I did interaction of chemical molecules with BSA. I confirmed with some studies and flourescence graph I plotted and i dont know plot the standard plot stern volmer plot. I request you to help me in plotting the standard graph
The Stern Volmer plot is the plot is that of Io/I vs Concentration as said . But it can be also expressed as (Io-I)/I=K[Q]. This is a particular case of the plot of (Io-I)/(I-I∞)=K[Q] where I∞ is the remaining intensity for an infinite concentration of quencher [Q]. It can be the blank fluorescence, if any, or the fluorescence of the quenched fluorophore, if any. This last relation is quite general since (Io-I) is proportional to [FQ] and (I-I∞) is proportional to [F] as long as the signal I, fluorescence, absorbance, scattering is proportional to concentration. Interestingly you can have a straight line in two cases : if you have a static quenching mechanism or a dynamic quenching mechanism. In your case a static quenching is the more probable. You will not get a straight line if both mechanisms act together. Indeed there is non more a linearity between concentrations and signals.
The y-axis is the ratio of the unquenched fluorescence intensity divided by the quenched fluorescence intensity. The x-axis is the quencher concentration. The y-intercept is 1. The slope is the Stern-Volmer constant.
The Stern Volmer plot is the plot is that of Io/I vs Concentration as said . But it can be also expressed as (Io-I)/I=K[Q]. This is a particular case of the plot of (Io-I)/(I-I∞)=K[Q] where I∞ is the remaining intensity for an infinite concentration of quencher [Q]. It can be the blank fluorescence, if any, or the fluorescence of the quenched fluorophore, if any. This last relation is quite general since (Io-I) is proportional to [FQ] and (I-I∞) is proportional to [F] as long as the signal I, fluorescence, absorbance, scattering is proportional to concentration. Interestingly you can have a straight line in two cases : if you have a static quenching mechanism or a dynamic quenching mechanism. In your case a static quenching is the more probable. You will not get a straight line if both mechanisms act together. Indeed there is non more a linearity between concentrations and signals.