TEST CODE
FORM TP 2013144
CARIBBEAN
02107020
MAY/JUNE 2013
E XAM I NAT I O N S
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®
BIOLOGY
UNIT 1 – Paper 02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULY.
1.
This paper consists of SIX questions in two sections. Answer ALL questions.
2.
For Section A, write your answers in the spaces provided in this booklet.
3.
For Section B, write your answers in the spaces provided at the end of each
question in this booklet.
4.
The use of silent non-programmable calculators is allowed.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright © 2011 Caribbean Examinations Council
All rights reserved.
02107020/CAPE/2013
- 2 SECTION A
Answer ALL questions.
Write your answers in the spaces provided in this booklet.
1.
(a)
Table 1 presents results of an investigation of the nutritional composition of various foods
from an average American diet.
TABLE 1: ANALYSIS OF SPECIFIC NUTRIENTS IN COMMON FOODS
USING KNOWN REAGENTS
{Note: For reagents which require a colour change to indicate the presence of a nutrient, results were
recorded based on a colour index (range ) of changes observed for any one reagent.}
FOOD TEST REAGENT
Food
Substance
Biuret
Solution
Lugol’s
Benedict’s Iodine (KI/I2)
Solution
Solution
Eggs
Purple
Blue
Milk
Pink
Pale yellow Dark blue
None seen.
Cheerios
cereal
Pale blue
(almost
clear)
Brick-red
Blue-black
Only seen when food sample is
dissolved in isopropyl alcohol and
then tested.
Blue
Yellow
Clearly seen when rubbed on paper.
Hamburger
Purple
patty
Black
Grease Spot
on Brown Paper*
None seen.
Carrot
Pale blue
(almost clear)
Orange
Dark blue
None seen.
Potato
chips
Pink
Brick-red
Black
Seen when rubbed on paper but not
very transparent.
Pepperoni
pizza
Purple
Yellow
Dark blue
Clearly seen when rubbed on paper.
Donut
Pink
Orange
Dark blue
Clearly seen when rubbed on paper.
*
Grease Spot on Brown Paper Test:
1.
Using a glass rod, a sample of the food being tested is rubbed on a section of brown
paper (labelled with the sample being tested). “Wet” spot appears on the paper.
2.
Any excess food which may stick to the paper is removed using a paper towel.
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02107020/CAPE 2013
- 3 3.
The sample paper is left to dry for about 10 minutes.
4.
A translucent spot indicates a positive result for the specific nutrient being tested.
5.
Isopropyl alcohol serves to dissolve the nutrient in the food substance.
(i)
State the nutrient being tested for, by EACH of the following tests:
Proteins
Biuret: _____________________________________________________________
Reducing and non-reducing sugars
Benedict’s: __________________________________________________________
Starch
Lugol’s Iodine: _______________________________________________________
Lipids
Grease spot: _________________________________________________________
[2 marks]
(ii)
Using the data presented in Table 1, determine TWO food substances which contain
all four nutrients being tested for in this investigation.
Potato chips, donuts and pepperoni pizza
___________________________________________________________________
[1 mark]
(iii)
With reference to the qualitative results given in Table 1, suggest which foods are
good sources of nutrients for cell structure and growth of the cell, and for supply of
energy. Justify your answer with a brief explanation.
Cell structure and growth of the cell:
Eggs, hamburger, pepperoni pizza, donuts
___________________________________________________________________
High levels of proteins for use in cells, e.g. to build phospholipid bilayer
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
Supply of energy:
Cheerios cereal and potato chips.
___________________________________________________________________
High levels of sugars to provide ATP, and lipids for energy reserves.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
[4 marks]
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02107020/CAPE 2013
-
(b)
4
-
Figure 1 is a diagrammatic representation of the molecular structure of a short section of a
glycogen molecule.
CH,OH
\ H
H
Xo\?H /
\ o
H
OH
O
c hh 22oo h
c
'h
V \O H
H
CH,OH
c h 2o h
CH,
ch2
V A ~'V / S \ H
\fH,A'
, / VXpH
V V X O H
/ V
OH
OH
OH
H
H
Figure 1. Short section of a glycogen molecule
(i)
(ii)
Using a labelled arrow on Figure 1, highlight the location of EACH of the following:
a)
An alpha 1,4-glycosidic linkage
b)
An alpha 1,6-glycosidic linkage
[2 marks]
In the space below, sketch the general shape of one glycogen molecule.
[Details of individual glucose residues are NOT required.]
Space for diagram
[2 marks]
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02107020/CAPE 2013
- 5 (iii)
Amylose (starch) and cellulose are both polymers of glucose molecules but they
differ in their structure and function in plant cells. With reference to ONE structural
property, briefly explain the functional differences of amylose and cellulose in plant
cells.
Amylose is comprised of long chains of alpha glucose units. It is insoluble.
___________________________________________________________________
Its main function is energy storage as it can be readily hydrolysed.
___________________________________________________________________
___________________________________________________________________
Cellulose is comprised of long, linear chains of beta glucose units.
___________________________________________________________________
Its function is for structural support.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
[4 marks]
2.
(a)
Total 15 marks
In a cross between a plant with purple flowers and a plant with white flowers, all the F1
plants had purple flowers. When the F1 offspring were crossed (selfed), 705 plants had purple
flowers and 224 plants had white flowers.
(i)
State the expected ratio for the cross of the F1 offspring.
3 purple : 1 white
___________________________________________________________________
[1 mark]
(ii)
State an appropriate null (H0) hypothesis and an appropriate alternative (H1) hypothesis
for a Chi-square test of the results.
There is no significant difference between the observed and expected ratios.
H0: ________________________________________________________________
________________________________________________________________
There is a significant difference between the observed and expected ratios.
H1: ________________________________________________________________
________________________________________________________________
[2 marks]
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02107020/CAPE 2013
- 6 (iii)
Complete Table 2 by calculating the missing values.
TABLE 2: DATA FOR CHI-SQUARE TEST
Phenotype Observed (O) Expected (E)
2í(
2í( 2/E
Purple
ÀRZHUV
705
696
9
0.12
White
ÀRZHUV
224
233
-9
0.35
Ȥð ‫( گ‬O í E) ð(
[
0.46
]
[4 marks]
(iv)
Determine the number of degrees of freedom. Show your calculation.
df = (Number of phenotypes) - 1 -- > 2 - 1 = 1
___________________________________________________________________
[1 mark]
(v)
Using the Chi-square values in Table 3, comment on the validity of the null hypothesis
stated on page 5.
The Chi square value at the critical p-value (0.05) is 3.84, which is less than
___________________________________________________________________
0.46. This indicates that there is no significant difference between the observed
___________________________________________________________________
and expected ratios. The null hypothesis is accepted.
___________________________________________________________________
[2 marks]
TABLE 3: CHI-SQUARE (Ȥð) VALUES
Degrees of
Freedom
Number of
ClassesChi-square Values
1
2
0.46
1.64
2.71
3.84
6.64
10.83
2
3
1.39
3.22
4.61
5.99
9.21
13.82
3
4
2.37
4.64
6.25
7.82
11.34
16.27
4
5
3.36
5.99
7.78
9.49
13.28
18.47
0.50
(50%)
0.20
(20%)
0.10
(10%)
0.05
(5%)
0.01 0.001
(1%) (0.1%)
Probability that chance
alone could produce
this deviation
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02107020/CAPE 2013
- 7 (b)
Figure 2 represents the elongation stage of translation in protein synthesis.
