1 Answers to end-of-chapter questions
1
A
[1]
2
C
[1]
3
B
[1]
4
B
[1]
5
C
[1]
6
B
[1]
7
D
[1]
8
A
[1]
Structured questions
9
a
i
α-carbon / central carbon
R group /
side group
carboxyl
group
amino
group
Each correct label [1]
[max 4]
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Original material © Cambridge University Press 2011
1
ii
H2O
condensation
Peptide bond /
linkage
Showing where condensation
would occur [1]
Correct dipeptide [1]
+ H2Opeptide bond [1]
Identifying
b
i
Level
Types of bond
Position of bonds
Effect of bond
formation
formed between the
–NH3 + group of one
amino acid and the
–COO– group of
another amino acid
formed between –CO
and –NH groups of
the peptide linkage
linear molecule,
determines the
sequence and number
of amino acids
Primary
structure
peptide /
covalent
Secondary
structure
hydrogen
Tertiary
structure
ionic
formed between R
groups / side chains
with –COO– and
–NH3+ groups
hydrogen
formed between –OH
groups, –OH and
–COOH groups, –OH
and –NH2 groups
disulphide
between the sulphur
groups of two
cysteine molecules
hydrophobic
formed between nonpolar groups
formed between side
chains of two or more
polypeptides
Quaternary
structure
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ionic
hydrogen
hydrophobic
disulphide
results in folding of
the protein to give
either an α-helix or βpleated molecule
results in the
polypeptide folding to
give a globular or 3D
structure
stabilises / holds the
polypeptide
monomers together
8–12 points [3]
4–7 points [2]
3 points [1]
< 3 points [0]
Original material © Cambridge University Press 2011
2
ii
c
[1]
[1]
Any three points (1 mark each):
•
•
•
•
•
•
•
•
10 a
Similarity: both have hydrogen bonds between –CO and –NH groups of the
peptide linkage groups
Difference:
α-helix: hydrogen bonding between –CO and –NH groups of amino acids four
places apart in a single chain
β-pleated sheet: hydrogen bonding between –CO and –NH groups of amino acids
of adjacent chains
transport molecules in cell membrane, i.e. carrier and channel proteins
oxygen-transport molecules, e.g. haemoglobin
structural proteins, e.g. collagen in skin, tendons, walls of blood vessels
enzymes, e.g. pepsin, amylase
defence, e.g. antibodies
as signalling molecules, e.g. hormones
contractile for movement, e.g. tubulin
storage, e.g. egg albumen
[max 3]
i
[3]
ii
b
In the β form of glucose, the –OH on carbon 1 is on the same side of the ring as
carbon 6 / it is above the plane
[1]
i
Procedure and test reagents
Biuret Iodine in Ethanol
Boil with
solution potassium
Benedict’s
dilute
iodide
and heat
acid and
neutralise
[tick]
[cross]
[cross]
[cross]
[cross]
reducing
sugar
[tick]
[tick]
[cross]
[cross]
[cross]
nonreducing
sugar
[cross]
[cross]
[cross]
[tick]
[cross]
starch
[cross]
[cross]
[tick]
[cross]
[cross]
protein
[cross]
[cross]
[cross]
[cross]
[tick]
lipid
Food
ii
4–5 correct [4]
3 correct [3]
2 correct [2]
1 correct [1]
Specimen contains a reducing sugar and protein
Biology for CAPE
Original material © Cambridge University Press 2011
[1]
3
c
Stage 1: calibration / colour standards prepared
(a) Make 0.1%, 0.5%, 1.0%, 2.0% concentrations of glucose by serial dilution.
(b) Label 5 test-tubes 0.1%, 0.5%, 1.0%, 2.0% and 4.0% glucose respectively.
(c) Add 5 cm3 of Benedict’s solution to each test-tube.
(d) Add 0.5 cm3 of each of the glucose solutions to the appropriate test-tube, using a
clean syringe each time.
(e) Shake each test tube in order to mix the contents.
(f) Place all five test tubes in a boiling water bath for 2 minutes.
(g) Carefully remove the test tubes and place in a test-tube rack.
