chapter 09 question

New Biology – a modern approach 2Chapter 9: Respiration

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Structured Questions

Core Section

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* John collected a grasshopper and a water beetle from a garden. He set up an experiment to investigate the

carbon dioxide release in these animals.

A

(grasshopper)

B

(water beetle)

C

(no animals)

Assume that the animals were equally active and the tubes were kept at the same temperature.

(i) State the indicator solution used. (1 mark)

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(ii) Describe and explain the colour change of the indicator solution in A after 2 days. (2 marks)

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(iii) State with reason whether tube B changed colour earlier than tube A. (2 marks)

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(iv) State the significance of the stopper. (1 mark)

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________________________________________________________________________________________© Aristo Educational Press Ltd 2007 29

stopper

indicator solution

wire gauze


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(v) If a pondweed is put in tube B at the beginning of the experiment and the tube is kept in bright light for

2 days, the colour change of the indicator solution is different from the previous result. State and

explain the colour change. (3 marks)

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(i) hydrogencarbonate indicator solution (1 mark)

(ii) The indicator will change from red to yellow (1 mark)

because the grasshopper will release carbon dioxide in respiration. (1 mark)

(iii) Yes, (1 mark)

it was because the water beetle released carbon dioxide to the indicator directly. (1 mark)

(iv) To prevent any gas from passing out or into the boiling tube (1 mark)

(v) The indicator solution becomes purple (1 mark)

because the pondweed undergoes photosynthesis at a faster rate than the respiration in bright light. (1

mark)

The tube thus contains less carbon dioxide. (1 mark)

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(9 marks)##

|!|EQA00900002|!|

The drawing below shows a set-up used to study anaerobic respiration in yeast:

An increase in dough volume was observed in both A and B after an hour.


dough(flour + yeast

+ glucose)

Set-up A(with water bath at 20℃)

Set-up B(with water bath at 30℃)


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(i) Account for the above observation. (2 marks)

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(ii) State and explain which container would show a greater increase in dough volume. (5 marks)

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(iii) If the water bath of container B were 100°C instead of 30°C, what would happen to the dough volume?

Explain. (2 marks)

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(iv) Apart from an increase in dough volume, suggest another possible observation of this experiment. (1

mark)

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(v) If a greater amount of yeast was used in container B, explain what would happen to the

(1) dough volume (3 marks)

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(2) rate of increase in dough volume (2 marks)

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(i) Carbon dioxide was produced when the yeast respires anaerobically, (1 mark)

causing the dough to rise. (1 mark)



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(ii) Container B. (1 mark)

The temperature of container B was higher than that of container A. (1 mark)

Thus, the enzymatic activity was more vigorous/ the enzymes were more active. (1 mark)

The rate of anaerobic respiration in yeast of container B was higher. (1 mark)

More carbon dioxide was produced within an hour,

causing greater swelling of the dough in container B. (1 mark)

(iii) The dough volume would remain unchanged/no swelling would be observed (1 mark)

because the yeast was killed at 100°C. (1 mark)

(iv) An alcoholic smell was detected. (1 mark)

(v) (1) The dough volume would remain unchanged since the amount of glucose was fixed. (2 marks)

Thus, the amount of carbon dioxide produced would be the same. (1 mark)

(2) The rate of the increase in dough volume would be faster (1 mark)

since more carbon dioxide would be produced per unit time when more yeast respire. (1 mark)

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(15 marks)##

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Sam set up two flasks to investigate one cell activity of yeast.

Flask A Flask B

(i) Suggest one external factor which should be kept constant to ensure that the results could be

compared. (1 mark)

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deflated balloon

mixture of yeast suspension and glucose solution

mixture of yeast suspension and distilled water


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(ii) (1) After two days, there is a change in the size of the balloon in A. State and explain the change. (4

marks)

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(2) Based on this result, state one application in food industry. (1 mark)

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(iii) (1) The solution in A contains a new compound. Name it. (1 mark)

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(2) Based on this result, state one application in food industry. (1 mark)

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(iv) If flask A is boiled for 3 minutes at the beginning of the experiment, it shows the same result as B.

