English: If we plot Km’ versus I, you would get the slope as Km/K_(I) and thus we could find From a plot of Km’ versus I as this, the equation was y = 0.012x + 0.12, y in this case is Km’, x is I, so The intercept Km = 0.12 from the equation $\therefore slope\ \frac{K_{m}}{K_{I}} = 0.012\ \ \ \ $ $\frac{K_{m}}{0.012} = \ K_{I}\ = \frac{0.12}{0.012} = 10mM\ \ $= the ∴ $\therefore slope\ \frac{K_{m}}{K_{I}} = 0.012$ $\therefore slope\ \frac{K_{m}}{K_{I}} = 0.012\ \ \ \ $
Telugu: Km’ = 0.012 I + 0.12. inhibitor constant Km' వర్సెస్ I యొక్క ఒక ప్లాట్ నుంచి, ఈక్వేషన్ y = 0.012x + 0.12, ఈ సందర్భంలో y అనేది Km', x అనేది I, అందువల్ల కి. మీ. = 0.012 I + 0.12. ఈక్వేషన్ నుంచి ఇంటర్సెప్ట్ Km = 0.12 $\therefore slope\ \frac{K_{m}}{K_{I}} = 0.012\ \ \ \ $కి= ఇన్హిబిటర్ కాన్స్టాంట్