SB6 - Plant Structures and their Functions Flashcards

1
Q

SB6a
1) Explain why photosynthetic organisms are producers of biomass.
2) Recall some substances produced from glucose and their roles in the plant.
3) Summarise what happens in photosynthesis.

A

1) Photosynthetic organisms are producers of biomass because they can make their own food using light, water and carbon dioxide.
2) Glucose is used for the following:
- It can be stored as starch
- It can be used for respiration (to release energy)
- It can be used to make other molecules for the plant (such as cellulose, lipids or proteins)
3) During photosynthesis: light energy is absorbed by chlorophyll - a green substance found in chloroplasts in the palisade cells in the leaf. Absorbed light energy is used to convert carbon dioxide (from the air) and water (from the soil) into a sugar called glucose. Oxygen is released as a by-product.

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2
Q

SB6a
1) Explain why photosynthesis is an endothermic reaction.
2) Explain how a leaf and its cells are adapted for photosynthesis.
3) What is the equation for photosynthesis (word and symbol equations)?

A

1) Photosynthesis is an endothermic reaction, meaning that energy is taken in during the reaction. It is endothermic because energy is transferred from the environment to the chloroplasts by light.
2) The palisade cells near the top contain a lot of chloroplasts, so that the leaf can absorb a lot of light. Leaves also have a large surface area so that they can absorb a lot of light.
3) Carbon dioxide + water —> (light above the arrow) oxygen + glucose
6CO₂+6H₂O→C₆H₁₂O₆+6O₂

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3
Q

SB6b
1) Recall what is meant by a rate of reaction.
2) Describe the effects of temperature, light intensity and carbon dioxide concentration on the rate of photosynthesis.
3) Explain the effects of limiting factors of photosynthesis

A

1) The rate of a reaction is a measure of how quickly a reactant is used up, or a product is formed.
2) Carbon dioxide concentration: As carbon dioxide concentration increases, the rate of photosynthesis also increases, until carbon dioxide is no longer the limiting factor and the rate stays the same.
Temperature: Increasing the temperature (up to the optimum) increases the rate at which the enzymes involved in photosynthesis work. After the optimum point, the temperatures are too high for the enzymes involved in photosynthesis to work. The enzymes start to denature, and the rate of photosynthesis will reduce.
Light intensity: As the light intensity increases, the rate of photosynthesis also increases (1), until light intensity is no longer the limiting factor and the rate stays the same (1).
3) A limiting factor is simply anything in short supply that prevents photosynthesis occurring at its maximum rate. If photosynthesis occurs more slowly in plant cells then a lower quantity of sugar will be produced and the quantity of chemical energy available for cell growth will be reduced.

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4
Q

SB6b
1) What are the three factors that affect photosynthesis?
2) Describe how light intensity and rate of photosynthesis are related.
3) Explain why the rate of photosynthesis is inversely proportional to the distance of a light source.

A

1) Light intensity, carbon dioxide concentration and temperature.
2) Increasing the light intensity increases the rate of photosynthesis, until some other factor – a limiting factor – becomes in short supply.
3) The inverse square law is light intensity is proportional to 1/distance squared. This means that as the square of the distance decreases, light intensity increases proportionally. In other words, if you halve the distance, the light intensity will be four times greater (2 squared).

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5
Q

SB6c
1) Explain how root hair cells are adapted to taking in water and mineral ions.
2) What are the three ways a substance can be transported by?
3) Describe what is meant by a concentration gradient.

A

1) Root hair cells are adapted for taking up water and mineral ions by having a large surface area to increase the rate of absorption. They also contain lots of mitochondria, which release energy from glucose during respiration in order to provide the energy needed for active transport.
2) Can be transported by diffusion, osmosis and active transport.
3) The concentration gradient is the difference in concentration of a substance between two areas.

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6
Q

SB6c
1) Explain why active transport is needed to transport some molecules in root hair cells
2) Explain how molecules move by osmosis.

A

1) The concentration of ions inside a root hair cell is greater than in the soil. Mineral ions cannot diffuse against this concentration gradient. So, proteins in the cell membrane pump the ions into the cell. This is an example of active transport.
2) Osmosis is when solvent molecules (such as water) diffuse through a semi-permeable membrane. They diffuse from where there are more solvent molecules (a dilute solution of solutes) to where there fewer solvent molecules (a more concentrated solution of solutes). Cell membranes are semi-permeable and so water passes into the cytoplasm of root hair cells by osmosis.

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7
Q

SB6d
1) Explain how xylem tissue is adapted to its function.
2) Explain how phloem tissue is adapted to its function.
3) Describe now transpiration occurs.

