CHAPTER 3 1

CHAPTER 3 : MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRANEConcept Savy

1. Substances from the outside of a cell can move into the cell, and from inside the cell can move out of the cell. The substances have to cross the plasma membrane for this type of movements.

2. The plant cell is enclosed by cell wall that is fully permeable and cell membrane that is partially permeable.

3. Animal cells have only partially permeable cell membranes.

Important because they separate the cell from the outside world. All membranes contain proteins and lipids.

Membrane consists of a lipid bilayer. The membrane is a fluid mosaic of phospholipids and proteins The phospholipids have hydrophilic polar heads pointing out and the

hydrophobic tails.Materials pass into or out of cells by the following processes:

By simple diffusion – directly through lipid bilayer: Lipid soluble molecules such as fatty acid,glycerol,steroid compounds and vitamins A, D,

E, and K can dissolve in the lipid layer and move across it. Non-polar molecules such as O2, H2O, CO2, rapidly diffuse across lipid bilayer

By simple diffusion through pores protein: Non-polar molecules such as O2, H2O, CO2

Water soluble molecule and ions such as K+, Na+, Ca2+

By osmosis , the net movement of freely moving water molecules from a region of low solute concentration to a region of high solute concentration through a semi-permeable membrane.The water molecules are small enough to move through pores of membrane.

The concentration of the solution has different effects on plant cells and animal cells. When a plant cells is immersed in hypotonic solution, the water diffuses into the cell by

osmosis. The cell becomes turgid. When a plants is immersed in hypertonic solution, the water diffuses out of the cell . The

cell becomes flaccid. When a plant cell is immersed in isotonic solution, the water diffuse into the cell and

diffuse out from the cell at the same rate. The cell remains the same. The animal cell did not have cell wall, so it will burst in hypotonic solution and remain

the same in isotonic solution.

By facilitated diffusion ,for larger compounds such as sugars and amino acids, substance binds to transport protein,and alters protein carrier, exposing the other surface. Particles move

Concept Savy

Cell Membrane

CHAPTER 3 2

from areas of high concentration to areas of low concentration.

By active transport , the movement of substances through membranes in living cells,against a concentration gradient,with the aid of carrier proteins and energy from ATP

PRACTICES:Activity 1:

a) Name transport P and Q.

P : ………………..……………………………

Q : ………………………………………………

b) Define

i) Transport P :

……………………………………………………………………………….

ii) Transport Q :

………………………………………………………………………………….

c) What is the main difference between transport P and transport Q?

…………………………………………………………………………………

d) What is process A and its definition ?

…………………………………………………………………………………

………………………………………………………………………………..

Activity 2

2. Diagram 1 shows model of a plasma membrane.

Transport P Transport Q

A Diffusion

Transport

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Diagram 1

Complete the text below to describe the structure of the plasma

membrane.

Consist of (1) -----------------------bilayer and (2) ------------------ molecules

- small non (3) --------------- molecules and lipid (4) ------------- molecules

- can diffuse (5) --------------- lipid bilayer

(6)----------------ions, for example, (7) -------------------- and bigger

molecules , e.g. (8)----------------- and (9) ----------------- ---------------can

move into the cell with the(10)-------------------of a (11)------------------- protein.

Hence, only certain substances can move across the (12)

……………….(permeable to this substance). Bigger molecules cannot

(13) …………………across, making the membrane (14)-------------------------

b) Name the processes that occur at P , Q and R.

i) P : …………………………………………………………………………

CHAPTER 3 4

ii) Q :

………………………………………………………………………………..

iii) R : …………………………………………………………………………………

Activity 3:

3) Figure 5 shows the condition of erythrocyte cells K , L and M after being soaked in 3

solutions at different concentration.

After being soaked After being soaked in After being soaked in

in solution K in solution L in solution M

Diagram 2 (a)

a) Name a physical process that causes the changes in erythrocyte cells L and M.

