A pH of 2 is optimal for specific digestive enzymes found in a pig's stomach. If these enzymes are placed into a neutral substrate solution, what do you predict will be the outcome? The enzyme reaction time would decrease. The enzyme reaction time would increase. The enzyme reaction time would stay the same. The optimal pH of the enzyme would change.
Question
A pH of 2 is optimal for specific digestive enzymes found in a pig's stomach. If these enzymes are placed into a neutral substrate solution, what do you predict will be the outcome? The enzyme reaction time would decrease. The enzyme reaction time would increase. The enzyme reaction time would stay the same. The optimal pH of the enzyme would change.
Solution
The enzyme reaction time would likely increase. Enzymes are very sensitive to their environment, including the pH level. If the pH is not at the optimal level, the enzyme may not function as efficiently, which could slow down the reaction time. The optimal pH of the enzyme would not change, as this is a characteristic of the enzyme itself. However, the enzyme's performance can change based on the pH of its environment.
Similar Questions
Which statement about the optimum pH for an enzyme is correct?It is the same for all enzymes.It is the pH at which the enzyme's activity occurs at the highest rate.It is the pH at which the enzyme cannot be damaged by heat.I DON'T KNOWSUBMIT ANSWER
7.What happens to the enzyme's active site above the optimum pH?
How does a decrease in pH typically affect enzyme activity?AIncreases activity by adding more substratesBAlters the enzyme's shape by affecting ionic bondsCPrevents denaturation by forming new peptide bondsDFacilitates the release of the product from the enzyme
b) Scientists investigated the effect of pH on the activity of the enzyme catalase in a fungus. The table below shows the scientists’ results.pH Enzyme activity in arbitrary units Test 1 Test 2 Test 3 Test 4 Test 5 Mean 3.0 0 0 0 0 0 0 4.0 6 5 8 4 7 6 5.0 38 65 41 42 39 5.5 80 86 82 84 88 84 6.0 100 99 96 103 102 100 6.5 94 92 90 93 91 92 7.0 61 63 61 62 63 62 8.0 22 22 21 24 21 22 Page 12 of 19The independent variables are the different parts of the bean seedling.The dependent variable is the amount of catalyst used The control variables are the temperature, concentration of hydrogen peroxide and the time for the investigationfirst get fresh bean seedling of the same age and sizethen get 5 test stubes labelled for each part of the bean seedling: seed, stem, root, leaf and seed coat and add the same volume of hydrogen peroxide to each test tube (25cm^3). Keep a separate 6th test tube with only hydrogen peroxide to compare against.cut a standardized piece of each part of the bean seedling seed, stem, root, leaf and seed coat and add them to their retrospective test tube.Observe and record the production of oxygen gas (which will be indicated by the amount of foam produced) after a certain time e.g. 30 seconds measure the height and volume of the foam and record it for each test tubeCompare the amount of foam produced for each test tube and that tells us the amounts of catalyst in the different parts of the bean seedling(i) Calculate the mean enzyme activity at pH 5.0. ______________________________________________________________ ______________________________________________________________ Mean = _____________ arbitrary units (2)
The enzyme pepsin becomes active when pH isMultiple Choicehigh.low.
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