Figure 2. Elongation stage of translation in protein synthesis
Source: http://www.frontiers-in-genetics.org/page.php?id=protein-synthesis_en
(i)
What is the term used to describe EACH group of three bases on the
codon
mRNA ____________________________________________________________
anticodon
tRNA? _____________________________________________________________
[1 mark]
(ii)
Using the mRNA strand below, identify the corresponding triplet bases on the tRNA
molecules. Write your answer in the boxes provided. Hint: Begin at the 5´ end.
mRNA 5´ UGG UUU GGC UCA 3´
ACC
AAA
CCG
AGU
[2 marks]
(iii)
Outline the nucleotide sequence on the DNA strand which serves as the template for
the mRNA strand in (b) (ii).
ACC AAA CGG AGT
___________________________________________________________________
[2 marks]
02107020/CAPE 2013
Total 15 marks
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- 8 3.
(a)
Table 4 is an incomplete table comparing the process of spermatogenesis with that of
oogenesis.
TABLE 4: COMPARISION OF THE PROCESS OF SPERMATOGENESIS
AND OOGENESIS
Feature
Number of meiotic daughter
cells which develop into
mature gametes
Duration of the process
(mitotic division of stem
cells and differentiated
spermatogonia/oogonia)
in the life span of the
individual
Spermatogenesis
Oogenesis
One.
Four.
Continuous process
from puberty to the
rest of the male's
adult life.
Produced in 28-day
cycles from puberty
until about 50
years old (menopause)
(i)
Complete Table 4 with respect to the features listed for spermatogenesis and
oogenesis in humans.
[3 marks]
Outline the roles of follicle-stimulating hormone (FSH) and luteinizing hormone
(LH) in the regulation of spermatogenesis.
(ii)
Promotes activity of Sertoli cells, which nourish developing sperm.
FSH: ______________________________________________________________
______________________________________________________________
______________________________________________________________
Binds to Leydig cells, which stimulate production of testosterone.
LH: _______________________________________________________________
_______________________________________________________________
[3 marks]
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02107020/CAPE 2013
- 9 (b)
Figure 3 is a representation of a photomicrograph of a stained section through a human ovary
showing developing ova.
Figure 3. Representation of a photomicrograph of a section through an ovary
In the box below, make a plan drawing of the ovary to include the stage of the developing
ovum just prior to release from the ovary.
[6 marks]
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02107020/CAPE 2013
- 10 (c)
Figure 4 is a diagram of a section through a carpel of an angiosperm plant showing the process
of double fertilization.
Sperm nucleus
Figure 4. Section through the carpel of an angiosperm plant
Comment on the fate of the structure labelled X and the structure labelled Y, which includes
the ovule walls and the embryo sac and its contents.
X:
The ovary wall becomes the pericarp of the fruit after fertilization.
______________________________________________________________________
______________________________________________________________________
Y:
The ovule walls become the testa (seed coat) and the ovule becomes the
______________________________________________________________________
seed. The fertilized egg forms the zygote and the polar nuclei (fused with
______________________________________________________________________
sperm nucleus) become the endosperm.
______________________________________________________________________
______________________________________________________________________
[3 marks]
02107020/CAPE 2013
Total 15 marks
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- 11 SECTION B
Answer ALL questions.
Write your answers in the spaces provided at the end of each question.
4.
(a)
Define the term ‘diffusion’, and distinguish between active transport and facilitated diffusion
in cells.
[4 marks]
(b)
(i)
With the aid of a simplified labelled diagram, describe the fluid mosaic model of
membrane structure.
[6 marks]
(ii)
With reference to the fluid mosaic model, discuss how polar molecules and ions are
transported.
[5 marks]
Total 15 marks
Write the answer to Question 4 here.
_______________________________________________________________________________________
(a) Diffusion is the net movement of molecules along a concentration gradient until uniformly
_______________________________________________________________________________________
distributed.
_______________________________________________________________________________________
_______________________________________________________________________________________
Two differences between 'facilitated diffusion' and 'active transport':
_______________________________________________________________________________________
1. Facilitated diffusion requires no ATP, while active transport does.
_______________________________________________________________________________________
2. Active transport moves molecules against a concentration gradient, while facilitated diffusion
_______________________________________________________________________________________
moves it down a concentration gradient.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 12 Write the answer to Question 4 here.
_______________________________________________________________________________________
(b) DIAGRAM:
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
- Fluid mosaic model consists of a phospholipid bilayer.
- Phospholipids have a hydrophilic phosphate head and two hydrophobic fatty acid tails. The tails
_______________________________________________________________________________________
face inward.
_______________________________________________________________________________________
- Proteins may span the width of the bilayer (integral) or adhere to the outer surface (peripheral).
_______________________________________________________________________________________
- Proteins may have carbohydrate chains, forming glycoproteins.
_______________________________________________________________________________________
- Cholesterol is present in the membrane between phospholipids, to maintain fluidity.
_______________________________________________________________________________________
_______________________________________________________________________________________
(ii) - Polar molecules and ions are transported via channel proteins or carrier proteins.
_______________________________________________________________________________________
- Protein channels span the membrane and allow flow through the membrane. They are shaped
_______________________________________________________________________________________
to accommodate particular molecules or ions.
_______________________________________________________________________________________
- Carrier proteins may use ATP to slightly change shape to shuttle across molecules.
_______________________________________________________________________________________
- Both channels and carriers are specific for particular ions and polar molecules.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 13 -
Space for diagram.
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02107020/CAPE 2013
- 14 5.
(a)
Mycobacterium tuberculosis, the bacterium that causes tuberculosis, has been virtually
eliminated by the widespread use of two effective antibiotics. Due to the development of
antibiotic-resistant strains of the bacterium, the disease has re-emerged as a major public
health problem.
(i)
Applying Darwin’s theory of evolution by natural selection, discuss how resistance
to antibiotics could have evolved in bacteria. Include in your discussion a concise
explanation of natural selection.
[6 marks]
(ii)
Briefly comment on why mutation is regarded as a driving force of evolution.
[2 marks]
(b)
Define the term ‘speciation’, and using examples, explain how geographical isolation may
lead to speciation.
[7 marks]
Total 15 marks
Write the answer to Question 5 here.
_______________________________________________________________________________________
(a)(i) Natural selection is the process by which organisms that are better adapted to their environment
_______________________________________________________________________________________
overcome selective pressures, survive and produce more offspring.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
1. Bacteria may have sensitive and resistant strains to antibiotics.
2. Mutations may contribute to genetic variation, increasing antibiotic resistance.
_______________________________________________________________________________________
3. Bacteria that are resistant to antibiotics are more likely to survive and reproduce than those
_______________________________________________________________________________________
that are susceptible to antibiotics.
_______________________________________________________________________________________
4. These resistant bacteria will produce offspring also resistant to antibiotics, thus increasing
_______________________________________________________________________________________
frequency of resistance.
_______________________________________________________________________________________
_______________________________________________________________________________________
(ii) Mutations are random and spontaneous, which give rise to new alleles. These new alleles
_______________________________________________________________________________________
may cause populations to be better able to adapt to changing environments.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 15 Write the answer to Question 5 here.
_______________________________________________________________________________________
(b)(i) Speciation is the process by which one species splits into two or more species.
_______________________________________________________________________________________
Reproductive isolation results in speciation.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
(ii) Geographical isolation creates physical barriers, e.g. mountains, that separate populations.
These barrier may result in environments that produce different selective pressures, such as
_______________________________________________________________________________________
new predators or abiotic obstacles. This isolation also prevents groups from mixing and thus
_______________________________________________________________________________________
preventing gene flow between individuals on both sides of the barrier.
_______________________________________________________________________________________
_______________________________________________________________________________________
Genetic differences accumulate over time, leading to reproductive isolation and could lead
to the emergence of a new species.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 16 6.
(a)
(b)
(i)
Give a concise description of the key stages in the fertilization of a human secondary
oocyte by a spermatozoon. Begin your account with the acrosome reaction.
