Diffusion osmosis and active transport

Question 1

Define Diffusion? (2)

Answer 1

Define Diffusion? (2)

net movement of molecules from an area of high concentration to an area of low concentration (1)

until equilibrium is reached (1)

Question 2

simple vs Facilitated Diffusion? (2)

Answer 2

simple vs Facilitated Diffusion? (2)

Simple = molecules move directly through the phospholipid bilayer (1)

Facilitated = molecules pass through transport proteins (large use carrier, charged use channel) (1)

Question 3

actors that affect rate of diffusion? (5)

Answer 3

actors that affect rate of diffusion? (5)

surface area (increase SA= increase rate of diffusion)

concentration gradient (increase concentration gradient = increase rate of diffusion)

thickness of the exchange surface = decrease diffusion distance = increase rate of diffusion

temperature (increase temperature = increase kinetic energy = molecules move faster = increase rate of diffusion)

size of molecules (smaller molecules = increase rate of diffusion)

Question 4

What is Ficks Law?

Answer 4

What is Ficks Law?

(Surface Area x Concentration Gradient)/Thickness

Question 5

Define Osmosis? (3)

Answer 5

Define Osmosis? (3)

movement of water molecules (1)

from an area of high water potential to an area of low water potential (1)

through a partially permeable membrane (1)

Question 6

Which liquid has the highest water potential?

distilled/pure water has a value of 0kPa

lower water potential by adding solutes (makes water potential negative)

water moves from less negative water potential (e.g. -35 kPa) to more negative water
potential (e.g. -75 kPa)

Answer 6

Which liquid has the highest water potential?

distilled/pure water has a value of 0kPa

lower water potential by adding solutes (makes water potential negative)

water moves from less negative water potential (e.g. -35 kPa) to more negative water
potential (e.g. -75 kPa)

Question 7

Surround animal cell with pure water? (1)

Answer 7

Surround animal cell with pure water? (1)

swells and burst (water enters by osmosis)

Question 8

Surround plant cell with pure water? (4)

Answer 8

Surround plant cell with pure water? (4)

swells but does not burst (1)

cell wall prevents it from bursting (1)

Cell wall made of cellulose – strong material (1)

the cell is Turgid (1)

Question 9

Surround animal cell with concentrated sugar/salt solution?

Answer 9

Surround animal cell with concentrated sugar/salt solution?

shrinks (water leaves by osmosis)

Question 10

Surround plant cell with concentrated sugar/salt solution? (4)

Answer 10

Surround plant cell with concentrated sugar/salt solution? (4)

water leaves by osmosis (1)

cell wall prevents cell from shrinking, keeps it rigid (1)

the protoplast (cell membrane plus contents) shrink (1)

the cell is Plasmolysed (1)

Question 11

Define Active Transport? (3)

Answer 11

Define Active Transport? (3)

Net movement of molecules (1)

from an area of low concentration to an area of high concentration (1)

using ATP and carrier proteins (against concentration gradient) (1)

Question 12

Describe the process of active transport? (5)

Answer 12

Describe the process of active transport? (5)

molecules in area of low concentration bind to carrier protein (1)

ATP breaksdown to ADP, Pi and Energy (1)

the Pi and Energy cause the carrier protein to change shape (1)

carrier protein releases molecules on opposite side (in area of high concentration)(1)

the carrier protein releases the attached Pi to return to its original shape (1)

Question 13

Adaptations of SI? (5)

Answer 13

Adaptations of SI? (5)

folded to form Villus (large surface area) (1)

cells lining SI have Microvilli (large surface area) (1)

wall of SI is thin (short diffusion distance) (1)

rich blood supply (maintains concentration gradient) (1)

cells lining SI have transport proteins and mitochondria (1)

Question 14

Active Transport of Glucose in SI?

Answer 14

Active Transport of Glucose in SI?

sodium ions are actively transported from the cells lining the SI into the blood (1)

lowers the sodium ion concentration in the cel (1)

therefore sodium ions move from the lumen of the SI into the cell (1)

this pulls in glucose via a cotransport protein (1)

therefore glucose builds up in the cell and moves into the blood by diffusion (1)

Question 15

What do we mean when we say diffusion is a passive process? (1)

Answer 15

What do we mean when we say diffusion is a passive process?

