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570220228-CAPE-Unit-1-Mark-Scheme-2021

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1
02107020/CAPE/KMS 2021
C A R I B B E A N
E X A M I N A T I O N S
C O U N C I L
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®
BIOLOGY
UNIT 1 - Paper 02
KEYS AND MARK SCHEME
MAY/JUNE 2021
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 1
(a)
Diagram
N.B.- Candidates are not expected to draw in the several less dense
areas within the cytoplasm in their diagram
1. Faithfulness of representation - Clear accurate line
representation of specimen- single line for cell membrane and
cell wall separately
2. Clean continuous lines of even thickness
3. Use of label lines that do not cross and must touch structure.
4. Accurate labelling (at least three correct labels)
5. Annotations are written beside the labels
6. Features correctly proportioned
7. Suitable title below the drawing- name of organism
5-7 points 4 marks
3-4 points 3 marks
2 points 2 marks
1 point 1 mark
4 marks XS
Magnification
Reasonable value (140,000-500,000). If magnification is placed beside
title, award mark for magnification, provided it is correct
1 mark XS
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Annotations
N.B.-Annotations should be written beside the labels. They are placed
below for clarity purposes. Annotations can be structural/functional.
•
•
•
•
•
•
Cell/plasma membrane - controls the movement of substances in
and out of cells/partially permeable/acts as a barrier/allows
only certain substances to pass through/composed of a
phospholipid bilayer.
Cell wall – composed of peptidoglycan/murein/gives structural
strength /provide protection/ stops cells from bursting if it
expands
Cytoplasm – contains all the enzymes needed for metabolic
reactions/composed of water, salts and proteins/ no membrane
bound organelles / site of metabolic reactions/cellular
reactions
Cytosol- aqueous component of cell/gel like
Nucleoid - region containing naked/circular DNA
Ribosomes - synthesizes proteins/translates mRNA/70S
Any 3 correct annotations given – 1 mark each
3 marks KC
[8 marks]
Question 1 cont’d
(b)
(i)
Similarities
•
•
•
•
•
•
•
•
(Have a) nucleus/ nuclear membrane
(Have a) cytoplasm
(Have a) cell/plasma membrane
(Contain) mitochondria
(Contain) ribosomes
(Contains) vacuoles
(Contains) endoplasmic reticulum/ER / Smooth ER / Rough ER
(Contains) Golgi body/apparatus
Any 3 similarities – 1 mark each
[3 marks KC]
(ii)
•
Plant cells have chloroplasts/ thylakoid (1) which contain a
pigment called chlorophyll which captures the energy from the
sun through a series of chemical reactions/photosynthesis, the
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
•
•
•
•
sun’s energy is converted into ATP to provide energy / food
(starch, sugars) / photosynthate for the plant/ Site of
photosynthesis/site of starch production/ starch storage. (1)
Plant cells, have a rigid cell wall (1) that gives the cell
strength and structure/ The cell wall prevents the cells from
bursting (in the presence of excess water)/resists turgor
pressure/provides protection. (1)
Central/large/permanent vacuole (which is surrounded by
tonoplast) (1) which stores cell sap (sugars/ pigments/enzymes/
waste products) / regulates the osmotic properties of cell (1)
Plant cells can have plasmodesmata (1) which allows movement of
substances between adjacent plant cells (1).
Plant cells can have an amyloplast (1), which is used for the
storage of starch granules (1).
For cell sap, accept either cell sap OR any two components
Any two
Identification of 2 structure =2mks, function of each structure= 2x
1mk =2mks
[4 marks KC]
Question 1 cont’d
(c)
Marking Instructions
For any THREE differences fully discussed, award 5 marks for EACH as
follows:
Award
mark
Award
-
1 mark for identifying the key difference (seen or implied) – 1
2 marks for expanding on the difference in starch
Describing the structure of starch – 1 mark each
Explaining the importance of the structural difference – 1 mark
Award 2 marks for expanding on the difference in cellulose
- Describing the structure of cellulose – 1 mark each
- Explaining the importance of the structural difference – 1 mark
Annotated drawing may be used to describe structure of
starch/cellulose for 1 mark each
Sample Responses
There is a difference in composition / monosaccharides from which it
is formed /type of polysaccharides that makes it (1). Starch is made
up of alpha glucose (monomers)(1). Cellulose is made up of beta
glucose (monomers) (1) / Starch is a mixture of two types
polysaccharides (Amylose and Amylopectin) (1). Cellulose is made up
of only one polysaccharide (1). Presence of amylose and amylopectin
makes starch a good energy storage molecule (1). Monomers that make
up the structure of cellulose allows it to give strength to the
cell/cell wall (1).