Explain. (1 mark)

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(i) temperature (1 mark)

(ii) (1) The balloon becomes bigger in size (1 mark)

because in anaerobic respiration, (1 mark)

yeast breaks down glucose (1 mark)

to evolve carbon dioxide. (1 mark)

(2) Baking bread (1 mark)

(iii) (1) alcohol (1 mark)

(2) brewing of wine (1 mark)

(iv) Yeast is killed by boiling and cannot carry out respiration. (1 mark)

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(9 marks)##



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The drawing below shows a set-up used to study respiration of germinating peas:

(i) Why was a thermos flask used instead of a plastic bottle? (1 mark)

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(ii) Why was dry cotton wool used instead of rubber bung? (1 mark)

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(iii) What was the purpose of the damp cotton wool? (1 mark)

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(iv) Draw a graph to show the temperature change in relation to time in flasks A and B. (3 marks)

(v) What is the purpose of setting up flask B? (1 mark)


thermometer

dry cotton wool

thermos flasks

sterilised peas of equal weights

damp cotton wool

Flask A,with germinating peas

Flask B,with boiled peas


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(vi) State and explain the result of flask A if the peas were not sterilised. (3 marks)

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(vii) If the germinating peas were replaced by a rat, state and explain what would happen to the temperature

change. (2 marks)

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(i) Thermos flask was a better heat insulator to prevent heat loss. (1 mark)

(ii) The cotton wool allowed gaseous exchange for the respiration of germinating peas. (1 mark)

(iii) It provides water for the germination of seeds. (1 mark)

(iv)

Correct title (1 mark)

Correct axis (1 mark)

Correct plotting of lines (1 mark)

(v) Flask B acted as a control. (1 mark)



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(vi) The temperature increase would be greater (1 mark)

since micro-organisms on the surface of the germinating peas respire, releasing heat energy which

causes greater overall temperature increase. (2 marks)

(vii) The temperature increase would be greater because a rat has greater respiratory rate/metabolic rate than

germinating peas. (2 marks)

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(12 marks)##

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STS Connections

* Read the passage below and answer the questions that follow:

LOW-ALCOHOL BEER

Most alcohol-free (less than 0.05% alcohol) and low-alcohol (usually 0.5 -1.0%) beers are

brewed in the conventional way with alcohol taken out. The idea is simple.

The brew is first put in a partial vacuum and warmed slightly. Then, a variation is to force

a beer through a membrane. Alcohol and water can go through the membrane more easily than

the larger molecules which are beer concentrate. Add water and you can have a low-alcohol

beer.

But the processes mean some change in flavour. The newer approach is to stop too much

alcohol from being produced in the first place.

There are three ways of doing this:

- by chilling the yeast before;

- by putting in less carbohydrate;

- by finding a new strain of yeast which converts little of the carbohydrate to alcohol.

It is the third avenue which brewing industry scientists believe to be most promising.

They hope to create a new strain of yeast to make a driver-friendly and liver-friendly brew

which does not have a great difference in taste.

(i) Explain why a low-alcohol beer can be obtained by putting the brew in a partial vacuum and warming

it slightly. (2 marks)

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(ii) Explain why each of the following practices can reduce the production of alcohol:

(1) Chilling the yeast before. (2 marks)

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(2) Putting in less carbohydrate. (2 marks)

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(iii) Name the technique to create a new strain of yeast. (1 mark)

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(iv) Based on the information given in the passage, state one example of the following, which people who

switch to low and alcohol-free beers can achieve.

(1) A social advantage (1 mark)

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(2) A health advantage (1 mark)

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(i) At low pressure, alcohol boils at a lower temperature. (1 mark)

After boiling, the alcohol becomes a vapour and escape. (1 mark)

(ii) (1) Fermentation of glucose to alcohol is sped up by the enzymes found in yeast cells. (1 mark)

The enzymes become inactive at low temperatures. (1 mark)

(2) The rate of reaction is controlled by the substrate concentration. (1 mark)

Carbohydrate is the substrate for anaerobic fermentation. A reduction in the amount of substrate

slows down the rate of fermentation. (1 mark)

(iii) genetic engineering (1 mark)

(iv) (1) Safe in driving, reduce the number of traffic accidents caused by drunk-driving. (1 mark)

(2) Reduction in damage to liver. (1 mark)

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(9 marks)##



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Henry was confused with the results showing that germinating seeds would give out heat. Miss Chan

instructed him to perform the following experiment.

