SB6 - Plant Structures and their Functions ✓ Flashcards Preview

Edexcel GCSE Biology (9-1) > SB6 - Plant Structures and their Functions ✓ > Flashcards

Flashcards in SB6 - Plant Structures and their Functions ✓ Deck (40)
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SB6a - How are leaves adapted for their purpose?

  • Broad and Flat - Large surface area to trap more energy
  • Palisade Cells -  These are packed with chloroplasts to trap sunlight energy
  • Leaf Stomata - Microscopic pores allow for gas exchange
  • Guard Cells - Controls the Opening/Closing of the Stomata
  • Thin Cells - Reduces diffusion pathway for gas exchange
  • Spongy Mesophyll Tissue - Contains air spaces which increase the rate of gas exchange


SB6a - What does photosynthesis do?

It traps energy from the sunlight and converts it to glucose


SB6a - What is a plant's biomass?

What are producers in a food chain?

The materials in an organism

Organisms that produce their own biomass and the produce the food for almost all other life on earth (PLANT AND ALGAE)


SB6a - What is the equation for photosynthesis?

Carbon dioxide + Water → Glucose + Oxygen


SB6a - What is the equation for respiration?

Glucose + Oxygen → Carbon dioxide + Water


SB6a - What type of reaction is photosynthesis?

  • Endothermic.
  • The products have more energy than the reactants.
  • This means they have taken in energy from the surroundings during the reactions


 SB6a - Where does photosynthesis occur?

In the chloroplasts of the plant cell


SB6a - Why are the stomata an example of a gas exchange system?

They let carbon dioxide diffuse in and let oxygen diffuse out


SB6a - Why is glucose necessary?

  • Glucose molecules are joined together to form a polymer of starch.
  • After photosynthesis stops, this is broken down to simple molecules which are used to form sucrose.
  • Sucrose is used to make:
    • Starch (In a storage organ such as a potato)
    • other molecules for the plant (cellulose, lipids etc.)
    • Glucose for respiration (to release energy)


SB6b - A graph showing how increasing rate of light intensity affects rate of photosynthesis eventually levels out. Why can't it get any higher despite light intensity increasing?

  • As the graph curves, light intensity is the limiting factor.
  • Once it levels out, something else is the limiting factor.


SB6b - Once the rate of photosynthesis can't increase anymore (due to light intensity in this case) how would you increase the rate of photosynthesis?

  • Something else is the limiting factor. 
  • Increasing the CO2 concentration or increasing the temperature will allow the rate of photosynthesis to continue to increase.
  • Eventually it will level out again as something else has become the limiting factor.


SB6b - What are the three main limiting factors that affect photosynthesis?

  • Carbon dioxide concentration 
  • Light Intensity 
  • Temperature 


SB6b - What is the inverse square law, and where does it apply to?

  • The inverse square law is used to find out how light intensity changes with distance from the source.
  • I: light intensity
  • d: distance  

 l (new) = l orig x d^2 (orig) / d^2 new   

light intensity is inversely proportionate to the square of the distance


SB6b - Why are plants less likely to grow higher up on a mountain?

Higher up, the air pressure is lower meaning the carbon dioxide concentration is lower


SB6b - Why is it that even if temperature is the limiting factor, you'll get to a point where increasing it won't increase the rate of photosynthesis?

  • At a temperature that is too high, the enzymes in the plant become denatured.
  • They can no longer bind to their substrate and therefore processes can't occur anymore


SB6b CP - Describe a method, using algae balls and hydrogen carbonate indicator, to investigate rates of photosynthesis at differing light intensities.

  1.  Add 20 algae balls and the same amount of indicator to as many glass bottles as you need
  2. Compare the colour of the bottle at the start to a key to work out its starting pH (they should all be the same)
  3. Place a tank of water between the light and the first glass to absorb the heat given off by the light
  4. Cover one with foil so it is in the dark and place it next to the one closest to the lamp
  5. Measure out the distances you place all of the bottles
  6. Turn on the light and wait till you see noticable changes in the pH
  7. Once you've decided to stop, work out the pH again by comparing to a key
  8. Work out the change in pH/hour to be your rate of reaction
  9. Plot a graph of rate of reaction vs distance from light


SB6c - How are roots adapted to absorb water?