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
There is a difference in type of chain formed / type of bond between
monomers (glucose units)/type of (glycosidic) bonding (1). Starch has
a mixture of branched and linear chains (1) whereas cellulose has
only linear/unbranched chains (1). Starch is composed of alpha-1,4glycosidic bonds (which makes it linear) and alpha-1,6-glycosidic
bonds (which allows for branching)(1). Cellulose is composed of beta1,4 glycosidic bonds (linkages), which allows the molecules to form
long and straight chains (1).
There is a difference in the orientation of monomers (1). In
cellulose alternate beta glucose molecules are rotated 180 degrees
(1). In starch the glucose molecules have the same orientation/are
not rotated 180 degrees (1). In cellulose, rotation of alternate
beta-glucose molecules allows glycosidic bonds (beta-1,4) to be
formed between alternate glucose molecules and gives the molecule its
strength/rigidity (1). Absence of rotation contributes to the
function of starch as a storage molecule as it can be packed into a
smaller area /uses less space in the cell (1).
There is a difference in overall shape of molecule (1). Amylose
molecules coil up into a spiral/helix and is compact (while
amylopectin molecule spirals to a lesser extent) (1). Cellulose
molecules do not coil but lie straight (linear)/parallel chains (1).
Hydrogen bonding occurs within the molecules of amylose and
amylopectin, which allows it to act as a storage molecule (1).
Hydrogen bonding occurs between molecules of cellulose/between -OH of
neighbouring cellulose molecules allowing it form fibers and fibrils
/Cellulose forms parallel chains, which are held together by hydrogen
bonds which gives the tensile strength (rigidity) (1).
Any 3, 5 marks each, [15 marks UK]
Total 30 marks
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 2
(a)
(i)
Labels to identify the parts of the DNA molecule
S
T
U
V
-
phosphate
deoxyribose
cytosine
thymine
Accuracy (with label lines)
3-4 correct labels – 3 marks
2 correct labels – 2 marks
1 correct label - 1 mark
1 mark for label lines not crossing (at least two drawn)
1 mark for no arrowheads on all line
IF NO LABEL LINES:
3-4 correct labels – 2 marks
1-2 correct labels - 1 mark
[5 marks XS]
(ii)
Identify a nucleotide.
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Any 1 shown on diagram – 1 mark – KC
Question 2 cont’d
(iii)
How two nucleotides in a DNA molecule are held together
1. Held together by hydrogen bonds between bases
2. A to T and C to G / purine to pyrimidine / base pairing
3. 2 hydrogen bonds between A and T/ 3 hydrogen bonds between C to
G
4. Phosphodiester bonds between nucleotides on a chain/backbone
DO NOT ACCEPT HYDROGEN BONDS OR PHOSPHODIESTER BONDS ONLY
(iv)
Any 1, 1 mark KC
Roles of messenger RNA (mRNA) and transfer RNA (tRNA) in the
production of polypeptides
mRNA
1. Provides instructions/information copied from the DNA (base
sequence)/transcription /gene/ genetic code for the formation
of a chain of amino acids in a polypeptide/protein (protein
synthesis)
2. Used to transfer genetic code (from nucleus (DNA)) to ribosomes
(in cytoplasm to make proteins / polypeptide)
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BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
1 mark each = 2 marks
tRNA
3. Deciphers/translates the codes / codons on mRNA sequence into a
sequence of amino acids in a polypeptide/protein /attaches
anticodon which are complementary to codons on mRNA and
specific amino acids
4. tRNA binds/attaches to a specific amino acid (on its amino acid
binding site according to anticodon and complementary codon)
5. tRNA transfers the specific/appropriate amino acid (to the end
of growing amino acid chain/polypeptide/protein) on the
ribosome
Any 2 points 1 mark each = 2 marks
4 marks KC
Question 2 cont’d
(b)
(i)
(ii)
Change in the gene which results in the production of abnormal
haemoglobin
• one base is changed/substituted in the sequence of DNA(coding
for normal Hb) / Adenine /A replaces thymine/T/ CTT is changed
to CAT /single base substitution
1 mark
How low oxygen saturation results in a change of the shape of red
blood cells
•
•
•
•
Valine forms a bond between themselves/hydrophobic amino acidcannot interact with water-but can interact with each other
This sticks the haemoglobin molecule together
Long fibres of stuck together haemoglobin molecules are
produced
As fibres are formed inside the red blood cells(erythrocytes),
they pull the cells out of the biconcave shape/ deform the
biconcave disc into sickle shaped cells
Any 3 points, 1 mark each
[3 marks]
(c)
(i)
Success of gene therapy of sickle cell anaemia (From data)
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BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
•
•
•
Gene therapy was effective in treating sickle cell anaemia
After 12 months, patient sickled red cell was much lower than
untreated people with the diseases (controls 3,4,5) /there was
15-19% lower sickled red cells than untreated people with the
disease.
