chapter 04 question

New Biology – a modern approach 1Chapter 4: Cell activities

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Structured Questions

Core Section

|!|EQA00400001|!|

Complete the following paragraph with suitable words selected from the list below:

diffusion hypertonic wall pressure osmosis water potential water pressure

hypotonic selectively randomly freely turgor pressure

Water is very important to life. It enters and leaves the living cells by (i)_______________. Plant cells

have a cell wall made of cellulose. The wall is porous and (ii) _______________ permeable. In contrast, the

cell membrane is made of protein and lipids and is (iii) _______________ permeable. Whereas water can

move freely through the cell membrane, dissolved solutes cannot. The tendency of water molecules to diffuse

from one place to another is called the (iv) _______________. When immersed in a (v) _______________

solution, the cells take in water. As a result, the cytoplasm enlarges until it presses against the cell wall with a

force called (vi) _______________.

(3 marks)

##

(i) osmosis (1/2 mark)

(ii) freely (1/2 mark)

(iii) selectively (1/2 mark)

(iv) water potential (1/2 mark)

(v) hypotonic (1/2 mark)

(vi) turgor pressure (1/2 mark)

__________

(3 marks)##

________________________________________________________________________________________© Aristo Educational Press Ltd 2007 126


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|!|EQB00400002|!|

* In an experiment to study the effect of temperature on the action of an enzyme for coagulating milk, milk and

the enzyme were first kept in separate test tubes in a water bath at ToC for 15 minutes before they were mixed.

The resulting mixture was kept in the same water bath. The time taken for milk to coagulate was noted and the

experiment was repeated by using water baths at different temperatures. The time taken for the milk to

coagulate at different temperature is shown in the table below.

Temperature of water bath (T o C) Time taken for coagulation ( m inutes)

10 No coagulation occurred.

20 7

30 4

40 2

50 4

60 No coagulation occurred.

(i) What is the purpose of keeping the milk and enzyme in the water bath for 15 minutes before mixing? (1

mark)

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(ii) At which temperature was the enzyme most active? (1 mark)

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(iii) State and explain whether there would be any change in the results if the mixture originally kept at

10oC was warmed to 30oC. (3 marks)

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(iv) What control experiment should be carried out to confirm that the coagulation of milk was due to the

action of an enzyme? What was the expected result? (3 marks)

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(v) State one factor, other than temperature, that can affect the activity of enzymes. (1 mark)

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##

(i) To allow sufficient time for milk and the enzyme to reach the required temperature. (1 mark)

(ii) 40℃ (1 mark)



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(iii) Coagulation will take place. (1 mark)

At 10oC, enzyme was inactive. However, at 30oC enzyme would become active and functional. Thus

coagulation would take place. (2 marks)

(iv) Replacing the enzyme solution with boiled enzyme solution / distilled water and repeating the whole

experiment. (2 marks)

The mixture should not coagulate at all temperatures. (1 mark)

(v) pH / enzyme concentration / substrate concentration (Choose any one) (1 mark)

_________

(9 marks)##

|!|EQA00400003|!|

Nancy performed an experiment to find out the effect of a strong salt solution on the growth of two potted

plants, A and B. She poured the salt solution to the pots daily for 15 days.

(i) Most parts of the survived plant B withered. Explain. (4 marks)

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(ii) Give a term used to describe the salt solution in relation to the cytoplasm of plant B. (1 mark)

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(iii) Draw a labeled diagram to show a root cell from plant B. (2 marks)

(iv) Name the processes in which plant A obtains salt from the soil. (2 marks)

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##

(i) The salt solution has a lower water potential than that of the root cells. (1 mark)

Water moves out of the root by osmosis. (2 marks)

The flaccid plant does not have enough water to support the body. (1 mark)

(ii) Hypertonic. (1 mark)

(iii)

(2 marks)

(iv) Diffusion and active transport. (2 marks)

_________

(9 marks)##


A plasmolysed root cell


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|!|EQB00400004|!|

* The following experiment was carried out to study the properties of a washing powder containing a

biologically active substance X.

