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New Senior Secondary Mastering Biology Book 1BPractical workbook answer p.1/32

Suggested answers to Practical Workbook

for SBA

Ch 7 Gas exchange in humans

Practical 7.1 Examination of the mammalian breathing system

Questions (p. 7-2)

1

A Nose B Pharynx

C Epiglottis D Trachea

E Cartilage F Right bronchus

G Right lung H Rib

I Intercostal muscle J Diaphragm

2 Nostrilsnasal cavitypharynxlarynxtracheabronchibronchioles

air sacs(in lungs)

3 The air is moistened, warmed and cleaner.

4 It closes the entrance to the larynx during swallowing, thereby preventing choking.

5 It protects the lungs and the heart.

Practical 7.2 Examination of the pig lungs

Results (p. 7-5)

1 There are three lobes in the left lung and two lobes in the right lung.

2 The trachea is hard. The lung tissue is soft and spongy.

3 The lungs increase in volume. / The lungs expand.

4 The piece of lung tissue floats in water.

Oxford University Press 2009




5

Questions (p. 7-5)

1 It is because they have a very rich supply of blood vessels.

2 The trachea, but not the lung tissue, is supported by cartilages.

3 Air.

4 The lungs tissue floats in water because the air in the air sacs gives the lung tissue a low density.



Practical 7.3 Examination of the mammalian air sacs

Results (p. 7-8)

Questions (p. 7-8)

1 In the air sacs, the oxygen concentration is higher than that in the capillaries. Oxygen in incoming air dissolves in the water film lining the air sacs, and then diffuses across the walls of the air sacs and the capillaries into the blood. In the capillaries, the carbon dioxide concentration is higher than that in the air sacs. Carbon dioxide in blood diffuses across the walls of the capillaries and the air sacs into the air in the air sacs.

2 The large number of air sacs provides a large surface area for gas exchange. The epithelium of the air sac is only one-cell thick. This provides a short distance for rapid diffusion of gases. The moist inner surface allows gases to dissolve in the water film for diffusion across the epithelium. The air sacs are richly supplied with blood. This allows rapid transport of gases to and from the air sacs so that a steep concentration gradient can be maintained for rapid diffusion.

Practical 7.4 Comparison of the composition of inhaled air and

exhaled air

Results (p. 7-11)

Inhaled air Exhaled air

Burning time of candle(s) 14 10

Final colour of hydrogencarbonate indicator Red / orange Yellow





Questions (p. 7-11)

1 The exhaled air contains less oxygen than inhaled air. The candle in exhaled air burns shorter.

2 Some oxygen in the inhaled air diffuses from the air sacs into the capillaries. Therefore, less oxygen is found in the exhaled air.

3 The exhaled air contains more carbon dioxide than inhaled air. The colour of hydrogencarbonate indicator turns yellow.

4 Some carbon dioxide diffuses from the capillaries to the air sacs. Therefore, more carbon dioxide is found in the exhaled air.

5 Lime water. The colour of lime water changes from colourless to milky.

Conclusion (p. 7-12)

The exhaled air contains less oxygen but more carbon dioxide than inhaled air.



Ch 8 Transport in humans

Practical 8.1 Examination of a blood smear

Results (p. 8-2)

1

2

Red blood cell White blood cell Blood platelet

ShapeBiconcavedisc shape

Irregular shape Irregular shape

Nucleus No nucleus Round or lobed No nucleus

Relative size Medium Large Small

Relative number Abundant Rare Occasional

Questions (p. 8-2)

1 a White blood cell is the largest. Blood platelet is the smallest.

b Red blood cell is the most abundant. White blood cell is the least abundant.

2 a It is biconcave disc shape. This provides a large surface area to volume ratio to facilitate the diffusion of gases.



b No. The absence of nucleus allows the accommodation of more haemoglobin.This increases the oxygen-carrying capacity of the red blood cells.

3 The body is infected with pathogens.

The body has abnormal cell growth.

Practical 8.2 Examination of the transverse sections of an

artery and a vein

Results (p. 8-4)

1

2

Artery Vein

Thickness of wall Thicker Thinner

Size of lumen Smaller Larger

Questions (p. 8-5)

1 Arteries have a thicker wall which contains a thick layer of muscles. The muscles contract and relax to regulate the blood flow to body cells. Veins have a larger lumen to reduce resistance to blood flow.