A

1) The xylem transports water and mineral ions.
Adaptations:
- No top and bottom walls between cells to form continuous hollow tubes through which water is drawn upwards towards the leaves by transpiration.
- Cells are dead, without organelles or cytoplasm, to allow free passage of water.
- Outer walls are thickened with a substance called lignin, strengthening the tubes, which helps support the plant.
- Xylem cells have thick cell walls
2) The phloem transports sucrose (dissolved sugars) and amino acids. Transport via the phloem is an active process that requires energy.
Adaptations:
- Made of living cells (as opposed to xylem vessels, which are made of dead cells) that are supported by companion cells
- Cells are joined end-to-end and contain holes in the end cell walls (sieve plates) forming tubes which allow sugars and amino acids to flow easily through (by translocation)
- Cells also have very few subcellular structures to aid the flow of materials
3) Transpiration is the movement of water from the root through the dead cells of the xylem in only one direction. This is driven by the evaporation of water from the leaves through the stomata, called the transpiration stream.
When the plant opens its stomata to let in carbon dioxide, water on the surface of the cells of the spongy mesophyll and palisade mesophyll evaporates and diffuses out of the leaf.
Water is drawn from the cells in the xylem to replace that which has been lost from the leaves. As water travels through the xylem in the stem and leaf, it is being replaced by water taken up by the roots. This is called the transpiration stream.

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8
Q

SB6d
1) Describe how translocation occurs.
2) Explain the effects of environmental factors on the rate of transpiration (light intensity, air movement, temperature, humidity).
3) Describe how to measure the rate of transpiration.

A

1) Translocation is the movement of sucrose and amino acids from the leaves through the plant. The sucrose travels through the living sieve cells of the phloem. The flow is in both directions, and travels from the sources to the sinks where the sucrose is needed. Translocation is an active process that requires energy.
2) If the air flow is high, the rate of transpiration is high. The greater the air flow around the plant, the greater the transpiration rate. Increasing air flow carries more water vapour water from the plant or reduces the concentration of water vapour outside the leaves. This sets up a concentration gradient between the leaf and the air, therefore increasing the rate of diffusion of water from the leaf cells to the air.
If the light intensity is high, the rate of transpiration is high. Guard cells are responsive to light intensity: when it is high they are turgid and the stomata open allowing water to be lost.
If the temperature is high, the rate of transpiration is high. At higher temperatures, particles have more kinetic energy so transpiration occurs at a faster rate as water molecules evaporate from the mesophyll and diffuse away faster than at lower temperatures.
If the humidity is high, the rate of transpiration is low. Humidity is a measure of moisture (water vapour) in the air. When the air is saturated with water vapour the concentration gradient weaker so less water is lost.
3) Rate of transpiration = mean volume of water uptake / time

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9
Q

SB6e
1) Identify the different tissues in a leaf.
2) Describe the functions of the different tissues in a leaf.
3) What is the function of the stomata?
4) What is the function of the air gaps in the spongy mesophyll?

A

1) Plant tissues from the top to bottom of the leaf: upper epidermis, the palisade mesophyll, spongy mesophyll and lower epidermis.
2) The palisade mesophyll layer of the leaf is adapted to absorb light efficiently. The cells: are packed with many chloroplasts; are column-shaped and arranged closely together; are towards the upper surface of the leaf.
Spongy mesophyll tissue has loosely packed cells and air spaces between them for efficient gas exchange. The spongy mesophyll cells are covered by a thin layer of water. Gases dissolve in this water as they move into and out of the cells.
The upper epidermis is thin and transparent to allow light to enter the palisade mesophyll layer underneath it. The lower epidermis contains guard cells and stomata.
3) The function of a stoma (plural stomata) is to allow gas exchange, and allow water vapour to leave the leaf.
4) The function of the air gaps in the spongy mesophyll is for gas exchange, so that carbon dioxide can enter the cell, and oxygen can leave the cell.

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10
Q

SB6e
1) Explain how leaf structure is adapted for photosynthesis and gas exchange.
2) Explain some ways in which plants are adapted to reducing water loss in extreme environments.

A

1) Photosynthesis adaptations:
Leaves are broad to catch lots of light, which is needed for photosynthesis. The upper epidermis is transparent to allow light to pass through to the palisade layer underneath. The palisade layer contains lots of chloroplasts to capture the light. Water is also needed for photosynthesis and is supplied to the leaves by xylem tissue. A waxy cuticle on the top layer of the leaf allows light to pass through, but helps to reduce water loss by evaporation.
Gas exchange adaptations:
Stomata allow carbon dioxide to enter the leaf for photosynthesis, but they can close to limit water loss when water availability is low. The spongy mesophyll tissue contains air spaces, to increase the rate of diffusion of carbon dioxide into and out of the cell’s leaves.
2) Plants can reduce water loss by trapping water vapour close to their leaves, which slows the rate of diffusion out of the leaves. Conifers achieve this by having stomata located in small pits, where water vapour collects because it is less exposed to air movement. Other plants use tiny hairs to trap water vapour.

