21: Respiration - Problems

Question 1

The electron micrograph below shows the cross section of a mitochondrion.

Name structures X and Y. (2)

Answer 1

X: crista
Y: outer mitochondrial membrane

Question 2

Mitochondria in animal cells can vary greatly in length and shape. However, their diameter is usually within 1μm. Suggest one advantage of having their diameter within 1μm. (1)

Answer 2

A smaller diameter decreases the diffusion distance. This allows molecules to diffuse rapidly into the centre of mitochondria.

Question 3

The electron micrograph below shows the cross section of a mitochondrion.

The ratio of the surface area of structure X to the surface area of structure Y from the muscle cells of mice P and Q were calculated. The ratio in mouse Q is higher than that in P.

Deduce which mouse is more likely to have carried out physical exercise for a longer period time and explain. (4)

Answer 3

Mouse Q.
The ratio of the surface area of structure X to the surface area of structure Y in mouse Q is higher. This indicates that the mitochondria in mouse Q contains more structure X.
Structure X is packed with enzymes involved in the reactions of respiration.
Mitochondria with more structure X can produce more ATP for muscle contraction during physical exercise.

Question 4

(AL 2009 I Q8)
An experiment was conducted other investigate the effects of chemical X on animal cell culture. After treating the cells with different concentrations of chemical X for 24 hours, the concentration of cellular ATP and percentage of cell death in each cell culture were measured. The table below shows the mean result obtained after repeated trials.
Explain which organelle might chemical X act on. (2)

Answer 4

Mitochondria,
since it is the main site of ATP production.

Question 5

(AL 2009 I Q8)
An experiment was conducted other investigate the effects of chemical X on animal cell culture. After treating the cells with different concentrations of chemical X for 24 hours, the concentration of cellular ATP and percentage of cell death in each cell culture were measured. The table below shows the mean result obtained after repeated trials.
Suggest how the change of concentration of cellular ATP would lead to the death of the cells. (1)

Answer 5

When ATP synthesis is inhibited, insufficient energy is provided to the cells, and they die since they have no energy to carry out metabolic activities.

Question 6

(AL 2009 I Q8)
An experiment was conducted other investigate the effects of chemical X on animal cell culture. After treating the cells with different concentrations of chemical X for 24 hours, the concentration of cellular ATP and percentage of cell death in each cell culture were measured. The table below shows the mean result obtained after repeated trials.
The cells used in the above experiment were taken from the epithelium of the small intestine rats. Suggest two ways the small intestine function would be affected if rats were fed with an animal feed containing a high concentration of chemical X. (2)

Answer 6

The absorption of food substances by active transport will be hindered,
and the production of intestinal juice will be hindered.

Question 7

The diagram below outlines the processes of aerobic respiration in a cell.
State the step in which ATP is used and state its function in this step. (2)

Answer 7

From glucose to triose phosphate,
ATP is used to activate glucose by phosphorylation.

Question 8

The diagram below outlines the processes of aerobic respiration in a cell.
Name two electron carriers that transfers electrons from glycolysis and Krebs cycle to the electron transport chain. (2)

Answer 8

NADH
and FADH

Question 9

The diagram below outlines the processes of aerobic respiration in a cell.
State the role of oxygen in the electron transport chain. (1)

Answer 9

Oxygen acts as the final electron acceptor in the electron transport chain.

Question 10

When studying aerobic respiration in germinating seeds, the seeds have to be sterilised. Then the seeds are placed in vacuum flask held in an inverted position. The temperatures inside the vacuum flasks are recorded.
Suggest the purpose of sterilising the seeds in the experiment. (3)

Answer 10

To kill the microorganisms on the surface of the seeds.
Otherwise, carbon dioxide released by them during respiration will cause an overestimation of the heat production by the germinating seeds.
Thus, the validity of the results decreases.

