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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
Nostrilsnasal cavitypharynxlarynx
tracheabronchibronchioles
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.
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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.
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Practical 7.3
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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)
Burning time of candle(s)
Final colour of hydrogencarbonate indicator
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Inhaled air
Exhaled air
14
10
Red / orange
Yellow
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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.
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Ch 8 Transport in humans
Practical 8.1
Examination of a blood smear
Results (p. 8-2)
1
2
Red blood cell
Biconcave
disc shape
No nucleus
White blood cell
Blood platelet
Irregular shape
Irregular shape
Round or lobed
No nucleus
Relative size
Medium
Large
Small
Relative number
Abundant
Rare
Occasional
Shape
Nucleus
Questions (p. 8-2)
1
2
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.
a
It is biconcave disc shape. This provides a large surface area to volume ratio to
facilitate the diffusion of gases.
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b
3
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No. The absence of nucleus allows the accommodation of more haemoglobin.This
increases the oxygen-carrying capacity of the red blood cells.
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).
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Practical 8.3
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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
The red blood cells are squeezing their way through the
capillaries.
Behaviour of blood cells
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.
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Practical 8.4
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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.
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3
A
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
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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.
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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.
C
Poor growth. Depending on the species, leaves may become dull green,
yellow or purple.
D
Poor 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.
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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)
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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.)
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A Identifying variables
Independent
Dependent variable
variable
(What will you
(What will you
measure?)
change?)
(Answer varies with (Answer varies with
the design.)
the design.)
Book 1B
p.13/26
Controlled
variables
(What will you keep
constant?)
(Answer varies with
the design.)
Control
(What is the control
in this experiment?)
(Answer varies with
the design.)
B Designing the set-up
1
(Answer varies with the design.)
2
(Answer varies with the design.)
C Collecting data
1
(Answer varies with the design.)
2
(Answer varies with Ss.)
D Risk assessment and safety precautions
1
(Answer varies with the design.)
2
(Answer varies with the design.)
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:
Method 2:
1 pair of forceps
1 stop-watch
1 potted plant
dry cobalt(II) chloride paper
sticky tape
1 pair of forceps
1 beaker of hot water
1 freshly-picked leaf from a terrestrial plant
Method 3:
1 microscope
2 microscope slides
2 cover slips
1 pair of forceps
1 microscope slide with a transparent grid
1 freshly-picked leaf (e.g. Zebrina)
distilled water
1 vaseline
1 electronic balance
2 freshly-picked leaves of the same
weight from a terrestrial plant
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Method 4:
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Procedure
Method 1:
Method 4:
1
1
2
Use sticky tape to stick a piece of dry
cobalt(II) chloride paper to the upper
side and underside of the leaf
respectively.
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.
2
3
4
Method 2:
1
2
5
Immerse a leaf from a terrestrial plant
quickly into a beaker of hot water.
6
Observe carefully and compare the
amount of bubbles coming out from each 7
side of the leaf.
Method 3:
1
2
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.
Compare the weight of the leaves.
8
Use a pair of forceps to peel off the lower
epidermis of a leaf. Put it on a slide.
Mount the epidermis with a drop of
distilled water.
Find a portion of the epidermis which
fills the microscope’s field of vision at
×100 magnification.
Count the number of stomata in the field
of vision.
Repeat step 4 for 3 times and take the
average value.
Repeat steps 1 to 5 for the upper
epidermis.
Use a slide with a transparent grid of a
stated grid size to estimate the dimension
of the field of vision.
Calculate the stomatal densities of the
upper and lower epidermis of the leaf
stomatal density
=
number of stomata
area of the microscope ' s
field of vision (mm 2 )
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.
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Method 4:
Area of the microscope’s
field of vision at ×100
magnification (mm2)
Number of
stomata
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.
5
(Answer varies with the design.)
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 condition
Colour 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.
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3
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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.
Put all the tubes in a water trough.
Conclusion (p. 9-18)
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)
Minimum
pressure (kPa)
0.39
Maximum
pressure (kPa)
2.81
Change in
pressure (kPa)
+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
Light condition
Bright light
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.
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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.
Conclusion (p. 9-22)
The higher of light intensity, the higher amount of oxygen is released by the plant.
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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 jar
A
A layer of moisture and drops
of liquid are formed on the wall.
B
The bell jar remains clear.
Cobalt(II) chloride paper
Original colour
Final colour
Blue
Pink
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.
Conclusion (p. 10-2)
Water vapour is released from plant A but not from plant B. Transpiration takes place in the
aerial parts of the plant.
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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 by
the air bubble (cm)
(Results vary with Ss.)
Rate of movement of the
air 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)
At start
After the practical
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(Results vary with Ss.)
Weight of the entire set-up
(g)
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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.
Conclusion (p. 10-8)
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.
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Light intensity: use a bench lamp. / Temperature: use a heater. /
Relative humidity: use a dehumidifier. /
Air movement: use a blowing fan.
3
(Answer varies with the design.)
A Identifying variables
Independent
Dependent variable
variable
(What will you
(What will you
measure?)
change?)
The environmental
The weight of water
factor being
lost in a weight
investigated.
potometer, the
distance travelled by
the air bubble in a
given time in a
bubble potometer,
etc.
Controlled
variables
(What will you keep
constant?)
The parameters and
conditions other than
the one being
investigated.
Control
(What is the control
in this experiment?)
The potometer that is
put in normal
conditions.
B Designing the set-up
(Answer varies with Ss.)
C Collecting data
1
(Answer varies with Ss.)
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.
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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 tubings
2 rubber tubings
2 retort stands
4 clamps
1 wash bottle with water
1 scalpel
1 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
Condition
Under an environmental
condition being investigated
Under normal conditions
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Amount of water absorbed
in 15 minutes (cm3)
Rate of water uptake
(cm3 / min)
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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)
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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.
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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)
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Conclusion (p. 10-20)
The xylem is the main tissue responsible for water transport in the herbaceous plant.
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