…………………………………………………………………………….

b) What are the characteristics of the solutions used to soak :

Erythrocyte K : ……………………………………………………………

Erythrocyte K

Erythrocyte L

Erythrocyte M

CHAPTER 3 5

Erythrocyte L : ……………………………………………………………

c) Explain the process that causes the changes of erythrocyte cell M

by completing the text below:

Water molecules move (1)----------- the cell by (2) ----------------., causing

the red blood cells to (3)----------- -----------. Eventually the built up of the osmotic pressure in the cells causes them to (4) ----------------.

d) At the start of the experiment at the end of the experiment

At the start of the experiment at the end of the experiment

Cell is soaked in 0.5% sucrose solution for 30 minutes

Cut surface

Cut surface

Cell is soaked in 0.1% sucrose solution for 30 minutes

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Diagram 2 (b)

On the diagram 2(b) above, draw a shape of balsam stem at the end of experiment.

d) Explain why there are no changes to the balsam stem at the :

i) epidermis:

…………………………………………………………………………………….

ii) cut surface :

………………………………………………………………………

Activity 4:

4. Diagram 3 shows a plant cell has been placed in salt solutions of different

concentrations. The figure shows the appearance of the cells after 20 minutes.

Diagram 3

a) Identify the structures labelled L to P :

L : …………………………….. O : ………………………………

M :…………………………….. P : ………………………………

N : …………………………….

b) Which cell was placed in

i) a concentrated salt solution ……………………………………

ii) a dilute salt solution …………………………………….

c) i) What happens to cell B ?

…………………………………………………………………………………..

ii) Cell B appears as shown in Diagram 3 . Complete the text below to explain the

CHAPTER 3 7

situation of cell B.

The (1)------------------------- concentration of (2)--------------------------- in the solution

((3…………………………solution) causes water to (4)…………………. the cell by

(5) ……………………………Both the (6) …………………….. and cytoplasm

(7)………………….. and the (8) …………………. membrane (9)……………...

away from the(10) ………... ………………… The cell becomes

(11)………………………..

d) How can cell B be restored to its normal condition ?

………………………………………………………………………………………

Activity 5

5. Diagram 4 shows the fluid-mosaic model of membrane structure.

M Y

X

Z

N

CHAPTER 3 8

Diagram 4

a) Name the structures M, N, X , Y and Z.

M : …………………………………………….

N: ……………………………………………..

X : ……………………………………………..

Y : ……………………………………………..

Z : ……………………………………………..

b)There are two various types of proteins:

i) ……………………………….: the protein molecule that forms a channel or pore

ii) ……………………………………: the protein molecule that acts as carrier.

c )Phospholipid bilayer also contains cholesterol which links the fatty acid

together. Complete the text below to state the roles of cholesterol in the plasma

membrane.

Cholesterol helps to (i)………………………… and (ii) ……………………..

the plasma membrane ,making it more (iii) …………………….

d ) Plasma membrane encircles the protoplasm of a cell. As a whole plasma

membrane exhibits fluidity as if it is a layer of oil in a beaker of water.

What is meant by plasma membrane exhibits fluidity ? ( 4 marks)

Explain the structure of plasma membrane that gives it a

characteristic of fluidity

Plasma membrane exhibits fluidity:

Consists of (1) ………………… …….. ………………. (phospholipids)

A polar head (2)……………………..property and a pair of non-

polar that tails that (3)………... ……………………property.

Various (4)……………………………. ……………are inserted within the two

layers of lipids.

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Form a membrane (5)………………………………… but very free moving/

dynamic.

Consists of (6)…………………………………// not easily broken because it

contains cholesterol.

e) Name two general factors that determine whether a molecule can pass through

the plasma membrane or not.

……………………………………………………………………………………

Activity 5:

Matching !!!:

b) Match the terms with their most suitable descriptions.

a) Simple

diffusion

i) Passive movement of polar molecules or ions down a

concentration gradient with the help of pore protein or

carrier protein.

b) Facilitated

diffusion

ii) Unassisted passive movement of molecules or ions

down a concentration gradient.

c) Osmosis iii) Movement of a solute across the plasma membrane

against its concentration gradient. The process requires

a carrier protein and expenditure of energy, usually

CHAPTER 3 10

from ATP .

d) Active

transport

iv) Movement of water across a partially permeable

membrane down a water concentration gradient.

Hypo = below

Hyper = above

Iso = same as

tonos = tension

Hydro = water

Philic =having an affinity for

Phobic = phobia,lacking affinity for

1. A hypotonic solutions is a solution that contains a lower concentration of solute molecules

( and higher concentration of water molecules)

2. A hypertonic solution contains a higher concentration of solute molecules and a lower

concentration of water molecules).

3. An isotonic solution has the same solute concentration (same concentration of water

molecules) as the other solution.

4. At equilibrium, water molecules in the two isotonic solutions move at the same rate in both

directions and there is no net movement of the water molecules.

Activity 6:

1) In hypertonic solutions, the red blood cells become (a) and plant cells become

(b) .