2 There are valves in veins but not in arteries (except in pulmonary artery and aorta).



Practical 8.3 Examination of the capillary flow in a fish tail fin

Results (p. 8-7)

1

Observations

Direction of blood flow One way

Speed of blood flow Slow

Diameter of blood vessels Similar to the diameter of red blood cells

Behaviour of blood cellsThe red blood cells are squeezing their way through the capillaries.

2

Questions (p. 8-8)

1 To provide a large surface area for rapid exchange of materials between the blood and the body cells.

To provide a large total cross-sectional area so that blood flows slowly in the capillaries. This allows a longer period of time for exchange of materials.

2 White blood cells can change their shape, so they can move along the narrower capillaries.



Practical 8.4 Dissection and examination of a pig heart

Results (p. 8-12)

1

2

Water run into What happens

venae cavae Water comes out from the pulmonary arteries.

pulmonary artery Water cannot enter and no water comes out from any vessels.

pulmonary vein Water comes out from the aorta.

aorta Water cannot enter and no water comes out from any vessels.



3A Anterior vena cava

B Right atrium

C Posterior vena cava

D Tricuspid valve

E Right ventricle

F Septum

G Pulmonary artery

H Aorta

I Pulmonary vein

J Semilunar valve

K Left atrium

L Bicuspid

M Heart tendon

N Left ventricle

Questions (p. 8-13)

1 When water is forced through the venae cavae and the pulmonary vein into the heart, it enters the heart and comes out as in the normal circulation. However, when water is forced through the pulmonary artery and the aorta, it cannot enter the heart because it is stopped by the semilunar valves.

2 The wall of the left ventricle is thicker than that of the right ventricle. It is because the left ventricle has to provide a greater force to pump blood to all parts of the body (except the lungs), whereas the right ventricle pumps blood only for a short distance to the lungs.

3 The ventricles have a thicker muscular wall. It is because the ventricles have to provide a greater force to pump blood to the lungs or other parts of the body, whereas the atria only pump blood to the nearby ventricles.

4 1 The ventricles have a thicker muscular wall to pump blood to all parts of the body.

2 Valves are present to prevent backflow of blood.

3 Heart tendons are present to prevent the valves from being turned inside-out when the

ventricles contract.

5 The septum prevents the oxygenated and deoxygenated blood from mixing. This ensures a high oxygen content in the blood in the aorta for the body cells.



Ch 9 Nutrition and gas exchange in plants

Practical 9.1 Investigation of the effects of different minerals on

plant growth

Results (p. 9-3)

1 (Answer varies with seedlings.)

2

Flask Appearance of seedlings

A Healthy growth of seedlings.

B Poor growth. Yellowing of older leaves.

CPoor growth. Depending on the species, leaves may become dull green, yellow or purple.

DPoor growth. Older leaves start to yellow at the edges, and then turn brown. Leaves may curl and dead spots appear.

E Poor growth. Yellowing of older leaves.

Questions (p. 9-4)

1 This prevents algal growth in the solutions. Algae take up the minerals in the solutions and affect the results.

2 It ensures the roots get enough oxygen for respiration. Respiration can provide energy for the root to absorb minerals by active transport.

3 All the major elements (e.g. nitrogen, phosphorus, potassium, magnesium, etc.) and trace elements (e.g. manganese, copper, zinc, etc.) needed by the plants.

4 It acts as a control to show that symptoms appear in the seedlings are due to the deficiency of a particular mineral.



5 Seedlings cannot use atmospheric nitrogen directly. Seedlings obtain nitrogen only in dissolved forms of nitrate or ammonium.



6 It leads to poor growth of the seedlings.

7 Older leaves. When the minerals are deficient, they are transported from the older leaves to the actively growing young leaves.

8 Yes. For each nutrient solution in the experimental set-ups (flasks B to E), only one variable (deficient in one mineral) is changed at a time, other variables are kept constant.

Conclusion (p. 9-5)

Plants need different minerals for growth. The deficiency of nitrogen, phosphorus, potassium or magnesium results in the development of deficiency symptoms.

Practical 9.2 Examination of the structure of roots

Results (p. 9-7)



Questions (p. 9-8)

1 Root hair. It provides a large surface area for absorption of water and minerals.

2 The epidermis consists of one layer of thin-walled cells only. It is not covered by cuticle. Therefore, water and minerals can easily pass into them.

Numerous root branches and root hairs provide a large surface area for absorption of water and minerals.

The root hairs are long and fine. They can easily grow between the soil particles to absorb water and minerals around them.

