osmosis w b 2 - thiacin...t r l j. version 2, 2018 3 the following points are what you need to know,...

MEMBRANE TRANSPORT

Osmosis

WORK BOOK 2

Name: ________________________ Option Group: _____________

T R L J. Version 2, 2018

2

Instructions

Regular revision throughout the year is essential. It’s vital you keep a track of what you understand and what you don’t understand. This booklet is designed

to help you do this. Use the following key to note how well you understand the work after your revision. Put the letter R, A or G in the table. If you place an

R or an A then you should make a note of what you are struggling with and the end of this book under the relevant section and seek help with this.

Key

• R = Red. I am not confident about my knowledge and understanding.

• A = Amber. I am fairly confident about my knowledge and understanding.

• G = green. I am very confident about my knowledge and understanding.

STUDY CHECKLIST AND ASSESSMENT OBJECTIVES


3

The following points are what you need to know, revise and answer questions on.

Place an R, A or G when you have revised and make notes of what you do not understand in the relevant section at the back of this booklet.

Cell Membrane Structure and Transport. Unit 1, Section 3

Cell Membrane Structure Topic a and b

1. Can you draw and fully label the cell membrane as shown by the fluid mosaic model.

2. Can you explain why the cell membrane is described as the fluid mosaic model.

3. Do you know the thickness of the cell membrane?

4. Can you explain the functions of all the components of the cell membrane?

5. Can you describe the function of the cell membrane?

6. Can you name the scientists that first described the fluid mosaic model of cell membrane structure?

Membrane Transport Topic c

7. Can you define: Diffusion, Facilitated Diffusion, Active Transport and Osmosis.

8. Can you relate the structure of the cell membrane to the transport of polar and non-polar substances?

9. Can you name specific substances that are polar and non-polar?

10. Can you state other names that mean polar and non-polar?

11. Can you calculate rate and the percentage change of rate from a graph?

Diffusion

12. Can you state the factors that affect the rate of diffusion?

13. Can you relate the structure of the cell membrane to the size and lipid solubility of the substances passing through the cell membrane to the rate of diffusion?

14. Can you recognise a graph that shows diffusion occurring?

Facilitated Diffusion

15. Can you state the factors that affect the rate of facilitated diffusion?

16. Can you relate the structure of the cell membrane to the facilitated diffusion?

17. Can you recognise a graph showing facilitated diffusion?

18. Do you understand the term saturation effect and where this occurs on a graph?

19. an you draw a diagram to show facilitated diffusion occurring.


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Active Transport

20. Can you state the conditions needed for active transport to occur?

21. Can you relate the structure of the cell membrane to active transport?

22. Can you recognise a graph showing active transport?

23. Can you recognise the keys in a question that tells you it’s about active transport?

24. Can you recognise a graph showing active transport?

25. Can you draw a diagram showing active transport?

26. Do you know how the Na+/K+ pump works and its importance to cellular function?

Osmosis

27. Do you understand the terms: Water Potential, Solute Potential and Pressure Potential?

28. Can you use the equation: ψw = ψs + ψp

29. Do you understand the terms: Isotonic, Hypotonic and hypertonic?

30. Can you describe and explain the effects of osmosis on animal cells, e.g. a red blood cell.

31. Can you use the terms in point 9.3 in relation to osmosis in animal cells.

32. Can you draw a red blood cell to show the effects of an isotonic, hypotonic and hypertonic solution?

33. Do you know what haemolysis and crenation is in animal cells. Do you know other words that could be used instead of crenation?

34. Can you describe the effects of an isotonic, hypertonic, and hypotonic solution in plant cells?

35. Do you understand the terms: turgid, plasmolysed and incipient plasmolysis in relation to plant cells?

36. Do you know the important relationships between ψw, ψs and ψp when a plant cell is plasmolysed, turgid and in incipient plasmolysis?

37. Can you use/plot a graph to explain the percentage change in mass that occurs in plant cells during osmosis when the external sugar concentration is increased.

38. From the graph in 37 can you determine the sugar concentration that would produce no net movement of osmosis.

39. Can you use/plot a graph to explain the percentage plasmolysed plant cells that occur when the external sugar concentration is increased. From this graph can you determine the solute potential of the plant cell.

40. Do you understand the importance of 50% plasmolysed cells from the graph in point 42.

41. Do you understand these important relationship in plant osmosis: 1 ψp = 0 so ψs= ψw. 42. 2 ψw = 0 so ψs= ψp

Endocytosis, Phagocytosis, Exocytosis and Pinocytosis

43. Can you draw diagrams to show endocytosis and exocytosis?

44. Can you link the function of the lysosome with endocytosis?


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Assessment Objective Description AO1 Demonstrate knowledge and understanding of scientific ideas, processes, techniques and procedures.

