2]/2D A, anisotropy; Af, anisotropy of the free DNA; Ab, anisotropy of the fully bound DNA; D, the DNA concentration; E, the enzyme concentration. Kd (FAM-AniI wt DNA) = 17.71 ± 2.33 nM Kd (FAM-SceI wt DNA) = 19.78 ± 3.34 nM

  2. Competitive titration X (Bound/Free)= (A-Af)/(Ab-A) C = {1+Ki/(KdX)}[E-D{X/(1+X)}-KdX] A, anisotropy; Af, anisotropy of the free DNA; Ab, anisotropy of the fully bound DNA; C, the unlabled DNA concentration; D, the labled DNA concentration; E, the enzyme concentration. Ki (Unlabled AniI wt DNA) = 51.56 ± 20.37 nM Ki (Unlabled SceI wt DNA) = 120.8 ± 23.18 nM

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