Henry recorded the temperature readings every hour.

Time in hour 0 1 2 3 4 5 6 7 8

Temperature in ℃ 20.0 20.2 20.5 21.0 21.5 22.4 23.5 24.5 26.0

(i) Plot a graph to show the change of temperature with time. (2 marks)



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(ii) In order to make the results valid, a control experiment has to be set up. State a difference between the

control and the test experiments. (1 mark)

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(iii) To which direction did the coloured water move in the U-tube? Explain. (3 marks)

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(iv) Predict the change in the amount of carbon dioxide in the air inside the container. (1 mark)

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(v) If the container is not covered with cotton wool, the results would be different. State and explain the

difference. (2 marks)

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(i)

Correct label for axes and title (1 mark)

Correct plotting of points and forming a curve (1 mark)

(ii) In the control experiment, there are no germinating seeds in the container. (1 mark)



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(iii) The coloured water on the right hand side rises. (1 mark)

As germinating seeds respire, heat is released. (1 mark)

The air inside the boiling tube will be heated up and expand and thus push the coloured water to the

right. (1 mark)

(iv) The amount of carbon dioxide is much larger than before. (1 mark)

(v) Without cotton wool, the increase in temperature is not as great as before. (1 mark)

Part of the heat released by the seeds is lost to the surroundings. (1 mark)

_________

(9 marks)##

|!|EQA00900007|!|

Read the passage below and answer the questions that follow:

RESPIRATION

Living organisms release energy by a process called respiration. A source of energy

(substrate) and oxygen are required for energy release which is controlled by enzymes.

Oxygen is required for the breakdown of substrate molecules. As these substrate molecules are

broken down, energy is released. The organism uses this energy for various life processes.

However, some of the energy is lost as heat. Carbon dioxide and water are formed during

energy release.

(i) According to the information given in the passage, complete the following table by using the word

‘decreases’ or ‘increases’ to show the various changes that take place during the energy release within a

living tissue. (2 marks)

Observation W ithin a living tissue

Oxygen concentration

Carbon dioxide concentration

Temperature

Mass

(ii) State TWO life processes in which living organisms use energy. (2 marks)

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(iii) Experiments to investigate energy release are often carried out in a water bath. Suggest reasons for this

practice. (3 marks)

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(iv) In muscle cells, the source of energy can be broken down in the absence of oxygen.



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(1) Compare the amount of energy released in the two types of respiration. (1 mark)

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(2) State the product formed in this type of respiration. (1 mark)

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(i)

Observation W ithin a living tissue

Oxygen concentration decreases

Carbon dioxide concentration increases

Temperature increases

Mass decreases

(1/2 mark each) (2 marks)

(ii) growth/movement/breathing (any two) (2 marks)

(iii) To keep the temperature constant. (1 mark)

Energy release is enzyme regulated and enzymes are sensitive to temperature changes. (1 mark)

A change in temperature will affect the rate of energy release. (1 mark)

(iv) (1) The amount of energy released in anaerobic condition is less than that in aerobic condition.

(1 mark)

(2) In anaerobic condition, lactic acid is formed. (1 mark)

_________

(9 marks)##

|!|EQA00900008|!|

The reaction of food burning can be described by the following word equation:

food + oxygen carbon dioxide + water + energy

(i) A physiological process in living cells has the same word equation. Name the process. (1 mark)

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(ii) Where does the process in (i) take place? (1 mark)

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(iii) Both food burning and the process in (i) release heat energy.