No energy required (1)

Question 16

What can diffuse directly through the membrane (1)

Answer 16

What can diffuse directly through the membrane (1)

Small, non-polar substances (e.g. oxygen) and water

Question 17

Exam tip

always say down the concentration gradient in the exam, not across or along – or you won’t get the marks

Answer 17

Exam tip

always say down the concentration gradient in the exam, not across or along – or you won’t get the marks

Question 18

Factors affecting the rate of facilitated diffusion (4)

Answer 18

Factors affecting the rate of facilitated diffusion (4)

Concentration gradient – the higher the concentration gradient, the faster the rate of facilitated diffusion. (1)
as equilibrium is reached the rate of facilitated diffusion will level off (1)

The number of channel or carrier proteins – once all the proteins in a membrane are in use, facilitated diffusion can’t happen any faster, (1) even if you increase the concentration gradient.(1)

Question 19

A photograph shows Ink diffusing through a beaker of water.

explain what is happening to the ink molecules (1)

Answer 19

A photograph shows Ink diffusing through a beaker of water.
explain what is happening to the ink molecules (1)

The ink molecules are moving from an area of higher concentration to an area of lower concentration

Question 20

Carbon dioxide is a waste product of respiration and must be removed from cells.
how will each of the following affect the rate of diffusion of carbon dioxide across a cell surface membrane?
Explain your answer in each case. (6)

A). Increasing the thickness of the cell membrane (2)

B).Increasing the number of folds In the cell membrane (2)

C). Reducing the concentration of carbon dioxide outside of the cell (2)

Answer 20

Carbon dioxide is a waste product of respiration and must be removed from cells.
how will each of the following affect the rate of diffusion of carbon dioxide across a cell surface membrane?
Explain your answer in each case.

A). Increasing the thickness of the cell membrane means the distance the particles have to travel is further (1)
so the rate of diffusion will decrease (1)

B). Increasing the number of folds In the cell membrane means the surface area of the cell will increase, (1)
so the rate of diffusion will increase (1)

C). Reducing the concentration of carbon dioxide outside of the cell Means that the concentration gradient will increase, (1)
so the rate of diffusion will increase (1)

Question 21

Simple diffusion and facilitated diffusion both move particles down their concentration gradient across a cell membrane.
suggest and explain how you could determine whether a particular particle is being transported by simple or facilitated diffusion in an experimental setting. (6)

Answer 21

Simple diffusion and facilitated diffusion both move particles down their concentration gradient across a cell membrane.
suggest and explain how you could determine whether a particular particle is being transported by simple or facilitated diffusion in an experimental setting. (6)

You could increase the concentration gradient of the particle and monitor the rate of diffusion.

facilitated diffusion requires proteins to transporter particles across the cell membrane. (1)
there are fixed number of proteins in the membrane (1)
once all the proteins are in use, increasing the concentration gradient won’t increase the rate of facilitated diffusion any further, (1)
whereas increasing the concentration gradient will continue to increase the rate of simple diffusion. (1)

Question 22

Briefly describe how a carrier protein transports molecules across a cell membrane (3)

Answer 22

Briefly describe how a carrier protein transports molecules across a cell membrane (3)

First, a large molecule attaches to a carrier protein in the membrane. (1)
then, the protein changes shape. (1)
this releases the molecule on the opposite side of the protein.(1)

Question 23

What is the channel protein (1)

Answer 23

What is the channel protein (1)

They are proteins within a cell membrane that form pores (1)

Question 24

Describe the role of channel proteins in the transport of particles across a cell membrane (1)

Answer 24

Describe the role of channel proteins in the transport of particles across a cell membrane (1)

They allow charged particles to pass through a cell membrane via facilitated diffusion (1)

Question 25

Explain how increasing the number of carrier and channel proteins in a membrane would affect the rate of facilitated diffusion (2)

