2.3- TRANSPORT ACCROSS MEMBRANES

Question 1

Q1.(a) Give two ways in which pathogens can cause disease. (2)

Answer 1

1. (Releases) toxins;
2. Kills cells / tissues.
3. Accept any reference to cell / tissue damage
   Ignore infecting / invading cells

Question 2

1(b) Putting bee honey on a cut kills bacteria. Honey contains a high concentration of
sugar.

Use your knowledge of water potential to suggest how putting honey on a cut kills
bacteria. (3)

Answer 2

1. Water potential in (bacterial) cells higher (than in honey) / water
   potential in honey lower (than in bacterial cells);
   Q candidates must express themselves clearly
2. Must be comparative e.g. high WP in cell and low WP in
   honey
3. Water leaves bacteria / cells by osmosis;
4. (Loss of water) stops (metabolic) reactions.
5. Needs a reason why lack of water kills the cell

Question 3

Q2.Read the following passage.

Low-density lipoprotein (LDL) is a substance found in blood. A high
concentration of LDL in a person’s blood can increase the risk of atheroma
formation. Liver cells have a receptor on their cell-surface membranes that LDL
binds to. This leads to LDL entering the cell. A regulator protein, also found in blood, can bind to the same receptor as LDL. This prevents LDL entering the
liver cell. People who have a high concentration of this regulator protein in their
blood will have a high concentration of LDL in their blood. Scientists have made
a monoclonal antibody that prevents this regulator protein working. They have
suggested that these antibodies could be used to reduce the risk of coronary
heart disease.
5
A trial was carried out on a small number of healthy volunteers, divided into two
groups. The scientists injected one group with the monoclonal antibody in salt
solution. The other group was a control group. They measured the concentration
of LDL in the blood of each volunteer at the start and after 3 months. They found
that the mean LDL concentration in the volunteers injected with the antibody was
64% lower than in the control group.
10
15
Use the information in the passage and your own knowledge to answer the following
questions.

(a) The scientists gave an injection to a mouse to make it produce the monoclonal
antibody used in this investigation (line 7).
What should this injection have contained? (1)

Answer 3

Regulator protein.
Accept regulator protein antigen
Reject regulator protein receptor
Ignore regular protein

Question 4

(b) LDL enters the liver cells (lines 3−4).
Using your knowledge of the structure of the cell-surface membrane, suggest how
LDL enters the cell. (2)

Answer 4

1. Lipid soluble / hydrophobic
2. Enters through (phospholipid) bilayer
   OR
3. (Protein part of) LDL attaches to receptor
4. Goes through carrier / channel protein.
5. Accept by facilitated diffusion or active transport
6. Reject active transport through channel protein

Question 5

(c) Explain how the monoclonal antibody would prevent the regulator protein from
working (lines 7−8). (2)

Answer 5

Any two from:
1. (Monoclonal antibody) has a specific tertiary structure / variable region /
is complementary to regulator protein

Do not award MP1 if reference to active site.
2. Binds to / forms complex with (regulator protein)
“It” refers to monoclonal antibody in MP1 and MP2

3. (So regulator protein) would not fit / bind to the receptor / is not
   complementary to receptor
4. Reject receptor on LDL

Question 6

(d) Describe how the control group should have been treated. (2)

Answer 6

1. Injection with salt solution
2. Accept inject placebo in salt solution
3. Otherwise treated the same.

Question 7

Q3.Scientists studied the rate of carbon dioxide uptake by grape plant leaves. Grape leaves
have stomata on the lower surface but no stomata on the upper surface.
The scientists recorded the carbon dioxide uptake by grape leaves with three different
treatments:
Treatment 1 − No air-sealing grease was applied to either surface of the leaf.
Treatment 2 − The lower surface of the leaf was covered in air-sealing grease that
prevents gas exchange.
Treatment 3 − Both the lower surface and the upper surface of the leaf were covered in
air–sealing grease that prevents gas exchange.
The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in
light and then for 20 minutes in the dark.
The scientists’ results are shown in the diagram below

(a) Suggest the purpose of each of the three leaf treatments.

(3)

Answer 7

1. (No grease)
 means stomata are open
 OR
 allows normal CO2 uptake;
Allow ‘gas exchange’ for CO2 uptake.
‘As a control’ is insufficient on its own. 

