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MEMBRANE STRUCTURE &

TRANSPORT

ANSWERS TO MEMBRANE

STRUCTURE QUESTIONS

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1. In the space below draw a labelled diagram of the cell membrane as described by the

fluid mosaic model.

2. Cells can be stained with chemical that dissolve in water. After being stained the

membrane was observed using the electron microscope. The membrane had the

appearance shown below.

Explain why the membrane has this appearance.

The two black lines represent the phosphate heads which are hydrophilic so the stain

which is also hydrophilic can bind to them. The white region represents the fatty acid tails

which are hydrophobic.

7.5nm

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3. Complete the following table by placing a tick to give the substance the correct

property.

4. The image below is of a phospholipid bilayer. The phospholipids are forming a

“sheet”.

Explain why phospholipids never form this “sheet” structure in organisms.

The fatty acid tails are exposed to water in this sheet arrangement. In organism the fatty

acid tails must not be in contact with water so the bilayer will form a continuous layer

with all the fatty acids covered by the phosphate heads.

Substance Hydrophobic Hydrophilic Amphiphilic Polar ionic

Phospholipid

Cholesterol

Glucose

Sodium chloride

Fatty acid

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5. The fluidity of the cell membrane can be altered by various factors, including the

cholesterol content, the type of phospholipid in the membrane and its composition.

People suffering from the genetic disorder abetalipoproteinemia have altered

amounts of the two phospholipids sphingomyelin and phosphatidylcholine in their

cell membranes. The structures of these phospholipids are shown below.

Abetalipoproteinemia results from a reduction in cell membrane fluidity. Using simple

diagrams for the above phospholipids and the information in this question draw an

annotated bilayer(s) to explain the reduction in membrane fluidity.

Sphingomyelin Phosphatidylcholine

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6. In an experiment to investigate the structure of the cell membrane Larry Frye and

Michael Edidin, at Johns Hopkins University, labelled the plasma membrane proteins

of a mouse cell and a human cell with two different fluorescent markers and fused

the cells. Using a microscope, they observed the markers on the hybrid cell.

Mouse

cell

Human

cell

Hybrid

cell

Hybrid cell after

one hour

The bilayer above contains both

sphingomyelin and phosphatidylcholine.

Phosphatidylcholine is an unsaturated

phospholipid so has a kink in its fatty

acid chain. There are sufficient

phosphatidylcholine in the bilayer to

make the bilayer fluid by causing the

phospholipids to be further apart and so

weakening the intermolecular forces

between the phospholipids.

The bilayer above is from a

Abetalipoproteinemia sufferer. There is

a reduced number of

phosphatidylcholine and a increased in

sphingomyelin. The reduced number of

unsaturated phospholipid allow the

phospholipids to pack closer together so

increasing the intermolecular forces so

causing the membrane to become less

fluid.

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Use your knowledge of the structure of the cell membrane to explain the results seen after

one hour.

The cell membrane from each cell are able to fuse together because of the fluid nature of

the phospholipids. The membrane proteins from both cells are able to mix suggesting

that proteins can move sideways within the plane of the cell membrane.

7. Proteins are an essential component of the cell membrane. These membrane

proteins are located in various positions within the phospholipid bilayer to allow

them to perform a specific function. The images below are a representation of the

general shapes of some membrane proteins.

(a) Based on their shapes and sizes only which of the above images can:

i. Can form an intrinsic protein.

A,B,C,D

ii. Can form a transmembrane protein.

B and C

iii. Can form an extrinsic protein.

A and D

A

B

C

D

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iv. Can form a glycoprotein.

A and D

v. Is a carrier protein.

C

vi. Is a channel protein.

B

(b) Proteins are polymers of amino acids. The structure of a general amino acid is

shown below.

The letter R is a chemical group, called a side chain, that gives the amino acid its specific

chemical property as shown in the table below.

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The image below is of a membrane protein. Amino acids are positioned at specific sites

within the protein polymer to allow it to do its function.

i. Using the table of amino acids on page 7, what type of amino acid R groups

would be found at positions 1 and 2 in the membrane protein above. Explain

your choices.

The protein is a channel protein. It interacts with the hydrophobic fatty acid tails via

amino acids that has a hydrophobic R group (the eight R’s within a square in the image).

Names of hydrophobic amino acids will be: glycine, alanine, valine, leucine, isoleucine,

proline, tryptophan and phenylalanine. A channel protein has a channel or pore that is

hydrophilic. So, the amino acid that form part of the pore must have hydrophilic R groups

(the nine R.s within circles in the image). Names of hydrophilic amino acids will be:

serine, threonine tyrosine, asparagine, glutamine, aspartic acid, glutamic acid, lysine and

histidine.

ii. Suggest, with an explanation, a function for this membrane protein

This protein will allow the transport of hydrophilic substances across the membrane as

hydrophilic substances cannot pass through the hydrophobic fatty acid tail region of the

bilayer.

