lq - 03 cell activity and organization

Certificate Biology New Mastering Basic Concepts Chapter 3Question Bank V2.0 Structured Questions

Ch 3 Cell activities and organization

[031001]

The following experimental set-up is used to investigate osmosis.

a Describe and explain the change in the liquid level in the capillary tube

after an hour. (4 marks)

b Does diffusion of sucrose molecules occur? Why? (2 marks)

c Which part of the cell is a differentially permeable membrane? (1 mark)

d If the distilled water in the set-up is replaced by dilute sucrose solution,

how will the results differ? (3 marks)

-- ans --

a The distilled water has a higher water potential than the sucrose solution. 1m

So water moleucules pass through the differentially permeable membrane from the

distilled water to the concentrated sucrose solution by osmosis. 2m

The liquid level rises. 1m

b No. Sucrose molecules are not small enough to pass through the holes on the

differentially permeable membrane. 2m

c Cell membrane. 1m

d The difference in water potentials of the two solutions is smaller. 1m

Oxford University Press 2005 / 37


The rate of osmosis will be slower, 1m

and so the change in water level will be smaller after the same period of time. 1m

-- ans end --



[031002]

To investigate the effect of pH on amylase activity, six shallow troughs were

made on a starch-agar plate and different mixtures of amylase and chemicals

were added to each trough as follows:

(1) fresh enzyme extract

(2) fresh enzyme extract + dilute HCl solution

(3) fresh enzyme extract + dilute NaOH solution

(4) boiled enzyme extract

(5) dilute HCl solution

(6) dilute NaOH solution

The plate was incubated at 35 ℃ for two hours. The plate was then treated with

iodine solution and washed. The appearance of the plate is shown below.

a Explain the results. (8 marks)

b What do the results of troughs 5 and 6 tell? (1 mark)

c Name the process by which iodine molecules get to the deeper region of

the starch agar. (1 mark)



-- ans --

a The fresh enzyme extract (1) contained enzyme which broke down some starch

around the trough. 1m

The clear area around the trough indicates that there is no starch left after

incubation. 1m

The enzyme in the mixture with dilute HCl (2) did not break down starch and the

starch was stained dark blue by iodine. 1m

The enzyme was inactivated by low pH. 1m

The enzyme in the mixture with dilute NaOH (3) digested much starch around the

trough. 1m

The enzyme had higher activity at high pH. 1m

The boiled enzyme extract remains dark blue because the enzyme is denatured at

high temperature, so cannot digest the starch 1m

The dilute HCl alone did not break down starch. 0.5m

The dilute NaOH did not break down starch. 0.5m

b They act as controls and are used to check whether HCl or NaOH has direct effect

on starch. 1m

c Diffusion 1m

-- ans end --



[031003]

The following diagram illustrates the ‘lock and key’ hypothesis for enzyme

actions.

a Which structures in the diagram represent substrate, product and enzyme

respectively? (3 marks)

b Is this a catabolic or anabolic reaction? (1 mark)

c How does the hypothesis explain the specificity of enzyme actions? (2

marks)

d State and explain how temperature affects enzyme actions. (4 marks)

-- ans --

a P—product 1m

Q—enzyme 1m

R, S—substrate 1m

b Anabolic reaction 1m

c The substrate molecules combine with the active sites on the enzyme molecules. 1m

The enzyme must have the right shape for the substrates to combine with the active

site. 1m

d Low temperature lowers enzyme activities because kinetic energy of enzyme and

substrate molecules decrease with temperature and collision of molecules is less

frequent. 1m

High temperature increases enzyme activities because high temperature increases



the chance of collision between substrate and enzyme molecules. 1m

When temperature is too high, the shape of enzyme molecules is changed. 1m

The active site no longer fits the substrate molecule and there is no reaction. 1m

-- ans end --



[031004]

David carried out an experiment to study the effect of temperature on enzyme

activity. He made 7 tubes of egg white of 6 cm long and put each one into a

petri dish containing water, digestive enzyme and hydrochloric acid. The petri

dishes were incubated at 0℃, 10℃, 20℃, 30℃, 40℃, 50℃ and 60℃

respectively for 4 hours. The table below shows the length of the egg white

tubes remaining after incubation.