2) In isotonic solutions, there is no ................ movement of water molecules.

3) In hypotonic solutions, red blood cells become (c) and plant cells become (d).

4) Adding excessive chemical fertilizers can cause (e) of root cells and (f) of the plant.

5) (a) is removed from food placed in (b) sugar or salt solution so that it can be preserved

Share

the

facts:

Checking concept

CHAPTER 3 11

for a longer period.

MCE: Name the specific physical process to describe the movement of water molecules

into or out of the red blood cell .

Answer: Passive transport. ( 0 mark ) . Give more specific answer. Correct answer

is: Osmosis .

Activity 7

Application of Osmosis

6. Diagram 5 shows the state experienced by food substance soaked in solution X .

Diagram 5

a) What is the most suitable content of solution X ?

………………………………………………………………………………………….

b) Explain the state that occurs in the diagram

………………………………………………………………………………………….

………………………………………………………………………………………….

c) State the process that occurs and the importance of its application in everyday life.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

d) What is the definition of that process?

………………………………………………………………………………………….

………………………………………………………………………………………….

e) State the importance of movement of substances into or out of a cell ? ( 3 marks)

Mind your Common Error

CHAPTER 3 12

f)

A few slices of mangoes is placed in a high concentration of salt solution

for 30 minutes. The slices of mangoes is then transferred into distilled

water for another 30 minutes.

Name the process that takes place in the cell of mangoes?

A Plasmolysis

B Deplasmolysis

C Crenation

D Turgidity

Activity 8

Part B ( Mini Essay)

7 ) Explain why fruits can be preserved using concentrated sugar solution ?

………………………………………………………………………………………..

8) What is the meaning of haemolysis and crenation respectively?

………………………………………………………………………………………...

…………………………………………………………………………………………

…………………………………………………………………………………………

9) a) What will happen to a plant if an excess of fertilizers is added to the soil?

Essay Exploration

CHAPTER 3 13

…………………………………………………………………………………………

………………………………………………………………………………………….

………………………………………………………………………………………….

c) Explain why tugor is important to plants.

…………………………………………………………………………………..

…………………………………………………………………………………….

10.) a Justify that osmosis is a special case of diffusion.

………………………………………………………………………………………….

………………………………………………………………………………………….

b) What do you understand by “ concentration gradient “ ?

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

Activity 9 : Skill in CCTS Compare and Contrast

a) Explain the similarities and differences between facilitated diffusion and

active transport in the movement of molecules across the cell membrane . [ 8 marks]

Practical Questions:

Q1. A group of students carried out an experiment to determine the concentration of external solution which is isotonic to the cell sap of a plant.

Diagram 1.1 shows the set up of apparatus for this experiment. By using a cork borer, a cylindrical potato strip was obtained. The potato strip was cut at 5 cm long.The final length of potato strip immersed in 0.1M after 30 minutes is shown in Diagram 1.2.

The experiment in diagram 1.1 was repeated by using different concentrations of sucrose solution.

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Table 1 shows the results of the experiment.

Concentrationof sucrose solution(M)

Final Length of cylindrical potato strips (cm)

Diagram 1.2

Potato strip(initial length – 5cm)

0.2M Sucrose solution

Diagram 1.1

CHAPTER 3 15

0.2

0.3

0.4

0.5

TABLE 1.1

(a) (i) Based on Table 1.1 state two observations that can be made from the experiment concerning the length of cylindrical potato strips.

1…………………………………………………………………………………………….……….

………………………………………………………………………………………………………………

………………………………………………………………………………………………

2…………………………………………………………………………………………….

………………………………………………………………………………………………………………

[ 3 marks](ii) State the inference which corresponds to the observation in 1(a)(i).

CHAPTER 3 16

1..…………………………………………………………………………………………….

………………………………………………………………………………………………………………

……………………………………………………………………………………………………

2…………………………………………………………………………………………….……….

………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

………………………………………………………………………………………

[ 3 marks](b) By using the information provided in Table 1.1, complete Table 1.2 by recoding the final length of cylindrical potato strips .

Concentrationof sucrose solution(M)

Final Length of cylindrical potato strips (cm)

0.2

0.3

0.4

0.5

[ 3 marks]TABLE1.2

(c) Complete Table 1.3 based on this experiment. State the type of variable of the experiment and method to handle the variable.

Variable Method to handle the variable

Manipulated variable

………………………………………………….

………………………………………………….

………………………………………………….

…………………………………………………

…………………………………………………

…………………………………………………

Responding variable

CHAPTER 3 17

………………………………………………….

