4: Enzymes and Metabolism - Problems Flashcards
A student carried out an experiment to study the activity of catalase. He added a piece of liver tissue to a boiling tube with hydrogen peroxide solution and collected the oxygen released.
With reference to the functions of the liver, suggest why liver tissue contains a large amount of catalase.
Many metabolic reactions take place in the liver. Therefore, a large amount of hydrogen peroxide may be produced in the liver.
A large amount of catalase is present to catalyse the breakdown of the toxic hydrogen peroxide into water and oxygen.
True or false: enzymes are denatured at extreme temperatures.
False
Fruits and vegetables like apples, bananas and lettuce turn brown after they are cut or dropped. This is because an enzyme called polyphenol oxidase (PPO) is found in their cells. The enzyme catalyses a reaction, which is the major cause of the browning in fruits and vegetables.
word equation: polyphenolic compounds + oxygen –PPO–> brown pigment
The enzyme is found in the cytoplasm of the cells, while polyphenolic compounds are usually present in the vacuoles. Explain why after dropping an apple, part of the fruit would become brown. (3)
Plant tissues may be damaged by mechanical force upon dropping, the vacuole membrane and cell membrane rupture.
The ruptured vacuole membrane allows the enzyme PPO in the cytoplasm and polyphenolic compounds in the vacuole to come together.
The ruptured cell membrane exposes polyphenolic compounds to another substrate oxygen, (allowing the formation of enzyme-substrate complexes and) producing brown pigment.
Fruits and vegetables like apples, bananas and lettuce turn brown after they are cut or dropped. This is because an enzyme called polyphenol oxidase (PPO) is found in their cells. The enzyme catalyses a reaction, which is the major cause of the browning in fruits and vegetables.
word equation: polyphenolic compounds + oxygen –PPO–> brown pigment
Explain how boiling lettuce for one minute after cutting prevents the browning of lettuce. (2)
Boiling brings the PPO to a high temperature that is above its optimum temperature, causing a conformational change in shape of the active site of PPO. The PPO is denatured.
It can no longer bind to the substrate molecules to form enzyme-substrate complexes. Thus it does not catalyse the conversion of polyphenolic compounds into brown pigments; the reaction does not occur.
Fruits and vegetables like apples, bananas and lettuce turn brown after they are cut or dropped. This is because an enzyme called polyphenol oxidase (PPO) is found in their cells. The enzyme catalyses a reaction, which is the major cause of the browning in fruits and vegetables.
word equation: polyphenolic compounds + oxygen –PPO–> brown pigment
Explain how putting an apple in the refrigerator after cutting prevents the browning of apple. (2)
At low temperature, PPO becomes inactive.
The chance of PPO and its substrate molecules colliding to form enzyme-substrate complex with each other is low, thus the reaction occurs only at low rate.
Fruits and vegetables like apples, bananas and lettuce turn brown after they are cut or dropped. This is because an enzyme called polyphenol oxidase (PPO) is found in their cells. The enzyme catalyses a reaction, which is the major cause of the browning in fruits and vegetables.
word equation: polyphenolic compounds + oxygen –PPO–> brown pigment
Explain how putting food products made from banana in vacuum packs prevents the browning of the food products. (2)
Oxygen is not available in a vacuum pack.
Oxygen is one of the substrates of the reaction. Without oxygen, the enzyme-substrate complex cannot be formed, and the enzymatic reaction cannot occur.
(DSE 2015 IB Q7)
Hydrogen peroxide is a by-product of some metabolic reactions inside our body. If it were allowed to accumulate, it would kill our cells. Fortunately, we have an enzyme called catalase which speeds up the breakdown of hydrogen peroxide into water and oxygen. An investigation is carried out to observe the effect of temperature on the activity of catalase. Below are the steps:
- Extract catalase from an animal organ.
- Mix 5mL 0.1% hydrogen peroxide solution with 1mL catalase extract.
- Place the mixture in a water bath set at 0ºC.
- Measure and record the volume of oxygen gas released in the first 5 minutes.
- Repeat steps 2 to 4 with the water bath set at 20ºC, 40ºC, and 60ºC.
Suggest an animal organ in which catalase is present in great abundance and from which the enzyme can be obtained. Explain why this organ has so much catalase.
Liver
It is the organ where many metabolic reactions take place; it is the organ for detoxification.
(Many metabolic reactions take place in the liver. Therefore, a large amount of hydrogen peroxide may be produced in the liver.
A large amount of catalase is present to catalyse the breakdown of the toxic hydrogen peroxide into water and oxygen.)
(DSE 2015 IB Q7)
Hydrogen peroxide is a by-product of some metabolic reactions inside our body. If it were allowed to accumulate, it would kill our cells. Fortunately, we have an enzyme called catalase which speeds up the breakdown of hydrogen peroxide into water and oxygen. An investigation is carried out to observe the effect of temperature on the activity of catalase. Below are the steps:
- Extract catalase from an animal organ.