2. (Grease on lower surface)
seals stomata
OR
stops CO2 uptake through
stomata
OR
to find CO2 uptake through
stomata
OR
shows CO2 uptake through cuticle / upper surface;

3. (Grease on both surfaces) shows sealing is effective
   OR
   stops all CO2 uptake.

Question 8

(b) (i) Describe the results shown for Treatment 1 (2)

Answer 8

1. (Mean rate of) carbon dioxide uptake was constant and fell after
   the light turned off;
   Ignore absence of arbitrary units in both marking points. Both ideas needed for mark.
   Accept ‘stayed at 4.5’ as equivalent to ‘was constant’.
2. Uptake fell from 4.5 to 0 / uptake started to fall at 60 minutes and
   reached lowest at 80 minutes / uptake fell over period of 20
   minutes;
   One correct use of figures required.
   Accept fell to nothing / no uptake for 0.

Question 9

(ii) The stomata close when the light is turned off.

Explain the advantage of this to the plant. (2)

Answer 9

1. (Because) water is lost through stomata;
2. (Closure) prevents / reduces water loss;
3. Maintain water content of cells.
   This marking point rewards an understanding of reducing
   water loss e.g. reduce wilting, maintain turgor, and is not
   related to photosynthesis.

Question 10

c) (i) Treatment 2 shows that even when the lower surface of the leaf is sealed
there is still some uptake of carbon dioxide.
Suggest how this uptake of carbon dioxide continues. (1) (1)

Answer 10

(Carbon dioxide uptake) through the upper surface of the leaf / through
cuticle.

Question 11

(ii) In both Treatment 1 and Treatment 2, the uptake of carbon dioxide falls to
zero when the light is turned off.
Explain why. (2)

Answer 11

1. No use of carbon dioxide in photosynthesis (in the dark);
2. No diffusion gradient (maintained) for carbon dioxide into leaf /
   there is now a diffusion gradient for carbon dioxide out of leaf (due
   to respiration).

Question 12

Q4.The figure below represents a capillary surrounded by tissue fluid.
The values of the hydrostatic pressure are shown.

Arteriole
end direction of blood flow
Venule
end
Hydrostatic pressure = 4.3 kPa Hydrostatic pressure = 1.6 kPa
Tissue fluid
Hydrostatic pressure = 1.1 kPa
(a) Use the information in the figure above to explain how tissue fluid is formed.  (2)

Answer 12

1. (Overall) outward pressure of 3.2 kPa;

2. Forces small molecules out of capillary

Question 13

) The hydrostatic pressure falls from the arteriole end of the capillary to the venule
end of the capillary. Explain why (1)

Answer 13

Loss of water / loss of fluid / friction (against capillary lining).

Question 14

(c) High blood pressure leads to an accumulation of tissue fluid. Explain how. (3)

Answer 14

1. High blood pressure = high hydrostatic pressure;
2. Increases outward pressure from (arterial) end of capillary / reduces
   inward pressure at (venule) end of capillary;
3. (So) more tissue fluid formed / less tissue fluid is reabsorbed.
   Allow lymph system not able to drain tissues fast enough

Question 15

The water potential of the blood plasma is more negative at the venule end of the
capillary than at the arteriole end of the capillary. Explain why. (3)

Answer 15

1. Water has left the capillary;
2. Proteins (in blood) too large to leave capillary;
3. Increasing / giving higher concentration of blood proteins (and thus wp)

Question 16

Q5.(a) Describe how you would test a piece of food for the presence of lipid. (2)

Answer 16

Dissolve in alcohol, then add water;

2. White emulsion shows presence of lipid.

Question 17

Q6.A group of students carried out an investigation to find the water potential of potato tissue.
The students were each given a potato and 50 cm3
of a 1.0 mol dm−3 solution of sucrose.
• They used the 1.0 mol dm−3 solution of sucrose to make a series of different
concentrations.
• They cut and weighed discs of potato tissue and left them in the sucrose solutions
for a set time.
• They then removed the discs of potato tissue and reweighed them.
The table below shows how one student presented his processed results.

