3. Exam Q

Question 1

A

Answer 1

6.1 m2
The area given is for a single layer – but in membranes, phospholipids form a bilayer, so the value must be halved.

Question 2

Answer 2

On the surface of the membrane
The charged groups are all around, so it must not be in contact with the hydrophobic portion and must be on the surface.

Question 3

label

Answer 3

A = phospholipid bilayer
B = extrinsic/surface protein/glycoprotein;
C = transmembrane/carrier/intrinsic protein;

Question 4

define function of the channel

Answer 4

allows passage of polar/charged/ionic/hydrophilic molecules/ facilitated/diffusion;

Question 5

b) The proteins are drawn to give some indication of their tertiary structure.
Explain the difference between secondary and tertiary structures of protein molecules including reference to the type of bonds involved.

Answer 5

(b) Secondary structure is folding of polypeptide chain/ref. to a helix or β pleated sheet;

held by hydrogen bonds;

tertiary is folding of a helix or secondary structure/correct reference to specific 3D shape;

held by bonds between R groups/name at least 2 from

covalent, disulphide, ionic, salt bridges, hydrophobic, hydrogen, van der Waals.

Question 6

c) (i) With reference to the fluid mosaic model explain how the distributions of charged and uncharged parts determine the position a protein will take up in a membrane.

Answer 6

Charged groups will associate with (hydrophilic) heads of lipids/layer/hydrophobic inside hydrophilic outside.

Uncharged groups will associate with (hydrophobic) tails.

Question 7

Answer 7

(ii) Will associate with heads only/attach to outside or inside of the membrane/correct use of extrinsic or would

be surface protein.
(not: would not be in the membrane unqual) [1]

Question 8

Question 9

Answer 9

X = protein (1)
Y = (phospho) lipid (1)
Z = carbohydrate/polysaccharide/glycocalyx/glycoprotein (1)

Question 10

Answer 10

(c) Labelled molecules/x/protein (1)
reference to mixing or equivalent (1)
[2]

Question 11

Answer 11

A. Cell/plasma membrane B. Cell Wall
C. Cytoplasm D. Tonoplast/vacuolar membrane
E. Vacuole F. Plasmodesma(ta)

Question 12

c) Cells X and Y are at incipient plasmolysis in 0.6 M sucrose solution. Describe one change that would be visible down a microscope if the cells were placed in a 1.0 M sucrose solution. [1]

Answer 12

(c) Cytoplasm/vacuole shrinks/gaps between wall and cytoplasm.
(not: plasmolysis/cell shrinks)

Question 13

Define the term water potential. [1]

Answer 13

The difference between the free energy of water molecules in a system and the free energy of molecules in pure water/the tendency for water molecules to leave/move out of a system.

Question 14

Answer 14

(iii) P – 700 kPA
Q – 600 kPA

Question 15

(ii) State the name of the condition shown by cell Y and explain how this condition could have arisen.
[3]

Answer 15

(ii) plasmolysed; [1]
cell in concentrated solution/low water potential; water passes out;
cytoplasm/vacuole shrinks.
(not: cell membrane comes away from the wall) [2 max]

Question 16

c) (i) Cell X has the higher pressure potential Wp. Explain how this pressure potential is built up in cell X. [3]

Answer 16

water passes into cell by osmosis; cytoplasm expands;
cell becomes turgid;
as cytoplasm/contents push against wall;
wall inelastic/resists further expansion. (not: rigid) [3 max]

Question 17

Suggest the effect on seedlings if all of their cells were in the condition as shown in plasmolysis. [1]

Answer 17

wilts

Question 18

Answer 18

1. water moves in by osmosis;
2. {higher water potential outside rbc/lower inside}/down water potential gradient into the cell;
   Accept: correct ref to water potential less negative outside cell
   Reject: ref water concentration
   Neutral: ref to soluble concentration/hypotonic/hypertonic
3. ref. no cell wall to prevent bursting/cell membrane unable to withstand pressure;
4. Most haemoglobin is released therefore lowest transmission of light/less light reaches sensor;

Question 19

Explain why there is a range of concentrations at which haemolysis occurs. [2]

Answer 19

Different cells have different {(solute) concentrations/solute potential/water potential};
Reject: water concentration
Each cell would require a different {external water potential/solute concentration}, before haemolysis/bursting;

Question 20

Answer 20

Any 3 from:
1. cell plasmolysis;
2. cytoplasm decreases in volume;
3. cell membrane pulls away from cell wall;
3. vacuole decreases in volume;

Question 21

(ii) State the pressure potential after it had been in the concentrated salt solution for 30 minutes.

hypertonic solution

Answer 21

0 kPa

The vacuole shrinks,
The cytoplasm pulls away from the cell wall (a process called plasmolysis),
The cell becomes flaccid, and the pressure potential becomes zero, as there is no internal pressure exerted on the cell wall.

