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Biology Book 1
Public examination questions
Suggested Answers
2 Molecules of Life
A.
Multiple choice
1.
B
2.
A
3.
C
4.
C
(1 mark each)
(Total: 4 marks)
B.
Short questions
1.
purine
pyrimidine
pyrimidine
purine
(Total: 3 marks)
2.
(a)
(b)
carbohydrates / sugars / polysaccharides
(1 mark)
vitamins
(1 mark)
(i)
(ii)
those that must be ingested
those that cannot be synthesised (by the human body)
(1 mark)
to make protein / polypeptide / named protein
(1 mark)
to make other / non-essential amino acids
(1 mark)
(Total: 5 marks)
1
C.
Essays
1.
Similar functions:
(1)
(max. 8 marks)
acting as respiratory substrates to release energy through respiration
•
lipid is suitable because triglyceride / fatty acid in lipid can be broken down
and enter Krebs cycle to yield ATP
lipids have high H:O ratio therefore easily oxidised.
•
protein is suitable because protein can be hydrolysed into amino acid
and amino acid can be broken down to carbon skeleton
and enter the Krebs cycle to yield ATP.
(2)
acting as structural components
•
phospholipids and proteins are components of plasma membrane of cells and
the components of cytoplasm
(3)
as hormones
•
lipid hormones like steroid hormones and
protein hormones like insulin regulate and coordinate body activities.
(4)
for protection
•
adipose tissue acts as shock absorber protecting internal organs from mechanical
injuries or
lipids form waterproof cuticle on the epidermis of plants and
oil on the skin of mammals preventing dehydration or
protein like keratin in nails hardens the nails for defence
(5)
as energy reserve and storage compounds
•
lipid stored as fat in animals and oil in plants and
proteins in body can be converted into respiratory substrates in extreme
starvation
(6)
contribute to differential permeability of cell membrane
•
lipid is suitable because phospholipids bilayer of membranes facilitates transport
of lipid soluble substances
•
protein is suitable because channel proteins selectively enhances only certain
molecules and ions to pass through the membrane
Different functions:
(1)
(max. 7 marks)
functions performed by proteins but not lipids
•
enzymes catalyses metabolic reactions or
•
haemoglobin transports oxygen or
•
antibodies for body defence or
•
carrier proteins and channel proteins help transport materials or
•
protein is suitable for the above function because they have specific 3-D
2
conformation
(2)
functions performed by lipids but not proteins
•
subcutaneous fat serves as an insulator to reduce heat loss
•
fat serves as a shock absorber protecting important organs against mechanical
injuries
•
lipid helps to transport lipid soluble substances like vitamins A, D, E and K
•
myelin sheath around nerve fibres help speeding up nerve impulses transmission
Communication
(max. 5 marks)
(Total: 20 marks)
3
3 Cellular Organisation
A.
Multiple choice
1.
D
2.
B
3.
B
4.
D
5.
B
6.
B
(1 mark each)
(Total: 6 marks)
B.
Short questions
1.
(a)
phospholipid
(b)
ions/polar or charged molecules such as glucose, amino acids, proteins, water (1 mark)
(c)
acting as ion channels/ trans-membrane carriers/ electron carriers/ enzymes/ receptors/
(d)
(1 mark)
maintaining structural integrity of membrane
(1 mark)
antigens/ recognition markers
(1 mark)
(Total: 4 marks)
2.
(i)
mitochondrion
(1 mark)
(ii)
respiration / aerobic respiration
(1 mark)
ATP production
(1 mark)
(iii) energy (ATP) needed for the movement of cilia
(1 mark)
(iv) 50 ±2 (mm) ÷ 10000
(1 mark)
= 5μm
(1 mark)
(Total: 6 marks)
3.
B:
6
C:
2
D:
5
E:
1
F:
4
(1 mark each)
(Total: 5 marks)
4
C.
Structured question
1.
(a)
eukaryotic cell
prokaryotic cell
sometimes present