Temperature (℃) Length of egg white tube (cm)

0 6

10 5

20 4

30 3

40 2

50 4

60 6

a Using the above data, draw a graph to show the relationship between

reduction in length of egg white tube and temperature. (3 marks)

b Describe and explain the graph drawn. (6 marks)

c What is the use of hydrochloric acid in the experiment? (1 mark)



-- ans --

a Correct labelling of x-axis and y-axis 1m

Correct shape of curve 2m

b The curve is bell-shaped. 1m

From 0 to 40℃, the reduction in length of egg white tube increases with temperature.

1m

At 40℃, the reduction in length of egg white tube reaches its maximum. 1m

From 40℃ to 60℃, the reduction in length of egg white tube decreases with

temperature. 1m

As temperature increases from 0 to 40℃, the digestive enzyme becomes more and

more active. 1m

At 40℃, the digestive enzyme begins to denature. 1m

c To provide an optimum pH for the digestive enzyme to work. 1m

-- ans end --



[031005]

The diagrams below show two experimental set-ups to investigate osmosis.

a Define the process of osmosis. (3 marks)

b One hour later, what are the changes in the sucrose solution levels inside

the capillary tubes? Explain the results. (4 marks)

c The dialysis tubing has one property that is very important for osmosis to

take place. What is that property and how can it make osmosis possible?

(3 marks)

-- ans --

a The movement of solvent (water) molecules 1m

from a region of higher water potential to a region of lower water potential 1m

through a selectively permeable membrane. 1m

b Sucrose solution level in set-up A rises. 1m

This is because water molecules moves from fresh water to 10% sucrose solution



through the dialysis tubing. 1m

Sucrose solution level in set-up B falls. 1m

This is because water molecules moves from 10% sucrose solution to 20% sucrose

solution through the dialysis tubing. 1m

c The dialysis tubing is selectively permeable. 1m

It allows free movement of small water molecules 1m

but not large sucrose molecules. 1m

-- ans end --



[031006]

A raw potato was cut into the shape of a cup. Concentrated sucrose solution was

put inside the well and the potato cup was immersed in a petri dish containing

distilled water.

a How did the sucrose solution level change after an hour? Explain your

answer. (4 marks)

b How will the sucrose solution level change if the potato is boiled? Explain

your answer. (4 marks)

c In another experiment, a potato cup containing distilled water was put into

a petri dish of sucrose solution. What was happened to the water level of

the well after an hour? Why? (2 marks)

-- ans --

a The sucrose solution level rose 1m

because water molecules moved from distilled water (higher water potential) 1m

to concentrated sucrose solution (lower water potential) 1m

through the living potato tissue which acted as a selectively permeable membrane.

1m

b The sucrose solution level will fall 1m

because the boiled potato is dead and loses its selective permeability. 1m

Both the water and sucrose molecules will leave the well 1m



through diffusion. 1m

c The water level fell 1m

because water molecules moved from distilled water to sucrose solution. 1m

-- ans end --



[031007]

a Suppose a differentially permeable tubing is filled with a 20% sugar

solution. What will happen to the volume of the solution if the tubing is

placed into the following solutions? Give a reason in each case.