Question 22

Answer 22

triglyceride

protection of vital organs (1)
{thermal/ electrical} insulation (1) NOT prevent heat loss
energy {storage / source}(1) NOT energy release
metabolic water (1)
buoyancy (1)
waterproofing (1)

Question 23

Describe how you would test for the presence of lipid in a tissue extract. [2]

Answer 23

mix sample thoroughly with ethanol (and water) (1)
emulsion (test) / goes {cloudy/ milky}(1)
Accept details of alternative tests, e.g. Sudan III - goes red,
brown paper test - translucent

Question 24

Describe how components Y and Z in the lipid differ from component X and explain
why it is recommended that humans eat a higher proportion of lipids containing Y and
Z.

Answer 24

X is saturated but {Y and Z are unsaturated /
Y monounsaturated + Z polyunsaturated} (1)
unsaturated fats decrease level of {LDL / low density
lipoprotein} / cholesterol/ ORA/ unsaturated fats increase
level of HDL (1)
reduces risk of{ heart disease / atherosclerosis/ atheroma
formation/ description of atheroma formation}/ ORA (1)

saturated = single bonds

Question 25

Answer 25

{phosphate/ head} is {hydrophilic/ polar} so attracted to
water/ owtte (1)
{fatty acids/ tails} are {hydrophic/ non-polar} so repelled from
water/ owtte (1)
Accept heads are hydrophilic, tails are hydrophobic (with no
explanation)

Question 26

The plasma membrane of a single human red blood cell has a surface area of
140 um 2
. Calculate the area of water covered by the lipids extracted
from the plasma membrane of a single red blood cell. [1]

WHY??

Answer 26

280 um

phospholipids arranged in bilayer in cell membrane (but in a
single layer on water)

Question 27

Answer 27

ref to fluid mosaic model (1)
proteins can {move / diffuse} within membrane / proteins are
arranged randomly(1)
{Fluorescence restored in the area exposed to the laser after
5 minutes / {Other/ non affected} proteins can move into the
area exposed to the laser after 5 minutes (1) (must in context
of protein movement)

Question 28

Answer 28

The total surface area of red blood cell is half the area of the
monofilm as phospholipids arrange themselves as a bilayer (1)
With hydrophilic heads facing the polar/ ionic molecules on both
sides of the membrane(1)

Question 29

Answer 29

proteins are embedded in the phospholipid bilayer as well as
being located outside phosphate head / there are extrinsic
(In the fluid mosaic model, proteins are embedded within the lipid bilayer and are not continuou)

and
intrinsic proteins not just extrinsic (1)

many proteins present in the membrane but not linked together
as in the diagram (1)
channel / pore is formed from a single protein not lined with
protein (1)

Question 30

Answer 30

Mosaic = protein molecules embedded in the membrane (1)
Fluid = protein and phospholipid molecules free to move (1)

Question 31

Answer 31

Change in temperature would change the kinetic energy of the
molecules and therefore the rate(1)
High temperature could denature proteins in the membrane(1)
Change in pH could change the tertiary structure of the protein
membranes(1)

Question 32

(ii) Using the results shown on the graph, what conclusions can be drawn
about the uptake of phosphate ions by the cells. Justify your
conclusions. [6]

Answer 32

Concentration of phosphate ions inside cell must be 30a.u.(1)
Because in the absence of oxygen and in concentrations less
than 30a.u. there is no uptake of phosphate(1)
Absence of oxygen uptake must be by facilitated diffusion/
diffusion(1)
In the presence of oxygen and in concentrations less than 30a.u.
must be active transport as phosphate uptake is against
concentration gradient and ATP is present(1)
At concentrations greater than 30 a.u. and oxygen present
uptake is by both active transport and facilitated diffusion/
diffusion (1)
At high concentration of phosphate all carrier proteins are in use
and become a limiting factor(1)

Question 33

Answer 33

“The concentration at which the line crosses the x-axis would be higher.”

To reach the isotonic point (no net water movement), the external sucrose solution must match the higher sucrose concentration of the sweet potato cells.
Therefore, the x-axis intercept (isotonic point) will occur at a higher sucrose concentration for sweet potatoes compared to regular potatoes.

Question 34

Answer 34

Increase the sucrose concentration (1)
Accept use a wider range
(Until a point is reached) where water will flow out of the potato
(1)

Question 35

which part is non-polar

Answer 35

S is non-polar/ hydrophobic (R groups) as it is adjacent to the
fatty acids (1)

Question 36

Explain why the water potential of the cell can be assumed to be equal to the solute
potential of the solution that causes 50% plasmolysis. [1]

Answer 36

At incipient plasmolysis pressure potential = 0 because cell
membrane is just pulled away from cell wall (1)

Question 37

State how the procedure could be modified to improve reliability and accuracy. [2]

Answer 37

Reliability:
{observe more pieces of onion (tissue from each solution) /
count more cells (in each piece of tissue)} AND calculate a mean
Accuracy:
narrower range {around 50% plasmolysis/ 0.4-0.6} (1)