AO2 Apply knowledge and understanding of scientific ideas, processes, techniques and procedures:

• In a theoretical context

• In a practical context

• When handling qualitative data

• When handling quantitative data

AO3 Analyse, interpret and evaluate scientific information, ideas and evidence, including in relation to issues, to:

• Make judgments and reach conclusions

• Develop and refine practical design and procedures

Application of Knowledge

45. Can you apply your knowledge of membrane structure and transport to unfamiliar scenarios?


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Below is a list of some key words and phrases you will need to learn and understand in this membrane

structure and transport section.

1. Active Transport 2. Amphiphilic 3. Antigen 4. Bilayer 5. Carrier protein 6. Channel protein 7. Cholesterol 8. Concentration gradient 9. Crenation 10. Diffusion 11. Endocytosis 12. Exocytosis 13. Extrinsic protein 14. Facilitated diffusion 15. Fatty acid tails 16. Fluid mosaic model 17. Glycocalyx 18. Haemolysis 19. Hydrophilic 20. Hydrophobic 21. Hydrophobic region 22. Intrinsic protein 23. Osmosis 24. Phospholipid 25. Phospholipid bilayer 26. Pinocytosis 27. Polar head 28. Pressure potential 29. Receptor 30. Solute potential 31. Surface area 32. Transmembrane protein 33. Water potential 34. Water potential gradient

WORD BANK


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1.0 WHAT IS OSMOSIS

1.0.1 Definition Osmosis is the passive net diffusion of water molecules from a region

of high water potential to a region of lower water potential through a semi permeable

membrane down a water potential gradient.

1.0.2 to 1.0.5 Important terminology and concepts 1

1.0.2 Concept Water potential is a measure of the tendency of water molecules to

move. It is based on the free energy which is the ability to do work. Water potential

has the unit of pressure expressed in mega or kilo pascals (MPa, KPa). The symbol for

water potential is the Greek letter psi (Ψ).

1.0.3 Concept Water potential is a relative measure, just like altitude on land is

measured relative to sea level, so water potential of a cell is relative to that of a

standard – this standard is pure water. Pure water has no solutes dissolved in it and has

a water potential arbitrary set at zero – this is considered the highest water potential.

1.0.4 Concept The difference in water potential is called a water potential gradient

and is the diving force for water movement.

1.0.5 Concept The water potential of a cell will be lowered by the addition of

solutes, e.g. glucose, sodium ions, sucrose. The water in cells have solutes dissolved in

it. This means, compared to pure water, the water potential of cells are negative values,

e.g. -100KPa. The degree to which the water potential is lowered by solutes is

dependent on the concentration of the solutes.

0KPa

• Pure water. No solutes added. Water has the greatest tendency to move so has the highest water potential and so the greatest free energy.

-100KPa

• Water with a low concentration of solutes added. The water potential has been lowered compared to pure water so has a negative value and a lower free energy.

-400KPa

• Water with a higher concentreation of solutes added. The water potential has been lowered further as has the free energy of the water molecules. The water potential value has become more negative.


8

1.0.6 to 1.07 Why does the addition of solutes lower the water potential

1.0.6 Concept Water is a polar molecule so has δ- and δ+ charges on the oxygen so

when solutes are added to water, they form bonds with water molecules. The water

molecules are then too big to fit through the semi permeable membrane – their free

energy has been reduced so their tendency to move and hence their water potential

have been reduced.

1.0.7 Concept The concentration of solutes is directly proportional to the solute

potential (ΨS). Because solutes reduce the tendency of water molecules to move by

reducing the number of free water molecules, so the greater the solute concentration

the greater the solute potential and so the greater the reduction is the tendency of

water to move. This is why all solute potential will be expressed as negative numbers,

e.g. -75KPa.

Free water molecules (2 small circle) are small enough to pass

through the semi permeable membrane. When solutes are added

(large circle) they bond to the water. The number of free water

molecules have been reduced from 2 to 1 in this example. The water

molecule attached to the solute is now too big to through the

semipermeable membrane.


9

1.0.8 to 1.0.10 Important terminology and concepts 2

1.0.8 Concept Isotonic solutions have the same water potentials, the same

concentration of solutes so the same solute potential, no water potential gradient and

no net movement of water by osmosis occurs. An equilibrium is present.