(1) Distilled water (3 marks)

(2) 20% sugar solution (3 marks)

(3) 40% sugar solution (3 marks)

b Deducing from the answer in a, what precaution should be taken in

injecting glucose solution to a patient? (1 mark)

-- ans --

a (1) The volume will increase. 1m

Water will enter the tubing by osmosis 1m

because the distilled water has a higher water potential than the solution

inside the tubing. 1m

(2) The volume will remain the same. 1m

There is no net water movement across the tubing 1m

because the water potential of the sugar solution inside and outside the tubing

is the same. 1m

(3) The volume will decrease. 1m

Water will leave the tubing by osmosis 1m

because the 40% sugar solution has a lower water potential than the solution

inside the tubing. 1m

b The water potential / concentration of the glucose solution for injection must be the

same as that of blood. 1m

-- ans end --



[031008]

a Complete the table below to show the applications of enzymes in daily

life.

(8 marks)

Enzyme Application Explanation

Amylases Making syrup and fruit juice

Washing clothes

Softening meat

b Why is the biological activity of enzyme-containing washing powder

reduced in hot water? Give an explanation with reference to the active

sites. (2 marks)

-- ans --

a

Enzyme Application Explanation

Amylases Making syrup and fruit juice The amylases digest starch

into sweet sugars.

2m

Proteases

1m

Washing clothes The proteases digest

protein stains such as

blood stains.

2m



Proteases

1m

Softening meat The proteases digest

certain proteins in the

meat.

2m

b The enzymes in the washing powder are denatured at high temperature. 1m

The shapes of active sites are changed and the substrates cannot fit in. 1m

-- ans end --



[031009]

The diagram below shows the action of an enzyme on its substrate.

a Identify the substrate(s), enzyme(s) and product(s) in the diagram. Give a

reason in each case. (6 marks)

b Enzymes are specific in their action. What does it mean? (1 mark)

c Why enzymes are specific in their action? (2 marks)

d Name a hypothesis that can explain enzyme specificity. (1 mark)

-- ans --

a X is the substrate. 1m

X is changed after the reaction. 1m

Q is the enzyme. 1m

Q remains unchanged after the reaction. 1m

Y and Z are the products. 1m

Y and Z are formed after the reaction. 1m

b Each enzyme acts on only one or a few chemically related substrates and catalyzes

only one kind of reaction. 1m

c The active sites of enzymes are of specific shapes. 1m

Each active site can only combine with one or a few chemically-related substrates

that fit into it. 1m

d The 'lock and key' hypothesis 1m



-- ans end --



[031010]

The table below shows the activity of an enzyme (a protease found in the

stomach) at different pH.

pH Activity of protease (% of maximum)

1.0 50

1.5 80

2.0 98

2.5 80

3.0 25

3.5 8

4.0 5

4.5 2

5.0 0

6.0 0

7.0 0

a Plot the data on a graph. (5 marks)

b What is the optimum pH for this enzyme? (1 mark)

c People suffering from "acid stomach" may find antacid tablets useful in

relieving the discomfort. Antacid tablets work by decreasing the acidity in

stomach. How will the digestion of proteins be affected by the antacid

tablets taken? (2 marks)

d Besides pH, give two factors that affect enzyme activity. (2 marks)



-- ans --

a

Correct title 1m

Correct labelling of axes 0.5m, 0.5m

Correct scale of axes 0.5m, 0.5m

Correct curve 2m

b pH 2 1m

c It will slow down the digestion of proteins 1m

because the pH in stomach exceeds the optimum pH value of 2. 1m

d Temperature / substrate concentration / enzyme concentration (any 2) 1m, 1m

-- ans end --



[031011]

The breakdown of hydrogen peroxide into water and oxygen can be catalyzed

by an enzyme found in various tissues. In an experiment, samples of liver

tissues or blood, either fresh or boiled, are added to test tubes containing

hydrogen peroxide.