………………………………………………….

………………………………………………….

…………………………………………………

…………………………………………………

………………………………………..............

Controlled variable

………………………………………………….

………………………………………………….

………………………………………………….

…………………………………………………

…………………………………………………

…………………………………………………

[ 3 marks]Table 1.3

(d) State the hypothesis for this experiment.

…………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………

…………………………………………………………………………………………………

[ 3 marks]

(f) Based on this experiment, what can you deduce about isotonic solution ?

…………………………………………………………………………………………………………………………

……………………………………………………………………………………

[ 3 marks]

Answer Scheme for Chapter 3 :

1 . a) Transport P : Active transport Q : Passive transport

b) i ) P : The movement of substances through membranes in living cells, against a concentration gradient, with the aid of carrier proteins and energy from ATP.

ii) Q : Transporting substances in and out of the cell membrane or intermediate material that does not involve energy usage.

c) Energy is required and used in active transport.

c) Osmosis: the net movement of freely moving water molecules from region of low d) solute concentration to a region of high solute concentration through a semi-permeable

membrane, until equilibrium is achieved.2) a)

1) lipid 7) mineral ions 13) move2) protein 8) glucose 14) semi-permeable3) polar 9) amino acid4) soluble 10) aid/help5) through 11) carrier6) charged ion 12) membrane

CHAPTER 3 18

Hence, only certain substances can move across the membrane (permeable to this substance.) Bigger molecules cannot move across, making the membrane to be semi-permeable.b) i) P : simple diffusion Q : simple diffusion through pores R : active transport

3 a) Osmosis b) K : isotonic L : Hypertonic c) 1 ) into 2) osmosis 3) swell up 4) burst

4) a) L : cytoplasm M : vacuole N : Nucleus O : Cell wall P : Plasma membrane

b) i ) B ii) A

c) i ) The cell has undergone plasmolysis ii) 1) higher 7) shrink2) solution 8) plasma3) hypertonic solution 9) pulls4)leave 10) cell wall5) osmosis 11) flaccid6)vacuole

e) by placing it in distilled water

5) a) M : phospholipids bilayer N : Hydrophilic head X : carrier protein Z : pore protein

b) i ) Pore protein ii) Carrier protein

c) i ) stabilize ii) strengthen iii) flexible

d) 1) two layer √ 2) give the hydrophilic √ 3) hydrophobic √

4) types of protein √ 5) that is not static √ 6) strong structure √

e) the size and the polarity of the moleculesActivity 7:Matching:b) a) (ii) b) (i) c) iv d ( iii) Activity 6:

Checking concept:

1 a) crenated b) plasmolysed 2 nett 3) a) haemolysed b) turgid 4) a) plasmolysis b) wilting/ death 5) a) water b) concentrated

6. a) Saturated salt or sugar solution

CHAPTER 3 19

b) Water content in fresh food diffuse out and enter the saturated solution. Dissolved substances like sugar and salt will diffuse into fresh food at the same time. After being left in the saturated solution, the size of the fresh food will shrink due to water diffusing out. c) The process of diffusion has taken place and this process can be applied to food preservation b using salt and sugar. d) Osmosis is the process of movement of water molecules from the fresh food into solution X.Application of Osmosis:

Q6 e: ) State the importance of movement of substances into or out of a cell ? ( 3 marks)

1. To supply cells with oxygen for respiration ( 1 m )2. To provide nutrients such as water, glucose, amino acids, lipids, vitamins and mineral salt for

growth and cellular respiration. ( 1 m)3. To regulate the solute concentration and pH for maintaining a stable internal environment for

enzymes to function optimally. ( 1 m)4. To remove nitrogenous waste products and carbon dioxide. ( 1 m )5. To allow useful substances, for examples , digestive enzymes hormone and mucus to be secreted.

7. The concentrated sugar solution is hypertonic to bacteria. Water will be drawn out of the Bacteria through the bacterial membrane, thus killing the microorganism.

8 a) Crenation is the shrinkage of animal cells that occurs when the surrounding solution is hypertonic to the cytoplasm.

Haemolysis is the bursting of red blood cells because the plasma membrane is too thin and delicate to withstand the osmotic pressure that develops within the cell.

9 a) The cells of the plant will lose water b osmosis. Eventually the plant will wilt. b) tugor plays a very important part in plants because it supports them and maintain their shape. Tugor is also responsible for holding the leaves in an opened-out position.