1.0.9 Concept A hypotonic solution has a low solute potential and a high water

potential.

Cell A Cell B

Cell A and B are separated by a semi permeable

membrane. Both cells have a water potential of 0KPa and

a solute potential of 0KPa so are isotonic. There is no

water potential gradient so there is no net movement of

water by osmosis as shown by the two arrows.

Cell A Cell B

Cell A and B are separated by a semi permeable

membrane. Both cells have a water potential of -200KPa

and a solute potential of -75KPa so are isotonic. There is

no water potential gradient so there is no net movement

of water by osmosis as shown by the two arrows.

Cell A Cell B

Cell A is hypotonic compared to cell B. Cell A has the lower

solute potential and the higher water potential. There is a

water potential gradient between cell A and cell B. Water

will move from cell A into cell B by osmosis down a water

potential gradient.


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1. The image below shows three cells all in contact with each other and all having a cell membrane.

The number in the cells are the water potential values of the cytoplasm. Draw arrows on the

image to show any movement of water by osmosis.

1.0.10 Concept A hypertonic solution has a high solute potential and a low

water potential. The diagram below is the same as the one from concept 1.0.9 but the

description is written in terms of cell B.

1.0.11 Concept It is the convention to relate the terms isotonic, hypotonic and

hypertonic to the solute potential not the water potential. The prefixes to these words

are describing the solute potential, i.e. hyper means high and hypo means low solute

potential. The water potential can be derived from the solute potential, i.e. high solute

potential means low water potential.

Cell B is hypertonic compared to cell A. Cell B has the

higher solute potential and the lower water potential.

There is a water potential gradient between cell A and cell

B. Water will move from cell A into cell B by osmosis down

a water potential gradient.

QUESTIONS ON SECTION 1.0

-257 kPa

B -321 kPa

C

-212 kPa

A


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2. The apparatus below was used to investigate osmosis. It consists of a capillary tube attached to a

visking tubing bag both containing solution A. Visking tubing is an artificial cell membrane. The

visking tubing is submerged into a beaker containing solution B. An air bubble is trapped in the

capillary tube. The composition of the solutions are as follows:

• Solution A: glucose 0.01mol dm-3, fructose 0.02 mol dm-3 and sucrose 0.3mol dm-3.

• Solution B: glucose 0.002 mol dm-3, fructose 0.06 mol dm-3 and maltose 0.005 mol dm-3.

The apparatus was left for 30 minutes after which the distance moved by the bubble was recorded

using the ruler.

The visking tubing is not permeable to disaccharides but is permeable to monosaccharides.

Capillary tube

Air bubble

Visking tubing

Solution B

Solution A

Ruler


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ai Suggest with an explanation appropriate solute potentials and water potentials for both

solution A and solution B.

aii Explain in which direction the bubble would move through the capillary tube.

b The starting position of the bubble in the capillary tube was 6.2 cm and at 30 minutes it had moved

to 14.8cm. The internal diameter of the capillary tube is 5.6mm.

bi Using the following equation calculate the volume of water that moved through the

capillary tube. Show your calculation steps and express your answer to 2 decimal places.

V = π x r2 x h π = 3.14

Answer______________________

bii Calculate the rate of water movement during the 30 minutes. Express your answer in

both minutes and hours.

Answers_____________________________________


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c. Name a solute that would:

i. Diffuse into solution A

ii. Diffuse out of solution A

3 A student sweetened some strawberries by sprinkling sugar on top of them one hour before

eating them. The student noticed that the sugar she sprinklered on them was no longer visible

and that there was juice at the bottom of the bowl.

The student thought that the juice was the sugar dissolved in water and that the water had come from the

fruit.

In order to test this hypothesis, she weighed some fresh strawberries and sprinkled them with sugar. One

hour later she rinsed off the juice and reweighed the strawberries. The mass of the strawberries before

adding the sugar was 77g. The mass after rinsing off the juice was 70g.

i. Calculate the percentage decrease in mass of the strawberries. Show your calculation steps.

Answer_____________________


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ii Suggest one possible source of error in the student’s procedure that could make this value for the

percentage decrease in the mass of the strawberries inaccurate. Explain how this source of error

would affect the value for the percentage decrease in the mass of the strawberries.

Source of error___________________________________________________________________________

Effect on value and explanation_____________________________________________________________

iii. Using your knowledge of cell transport mechanisms and the properties of water, explain how the

juice is formed from the water that came form the fruit.