(1) State another form of energy released by food burning. (1 mark)

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(2) Heat does not comprise all the energy released by the process in (i). What happens to the energy

that is not released as heat? (3 marks)

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(iv) State THREE differences, which have not been discussed above, between food burning and the process

in (i). (3 marks)

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(v) Suggest TWO examples of the use of the energy released by the process in (i) in humans. (2 marks)

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(i) aerobic respiration (1 mark)

(ii) mitochondria (1 mark)

(iii) (1) light energy (1 mark)

(2) The energy is used in bond formation between ADP molecules and phosphates to produce ATP

molecules, which are high energy compounds storing energy temporarily inside cells. (3 marks)

(iv) - Food burning is a single-step process while aerobic respiration is a series of chemical reactions.

(1 mark)

- Food burning is not controlled by enzymes while aerobic respiration is controlled by enzymes.

(1 mark)

- The rate of energy release in food burning is vigorous and not under control while the energy release

in aerobic respiration is slower and under control. (1 mark)

(v) - muscle contraction

- nerve impulse transmission

- protein synthesis

- maintaining body temperature

- active transport

or any other possible answers (any two) (2 marks)

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(11 marks)##

|!|EQB00900009|!|

* Peter rushed to catch a bus and then he panted heavily right after he got on the bus.



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(i) Name the process occurring in his muscle cells during the rush. (1 mark)

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(ii) Write a word equation for the process in (i). (2 marks)

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(iii) Explain, in detail, why he panted so heavily. (4 marks)

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(iv) Did all the energy for the rush come from the process in (i)? Explain your answer. (2 marks)

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(i) anaerobic respiration (1 mark)

(ii) enzyme

glucose lactic acid + energy (2 or 0 marks)

(iii) To take in a large amount of oxygen to pay for the oxygen debt, (1 mark)

so as to convert the harmful lactic acid to carbon dioxide and water, or to glycogen. (2 marks)

Otherwise the lactic acid accumulated would cause muscle fatigue or muscle cramp. (1 mark)

(iv) No. The energy produced by anaerobic respiration alone is too little to support the motion. Aerobic

respiration proceeded at the same time to provide sufficient energy. (2 marks)

__________

(9 marks)##

|!|EQB00900010|!|

* The following diagram shows an experimental set-up to investigate respiration in seeds:



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(i) What is the function of

(1) the screw clip? (2 marks)

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(2) the thermos flasks? (1 mark)

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(3) the test tubes in this experiment? (2 marks)

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(ii) What treatment should be done to the seeds before putting them in the thermos flasks? State the

purpose of this treatment. (2 marks)

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(iii) (1) What will happen to the levels of coloured water in the U-tube of the manometer? (1 mark)

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(2) Give an explanation for your answer in (1). (3 marks)

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(iv) Would the same result be obtained if a living mouse were used instead of seeds in flask A?

(1 mark)

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(v) What conclusion can be drawn from this experiment? (1 mark)

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(vi) What was the function of flask B? (1 mark)

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(i) (1) to adjust the air pressure in the U-tube and the test tubes so that the levels of coloured water in the

two arms of the U-tube are the same (2 marks)

(2) to prevent heat loss (1 mark)

(3) to keep a small amount of air which can easily expand when warmed to push the coloured water in

the manometer (2 marks)

(ii) Wash the seeds with a dilute disinfectant. This can kill the bacteria and fungi on the seed surfaces. (2

marks)

(iii) (1) The level of coloured water in the left arm of the U-tube of the manometer rises. (1 mark)

(2) The seeds in flask A release a large amount of heat in respiration (1 mark)

while those in flask B are dead seeds and do not release heat. (1 mark)

Expansion of the air in the test tube in flask A will press down the water level in the right arm of

the manometer. (1 mark)

(iv) The same result would be obtained. (1 mark)

(v) Only living organisms can carry out respiration and release heat energy (1 mark)

(vi) To minimise errors caused by changes in the pressure of the environment.

(To serve as a control – No mark) (1 mark)

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(14 marks)##

|!|EQB00900011|!|

* In the set-up below, the bell jar could be lowered as deeply as possible after the rubber tubing connecting the



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air inside and outside the bell jar has been removed.

