1. **Problem Statement:** a) Define immunoassay. b) Explain competitive ELISA. c) Name and explain two other types of ELISA. d) Use the provided standards absorbance data to graphically determine the aflatoxin concentration for samples 1, 2, and 3. 2. **Definitions and Explanations:** a) An immunoassay is a biochemical test that measures the presence or concentration of a substance, commonly using the reaction of an antibody or antibodies to its antigen. b) Competitive ELISA is a type of enzyme-linked immunosorbent assay where sample antigen competes with a labeled antigen for binding to a specific antibody. The intensity of the signal inversely correlates with the antigen concentration. c) Other types of ELISA include: - **Direct ELISA:** An antigen is immobilized on the plate and detected by a labeled antibody, providing a direct measure of antigen presence. - **Sandwich ELISA:** Uses two antibodies; the first antibody captures the antigen, and the second labeled antibody reacts with another epitope on the antigen, allowing sensitive and specific detection. 3. **Graphical Determination of Unknown Concentrations:** - The calibration curve plots absorbance (y-axis) against concentration (x-axis) with points: $$(0,1.85), (4,1.6), (10,1.4), (20,0.9), (40,0.25)$$ - Because this is competitive ELISA, absorbance decreases as concentration increases. - Fit a line between the points (approximate linear fit for interpolation): Calculate slope $m$: $$m = \frac{1.85 - 0.25}{0 - 40} = \frac{1.6}{-40} = -0.04$$ Using point-slope form for the line: $$y - 1.85 = -0.04(x - 0) \implies y = -0.04x + 1.85$$ - To find concentration $x$ for a given absorbance $y$: $$x = \frac{1.85 - y}{0.04}$$ Calculate for each sample: - Sample 1: absorbance = 0.55 $$x = \frac{1.85 - 0.55}{0.04} = \frac{1.3}{0.04} = 32.5\,\text{ppb}$$ - Sample 2: absorbance = 0.30 $$x = \frac{1.85 - 0.30}{0.04} = \frac{1.55}{0.04} = 38.75\,\text{ppb}$$ - Sample 3: absorbance = 0.80 $$x = \frac{1.85 - 0.80}{0.04} = \frac{1.05}{0.04} = 26.25\,\text{ppb}$$ 4. **Enzyme Activity Questions:** 1. Reaction rate increases with temperature because higher temperature increases molecular kinetic energy, leading to more frequent and energetic collisions between enzymes and substrates. 2. At 70 °C, enzyme activity will likely decrease or cease because high temperatures cause denaturation, altering enzyme structure and active site, thus preventing substrate binding and catalysis.