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2.0 OSMOSIS IN ANIMAL CELLS

2.0.1 to 2.0.3 The affect of osmosis on a typical animal cell – the red blood cell

2.0.1 Concept If red blood cells are placed into an isotonic solution the red blood

cell will not change shape because:

2.0.2 Concept If red blood cells are placed into a hypotonic solution the red blood

cell will swell and burst because:

2.0.3 Concept If red blood cells are placed into a hypertonic solution the red blood

cell will shrivel up because:

1. The solute potential of the red blood cell and the surrounding

solution are equal.

2. The water potential of the red blood cell and the surrounding

solution are equal.

3. There is no water potential gradient.

4. There is no net movement of water by osmosis (black arrow), so the

red blood cell does not gain or loss water by osmosis so does not

change shape as the volume of water in the red blood cell remains

the same.

1. The solute potential of the red blood cell is greater than the

surrounding solution.

2. The water potential of the red blood cell is lower than that of the

surrounding solution.

3. There is a water potential gradient.

4. There is net movement of water by osmosis into the red blood cell

(black arrow) from the surrounding solution. The red blood cell will

gain water and start to swell. The cell membrane cannot withstand

the increased pressure caused by the greater volume of water

entering the cell, so it will break causing the cell to burst. The cell

membrane has no strength or rigidity.

1. The solute potential of the red blood cell is lower than the surrounding

solution.

2. The water potential of the red blood cell is higher than the surrounding

solution.

3. There is a water potential gradient.

4. There is a net movement of water out of the red blood cell (black

arrow) into the surrounding solution. Because the cell membrane is

flexible it is able to shrivel up (or become crenated) as water leaves the

cell.


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2.0.4 to 2.0.6 How unicellular fresh water aquatic organisms prevent an increase in

their cell volume caused by osmosis. A research project.

2.0.4 Concept The intracellular contents of a fresh water unicellular aquatic

organism, like a paramecium, has a higher solute potential compared to the water they

live in.

2.0.5 Concept

2.0.6 Concept

Drawing of a paramecium Electron micrograph of a paramecium


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1. Isolation of mitochondria from liver cells was needed to investigate the effects of different

substrates on the rate of respiration. Images of the isolated mitochondria are shown below.

Explain the appearance of the three mitochondria.



18

2


19


20


21


22

3


23

4


24


25

3.0 OSMOSIS IN PLANTS

3.0.1 to 3.0.5 Plant cells have a pressure potential

3.0.1 Concept In plants water potential can be calculated by:

𝛹 = 𝛹𝑠 + 𝛹𝑝

Where: Ψ = water potential, ΨS = solute potential, ΨP = pressure potential 3.0.2 Concept Plant cells become turgid when placed into a hypotonic solution

because water enter the cell by osmosis. The cell contents are pushed against the cell

wall as the vacuole enlarges as it gains water. The cell wall is made of cellulose and is

inelastic so will not stretch so plant cells do not burst.

3.0.3 Concept The inelastic nature of the cell wall is responsible of creating a

pressure potential when the cell becomes turgid. When turgid the pressure potential of

the cell is equal, but a positive value, to the solute potential so the water potential is

zero. For example, if the pressure potential is +400KPa and the solute potential is

-400KPa then: Ψ = -400 + 400. So Ψ = 0KPa.

Cell wall

Cell membrane

Cytoplasm

Tonoplast

Vacuole

Turgid onion cells Drawing of a turgid cell

In a turgid cell this relationship exists:

Ψ = 0

ΨS = ΨP


26

3.0.4 Concept Plant cells become plasmolysed when placed in a hypertonic

solution. Water will move out of the cell and the cell contents will shrink and the cell

membrane will pull away from the cell wall. The cell wall, however, due to its inelastic

properties will not change shape.

3.0.5 Concept When a plant cell is plasmolysed its pressure potential is zero, but

the water potential is equal to the solute potential.

Vacuole

Tonoplast

Cytoplasm

Cell membrane

Cell wall

In a plasmolysed cell this relationship exists:

ΨP= 0

Ψ = ΨS

Drawing of a plasmolysed cell plasmolysed onion cells

Cell membrane (arrow 1) pulled away

from the cell wall (arrow 2).

1 2


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1. A plant cell has a solute potential of -1240KPa and a pressure potential of 350KPa. What is the

water potential of the cell?

Answer_____________________

2. A plasmolysed cell is found to have a solute potential of -960KPa. What is the water potential of

the cell? Explain your answer.

Answer_____________________

3. A plant cell, after being immersed in pure water for several hours, has a solute potential of -

800KPa. What is the water potential and the pressure potential of the cell? Explain your answer.