(i) What would happen to the water levels inside and outside the bell jar at the beginning of the

experiment if a rubber tubing was not used? (1 mark)

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(ii) (1) After moving the tubing, what change would be observed after a few hours? (1 mark)

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(2) Account for the change in (1). (5 marks)

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(iii) How can you set up a control for this experiment? (1 mark)

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(i) The water level inside the bell jar would be lower than that outside. (1 mark)

(ii) (1) After a few hours, the water level inside the bell jar slightly rose. (1 mark)

(2) There were some living organisms in the fresh soil. They carried out respiration. They absorbed

oxygen in the jar and released carbon dioxide which was absorbed by the caustic soda solution.

The pressure inside therefore decreased and became lower than that outside and the water level

rose. (5 marks)

(iii) The apparatus is the same except that the fresh soil is replaced by baked soil. (1 mark)

_________

(8 marks)##

|!|EQA00900012|!|

The drawing below shows the apparatus used in an experiment to investigate anaerobic respiration in yeast:



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(i) Write a word equation for anaerobic respiration in yeast. (2 marks)

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(ii) Why should the glucose solution be boiled and then cooled down? (3 marks)

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(iii) Would there be any change in the reading of the thermometer? Explain your answer. (2 marks)

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(iv) State how the anaerobic respiration of yeast can be applied in industry. (4 marks)

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(v) The indicator in the small tube was used to test for the by-product of respiration.

(1) Name the by-product. (1 mark)

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(2) Give the name of the indicator and describe its colour change. (2 marks)

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(i) glucose alcohol + carbon dioxide + energy (2 or 0 marks)

(ii) to exclude air from the glucose solution so as to make sure that the respiration of yeast cells was

anaerobic (2 marks)

If hot glucose solution were used, the yeast would be killed. (1 mark)

(iii) The temperature rose because heat was released by yeast cells during anaerobic respiration.

(2 marks)

(iv) for brewing/for making wine by the production of alcohol from anaerobic respiration (2 marks)

for baking/for making bread because the production of carbon dioxide from anaerobic respiration

makes the dough rise/spongy (2 marks)

(v) (1) carbon dioxide/CO2 (1 mark)

(2) hydrogencarbonate indicator solution -- turns from red to yellow

OR lime water -- turns from colourless to milky (2 marks)

_________

(14 marks)##

|!|EQA00900013|!|

Three test tubes were set up as shown in the following diagram. They were kept at 37°C for two hours.

(i) What changes in the size of the balloons would be expected in test tubes A, B and C, respectively?

(3 marks)

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(ii) What was the common gas collected in the balloons? (1 mark)

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(iii) How would you test for the gas collected? (1 mark)

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(iv) What was the other product formed in one of the test tubes? (1 mark)

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(v) Which test tube(s) serve(s) as the control(s)? (1 mark)

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(vi) Name the biological process demonstrated in this experiment. (1 mark)

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(i) Test tube A - size of balloon increased (1 mark)

Test tube B - size of balloon remained unchanged (1 mark)

Test tube C - size of balloon remained unchanged (1 mark)

(ii) carbon dioxide/CO2 (1 mark)

(iii) Carbon dioxide turns lime water milky (or turns hydrogencarbonate indicator solution yellow).

(1 mark)

(iv) Another product formed in test tube A was alcohol. (1 mark)

(v) Test tubes B and C are the controls. (1 or 0 mark)

(vi) fermentation/anaerobic respiration of yeast (1 mark)

__________

(8 marks)##



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|!|EQA00900014|!|

Some yeast was mixed with water and flour to make a dough which was kept at a constant temperature of

50℃ in an oven. The volume of the dough was measured every 5 minutes. The results are shown in the

following graph:

(i) From the graph, calculate

(1) the increase in volume of the dough during the first 20 minutes of the experiment. (2 marks)

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(2) the percentage increase in the volume of the dough between the 40th minutes and 50th minute.

(2 marks)

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(ii) Explain why there was an increase in volume of the dough. (3 marks)

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(i) (1) (14 - 10)cm3 = 4 cm3 (2 or 0 marks)

(2) x 100% = 11.9% OR x 100% = 13% (2 or 0 marks)

(ii) The yeast carried out anaerobic respiration to produce carbon dioxide which expanded on heating. (3

marks)

__________

(7 marks)##

|!|EQB00900015|!|

* The following diagrams show four sets of experimental apparatus:

The four sets of apparatus A, B, C and D were kept under identical conditions. The above diagrams show their

appearances after 3 days. The dotted lines indicate the initial levels of coloured water.