[5 marks]
(ii)
Briefly comment on the significance of the process of fertilization.
[2 marks]
(i)
Describe TWO major functions of the placenta.
[4 marks]
(ii)
Discuss TWO ways in which maternal behaviour can affect foetal development.
[4 marks]
Total 15 marks
Write the answer to Question 6 here.
_______________________________________________________________________________________
6. (a)(i) During the acrosome reaction, the outer surface of the sperm membrane and acrosome
_______________________________________________________________________________________
membrane rupture to release hydrolytic enzymes.
_______________________________________________________________________________________
These enzymes digest a path through the corona radiata that surrounds the oocyte.
_______________________________________________________________________________________
The spermatozoa swim towards the oocyte's zona pellucida and enzymes digest a path through it.
_______________________________________________________________________________________
The sperm disconnects its tail and moves to the surface of the oocyte. The head of the sperm
_______________________________________________________________________________________
fuses with microvilli on the oocyte. The impermeable fertilization membrane forms around the oocyte
_______________________________________________________________________________________
just before the male nucleus fuses with the female nucleus to form a zygote.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
(ii) Fertilization restores the normal diploid number of chromosomes (46) for the zygote
_______________________________________________________________________________________
and results in variation as maternal and paternal chromosomes intermingle. Sex is determined.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 17 -
Write the answer to Question 6 here.
_______________________________________________________________________________________
(b)(i) The placenta:
_______________________________________________________________________________________
1. Allows production of hormones (e.g. hCG) to maintain endometrium during pregnancy.
_______________________________________________________________________________________
2. Allows exchange of gases.
_______________________________________________________________________________________
3. Provides nutrients for developing embryo.
_______________________________________________________________________________________
4. Removes excretory products from embryo.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
(ii) Maternal behaviour that influences foetal development:
_______________________________________________________________________________________
1. Proper nutrition, such as intake of folic acid, is important for foetal growth and prevention
_______________________________________________________________________________________
of diseases such as spina bifida.
_______________________________________________________________________________________
_______________________________________________________________________________________
2. Intake of drugs, alcohol and nicotine can all impede development of the foetus. Smoking can
_______________________________________________________________________________________
contribute to low birth weights and pre-term deliveries, while alcohol can contribute to foetal
_______________________________________________________________________________________
alcohol syndrome and birth defects. Symptoms of withdrawal can also appear at birth.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
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02107020/CAPE 2013
- 18 -
Write the answer to Question 6 here.
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________
END OF TEST
IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK ON THIS TEST.
02107020/CAPE 2013
TEST CODE
FORM TP 2014141
CARIBBEAN
02107020
MAY/JUNE 2014
E XAM I NAT I O N S
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®
BIOLOGY
UNIT 1 – Paper 02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULY.
1.
This paper consists of SIX questions in two sections. Answer ALL questions.
2.
For Section A, write your answers in the spaces provided in this booklet.
3.
For Section B, write your answers in the spaces provided at the end of each
question in this booklet.
4.
You may use a silent non-programmable calculator.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright © 2012 Caribbean Examinations Council
All rights reserved.
02107020/CAPE/2014
- 2 SECTION A
Answer ALL questions.
Write your answers in the spaces provided in this booklet.
1.
(a)
Using haemoglobin as an example, explain EACH of the following levels of structural
organization of proteins:
(i)
Primary structure
The order or sequence of amino acids in a polypeptide chain.
________________________________________________________________
________________________________________________________________
[1 mark]
(ii)
Secondary structure
The folding of polypeptide chains due to hydrogen bonding.
________________________________________________________________
Haemoglobin consists of alpha helices.
________________________________________________________________
________________________________________________________________
________________________________________________________________
[2 marks]
(iii)
Tertiary structure
Polypeptides fold in a 3D configuration, held together by ionic, sulphuric
________________________________________________________________
and hydrogen bonds. This gives a haemoglobin a globular shape.
________________________________________________________________
Prosthetic haem groups are found in the molecule.
________________________________________________________________
________________________________________________________________
[2 marks]
(iv)
Quaternary structure
________________________________________________________________
Multiple secondary or tertiary structures are stabilized by hydrogen and ionic bonds.
________________________________________________________________
Haemoglobin consists of 4 polypeptide chains.
________________________________________________________________
________________________________________________________________
[2 marks]
02107020/CAPE 2014
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- 3 (b)
With reference to its protein structure, explain how the haemoglobin molecule functions
in its essential role.
______________________________________________________________________
Haemoglobin functions to transport oxygen. It does this by binding an oxygen
______________________________________________________________________
molecule to each haem group. There are four haem groups in the quaternary structure.
______________________________________________________________________
The protein changes structure when first oxygen binds to allow more oxygen molecules
______________________________________________________________________
to bind more readily.
______________________________________________________________________
______________________________________________________________________
[3 marks]
02107020/CAPE 2014
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- 4 (c)
Figure 1 is a drawing of an electron micrograph of a plant cell.
Ramesar, Jones and Jones 2011, Fig. 2.15, page 41
Figure 1. Drawing of an electron micrograph of a plant cell ( × 5600)
(i)
Identify the organelles labelled A, B and C in Figure 1.
Mitochondrion
A: _____________________________________________________________
Chloroplast
B: _____________________________________________________________
C: _____________________________________________________________
Golgi apparatus (body)
[3 marks]
Calculate the actual maximum length of the organelle labelled X to the nearest
micrometre (μm). Show your working.
(ii)
Magnification = Length of image / Length of cell
Length of cell = Length of image / mag = 40000 micrometers / 5600
= 7.1 micrometers
Length: _________________________________________________________
[2 marks]
02107020/CAPE 2014
Total 15 marks
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- 5 2.
(a)
(i)
Protein synthesis requires two steps, transcription and translation. Table 1 is
an incomplete comparison of some features of transcription and translation in
eukaryotes. Complete Table 1 by writing the correct answers in the relevant
spaces in the table.
TABLE 1: COMPARISION OF TRANSCRIPTION AND TRANSLATION
Feature
Site
Precursor
molecule
Transcription
Generally in the nucleus
Occurs in the cytoplasm / RER
DNA
mRNA
Enzymes and/or RNA polymerase and other associated
factors
proteins
Function
Translation
Produces mRNA from DNA
template
rRna, tRNA transferase
Produces the peptide sequence
which is complementary to the
mRNA
[4 marks]
(ii)
Figure 2 is a diagrammatic representation of the elongation phase of translation. In
the box labelled A in Figure 2, sketch a diagrammatic representation of the tRNA
molecule carrying the next amino acid to be added to the growing polypeptide
chain.
Source: http://www.motifolio.com/1021138.html
Figure 2. Diagrammatic representation of the elongation phase of translation
[3 marks]
02107020/CAPE 2014
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- 6 (b)
In humans, the A, B, O blood groups are determined by multiple alleles of a single gene.
The gene locus is usually represented by the symbol I and the blood genotypes may be
represented as follows:
–
–
–
–
IAIA or IAi = blood group A
IBIB or IBi = blood group B
ii
= blood group O
IAIB
= blood group AB
(i)
Briefly explain the nature of the relationship between the alleles in the AB blood
group.
________________________________________________________________
They are codominant, where both alleles are expressed in the phenotype.
________________________________________________________________
Neither are dominant.
________________________________________________________________
[2 marks]
(ii)
In a paternity suit, a female with blood type O has accused a male with blood type
B of being the father of her child. The child has blood type O.
a)
Deduce the blood genotype of the accused male which will clearly prove
that he is NOT the father of the child. Give a brief explanation to justify
your answer.
Homozygous for B.