Stage 2: Benedict’s test on fruit juices
(a) Repeat steps (b) to (g) using the fruit juices A, B and C.
(b) Stir the colour standards and fruit juices.
(c) Compare the colours of the fruit juices with the known glucose concentrations.
(d) Estimate the % concentration of glucose of the fruit juices using the colour
standards.
Stage 3: Conversion of % glucose to mg glucose
4% glucose = 4000 mg of glucose in 100 cm3 of water
Therefore 0.5 cm3 of 4% glucose contains 4000 / 100 × 0.5 mg of glucose = 20 mg
glucose.
Logical sequence of method [4]
0.5 cm3 of 2.0% glucose = 10 mg
Same volumes, time used
0.5 cm3 of 1.0% glucose = 5 mg
for both Stage 1 and 2 [2]
0.5 cm3 of 0.5% glucose = 2.5 mg
Conversion of % to mg [1]
0.5 cm3 of 0.1% glucose = 0.5 mg
[max 7]
11 a
i
Correct structure for glucose [1]
Correct structure for fructose [1]
Identification of where bond will be
formed [1]
Correct structure of sucrose [1]
ii
Condensation reaction
iii
Function: Sucrose is the main form in which
carbohydrates are transported in plants.
Relation to structure: It is soluble and less reactive than
glucose.
Function: It is an important source of energy.
Relation to structure: It can be hydrolysed to give
glucose and fructose. Fructose isomerises to give glucose.
Hence two/one molecule of sucrose produces two
Any function [1]
molecules of glucose for respiration.
Corresponding relation to function [1]
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[1]
Original material © Cambridge University Press 2011
4
iv
Sucrose is a non-reducing sugar because its reactive –OH groups on carbon 1 in
glucose and carbon 2 in fructose are used to form the glycosidic bond.
Therefore, these groups are not available to reduce the Cu2+ ions in
Benedict’s solution
[1]
[1]
b
Form of
glucose
Bonds
between
monomers
Features
of
molecule
Function
Starch
α-glucose
Cellulose
β-glucose
Glycogen
α-glucose
amylose:
α 1–4;
amylopectin:
α 1–4 and
α 1–6
compact,
insoluble and
easily
hydrolysed to
glucose
β 1–4
α 1–4 and α 1–6
straight chain,
with –OH
projecting
outwards;
hydrogen bonding
between
molecules to form
fibrils
structural in
plants, cell wall
formation in
plants
Compact,
insoluble
molecule, easily
hydrolysed to
glucose
energy
storage
molecule in
plants
energy storage
in animals
Any 2 points [1]
[max 6]
Essay questions
12 a
Structure of water can be shown by a diagram as well as described in words.
In a water molecule, the two hydrogen atoms are found to
be on one side of the oxygen atom.
The oxygen atom pulls the bonding electrons towards it,
which makes the oxygen slightly negatively charged. The
hydrogen atoms have small positive charges. This unequal
distribution of charge is called a dipole.
• States that hydrogen bonding is responsible for properties of water.