1 1

2

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iii. Which group(s) of amino acids from the table on page 7 will be needed to

produce an extrinsic protein, explain your answer.

Amino acids with polar side chains and charged side chains. These are needed as an

extrinsic protein will interact with the phosphate heads on the bilayer and water, so the

protein must be hydrophilic all around it surface.

8. An ionic molecule is one that has a true electrical charge, for example Na+ and Cl- that

is caused by either a gain or loss of electron(s). A polar molecule, however, has a

partial electrical charge caused by a difference in electronegativity of the atoms that

make up the molecule causing an even distribution of electron density. Non- polar

molecules have no partial charge so the atoms have equal electronegativity.

Glucose and water are polar, and non-polar molecules would be lipids, oxygen and

carbon dioxide. The image below is of a phospholipid bilayer showing what

substances can cross the bilayer and those that cannot cross the bilayer

Non-Polar

Molecules O2 & CO2

Small polar

molecules H2O

Large

polar molecules Glucose

Ions

Na+ & K

+

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(ai) In relation to the structure and properties of the bilayer, explain the transport or

non-transport of the molecules shown in the image on page 9.

The bilayer has a hydrophobic region produced by the fatty acid tails of the

phospholipids. This hydrophobic layer is one major factor that determines what can pass

through, or diffuse, through the bilayer. Any substance that is hydrophobic can pass

through the hydrophobic region as they can dissolve in the fatty acids. This is why oxygen

and carbon dioxide can diffuse across the membrane. Water is a polar molecule, but

small amounts can move across the bilayer very slowly due to its small size. Large polar

molecules and ions cannot pass across the bilayer as they cannot dissolve in the fatty acid

layer.

(aii) Explain what modification(s) to the bilayer structure would need to be made to allow

the transport of large polar molecules and ions.

The bilayer would need transmembrane proteins like channel and carrier proteins. These

proteins able to transport ions and large polar molecules across the membrane without

them coming into contact with the hydrophobic fatty acids.

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(b) Explain how phospholipids could arrange themselves as shown below.

By being added to oil and then shaken. The hydrophilic fatty acids will interact with the

oil and the phosphate heads would be forced inwards away from the oil.

9. The image below shows the entry of the HIV virus into a helper T cell.

(a) Describe how HIV infects helper T cells.

HIV has two antigens on the protein coat. These antigens bind to two receptors on the

surface of helper T-cells because their shapes are complementary. The virus plastid will

then enter the cell when the two membranes fuse by endocytosis.

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(b) Explain how scientists could prevent entry of HIV into the helper T cell.

Scientists could develop a drug that had a complementary in shape to one of the

receptors on the surface of the helper T cell. This drug would bind to the receptor and

prevent the viral antigen from binding to it. The virus would be prevented from entering

as it needs to bind to both receptors on the T cell to enter the t cell.

10. Cell membrane proteins have carbohydrates attached to them in the ER and Golgi

body, then are transported in vesicles to the cell surface. On which side of the vesicle

membrane are the proteins and carbohydrate? Explain your answer with an

annotated diagram.

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11. An experiment was conducted to investigate the effect of temperature on the

permeability of the beetroot cell membranes. Beetroot disks were cut and added to

water and incubated at different temperatures and the colour of the water was

measured using a colorimeter with blue light. The results are shown in the table

below. The lower the percentage transmission reading the darker the red colour.

(ai) Plot the above data on the graph paper on page 14

Temperature (oC) 10 20 30 40 50 60 70

Percentage transmission

of blue light (%)

100 90 80 7 4 1 0

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(a ii) Describe the trend of the graph. Explain if you have confidence in the trend?

As the temperature increases the percentage transmission of blue light decreases. I have

no confidence in the trend as there are no replicate readings or range bars to allow

assessment of the consistency of the replicate results.

(a iii) State the controlled variable for this experiment.

Volume of water, diameter of beetroot disk, use of the same beetroot.

(a iv) Can the reliability of this data be determined, explain your answer.

No, because there are no repeat readings to determine if the results are repeatable.

(a v) Discuss the precision and accuracy of the percentage transmission results.

The colorimeter can only give results to whole numbers. This mean that its precision is

low, so the results will not be accurate.

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(bi) With reference to the structure of the cell membrane, explain the results.

The cell membrane is made of phospholipids and membrane proteins. At low

temperatures the cell membrane is functioning by preventing beetroot pigment from

leaking out. As the temperature increases the phospholipids gain kinetic enery and bein

to move further apart. This creates small holes in the membrane that allow more

pigment to diffuse out. As the temperature increases further the membrane proteins

become denatured and fall out of the bilayer producing large hols in the bilayer causing

far more pigment to leak out.

(bii) Explain the difference in the results if the cell membrane was initially treated with

ethanol.

The ethanol will dissolve the phospholipids so the beetroot pigment would instantly leak

out even at 10oC causing a very low percentage transmission reading.