Blood genotype of male (no symbols required): ____________________
Justification: _______________________________________________
B alleles are dominant to O alleles. A child with blood type O would have
__________________________________________________________
__________________________________________________________
received an O allele for each parent. If the father has two B alleles,
this would be impossible.
__________________________________________________________
[3 marks]
b)
If the male parent in (b) (ii) a) above has blood type B, demonstrate the
inheritance of the blood type (O) of the child. Use the given symbols and
a Punnett square.
[3 marks]
02107020/CAPE 2014
Total 15 marks
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- 7 3.
(a)
Figure 3 is a photomicrograph of a cross section of a seminiferous tubule, and Figure 4
shows a part of the tubule.
Box X
http://www.pmrc.org.pk
A
Figure 3. Photomicrograph of a
section of a seminiferous tubule
(i)
02107020/CAPE 2014
B
Figure 4. Part of Tubule A
Make a detailed labelled drawing of the region highlighted by Box X in
Figure 4 B.
[6 marks]
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- 8 (ii)
Using Figure 3 or Figure 4 as a guide, outline the key development stages of
spermatozoa within the seminiferous tubule.
________________________________________________________________
- Germinal epithelial cells differentiate to form spermatogonia.
________________________________________________________________
- Spermatogonia form primary then secondary spermatocytes.
________________________________________________________________
- Secondary spermatocytes undergo meiosis to become haploid spermatids.
________________________________________________________________
- Spermatids undergo growth of flagella (tails) to form spermatozoa.
________________________________________________________________
________________________________________________________________
[3 marks]
(b)
An experiment is conducted to investigate the effect of sucrose concentration on the
germination of pollen grains for a particular plant species. The results are shown in
Figure 5.
50
Pollen germination %
40
30
20
10
0
0
2
4
6
8
10
Sucrose concentration (%,w/v)
Figure 5. Effect of sucrose concentration on germination of pollen grains
02107020/CAPE 2014
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- 9 (i)
Briefly describe the results of this experiment shown in Figure 5.
________________________________________________________________
No germination occurs at 0% sucrose and percentage germination increases
________________________________________________________________
as sucrose concentration increases (up to 6%). After 6% concentration,
________________________________________________________________
there is a slight decrease before remaining constant.
________________________________________________________________
[2 marks]
(ii)
Explain the significance of this response for the pollination process.
________________________________________________________________
________________________________________________________________
Sucrose is secreted by stigma for pollen germination. The sucrose
helps to nourish growth of pollen tube and facilitates the pollen grains
________________________________________________________________
sticking to the stigma.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
[4 marks]
02107020/CAPE 2014
Total 15 marks
GO ON TO THE NEXT PAGE
- 10 SECTION B
Answer ALL questions.
Write your answers in the spaces provided at the end of each question.
4.
(a)
Outline the structure of a molecule of water and explain how this structure allows water
[5 marks]
to be an excellent solvent.
(b)
Discuss TWO major roles of water in cell function.
(c)
Distinguish between endocytosis and exocytosis, and briefly comment on ONE cellular
[6 marks]
process that involves exocytosis.
[4 marks]
Total 15 marks
Write the answer to Question 4 here.
___________________________________________________________________________________
(a) - Water contains two hydrogen atoms covalently bonded to an oxygen molecule.
___________________________________________________________________________________
The bonding is a non-linear arrangement. Oxygen has a slightly negative charge
___________________________________________________________________________________
and hydrogen has a slightly positive charge. Water is therefore a dipole molecule.
___________________________________________________________________________________
___________________________________________________________________________________
Its dipole structure attracts other molecules or ions that are charged. When other polar
___________________________________________________________________________________
compounds enter water, the water molecules surround it, acting as an excellent solvent.
___________________________________________________________________________________
___________________________________________________________________________________
(b) 1. Water is used as a solvent or transport medium. Substances dissolve readily in
___________________________________________________________________________________
water and can be transported through the bloodstream or the xylem to cells.
___________________________________________________________________________________
2. Water acts as a buffer for rapid temperature and pH changes. It helps to keep cells
___________________________________________________________________________________
at optimum temperature due to its high specific heat capacity.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 11 Write the answer to Question 4 here.
___________________________________________________________________________________
___________________________________________________________________________________
(c) Endocytosis facilitates entry of molecules and larger material into cells.
The material is engulfed by the folding of the cell membrane.
___________________________________________________________________________________
___________________________________________________________________________________
Exocytosis is the process that moves materials out of a cell. Vesicles from within
___________________________________________________________________________________
the cell cytosol fuse with the cell membrane. An example of this is the removal of waste particles
___________________________________________________________________________________
by lysosomes.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 12 Write the answer to Question 4 here.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 13 5.
(a)
State FOUR observations and THREE deductions that formed the basis of Darwin’s
theory of natural selection.
[5 marks]
(b)
Use the theory of natural selection to explain how a new species can evolve from an
existing one by allopatric (geographical) speciation.
[4 marks]
(c)
Discuss THREE potential threats to humans, and other organisms, of the use of genetically
modified crops. Include a definition of the term ‘genetically modified organism’.
[6 marks]
Total 15 marks
Write the answer to Question 5 here.
___________________________________________________________________________________
___________________________________________________________________________________
FOUR OBSERVATIONS from Darwin's theory:
1. All organisms produce much more offspring than are required.
___________________________________________________________________________________
2. Organisms within a species experience variation.
___________________________________________________________________________________
3. Some of these variations are inherited.
___________________________________________________________________________________
4. Population numbers tend to be fairly constant / stable over long periods of time.
___________________________________________________________________________________
___________________________________________________________________________________
DEDUCTIONS from Darwin's theory
___________________________________________________________________________________
1. Among populations in habitats, there is a struggle for existence.
___________________________________________________________________________________
2. Only the best adapted individuals can survive and reproduce.
___________________________________________________________________________________
3. The offspring inherit traits from these surviving, well-adapted individuals.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 14 Write the answer to Question 5 here.
___________________________________________________________________________________
(b) Allopatric or geographical separation occurs when there are geographical or spatial
___________________________________________________________________________________
barriers present, such as large rivers or mountain ranges. Using a valley as an example,
___________________________________________________________________________________
these would prevent gene flow between organisms living on both sides of the valley.
___________________________________________________________________________________
___________________________________________________________________________________
There may be different selective pressures and environmental conditions on each side of
___________________________________________________________________________________
the valley (e.g. different predators, water availability). Organisms on each side adapt to these
___________________________________________________________________________________
selective pressures, which lead to changes in genotype frequencies. New species may
___________________________________________________________________________________
develop when the gene pools become varied enough, and populations can no longer
___________________________________________________________________________________
interbreed to produce fertile offspring.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 15 Write the answer to Question 5 here.
___________________________________________________________________________________
(c) A genetically modified organism is one that has had its genome modified due to
___________________________________________________________________________________
genetic engineering. It may have had genes added or altered.
___________________________________________________________________________________
___________________________________________________________________________________
THREATS:
___________________________________________________________________________________
1. GMOs may invade natural habitats and out-compete native species, thus affecting
___________________________________________________________________________________
the food webs and biodiversity.
___________________________________________________________________________________
___________________________________________________________________________________
2. Added genes that produce toxins for pest control can have adverse effects on
beneficial organisms, such as pollinators.
___________________________________________________________________________________
___________________________________________________________________________________
3. GMO crops may be unsafe for human consumption and may cause allergic reactions.
___________________________________________________________________________________
(e.g. GMO soybeans with Brazil nut genes)
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 16 6.
(a)
(i)
Explain why vegetative propagation is NOT considered a form of sexual
reproduction. Include in your explanation a brief definition of ‘vegetative
propagation’.