Question 11

When studying aerobic respiration in germinating seeds, the seeds have to be sterilised. Then the seeds are placed in vacuum flask held in an inverted position. The temperatures inside the vacuum flasks are recorded.
Suggest three reasons of placing the vacuum flasks in an inverted position. (3)

Answer 11

1. Warm air rises as it is less dense than cold air. Putting the vacuum flasks in an inverted position can trap the warm air and minimize heat loss.
2. Readings of the thermometer can be taken more easily.
3. As elevated concentration of carbon dioxide can reduce the rate of respiration, the inverted vacuum flask that is closed by dry cotton wool allows the escape of carbon dioxide released by the seeds from the flask because carbon dioxide is more dense than air.

Question 12

A researcher extracted mitochondria from muscle cells. He then put the mitochondria into a buffer solution containing a substrate, ADP and phosphate ions. He monitored the consumption of oxygen, the substrate and phosphate ions in the solution. The table below shows the results.
Suggest one reason to explain why the amount of ATP produced in the experiment is much lower than that in an intact cell. (1)

Answer 12

Mitochondria may be damaged in the extraction process.

Question 13

A researcher extracted mitochondria from muscle cells. He then put the mitochondria into a buffer solution containing a substrate, ADP and phosphate ions. He monitored the consumption of oxygen, the substrate and phosphate ions in the solution. The table below shows the results.
Explain how the amount of ATP produced would be different if the experiment was repeated using glucose as the substrate. (2)

Answer 13

No ATP would be produced,
because mitochondria cannot use glucose as the substrate directly.

Question 14

In plants, both chloroplasts and mitochondria are involved in the production of ATP. Discuss the similarities and differences between the production of ATP in these organelles. Explain why plants cannot rely on chloroplast as the only source of ATP. (8+3)

Answer 14

Similarities:
In both organelles, ATP is produced from the electron transport chain,
which is embedded in membranes.
Differences:
In chloroplasts, light energy is captured to form ATP. In mitochondria, chemical energy stored in the food is released to form ATP.
In chloroplasts, ATP is formed from photophosphorylation. In mitochondria, ATP is formed from Krebs cycle and oxidative phosphorylation.
In chloroplasts, chlorophyll donates electrons to the electron transport chain. In mitochondria, the electrons are donated by NADH and FADH.
In chloroplasts, NADP is the final electron acceptor in the electron transport chain. In mitochondria, oxygen is the final electron acceptor.
Why plants cannot rely on chloroplast as the only source of ATP:
Chloroplasts cannot produce ATP in the dark. Therefore, plants need mitochondria to provide a steady supply of ATP in the dark.
Some plant cells do not have chloroplasts and ATP cannot be transported from cell to cell. Therefore, these plant cells cannot survive without mitochondria.

Question 15

The diagram below shows the change in concentration of lactic acid in the blood of two individuals (X and Y) when they ran a 200-m race. One of the individuals is an athlete.
Explain why the concentration of lactic acid in the blood increases in both individuals during running. (2)

Answer 15

During running, skeletal muscles also respire anaerobically to provide additional energy for muscle contraction.
Pyruvate formed in glycolysis is reduced to lactic acid by NADH. Therefore, the lactic acid concentration in the blood increases.

Question 16

The diagram below shows the change in concentration of lactic acid in the blood of two individuals (X and Y) when they ran a 200-m race. One of the individuals is an athlete.
Explain which individual is an athlete. (3)

Answer 16

Individual Y
The change in lactic acid concentration in the blood of individual Y is smaller than that of individual X.
This indicates that individual Y can remove lactic acid more efficiently.