(a) Explain why the data in the table above are described as processed results. (1)

Answer 17

Calculations made (from raw data) / raw data would have recorded initial and final
masses

Question 18

Describe how you would use a 1.0 mol dm−3 solution of sucrose to produce 30 cm3
of a 0.15 mol dm−3 solution of sucrose. (2)

Answer 18

Add 4.5 cm3
 of (1.0 mol dm–3) solution to 25.5 cm3
 (distilled) water.
If incorrect, allow 1 mark for solution to water in a proportion
of 0.15:0.85

Question 19

(c) Explain the change in mass of potato tissue in the 0.40 mol dm−3 solution of sucrose. (2)

Answer 19

1. Water potential of solution is less than / more negative than that of
   potato tissue;
   Allow Ψ as equivalent to water potential
2. Tissue loses water by osmosis.

Question 20

(d) Describe how you would use the student’s results in the table above to find the
water potential of the potato tissue. (3)

Answer 20

1. Plot a graph with concentration on the x-axis and percentage change in
   mass on the y-axis;
2. Find concentration where curve crosses the x-axis / where percentage
   change is zero;
3. Use (another) resource to find water potential of sucrose concentration
   (where curve crosses x-axis).

Question 21

Q1.(a) Describe how phospholipids are arranged in a plasma membrane. (2)

Answer 21

1. Bilayer;

Accept double layer
Accept drawing which shows bilayer
2. Hydrophobic / fatty acid / lipid (tails) to inside;
3. Polar / phosphate group / hydrophilic (head) to outside;

2. & 3. need labels
3. & 3. accept water loving or hating

Question 22

(b) Cells that secrete enzymes contain a lot of rough endoplasmic reticulum (RER) and a
large Golgi apparatus.

(i) Describe how the RER is involved in the production of enzymes. (2)

Answer 22

1. (Rough endoplasmic reticulum has) ribosomes;

accept “contains / stores”
2. To make protein (which an enzyme is);
Accept amino acids joined together / (poly)peptide
Reject makes amino acids
Ignore glycoprotein

Question 23

(ii) Describe how the Golgi apparatus is involved in the secretion of enzymes. (1)

Answer 23

(ii) (Golgi apparatus) modifies (protein)

Question 24

(a) Explain why the rate of diffusion is more rapid at higher temperatures. (2)

Answer 24

(a) More (kinetic) energy;
Molecules are moving faster;

Ignore references to collisions

Question 25

(c) The graph shows how the concentration of a substance affects its rate of absorption
into a cell.

(i) Substance A enters the cell by simple diffusion. Use Fick’s law to explain the
shape of the curve. (1)

Answer 25

(c) (i) Greater the concentration difference / gradient, faster rate of entry /

diffusion;

Question 26

Substance B enters the cell by facilitated diffusion. Explain the evidence from the
graph which supports this. (2)

Answer 26

(ii) Curve flattens out;

Channel / carrier proteins / carriers become limiting;

Question 27

(a) An optical microscope cannot be used to see a plasma membrane. Explain why. (2)

Answer 27

(a) Does not have the resolution / cannot distinguish between points this close

together;
As light has longer wavelength;
The key ideas in marking this part of the question are resolution and wavelength.

Question 28

(b) Give one property of the molecules of substance X which allows them to diffuse
through the membrane at the position shown. (1)

Answer 28

Lipid soluble / small / non-polar / not charged;

Question 29

The effect of the concentration of sodium ions in the surrounding solution on their rate
of diffusion across the membrane was investigated. The graph shows the results.
(i) What limits the diffusion of sodium ions across the membrane between
A and B on the graph? Give the evidence for your answer.

(2)

Answer 29

Concentration of sodium ions (outside cell);

As concentration / independent variable increases so does
the rate of diffusion;

Question 30

(ii) Explain the shape of the curve between C and D.