[3 marks]
(ii)
Comment on why vegetative propagation is especially beneficial to agriculturists
and horticulturists. Limit your commentary to FOUR main points. [4 marks]
(i)
Give a concise explanation of how combined oral contraceptives work to prevent
[3 marks]
pregnancy.
(ii)
A young, recently married couple seeks advice at a family planning clinic. The
23-year-old female explains that her last normal menstrual cycle started two weeks
ago and they both confirm that they have not engaged in sexual intercourse since
then. She would like to begin using combined oral contraceptives immediately,
as she has been advised that there are no medical reasons which prevent her from
using this form of contraception.
Suggest what advice should be given to the couple about the most appropriate
way to use combined oral contraceptives. Include in your account, justification
of your advice in relation to the physiological details provided by the couple and
[5 marks]
your knowledge of the menstrual cycle.
(b)
Total 15 marks
Write the answer to Question 6 here.
___________________________________________________________________________________
(a)(i) Asexual reproduction refers to the production of offspring where only one
___________________________________________________________________________________
parent is involved. No fertilization of gametes occur, and offspring are genetically identical
___________________________________________________________________________________
to the parent.
___________________________________________________________________________________
___________________________________________________________________________________
Vegetative propagation refers to the ability of plants to asexually reproduce by using
___________________________________________________________________________________
vegetative structures, such as runners, rhizomes and specialized roots.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 17 Write the answer to Question 6 here.
___________________________________________________________________________________
___________________________________________________________________________________
(ii) Four benefits of vegetative propagation
1. Due to offspring being genetically identical, desirable characteristics from
___________________________________________________________________________________
the parent can be retained.
___________________________________________________________________________________
2. Can be used to grow seedless varieties of fruits, such as bananas.
___________________________________________________________________________________
3. Can be done on a large scale since only one parent is needed and some
___________________________________________________________________________________
methods are relatively inexpensive (e.g. cuttings).
___________________________________________________________________________________
4. Allows the ability to combine desirable traits of different crops, e.g. grafting.
___________________________________________________________________________________
5. Can propagate plants that are difficult to grow via seed germination.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
(b)(i) Oral contraceptives work by changing the body's hormonal balance, resulting in
___________________________________________________________________________________
no secondary oocyte being released. As a result, no fertilization by sperm can occur.
___________________________________________________________________________________
___________________________________________________________________________________
It can also cause cervical mucus to thicken, making entry of sperm difficult. It may also
___________________________________________________________________________________
thin the endometrium, therefore preventing implantation of a fertilized egg.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
(ii) The client is most likely near day 14 of the menstrual cycle, where the ovarian
___________________________________________________________________________________
follicle has begun developing.
___________________________________________________________________________________
Oral contraceptives may not stop ovulation at this point, as the best time to begin taking
___________________________________________________________________________________
the contraceptives is within the first 5 days of the start of the cycle. The contraceptive must be
___________________________________________________________________________________
taken consistently every day.
___________________________________________________________________________________
It is recommended that the client use some other kind of protection, e.g. condoms
___________________________________________________________________________________
02107020/CAPE 2014
GO ON TO THE NEXT PAGE
- 18 Write the answer to Question 6 here.
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
END OF TEST
IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK ON THIS TEST.
02107020/CAPE 2014
%.
FORM TP 2015147
rESr coDE 02107020l|
CARIBBEAN EXAMINATION S
MAY/JLTNE 2015
C OUNC IL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION@
BIOLOGY
UNITl-PaPer02
2 hours 30 minutes
l
I
R.EAD THE FOLLOWING INSTRUCTIONS CAREFULLY.
l.
This paper consists of SIX questions in TWO sections. Answer ALL questions.
2.
Write your answers in the spaces provided in this booklet.
3.
Do NOT write in the margins'
4.
You may use a silent, non-programmable calculator to answer questions.
5.
You are advised to take some time to read through the paper and plan your answers.
6.
If you need to rewrite any answer and there is not enough space to do so on the
7.
If you use the extra page(s) you MUST write the question number clearly in
original page, you must use the extra lined page(s) provided at the back of this
booklet. Remember to draw a line through your original answer.
the box provided at the top of the extra page(s) and, where relevanto include
the question part beside the answer.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright @ 2Ol3 Caribbean Examinations Council
All riehts reserved.
lorroro2o.*P'
2ors
I
tililt lllll lllll llll lllll lllll lllll llll lllll lllll lil llll
0210702003
-2SECTION A
AnswerALL questions.
Write your answers in the spaces provided in this booklet.
(a)
1.
Table 1 summarizes the results of an experiment comparing the effect of temperature on
two protease enzymes, one from a mammal that lives in the tropics and the other from a
prokaryote living in hot springs. The enzyme activity is expressed as a percentage of its
maximum efficiency.
TABLE 1: EFFECT OF TEMPERATURE ON
TWO PROTEASE ENZYMES
7o Maximum EfficiencY
Temperature
oc
Prokaryote Protease
Mammalian Protease
0
0
0
5
0
10
10
0
25
20
8
35
30
l5
48
40
30
95
50
45
35
60
65
0
70
100
0
90
40
0
GO ON TO THE NEXT PAGE
02l07020lcAPE 201s
t-
I
|
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0210702004
-l
-3(D
On the grid in Figure 1, plot the data in Table I for the mammalian protease as a
[3 marksl
line graph.
-ii t-i
'l
-t" r"1- f'
..i....i...1...1..
...i...i...i.,.i...
...i....i...i...i...
100
...i....i...i...i...
...i....i...:...1...
-i-l-'!-'l
!!!!
...i...i...i-..i...
iiii
--l :.-':" 1'
'"i i-i-i
[..i...i.'i'
90
ii:i
80
-i"i-! "+..i-i-
(D
<)
60
"i" i-:- i
€)
{
--i +--i :
i:
i-i i-i
.;.i
ii i-i
t
ti.-+".r- r'
t..1...i....i.-
i i-i-i
I+ii
-f i-i_r'
-i -I-l-l '
..i.. ..ii.
-j-i-r"
i"'i ! .l-i i-i"!'ll-:'"r-i:'
,....i...+...i..
-i
E
1_1- f
40
l..
!
J
TProkaryote -i-t " i-i'
...i....i...!...i.
/
-i:i:ii "?-:
i" l
0!
i:i i iti:iii::l t'i-i i i'
,i-i',,il ;'i-i-i-il"!-r'i-i r'i-i-i-i
ftiil.
"i i-:'
.:...i:...i} 'i !-i
-i i-i-i
6)
q)
.\
't-
"i i"i i litii
'' !
50
6l
-t i-i-i
-i- + -i- i
-i" i-i i
IE
x6l
.i.
..:,..:-..i,,.i..
'':" 1" 1 _r_
e
..i...!...!...)..
li:::r:l: i.../.i.
-i- +-.+-.i
-i':_ i' : _
70
-{ -+. } -f.
..i i.::i i.
iiii
-j-i-i
...
i
-.i i-
protease
iai-i" i1- i i-i i
i::
i .i::i.
30
t,
;:::;;;
q)
20
Ii-i.-i
.i::fi
-r'-t-1-i'
10
0
1
-t -i-j-i'
,t
:y
ii:i
-i-1-i
'':
..i.,i-i-i.
-1".1-l" l_
...i...;...i....i...
-f_i "i- i
-r'i-.i-.1.i...i...1...i...
i:ii
'ii i- i"
ili; i" i'i-i_
.i...,i...!,,,i,..
,i.,,.i-i i -
-i-i -i.-+
40
20
.i...i..,i...i...
i"i i-i
60
100
80
Temperature (oC)
Figure 1. Effect of temperature on protease enzymes
(iD
temperature
Using affows labelled Pl andP2,indicate, on the.r-axis of Figure 1, the
What
efficiency'
at which EACH protease enzyme functions at its maximum
accounts for any difference observed?