Question 17

(DSE 2012 IB Q9)
Drugs X and Y may inhibit enzymes involved in glycolysis, the Krebs cycle or oxidative phosphorylation. To study the effects of the drugs, some muscle cells were isolated and treated with these two drugs separately in the presence of oxygen. The cellular levels of ATP, NADH, and pyruvate were determined and recorded in the table below.
The data for the control are set as 100% for comparative purpose.
Suggest and explain the key process that is inhibited by drug X. (3)

Answer 17

Drug X inhibited glycolysis.
As glycolysis is the first step in the respiratory pathway, the inhibition of glycolysis will halt the processes that follow, which are Krebs cycle and oxidative phosphorylation.
Hence, the overall production of pyruvate, ATP and NADH are greatly reduced, showing that the whole respiratory pathway was jeopardised.
OR:
Pyruvate is the product of glycolysis.
As the production of pyruvate is greatly reduced after treating with drug X, glycolysis was inhibited in this case.

Question 18

(DSE 2012 IB Q9)
Drugs X and Y may inhibit enzymes involved in glycolysis, the Krebs cycle or oxidative phosphorylation. To study the effects of the drugs, some muscle cells were isolated and treated with these two drugs separately in the presence of oxygen. The cellular levels of ATP, NADH, and pyruvate were determined and recorded in the table below.
The data for the control are set as 100% for comparative purpose.
Suggest the process that is inhibited by drug Y, and explain why there is an accumulation of pyruvate in the muscle cells after treatment with drug Y. (3)

Answer 18

Drug Y inhibited Krebs cycle.
When the respiratory pathway is halted at Krebs cycle, pyruvate will not be metabolised and the rate of consumption of pyruvate is greatly reduced.
On the other hand, glycolysis still proceeds as usual and produces pyruvate. The rate of pyruvate production is higher than the rate of pyruvate consumption, hence pyruvate accumulates.

Question 19

(DSE 2012 IB Q9)
Drugs X and Y may inhibit enzymes involved in glycolysis, the Krebs cycle or oxidative phosphorylation. To study the effects of the drugs, some muscle cells were isolated and treated with these two drugs separately in the presence of oxygen. The cellular levels of ATP, NADH, and pyruvate were determined and recorded in the table below.
The data for the control are set as 100% for comparative purpose.
The untreated muscle cells were incubated under anaerobic conditions. Predict the change in cellular ATP, NADH and lactate levels. (3)

Answer 19

In anaerobic conditions, muscle cells undergoes anaerobic respiration which produce less ATP,
and less NADH than that of aerobic respiration.
Lactic acid and lactate levels rise as it is a product of anaerobic respiration of muscle cells.

Question 20

(DSE 2012 IB Q9)
Drugs X and Y may inhibit enzymes involved in glycolysis, the Krebs cycle or oxidative phosphorylation. To study the effects of the drugs, some muscle cells were isolated and treated with these two drugs separately in the presence of oxygen. The cellular levels of ATP, NADH, and pyruvate were determined and recorded in the table below.
The data for the control are set as 100% for comparative purpose.
A student would like to study the enzymes involved in glycolysis, the Krebs cycle, and oxidative phosphorylation respectively. Suggest which cellular components they need to isolate for the investigation. (3)

Answer 20

Glycolysis: cytoplasm
Krebs cycle: Mitochondrial matrix
Oxidative phosphorylation: Inner mitochondrial membrane

Question 21

(AL 2006 I Q11)
To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC.
For both species, suggest and explain which process (photosynthesis or respiration) is more sensitive to heat stress. (2)

Answer 21

Photosynthesis,
since the photosynthetic rate for bot species dropped at a lower temperature than respiration rate.

Question 22

(AL 2006 I Q11)
To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC.
Give two clauses that may lead to the inhibition of photosynthesis and respiration at high temperatures. (2)

Answer 22

1. Heat denatures enzymes required in the dark reactions of photosynthesis and in the reactions of respiration.
2. The high temperature destroys the membrane of chloroplasts and mitochondria.

Question 23

(AL 2006 I Q11)
To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC.
The two plant species are native to Hong Kong and the Sahara desert respectively. Identify and explain the species which is native to the Sahara desert. (3)

Answer 23

A is from the Sahara desert.
The rate of photosynthesis and respiration of plant A starts to drop at a higher temperature compared to plant B,
showing that it is more tolerant to high temperatures typical of the desert area.

