1. **Problem:** Calculate the rate of reaction at [S] = 3 mM with and without inhibitor for an oxidase enzyme with $V_{max} = 20$ μmol/min, $K_m = 2$ mM without inhibitor and apparent $K_m = 6$ mM with inhibitor. Identify inhibition type and discuss usefulness in browning inhibition. 2. **Problem:** Explain decreased protease activity beyond 15 mM substrate in soy sauce fermentation, suggest substrate inhibition, experimental confirmation, and importance in fermentation. 3. **Problem:** Calculate specific activity, purification fold, and percentage yield for pectinase enzyme purification data. 4. **Problem:** Compare enzyme activity at two pH conditions for β-galactosidase given $V_{max}$ and $K_m$ values; discuss pH effects and optimal conditions. 5. **Problem:** Explain pectinase activity variation with pH and its application in fruit juice processing. 6. **Problem:** Using lactose hydrolysis data, determine $V_{max}$ and $K_m$ by Lineweaver-Burk plot, calculate rate at $[S] = 8$ mM, and interpret high $K_m$ implication. --- ### 7. Oxidase Enzyme Inhibition 1. **Identify inhibition type:** Given that $V_{max}$ stays constant and $K_m$ increases, this is **competitive inhibition**. 2. **Calculate rates:** Using Michaelis-Menten equation $$v = \frac{V_{max}[S]}{K_m + [S]}$$ - Without inhibitor: $$v = \frac{20 \times 3}{2 + 3} = \frac{60}{5} = 12 \text{ μmol/min}$$ - With inhibitor ($K_m = 6$ mM): $$v = \frac{20 \times 3}{6 + 3} = \frac{60}{9} = 6.67 \text{ μmol/min}$$ 3. **Inhibition usefulness:** Competitive inhibition by ascorbic acid reduces browning by slowing oxidase activity, preventing enzymatic browning in apples and food products, thus preserving quality. ### 8. Soy Sauce Protease Activity 1. **Explanation:** Rate decreases after 15 mM possibly due to **substrate inhibition**, where excess substrate hinders enzyme activity. 2. **Experimental confirmation:** Measure reaction rates at increasing [S], plot data, look for rate decline after a substrate threshold. 3. **Importance:** Understanding substrate inhibition allows optimizing substrate concentration to maximize yield and prevent efficiency loss in fermentation. ### 9. Pectinase Purification Given: - Crude: Volume = 100 mL, Total activity = 500 U, Protein = $100 \times 100 = 10,000$ mg - Purified: Volume = 20 mL, Total activity = 200 U, Protein = $5 \times 20 = 100$ mg 1. Specific activity (U/mg): - Crude: $$\frac{500}{10,000} = 0.05$$ U/mg - Purified: $$\frac{200}{100} = 2$$ U/mg 2. Purification fold: $$\frac{2}{0.05} = 40$$ fold 3. Percentage yield: $$\frac{200}{500} \times 100 = 40\%$$ ### 10. β-Galactosidase pH Effects 1. Lower pH (4.5) reduces $V_{max}$ but decreases $K_m$, increasing substrate affinity. 2. Optimal: pH 6.5 has higher $V_{max}$ for process efficiency; pH 4.5 favors substrate binding but lower activity. Selection depends on sweetness (more product formation) and efficiency. ### 5. Pectinase Activity vs pH Enzyme activity varies due to ionization state changes affecting active site shape; maximum at pH 5.0. This guides processing pH to maximize enzyme efficiency, avoiding conditions where activity drops (pH 3.0, 7.0). ### 6. Lactase Kinetics Using Lineweaver-Burk: 1/Rate = (Km/Vmax)(1/[S]) + 1/Vmax From data, estimate: - $V_{max} \approx 13$ μmol/min - $K_m \approx 6$ mM Calculate rate at $[S] = 8$ mM: $$v = \frac{13 \times 8}{6 + 8} = \frac{104}{14} = 7.43\text{ μmol/min}$$ High $K_m$ means lower affinity: enzyme requires higher substrate concentration for activity, implying less efficient lactose hydrolysis. --- "slug": "enzyme kinetics", "subject": "biochemistry", "desmos": {"latex": "y= \frac{20 x}{2 + x}","features": {"intercepts": true,"extrema": true}}, "q_count": 7