(i) Describe the action of substance X on the protein coat. (2 marks)

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(ii) What is the effect of boiling on substance X? (1 mark)

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(iii) Name substance X. Explain why this washing powder can remove egg stain from clothes.

(3 marks)

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(iv) It is not advisable to use this washing powder with water at temperatures above 60oC. Explain why. (1

mark)

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(v) It is also suggested not to use this washing powder together with bleaching solution. Explain briefly. (2

marks)

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film with a layer of protein coat

washing powder solution

boiled washing powder solution

protein coat

absent

protein coat

present

A B

After 2 hours


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##

(i) Substance X breaks down / digests the solid protein coat into soluble substances. (2 marks)

(ii) Substance X loses its function upon boiling (making it unable to break down the protein coat). (1

mark)

(iii) X is an enzyme / protease. (1 mark)

This washing powder contains enzyme / protease that digests the protein in egg stains to small

molecules which are soluble in water / can be removed by water. (2 marks)

(iv) The enzyme would be denatured at temperatures above 60oC. (1 mark)

(v) Bleaching solution is alkaline. The activity of the enzyme in the washing powder may be inhibited at

this high pH value. (2 marks)

_________

(9 marks)##

|!|EQA00400005|!|

Tom prepared an experiment to investigate the action of an enzyme. The test tubes were placed in a water bath

at 37°C.

At the beginning 3 minutes later 30 minutes later

cloudy cloudy clear

(i) Name the enzyme. (1 mark)

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(ii) Explain why the suspension became clear. (1 mark)

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(iii) Suggest with reason one method to speed up the action. (3 marks)

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10 cm3 of egg white

2 cm3 of enzyme suspension was added and stirred


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(iv) Predict and explain the result if the test tube is placed in a water bath at

(1) 80°C. (2 marks)

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(2) 0°C. (2 marks)

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##

(i) protease (1 mark)

(ii) The protein in egg white was broken down in the presence of protease. (1 mark)

(iii) Add two drops of hydrochloric acid into the test tube. (1 mark)

This brings the pH of the suspension to a value close to 2 (1 mark)

which is the optimal pH of protease. (1 mark)

(iv) (1) The suspension would remain cloudy because (1 mark)

at a temperature higher than 37°C, the enzyme would be denatured and cannot catalyse the

hydrolytic reaction. (1 mark)

(2) The suspension would remain cloudy because (1 mark)

at a low temperature, the enzyme would become inactive. (1 mark)

_________

(9 marks)##

|!|EQA00400006|!|

Helen set up an experiment to investigate the action of amylase on starch:


Boiling tube A Boiling tube B Boiling tube C

dialysis tubing

water bath at 20℃

distilled water


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Boiling tube C ontent inside the dialysis tubing pH of the content

A starch + amylase 2

B starch + amylase 7.5

C starch + amylase 11

(i) State the action of amylase on starch. (2 marks)

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(ii) State the tube in which the action was the fastest. Explain your answer. (2 marks)

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(iii) Design a control to avoid the argument that the result is caused by other non-pH factors. (1 mark)

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(iv) Suggest with reason a method to speed up the action. (2 marks)

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(v) Predict, if the amylase can act on starch, whether the distilled water contains the end product. Explain

your answer. (2 marks)

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##

(i) Starch is broken down (1 mark)

in the presence of amylase to reducing sugars, either glucose or maltose. (1 mark)

(ii) B, because amylase works best at pH 7.5. (2 marks)

(iii) Same set-up but the dialysis tubing contains starch and distilled water. (1 mark)

(iv) The temperature of the water bath is kept at 37℃. (1 mark)

This is the optimal temperature at which the enzyme works most efficiently. (1 mark)

(v) Yes, (1 mark)

because the end product inside is small enough to diffuse through the dialysis tubing to the distilled

water outside. (1 mark)

_________

(9 marks)##

|!|EQA00400007|!|



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STS Connections

Read the passage below and answer the questions that follow:

Tenderisation of Meat

Enzymes have been utilized by man for thousands of years. One example is the use of

enzyme by some Africans. Much of the meat eaten in these areas is very fibrous and tough.