Mammal - 40 Celsius, Prokaryote - 70 Celsius
Mammal's optimum temperature is determined by its body temperature,
while the higher prokaryotic optimum is determined by its environment (hot spring).
Maximum efficiency of the enzyme occurs at the optimum temperatures.
[3 marks]
GO ON TO THE NEXT PAGE
02r07020lcAPE 2015
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0210702005
Increasing temperature also increases the kinetic energy of the enzyme
molecules.
This KE results in vibrations and eventual breaking of hydrogen bonds that
hold the tertiary structure of the proteins together.
This changes the shape of the active site, which prevents the substrate from
binding, thus decreasing or stopping enzyme activity.
Nucleus
Mitochondrion
ER
t-
-5(D
Identiff the organelles labelled I, II and III.
I:
II:
III:
(iD
Mitochondrion
ER
Nucleus
[3 marksl
If the actual size of the structure labelled Y is 1.5 pm, calculate the magnification
of the electron micrograph in Figure 2'
Size of drawing = 11mm = 11000 micrometers
Magnification: ............
Mag = 11000/1.5 = 7333
(iiD
[l mark]
Suggest a reason which may account for the extensive network of membranes
seen
in this tyPe of animal cell.
Allows for efficient transport of metabolic products, such as bile salts or lipids.
OR allows for protein and lipid synthesis by rough and smooth ER, respectively.
[2 marks]
Total 15 marks
GO ON TO THE NEXT PAGE
02r07020/cAPE 2015
I
t-
ililillllmllllllllllllllllllllllllllllllllllllllllllllllll
0210702007
-6)
(a)
Figure 3 is a diagram of a cell with four chromosomes (haploid number
process of meiosis.
: 2) during the
Figure 3. Diagram of a cell with four chromosomes
(D
Name the stage of meiosis illustrated in Figure 3'
Prophase I
(iD
Name of stage:
ii;;;k;
key stages
In the boxes below, make labelled sketches to illustrate the next TWO
below
of meiosis in this cell. Write the name of the stages on the lines provided
the boxes.
Name of stage:
Metaphase I
Anaphase I
[6 marksl
GO ON TO THE NEXT PAGE
o2l07020lcAPE 2015
|
I
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0210702008
-7 -
(iii)
Explain how the key stages identified in (a) (i) and (ii) contribute to genetic variation'
Chiasmata (regions of crossing over) occur between chromatids in Prophase I.
Homologous chromosomes undergo independent assortment in Anaphase I.
Due to random alignment, there can be many different arrangements of alleles.
[3 marks]
(b)
Explain how 'substitution'and 'insertion'result in genetic variation. Discuss how gene
mutation causes sickle-cell anaemia.
Substitution occurs when a nucleotide is replaced with another (e.g. A instead of T) and
thus, a change in base pairs occur. This results in a different amino acid sequence and a
different protein can be made during translation.
Insertion is the addition of a nucleotide (or base pair) into the DNA code. This causes a
frame shift, which can highly influence protein structure during translation.
Sickle cell anaemia is caused by a substitution (where A is replaced by T in one instance).
As a result, the polypeptide chain in haemoglobin is changed and thus, the tertiary (3D)
structure of haemoglobin also changes. Red blood cells become sickle-shaped and their
oxygen-carrying capacity is reduced.
[5 marks]
Total 15 marks
GO ON TO THE NEXT PAGE
02l07020lcAPE 2015
l
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0210702009
I
r
(a)
3.
-l
-8Table 2 is an incomplete comparison of some features of a human ovum
with some features
jijij-i:[-j
----.--t'-
-'-V_--F
i1'j-'i----ljj
i:-l*F:-F
of a spermatozoon.
[.f;ti+r
lj-:-*-:+l
TABLE 2: COMPARISON OF FEATURES OF
OVUMAND SPERMATOZOON
Ovum
Feature
Spermatozoon
l-r.t i-r+i
l"l\,i:-I.
L-5i+i.I1
t.--f|tr.--l.
I..tr+:.l.
[,.k+j
j-r-\+---tj
I
f-:-.'-'---'f-
F.t*lr-+.
l'jl*I:+l
Size
Can be seen with naked eye;
about 0.1 mm to 0.2 mm in
width.
li-f!*i.+
Cannot be seen with
naked eye
(about 0.05mm long)
L-,s*
t---ri\-.'-f
l---r\--alf
t-:-hij..-1:
f..trtnt
l:'Eit
f.-.-.-.T.'.-&
[:-$.:l
r.-.n--'.'t
r---ha'-'t
I '_'_'_l-'-_ t
f----.--t.
Shape
Elongated.
[...J
t_---,-__t
Round
t-------t
I:':'n-t
r-----_.'.'.'.7
t--------t
[-]-----:-f
t::!ii+
l.-i;T{
l---1*f-:1
l.:#:1
ttxJ
lr:Fe-+
t----E---l
Overall structure
Not divided into distinct
regions, surrounded by zona
pellucida and follicle cells.
Differentiated into head,
neck, middle piece
and tail.
l---ti.'---1
I-if'li('--l
li.*iil
f:ftj
l.+iij
l-trt:l
f
:srj
l.'.\t:-l
L.-$4J
l-:-E-rl
f-'-*-:-l
li:lf*iJ
t-'.'a=---.1
Motility
l-'-'Li.---l
Cannot move by itself.
Motile (can swim)
[.:El
l:-:txi
f--y_--_l
Fj$l
No food reserves.
Food reserves
r.-iaa:
L---iil
['--f1.:
Contains food reserves.
F---'iral:
ll:fii-r.
F..-is-:1
r-'i'Fr:
li:?Fii'
I..Ef..
f.:*.:i
L-:l!i--:
Metabolic activity
Active, substances
are absorbed and released
Inactive, does not absorb
or release substances.
F,S+t
l'jH-.i
r--xt:
l.:-td-::
t-'-fa---,
trl+:'j
t-:-!!(-j
l:'.C!;r
r---'J-:
f:iR.
llrl':.
l---a:ti.
[:.ftr
02r07020lcAPE 20ls
L
l------||
illlll llll lllll llll lilll lllil lllll lllll lilll llll llll llll
0210702010
"'-'-'---'.
-9(D
in the
Complete Table2by writing the missing information in the relevant spaces
marks]
[6
table.
(ii)
With reference to a specific function of the human ovum, discuss how its structure
is suited to the stated function.
The ovum, when fertilized by a sperm cell, becomes a zygote. It facilitates the
reception/entry of a sperm cell.
Its outer layer, which contains cell surface proteins, only allows entry of one sperm cell.
Its large size also facilitates easier contact with sperm cell.
[3 marksl
GO ON TO THE NEXT PAGE
02l07020lcAPE 2015
l
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0210702011
-l
- 10(b)
Figure 4 is a diagram of a cross section of a fertilized ovule of a flowering plant'
Pollen tube
X
X is the fertilized egg cell.
Mic
Figure 4. Cross section of a fertilized ovule
(i)
On Figure 4, indicate, using an affow labelled X, the structure which will develop
[1 mark]
a radicle and plumule.
(iD
Identiff the structure labelled Y and explain its role in the developing embryo'
Endosperm nucleus (3n)
Identity of Y:
Role:
(iiD
Develops into endosperm, which acts as a food reserve for embryo
[2 marksl
An orange is described as a fleshy fruit. Name the part of a flowering plant which
develops into a fleshy fruit after the ovule has been fertilized. Comment on the
importance of the fruit, such aS an orange, in the life cycle of a flowering plant.
The ovary becomes the fruit.