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11
Q

SB6f
1) Recall the names of three types of plant hormone
2) Define the term tropsim.
3) Identify negative and positive photo- and geo-/gravitropisms.

A

1) Auxins, gibberellin, and ethene.
2) A tropism is when there is a growth that is responding to a stimulus. Plants tend to grow towards water and sunlight, as these are 2 essential things that plants require in order to survive.
3) Shoots are positively phototropic (they grow towards the light) and negatively geotropic (they grow away from the ground).
The roots are the opposite. They are negatively phototropic (they grow away from the light) and positively geotropic (they grow towards the ground).

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12
Q

SB6f
Explain how auxins cause phototropism in plant shoots and roots.

A

Auxin accumulates on the shaded side of the shoots/ the side of the shoots away from the light. The auxin made the cells elongate (grow) faster on the shaded side, so the shoots bent and grew towards the light.
In plant roots, auxins inhibit growth. If a root is growing sideways, auxins will accumulate in the lower side of the root tip due to gravity. These cells therefore grow and elongate less than the cells on the upper side, causing the shoot to bend downwards, in the direction of gravity.
In the shoots of a plant, auxin promotes cell elongation, which is positive phototropism or negative geotropism/ gravitropism. (growth towards light). Whereas, in the roots of a plant, auxin inhibits (prevents) cell elongation, which is positive geotropism/ gravitropism and negative phototropism (growth towards gravity).

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13
Q

SB6g
1) Describe the uses of auxins by plant growers.
2) Describe the uses of gibberellins by plant growers and fruit farmers.

A

1) Auxins are used in selective weed killers. This hormone can kill the weed but does not affect the crop because weeds have broad leaves, and crops have narrow leaves. The auxin/weed killer only affects broad-leaved plants.
2) Gibberellin are used to start the process of germination in plants. This is the process where seeds sprout and begin to grow, the first stage of plant growth. Gibberellin can also grow more fruit, and prevent the seed formation in fruit.

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14
Q

SB6g
1) Describe how fruit is artificially ripened using plant hormones.
2) Compare the advantages and disadvantages of using plant hormones (gibberellins) in fruit farming.

A

1) Ethene is the hormone that causes fruit to ripen. Fruit is often picked before they are ripe. This is because transporting fruit will take time. By picking them before they are ripe, they will still be fresh when they reach the supermarket. They will be firm and not as easily damaged while being transported.
2) Advantages:
- Gibberellins can promote seed germination at times of the year when it wouldn’t normally happen. This would allow grapes to be grown all year round.
- Gibberellins can be used to promote stem growth in the plant’s stem, so that more fruit can be grown, and a greater yield can be produced in a shorter time period.
- Gibberellin can be used to stimulate flower formation, so that the fruit quality and size increases
- Applying gibberellins to unpollinated flowers causes the fruit to grow but seeds will not.
Disadvantages:
- The potential for harm to the environment and non-target species
- Overuse of plant hormones can also lead to resistance in plants and reduce the effectiveness of the hormones over time.

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15
Q

SB6b - Core Practical
1) What is the aim of the photosynthesis core practical?
2) What is the method for the photosynthesis core practical?

A

1) Investigate the effect of light intensity on the rate of photosynthesis.
2) A. Decide on the different distances between the algae and the lamp you are going to use.
B. For each distance you will need one clear glass bottle. You will also need one extra bottle.
C. Add 20 of the algal balls to each bottle.
D. Add the same amount of indicator solution to each bottle, and put on the bottle caps.
E. Your teacher will have a range of bottles showing the colours of the indicator at different pHs. Compare the colour in your tubes with this pH range to work out the pH at the start.
F. Set up a tank of water between the lamp and the area where you will place your tubes. Take extreme care not to spill water near electrical apparatus (such as a lamp).
G. Cover one bottle in kitchen foil, so that it is in the dark.
H. Measure the different distances from the lamp. Place your bottles at those distances. Put the bottle covered in kitchen foil next to the bottle that is closest to the lamp.
Turn on the lamp and wait until you can see obvious changes in the colours in your bottles. The longer you can wait, the more obvious your results are likely to be.
J. Compare the colours of all your bottles with the pH range bottles. Write down the pHs of the solutions in your bottles.
K. For each bottle, calculate the ‘change in pH/hour’. Plot a suitable graph or chart of your results.

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16
Q

SB6b - Core Practical
1) What are the independent, dependent and 4 control variables for the photosynthesis core practical?
2) What are safety considerations for the investigating photosynthesis core practical (state 3)?
3) What gas is produced during the experiment?