After 12 months, sickled red cell were not significantly
different from his asymptomatic mother /there was only 1-4%
difference in sickled red cells compared to the asymptomatic
mother.
1 mark each
3 marks
Question 2 cont’d
(c)
(ii)
Marking Instructions
For any THREE challenges fully discussed, award 4 marks for EACH as
follows:
Award 1 mark for stating a relevant challenge
Award 3 marks for expanding on the challenge
- Award 3 marks for a complete discussion with at least THREE
relevant and accurate supporting points.
- Award 2 marks for a partial discussion with at least TWO
relevant and accurate supporting points.
- Award 1 mark for a limited discussion with at least ONE
relevant and accurate supporting point.
Sample Responses:
Delivering the gene to the right place (1). Gene therapies only work
if a normal gene is delivered to a large number of cells in the correct
tissue/crucial new gene reaches correct cells for success of gene
therapy (1). There may be challenges with the vectors which results in
the gene being delivered to the wrong cells (1). If delivered to the
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
wrong cells, it would be inefficient and cause health problems for
patient (1).
Activating the gene/Switching on the gene (1). Gene therapies require
the inserted gene results in the production of new proteins (1). Once
the gene is delivered to the right cell, it must be activated or turned
on to make required proteins (1). Gene once activated must remain on
to shut down genes that are too active or show unusual activity (1).
Avoiding the immune response/delivering gene without triggering immune
system (1). The role of the immune system is to fight off pathogens
such as bacteria and viruses (1). Gene delivery agents are usually
viruses which may trigger an immune response in patients (1) this could
cause serious illness or even death (case in USA of death cause by
inflammatory response to adenovirus used to transfer gene) (1).
Making sure the new gene does not disrupt the function of other genes
(1). Good gene therapy involves the new gene introduced will integrate
into patient’s genome and work for rest of their lives (1). There is
a risk that the gene will insert itself into path of another gene
disrupting its activity (1) this could interfere with important genes
involved in regulating cell division which leads to cancer (tumour
formation)/ leukaemia formation in SCID patients when gene stitched
itself to a gene regulating cell division / insertion of gene into
protooncogene/cancer causing gene (1).
The effects of gene therapy are usually short lived (1). Cells
containing therapeutic DNA should be long lived and stable in order to
have a permanent cure (1). Only treatment of somatic cells is allowed
and somatic cells die and are replaced frequently (e.g lung cells) (1).
Patients therefore need to undergo multiple treatments to achieve
consistent relief (1).
Type of genetic diseases that can be treated (1). Diseases that are
caused by a (single) recessive gene are usually treated
because it
does not have to be removed/inactivated/switched off (1) The dominant
allele can mask defective recessive allele when added to the genome
(1) however a disease caused by a dominant allele would have to be
inactivated/turned off which is difficult to do (since many processes
have to be turned off)(1).
Cost of gene therapy/commercial viability/ Ethical issue of equal
access to treatment (1). Many genetic disorders that can be targeted
by gene therapy are extremely rare and requires an individual case by
case approach (1). Treatment becomes costly since there is a need for
repeated treatments/ many patients not able to afford treatment (1).
Developing new therapy is very expensive (1) pharmaceutical firms may
not want to develop treatment because they cannot make a profit (1).
Any 3, 4 marks each
[12 marks UK]
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 3
(a)
(i)
1.
2.
3.
4.
Plot points correctly for each hormone - 2 marks
Correct labelling for both X- and Y-axes with units - 1 mark
Appropriate key (legend) labelled/curves identified - 1 mark
Appropriate scale used (25% to 100% of area used) - 1 mark
[5 marks XS]
(ii)
1. Increased FSH (around Day 5) stimulate maturation of follicle
(secondary follicle to graafian follicle/ dominant follicle)
2. LH/FSH stimulate dominant follicle to secrete oestrogen,
3. oestrogen has negative feedback on LH and FSH/surge in LH and
FSH as oestrogen continues to increase.
4. LH builds up pressure in the antrum/weakens follicular walls
leading to bursting of the follicle.