The fruit is necessary for seed dispersal by animals (which scatter the seeds
via defecation). The fruit also protects the seeds from damage.
[3 marksl
Total 15 marks
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02t07020lcAPE 2015
I
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0210702012
r
-11
--l
SECTION B
AnswerALL questions.
Write your answers in the spaces provided in this booklet'
(a)
(i)
With the aid of an annotated diagram, describe the structure of phospholipids.
[4 marks]
Space for annotated diagram of phospholipids
Phospholipids consist of hydrophilic (water-loving) phosphate heads and
hydrophobic (water-hating) fatty acid tails. These are connected by glycerol
molecules through ester bonds.
GO ON TO THE NEXT PAGE
02107020lcAPE 2015
I
-':---'-'f--'------:
jji-'I-:ji a
,t-----_r--_-:
t-
|
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o210702013
t-
-12(ii)
With reference to the fluid mosaic model, discuss the role of phospholipids and
[5 marks]
proteins in cell membrane functioning.
The fluid mosaic model is represented as a sea of phospholipids, with
proteins floating throughout (like icebergs).
The phospholipids' role is to form a barrier that separates the cell interior (cytosol) and
exterior. They also prevent the entry/exit of large, polar molecules and facilitates
movement of non-polar molecules (e.g. carbon dioxide).
'€={'--l
:S.-tj{
iR.-.t
i:'----::':-:l--;
Proteins can span the entire width of the bilayer and can affect as channels
:'-_-_-_---_---Ti.'-'-
(allowing facilitated diffusion for molecules, e.g. glucose) or act as carrier proteins
(which act as gated channels with binding sites).
Proteins can also assist with cell recognition and signalling, as well as
carry out metabolic reactions (enzymes).
GO ON TO THE NEXT PAGE
02r07020lcAPE 201s
I
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0210702014
n:i+j-:1i
1-l'-;
-l
-13(b)
Using a generalized structure for amino acids, outline the process of formation of a peptide
bond between two amino acids. Explain, using an example, how amino acids differ from
[6 marks]
each other.
Space for formation of peptide bond between two amino acids
Amino acids vary from each other based on the residual (R) group attached.
R groups can vary in size, polarity and may even contain aromatic rings.
Total 15 marks
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02107020lcAPE 2015
t-
I
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0210702015
J
r
-14(a)
f,.
Explain what is meant by the term'gene'. Outline the principles of using restriction
enzymes to cut a gene from a length of DNA.
Note: Details of the steps involved in recombinant DNAtechnology are not required.
[5 marks]
A gene is a basic unit of inheritance that determines traits. It is defined as a segment of
DNA on a chromosome that codes for a sequence of amino acids.
A restriction enzyme is used to cut across DNA strands at specific sites, due to having
specific nucleotide sequences (which can usually be read 5' to 3' and 3' to 5').
They are usually made by bacteria to 'cut out' viral DNA. Think of restriction enzymes
as DNA scissors, thus. Usually, when they make these DNA cuts, they leave free
unpaired ends, known as 'sticky ends', used for easy attachment.
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02107020lcAPE 201s
I
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0210702016
I
- 15 (b)
Gene therapy is the application of the principle of genetic engineering in the treatment of
diseases.
(i)
Give a brief description of one human disease for which the use of somatic gene
[2 marksl
therapy has been shown to be effective.
An example is cystic fibrosis (CF), caused by a substitution mutation.
This mutation allows a channel protein (CFTR) to be defective. As a result,
the channel cannot release chloride ions, which results in a salt imbalance and
build-up of thick mucus that clogs the lungs.
Somatic gene therapy can replace
the 'incorrect' nucleotide, thus allowing formation of normal channel proteins.
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o2r07020lcAPE 2015
I
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0210702017
lllll llll llll
J
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-16-
(ii)
Discuss FOUR reasons why somatic gene therapy, despite its potential, is still not
used as an effective treatment for human diseases.
[8 marksl
1. Viral vectors may be used to transport the new DNA, but can cause
inflammatory responses once inside the patient. The virus may also become a
a pathogen while within the patient.
2. It is limited to diseases caused by a single gene mutation. Diseases that
affect multiple genes cannot be treated effectively.
3. Since it requires repeated treatments over the course of the patient's life,
it may be too costly.
4. The effects may not be permanent due to the 'repaired' cells ultimately dying
and replaced by new cells. The vector may not always target the correct set of
tissues to deliver the DNA.
Total 15 marks
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02107020/CAPE 2015
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0210702018
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6.
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-17(a)
(i)
Explain TWO mechanisms for the promotion of cross-fertilization in plants.
[4 marksl
1. Dioecious plants contain male and female flowers on separates, thus
preventing self-fertilization.
2. Protandry and protogyny ensure that male and female reproductive organs
(stamens and pistils) do not open or mature at the same time.
Others: Self-incompatibility, male sterility, heterostyly.
(ii)
Evaluate the genetic consequences of self-fertilization and cross-fertilization in
[7 marksl
Self-fertilized plants are those that are self-pollinated. Cross-fertilized plants
plants.
are those that are cross-pollinated (two different plants of the same species).
Self-fertilization reduces genetic variation in flower populations and thus leads
to reduced vigour and decreased chances of individuals overcoming selective
pressures such as pests and disease. However, advantageous alleles and traits
can be passed down to offspring reliably.
Cross-fertilization increases genetic variation and vigour in a population.
Organisms are more likely to overcome selective pressures and there is a
greater evolutionary potential and chance for speciation to occcur.
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02r07020/cAPE 2015
tI
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0210702019
I
-18(b)
Starting with contact of sperm cells with follicle cells of the oocyte, outline the process
of fertilization in the human reproductive system.
[4 marksl
- Sperm undergoes capacitation. Acrosome releases enzymes.
- The enzymes digest a pathway through follicle cells and zona pellucida in secondary
oocyte.
- Sperm cell receptors bind to zona pellucida and sperm nucleus enters secondary
oocyte.
- The zona pellucida becomes impermeable to other sperm cells, and prevents them from
binding.
- The secondary oocyte completes meiosis II, just before sperm and oocyte nuclei
fuse to become a (diploid) zygote.
Total 15 marks
END OF TEST
IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORI( ON THIS TEST.
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0210702020
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-19EXTRA SPACE
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0210702022
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FORM TP 2016150
%
TEST CODE O2IO7O2O
MAY/JUNE 2OI6
CARIBBEAN EXAMINATIONS COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONO
BIOLOGY
UNITl-Paper02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULLY.
This paper consists of SIX questions in TWO sections. Answer ALL
I
questions.
2.
Write your answers in the spaces provided in this booklet.
J.
Do NOT write in the margins.
4.
You may use a silent, non-programmable calculator to answer questions.
5.
You are advised to take some time to read through the paper and plan your
answers.
6.
If you need to rewrite any answer and there is not enough space to do so
on the original page, you must use the extra lined page(s) provided at the
back of this booklet. Remember to draw a line through your original
answer.
7
If you use the extra page(s), you MUST write the question number
clearly in the box provided at the top of the extra page(s) and, where
relevant, include the question part beside the answer.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright @ 2014 Caribbean Examinations Council
All rights reserved.
L
02107020/CAPE 2016
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0210702003
I
r
-l
-4SECTION A
AnswerALL questions.
Write your answers in the spaces provided in this booklet.
1.
(a)
Figure I shows the formation of sucrose
Glucose
Fructose
+
R
x
Sucrose
Figure 1. Formation of sucrose
(i)
Name the reaction labelled R and the reaction product labelled X in Figure I
Condensation
R.......
x
Water
[2 marksl
(ii)
In the box below, illustrate the full molecular ring structure ofthe glucose molecule
involved in the reaction in Figure 1.
