1. **Problem:** Calculate the rate of reaction at substrate concentration $[S]=3$ mM with and without inhibitor for an oxidase enzyme with $V_{max}=20$ $\mu$mol/min and $K_m=2$ mM without inhibitor, and apparent $K_m=6$ mM with inhibitor, $V_{max}$ unchanged. 2. The rate of reaction follows Michaelis-Menten kinetics: $$v = \frac{V_{max}[S]}{K_m + [S]}$$ 3. **Without inhibitor:** $$v = \frac{20 \times 3}{2 + 3} = \frac{60}{5} = 12\ \mu mol/min$$ 4. **With inhibitor:** $$v = \frac{20 \times 3}{6 + 3} = \frac{60}{9} \approx 6.67\ \mu mol/min$$ 5. **Inhibition type:** Since $K_m$ increases and $V_{max}$ is unchanged, this is competitive inhibition. It competes with substrate for active site. 6. **Usefulness:** Competitive inhibition by ascorbic acid slows browning by reducing enzymatic oxidation of phenolics that lead to brown pigments. --- 7. **Problem:** Explain substrate inhibition observed during soy sauce fermentation where beyond 15 mM substrate concentration, rate decreases. 8. **Explanation:** High substrate concentration causes substrate molecules to bind to inhibitory sites on the enzyme, reducing activity (substrate inhibition). 9. **Experimental confirmation:** Measure reaction rates at varying substrate concentrations and fit data to substrate inhibition kinetics equations or perform enzyme assays with and without substrate analogs. 10. **Importance:** Understanding substrate inhibition helps optimize substrate levels in fermentation to maximize product yield and prevent enzyme inactivation. --- 11. **Problem:** Calculate specific activity, purification fold, and percentage yield of pectinase enzyme purification. Given: - Crude extract: volume = 100 mL, total activity = 500 U, protein conc. = 100 mg/mL - Purified: volume = 20 mL, total activity = 200 U, protein conc. = 5 mg/mL 12. **Calculations:** - Total protein crude = $100 \times 100 = 10000$ mg - Specific activity crude = $\frac{500}{10000} = 0.05$ U/mg - Total protein purified = $20 \times 5 = 100$ mg - Specific activity purified = $\frac{200}{100} = 2$ U/mg - Purification fold = $\frac{2}{0.05} = 40$ - Percentage yield = $\frac{200}{500} \times 100 = 40\%$ --- 13. **Problem:** Discuss effect of pH on lactase enzyme activity; two conditions with data: | Condition | pH | $V_{max}$ ($\mu$mol/min) | $K_m$ (mM) | |---|---|---|---| | A | 6.5 | 15 | 5 | | B | 4.5 | 12 | 2 | 14. **Discussion:** - Higher $V_{max}$ at pH 6.5 means faster maximum reaction rate. - Lower $K_m$ at pH 4.5 means higher substrate affinity. - Optimal conditions balance high $V_{max}$ and suitable affinity; pH 6.5 offers better efficiency and sweetness due to more product formed. --- 15. **Problem:** Explain enzyme activity variation with pH for pectinase (max at pH 5.0, drops to 40% at pH 3.0 and 60% at pH 7.0). 16. **Explanation:** Enzyme structure and charge distribution change with pH affecting active site shape and substrate binding. Extremes cause denaturation or reduced activity. 17. **Application:** pH optimization in juice processing maintains enzyme activity for efficient clarification. --- 18. **Problem:** Determine $V_{max}$ and $K_m$ of lactase from data using Lineweaver-Burk plot. Data: $[S]$: 2,4,10,20,40 mM Rate: 4,6.4,9,10.5,11.5 $\mu$mol/min 19. **Calculation:** Lineweaver-Burk plot uses $1/v$ vs $1/[S]$: - Calculate reciprocals. - Linear regression gives intercept = $1/V_{max}$, slope = $K_m/V_{max}$. - Approximate $V_{max} \approx 13$ $\mu$mol/min, $K_m \approx 4.8$ mM. 20. **Rate at $[S]=8$ mM:** $$v = \frac{13 \times 8}{4.8 + 8} = \frac{104}{12.8} \approx 8.13\ \mu mol/min$$ 21. **High $K_m$ implication:** Lower substrate affinity, less efficient lactose hydrolysis. --- 22. **Answers to ELISA questions:** - Immunoassay: technique to detect/quantify substances using antibody-antigen reaction. - Competitive ELISA: sample antigen competes with labeled antigen for antibody binding. - Other ELISA types: Sandwich ELISA (captures antigen between two antibodies), Indirect ELISA (detects antibody with labeled secondary antibody). - Concentrations for samples from standard curve: Sample 1 (~0.55 absorbance) ~~ 15 ppb Sample 2 (~0.30 absorbance) ~~ 35 ppb Sample 3 (~0.80 absorbance) ~~ 10 ppb (approximate from interpolation). --- 23. **Temperature effect on amylase rate:** - Increase from 37°C to 50°C raises rate due to faster molecular collisions. - At 70°C, denaturation likely reduces activity due to enzyme structural damage. --- 24. **Industrial amyloglucosidase kinetics (Lineweaver-Burk for given data):** - $V_{max} \approx 40$ mM/min, $K_m \approx 25$ mM. - Turnover number $k_{cat} = \frac{V_{max}}{[E]} = \frac{40}{1 \times 10^{-4}} = 4\times10^{5}$ min$^{-1}$. 25. **Eadie-Hofstee plot data:** - $V_{max} \approx 33$ uM/min, $K_m \approx 8$ mM. - Turnover number $k_{cat} = \frac{33}{1 \times 10^{-4}} = 3.3 \times 10^{5}$ min$^{-1}$. ---