Test Tube Content Gas evolved

A Hydrogen peroxide No oxygen evolved

B Hydrogen peroxide + fresh manganese dioxide Oxygen evolved

C Hydrogen peroxide + boiled manganese dioxide Oxygen evolved

D Hydrogen peroxide + fresh liver tissue Oxygen evolved

E Hydrogen peroxide + boiled liver tissue No oxygen evolved

F Hydrogen peroxide + fresh blood Oxygen evolved

G Hydrogen peroxide + boiled blood No oxygen evolved

a How to test for oxygen on the gas evolved? (2 marks)

b What is the purpose of setting up test tube A? (1 mark)

c Explain why the results are the same for test tubes B and C. (2 marks)

d Explain the results of test tubes D and F. (1 mark)

e Explain the results of test tubes E and G. (2 marks)

f State two parameters that should be kept constant in all the test tubes.

(2 marks)

--ans --

a Bring a glowing splint near the gas evolved. 1m

If the splint lights up again, the gas evolved is oxygen. 1m

b It is a control to show that no oxygen is released from hydrogen peroxide alone. 1m



c Manganese dioxide is a chemical catalyst but not an enzyme. 1m

Its catalytic property is not affected by boiling. 1m

d Fresh liver tissue and blood contain the enzyme that catalyzes the breakdown of

hydrogen peroxide into water and oxygen. 1m

e Biological enzymes are proteins in nature. 1m

They are denatured after boiling and their catalytic property is lost. 1m

f Temperature of the reaction materials / pH of the reaction materials / volume of the

hydrogen peroxide (any 2) 1m, 1m

-- ans end --



[031012]

a For the following statements, state whether they are correct or incorrect.

Give reasons if you think they are incorrect.

i Enzymes are killed at high temperatures.

ii The optimum temperature for all enzyme is 37 oC because this is the

normal body temperature of humans.

iii Enzymes are denatured at very low temperatures.

iv The reaction rate of enzymes decreases with time as enzyme

molecules are used up during the reaction. (6 marks)

b Explain why enzyme actions are specific. (2 marks)

c How do you think human life would change if all our enzymes worked at

half their normal rates? (2 marks)

-- ans --

a i Incorrect. At high temperatures, enzymes are denatured but not killed. 1m

ii Incorrect. Enzymes in other organisms may have different optimum

temperatures. 1m

iii Incorrect. Enzymes are only inactive at very low temperatures. 1m

Their shapes are not changed at low temperatures. 1m

iv Incorrect. Enzyme molecules are not used up during a reaction. 1m

The reaction rate decreases because fewer substrate molecules are available

for reaction as time goes by. 1m

b During an enzymic reaction, the substrate molecule combines with the enzyme

molecule at the active site. 1m

The active site of a particular enzyme is of a specific shape and so can only combine

with a substrate of the right shape. 1m



c All metabolic reactions would slow down. 1m

And so all our daily activities, like movement, growth and thinking would slow down

as well. We would not be as active as we are now. 1m

-- ans end --



[031013]

An experiment was set-up as shown in the following diagram to study diffusion.

Both water molecules and iodine molecules can pass through the dialysis

tubing. The set-up was left to stand at room temperature for one hour and

observation on any changes was made.

a Would there be any changes in the distribution of iodine molecules?

Explain. (2 marks)

b What would be the movements of water molecules? How could you

observe the change? (2 marks)

c If the whole set-up was put in a water bath at 40 oC, do you think there would be any differences in the experimental results? Explain. (2 marks)

d State and explain whether net movement of molecules would occur in the

following cases:

i Iodine solution of a lower concentration than the one outside,

instead of pure water, was put inside the dialysis tubing.

ii Iodine solution of the same concentration as the one outside, instead

of pure water, was put inside the dialysis tubing. (4 marks)

-- ans --

a Some iodine molecules would be found inside the dialysis tubing. 1m

Iodine molecules moved from a region of higher concentration (outside the tubing) to

a region of lower concentration (inside the tubing). 1m

b Water molecules would move from the inside of the dialysis tubing to the outside



down the concentration gradient. 1m

The volume of the dialysis tubing would decrease. 1m

c The changes would become more rapid. 1m

Higher temperature results in faster movements of all molecules. 1m

d i Net movement of iodine into the dialysis tubing and net movement of water out

of it would occur 1m

because there was a concentration difference of iodine and water molecules

across the differentially permeable tubing. 1m

ii There is no net movement of molecules across the membrane because there

was no concentration difference of water molecules and iodine molecules. 2m

-- ans end --



[031014]

In a hen's egg, there is a thin and differentially permeable membrane inside the

shell. It can be used to study osmosis. Two hen's eggs were immersed in dilute

hydrochloric acid until the shells had dissolved. The eggs were washed. One

was put into distilled water and the other into a concentrated salt solution.

