transport in multicellular plants cambridge bio lesson 1

Transport in multicellular plants Part 1 Lesson 1

Redington College Kajana Sivarasa 1

Transport in Multicellular plants

Explain the need for transport system in multicellular plants and animals in terms of size and surface area to volume ratios All cells need to take in substances from their environment, and get rid of unwanted substances. For example, a cell that is respiring aerobically has to take in oxygen and get rid of carbon dioxide. v In single-‐ celled organism, this can happen quickly enough by diffusion alone. This is because:

• No point in the cell is very far from the surface, so it does not take long for gases to diffuse from the cell surface membrane to the center of the cell, or vice versa;

• The surface area to volume ratio of the cell is relatively large – that is, it has a large amount of surface area compared to its total volume.

v In a large organism, diffusion is no longer sufficient. This is because:

• The center of the organism may be a long way from the surface, so it would take too long for substances to diffuse all that way;

• The surface area to volume ratio is much smaller – that is, it has a small amount of surface area compared to its total volume. v Large organisms solve these difficulties in two ways:

• They have transport systems that carry substances by mass flow from one part of the body to another, rather than relying solely on diffusion.

• They increase the surface area of parts of the body involved in exchange with the environment, for example by having thin, flat leaves or by having a highly folded gas exchange surface.

Transport in plants

v Their energy needs are generally small compared with those of animals v Their leaves are very thin and have a large surface area inside them in contact with

the air spaces. v Therefore diffusion is sufficient to supply the mesophyll cells with carbon dioxide

for photosynthesis, and to remove oxygen.



v Plants have two transport systems: • Xylem, which transports water and inorganic ions from the roots to all other

parts of the plant • Phloem,which transports substances made in the plant, such as sucrose and

aminoacid to all parts of the plant.

Transport in Xylem

• The driving force that causes this movement is the loss of water vapour from the leaves. This is called transpiration.

• Transpiration is the loss of water vapour from a plant. • Most transpiration happens in the leaves. • A leaf contains many cells in contact with air spaces in the mesophyll layers. • Liquid water in the cell walls changes to water vapour, which diffuses in to the air

spaces. • The water vapour then diffuses out of the leaf through the stomata, down a water

potential gradient, into the air surrounding the leaf.

• A pair of guard cells surrounds each stoma. These can change shape to open or close the stoma.

• When the pair of guard cells is turgid (full of water), the bow in such a way as to increase the gap – stoma – between them.



• When potassium ions move in from the surrounding cells, it decreases water potential in guard cells.so Water moves in to the guard cells creating pressure on walls, which open stomata.

• In order to photosynthesis, the stomata must be open so that carbon dioxide

can diffuse in to the leaf. Plants cannot therefore avoid losing water vapour by transpiration.

Factors that affect transpiration rate

• Temperature – At higher temperature water molecules have more kinetic energy. Evaporation from the cell walls inside the leaf therefore happens more rapidly, and diffusion also happens more rapidly.

• Light – High light intensity increases the rate of transpiration. This is because the plant may be photosynthesizing rapidly, requiring rapid supply of carbon dioxide. This means that more stomata are likely to open, through which water vapour can diffuse out of the leaf.



Example Q:



Investigating experimentally the factors that affect the transpiration rate.

Ø The potometer is used as follows:

• The potometer is filled under water to get rid of all air inside it • A shoot is cut and the end is quickly put into water to prevent any air

bubbles in the xylem. Using s sharp blade, make a slanting cut across the stem.

• The shoot is inserted into the apparatus with everything submerged in water to avoid any air bubbles.

• Make sure the apparatus is air tight and there are no bubbles inside. • Wait atleast 10 minutes for it to dry out after taking the whole apparatus out

of the water. • By using a stop clock measure the distance moved over a set time. • Repeat measurements over the same time.

• The amount of water taken up by the shoot per unit time is measured by

multiplying the distance travelled by the bubble and the cross sectional area of the tube.

• The shoot can be tested under various conditions such as temperature, humidity, light and wind

v A potometer actually measures the rate of water uptake by the cut stem, not the rate

of transpiration; and these two are not always the same. v During the day plants often transpire more water than they take up (i.e. they lose

water and may wilt), and during the night plants may take up more water than they transpire (i.e. they store water and become turgid).

v The difference can be important for a large tree, but for a small shoot in a potometer the difference is usually trivial and can be ignored.



Example Q1:



Example Q2:

Transpiration is the loss of water from plants by evaporation. The diagram below shows a potometer, an apparatus used to estimate transpiration rates.

(a) Transpiration itself is not measured directly by a potometer.

State what is measured by this apparatus.

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(b) Describe how the apparatus should be set up to ensure that valid measurements can be made.

In your answer, you should make clear how the steps in the process are sequenced. [7]

[Total 8 marks]

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water reservoir

leafy shoot

screw clip

air bubble

scale

transport in multicellular plants cambridge bio lesson 1

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