1. Enzyme activity variation with pH: Enzymes have an optimal pH at which their active site is most effective. Deviations from this pH can alter the enzyme's structure or the ionization state of amino acids at the active site, reducing activity. For pectinase, maximum activity at pH 5.0 means the enzyme structure and substrate binding are optimal there. At pH 3.0 and 7.0, activity drops to 40% and 60% because the enzyme likely denatures or its active site charge changes, decreasing affinity for substrate. This information guides fruit juice processing to maintain pH near 5.0 for efficient pectinase function. 2. Lactase enzyme data analysis: Given substrate concentrations [S] and initial rates v: | [S] (mM) | Rate (µmol/min) | |---------|-----------------| | 2 | 4 | | 4 | 6.4 | | 10 | 9.0 | | 20 | 10.5 | | 40 | 11.5 | 1. Calculate Lineweaver-Burk plot values: Lineweaver-Burk plot is $$\frac{1}{v} = \frac{K_m}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}}$$ Compute $$\frac{1}{[S]}$$ and $$\frac{1}{v}$$: | [S] | v | 1/[S] | 1/v | |-----|----|-------|------| | 2 | 4 | 0.5 | 0.25 | | 4 |6.4 | 0.25 | 0.15625| |10 |9.0 | 0.1 | 0.1111| |20 |10.5| 0.05 | 0.0952| |40 |11.5| 0.025 | 0.087| Use linear regression or approximate two points to find slope and intercept. Using points (0.5,0.25) and (0.025,0.087): $$m = \frac{0.25-0.087}{0.5-0.025} = \frac{0.163}{0.475} \approx 0.343$$ $$b = y - mx = 0.25 - 0.343\times 0.5 = 0.25 - 0.1715 = 0.0785$$ Therefore, $$V_{max} = \frac{1}{b} = \frac{1}{0.0785} \approx 12.74\ \mathrm{\mu mol/min}$$ $$K_m = m \times V_{max} = 0.343 \times 12.74 \approx 4.37\ \mathrm{mM}$$ 2. Rate at $$[S]=8\ \mathrm{mM}$$: Using Michaelis-Menten: $$v = \frac{V_{max} [S]}{K_m + [S]} = \frac{12.74 \times 8}{4.37 + 8} = \frac{101.92}{12.37} \approx 8.24\ \mathrm{\mu mol/min}$$ 3. High $$K_m$$ implies low substrate affinity; thus, fatty enzymatic efficiency at low lactose concentrations would be reduced. 3. Oxidase enzyme inhibition: 1. Inhibitor raises apparent $$K_m$$ from 2 mM to 6 mM; $$V_{max}$$ remains 20 µmol/min. This indicates competitive inhibition, where inhibitor competes with substrate for active site, increasing $$K_m$$ but not affecting $$V_{max}$$. 2. Rate without inhibitor at $$[S]=3\ \mathrm{mM}$$: $$v = \frac{20 \times 3}{2 + 3} = \frac{60}{5} = 12\ \mathrm{\mu mol/min}$$ Rate with inhibitor (apparent $$K_m=6\ \mathrm{mM}$$): $$v = \frac{20 \times 3}{6 + 3} = \frac{60}{9} = 6.67\ \mathrm{\mu mol/min}$$ 3. Competitive inhibition reduces browning by lowering reaction rate when ascorbic acid is present. 4. Soy sauce protease inhibition: The rate decreases beyond 15 mM substrate; this suggests substrate inhibition where excessive substrate binds additional inhibitory sites, decreasing enzyme activity. This affects process control in fermentation ensuring substrate concentration remains optimal for maximum protease activity.