Answer 25

Explain how increasing the number of carrier and channel proteins in a membrane would affect the rate of facilitated diffusion (2) It would increase the rate of facilitated diffusion (1) as it would allow more particles to be transported across the membrane at the same time (1)

Question 26

You should always use the term water potential in the exam – never say water concentration

Answer 26

Explain how increasing the number of carrier and channel proteins in a membrane would affect the rate of facilitated diffusion (2)

Question 27

Describe the net movement of water molecules in each of the following situations: (3) A). Human cheek cells with the water potential of -300 kPa are placed in a salt solution with the water potential of -325 kPa B). Apple slices with the water potential of -750 kPa are placed in a beaker of pure water C). Orange squash with the water potential of -450 kPa is sealed in a length of visking tubing and suspended in a solution of equal water potential

Answer 27

Describe the net movement of water molecules in each of the following situations: (3) A). Human cheek cells with the water potential of -300 kPa are placed in a salt solution with the water potential of -325 kPa Water molecules will move from the cheek cells into the salt solution B). Apple slices with the water potential of -750 kPa are placed in a beaker of pure water Water molecules will move into the apple slices out of the beaker of water C). Orange squash with the water potential of -450 kPa is sealed in a length of visking tubing and suspended in a solution of equal water potential The solutions are isotonic, this means there will be no net movement of water molecules as the water potential in both solutions is the same

Question 28

Potato chips with the water potential of -350 are placed in sucrose solution is with varying water potentials solution 1 -250 kPa solution 2 -500 kPa solution 3 -1000 kPa A). After 15 minutes, the potato cells and solution one have increased in volume. explain why this is the case. B). Predict whether the cells in solutions two and three will increase or decrease in volume. explain your answers.

Answer 28

Potato chips with the water potential of -350 are placed in sucrose solution is with varying water potentials solution 1 -250 kPa solution 2 -500 kPa solution 3 -1000 kPa A). After 15 minutes, the potato cells and solution one have increased in volume. explain why this is the case. The potato cells have a lower water potential than the sucrose solution, so they gain water by osmosis (1) B). Predict whether the cells in solutions two and three will increase or decrease in volume. explain your answers. The cells and both solutions will decrease in volume, this is because they have a higher water potential than the sucrose solution is, so will lose water by osmosis

Question 29

A scientist has a 1.5 M saline solution. For her experiment she needs 30 cm³ of 125 mM solution (1M = 1000 mM) A).Calculate the volume of the original solution and distilled water that she needs to make the new solution B). From her 30 cm³ of 125 mM solution, she needs to make two more solutions. she will make serial dilutions, diluting by a factor of 5 each time. describe fully how she would do this.

Answer 29

A scientist has a 1.5 M saline solution. For her experiment she needs 30 cm³ of 125 mM solution A).Calculate the volume of the original solution and distilled water that she needs to make the new solution 1.5 M = 1500 mM Scale factor = 1500 mM / 125 mM = 12 30 / 12 = 2.5 So she needs to use 2.5 cm³ of the original solution and 30 - 2.5 = 27.5 cm³ of distilled water B). From her 30 cm³ of 125 mM solution, she needs to make two more solutions. she will make serial dilutions, diluting by a factor of 5 each time. describe fully how she would do this. She is starting with 30 cm³ and diluting by a factor of 5. 30 cm³ / 5 = 6 cm³, so she would need to add 30 - 6 = 24 cm³ of distilled water to two test tubes. she would then use a pipettes to take 6 cm³ from the 125 mM solution and add it to the one of the test tube containing 24 cm³ of distilled water. she would then mix the contents of this test tube thoroughly to create one solution before taking 6 cm³ of the solution and adding to it to the distilled water In the remaining test tubes and mixing thoroughly to create the second solution