Practical 9.3 Design an investigation of the distribution of

stomata on both sides of a leaf

Design and perform an experiment (p. 9-10)

1 The upper side of the leaf has a lower stomatal density.

2 (Answer varies with Ss.)



A Identifying variables

Independent variable

(What will you change?)

Dependent variable(What will you

measure?)

Controlled variables

(What will you keep constant?)

Control(What is the control in this experiment?)

(Answer varies with the design.)




B Designing the set-up

1 (Answer varies with the design.)


C Collecting data



D Risk assessment and safety precautions



Write an experimental report (p. 9-12)

Objective

To compare the distribution of stomata on both sides of a leaf.

Apparatus and materials

Method 1:

1 pair of forceps1 stop-watch1 potted plantdry cobalt(II) chloride papersticky tape

Method 3:

1 vaseline1 electronic balance2 freshly-picked leaves of the same

Method 2:

1 pair of forceps1 beaker of hot water1 freshly-picked leaf from a terrestrial plant

Method 4:

1 microscope2 microscope slides2 cover slips1 pair of forceps1 microscope slide with a transparent grid



weight from a terrestrial plant 1 freshly-picked leaf (e.g. Zebrina)distilled water

Procedure

Method 1:

1 Use sticky tape to stick a piece of dry cobalt(II) chloride paper to the upper side and underside of the leaf respectively.

2 Measure the time taken for the dry cobalt(II) chloride paper to reach the same colour as a piece of moist cobalt(II) chloride paper used as a control.

Method 2:

1 Immerse a leaf from a terrestrial plant quickly into a beaker of hot water.

2 Observe carefully and compare the amount of bubbles coming out from each side of the leaf.

Method 3:

1 Smear vaseline to cover the upper side of one of the leaf. Smear vaseline to cover the underside of the other leaf.

Weigh the leaves after one hour.

2 Compare the weight of the leaves.

Method 4:

1 Use a pair of forceps to peel off the lower epidermis of a leaf. Put it on a slide.

2 Mount the epidermis with a drop of distilled water.

3 Find a portion of the epidermis which fills the microscope’s field of vision at ×100 magnification.

4 Count the number of stomata in the field of vision.

5 Repeat step 4 for 3 times and take the average value.

6 Repeat steps 1 to 5 for the upper epidermis.

7 Use a slide with a transparent grid of a stated grid size to estimate the dimension of the field of vision.

8 Calculate the stomatal densities of the upper and lower epidermis of the leaf stomatal density

=

Results

Method 1:

The dry cobalt(II) chloride paper on the underside of the leaf changes to pink faster than the one on the upper side.

Method 2:

More bubbles come out from the underside of the leaf.

Method 3:

The decrease in weight of the leaf with the upper side smeared with vaseline is larger than the one with the underside smeared with vaseline.



Method 4:

Area of the microscope’s field of vision at ×100magnification (mm2)

Number ofstomata

Stomatal density (number of stomata/mm2)

Upper epidermis

Lower epidermis

Analysis and discussion

1 The upper side of the leaf has fewer stomata. The upper side is directly illuminated by sunlight. Fewer stomata on the upper side can reduce water loss by evaporation.

2 Stomatal densities of both sides of a monocotyledonous leaf are about the same because both sides are exposed to more or less the same amount of sunlight.

3 Stomata are absent.

4 Stomata are present only on the upper side which is in contact with the air.


Conclusion

The upper side of a dicotyledonous leaf has fewer stomata than the underside.

Practical 9.4 Investigation of the effect of light intensity on gas

exchange in plants using hydrogencarbonate

indicator

Results (p. 9-16)

Tube Light conditionColour of hydrogencarbonate indicator

Before experiment After experiment

A Bright light Red / orange Deep purple

B Moderate light Red / orange Light purple

C Dim light Red / orange Red / orange

D Dark Red / orange Yellow

Questions (p. 9-17)

1 It is used as a control. It shows that any colour changes in the indicator are due to light.



2 a The colour changes in the indicator of tubes A and B show that carbon dioxide is absorbed by leaves in bright or moderate light. Leaves in light carry out both photosynthesis and respiration. Since photosynthesis proceeds at a much faster rate than respiration, the amount of carbon dioxide absorbed by photosynthesis is much more than that released by respiration under the two conditions.

b The colour change in the indicator of tube D shows that carbon dioxide is released by leaves in the dark. Leaves in the dark carry out respiration only.