Answer 30

Sodium ions are passing through the channels / pores at their maximum
rate;
Rate is limited by the number of sodium channels / another limiting factor;

Question 31

Q4. A student investigated the effect of putting cylinders cut from a potato into sodium
chloride solutions of different concentration. He cut cylinders from a potato and weighed each
cylinder. He then placed each cylinder in a test tube. Each test tube contained a different
concentration of sodium chloride solution. The tubes were left overnight. He then removed
the cylinders from the solutions and reweighed them.
(a) Before reweighing, the student blotted dry the outside of each cylinder. Explain why. (2)

Answer 31

(a) Water will affect the mass / only want to measure water taken up or lost;

Amount of water on cylinders varies / ensures same amount of water on outside;

Neutral: removes water
Accept: ‘(sodium chloride) solution’ for water
Do not accept ‘sodium chloride’
Neutral: refs. to fair testing

Question 32

(c) The student calculated the percentage change in mass rather than the change in mass.
Explain the advantage of this. (2)

Answer 32

Allows comparison / shows proportional change;
Idea that cylinders have different starting masses / weights;

Reject: if comparison is in context of the start and final mass of
the same cylinder
Neutral: different masses
Neutral: different starting sizes

Question 33

(d) The student carried out several repeats at each concentration of sodium chloride
solution. Explain why the repeats were important. (2)

Answer 33

(Allows) anomalies to be identified / ignored / effect of anomalies to be reduced /
effect of variation in data to be minimised;
Makes the average / mean / line of best fit more reliable / allows concordant
results;

Question 34

(ii) The students gave their results as a ratio. What is the advantage of giving the
results as a ratio? (2)

Answer 34

(ii) Allows comparison / shows proportional change;

Neutral: sizes / amounts
Idea that discs had different starting masses / weights;
Neutral: different masses

Question 35

(iii) The students were advised that they could improve the reliability of their results
by taking additional readings at the same concentrations of sodium chloride.
Explain how. (2)

Answer 35

(iii) (Allows)

Accept: outliers instead of anomalies
Anomalies to be identified / effect of anomalies to be reduced / effect of
variation in data to be minimised;
Reject: idea of not recording anomalies / preventing anomalies
from occurring
A mean to be calculated;
Neutral: average

Question 36

(b) (i) The students used a graph of their results to find the sodium chloride solution
with the same water potential as the apple tissue. Describe how they did this. (2)

Answer 36

Plot (sodium chloride) concentration against ratio / draw line of best fit;

Reject: if wrong axes or type of graph

Find (sodium chloride concentration from the graph) where the ratio is 1 /
there is no change in mass;

Question 37

(ii) The students were advised that they could improve their graph by taking
additional readings. Explain how. (2)

Answer 37

(ii) Line / curve of best fit is more reliable / precise;

Neutral: graph
Intercept / point where line crosses axis is more reliable / precise;
Reject: references to ‘more accurate’
OR
Can plot SD values / error bars;
(To show) variability about the mean / how spread out the results are;

Question 38

(a) A plant cell was observed with an optical microscope.

Describe how the length of
the cell could be estimated.

Answer 38

(a) Measure diameter of field with ruler; And proportion taken up by the cell; or

Measure length with (eyepiece) graticule / eyepiece scale;
Calibrated against stage micrometer / something of known length;
Reject divide apparent length by magnification

Question 39

(b) The water potential of a plant cell is –400 kPa. The cell is put in a solution with a water
potential of –650 kPa. Describe and explain what will happen to the cell. (3)

Answer 39

Membrane / cytoplasm shrinks / pulls away from cell wall / cell plasmolysed / goes
flaccid; Water moves down water potential gradient / to lower / more negative water
potential; By osmosis;

Question 40

(i) The carrot cylinders were left for 18 hours in the sucrose solutions. Explain why
they were left for a long time. (1)

Answer 40

(i) Reaches equilibrium / no further / maximum change in length;

Reject osmosis takes time

Question 41

(ii) Explain how you would use a graph to predict the concentration of sucrose that
would result in no change in length of the carrot cylinders. (2)

Answer 41

(ii) Line / curve of best fit; Extrapolate (and read off) / find where it crosses x-axis;

Question 42

Young carrots store sugars in their tissues but, in older carrots, some of this is
converted to starch. How would using cylinders of tissue from older carrots affect
the results obtained for a sucrose solution of 0.6 mol dm–3? Give a reason for your
answer. (2)

Answer 42

Greater decrease / length smaller; More water removed;
Greater difference in water potential / cell with higher / less negative water
potential; Starch is insoluble / has no effect on osmosis

Question 43

Q7.A scientist investigated the uptake of sodium ions by animal tissue.
To do this, he:
• used two flasks, F and G
• put equal masses of animal tissue into each flask
• added equal volumes of a solution containing sodium ions to each flask
• added to flask F a solution of a substance that prevents the formation of ATP by cells
• measured the concentration of sodium ions remaining in the solution in each flask.
The graph below shows his results.