This type of meat would need to be cooked for hours to break down the fibrous protein, and

so make the meat edible. The people in these areas have learned that after covering the meat

with fruit extracts from the paw-paw tree for several hours, the meat is tenderised. They also

found that after cooking, the process of tenderisation stopped and the meat is no longer

tender.

(i) Name the substrate on which the enzyme in paw-paw works. (1 mark)

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(ii) State the chemical nature of the enzyme. (1 mark)

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(iii) Describe the action of the enzyme on the specific substrate. (2 marks)

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(iv) Explain why cooking stops tenderisation. (2 marks)

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(v) Complete the following table by filling either ‘more’ or ‘less’: (2 marks)

U ntreated meat Tenderi s ed meat

Toughness of meat

Cooking time

(vi) Suggest another fruit which has the same effect on meat. (1 mark)

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##

(i) The enzyme works on protein. (1 mark)

(ii) The enzyme is protein in nature. (1 mark)

(iii) The enzyme speeds up the breakdown of protein to amino acids. (2 marks)

(iv) Enzyme is denatured when cooked at a high temperature. (1 mark)

The denatured enzyme cannot carry out the breakdown process. (1 mark)



___________________________________________________________________________________

(v)

U ntreated meat Tenderi s ed meat

Toughness of meat more less

Cooking time more less

(½ mark each) (2 marks)

(vi) pineapple (1 mark)

_________

(9 marks)##

|!|EQB00400008|!|

* Four potato cylinders with equal length and mass were freshly prepared from a peeled potato. They were

placed in four beakers as shown below:

Their masses were recorded and presented graphically as shown below:

mass (g)

A

C

D

B

0 1 2 3 4 5 6 7 8 9 10 11 12

1

2

3

4

5

6

(i) What was the independent variable in this experiment? (1 mark)

_______________________________________________________________________________

(ii) What process did the experiment demonstrate? What molecule(s) was / were involved in the movement

of substances in the process? (2 marks)

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time (hour)


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(iii) What relationship was studied in the experiment? Describe the relationship according to the results.

(2 marks)

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(iv) Why did all potato cylinders show no change in mass after the first three hours? (2 marks)

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(v) Name another feature other than change in mass, which can be studied to obtain similar results. (1

mark)

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(vi) Describe and explain curves B and D. (4 marks)

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(vii) Another identical potato cylinder was put into a 10% sucrose solution (not shown in the diagram).

Sketch a graph to show the results for this potato cylinder. (2 marks)

##



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(i) the concentrations of sucrose solution (1 mark)

(ii) osmosis (1 mark)

water molecules (1 mark)

(iii) To study how sucrose solutions of different concentrations affected the mass of the potato cylinders. (1

mark)

When the sucrose concentration was higher than a particular value, the mass of the potato cylinder

decreased. When the sucrose concentration was lower than a particular value, the mass of the potato

cylinder increased. (1 mark)

(iv) After the first three hours, the water potential on both sides (within the cylinder and outside the

cylinder) became equal. (1 mark)

No osmosis occurred and there was no change in mass. (1 mark)

(v) the length of potato cylinders (1 mark)

(vi) The mass of the potato cylinder in set-up B did not change (1 mark)

since the water potential within the cylinder and that outside the cylinder were almost the same and

hence no osmosis occurred. (1 mark)

The mass of the potato cylinder in set-up D decreased since the water potential within the cylinder was

higher than that outside. (1 mark)

Water left the cylinder via osmosis and its mass was reduced. (1 mark)

(vii)