Question 30

How to make a desired solution If if you want to make 15 cm³ of 0.4 M sucrose solution 1). Start with a solution of a known concentration, E.g. 1M 2). Find the scale factor by dividing the concentration of the solution by the concentration of the solution you want to make. the scale factor is 1M / 0.4M =2.5 3). This means that the solution you want to make is 2.5 times weaker than the one you have got to make the solution 2.5 times weaker, use 2.5 times less of it, I.e. 15 cm³/ 2.5 = 6 cm³ transfer this amount to a clean test tube. 4). Top up the test tube with distilled water to get the volume you want to make in this case you want to make 15 cm³ of solution, so you need to add: 15 - 6 = 9 cm³ of distilled water

Answer 30

How to make a desired solution If if you want to make 15 cm³ of 0.4 M sucrose solution 1). Start with a solution of a known concentration, E.g. 1M 2). Find the scale factor by dividing the concentration of the solution by the concentration of the solution you want to make. the scale factor is 1M / 0.4M =2.5 3). This means that the solution you want to make is 2.5 times weaker than the one you have got to make the solution 2.5 times weaker, use 2.5 times less of it, I.e. 15 cm³/ 2.5 = 6 cm³ transfer this amount to a clean test tube. 4). Top up the test tube with distilled water to get the volume you want to make in this case you want to make 15 cm³ of solution, so you need to add: 15 - 6 = 9 cm³ of distilled water

Question 31

How do we find the scale factor in serial dilutions?

Answer 31

How do we find the scale factor in serial dilutions? scale factor = concentration of the solution / concentration of the solution you want to make.

Question 32

To make a solution weaker what do we do

Answer 32

To make a solution weaker what do we do Use less of the solution E.g. to make a solution 2.5 times weaker, use 2.5 times less of it, I.e. 15 cm³/ 2.5 = 6 cm³

Question 33

Define water potential (1)

Answer 33

# Define water potential (1) water potential is the potential of water molecules to diffuse out of or into a solution

Question 34

Give three factors that affect the rate of osmosis

Answer 34

Give three factors that affect the rate of osmosis The water potential gradient (the higher the water potential gradient, the faster the rate of osmosis). The thickness of the exchange surface (the thinner the exchange surface the faster the rate of osmosis) The surface area of the exchange surface (the larger the surface area, the faster the rate of osmosis)

Question 35

Describe an investigation that you could do to find the water potential of potato cells

Answer 35

Describe an investigation that you could do to find the water potential of potato cells Firstly cut equal size chips from a potato. divide the chips into groups of three and measure the mass of each group and make up several different sucrose concentrations. place each group of chips into a different sucrose solution and leave all the chips for the same length of time. remove the chips and measure the mass of each group again. record each groups percentage change in mass, then make a calibration curve by putting the percentage change in mass against a concentration of the sucrose solution the group was in. Read off the concentration where the curve crosses the X axis look up the water potential for that concentration of solution in, For example, a textbook to give you the water potential of the potatoes cells

Question 36

ATP is produced by mitochondria during aerobic respiration. the equation for this process is glucose + oxygen -> carbon dioxide + Water + ATP the graph shows the relationship between the relative rates of oxygen consumption and the active transport of sodium ions across epithelial cells as a proportional relationship Q1. a) Describe the relationship shown by the graph (1) b) Suggest an explanation for this relationship (3) c) Suggest one other factor that may affect the rate of sodium ion active transport (1)

Answer 36

ATP is produced by mitochondria during aerobic respiration. the equation for this process is glucose + oxygen -> carbon dioxide + Water + ATP the graph shows the relationship between the relative rates of oxygen consumption and the active transport of sodium ions across epithelial cells as a proportional relationship Q1. a) Describe the relationship shown by the graph As the rate of sodium ion active transport increases, so does the rate of oxygen consumption b) Suggest an explanation for this relationship Sodium ion active transport requires energy from ATP. as the rate of active transport increases, the rate of aerobic respiration must also increase in order to produce more ATP, which means the rate of oxygen consumption must increase too. c) Suggest one other factor that may affect the rate of sodium ion active transport The rate of glucose consumption