Changes were observed after a day.

a Explain the term 'differentially permeable membrane'. (1 mark)

b What changes occurred to the eggs in distilled water and in concentrated

salt solution? Explain. (6 marks)

c State two factors that affect the rate of osmosis. (2 marks)

d In natural circumstances, hen's eggs are not immersed in water. What

substances do you think normally diffuse through the membrane around

the hen's egg? (1 mark)

-- ans --

a A membrane with holes that only allow small molecules to pass through. 1m

b For the egg in distilled water, there was a net movement of water molecules from the

outside into the egg 1m

as the concentration of solution was high inside the egg. 1m

Therefore the volume of the egg increased. 1m

For the egg in concentrated salt solution, there was a net movement of water

molecules from the egg to the salt solution 1m

as the concentration of salt solution was higher than the solution inside the egg. 1m

The volume of the egg thus decreased. 1m

c The degree of the concentration gradient across the differentially permeable

membrane 1m

The temperature of the environment. 1m

d Gases 1m



-- ans end --



[031015]

A plant is divided into three parts A, B and C. The following diagram on the

right shows the cross section of part A.

a What are the names of parts A, B and C? (3 marks)

b With reference to the structural diagram, give an example of a cell, a

tissue and an organ. (3 marks)

c Give one advantage and three disadvantages for an organism being multi-

cellular. (4 marks)

-- ans --

a A: leaf

1m

B: stem 1m

C: root 1m

b Cell: palisade cell 1m

Tissue: palisade layer 1m

Organ: leaf 1m

c Advantage:

Enable division of labour. 1m



Disadvantages:

Problem of support. 1m

Problem of transportation of materials inside the body. 1m

Problem of coordination of different parts of the body. 1m

-- ans end --



[031101]

In order to show the movement of water in and out of a human cheek cell, a

model is set up using the following apparatus:

dialysis tubing 1 piece

thread 2 pieces

20% glucose solution 200 ml

distilled water 100 ml

beaker 1

a Draw and label the model in the space below. (3 marks)

b What will you observe in a after one hour? Give an explanation to your

observation. (3 marks)

c Which part of a human cheek cell does the dialysis tubing represent? (1

mark)



-- ans --

a Correct drawing 0.5m

Correct labelling 0.5m x 5

b The tubing will shrink. 1m

The water potential of the distilled water inside the dialysis tubing is higher than

that of the glucose solution outside the tubing. 1m

Therefore, there is a net movement of water molecules from distilled water to

glucose solution. The volume of the distilled water inside the tubing decreases.

1m

c Cell membrane 1m

-- ans end --



[031102]

Catalase is an enzyme present in most living tissues. To show that catalase is

present in a tissue, hydrogen peroxide can be used. In the presence of catalase

and hydrogen peroxide, bubbles of oxygen are given off. An experiment as

shown in the diagram below was carried out to study the activity of catalase.

Discs of tissues were prepared from potato, celery, turnip and carrot. Two drops

of detergent and 2 ml of dilute hydrogen peroxide were added to each of the

four test tubes. Then a disc of living tissue of each type was added. Another set

of test tubes using the boiled tissues was included as the controls. The maximum

height of the froth produced in each test tube was recorded in the table below.