(b) The scientist concluded that the cells in flask G took up sodium ions by active transport.
Explain how the information given supports this conclusion. (4)

Answer 43

1. Uptake in flask G much greater than in flask F;
2. Showing use of ATP in flask G;
3. Sodium ion concentration in flask G falls to zero;
4. Showing uptake against a concentration gradient;

Question 44

(c) The curve for flask F levelled off after 20 minutes. Explain why. (2)

Answer 44

1. (Uptake of sodium ions occurring by) facilitated diffusion;
2. Equilibrium reached / sodium ion concentrations in solution and in cells the
   same;

Question 45

Read the following passage.

Human milk contains all the nutrients a young baby needs in exactly the right
proportions. It is formed in the mammary glands by small groups of milk-producing cells.
These cells absorb substances from the blood and use them to synthesise the lipids,
carbohydrates and proteins found in milk. Milk-producing cells are roughly cube-shaped
5 and have a height to breadth ratio of approximately 1.2 : 1.
The main carbohydrate in milk is lactose. Lactose is a disaccharide formed by the
condensation of two monosaccharides, glucose and galactose. (A molecule of galactose
has the same formula as a molecule of glucose – the atoms are just arranged in a different
way.)
10 Lactose is synthesised in the Golgi apparatus and transported in vesicles through the
cytoplasm. Because lactose is unable to escape from these vesicles, they increase in
diameter as they move towards the plasma membrane. The vesicle membranes fuse with
the plasma membrane and the vesicles empty their contents out of the cell.
Use the information from the passage and your own knowledge to answer the following
questions.

(ii) Describe and explain how you would expect the height to breadth ratio of an
epithelial cell from a lung alveolus to differ from the height to breadth ratio of a
milk-producing cell. (2)

Answer 45

Ratio would be less / smaller;
Cell is thin / has large surface area / (adapted) for diffusion;
Accept converse. Must relate to concept of ratio.

Question 46

How many oxygen atoms are there in a molecule of galactose; ?
lactose? (2)

Answer 46

(i) 6;

(ii) 11;

Question 47

The lactose-containing vesicles increase in diameter as they move towards the plasma
membrane of the milk-producing cell (lines 11-12). Use your knowledge of water
potential to explain why. (2)

Answer 47

Water potential inside vesicle more negative / lower;

Water moves into vesicle by osmosis / diffusion;

Question 48

Suggest one advantage of milk-producing cells containing large numbers of
mitochondria. (2)

Answer 48

Mitochondria supply energy / ATP;
For active transport / absorption against concentration
gradient / synthesis / anabolism / exocytosis / pinocytosis;

Do not credit references to making,
creating or producing energy.

Question 49

(e) Some substances pass through the plasma membrane of a milk-producing cell by
diffusion.

Describe the structure of a plasma membrane and explain how different
substances are able to pass through the membrane by diffusion. (6)

Answer 49

1 Phospholipids forming bilayer / two layers;
2 Details of arrangement with “heads” on the outside;
3 Two types of protein specified;
e.g. passing right through or confined to one layer /
extrinsic or intrinsic /
channel proteins and carrier proteins /
two functional types
4 Reference to other molecule e.g. cholesterol or glycoprotein;
5 Substances move down concentration gradient / from high to low
concentration;
Reject references to across or along a gradient
6 Water / ions through channel proteins / pores;
7 Small / lipid soluble molecules / examples pass between phospholipids /
through phospholipid layer;
8 Carrier proteins involved with facilitated diffusion;
Ignore references to active transport.
Credit information in diagrams.

Question 50

Q10. Read the following passage.