nuclear membrane


ribosomes


carries out respiration


sometimes present

cell wall
chloroplast
(1 mark each)
(b)
(i)
a group of similar cells specialised to perform a particular function
(ii)
Any one of the following:
(2 marks)
(1 mark)
xylem
phloem
epidermis
palisade mesophyll
spongy mesophyl
(iii) Any one of the following:
(1 mark)
muscle
bone
epithelium
nervous tissue
connective tissue
blood
(Total: 9 marks)
5
4 Movement of Substances across Membrane
A.
Multiple choice
1.
B
2.
B
3.
C
4.
C
5.
B
6.
D
7.
C
(1 mark each)
(Total: 7 marks)
B.
Short questions
1.
(a)
The water outside has a higher water potential than the cells of pears
(1 mark)
water moves into the cells
(1 mark)
increasing the volume of the cell content/ water content of the cell.
(1 mark)
As a result, more fruit juice can be extracted.
(b)
(c)
The skin of the fruit acts as a barrier to water.
(1 mark)
Little/ no water movement will occur.
(1 mark)
The water/ pear may be contaminated with pathogens/ lead to food poisoning. (1 mark)
(Total: 6 marks)
2.
(a)
plasmolysed
(1 mark)
(b)
As the plasma membrane is permeable to glycerol, glycerol can diffuse into the cell.
(1 mark)
This lowers the water potential of the cell
(1 mark)
leading to the entry of water into the cell by osmosis.
(1 mark)
The cell regains turgidity/ restores its original appearance.
(1 mark)
Sucrose molecule is large, it cannot diffuse into the cell easily,
(1 mark)
thus the cell remains plasmolysed.
(Total: 6 marks)
3.
(i)
K
(ii)
vacuole in cell K has less water in it than cell L.
(1 mark)
Vacuole in cell K has lost more water than cell L.
The water potential outside cell K is lower. (Any one)
(iii) Cell wall is freely permeable / permeable to salt
(1 mark)
(1 mark)
(Total: 3 marks)
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4.
(a)
(b)
diffusion
(1/2 mark)
facilitated diffusion
(1/2 mark)
active transport
(1/2 mark)
energy / ATP
(1/2 mark)
carrier
(1/2 mark)
phagocytosis
(1/2 mark)
Similarity:
Both are passive / do not require energy.
Both of them move molecules down a concentration gradient. (Any 1)
(1 mark)
Difference:
Osmosis involves transport of water only or vice versa.
Osmosis always involves movement of molecules through a partially permeable
membrane or vice versa. (Any 1)
(1 mark)
(Total: 5 marks)
C.
Structured question
1.
(a)
Facilitated diffusion:
Proteins provide channels/ pores/ carriers in membrane.
(1 mark)
Channels/ pores can open or close; carriers can change shape
(1 mark)
for ions/ large molecules/ molecules insoluble in lipids to pass through down the
concentration/ diffusion gradient
(1 mark)
Active transport:
(b)
Energy is used to change the shape of the carrier protein.
(1 mark)
Transport substances up the concentration gradient / for faster uptake.
(1 mark)
(i)
(ii)
Increase in concentration (in cytoplasm) for first (2.3 hours /2.4 hours / 2h 18
min. / 2h 24 min.).
(1 mark)
Then concentration remains constant/ plateau/ levels off.
(1 mark)
At first, X diffuses in/ can enter down concentration gradient.
(1 mark)
Once concentration of X in cytoplasm equals concentration outside, no more X
can be taken up.
(1 mark)
(Total: 9 marks)
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5 Cell Cycle and Cell Division
A.
Multiple choice
1.
D
2.
C
3.
A
4.
A
5.
D
6.
B
(1 mark each)
(Total: 6 marks)
B.
Short questions
1.
During the first division of meiosis, homologous chromosomes form pairs.
(1 mark)
Homologous chromosomes orient/align themselves randomly in the middle of the cell.
(1 mark)
The orientation of one homologous pair of chromosomes is independent of others. (1 mark)
Then the members of each homologous pair separate/move to opposite poles;
(1 mark)
The random distribution and subsequent independent assortment of the chromosomes lead to
great genetic variety among gametes.
(1 mark)
(Total: 5 marks)
2.
(a)
centromere
(b)
membrane
(c)
chromosomes
(d)
poles / ends
(e)
cytokinesis
(f)
genetically
(1 mark each)
(Total: 6 marks)
C. Structured Question
1
(a)
(i)
Label any one of the pairs in the diagram:
(1 mark)
8
homologous chromosome pair 1
homologous chromosome pair 2
(1 mark)
(ii)
(b)
2
(1 mark)
46
(1 mark)
(c)
Cells produced by mitosis
Cells produced by meiosis
- diploid number of chromosome
- haploid number of chromosome
(2 marks)
- genetically identical to parent
- genetically different to parent
(2 marks)
cells
cells
(Total: 7 marks)
9
6 Metabolism and Enzymes
A.
Multiple choice
1.
B
2.
D
3.
A
4.
B
5.
C
6.
C
(1 mark each)
(Total: 6 marks)
B.
Short questions
1.
Example 1:
•
non-competitive inhibitor/ allosteric inhibitor
•
inhibitory action of the inhibitor cannot be overcome by increasing substrate
(1 mark)
concentration
OR
•