5. surge of the hormones stimulates release of secondary oocyte
from follicle into oviduct)
Any 4 points = 1 mark each = [4 marks KC]
(b)
1. progesterone stimulates thickening of the uterine lining
(endometrium),
2. cells on surface of blastocyst (trophoblasts) secrete enzymes
to partially digest cells in endometrium /cause tissues in the
endometrium to break down,
3. blastocyst burrows into the endometrium,
4. trophoblast cells produce tiny/finger-like projections
(villi)/trophoblast cells differentiate (into an outer layer,
the chorion), forming the chorionic villi,
5. villi project/grow into endometrium,
6. trophoblast secretes hCG, which stimulates corpus luteum to
continue producing progesterone,
7. endometrium also forms blood spaces (sinuses) around villi,
(whole structure called placenta)
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BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Any 4 points [4 marks KC]
Question 3 cont’d
(c)
(i)
•
•
A/Ovary wall develops into pericarp / fruit wall of fruit
B/Ovule/embryo sac develops into seed
1 mark for each correct description
[2 marks KC]
(ii)
Marking Instructions
For any THREE advantages fully discussed, award 5 marks for EACH as
follows:
Award 1 mark for stating a relevant advantage
Award 3 marks for expanding on the challenge
- Award 3 marks for a complete discussion with at least THREE
relevant and accurate supporting points.
- Award 2 marks for a partial discussion with at least TWO
relevant and accurate supporting points.
- Award 1 mark for a limited discussion with at least ONE
relevant and accurate supporting point.
Award 1 mark for providing a limitation linked to the advantage
discussed (Any one of the limitations underlined)
Sample Responses:
Maintenance of cultivar characteristics/preservation of useful
parental characteristics in next generation/Desirable characteristics
are maintained in the offspring (an example: flowers of same colour,
tea leaves, apples, mangoes) (1). New plants contain genetic material
of only one parent /clones of parent plant (which is not the case in
sexual reproduction) (1). This is especially important for commercial
growers who want to reproduce the highest quality plants and ensure
consistency of plant or crop for sale (1). It helps maintain
consistent quality and taste in produces made from plants or crops/
batch after batch of standard plants can be produced ensuring
reliability and quality control (1). There is an increased risks due
to pest and diseases due to limited genetic diversity OR if a
particular plant clone is susceptible to certain diseases there is a
potential to lose the entire crop OR The method cannot be applied to
all plant species OR Decreased genetic diversity that may have
advantageous characteristics OR Decreased ability to adapt in
changing environment (1)
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Rapid propagation/large number of plants in a short period of time
(1). Plantlet formation is induced from cultured tissue cells,
resulting in rapid multiplication by tissue culture (1). This allows
rapid production of a large numbers of plants from just one or a few
stocks plants (1). Plants can be produced anytime of the year/ allyear round availability of planting material OR It allows the
balancing of crop losses as asexual reproduction makes it possible to
rapidly regenerate a current generation of crops and minimize
losses(1). There is a high cost of tissue culture facilities and
training / must be carried out in sterile environment / labour
intensive OR There is the Possibility of off-types due to mutation
during cell culture / possibility of plants being sterile OR There is
competition for resources between parent plant and offspring (1).
Propagation of species that do not produce seeds/non-viable seeds/
too few seeds/ where seed productions is uncommon / process makes
possible production of seedless varieties(1). The seedless plants
(examples: banana, orange, rose, jasmine) can be propagated by
several natural/artificial methods (tissue culture, fragmentation,
runners, stolons, suckers, grafting and budding) without germination
of seeds/does not need to depend on fertilization to reproduce (1).
These methods reduce the complexity of obtaining offspring for
market, as the difficult processes of seed harvesting and germination
are avoided (1). Plants can be produced artificially by cuttings
(e.g. sugarcane, African Violets) / Exotic plants such as orchids
(that are hard to produce by seeds) can be cloned in large numbers by
tissue culture (1). The methods are Labour intensive to plant large
numbers of plants needed compared to mechanical drilling of a large
field of seeds OR Training required for methods like grafting/tissue
culture (1).
Easier management of plants (1). All plants have identical genetic
makeup therefore need the same environmental conditions/all plants of
the same cultivar can be taken care of under the same environmental
conditions (1) Therefore, field operations such as pest and disease
management and fertilizers can be synchronized / Disease free
plants/virus free plants can be produced from parental plants (1).
Plants grown using tissue culture are light weight and small in size
mean the plants can be air-freighted easily/takes up very little
space compared to fields (1). The root systems may not be as robust
as plants propagated from seeds OR The method cannot be applied to
all plants OR Difficulty in adapting to change in environmental
conditions (1).
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02107020/CAPE/KMS 2021
BIOLOGY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 3 cont’d
Faster maturity of crops (1). Plants reach the mature phase/flowering
sooner than using seeds (1). This shortened growing time makes it
possible for multiple yields in some environments (1). Faster
maturity of crops save a lot of time and money for commercial plant
production (1). The root systems may not be as robust as plants
propagated from seeds OR Faster maturation leads to nutrient
deficiency similar to the parent plant (1).
**any other plausible advantages discussed.
Any 3, 5 marks each
15 marks UK
Total 30 marks
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