493. In the industrial production of clarified apple juice, pectinases are added to break down pectin. If a 500 mL batch of cloudy apple juice contains 10 g of pectin, and the enzyme has an activity of 1000 units per mg, how many milligrams of pectinase would be needed to fully degrade the pectin if each unit degrades 1 $\mu$g of pectin per minute over 30 minutes?
Key Concept: Enzyme Action, Industrial Applications
c) 333.33 mg
[Solution Description]
First, calculate the total amount of pectin to be degraded: 10 g = 10,000 mg. Each unit of enzyme degrades 1 $\mu$g (0.001 mg) of pectin per minute. Over 30 minutes, one unit degrades $0.001 \, \text{mg/min} \times 30 \, \text{min} = 0.03 \, \text{mg}$. To degrade 10,000 mg of pectin, the number of units required is $\frac{10,000 \, \text{mg}}{0.03 \, \text{mg/unit}} = 333,333.33 \, \text{units}$. Since the enzyme has an activity of 1000 units per mg, the mass of enzyme needed is $\frac{333,333.33 \, \text{units}}{1000 \, \text{units/mg}} = 333.33 \, \text{mg}$.
Your Answer is correct.
c) 333.33 mg
[Solution Description]
First, calculate the total amount of pectin to be degraded: 10 g = 10,000 mg. Each unit of enzyme degrades 1 $\mu$g (0.001 mg) of pectin per minute. Over 30 minutes, one unit degrades $0.001 \, \text{mg/min} \times 30 \, \text{min} = 0.03 \, \text{mg}$. To degrade 10,000 mg of pectin, the number of units required is $\frac{10,000 \, \text{mg}}{0.03 \, \text{mg/unit}} = 333,333.33 \, \text{units}$. Since the enzyme has an activity of 1000 units per mg, the mass of enzyme needed is $\frac{333,333.33 \, \text{units}}{1000 \, \text{units/mg}} = 333.33 \, \text{mg}$.