Mass (g)

0 1 2 3 4 5 6 7 8 9 10 11 12

1

2

3

4

5

6

(2 marks)

__________

(14 marks)##


Time (hour)


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|!|EQB00400009|!|

* Four different jelly blocks are shown in the diagram below:

(i) The experiment demonstrated a mechanism of the movement of substances. Name and define this

mechanism. (3 marks)

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_______________________________________________________________________________

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(ii) The rate of the mechanism that you answered in (i) was studied in the experiment. Suggest how this

rate could be determined. (1 mark)

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(iii) Several factors affecting the rate of the mechanism were studied in the experiment. Fill in the table

below: (8 marks)

Factor studied Set-ups compared

The set-up that w ould

have a deeper colour

penetration after 30

minutes.

Implication

Surface area to

volume ratio

The larger the ratio, the

faster the rate of the

mechanism studied.

A and B

Total surface area

##

(i) Diffusion (1 mark)

Diffusion is the process of net movement of molecules from a region of higher concentration to a

region of lower concentration (down the concentration gradient). (2 marks)

(ii) Examine the depth of colour penetration after a certain period of time. (1 mark)



___________________________________________________________________________________

(iii)

Factor studied Set-ups compared

The set-up that w ould

have a deeper colour

penetration after 30

minutes.

Implication

Surface area to

volume ratio

C and D

(or any reasonable answers)

C

(or any reasonable answers)

The larger the ratio, the

faster the rate of the

mechanism studied.

Dye concentration A and B B

The higher the dye

concentration, the faster

the rate of the mechanism

studied.

Total surface area B and C B

The larger the total surface

area, the faster the rate of

the mechanism studied.

(1 mark each) (8 marks)

__________

(12 marks)##

|!|EQA00400010|!|

The diagrams below show an experiment on osmosis:

The dialysis tubing in each set-up was filled with different liquids as shown below:

Dialysis tubing Liquid filled

A distilled water

B 5% glucose solution

C 5% sucrose solution



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D 10% glucose solution

E 10% sucrose solution

The set-ups were left undisturbed for 30 minutes.



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(i) Mark on the diagram below, the expected liquid level in the glass tubing of each set-up after 30

minutes. (3 marks)

(ii) Long and thin dialysis tubings and glass tubings were preferred in this experiment. Why?

(3 marks)

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(iii) When the set-ups were prepared, the tied dialysis tubings were put under a stream of running tap water

for a while after they had been filled with the liquids. Explain why this step was necessary.

(2 marks)

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(iv) Compare the results of set-ups B and C. Explain their difference(s). (4 marks)

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(i)

A: remained unchanged (1 mark)

B and D: remained unchanged (1 mark)

C and E: the liquid levels were higher than the original one and the liquid level in E was higher than

that in C. (1 mark)

(ii) A long and thin dialysis tubing provided a greater total surface area and a shorter distance for faster

diffusion / osmosis to occur. Results can be obtained in a shorter time. (2 marks)

A long and thin glass tubing enables clear observation of the changes in liquid level. (1 mark)

(iii) To wash away the liquid left on the outer surface of the dialysis tubing. (1 mark)

The liquid left could contaminate the distilled water and affect the diffusion / osmosis rate. (1

mark)

(iv) In set-up B, there was no change in the liquid level. Since glucose molecules could pass through the

dialysis tubing, they diffused from the dialysis tubing to the distilled water. The glucose concentrations

inside and outside the dialysis tubing were eventually equal and osmosis did not occur. Therefore, there

was no change in the liquid level in set-up B. (2 marks)

In set-up C, the liquid level rose. As the sucrose molecules were too large to leave the dialysis tubing,

the water potential outside the dialysis tubing was higher than that inside. Water entered the dialysis

tubing by osmosis and the liquid level rose. (2 marks)

__________

(12 marks)##



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|!|EQA00400011|!|

A student set up an apparatus as shown below to study how temperature affected the rate of diffusion:

Six identical set-ups were separately left at six different temperatures: 10℃, 20℃, 30℃, 40℃, 50℃ and

60℃. A few drops of liquid outside the dialysis tubing were removed and tested by Benedict’s test at 30-

second intervals until positive results were obtained. The time required to obtain positive results was

recorded. The student then presented his data in a graph as follows:

(i) Based on the graph, explain the relationship between temperature and diffusion rate. (2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

(ii) Explain why the liquid outside the dialysis tubing had to be stirred with a glass rod before removal for

Benedict’s test. (1 mark)

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(iii) Sketch a simple graph to show how the activity of a typical enzyme varies with temperature.

(2 marks)



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##

(i) Diffusion rate increases with temperature (1 mark)

because molecules have more kinetic energy at higher temperatures. (1 mark)

(ii) To allow even distribution of glucose molecules before extraction. (1 mark)

(iii)

(2 marks)

_________

(5 marks)##

|!|EQA00400012|!|

The drawing below shows how a certain biochemical reaction takes place inside the small intestine of a

mammal:

(i) Which molecule is an enzyme? (1 mark)

_______________________________________________________________________________

(ii) Which are the substrate(s) and product(s) of this reaction? (4 marks)

_______________________________________________________________________________

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(iii) What is the name of the part of the enzyme that determines the specificity of the reaction?

(1 mark)

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___________________________________________________________________________________

(iv) Draw a simple sketch to show the relationship between enzyme activity and

(1) temperature.

(2) pH.

(1) (2)

(4 marks)

(v) Give an example of this type of reaction in the small intestine and state the products of the reaction. (2

marks)

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##

(i) molecule B (1 mark)

(ii) substrates: -molecule A and water (2 marks)

products: -molecules D and E (2 marks)

(iii) active site (1 mark)

(iv)

(1)

(2 marks)

(2)

(2 marks)

(v) Digestion of sucrose by sucrase. (1 mark)

The products are glucose and fructose. (1 mark)

OR

Digestion of lipid by lipase.

The products are fatty acids and glycerols.


Enzyme activity

Enzyme activity

Temperature (℃)


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__________

(12 marks)##

|!|EQB00400013|!|

* The graph below shows the results of an experiment demonstrating the breakdown of starch by amylase:

(i) Describe the changes in the activity of the enzyme when temperature increased from 0℃ to 60℃? (3

marks)

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_______________________________________________________________________________

(ii) Why did the rate of reaction drop rapidly at temperatures above 40℃? (1 mark)

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(iii) What is the product of digestion of starch? (1 mark)

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(iv) How would you test for this product? (3 marks)

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(v) State one factor, other than temperature, that affects the activity of an enzyme. (1 mark)

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(vi) State where amylase can be found in the human body. (2 marks)

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##

(i) The enzyme activity increased from 0℃to 40℃. (1 mark)

The optimum temperature of the enzyme was 40℃. (1 mark)

The enzyme activity decreased from 40℃to 60℃. (1 mark)



___________________________________________________________________________________

(ii) High temperatures caused denaturation of enzymes. (1 mark)

(iii) maltose (1 mark)

(iv) Heat it with Benedict’s solution. If an orange or a brick red precipitates is formed, it indicates that the

product is maltose (reducing sugar). (3 marks)

(v) pH (1 mark)

(vi) It can be found in the mouth cavity and also in the small intestine. (2 marks)

__________

(11 marks)##

|!|EQB00400014|!|

* The graph below shows the effect of pH on the activities of three digestive enzymes found in the human body:

(i) What conclusion can be drawn from the graph on the relationship between pH value and enzymatic

activity? (2 marks)

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_______________________________________________________________________________

(ii) Which curves would best represent the actions of salivary amylase and pepsin, respectively?

