Download Year 12 Biology: Intervention booklet. Name

Year 12 Biology: Intervention booklet.
Name ……………………………………….
This booklet needs to be fully completed and brought to the intervention session at 2:45 on
Friday 11th November. The session will be 1 hour. You should attempt the booklet open
book using your notes and textbook to help. Answer in bullet points 1 point for each mark.
1.
The diagram shows a triglyceride molecule.
H
O
Fatty acid 1
H
C
O
C
O
Fatty acid 2
H
C
O
C
O
Fatty acid 3
H
C
O
C
H
A
(a)
Name part A.
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(1)
(b)
With reference to named parts of the diagram, explain the difference between the terms:
(i)
triglyceride and phospholipid;
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(2)
(ii)
saturated and unsaturated.
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(2)
(Total 5 marks)
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2.
Read the following passage.
During the course of a day, we come into contact with many poisonous substances. These
include industrial and household chemicals. The skin acts as a barrier and prevents many
of these substances entering and harming the body.
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The skin is one of the largest organs in the body. It is composed of several layers of
tissue. The outer layer consists of dead cells packed with keratins. Keratins are a group of
proteins that differ from each other in their primary structure. Each keratin molecule
consists of several polypeptide chains, each individual chain wound into a spiral or helix.
The polypeptide chains include many sulphur-containing amino acids and these help to
give the keratin molecules their characteristic strength.
Use information from the passage and your own knowledge to answer the questions.
(a)
What is the evidence from the passage that keratin molecules have a quaternary structure?
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(1)
(b)
Explain how sulphur-containing amino acids help to give keratin molecules their
characteristic strength (lines 8–9).
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(2)
(c)
Explain why differences in primary structure result in keratins with different properties
(line 6).
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(2)
(d)
The skin prevents poisonous substances entering and harming the body (line 3). Explain
why these substances are unable to pass through the outer layer of skin cells by active
transport.
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(3)
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(e)
Skin cells may be studied with a transmission electron microscope or an optical
microscope. Explain the advantages and limitations of using a transmission electron
microscope to study cells.
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(6)
(Total 14 marks)
3.
(a)
Cylinders of potato were cut using a cork borer. Each cylinder was placed into one of a
range of sucrose solutions of different concentrations. The cylinders were left for 6 hours
and then removed from the solutions. The mass of each cylinder was recorded before and
after immersion. The graph shows the results of this investigation.
+10
Increase
Percentage
change in 0
mass
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Concentration of
sucrose / mol dm –3
Decrease
–10
(i)
Explain why the change in mass was given as a percentage change.
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(2)
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(ii)
Explain the shape of the curve as the concentration of sucrose decreases from
0.3 mol dm–3 to 0.1 mol dm–3.
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(4)
(iii)
What concentration of sucrose solution is equivalent to the mean water potential of
the potato cells? Explain your answer.
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(2)
(b)
The diagram represents three plant cells and shows the water potential of two of these
cells. The arrows show the direction of water movement between these three cells.
A
–510 kPa
B
C
–550 kPa
Suggest the range of possible values for the water potential of cell B. Explain your
answer fully.
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(2)
(Total 10 marks)
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4.
The diagram shows part of a plant cell as seen through an electron microscope.
(a)
Name organelles A, B and C.
A ...................................................................
B ...................................................................
C ...................................................................
(3)
(b)
Give the function of
(i)
structure X;
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(ii)
structure Y.
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(2)
(c)
Calculate the width of the structure labelled X in micrometres.
Show your working.
Width ............................ micrometres (µm)
(2)
(Total 7 marks)
5.
Mitochondria were isolated from the liver tissue using differential centrifugation. The tissue was
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chopped in cold, isotonic buffer solution. A buffer solution maintains a constant pH. The first
stages in the procedure are shown in the diagram.
Tissue chopped
in cold isotonic
buffer solution
Homogenised
Stage 1
Stage 2
(i)
Centrifuged
al low speed
for 10 minutes
Stage 3
The tissue was chopped in cold, isotonic buffer solution. Explain the reason for using
a cold solution; ............................................................................................................
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an isotonic solution; ....................................................................................................
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a buffer solution. .........................................................................................................
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(3)
(ii)
Why is the liver tissue homogenised?
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(1)
(iii)
Describe what should be done after Stage 3 to obtain a sample containing only
mitochondria.
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(2)
(Total 6 marks)
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6
The diagram shows the arrangement of protein molecules in part of a cell surface membrane.
Key
Protein
(a)
Explain how amino acid molecules may be linked to form a polypeptide chain which is
folded into a specific tertiary shape.
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(6)
(b)
Describe the role of proteins in the transport of molecules and ions across cell surface
membranes.
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(7)
(c)
The hormone glucagon is a protein. It targets liver cells but does not affect other cells in
the body. Explain why.
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(4)
(Total 17 marks)
West Bridgford School
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7.
Six cylinders of a standard size were cut from a single large potato. One cylinder was placed in
distilled water and the others were placed in sucrose solutions of different concentrations. The
length of each cylinder was measured every 5 minutes for the next 50 minutes.
The graph shows the changes in length at each sucrose concentration.
+5
Distilled water
+4
Decrease +3
+2
0.2 mol dm –3
+1
Change in
length /mm 0
10
20
30
40
50
Time / minutes
–1
0.4 mol dm –3
–2
–3
0.6 mol dm –3
Increase
–4
0.8 mol dm –3
–5
(a)
1.0 mol dm –3
Explain why
(i)
the potato cylinder in distilled water increased in length;
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(2)
(ii)
the potato cylinder in the 1.0 mol dm–3 sucrose solution showed no further decrease
in length after 40 minutes.
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(2)
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(b)
(i)
Describe the difference in the rate of decrease in length during the first 10 minutes
between the cylinder in the 0.4 mol dm–3 and the cylinder in
the 0.8 mol dm–3 solution.
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(1)
(ii)
Use your knowledge of water potential to explain this difference.
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(1)
(c)
After 45 minutes the potato cylinder in the 0.8 mol dm–3 solution was removed and blue
dye added to this solution. Some of this blue-stained solution was drawn into a syringe. A
drop was then released, slowly, halfway down a test tube of fresh 0.8 mol dm–3 sucrose
solution as shown in the diagram. The blue drop quickly moved to the surface of the
liquid in the test tube.
Syringe needle
Blue drop
0.8 mol dm –3 sucrose solution
(i)
The density of a solution depends on its concentration. The more concentrated the
solution the greater its density. Explain why the blue drop had a lower density and
therefore moved up.
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(2)
(ii)
A sucrose solution of concentration 0.3 mol dm–3 has a water potential which is
equivalent to that of the potato cells. Describe and explain what would happen to
the blue drop from this solution.
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(2)
(Total 10 marks)
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