lower enzymatic activity irrespective of substrate concentration
(1 mark)
because inhibitor does not bind to the active site.
(1 mark)
Example 2:
•
irreversible inhibitor
(1 mark)
•
lower enzymatic activity irrespective of substrate concentration
(1 mark)
•
binds to enzyme permeability/ forms a covalent bond with enzyme
(1 mark)
(Total: 3 marks)
2.
(a)
(i)
Advantages (any two of the following):
•
(2 marks)
speed up chemical reactions in industrial process for mass production of
products
•
enzymes are specific in action, thus they can catalyse specific processes /
less likely to generate undesirable products
•
enable artificial manipulation of the rate of industrial process by controlling
reaction temperature or pH
(accept other correct / reasonable alternatives)
(ii)
Limitations (any two of the following):
•
(2 marks)
the industrial process will be sensitive to changes in temperature as enzyme
works in specific temperature range (accept similar concept such as enzyme
denatured by high temperature)
10
•
the industrial processes will be sensitive to pH changes as enzyme works
within specific pH range
•
vessels / utensils used in the industrial processes have to be clean so as to
avoid degradation of the enzyme / easily affected by inhibitors
(accept other correct / reasonable alternatives)
(b)
(i)
(ii)
protease and lipase may damage the skin / eye / mucous membrane as these body
parts consist of protein and lipid
(1 mark)
wear gloves (accept other correct / reasonable alternatives)
(1 mark)
(Total: 6 marks)
3.
(a)
the site (on the surface of an enzyme) to which substrate(s) bind(s) / the site (on the
enzyme) where it catalyses a chemical reaction
(b)
(c)
(1 mark)
Any two of the following:
(2 marks)
•
brings substrates close to an active site / in correct orientation
•
forms enzyme-substrate complex / substrate(s) bind(s) to the active site
•
lowers the activation energy for the reaction
•
weakens the bonds in the substrate
Any three of the following:
(3 marks)
•
enzymes have an optimal pH
•
lower activity above and below optimum pH
•
too acidic / basic pH can denature an enzyme
•
changes shape of the active site / tertiary structure altered
•
substrate cannot bind to active site / enzyme-substrate complex cannot form
•
hydrogen / ionic bonds in the enzyme / active site are broken / altered
(Total: 6 marks)
4.
enzymes have an optimum temperature
(1 mark)
increase in temperature increases motion/energy of enzyme and substrate molecules;
increases the frequency of contact between substrates and active sites;
reaction rate increases
(accept converse i.e. effect of lowering temperature)
(1 mark)
high temperature disrupts intermolecular forces / denatures enzymes
(1 mark)
(Total: 3 marks)
C.
Structured questions
1.
(i)
The catalase in the paper disc broke down hydrogen peroxide, releasing oxygen.
(1 mark)
When the amount of oxygen evolved reached a certain level, the oxygen bubbles
buoyed up the disc to the surface of the hydrogen peroxide solution.
(1 mark)
11
(ii)
(1)
pH of hydrogen peroxide solution
Rate of reaction (s-1)
3
＜0.01
5
0.02
7
0.05
9
0.05
11
0.04
(2 marks)
(2)
Effect of pH on catalase activity
0.05
-1
Rate of reaction (s )
0.04
0.03
0.02
0.01
0
3
5
7
pH
9
11
(3 marks)
(iii) (1)
(2)
any value between 7 and 9 (or any value read from the graph drawn)
(1 mark)
Repeat the experiment using hydrogen peroxide solutions with smaller pH
intervals between 7 and 9.
(1 mark)
(iv) The catalase activity would be similar to that of the previous investigation
(1 mark)
because catalase is not denatured at 4ºC and it can become active again at room
temperature.
(1 mark)
(Total: 11 marks)
12
7 Photosynthesis
A.
Multiple choice
1.
A
2.
D
3.
A
(1 mark each)
(Total: 3 marks)
B.
Structured questions
1.
(i)
(ii)
The bacteria distributed evenly in I,
(1 mark)
but concentrated on the chloroplast in II.
(1 mark)
In the presence of light, the chloroplast produced oxygen during photosynthesis.
(1 mark)
This led to the movement of bacteria toward the chloroplast.
(1 mark)
(iii) To study the effect of light of different colours on (the rate of) photosynthesis. (1 mark)
(iv) Photosynthesis occurs at similar rates in red and blue lights
(1 mark)
which are higher than that in green light.
(v)
(1)
(1 mark)
Workable set-up: (must include light source, water, waterweed and colour filters)
(1 mark)
Labels: [1] colour filters, [2] waterweed, [3] water / lamp / gas or oxygen
(1 mark)
Title
(1 mark)
Set-up for studying the effect of light colours on photosynthesis
(2)
Measure the volume of oxygen produced / number of oxygen bubbles released
per unit time / over a period of time.
(1 mark)
(Total: 11 marks)
2.
(a)
Concept for mark award:
Any two for the following:

(2 marks)
why CO2 concentration can affect the rate of GP formation
13

why temperature can affect the rate of GP formation

why light intensity can affect the rate of GP formation

why water in soil can affect the rate of GP formation
Example 1:

CO2 is raw material for GP

temperature will affect the kinetic energy of the reactants
Example 2:

temperature will affect the rate of enzyme catalysis

light for ATP synthesis which is necessary for the resynthesis of RuBP, which in
turn forms GP/ light opens up stomata and increases CO2 availability
(b)
(c)
(d)
ATP provides the energy for the conversion of GP to TP
(1 mark)
NADPH reduces GP to TP/ supply H
(1 mark)
Calvin cycle will stop
(1 mark)
no regeneration of RuBP from TP/A
(1 mark)
RuBP is necessary to fix CO2
(1 mark)
Concept for mark award:

name and relation of the properties of two different polysaccharides with their
roles (accept tabulated answer)
(2 marks)
Example:
Polysaccharide
Relationship between property and role
starch
insoluble, osmotically inactive, thus good for storage
has mechanical strength, thus good for support/
cellulose
(e)
(i)
high tensile strength, can maintain shape of cells
Concept for mark award:

TP/ GP/ compound A feeds to glycolysis/ provides carbon skeleton for
amino acid synthesis

formation of amino acid from pyruvate/ intermediates of Kreb cycle 
amino acids

glycolysis  Krebs cycle

polymerisation of amino acids  protein

concept of amination/ concept of transamination
14
Example of flowchart:
Compound A
or
Triose phosphate (also accept GP)
pyruvate
or
glycolysis
+ NH4+
amino acid
polymerization
protein
Krebs cycle
intermediates
+ NH4+
amino acid
(4 marks)
(ii)
(f)
Any two of the following:

ammonium/ nitrate

sulphate

phosphate
(2 marks)
use tracer technique
label the O of CO2 with
(1 mark)
18O/
the isotope of O
(1 mark)
let photosynthesis take place/ exposure the plant to light and collect the gas (O2)
evolved
(1 mark)
in a separate experiment, label the O of H2O with 18O/ the isotope of O
(1 mark)
interpretation of result: whichever experiment releases the labelled O2 will determine
the source
(1 mark)
(Total: 20 marks)
3.
(a)
(b)
(i)
light intensity
(1 mark)
(ii)
Some other factors becomes limiting.
(1 mark)
e.g. carbon dioxide or temperature
(1 mark)
(i)
experiment 1
(1 mark)
(ii)
If temperature increases from 15ºC to 25ºC, the rate of photosynthesis increases.
(1 mark)
If carbon dioxide concentration increases from 0.04% to 0.4%, the rate of
photosynthesis increases.
(c)
(1 mark)
denaturing of enzyme / change in shape of active site
(1 mark)
photochemical reactions and the reactions in Calvin cycle affected
(1 mark)
increase in the rate of respiration, which occurs at faster rate than photosynthesis
(1 mark)
15
increased rate of transpiration leads to stomatal closure; less carbon dioxide uptake
(1 mark)
(Total: 10 marks)
16
8 Respiration
A.
Multiple choice
1.
A
2.
C
3.
D
4.
D
5.
A
6.
D
7.
D
8.
C
(1 mark each)
(Total: 8 marks)
B.
Structured questions
1.
(a)
(b)
130
(1 mark)
(1/2
correct title
mark)
(1/2 + 1/2mark)
correct labeling of axes
(11/2marks)
correct drawing and labeling of bars
Title: Activity of three brands of yeast
160
Percentage change in the
volume of mixture (%)
140
120
100
80
60
40
20
0
Brand A
Brand B
Brand C
Different brands of yeast
(c)
(d)
Anaerobic respiration of yeast
(1 mark)
produces carbon dioxide
(1 mark)
which is trapped inside the dough.
(1 mark)
Brand B
(1 mark)
17
(e)
(f)
To ensure the temperature of the three mixtures are the same / maintain the temperature
at 30ºC throughout the investigation
(1 mark)
use only one brand of yeast
(1 mark)
put the measuring cylinder in water baths at different temperatures
(1 mark)
(Total: 11 marks)
2.
(a)
(i)
(ii)
Animal cells: lactic acid
(1 mark)
Plant cells: ethanol and carbon dioxide
(1 mark)
because the enzyme(s) involved in the breaking down of pyruvic acid in animal
cells is/are different from those in plant cells
(1 mark)
(b)
NADH / FADH
(1 mark)
(c)
There is no oxygen in anaerobic conditions. Oxygen is required as the final electron
acceptor.
(d)
(1 mark)
On the diagram: CO2 is released from the steps of pyruvic acid → acetyl-CoA
(1 mark)
citric acid → 5-carbon compound
(1 mark)
(Total: 7 marks)
3.
(a)
A
(1 mark)
C
(1 mark)
C
(1 mark)
B
(1 mark)
(b)
cytoplasm/ cytosol
(1 mark)
(c)
2
(1 mark)
(d)
conversion to lactate / lactic acid
(1 mark)
by addition of hydrogen ; pyruvate acting as a hydrogen acceptor from reduced NAD
(1 mark)
recycle NAD for glycolysis
4.
(a)
accept labelled sketch diagram for marking the points below:

nitrogenous base / purine/ adenine

pentose / 5-carbon sugar / ribose

three phosphate groups / Pi

phosphorylated nucleotide
(1 mark)
(4 marks)
(adenosine as an alternative to adenine plus ribose)
(b)
Any three of the following:

biosynthesis / anabolism

protein synthesis

DNA replication

glycolysis (initial step)
(3 marks)
18
(d)

muscle contraction

cell division

active transport

electrical conduction
NAD / FAD involved in respiration:
(max. 4 marks)

associated with dehydrogenation

2 molecules of reduced NAD in glycolysis

link reaction produces 1 molecule of reduced NAD (per turn of cycle)

Krebs cycle produces 3 reduced NAD (per turn of cycle)

Krebs cycle produces 1 reduced FAD (per turn of cycle)

carries / transfers, hydrogen to, inner mitochondrial membrane / cristae /
cytochromes

mitochondrial shuttle (bringing NAD reduced from glycolysis into matrix)
NADP involved in photosynthesis:
(max. 4 marks)

produced in non-cyclic (photo)phosphorylation

hydrogen comes from, water / photolysis

(used in) Calvin cycle / light independent stage

glucose to triose phosphate step

e.g. NADP involved in transporting hydrogen from grana to stroma

e.g. hydrogen split into electrons and protons at electron transport chain
(Total: max. 14 marks)
C.
Essay
1.
Content:
(max. 8 marks)
Any eight of the following:

glucose is broken down to pyruvate in the cytoplasm

with a small yield of ATP/net yield of 2 ATP

and NADH / NADH + H+

aerobic respiration in the presence of oxygen

pyruvate is converted to acetyl CoA

acetyl CoA enters Krebs cycle

Krebs cycle yields a small amount of ATP/one ATP per cycle

and FADH / FADH + H+, NADH / NADH + H+ molecules

these molecules pass electrons to electron transport chain

oxygen is the final electron acceptor / water produced

oxidation of FADH / FADH + H+ and NADH / NADH ATP yields further ATP
Communication
(max. 3 marks)
(Total: 11 marks)
19