(2 marks)

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(iii) Which parts of the digestive system secrete salivary amylase and pepsin, respectively? Briefly describe

the functions of these two enzymes. (6 marks)

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##

(i) Each type of enzyme works best at its own optimum pH. (2 marks)

(ii) Action of salivary amylase: -curve B (1 mark)

Action of pepsin: -curve A (1 mark)

(iii) Salivary amylase is secreted by the salivary gland. It can digest starch into maltose. (3 marks)

Pepsin is secreted by the stomach. It can digest protein into peptides. (3 marks)

__________

(10 marks)##

|!|EQB00400015|!|

* In an experiment investigating the action of protease on protein, a capillary tube of solid egg white was placed

in each of the following labelled test tubes:

Test tube A – with 4 cm3 1% protease

Test tube B – with 4 cm3 1% protease and several drops of dilute hydrochloric acid

Test tube C – with 4 cm3 boiled 1% protease and several drops of dilute hydrochloric acid

Test tube D – with 4 cm3 dilute hydrochloric acid

The test tubes were put into a water bath at 37℃. The original lengths of egg white were all 30 mm. After one

hour, the length of egg white in each test tube was measured again.

The length of egg white in each tube after one hour is shown below:



___________________________________________________________________________________

A: 29 mm

B: 25 mm

C: 30 mm

D: 30 mm

(i) What conclusion can be drawn from the results of the experiment? (2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

(ii) Explain the results in tubes A, C and D. (3 marks)

_______________________________________________________________________________

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_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(iii) Explain the following applications of protease in our daily lives.

(1) Biological washing powder contains protease which can wash away the egg stains on clothes.

(3 marks)

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(2) When pineapple juice, which contains protease, is added to a chunk of beef and the beef is left for

30 minutes, it will become tenderer. (3 marks)

_______________________________________________________________________________

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##

(i) Protease can digest protein in the presence of dilute HCl but boiled protease is ineffective in the

digestion of protein. (2 marks)

(ii) Tube A – Protease was the most effective in acidic condition. It alone could not digest protein

effectively. (1 mark)

Tube C – Protease was denatured by boiling. It could not digest protein though dilute HCl was added.

(1 mark)

Tube D – Dilute HCl alone could not digest protein. (1 mark)

(iii) (1) Protease in the biological washing powder digests protein in egg stain into simpler forms which are

soluble in water and can be washed away by water. (3 marks)

(2) Beef is rich in protein. Protease in pineapple juice digests the protein in beef into simpler forms,

making it tenderer. (3 marks)



___________________________________________________________________________________

__________

(11 marks)##

|!|EQA00400016|!|

The following experiment was set up to investigate the action of saliva on starch:

The dialysis tubings contained the following substances:

Tube A – starch and saliva (pH 5)

Tube B – starch and saliva (pH 7.5)

Tube C – starch and saliva (pH 11)

Tube D – starch and distilled water (pH 7)

At regular intervals, the water in each of the test tubes surrounding the dialysis tubing was tested for the

presence of reducing sugars.

(i) Name a reagent which could be used to test for reducing sugars. (1 mark)

_______________________________________________________________________________

(ii) In which dialysis tubing would the breaking down of starch be the most rapid? (1 mark)

_______________________________________________________________________________

(iii) Which tube acted as a control in this experiment? Give one reason why the control was necessary. (2

marks)

_______________________________________________________________________________

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_______________________________________________________________________________

(iv) Without making any changes to the contents of the tubings, how could the maximum rate of starch

digestion be achieved? Explain your answer. (2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(v) Why were the test tubes put in a water bath? (1 mark)

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##


dialysis tubing

distilled waterwater bath at 25℃

A B C D


___________________________________________________________________________________

(i) Benedict’s solution (1 mark)

(ii) tube B (1 mark)

(iii) tube D (1 mark)

A control was set up to avoid arguments that the result was produced because of some experimental

conditions / produced by factors other than the one under study. (1 mark)

