With substrate in excess, 2 cm3 of an enzyme solution makes 12 mg of product per minute. What rate is expected with 5 cm3 of the same enzyme solution, all other conditions the same?
- Substrate is in excess, so rate is directly proportional to the amount of enzyme.
- Rate per cm3 of enzyme = 12 ÷ 2 = 6 mg per minute.
- Rate with 5 cm3 = 6 × 5 = 30 mg per minute.
Answer: 30 mg of product per minute.