Certain environmental changes can denature proteins. denatured proteins lose their three dimensional structure and when the protein is an enzyme, they lose their ability to function. ph is just one factor that can change the effectiveness of an enzyme. what other factor might the students test regarding catalase specificity?
a) students could test various other enzymes to see if all enzymes were denatured in a basic environment (ph = 13).
b) after determining the optimum ph, they could vary the temperature of the environment to see if catalase is temperature specific.
c) after determining the optimum ph, the students could vary the amount of catalase used while keeping the amount of substrate constant.
d) students could test an assortment of food items besides liver, raw potato, apple, crackers, to see if catalase is consistently ph specific.

B) After determining the optimum pH, they could vary the temperature of the environment to see if catalase is temperature specific.

Explanation:

As already mentioned in the question above, pH can alternate the efficiency of the enzyme. However, pH is not the only factor capable of altering this efficiency, temperature and time can also cause problems for the work of an enzyme. For this reason, we can say that in relation to the question shown above, after checking the pH, students can vary the temperature of the environment to see if the catalase is specific to the temperature.

The speed of a chemical reaction, in which the enzyme is involved, increases or decreases with the interference of the temperature. In addition, the enzymes have an ideal temperature so that they can maintain their three-dimensional structure and so that they can work correctly. When the temperature of the environment is above or below the ideal temperature of the enzyme, it will begin to break its structure, losing its three-dimensional shape and failing to properly fit the substrate necessary for its reaction.

9.9 years

a = p(1 + rt) . . account balance after time t at rate r starting with principal p

3000 = 2500(1 + 0.0202t) . . filling in the given numbers

1.2 = 1 + 0.0202t . . divide by 2500

0.2 = 0.0202t . . . subtract 1

0.2/0.0202 = t ≈ 9.901

it will take about 9.9 years for the account balance to reach $3000.