The plasma membrane plays a vital role in microorganisms. It forms a barrier between the
cell
and its environment, controlling the entry and exit of solutes. This makes bacteria vulnerable
to a range of antiseptics and antibiotics
When bacteria are treated with antiseptics, the antiseptics bind to the proteins in the
5 membrane and create tiny holes. Bacteria contain potassium ions at a concentration many
times that outside the cell. Because of the small size of these ions and their concentration in
the cell, the first observable sign of antiseptic damage to the plasma membrane is the
leaking
of potassium ions from the cell. Some antibiotics damage the plasma membrane in a similar
way. One of these is tyrocidin. This is a cyclic polypeptide consisting of a ring of ten amino
10 acids. Tyrocidin and other polypeptide antibiotics are of little use in medicine.
Other antibiotics also increase the rate of potassium movement from cells. It is thought that
potassium ions are very important in energy release and protein synthesis, and a loss of
potassium ions would lead to cell death. Gramicidin A coils to form a permanent pore
passing
through the plasma membrane. This pore enables potassium ions to be conducted from the
15 inside of the cell into the surrounding medium. Vanilomycin also facilitates the passage of
potassium ions from the cell. A molecule of vanilomycin forms a complex with a potassium
ion and transports it across the membrane. The potassium ion is released on the outside and
the vanilomycin is free to return and pick up another potassium ion. Vanilomycin depends on
the fluid nature of the plasma membrane in order to function.
20 Polyene antibiotics have flattened ring-shaped molecules. The two sides of the ring differ
from
each other. One side consists of an unsaturated carbon chain. This part is strongly
hydrophobic and rigid. The opposite side is a flexible, strongly hydrophilic region. It has been
shown that polyene antibiotics bind only to sterols. Sterols are lipids found in the membranes
of eukaryotes but not in the membranes of prokaryotic organisms. It is thought that several
25 sterol-polyene complexes come together. The plasma membranes of eukaryotic cells treated
with these polyene antibiotics lose the ability to act as selective barriers and small ions and
molecules rapidly leak out

Use information in the passage and your own knowledge to answer the questions.
(a) (i) By what process do potassium ions normally enter a bacterial cell? Explain the

evidence for your answer.

Answer 50

(i) Active transport;

Low to high concentration / against concentration gradient;
Reject answers relating only to high concentration in cell

Question 51

Read the following passage.

The plasma membrane plays a vital role in microorganisms. It forms a barrier between the
cell
and its environment, controlling the entry and exit of solutes. This makes bacteria vulnerable
to a range of antiseptics and antibiotics
When bacteria are treated with antiseptics, the antiseptics bind to the proteins in the
5 membrane and create tiny holes. Bacteria contain potassium ions at a concentration many
times that outside the cell. Because of the small size of these ions and their concentration in
the cell, the first observable sign of antiseptic damage to the plasma membrane is the
leaking
of potassium ions from the cell. Some antibiotics damage the plasma membrane in a similar
way. One of these is tyrocidin. This is a cyclic polypeptide consisting of a ring of ten amino
10 acids. Tyrocidin and other polypeptide antibiotics are of little use in medicine.
Other antibiotics also increase the rate of potassium movement from cells. It is thought that
potassium ions are very important in energy release and protein synthesis, and a loss of
potassium ions would lead to cell death. Gramicidin A coils to form a permanent pore
passing
through the plasma membrane. This pore enables potassium ions to be conducted from the
15 inside of the cell into the surrounding medium. Vanilomycin also facilitates the passage of
potassium ions from the cell. A molecule of vanilomycin forms a complex with a potassium
ion and transports it across the membrane. The potassium ion is released on the outside and
the vanilomycin is free to return and pick up another potassium ion. Vanilomycin depends on
the fluid nature of the plasma membrane in order to function.
20 Polyene antibiotics have flattened ring-shaped molecules. The two sides of the ring differ
from
each other. One side consists of an unsaturated carbon chain. This part is strongly
hydrophobic and rigid. The opposite side is a flexible, strongly hydrophilic region. It has been
shown that polyene antibiotics bind only to sterols. Sterols are lipids found in the membranes
of eukaryotes but not in the membranes of prokaryotic organisms. It is thought that several
25 sterol-polyene complexes come together. The plasma membranes of eukaryotic cells treated
with these polyene antibiotics lose the ability to act as selective barriers and small ions and
molecules rapidly leak out

(ii) Use Fick’s law to explain why leakage of potassium ions occurs following
antiseptic damage to the plasma membrane (lines 7 - 8). (2)

Answer 51

Rate of movement / diffusion proportional to concentration gradient /
difference in concentration;
High concentration of potassium ions inside cell compared to outside;
Must mention high concentration. Ignore reference to other factors if
reasoning is appropriate.