A

1) Independent variable: the light intensity/ the distance from the lamp
Dependent variable: the change in pH
Control variables:
- Temperature of water
- Amount of hydrogen carbonate indicator
- Number of algal balls
- The type of bottles used
2) - Wash hands after contact with algal balls, water and indicator
- Lamp may get hot
- Keep electrical equipment dry and do not handle if hands are wet
3) Oxygen bubbles are produced.

17
Q

SB6b - Core Practical
1) Explain a reason for using hydrogen carbonate indicator?
2) Why must the same type of plant be used for the experiment?
3) Why is there a tube wrapped in foil?

A

1) The hydrogen carbonate indicator also ensures that there is a lot of carbon dioxide dissolved in the solution around the agal balls. It is important that there is a good supply of carbon dioxide is so that carbon dioxide does not become a limiting factor.
2) - Plants may have different number of leaves/surface area/number of cells etc.
- Plants may have different amount of chloroplasts/less light absorbed
3) This tube is the control that ensures that the change in pH is due to differences in light intensity and not something else.

18
Q

SB6b - Core Practical
1) Suggest 3 pieces of equipment that could be used to accurately measure the volume of gas in the photosynthesis practical that involves pondweed
2) Explain the possible sources of error in the photosynthesis core practical and suggest how the experiment could be adapted to reduce them (improve the accuracy)

A

1) A measuring cylinder, granulated gas syringe, or burette.
2) Bubbles on plant before experiment starts: gently shake the plant to dislodge any bubbles before experiment starts.
Temperature changes: measure with a thermometer to watch for temperature changes. Do not leave lamp on close to the algal balls. Place a heat shield or water bath in front of the lamp.
Light intensity: ensure same light source is used and distance is measured from the filament to the algal balls – not from the bulb to the edge of the beaker.

19
Q

SB6b - Core Practical
1) How would your deal with anomalous results in the photosynthesis core practical?
2) Give methods of measuring light intensity in the photosynthesis core practical

A

1) To deal with anomalies, repeat the reading to get consistent results or calculate the mean without the anomalous results.
2) Use a light meter or lux meter, or count the number of bubbles produced and use this to calculate the light intensity.

20
Q

SB6b - Core Practical
1) Explain how you could improve the method to measure the volume of gas released more accurately in the photosynthesis core practical
2) Explain the inverse square law

A

1) Collect the gas or oxygen produced in a granulated gas syringe. This is to reduce the errors generated when bubbles may be different sizes.
2) As the distance increases the light is spread out over a larger area. Area is larger in two dimensions. If the distance is doubled, the area is increased by 2 squared. Light intensity is therefore divided by 2 squared when the distance doubles.

21
Q

SB6e
1) Describe some adaptations that plants have to living in dry environments (state 4)
2) Describe some adaptations that marram grass have to living in dry environments (state 4)

A

1) Adaptations that desert plants have to allow them to survive in dry environments:
- Desert plants have spines and small leaves so that they have a reduced surface area, they deter animals from eating the flesh for water. The spines also mean that there is less water lost by evaporation. The plant has moist air trapped in curled leaves.
- They have a thick waxy cuticle to reduce water lost by evaporation.
- The plants have a fleshy or swollen stem which collects and stores water.
- The stomata only open at night. Carbon dioxide is taken in at night and is stored for use during the day.
2) Adaptations that marram grass has to survive in dry conditions:
- To reduce wind damage: the grass is flexible, there is a good root structure, there are long thin leaves and the leaf is rolled up.
- Thick waxy cuticle to prevent water loss by transpiration.
- The inner surface has a large number of hairs to trap air and reduce air movement. The leaf is rolled up to trap air inside as well.
- No stomata on the upper surface to prevent water loss.

22
Q

SB6e
1) Describe some adaptations that plants have to living in cold environments
2) Describe some adaptations that plants have to living in humid environments
3) Explain why a waxy cuticle is important for a pine leaf.

A

1) Adaptations in cold environments:
In winter, many broad-leaved deciduous plants lose all their leaves, preventing water loss when soil water may be frozen. However, most conifers (e.g. pine trees) do not do this. Conifers have needle-shaped leaves with a much smaller surface area and a very thick cuticle. Additionally, this shape creates less wind resistance than broad leaves, allowing conifers to withstand high winds. It also means that they collect less snow.
2) Plants that live in conditions with a plentiful supply of freshwater have leaves with a short diffusion distance through the stomata and a large surface area provided by the air spaces in the spongy mesophyll. These factors make them vulnerable to water loss.
3) The waxy cuticle surrounds the pine leaf, so it prevents water loss from the pine leaf.

23
Q

Describe how stomata open

A

Guard cells take in water by osmosis, so that the guard cells become turgid and swell.