3 Put all the tubes in a water trough.


In bright or moderate light, there is a net absorption of carbon dioxide by the plant.In the dark, there is a net release of carbon dioxide by the plant.

Practical 9.5 Investigation of the effect of light intensity on gas

exchange in plants using a data logger

Results (p. 9-21)

Light conditionMinimum

pressure (kPa)Maximum

pressure (kPa)Change in

pressure (kPa)

Bright light 0.39 2.81 +2.42

Moderate light 0.39 2.25 +1.86

Dim light 0.35 1.89 +1.54

Dark 0.04 0.44 – 0.4

Questions (p. 9-21)

1 The water prevents the plant from heating up by the lamp.

2 To allow the rate of photosynthesis to become steady.

3 a The larger increases in pressure show that more oxygen is released by leaves in bright or moderate light. Leaves in light carry out both photosynthesis and respiration. Since photosynthesis proceeds at a faster rate than respiration, the amount of oxygen released by photosynthesis is more than that absorbed by respiration under the two conditions.



b The decrease in pressure shows that oxygen is absorbed by leaves in the dark. Leaves

in the dark carry out respiration only.

4 Change the distance between the bench lamp and Hydrilla.

Wrap the boiling tube with different numbers of layers of fine muslin or aluminium foil.

5 The volume of oxygen released per unit time. Oxygen is produced as a by-product in photosynthesis. The volume of oxygen released per unit time can act as an indicator of the rate of photosynthesis.


The higher of light intensity, the higher amount of oxygen is released by the plant.



Ch 10 Transpiration, transport and support in plants

Practical 10.1 Demonstration of the occurrence of transpiration

Results (p. 10-2)

Plant Changes in the bell jarCobalt(II) chloride paper

Original colour Final colour

AA layer of moisture and drops of liquid are formed on the wall.

Blue Pink

B The bell jar remains clear. Not applicable Not applicable

Questions (p. 10-2)

1 To prevent the respiration of soil organisms and the evaporation of soil water from affecting the results.

2 Water. (Water turns dry cobalt(II) chloride paper from blue to pink.)

3 Set-up A. The liquid is water.

4 No. Little or no transpiration will take place in plant A because the stomata are only slightly open or even closed in the dark.


Water vapour is released from plant A but not from plant B. Transpiration takes place in the aerial parts of the plant.



Practical 10.2 Measurement of the rate of transpiration using a

bubble potometer

Results (p. 10-4)

Reading 1 Reading 2 Reading 3

Time period (min)

Distance travelled bythe air bubble (cm)

Rate of movement of theair bubble (cm / min)

Questions (p. 10-4)

1 To prevent air bubbles from entering the xylem vessels of the plant and blocking water uptake.

2 When the plant transpires and absorbs water, water is drawn from the capillary tube.

The air bubble therefore moves towards the shoot along the tube.

3 No. The rate of movement of the air bubble indicates the rate of water uptake.

4 The water absorbed is to replace an equal amount of water lost by transpiration.

5 The movement of the air bubble may be affected by the friction between the air bubble and the wall of the capillary tube.

Practical 10.3 Measurement of the amount of water absorbed

and lost by a plant using a weight potometer

Results (p. 10-7)

Water level in the burette (cm3)

Weight of the entire set-up (g)

At start (Results vary with Ss.)

After the practical


(Results vary with Ss.)




Questions (p. 10-7)

1 To prevent the evaporation of water in the burette which will affect the results.

2 (Answer varies with Ss. The amount of water absorbed is the difference between the water levels in the burette before and after the practical.)

3 (Answer varies with Ss. The amount of water lost is the difference between the weights of the whole set-up before and after the practical.)

4 The amount of water lost is slightly less than the amount of water absorbed. It is because some of the absorbed water is used in photosynthesis and other metabolic activities.

5 No. As some water remains in the plant, the rate of water uptake is slightly higher than the rate of transpiration.

6 Error: Water may be present on the leafy shoot when the plant is removed from water.

Improvement: Blot the plant with tissue paper before the experiment.

7 A weight potometer is easier to handle and has a higher accuracy in measurement. It is because the weight potometer can measure the rate of transpiration directly, but the bubble potometer can only measure the rate of water uptake of plants.


The amount of water lost is slightly less than the amount of water absorbed by the plant.

Practical 10.4 Design an investigation of the effects of

environmental factors on the rate of transpiration

Propose a hypothesis (p. 10-10)

Higher light intensity / higher temperature / lower relative humidity / higher wind speed increases the rate of transpiration.