Question 52

Read the following passage.

The plasma membrane plays a vital role in microorganisms. It forms a barrier between the
cell
and its environment, controlling the entry and exit of solutes. This makes bacteria vulnerable
to a range of antiseptics and antibiotics
When bacteria are treated with antiseptics, the antiseptics bind to the proteins in the
5 membrane and create tiny holes. Bacteria contain potassium ions at a concentration many
times that outside the cell. Because of the small size of these ions and their concentration in
the cell, the first observable sign of antiseptic damage to the plasma membrane is the
leaking
of potassium ions from the cell. Some antibiotics damage the plasma membrane in a similar
way. One of these is tyrocidin. This is a cyclic polypeptide consisting of a ring of ten amino
10 acids. Tyrocidin and other polypeptide antibiotics are of little use in medicine.
Other antibiotics also increase the rate of potassium movement from cells. It is thought that
potassium ions are very important in energy release and protein synthesis, and a loss of
potassium ions would lead to cell death. Gramicidin A coils to form a permanent pore
passing
through the plasma membrane. This pore enables potassium ions to be conducted from the
15 inside of the cell into the surrounding medium. Vanilomycin also facilitates the passage of
potassium ions from the cell. A molecule of vanilomycin forms a complex with a potassium
ion and transports it across the membrane. The potassium ion is released on the outside and
the vanilomycin is free to return and pick up another potassium ion. Vanilomycin depends on
the fluid nature of the plasma membrane in order to function.
20 Polyene antibiotics have flattened ring-shaped molecules. The two sides of the ring differ
from
each other. One side consists of an unsaturated carbon chain. This part is strongly
hydrophobic and rigid. The opposite side is a flexible, strongly hydrophilic region. It has been
shown that polyene antibiotics bind only to sterols. Sterols are lipids found in the membranes
of eukaryotes but not in the membranes of prokaryotic organisms. It is thought that several
25 sterol-polyene complexes come together. The plasma membranes of eukaryotic cells treated
with these polyene antibiotics lose the ability to act as selective barriers and small ions and
molecules rapidly leak out

Experiments have shown that vanilomycin is unable to transport potassium ions across
a membrane when it is cooled. Gramicidin A continues to facilitate the movement of
potassium ions at these low temperatures. Explain these results. (3)

Answer 52

Action of vanilomycin depends on fluidity of membrane;
Fluidity reduced / not fluid at low temperatures;
Pore formed by gramicidin A remains in place / permanent;

Question 53

Contrast the processes of facilitated diffusion and active transport.
[3 marks]

Answer 53

1. Facilitated diffusion involves channel or carrier
   proteins whereas active transport only involves
   carrier proteins;
2. Facilitated diffusion does not use ATP / is
   passive whereas active transport uses ATP;
3. Facilitated diffusion takes place down a
   concentration gradient whereas active transport
   can occur against a concentration gradient;

Question 54

P2017 Students investigated the uptake of chloride ions in barley plants. They divided the
plants into two groups and placed their roots in solutions containing radioactive
chloride ions.

• Group A plants had a substance that inhibited respiration added to the solution.
• Group B plants did not have the substance added to the solution.
The students calculated the total amount of chloride ions absorbed by the plants every
15 minutes. Their results are shown in Figure 4.

Explain the results shown in Figure 4.
[4 marks]

Answer 54

1. Group A – initial uptake slower because by
   diffusion (only);
2. Group A – levels off because same
   concentrations inside cells and outside cells /
   reached equilibrium;
3. Group B – uptake faster because by diffusion
   plus active transport;
4. Group B fails to level off because uptake
   against gradient/no equilibrium to be reached;
5. Group B – rate slows because few/fewer
   chloride ions in external solution/respiratory
   substrate used up;

Question 55

During early pregnancy, the glycogen in the cells lining the uterus is an important
energy source for the embryo.
Suggest how glycogen acts as a source of energy.
Do not include transport across membranes in your answer.
[2 marks]

Answer 55

1. Hydrolysed (to glucose);

2. Glucose used in respiration;

Question 56

Describe the structure of glycogen.