Answer_____________________

4. Two plant cells, A and B, are next to each other in a tissue. Their solute and pressure potentials

are: Cell A ΨS -630KPa and ΨP 380 KPa, Cell B Ψs -650 KPa and Ψp 320 KPa. In which direction

will water move, from cell A to cell B or from cell B to cell A?

Answer_____________________



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5. The diagram below shows two plant cells, X and Y. the figures show the solute and pressure

potential of for both cells and the water potential for cell Y.

i. Calculate the water potential for cell X. Show your calculation steps.

Answer_____________________

ii State the name of the condition shown by cell Y and explain how this condition could

have arisen.

iii Cell X has the higher pressure potential. Explain how this pressure potential is built up in

cell X.

iv Suggest the effect on a house plant if all their cells were in the condition as shown in cell

Y.

Cell X

ΨP = 1000KPa

ΨS = -1800KPa

Cell Y

Ψp = 0KPa

Ψs = -1000KPa

Ψ = -1000KPa


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6


30


31


32

4.0 OSMOSIS EXPERIMENTS IN PLANTS

4.0.1 to 4.0.9 Percentage plasmolysed cells and Incipient plasmolysis

4.0.1 Concept When a turgid plant cell is placed into increasing concentration of

solute the cell will lose water by osmosis and enter a condition of incipient plasmolysis

and then a condition of plasmolysis.

4.0.2 Concept A cell in incipient plasmolysis has its cell membrane just slightly

starting to pull away from the cell wall as represented in the diagram below:

4.0.3 Concept When a plant cell is in incipient plasmolysis its pressure potential is

zero, but the water potential is equal to the solute potential. See also concept 3.0.5.

4.0.4 Concept A cell in incipient plasmolysis cannot be seen under the microscope

and often just looks like a turgid cell.

4.0.5 Concept The concentration of sucrose that causes incipient plasmolysis can be

determined from a graph of percentage plasmolysed cells against sucrose concentration

see prac concept 4.0.7 and 4.0.8.

A

B

A plant cell showing incipient plasmolysis. The

cell membrane has just started to pull away

from the cell wall at points A and B

In a incipient plasmolysed cell this relationship exists:

ΨP= 0

Ψ = ΨS


33

4.0.6 Prac Concept Strips of red onion membrane are placed in small containers

each with a different concentration of sucrose. They are left for 20 minutes and

the cells are viewed under the microscope. The red pigment in the onion allows

the vacuole to be seen. The total number of cells in the field of view are counted

as are the number of cells that are plasmolysed. The percentage plasmolysed

cells are calculated by the following equation:

Diagram of the incipient plasmolysis practical.

𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑙𝑎𝑠𝑚𝑜𝑙𝑦𝑠𝑒𝑑 𝑐𝑒𝑙𝑙𝑠 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑠𝑚𝑜𝑙𝑦𝑠𝑒𝑑 𝑐𝑒𝑙𝑙𝑠

𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑒𝑙𝑙𝑠 𝑖𝑛 𝑓𝑖𝑒𝑙𝑑 𝑜𝑓 𝑣𝑖𝑒𝑤 × 100

Image of plasmolysed red onion cells. The

percentage plasmolysed cells in this image is 100%

as there are no turgid onion cells. The red pigment

in these cells allows the visualisation of the shrinking

of the cell’s contents.

Key to diagram

Red onion in sucrose solution Microscope field of view Plasmolysed cell Turgid cell


34

4.0.7 Prac Concept A graph is plotted with percentage plasmolysed cells on the Y

axis and sucrose concentration on the X axis as shown below.

4.0.8 Prac Concept On the graph above a line is drawn from 50% plasmolysed cells

across to the line drawn between the data points and then down to the sucrose

concentration on the x axis. So, the concentration that causes incipient

plasmolysis from the above graph is 0.47 mol dm-3.

4.0.9 Interpretation of 50% plasmolysed cells being the point of incipient

plasmolysis. This is an arbitrary value where it is assumed that the sucrose

concentration has caused 50% of the cells to be plasmolysed while the other

50% are assumed to be turgid. So, the sucrose concentration that causes

incipient plasmolysis can only be determined from a graph like the one above.

At incipient plasmolysis:

1. The pressure potential of the plant cell will be zero. When this happens the

solute potential of the sucrose will be equal to the water potential of the

cell.