Design and perform an experiment (p. 10-10)

1 Light intensity, temperature, humidity or air movement, etc.





2 Light intensity: use a bench lamp. / Temperature: use a heater. /

Relative humidity: use a dehumidifier. /

Air movement: use a blowing fan.


A Identifying variables

Independent variable

(What will you change?)

Dependent variable(What will you

measure?)

Controlled variables

(What will you keep constant?)

Control(What is the control in this experiment?)

The environmental factor being investigated.

The weight of water lost in a weight potometer, the distance travelled by the air bubble in a given time in a bubble potometer, etc.

The parameters and conditions other than the one being investigated.

The potometer that is put in normal conditions.

B Designing the set-up

(Answer varies with Ss.)

C Collecting data


2 Use a shoot with more leaves. / Use a capillary tube with a narrower bore in the bubble potometer.

3 Allow a few minutes for the shoot to equilibrate before taking any readings or ignore the first few readings. / Take the average of several readings under the same condition.

D Risk assessment and safety precautions

1 The scalpel used to cut the plant is very sharp and may cut our fingers.

2 Handle the scalpel with care.



Write an experimental report (p. 10-12)

Objective

To investigate the effect of light intensity / temperature / relative humidity / air movement on the rate of transpiration.

Hypothesis

Higher light intensity / higher temperature / lower relative humidity / higher wind speed increases the rate of transpiration.

Apparatus and materials

2 pipettes (1 cm3)2 glass tubings2 rubber tubings2 retort stands

4 clamps1 wash bottle with water1 scalpel1 plant with leafy shoots

Procedure

1 Set up the apparatus as shown on the right.

2 Put the U-shaped potometer in one of the following places, depending on the environmental factor being investigated:

Light intensity – near a bench lamp

Temperature – near a heater

Relative humidity – near a dehumidifier

Air movement – near a blowing fan

3 Put the set-up in a laboratory with normal conditions to act as the control.

4 Allow 5 minutes for equilibration.

5 Adjust the water levels in the glass tubing and the pipette to the same level by raising or lowering

the2 arms of the U-shaped potometer.

6 Record the initial water level in the pipette.

7 Record the water level again after a certain time (e.g. 15 minutes).

8 Readjust the water levels and repeat with 2 more readings.

Results

ConditionAmount of water absorbed

in 15 minutes (cm3)Rate of water uptake

(cm3 / min)

Under an environmental



condition being investigated

Under normal conditions



Analysis and discussion

1 To allow the rate of transpiration to become steady.

2 a Higher light intensity causes the stomata open wider. More water vapour in the air space diffuses out through the stomata. Hence, the rate of transpiration increases.

b Higher temperature increases the rate of evaporation from the water film on the cell surfaces and the diffusion rate of water vapour out of the stomata. It also

lowers the relative humidity of the air. Hence, the rate of transpiration increases.

c Lower relative humidity in the surrounding air increases the concentration gradient ofwater vapour between the air spaces in the leaves and the atmosphere. Hence, water vapour diffuses out of the leaves more rapidly and the rate of transpiration increases.

d Wind blows away the water vapour and prevents the decrease in the concentration gradient of water vapour between the air space in the leaves and the surrounding air. The rate of diffusion and therefore the rate of transpiration increases in windy conditions.

3 Error: Changing of one environmental condition may have changed another, e.g. the use of the bench lamp to increase the light intensity may also have increased the temperature of the surrounding air.

Improvement: When investigating the effect of light intensity, put a beaker of water in front of the plant to prevent the plant from heating up by the lamp.

Conclusion

The rate of transpiration increases at higher light intensity / higher temperature / lower relative humidity / in windy conditions.

Practical 10.5 Examination of the vascular tissues of a young

dicotyledonous plant

Results (p. 10-16)



Questions (p. 10-17)

1 Xylem transports water and minerals in the plant.

Phloem transports organic nutrients in the plant.

2 The vascular tissues in the stem, the root and the leaf are found on the periphery, at the centre and in the midrib vein respectively.



Practical 10.6 Investigation of the plant tissue responsible for

water transport

Results (p. 10-19)

Questions (p. 10-20)

1 Yes. The xylem.

2 Put the plant near a blowing fan. / Put the plant in bright light or near a bench lamp. / Put the plant near a heater. / Put the plant near a dehumidifier. / Blot dry the leaves. (Any two)




The xylem is the main tissue responsible for water transport in the herbaceous plant.


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