Question 24

(AL 2006 I Q11)
To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC.
Explain why ion leakage can be used as a parameter for measuring the heat stress response of the plants. (2)

Answer 24

High temperatures denature the channel proteins and destroys membrane integrity.
Therefore, high temperature increases the membranes’ permeability to ions, hence resulting in ion leakage detected.

Question 25

(AL 2006 I Q11) To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC. State and explain the critical heat-killing temperature for species B. (2)

Answer 25

52ºC, At this temperature, the rate of respiration of the cell drops to 0, indicating cell death.

Question 26

(AL 2006 I Q11) To study the responses of two plant species A and B to heat stress, the rates of photosynthesis and respiration, and the ion leakage of the leaves of these plants were measured after they had been exposed to different temperature treatments. The control experiments were carried out at 30ºC and the rates of photosynthesis and respiration at each temperature treatment were expressed as a percentage of respective rate at 30ºC. Fruits and vegetables that are grown in high but sub-lethal temperatures for a prolonged period during the growing season are less sweet. With reference to the findings of this study, explain the physiological basis underlying such a phenomenon.

Answer 26

Photosynthetic rate decreases at a lower temperature compared to the respiration rate. At high but sub-lethal temperatures, the respiration rate remains high but the photosynthetic rate is much lower. Hence, the net production of sugar in the plant decreases.

Question 27

(2008 AL Q2) The pathway of glycolysis is outlined in the diagram below. In the pathway, the conversion of intermediate A to intermediate B is catalysed by an enzyme E, which is inhibited by a high level of ATP. Based on this information, suggest and explain whether the rate of glycolysis would be higher under aerobic or anaerobic conditions. (4)

Answer 27

The rate of glycolysis would be higher under anaerobic conditions, because ATP production is lower in anaerobic conditions. The **inhibitory effect** of ATP on enzyme E is lower, hence the conversion of intermediates A to B would be faster, resulting in a higher rate of product formation and hence a higher rate of glycolysis.

Question 28

Aerobic respiration consists of 3 main stages of which one takes place outside the mitochondrion. Name this stage and state its importance. (3)

Answer 28

Glycolysis, which is the initial breakdown of glucose to release energy without oxygen required. It produces **pyruvate** which enters the remaining 2 stages of aerobic respiration, for **complete oxidative breakdown** of the organic compound to release more **energy** in the form of **ATP**.

Question 29

Describe the relationship between Krebs cycle and oxidative phosphorylation. (5)

Answer 29

Each acetyl-CoA molecule enters the Krebs cycle first to have further oxidative breakdown, producing 3 NADH, 1 FADH, and 1 ATP in the process. During oxidative phosphorylation in the inner membrane of the mitochondrion, NADH and FADH molecules pass their hydrogen atoms to the **electron transport chain**. This **regenerates NAD and FAD**, which allows them to receive hydrogen from the Krebs cycle so the Krebs cycle can **proceed continuously**.

Question 30

State the importance of oxygen in oxidative phosphorylation. (3)

Answer 30

Oxygen is the **final hydrogen acceptor** so it can receive hydrogen atoms from NADH and FADH, for the redox reactions to proceed in the electron transport chain to **produce ATP**. It causes **regeneration of NAD and FAD** which allows them to receive hydrogen from the Krebs cycle so the Krebs cycle can **proceed continuously**.

Question 31

Explain the importance of anaerobic respiration during vigorous activity. (4)

Answer 31

Vigorous activity requires **large amounts of energy** to support powerful and/or frequent muscle contractions. The increase in rate of breathing and heart rate is **insufficient** to supply enough oxygen to skeletal muscles for them to carry out **aerobic respiration** rapid enough to meet the high energy requirement. Anaerobic respiration can produce extra energy in a short amount of time to meet the energy requirement.