100

80

60

40

0

20

0 0.2 0.4 0.6 0.8 0

Pe

rcen

tage

pla

smo

lyse

d c

ells

/ %

Concentration of sucrose/ mol dm-3


35

2. Point 1 describes an equilibrium where there is no water potential gradient and so no net movement of water by osmosis. To understand this, we need to convert concentration in mol dm-3 in the units of Kpa. So, a sucrose concentration of 0.47mol dm-3 is equal to a solute potential of -1380KPa. Therefore, the cell has a solute potential of -1380KPa as well as a water potential of -1380KPa. This is summarized in the diagram below:

3. The results of the incipient plasmolysis experiment are summarized on graph below

Cell in incipient plasmolysis

Sucrose solution of

concentration 0.47mol dm-3

as determined from the

graph so this has a solute

potential of -1380KPa.

Pressure potential inside cell is

0KPa. Solute potential inside the

cell is equal to the solute potential

of the sucrose solution outside

which is -1380KPa.

No net movement of

water by osmosis

because the water

potential is equal to the

solute potential, so the

water potential is

-1380KPa.

Turgid Cell

ΨP = Ψs

Ψ = 0

Ψ = 0

Increasing concentration of sucrose up to 0.47mol dm-3 Incipient

plasmolysis

Ψ = Ψs

ΨP = 0

Ψ = -1380KPa

Ψp= 0KPa

Ψs= -1380KPa


36

4.0.10 to 4.0.15 Percentage change in mass

4.0.10 Concept When a cylinder of potato is placed into a hypotonic solution for several hours it will increase in mass because the potato cells will gain water by osmosis. When a cylinder of potato is placed into a hypertonic solution is will loss mass because water will leave the potato cells. 4.0.11 Concept The percentage change in mass of the potato cylinder is

calculated by:

% 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑎𝑠𝑠 = 𝑓𝑖𝑛𝑎𝑙 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 × 100

4.0.12 Prac Concept Potato cylinders are weighed and then placed in tubes

each with a different concentration of sucrose. They remain in the tubes for 2 hours

after which they are removed dried and the final mass is measured. The percentage

change in mass is then calculated.

4.0.13 Prac Concept The percentage change in mass is plotted against sucrose concentration.

4.0.14 Prac Concept From the graph stated in 4.0.13 the solute potential of

the potato cells can be determined.

Potato cylinder in

Hypotonic solution.

Mass of cylinder is

3.8g.

Potato cylinder in

Hypertonic

solution. Mass of

cylinder is 0.85g

Original mass of potato cylinder is 2g


37

4.0.15 Prac Concept From the graph below a zero percentage change in mass is caused by a sucrose concentration of 0.425 mol dm-3. When there is no change in mass there must be no net movement of water into or out of the potato cells by osmosis. This is due to the solute potential of the sucrose solution being equal to the solute potential of the potato cell, i.e. they are isotonic.

20

10

0

-10

-20

0 0.2 0.4 0.6 0.8 0 0.1 0.3 0.5 0.7

Pe

rcen

tage

ch

ange

in m

ass/

%

Concentration of sucrose/ mol dm-3


38

QUESTIONS ON SECTION 4

1


39


40

2


41


42

3


43


44

4


45


46

1. Red blood cells (erythrocytes) can have abnormalities of their cell membrane. The abnormalities are

hereditary. Two examples of abnormal red blood cell membranes are: hereditary spherocytosis and

hereditary elliptocytosis both of which cause haemolytic anemias. With these conditions the red

blood cells have altered cell membrane strength. One investigation to diagnose these conditions is

the membrane fragility test. The test involves adding different concentrations of sodium chloride

along with a sample of the patient’s blood to a tube. The tubes were left for 20 minutes, then the

red blood cells we removed by filtration and the percentage transmission of the filtrate was

measured with a colorimeter using light of 470nm (blue light). The images below are of the filtrates

taken from two different people.

a. The concentration of sodium chloride used in the membrane fragility test for both people

were: 1%, 0.85%, 0.66%, 0.55%, 0.35% and 0%. Complete the table below by assigning a

concentration to each of the 6 tubes.

Tube number Concentration of sodium chloride/%

1

2

3

4

5

6

Person 1

Person 2

Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Tube 6

Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Tube 6

APPLICATION & EXTENSION


47

b. The percentage transmission of blue light passing through the filtrate for each patient are

shown in the table below:

i. Using your answer to part a and the information in the table above plot a suitable

graph of the results on the graph paper provided on the next page.

ii. A medical student hypothesised that the solute potential of the red blood cell can be

determined from a membrane fragility experiment by the concentration of sodium

chloride that caused a 50 percent transmission reading. Using your own knowledge

and all the information in this question, evaluate this hypothesis.