(i) Name chemical X and state its function. (2 marks)

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(ii) What was the physiological process investigated? (1 mark)

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(iii) Explain the results of experimental set-up A. (5 marks)

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(iv) Explain the results of experimental set-up B. (3 marks)

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(v) Why did the level of coloured water rise in experimental set-up D? (2 marks)

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(vi) Why did the level of coloured water remain unchanged in experimental set-up C? (1 mark)

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(i) concentrated sodium hydroxide solution/caustic soda solution (1 mark)

It absorbed carbon dioxide. (1 mark)

(ii) respiration (1 mark)

(iii) The peas took in oxygen and released carbon dioxide which was absorbed by the concentrated sodium

hydroxide solution. It resulted in the reduction of pressure so that the coloured water was sucked

towards the flask when the pressure inside the flask became lower than the atmospheric pressure.

(5 marks)

(iv) The peas took in oxygen and released an equal amount of carbon dioxide. The pressure did not change.

Therefore, the levels of the coloured water did not change. (3 marks)

(v) On the surfaces of the peas there were some microorganisms which carried out respiration / gaseous

exchange. (2 marks)

(vi) Both the peas and microorganisms were killed and no respiration took place. (1 mark)

__________

(14 marks)##

|!|EQB00900016|!|

* The following diagram shows an experiment to investigate the respiration of a small mammal:



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(i) The above set-up failed to give conclusive results. Suggest and explain how you would correct it.

(2 marks)

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(ii) What solution should be used in flasks A and B? (1 mark)

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(iii) What was the purpose of using the solution in each flask? (2 marks)

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(iv) After your correction, what would you observe in flasks A and B at the end of this experiment? (2

marks)

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(v) Suggest, with reason(s), the alteration(s) you would make in order to investigate the respiration of a

potted plant. (5 marks)

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(i) Replace the calcium chloride with caustic soda/sodium hydroxide so as to absorb carbon dioxide.

(2 marks)

(ii) lime water / hydrogencarbonate indicator solution (1 mark)

(iii) Flask A - to ensure that the incoming air was free of CO2 (1 mark)

Flask B - to show that CO2 came from the bell jar/ the mammal (1 mark)

(iv) - The lime water in flask A remained clear OR hydrogencarbonate indicator solution remained red / no

change in colour. (1 mark)

- The lime water in flask B turned milky OR hydrogencarbonate indicator solution turned yellow.

(1 mark)

(v) - Put the potted plant in the bell jar. Cover the bell jar with black paper to exclude light / prevent

photosynthesis. (2 marks)

- Cover the pot with a plastic bag to prevent the release of carbon dioxide produced by the

microorganisms in soil, otherwise this may interfere with the results of the experiment. (3 marks)

_________

(12 marks)##



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|!|EQA00900017|!|

The following graph shows the rate of three processes in a terrestrial plant within a day:

(i) State, with a reason, which curve represents

(1) photosynthesis. (2 marks)

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(2) respiration. (2 marks)

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(3) transpiration. (2 marks)

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(ii) (1) State at which time(s) of the day the respiratory rate is equal to the photosynthetic rate.

(2 marks)

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(2) Name the gas, if any, liberated by the plant at this/these time(s). (1 mark)

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##

(i) (1) Curve A represents photosynthesis because it takes place only in the presence of light. (2 marks)

(2) Curve C represents respiration because the respiratory rate is constant all the time. (2 marks)

(3) Curve B represents transpiration because its rate is higher in the daytime and occurs all the time.

