1. **Problem Statement:** We are given concentration values and corresponding absorbance (T1, T2, T3) readings. We need to find the IC50 value, which is the concentration at which the response (absorbance) is reduced by 50% compared to the control (0 concentration). 2. **Understanding IC50:** IC50 is the concentration of an inhibitor where the response (or binding) is reduced by half. We first calculate the average absorbance for each concentration and normalize it relative to the control (0 concentration). 3. **Calculate average absorbance for each concentration:** - For 960ug/ml: $\frac{0.869 + 0.867 + 0.649}{3} = 0.795$ - For 480ug/ml: $\frac{0.911 + 0.943 + 0.836}{3} = 0.897$ - For 240ug/ml: $\frac{0.962 + 1.026 + 0.974}{3} = 0.987$ - For 120ug/ml: $\frac{0.948 + 0.952 + 0.992}{3} = 0.964$ - For 60ug/ml: $\frac{0.909 + 0.945 + 0.894}{3} = 0.916$ - For 0ug/ml (control): $\frac{0.962 + 0.805 + 0.937}{3} = 0.901$ 4. **Normalize absorbance values relative to control:** $$\text{Normalized} = \frac{\text{Average absorbance at concentration}}{\text{Average absorbance at 0 concentration}}$$ - 960ug/ml: $\frac{0.795}{0.901} = 0.882$ - 480ug/ml: $\frac{0.897}{0.901} = 0.995$ - 240ug/ml: $\frac{0.987}{0.901} = 1.096$ - 120ug/ml: $\frac{0.964}{0.901} = 1.070$ - 60ug/ml: $\frac{0.916}{0.901} = 1.017$ - 0ug/ml: $1$ 5. **Determine IC50:** IC50 corresponds to 50% inhibition, so normalized absorbance should be 0.5. From the data, the lowest normalized value is 0.882 at 960ug/ml, which is still above 0.5. This suggests the IC50 is higher than 960ug/ml or not reached within the tested concentrations. 6. **Conclusion:** Based on the data, the IC50 value is greater than 960ug/ml since the response does not drop to 50% at the highest tested concentration. **Final answer:** IC50 > 960ug/ml