TissueMaximum height of froth (cm)

Experimental set-up Control

Potato 5 0

Celery 3 0

Turnip 3 0

Carrot 1 0

a How do you show that the gas released is oxygen? (2 marks)



b Write a word equation to represent the action of catalase on hydrogen

peroxide. (1 mark)

c From the results, which tissue has the highest activity of catalase? (1 mark)

d Suggest two possible changes in the procedure so as to distinguish the

results between celery and turnip. (2 marks)

e What is the significance of using boiled tissue in the control? (2 marks)

f A student said that measuring the height of froth was not accurate enough

to indicate the activity of catalase. Briefly suggest one modification of the

set-up to improve the accuracy. (2 marks)

-- ans --

a Collect the gas released and put a glowing sprint in it. 1m

Oxygen can relight a glowing splint. 1m

b catalase

hydrogen peroxide -----------------------------> oxygen + water 1m

c Living potato tissue 1m

d Add more discs or thicker discs to the test tubes.

Add more detergent to the test tubes.

Increase the amount or concentration of hydrogen peroxide.

(any 2) 1m x 2

e Boiling causes denaturation of catalase so that no froth should be produced. 1m

The controls enable us to compare the results of the experimental set-ups and

show that only enzyme in living tissues can break down hydrogen peroxide. 1m

f Collect the gas released 1m

and measure its volume under atmospheric pressure. 1m

-- ans end --



[031103]

The table below shows the effect of temperature on the mass of a potato cube

placed in distilled water.

Temperature of

water surrounding

the potato cube

Percentage change in mass (%)

Time (hours)

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

20 oC 0 +5 +7 X +12 +14 +16 +17 +17

50 oC 0 +7 0 Y -8 -10 -10 -10 -10

a Given that the initial mass of both potato cubes is 50 g. If the potato cubes

at 20 oC and 50 oC weigh 54.5 g and 47.5 g respectively after 1.5 hours,

calculate the values of X and Y. (2 marks)

b From the results of the first 0.5 hour, what appears to be the effect of

temperature on the mass of the potato cubes? (1 mark)

c i Describe the changes in the mass of the potato cube at 20 oC during

the 4-hour period. (3 marks)

ii Why was there no further increase in mass after 3.5 hours? (2 marks)

d What happened to the potato cube at 50 oC after 0.5 hour? Explain your

answer. (3 marks)

-- ans --

a At 20 oC: [(54.5 - 50) / 50] x 100 % = +9 % 1m

At 50 oC: [(47.5 - 50) / 50] x 100 % = -5 % 1m

b Increasing the temperature increased the mass of the potato cubes. 1m

c i There was a rapid increase in the mass of the potato cube during the first

0.5 hour. 1m

Then there was a gradual increase from 0.5 to 3.5 hours. 1m

The mass remained unchanged after 3.5 hours. 1m



ii The potato cells were fully turgid 1m

and hence, there is no net movement of water molecules into the cells by

osmosis. 1m

d The mass of the potato cube dropped rapidly and remained unchanged after 2.5

hours. 1m

The cell membranes of the potato cells were destroyed at 50 oC 1m

so that their differential permeability could not be retained. 1m

-- ans end --



[031104]

A student did the following demonstration to simulate an enzyme-controlled

reaction.

a Was the reaction a catabolic or anabolic one? (1 mark)

b What did the pin represent? How do you know? (2 marks)

c The student then used the same pin to burst another balloon. Which

property of enzyme was demonstrated? (1 mark)

d State two ways in which the demonstration cannot represent an enzyme-

controlled reaction accurately. (2 marks)

-- ans --

a A catabolic reaction 1m

b Enzyme 1m

It is not consumed at the end of the reaction. 1m

c Enzyme is reusable. 1m

d The enzyme-controlled reaction is reversible.

The enzyme-controlled reaction can also be an anabolic one.



Rate of the enzyme-controlled reaction depends on temperature.

Rate of the enzyme-controlled reaction depends on pH.

(any 2) 1m x 2

-- ans end --


lq - 03 cell activity and organization

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