Question 37

The thyroid gland needs iodide ions to make hormones, so there is a higher concentration of iodide ions inside the thyroid soils than in the blood plasma. the sodium iodide co-transporter is involved in transporting iodide ions into the thyroid gland. the concentration of sodium ions is higher in the blood plasma then in the thyroid gland a) Which ion needs to be actively transported by the sodium iodide co-transporter. explain your answer. b) Using your knowledge of co-transporters, describe and explain how active transport is carried out by the sodium iodide co-transporter

Answer 37

The thyroid gland needs iodide ions to make hormones, so there is a higher concentration of iodide ions inside the thyroid soils than in the blood plasma. the sodium iodide co-transporter is involved in transporting iodide ions into the thyroid gland. the concentration of sodium ions is higher in the blood plasma then in the thyroid gland a) Which ion needs to be actively transported by the sodium iodide co-transporter. explain your answer. The iodide ions because it needs to move from an area of lower concentration to an area of higher concentration b) Using your knowledge of co-transporters, describe and explain how active transport is carried out by the sodium iodide co-transporter The co-transporter binds to an Iodide ion and a sodium ion. the sodium ion moves across the membrane into the thyroid gland down its concentration gradient. this moves iodide across the membrane into the cell too, against its concentration gradient.

Question 38

Describe the chemical reaction that occurs to release energy from ATP (1)

Answer 38

Describe the chemical reaction that occurs to release energy from ATP (1) Hydrolysis reaction occurs which splits ATP into ADP and P (an inorganic phosphate)

Question 39

Describe how the following are used to transport substances across the cell membrane during active transport: a) Carrier proteins (3) b) Co-transporters (2)

Answer 39

Describe how the following are used to transport substances across the cell membrane during active transport: a) Carrier proteins - The molecule attaches to a carrier protein in the membrane. the protein changes shape and releases the molecule on the opposite side of the membrane. the process requires energy b) Co-transporters - co-transporters bind two molecules at a time. the concentration gradient of one of the molecules is used to move the other molecule against its own concentration gradient.

Question 40

Why are sodium ions important in the transport of glucose from the ileum into the blood? (3)

Answer 40

Why are sodium ions important in the transport of glucose from the ileum into the blood? (3) Because sodium ions diffuse from the lumen of the Ileum into the intestinal epithelium cells down the concentration gradient, through a sodium glucose co-transporter protein. at the same time, the co-transporter carries glucose into the epithelium cells against a concentration gradient. glucose is then able to diffuse into blood from the epithelial cell.

Question 41

Will the rate of active transport increase or decrease with an increasing number of carrier proteins

Answer 41

Will the rate of active transport increase or decrease with an increasing number of carrier proteins Increase

Question 42

State the type of molecule that actively transports glucose across the cell surface membranes of the ileum and briefly describe how it does so.

Answer 42

State the type of molecule that actively transports glucose across the cell surface membranes of the ileum and briefly describe how it does so. It is a carrier protein/co-transporter, which binds glucose and sodium ions at the same time

Question 43

Another stage of the absorption process happens by facilitated diffusion. suggest why glucose must use facilitated diffusion rather than simple diffusion to cross the cell surface membranes of the Ileum. (3)

Answer 43

Another stage of the absorption process happens by facilitated diffusion. suggest why glucose must use facilitated diffusion rather than simple diffusion to cross the cell surface membranes of the Ileum. (3) The centre of the phospholipid bilayer is hydrophobic. (1) it forms a barrier to the diffusion of water soluble substances including most polar molecules. (1) glucose is a polar molecule that can’t diffuse directly across the membrane (1)

Question 44

Percentage change in mass

Answer 44

Percentage change in mass Change in mass / initial mass x 100 Example A piece of potato has a mass of 2.5 g at the start and 3.0 g at the end. percentage change in mass = (3.0 – 2.5) ÷ 2.5 × 100 = 0.5 ÷ 2.5 × 100 = +20%

Question 45

Suggest how the students could make the results more precise

Answer 45

Suggest how the students could make the results more precise They could do repeats of the experiment for each concentration of sucrose solution and calculate a mean percentage change in mass

Question 46

How does surface area affect the diffusion rate (1)

Answer 46

How does surface area affect the diffusion rate (1) More space (area) for absorption per given size of cell