Tube Number Percentage transmission of blue light passing through sodium chloride filtrate/%

Patient 1 Patient 2

1 100 100

2 100 90

3 90 80

4 85 65

5 35 20

6 10 5


48

iii For person 1 and 2 state precisely what concentration range of sodium chloride will cause haemolysis of the red blood cells.


49

iv Explain which person, 1 or 2, has a higher fragility of their red blood cell membrane.

2. The cytoplasm of three cells, A, B and C have a concentration of sucrose of: 0.2mol dm-3, 0.4mol

dm-3 and 0.65 moldm-3 respectively. Solute potential can be calculated using the equation:

Calculate the solute potential of cell A, B and C at a temperature of 15oC and 37oC. Show your calculation

steps and quote your answer to the nearest hundredth.

Answer for 15oC____________________

Answer for 37oC____________________

3. A cell is submerged in solution A which has a concentration of 250 mmol dm-3. In solution A the

cell has a volume of 1000µm3 and a concentration of 250 mmol dm-3. Water was added to

solution A to form solution B which had a concentration of 200 mmol dm-3. In solution B the cell

had a concentration of 200 mmol dm-3.

i. Using the equation V1C1=V2C2 calculate the volume of the cell in solution B. show your

calculation steps.

V1 = volume of cell in solution A. C1 = concentration of the cell. V2 = volume of the cell in

solution B. C2 = Concentration of the cell.

Ψs = -iCRT Where: -i = 1 C = concentration in mol dm-3 R = Pressure constant = 8.31 T = temperature in oKelvin = 273 + the temperature of the cell in oC.


50

Answer________________

4. Plant cell A has a solute concentration of 0.2 mol dm-3 and plant cell B has a solute concentration

of 0.4 mol dm-3. In terms of water potential, explain if these two cells can be in equilibrium with

each other.

5. For a protein to become functional it must be modified after its synthesis. One such modification

is called glycosylation and occurs when proteins have sugar chains added to key amino acids that

make up the primary structure. The drawings below represent red blood cells with a membrane

protein attached to which are different sugar chains. Red Blood cell 1 has been newly synthesised

in the bone marrow, while red blood cell 2 is three month old. The only difference between the

two cells is the glycosylation of the membrane protein with regard to the terminal sugar unit.

Terminal sugar unit

sialic acid

RED BLOOD CELL 1

PROTEIN

GAL GAL

SIA SIA

RED BLOOD CELL 2

PROTEIN

3 1

2

GAL GAL

Terminal sugar unit

galactose

Cell membrane


51

Due to wear and tear of the red blood cell the terminal sialic acid sugar breaks off, this will target the cell

for destruction by the liver. A single liver cell is shown below.

During the life cycle of a red blood cell the haemoglobin inside the red blood cell becomes glycosylated by

glucose present in the blood. The concentration of glucose bound to haemoglobin forms the test for

diabetes called the HBA1c test.

Over a 12 month period a patient had her HBA1c measured and her results are shown below. Diabetes is

diagnosed with a HBA1c reading of >55 moldm-3 and pre-diabetes is diagnosed with a HBA1c reading of 30-

40 moldm-3.

LIVER CEll

Protein Receptor Protein

Binding site

Co

nce

ntr

atio

n o

f gl

uco

se b

ou

nd

to

hae

mo

glo

bin

/mo

ldm

-3

30

50

70

0

Time/Months

90

110

130

0

150

3 6 9 12 15

170


52

The following questions require you use all the information in this questions as well as your own

knowledge.

(a) i. State the correct name given to the chain of sugars attached to membrane bound proteins as

shown with red blood cell 1 and 2.

ii. On red blood cell 2 the triangle and the circle represent hexose sugars. Suggest the name of

the bond between circles 1 and 2 and circle 1 and triangle 3.

iii. The terminal sugar in red blood cell 2 is galactose. Compare the structure of galactose with

β-glucose.

(b) Red blood cells are unusual in that they do not have any organelles inside them, just

haemoglobin. However, they do have a cell membrane which has the fluid mosaic structure.

i. State the names of the scientists who first describe the cell membrane as having a fluid

mosaic structure.

ii. State the name of the biological molecule that forms the bilayer of a cell membrane.

iii. Explain how the structure of the biological molecule you stated in your answer to bii gives it

the physical properties required to form a bilayer.

iv. The circular and diamond shaped proteins in the images above represent cell membrane

proteins. In relation to their positioning in the membrane of the red blood cell and the liver

cell, evaluate the physical properties required by these two proteins to explain how they have

come to be positioned the way they have in the two cell membranes.