[2 marks]

Answer 56

1. Polysaccharide of α-glucose;
OR
polymer of α-glucose;
2. (Joined by) glycosidic bonds
 OR
Branched structure;

Question 57

2 . 3 Suggest and explain two ways the cell-surface membranes of the cells lining the
uterus may be adapted to allow rapid transport of nutrients.
[2 marks]

Answer 57

1. Membrane folded so increased/large surface
   area;
   OR
   Membrane has increased/large surface area
   for (fast) diffusion/facilitated diffusion/active
   transport/co-transport;
2. Large number of protein channels/carriers (in
   membrane) for facilitated diffusion;
3. Large number of protein carriers (in membrane)
   for active transport;
4. Large number of protein (channels/carriers in
   membrane) for co-transport;

Question 58

Mammals have some cells that produce extracellular proteases. They also have
cells with membrane-bound dipeptidases.
Describe the action of these membrane-bound dipeptidases and explain their
importance.
[2 marks

Answer 58

1. Hydrolyse (peptide bonds) to release amino
   acids;
2. Amino acids can cross (cell) membrane;
   OR
   Dipeptides cannot cross (cell) membrane;
   OR
   Maintain concentration gradient of amino acids
   for absorption;
   OR
   Ensure (nearly) maximum yield from protein
   breakdown;

Question 59

3 . 1 Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some
of these cells have membranes with a carrier protein called NHE3.
NHE3 actively transports one sodium ion into the cell in exchange for one proton
(hydrogen ion) out of the cell.
Use your knowledge of transport across cell membranes to suggest how NHE3
does this.
[3 marks]

Answer 59

1. Co-transport;
2. Uses (hydrolysis of) ATP;
3. Sodium ion and proton bind to the protein;
4. Protein changes shape (to move sodium ion
   and/or proton across the membrane)

Question 60

0 3 . 2 Scientists investigated the use of a drug called Tenapanor to reduce salt absorption
in the gut. Tenapanor inhibits the carrier protein, NHE3.
The scientists fed a diet containing a high concentration of salt to two groups of
rats, A and B.
• The rats in Group A were not given Tenapanor (0 mg kg–1).
• The rats in Group B were given 3 mg kg–1 Tenapanor.
One hour after treatment, the scientists removed the gut contents of the rats and
immediately weighed them.
Their results are shown in Table 2.
Table 2

The scientists carried out a statistical test to see whether the difference in the
means was significant. They calculated a P value of less than 0.05.
They concluded that Tenapanor did reduce salt absorption in the gut.
Use all the information provided and your knowledge of water potential to explain
how they reached this conclusion.
[4 marks]

Answer 60

1. Tenapanor/(Group)B/drug causes a significant
   increase;
   OR
   There is a significant difference with
   Tenapanor/drug/between A and B;
2. There is a less than 0.05 probability that the
   difference is due to chance;
3. (More salt in gut) reduces water potential in gut
   (contents);
4. (so) less water absorbed out of gut (contents)
   by osmosis
   OR
   Less water absorbed into cells by osmosis
   OR
   Water moves into the gut (contents) by
   osmosis.
   OR
   (so) water moves out of cells by osmosis;

Question 61

0 3 . 3 High absorption of salt from the diet can result in a higher than normal
concentration of salt in the blood plasma entering capillaries. This can lead to a
build-up of tissue fluid.
Explain how.
[2 marks]

Answer 61

1. (Higher salt) results in lower water potential of
   tissue fluid;
2. (So) less water returns to capillary by osmosis
   (at venule end);
   OR
3. (Higher salt) results in higher blood
   pressure/volume;
4. (So) more fluid pushed/forced out (at arteriole
   end) of capillary

Question 62

give 1 way in which the student could ensure the 1st three beetroot cylinders were kept at 25 in her experiment (1)

Answer 62

Measure the temperature and use appropriate corrective measures e.g. thermomete

Question 63

What can you conclude about the damage caused by beetroot cells by water, ethanol , hydrochloric aci an different temperatures? (4)

Answer 63

• Water caused no damage at 25 degrees
• High temperature means membrane becomes more fluid due to more kinetic energy
• Damage to cell membrane
• ethanol and acid caused similar types of damage