- Mix 5mL 0.1% hydrogen peroxide solution with 1mL catalase extract.
- Place the mixture in a water bath set at 0ºC.
- Measure and record the volume of oxygen gas released in the first 5 minutes.
- Repeat steps 2 to 4 with the water bath set at 20ºC, 40ºC, and 60ºC.
Suggest and explain the importance of an important procedure that was missed out.
The hydrogen peroxide and catalase solution should be incubated at the set temperatures for 10 minutes before mixing.
This is to ensure that the mixture has reached the set temperature at the beginning of the reaction.
Three test tubes (W, X, and Y) are set up to measure the rate of an enzymatic reaction. The enzyme catalyses the following reaction:
substrate P —-> product Q
In each test tube, either the concentration of substrate P or product Q was measured. The graph below shows the results.
State in which of the test tubes is the concentration of substrate P measured. (1)
X and Y
Three test tubes (W, X, and Y) are set up to measure the rate of an enzymatic reaction. The enzyme catalyses the following reaction:
substrate P —-> product Q
In each test tube, either the concentration of substrate P or product Q was measured. The graph below shows the results.
Compare and explain the difference in rate of reaction at the 2nd minute and the 20th minute in tube W. (4)
The rate at the 2nd minute is higher than that at the 20th minute since the steepness of the curve was larger at the 2nd minute than the 20th minute.
This is because the concentration of substrate at the 2nd minute is higher than that at the 20th minute.
Therefore, the enzyme molecules collide with the substrate molecules more frequently at the 2nd minute.
The chance of forming enzyme-substrate complexes is higher, thus the rate of reaction is higher.
Three test tubes (W, X, and Y) are set up to measure the rate of an enzymatic reaction. The enzyme catalyses the following reaction:
substrate P —-> product Q
In each test tube, either the concentration of substrate P or product Q was measured. The graph below shows the results.
The experimental condition of one of the test tubes is different from the others. Identify the test tube and give one piece of evidence from the graph to support the answer. (2)
X
The sum of the substrate concentration in tube Y and the product concentration in tube W is always 100%, indicating that the experimental conditions of these two tubes are the same.
Three test tubes (W, X, and Y) are set up to measure the rate of an enzymatic reaction. The enzyme catalyses the following reaction:
substrate P —-> product Q
In each test tube, either the concentration of substrate P or product Q was measured. The graph below shows the results.
Suggest one possible difference in experimental condition between test tube X and the other test tubes. (2)
Lower temperature
Jelly was used to investigate the effect of pH on the activity of a protease. Jelly contains a protein called gelatin; the protease catalyses the breakdown of gelatine into soluble products. Four petri dishes containing 10cm3 buffer solutions at different pH. Four jelly blocks of the same size were cut and one was put into each petri dish. Then, 10cm3 protease solution was added to each petri dish. After leaving the petri dishes at room temperature for two hours, the area of the upper face of each jelly block was measured. The results are shown below.
Describe and explain the effect of pH and the activity of the protease. (4)
When pH increases from 4 to 6, the area of the remaining jelly block decreases. This shows the activity of the protease increases as pH increases from 4 to 6.
When pH increases from 6 to 10, the area of the remaining jelly block increases. This shows the activity of the protease decreases as pH increases from 6 to 10.
The protease denatures when pH is lower or higher than the optimum pH. The substrate can no longer fit into the active site of the protease to form the enzyme-substrate complex. Therefore, the rate of the reaction is greatly lowered.
Jelly was used to investigate the effect of pH on the activity of a protease. Jelly contains a protein called gelatin; the protease catalyses the breakdown of gelatine into soluble products. Four petri dishes containing 10cm3 buffer solutions at different pH. Four jelly blocks of the same size were cut and one was put into each petri dish. Then, 10cm3 protease solution was added to each petri dish. After leaving the petri dishes at room temperature for two hours, the area of the upper face of each jelly block was measured. The results are shown below.
Describe and explain the effect of pH and the activity of the protease. (4)
When pH increases from 4 to 6, the area of the remaining jelly block decreases. This shows the activity of the protease increases as pH increases from 4 to 6.
When pH increases from 6 to 10, the area of the remaining jelly block increases. This shows the activity of the protease decreases as pH increases from 6 to 10.
The protease denatures when pH is lower or higher than the optimum pH. The substrate can no longer fit into the active site of the protease to form the enzyme-substrate complex. Therefore, the rate of the reaction is greatly lowered.
An investigation was carried out about how pH affects the activity of an enzyme used in the extraction of fruit juice. The table below shows the content of four beakers in the experiment.
The beakers were allowed to stand for 10 minutes. The contents were then filtered and the volumes of juice collected were measured:
Volume of juice collected (W, X, Y, Z) = (10, 19, 11, 11)
Explain why water was added instead of enzyme solution in some of the beakers. (2)
The beakers were the control set-ups with no enzymes added.
Water was added to these beakers to keep the total volume the same, so that the results of different beakers can be compared.