53

v. Using the diagrams on page 50 and 51, and your own knowledge, explain why the removal

of sialic acid causes the red blood cells to be targeted for destruction by liver cells.

vi. Describe how organelles in liver cells could lead to the destruction of red blood cells.

(c) Diabetes is described as a metabolic disorder. It’s a complex condition with several different

causes, for example, type II diabetes can be caused by insulin being less effective at lowering

blood glucose concentration. Whatever the cause of diabetes the result is the same – high

blood glucose concentrations. This can be very dangerous as glucose can cause damage to

nerves and small blood capillaries. To achieve an accurate measure of blood glucose

concentration scientists have exploited the life cycle of a red blood cell and the glycosylation of

the haemoglobin within it.

Using the graph on page 51 answer the following questions.

i. Suggest how long a red blood cell life cycle is.

ii. Over which time period was there the smallest drop in glucose bond to haemoglobin

(HBA1c)?

iii. When did the patient become pre-diabetic.

iv. Calculate the percentage decrease in glucose bound to haemoglobin between one and

three months. Show your calculation steps.


54

v. Which time period showed the slowest rate of decrease of glucose. Explain your answer

mathematically. Show your calculation steps.

vi. Calculate the percentage decrease in glucose bound to haemoglobin between 1-3 months

and 6-9 months. Also express your answer as rate. Show all your calculation steps.

vii. The patient had further HBA1c tests at 18 and 21 months. The concentration of glucose

bound to haemoglobin remained at 30 moldm-3. The rate of glucose uptake to haemoglobin

from 12 to 18 months was calculated to be 0mol dm-3 month-1. Evaluate this answer.


55

(d) The mammalian kidney has many roles, these include regulating water levels in the body,

excreting waste products like urea into the urine and reabsorbing important substances like

glucose, amino acids and water that have been filtered out of the blood. As the kidney filters

the blood under high pressure a fluid called the filtrate enters the many kidney tubules through

the small gapes in the capillary. Once in the kidney tubules the filtrate composition is altered

by reabsorption of substances along the whole length of the tubule through the cells lining the

tubule. The filtrate now becomes urine which enters the bladder. These events are

summarised in the diagram below.

a. i. On the above diagram place an “*” where filtration is occurring and a “#” where

reabsorption is occurring.

Blood from body to

kidney

Blood from kidney

back to body Urine to

the bladder

Start of kidney tubule

Blood capillary 1

Blood capillary 2

End of kidney tubule

Cells lining

the kidney

tubule

Filtrate


56

ii If the concentration of urea along the kidney tubule is plotted, the line would follow the

equation

Where: 𝑚 is the gradient and C is the value of y when 𝑥 is 0.

The concentration of urea in the filtrate at the start of the kidney tubule is 20 arbitrary units

(AU) and the value of m is 0.2 AU/%. Use these values provided to calculate the

concentration of urea at the end of the kidney tubule and draw a line on the graph below for

the concentration of urea along the length of the kidney tubule. The concentration at the

start of the kidney tubule has been plotted for you.

𝑦 = 𝑚𝑥 + 𝑐

Distance along the kidney tubule (% of total distance)

Start of kidney tubule

End of kidney tubule

Ure

a co

nce

ntr

atio

n (

AU

)


57

The diagram below is a simplified view of the cell membrane of the cells that line the kidney tubule. It

shows three membrane proteins along with the substances they are involved in transporting across the

membrane and into the tubule cell.

iii. Using the information about the kidney tubule cell membrane and your own knowledge

explain your answer to aii.

An experiment was undertaken to investigate the membrane transport of glucose. The concentration of

glucose in capillary 1 was increased while the concentration of glucose in capillary 2 was measured. The

graph below shows the results of this investigation.

Glucose

Water

Amino acid

Concentration of glucose in capillary 2 and urine/mg cm-3

Co

nce

ntr

atio

n o

f gl

uco

se in

cap

illar

y 1

/mg

cm-3

12

10

8

6

0 2 4 6 8 10 12 0

2

4

Urine

Capillary 2


58

iv State the concentration of glucose in capillary 1 above which glucose will be found in the

urine.

Answer________________

v Using all the information in this question and our own knowledge explain why glucose

concentrations above the answer you stated in part iv would result in glucose being

present in the urine.

osmosis w b 2 - thiacin...t r l j. version 2, 2018 3 the following points are what you need to know,...

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