(2 marks)

(ii) (1) 9 a.m., 6 p.m. (2 marks)

(2) water vapour (1 mark)



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_________

(9 marks)##

|!|EQB00900018|!|

* The following diagram shows an experiment to investigate the respiration of some small animals:

(i) Name chemical X in the test tube. State its function in this experiment. (2 marks)

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(ii) How can you set up a control for this experiment? (1 mark)

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(iii) The heights of the coloured water in the capillary tube of the above set-up were 3 cm at 9:00 a.m.

and 6.5 cm at 12:00 noon. The heights of the coloured water of the control were 3 cm at 9:00 a.m.

and 2 cm at 12:00 noon. Calculate the rate of the respiration of the woodlice in the set-up. (It is

given that the cross-sectional area of the capillary tube = 1 cm2) (2 marks)

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(iv) Explain the increase in height in the above set-up and the decrease in height in the control. (6 marks)

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(i) caustic soda / soda lime / sodium hydroxide (1 mark)

to absorb carbon dioxide (1 mark)

(ii) the same set-up without woodlice (1 mark)

(iii) 1 × = = 1.5 cm3/ hour (2 marks)

(iv) Oxygen was used up by the woodlice for respiration. (1 mark)

The carbon dioxide produced was absorbed by soda lime. (1 mark)

The pressure decreased. Coloured water was sucked upwards when the pressure inside was lower than

the atmospheric pressure. (1 mark)

In the control set-up no respiration took place. (1 mark)

The volume increase was due to gas expansion caused by the increase in temperature at noon / a drop

in the atmospheric pressure. (2 marks)

__________

(11 marks)##



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|!|EQB00900019|!|

* An experiment was set up to demonstrate the process of respiration in germinating seeds. Test tube B served

as a control.

(i) Which type of respiration was carried out by the germinating seeds in tube A? (1 mark)

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(ii) What results would be observed in tube B? Explain your answer. (3 marks)

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(iii) (1) What gas was present in space X? (1 mark)

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(2) How can you test for the presence of this gas? (2 marks)

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(3) Will the seeds germinate after several days? Explain your answer. (2 marks)

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(i) anaerobic respiration (1 mark)

(ii) The level of liquid paraffin in the test tube decreased slightly due to the presence of

micro-organisms/fungi on the surfaces of dead seeds and they carried out anaerobic respiration to

release CO2. (3 marks)

(iii) (1) carbon dioxide (1 mark)

(2) Introduce some caustic soda pellets into the test tube. (1 mark)

The liquid paraffin level will rise. (1 mark)

OR

Collect the gas sample with a J-tube and then pump the gas into lime water /

hydrogencarbonate indicator solution which will turn milky/ yellow respectively.

(3) no (1 mark)

because the seeds were killed by one of the by-products, alcohol. (1 mark)

_________

(9 marks)##

|!|EQB00900020|!|

* The diagrams below show the set-up to investigate if heat is given out by germinating seeds:

The table below shows the temperature inside the flasks over a period of 24 hours:

Time (hr) 0 3 6 9 12 15 18 21 24

Temperature in flask A (°C) 20 20 25 30 35 35 35 35 35

Temperature in flask B (°C) 20 20 20 20 25 30 35 40 45



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(i) On the same graph, draw TWO curves to present the results. (3 marks)

(ii) The temperature change was lower than expected. How can you improve set-up A and set-up B to

obtain a more obvious result? (3 marks)

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(iii) State the environmental conditions needed for germination of seeds. (3 marks)

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(iv) (1) Suggest a reason for the increase in temperature in flask B. (2 marks)

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(2) How can the increase in temperature in flask B be prevented? (1 mark)

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##

(i)



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Title (1/2 mark)

Correct labels for axes with correct choice (1/2 mark)

Correct plotting of any 6 points for each curve (1 mark)

Joining the points to form a curve (1 mark)

(ii) The flasks should be fixed upside down to trap the warm air. (1 mark)

The flasks should be plugged with cotton wool to trap the warm air. (1 mark)

The number of seeds/mass of seeds should be increased (1 mark)

(iii) warmth, moisture, oxygen (3 marks)

(iv) (1) There were microorganisms e.g. bacteria. (1 mark)

They respired to give out heat energy. (1 mark)

(2) The seeds should be sterilised by washing them with a disinfectant. (1 mark)

__________

(12 marks)##


chapter 09 question

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