(iv) It could be achieved by placing the test tubes in a water bath at about 37℃ because the enzyme in

saliva worked better at this temperature. (2 marks)

(v) To provide a relatively constant temperature. (1 mark)

__________

(7 marks)##

|!|EQB00400017|!|

* State THREE differences between active transport and diffusion. Give ONE example of each of the two

processes. (8 marks)

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##

A ctive transport Diffusion

Energy needed Yes No (2 marks)

Direction of net

movement of

molecules

From low concentration to high

concentration

From high concentration to low

concentration (down the

concentration gradient)

(2 marks)

Presence of

carrier

Yes No (2 marks)

Example Absorption of mineral salts by

the root hairs from the soil

Carbon dioxide gas diffuses

from the capillaries to the

alveolar space inside the lungs

(2 marks)

__________

(8 marks)##

|!|EQB00400018|!|

* STS Connections

Wendy had prepared some fresh pineapples and was ready to use them to make a bowl of pineapple jelly for a

party. However, the jelly solution failed to coagulate and remained in the liquid state even after it had cooled

down.

She then searched for information on the Internet and found this sentence: For the jelly to turn into a semi-

solid, the proteins inside need to form bonding with each other.

(i) Which enzyme did fresh pineapple contain that prevented the jelly solution from solidifying? Explain

how it works. (3 marks)

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(ii) State TWO characteristics of the enzyme involved in (i). (2 marks)

_______________________________________________________________________________

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(iii) Her classmate suggested that she should use canned pineapple instead of fresh pineapple. Since the

pineapple was treated at a high temperature during the canning process and bacteria were killed, the

jelly solution could coagulate successfully.

Comment on this method. (4 marks)

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(iv) Based on (i), suggest ONE application of fresh pineapples on food processing. (1 mark)

_______________________________________________________________________________

##

(i) protease (1 mark)

The proteases from the pineapple broke down the bonds formed between the protein molecules in the

jelly solution. (1 mark)

Without the bonds to hold the protein molecules in a regular pattern, the jelly did not turn from the

liquid state into a semi-solid. (1 mark)

(ii) They are specific in nature. (1 mark)

They are reusable. (or any reasonable answers) (1 mark)

(iii) The method of using canned pineapple instead of fresh pineapple could really make the jelly

coagulate. (1 mark)

However, the reason was incorrect. (1 mark)

The proteases were denatured at the high temperature used in the canning process and they could not

break the bonds formed between the protein molecules. Therefore the jelly was able to solidify. (2

marks)

(iv) We can use fresh pineapples as meat tenderizers. (or any reasonable answers) (1 mark)

__________

(10 marks)##



___________________________________________________________________________________

|!|EQB00400019|!|

* The table below shows the contents in 5 different test tubes:

Test tubes

Test tube contents A B C D E

Whole fat milk (ml) 5 5 5 5 5

Sodium bicarbonate (ml) 1 1 1 1 1

Bile salt (ml) 0 1 0 1 1

Lipase (ml) 0 0 1 1 1 (boiled)

Water (ml) 2 1 1 0 0

After 1 hour, pH paper was used to test the pH of the solutions in the 5 test tubes. It was found that only

solution in test tubes C and D was acidic.

(i) Which substance present in the milk was the substrate in this experiment? (1 mark)

_______________________________________________________________________________

(ii) What was the main purpose of adding sodium bicarbonate solution into each test tube? (1 mark)

_______________________________________________________________________________

(iii) State the final product which made the solutions in test tubes C and D acidic. (1 mark)

_______________________________________________________________________________

(iv) Compare the acidity of the solutions in test tubes C and test tube D. Explain briefly. (3 marks)

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

(v) Explain the results in test tube E at the molecular level. (2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

(vi) Another test tube, F, contained a mixture of the following:

Whole fat milk (ml) 5

Sodium bicarbonate (ml) 1

Bile salt (ml) 1(boiled)

Lipase (ml) 1

Water (ml) 0

The solution was found to be acidic after 1 hour.