  • Roots have root hair cells
  • The hairs make the surface area larger meaning there is more area for mineral ions to be quickly absorbed through water


SB6c - How can some water enter the root hair cells if not through openings?

  • The root hair cells have a semi-permeable membrane meaning that osmosis can take place with the water moving down the concentration gradient into the cytoplasm of the cell


SB6c - How do plants take in mineral ions?

  • Through the water they absorb.
  • However as there is a higher concentration of these in the plant than in the soil, they can't absorb it through diffusion but rather through active transport which takes up energy


SB6c - What do root hair cells and root cells have between them and why?

They have a little tube allowing diffusion of fluids between cells


SB6c - What is a concentration gradient and what can it cause to occur?

  • When two areas are connected in some way and having differing levels of concentration of a substance, they have a concentration gradient
  • If this is in a fluid, diffusion can occur, where the substance moves from the area of higher to lower concentration


SB6c - What are the 4 uses of water in plants?

To be used in/to:

  • Carrying dissolved mineral ions
  • Keeping cells rigid so plants don't wilt (droop)
  • Cooling leaves (when it evapourates)
  • Photosynthesis


SB6d - Describe how you can investigate rates of transpiration

  • A potometer is used for this. It involves using a plant attached to a rubber stopper connected to a reservoir of water and a capillary tube.
  • The capillary tube should have at least one bubble in it and should have a scale .
  • As the plant uses up water it will draw water from the tube moving the air bubble.
  • The speed of the bubble will allow you to calculate the rate of transpiration


SB6d - Describe the adaptations of the phloem

  • No nucleus or cytoplasm as they aren't needed and would be a waste of energy. They create more room for the central channel
  • Companion cells to pump have many mitochondria so they have energy to actively transport sucrose
  • Pore through whcih sucrose solution can be pumped


SB6d - Describe the adaptations of the xylem

  • Found in the plant stem, form long tubes. These tubes carry water and other dissolved minerals
  • Thick cellulose cell walls strengthened by lignin prevents it from bursting due tot eh water pressure.
  • Lignin kills the xylem cells, this ensures no water is used up in cell processes.
  • End walls are also broken to create one long tube, so water flow isn't impeded.
  • Xylem cells have no subcellular structures e.g cytoplasm, nucleus, vacuole or chlorplasts so the path of water isn't obstructed. 


SB6d - Describe the factors that affect transpiration.

  • Light intensity: stomata opens wider allowing water in quicker
  • Temperature: More evapouration means a greater concentration gradient on the leaf surface
  • Wind speed: Less water on leaf surface steepens the concentration gradient, increasing rate
  • Low humidity: As the air becomes less saturated, water vapour evaporates more easily steepening the concentration gradient, increasing rate


SB6d - Describe the process of translocation

  • Translocation is the movement of food substances (such as sucrose) made in the leaves up or down the phloem, for use immediately or storage.
  • Plants make sucrose from glucose and starch made by photosynthesis. It is then translocated in the sieve tubes of the phloem tissue. The large central channel in each sieve cell is connected to its neighbours by pores, through with sucrose solution flows.
  • Companion cells actively pump sucrose into or out of the seive cells that form the sieve tubes. As sucrose is pumped into sieve tubes, the increased pressure causes the sucrose solution to flow up to the growing shoots or down to storage organs.


SB6d - Describe the process of transpiration

Transpiration is the loss of water vapour from the leaves and stems of the plant. It is a consequence of gaseous exchange, as the stomata are open so that this can occur.

● Water also evaporates at the open stomata (pores) on the leaf surfaces

● As water molecules are attracted to each other, when some molecules leave the plant the rest are pulled up through the xylem

● This results in more water being taken up from the soil resulting in a continuous transpiration stream through the plant


SB6e - How do plants reduce water loss?

  • Having stomata located inside small pits - where water vapour collects because it is less exposed to air movement
  • By losing leaves in winter (deciduous), preventing water loss when soil water may be frozen
  • By opening stomata at night, CO2 is taken i nat night and stored for use during the day. 


SB6e - Describe how specialised guard cells open/close stomata

  • When lots of water is available to the plant, the cells fill and change shape, opening stomata (they are also light sensitive)
  •  This allows gases to be exchanged and more water to leave the plant via evaporation
  •  More stomata are found on the bottom of the leaf, allowing gases to be exchanged whilst minimising water loss by evaporation as the lower surface is shaded and cooler.