[2]
[1]
[1]
Any five features with effect (1 mark each point):
• solvent properties: can dissolve ionic / polar substances by isolating
and dissolving them
water can act as a transport medium / allows for metabolic reactions
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Original material © Cambridge University Press 2011
5
• specific heat capacity: requires a large amount of heat to increase
temperature
this reduces fluctuations in temperature / more stable habitat /
maintains temperature for enzyme activity / metabolic reactions
• high heat of vaporisation: requires a large amount of energy to
change from liquid to vapour
helps to maintain body temperature by sweating / cooling mechanism /
cooling in plants
• high heat of fusion: requires a large amount of energy to change state
from liquid to solid and solid to liquid
makes it difficult for water to freeze / cells contain cytoplasm which is
mostly water / ice crystals would not form inside cells
• density and freezing properties: maximum density at 4 oC / ice less
dense than liquid water / ice floats acts as an insulator / allows for
organisms to survive in freezing conditions below ice / circulation of
nutrients
• high cohesion and surface tension: hydrogen bonding holds water
molecules together
allows for mass flow in plants / small animals can walk across water
for food
• pH: measure of concentration of hydrogen ions in a solution works
with other substances to act as buffers
• transparency: transparent to visible wavelengths of light allows for
photosynthesis to take place by algae / plants in water
• reactivity: water takes part in many metabolic reaction
[1 mark each, max 5]
b
i
Globular – formed by the folding of polypeptide by interaction of R groups (ionic,
hydrophobic, disulphide and hydrogen bonding) into complex shapes / compact
spherical protein
[2]
ii
4 subunits (2 α and 2 β) interlock
haem group which carries oxygen
globular protein
Diagram with four subunits [1]
Labels [1]
•
•
•
•
13 a
i
Hb molecule has 4 haem groups
oxygen binds to haem group
4 oxygen molecules carried by each Hb molecule
when first oxygen molecule binds to first haem group, there is a
small change in the four polypeptides’ tertiary structure, making it
easier for an oxygen molecule to bind to the other three
Starch: made up of amylose and amylopectin
Monomer: α-glucose
In amylose: α 1–4 glycosidic bond, unbranched / helix / spiral
In amylopectin: α 1–4 and α 1–6 glycosidic bond / branched
Biology for CAPE
Original material © Cambridge University Press 2011
[1]
[1]
[1]
[1]
[2]
6
Features: compact; takes up little space; does not interfere with metabolic reactions
/ inert; insoluble; does not affect water potential of cell
[2]
ii
Starch
• made up of α-glucose
monomers
• α 1–4 glycosidic bond
• no rotation of monomer
units
• spiral / helix; amylopectin
branches
• no hydrogen bonding
between molecules
b
Cellulose
• made up of β-glucose
monomers
• β 1–4 glycosidic bond
• alternate units rotated 180o
• linear / straight chain /
unbranched
• hydrogen bonding between
molecules to form
microfibrils
Any 4 comparative
points [4]
Any two points 1 mark:
• alternate monomer units rotated 180o
• many –OH groups sticking above and below the plane of molecule
• many hydrogen bonds within molecule
• many hydrogen bonds between molecules
• linear / unbranched chains
• 60–70 chains link by hydrogen bonding to form microfibrils
• arranged in bundles to form fibrils
c Any point 1 mark:
• A single molecule made up of three single polypeptide chains twisted around
each other / Each chain has a shape of a helix (due to hydrogen bonding and
the secondary structure) / Molecule is described as a triple helix
• Every third amino acid is the smallest amino acid, glycine / glycine allows
the chain to coil tightly
• There are covalent bonds formed between the triple helix. This occurs
between the NH2 group of one chain and the COOH of another
• Ends of the parallel molecule are staggered and hence there are no weak
spots. It is strong because the fibrils overlap
• Each complete three-stranded molecule cross-links with other collagen
molecules running parallel to it to form fibrils
[max 3]
[max 4]
14 a
ester linkage – bond formed by
condensation of carboxyl and hydroxyl
groups
2 saturated fatty
acids (non-polar
tails)
unsaturated fatty acid –
causes kinks in the
molecule – nonpolar tail
Biology for CAPE
provides –OH groups,
forms nonpolar head of
molecule
Drawing [2]
Original material © Cambridge University Press 2011
7
4 labels [3]
b
•
•
•
•
•
phospholipid – consists of a glycerol residue with two fatty acid tails and a
phosphate group
phosphate group is hydrophilic
the fatty acid tails are hydrophobic
5 points [3]
triglyceride – consists of a glycerol residue with 3 fatty acids tails
3–4 points [2]
the molecule is nonpolar / hydrophobic
2 points [1]
c Any point 1 mark:
•
•
•
•
•
•
•
•
phospholipids: consist of hydrophilic head and 2 hydrophobic tails
hydrophilic head attracted by water so orients towards the aqueous medium
hydrophobic tails repelled by water and so oriented away from the aqueous
medium
forms a bilayer
hydrophobic tails prevent movement of water-soluble / polar molecules
allows the movement of lipid-soluble / nonpolar molecules
weak interaction between phospholipid molecules makes membrane fluid
the fluid nature of the membrane allows it to unfold, break and reconnect easily /
exocytosis / endocytosis
max [7]
Biology for CAPE
Original material © Cambridge University Press 2011
8