10 a) Osmosis is a special case of diffusion because it involves the diffusion of water molecules from a region where they are more concentrated to a region where they are less concentrated across a semi-permeable membrane.

b) Concentration gradient is the difference in concentration of the substance which causes diffusion. The greater the concentration gradient the greater the rate of diffusion.

Activity 9

Answer for Activity 6 CCTS : Answer :

a) Mini Essay ( 8 marks ): S= similarities E= Explanation

Able to explain the similarities between facilitated diffusion and active transport

S1: Both ways of transportation need carrier protein in the cell membrane

E1: to bind with molecules/ions/substances

CHAPTER 3 20

S2: Both processes occur in living cell

E2: because carrier proteins need/can change shape to (allow substances to move across)

S3: Both transport specific molecules only/ examples of molecules/ions/names

ions/molecules

E3: because the carrier protein have specific sites

[ any 2 S and suitable Explanation ]

Able to explain the differences between facilitated diffusion and active transport

CHAPTER 3 21

Facilitated Diffusion Active Transport

D1: Down concentration gradient//

E1: Molecules move from higher

concentration(region) to lower

concentration

Against concentration gradient

Molecules move from lower

concentration(region) to higher

concentration

D2: Molecules move in both

directions across the plasma

membrane//

E2: Molecule can move through

pore proteins /carrier protein

Molecules move in one direction across the

plasma membrane//

Molecule moves through carrier protein

D3: No ATP /energy used //

E3: Molecules can move through

pore protein

ATP / energy used

Energy needed for binding/bind with active

site

D4: molecules need carrier

protein and pore protein

Molecules need carrier protein

***** any suitable answer

Answer for Practical Questions

(a) (i) Based on Table 1.1 state two observations that can be made from the experiment concerning the length of cylindrical potato strip.

1. The final length of potato strip immersed in 0.2M/0.3M sucrose solution is

5.6 cm / 5.3cm

2. The final length of potato strip immersed in 0.4M/0.5M sucrose solution is

4.5cm /4.2cm

[ 3 marks]

(ii) State the inference which corresponds to the observation in 1(a)(i).

1. For the 0.2M / 0.3M sucrose solution, the increase in length is caused by the

Diffusion / movement of water molecules into the cell sap by osmosis.

2. For the 0.4M / 0.5M sucrose solution, the decrease in length is caused by the

diffusion / movement of water molecules out of the cell sap by osmosis.

[ 3 marks]

CHAPTER 3 22

(b) Using the information provided in Table 1.1, complete Table 1.2 by recoding the final

length of potato strips .

Concentration of

sucrose solution(M)

Final Length of potato

strips (cm)

0.2 5.6

0.3 5.3

0.4 4.6

0.5 4.2

[ 3 marks]

TABLE1.2

(c) Complete Table 1.3 based on this experiment

Variable Method to handle the variable

Manipulated variable

Concentration of sucrose solution

Use different concentration of sucrose

solutions: 0.1M, 0.2M,0.3M and 0.4M

Responding variable

Final length of potato strips/ change in

length of potato strips/ percentage

change in length of potato strips.

Measure the final length / change in

length of potato strips by using ruler

// calculate the percentage in length by

using formula :

Final length – Initial length X 100%

Initial length

Controlled variable

Duration of immersion/

Size /length of potato strips/ volume of

sucrose solution

Fix the time ( 30 min) duration for the

immersion of the potato strips./

Fix the diameter of the potato strips by

using the same cork borer for all the

strips / fix the initial length ( 5 cm) of

potato strips /measure the same volume

of sucrose solutions using a measuring

cylinder,

CHAPTER 3 23

[ 3 marks]

Table 1.3

(d) State the hypothesis for this experiment.

The concentration of an external solution which is isotonic to the cell sap is the

concentration of the sucrose solution that does not change the length of the

potato strip

[ 3 marks]

(f) Based on this experiment, what can you deduce about osmosis ?

A process in which water molecules entering or leaving the potato discs across the

plasma membranes of the potato cells occur when there is a difference in osmotic

concentration/ concentration gradient between the cell sap and the sucrose

solutions.

[ 3 marks]

(g) The experiment using 0.7M sucrose solution was repeated with spinach stem strips.

Predict the changes that occur to the strips?

The spinach stem strips will become shrunken and soft/flaccid and curved

inwards with the epidermis on the outside as water leaves the exposed tissues

and not through the impermeable epidermis

[ 3 marks]

CHAPTER 3 24

CHAPTER 3 25

CHAPTER 3 26