___________________________________________________________________________________

(1) Which test tube could act as a control to test tube F? (1 mark)

____________________________________________________________________________

(2) Is bile salt an enzyme? Explain your answer. (2 marks)

____________________________________________________________________________

____________________________________________________________________________

##

(i) fats (1 mark)

(ii) To provide a slightly alkaline condition for the lipase to act on. (1 mark)

(iii) fatty acids (1 mark)

(iv) Solution in test tube D is more acidic than that in test tube C. (1 mark)

The bile salts in D emulsified (broke down the fat into small droplets) the fat and increased the surface

area for the action of lipase. (1 mark)

With a larger surface area, the enzymatic rate was faster in D and more fatty acids were formed. (1

mark)

(v) The shape of the active site of the lipases was altered after boiling. (1 mark)

The substrates could not fit into the active site and the enzyme could not function. (1 mark)

(vi) (1) test tube D (1 mark)

(2) no (1 mark)

It will not denature after boiling. (1 mark)

__________

(11 marks)##

|!|EQB00400020|!|

* The experimental set-up below demonstrates a certain process in the living cells of a carrot:

A large carrot was cut into many small discs of similar size. Then the initial weight of every piece of carrot

was measured and recorded. Five pieces of carrot were separately added to five different test tubes containing

sucrose solution of different concentrations. After four hours, the weights of the carrots were measured again.

The experimental results are shown in the table below:

Test tube A B C D E

Sucrose concentration (M) 0.00 0.10 0.20 0.30 0.40

Initial weight (g) 21.4 23.3 19.6 19.9 22.1

Final weight (g) 21.9 23.5 19.5 19.5 21.6

The percentage change in

weight (%)



___________________________________________________________________________________

(i) Complete the table. (2½ marks)

(ii) Plot the graph of the percentage change in weight against the sucrose concentrations. (4 marks)

(iii) From the graph, estimate the water potential of the carrot cells in terms of the sucrose concentrations.

(1 mark)

_______________________________________________________________________________

(iv) Some cells were then taken from the carrot discs in tubes C and E and put onto a slide for observation

under a microscope.

(1) State TWO observable differences between the cells in tubes C and E. (2 marks)

_____________________________________________________________________________

_____________________________________________________________________________

(2) Name the process that caused the differences. (1 mark)

____________________________________________________________________________

(v) Explain why we had to cut the carrot into many small pieces. (1 mark)

_______________________________________________________________________________

(vi) If animal cells were put into solution A, what would happen to the cells? (2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

(vii) Suggest an improvement when performing this experiment in order to yield more reliable results.

(2 marks)

_______________________________________________________________________________

_______________________________________________________________________________

##

(i)



___________________________________________________________________________________

Test tube A B C D E

Percentage change

in weight (%)+2.34 +0.85 -0.51 -2.01 -2.26

(½mark each) (+ / - should be included) (2½ marks)

(ii)

X- and Y-axes with correct names, units and scales (2 marks)

Correct plotting and joining of the 5 points on the graph (1 mark)

Correct title (1 mark)

(iii) Accept answers within the range of 0.15-0.18 M. (1 mark)

(iv) (1) The plasma membrane was detached from the cell wall in cells from test tube E but remained

attached to the cell wall in cells from test tube C. (1 mark)

The cytoplasm shrank in cells from test tube E but not in those from test tube C. (1 mark)

The vacuole shrank in the cells from test tube E but not in those from test tube C.

(any two)

(2) osmosis (1 mark)

(v) to increase the surface area so that osmosis could proceed more rapidly (1 mark)

(vi) The animal cells would swell (1 mark)

and eventually burst. (1 mark)

(vii) Several carrot discs instead of one can be added into each test tube. (1 mark)

This allows us to obtain a mean value of change in weight. (1 mark)

__________

(15½ marks)##


chapter 04 question

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