1. **State the problem:** We have ELISA data of absorbances vs concentrations, and samples absorbances. We need to find sample concentrations by interpolation. 2. **What is immunoassay?** It is a biochemical test that measures the presence or concentration of a substance using antibodies or antigens. 3. **What is competitive ELISA?** In competitive ELISA, the sample antigen competes with a labeled antigen for binding sites on antibodies, so a higher analyte concentration leads to lower signal. 4. **Other 2 types of ELISA:** a) **Sandwich ELISA:** Antibody is coated on plate, captures antigen from sample, then a second antibody detects it, giving signal proportional to antigen. b) **Direct ELISA:** Antigen is fixed on plate, enzyme-labeled antibody binds directly for detection. 5. **Determining sample concentrations:** a) Plot standards: Concentration $(x)$ vs Absorbance $(y)$: Points: $(0,1.85), (4,1.6), (10,1.4), (20,0.9), (40,0.25)$ b) Perform linear interpolation between pairs to find the concentration corresponding to each sample absorbance: For $Sample\ 1$ with $A=0.55$: Between $(20,0.9)$ and $(40,0.25)$ Slope $m=\frac{0.25-0.9}{40-20} = \frac{-0.65}{20} = -0.0325$ Equation: $y=m(x-20)+0.9$ so $0.55 = -0.0325(x-20)+0.9$ $-0.0325(x-20) = 0.55 - 0.9 = -0.35$ $x-20=\frac{-0.35}{-0.0325} = 10.77$ $x=30.77$ ppb For $Sample\ 2$ with $A=0.30$: Also between $(20,0.9)$ and $(40,0.25)$ $0.30 = -0.0325(x-20) + 0.9$ $-0.0325(x-20) = 0.30-0.9 = -0.6$ $x-20= \frac{-0.6}{-0.0325} = 18.46$ $x=38.46$ ppb For $Sample\ 3$ with $A=0.80$: Between $(10,1.4)$ and $(20,0.9)$ Slope $m= \frac{0.9-1.4}{20-10} = \frac{-0.5}{10} = -0.05$ $0.80 = -0.05(x-10) + 1.4$ $-0.05(x-10) = 0.80-1.4 = -0.6$ $x-10= \frac{-0.6}{-0.05} = 12$ $x=22$ ppb 6. **Explain rate and temperature for amylase:** 1. Rate increases because temperature increases molecular collisions and enzyme activity until an optimum. 2. At 70 °C, enzyme might denature, losing structure and activity, so rate decreases.