[2 marksl
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02107020/cAPE20t6
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0210702004
J
la
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5
(iii)
With reference to the molecular structure of sucrose, explain why sucrose has an
advantage over glucose as a transport sugar in plants.
Sucrose is less reactive than glucose, as it is a non-reducing sugar and
disaccharide. It is a more efficient energy transfer and storage molecule.
[3 marks]
(b)
(i)
Draw a simple labelled diagram to illustrate the fluid mosaic model of plasma
membrane structure.
[5 marksl
(ii)
With reference to the fluid mosaic model, explain how large polar molecules can
pass through the membrane.
Polar molecules cannot easily cross the lipid layers.
- Polar molecules cannot interact with the hydrophobic regions.
- Movement has to occur through channel proteins or carrier proteins.
- Channel proteins allow facilitated diffusion. Carrier proteins are gated
channels that allow active transport (use of ATP to allow passage of molecules
against a concentration gradient)
[3 marksl
Total 15 marks
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02t07020lcAPE 2016
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0210702005
J
r
2.
-l
6
(a)
(i)
Using a simplified diagram of a small section of ribonucleic acid (RNA), describe
the structure of RNA. Note: Details of the chemical composition of components
are NOT required.
- RNA is single-stranded.
- Comprised of nucleotides (adenine, cytosine, guanine and uracil).
- Ribose sugar is linked to phosphates via phosphodiester bonds.
[5 marks]
(ii)
Comment on the role of m-RNA in eukaryotes
mRNA is sent to ribosomes in order to synthesize polypeptide chains and
proteins from genetic instructions. It does this by copying a complementary
code from a DNA template and using ribonucleotides (A, U, C and G).
[2 marks]
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02107020/0APE 2016
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0210702006
J
r
-l
-7 (b)
A plant geneticist is investigating the inheritance of genes for bitter taste (,Sz) and explosive
rind (e) in watermelon. Explosive rind is recessive and causes the watermelon to burst
when cut. Non-bitter watermelons are as a result of the homozygous recessive allele (sz).
The geneticist wishes to determine if the genes assort independently. A test cross is done
between a bitter, non-explosive plant and a plant homozygous recessive for both traits.
Table I shows the offspring produced.
TABLE 1: CATEGORIES OF OFFSPRING AND RESPECTIVE NUMBERS RECORDED
Phenotype
Number
Genotype
SusuEe
Bitter, non-explosive
88
Bitter, explosive
68
Non-bitter, non-explosive
62
Susuee
susuEe
Non-bitteq explosive
81
susuee
(i)
Determine the genotype of EACH phenotypic category in Table l, using the symbols
given. Write your answers in Table l.
(ii)
[2 marksl
Suggest a null hypothesis for the test cross in (b).
There is no significant difference between the ratio of the observed results and those
expected from the genetic ratio.
[2 marksl
(iii)
A Chi-square test is conducted on the results ofthe test cross. Table 2 is incomplete
for calculated values.
Complete Table 2 by writing the missing values in the relevant spaces.
TABLE 2z CALCULATED VALUES FOR CHI-SQUARE
TEST FOR DATA GIVEN IN TABLE T
Phenotype
Observed
Expected
Bitter, non-explosive
88
Bitter, explosive
68
Non-bitter, non-explosive
62
Non-bitter, explosive
8l
75
75
75
75
(Observed - Expected )z
Expected
2.25
0.6s
2.25
0.6s
)K2 =
TOTAL = 299
5.8
[2 marksl
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02107020/CAPE 2016
L
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02't0702007
I
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-l
-8(iv)
Using the table of probabilities provided below, and with reference to the calculated
Chi-square value from Table 2, evaluate the validity of the hypothesis.
Chi square value (5.8) is less than the critical value for 3 degrees of freedom.
The critical value at 0.05 is 7.82.
Thus, the null hypothesis will be accepted (or fail to reject) since there
is no significant difference between observed and expected values.
[2 marks]
TABLB 3: CHI-SQUARE DISTRIBUTION
Probability
Degrees
of
Freedom
0.95
0.90
0.80
0.75
0.50
0.30
0.20
0.10
0.05
0.01
0.001
I
0.004
0.02
0.06
0.1 5
0.46
1.07
1.64
2.71
3.84
6.64
r 0.83
2
0.10
0.21
0.45
0.71
r.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.3s
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.28
18.47
5
1.14
r.6l
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
6
r.63
2.20
3.07
3.83
s.35
7.23
8.s6
10.64
12.59
16.81
22.46
7
2.17
2.83
3.82
4.67
6.35
8.38
9.80
12.02
14.07
18.48
24.32
8
2.73
3.49
4.59
5.53
7.34
9.52
I 1.03
13.36
15.51
20.09
26.12
9
3.32
4.17
s.38
6.39
8.34
r 0.66
12.24
r4.68
16.92
21.67
27.88
r0
3.94
4.86
6.18
7.27
9.34
11.78
13.44
15.99
t 8.31
23.21
29.59
Nonsignificant
Significant
Total 15 marks
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02107020/CAPE 2016
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0210702008
I
r
3.
-l
-9(a)
Figure 2 is a photomicrograph showing a cross section of a mature anther.
Figure 2. Photomicrograph of a mature anther
Sourc e : http : //www. m i uo s copyvi ew. c om/MENU/M 1 4 - B OTA/5 1 4 A- 0 1 - 1 6/ 5 1 4 A- 0 2 A. htm
l
In the box below, make a plan drawing of the anther in Figure 2. Use annotated labels to
identify TWO tissues.
1. Epidermis - outer protective tissue
2. Vascular bundle - made up of
xylem and phloem
3. Fibrous layer - allows dehiscence
of anther
4. Tapetum - nourishment for
pollen grains
5. Pollen grain - released by anther,
contains male sperm nuclei
6. Connective tissue - keeps
support and shape of anther
[6 marksl
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02t07020lcAPE 2016
L
I
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021 0702009
I
r
-l
-10-
(b)
Figure 3 is a diagram of a section through the placenta with the umbilical vein.
A
B
C
Umbilical vein
Figure 3. Diagram showing a section through a placenta
Source: Biologt Unit I for CAPE Exominations by Ramesari Jones and Jones, 2011
(i)
Identify the structures labelled A, B and C and indicate whether EACH is of foetal
or maternal origin. Write your answers in the following table.
Label
Name of Structure
Foetal or Maternal Origin
A
Chorionic villus
Foetal
B
Intervillous space / sinus
Maternal
C
Endometrium / uterine lining
Maternal
[3 marks]
(ii)
Outline TWO functions of the placenta.
1. Exchange of substances (e.g. gases, nutrients, removal of waste products)
2. Delivery of maternal antibodies to foetus
3. Acts as a barrier for larger materials and allows for regulation of maternal and
foetal blood pressure.
[2 marks]
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021070201cAP8 2016
L
ilililtilt il!iltil ilfl ll]t illl tilt t]] tilt ilillll
0210702010
I
r
-l
- lt (i ii)
comment on the importance of the amnion during foetal development. In your
comment, give TWo effects of reduced levels of the amniotic fluid.
1. The amnion supports developing foetus against gravity. It also acts as a shock
absorber.
Low levels leads to increased risk of physical injury occurring in foetus.
2. Amniotic fluid acts as a temperature buffer, helping to regulate temperature.
Reduced levels can lead to higher temperature fluctuations in foetus.
[4 marksl
Total 15 marks
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02t07020tcAPE 2016
L
r
iltil til flil il]t ilil fl!il ililt ll]r ilil ilil flIlril
0210702011
J
r
-l
-12SECTION B
AnswerALL questions.
write your answers in the spaces provided in this booklet.
4.
(a)
According to the endosymbiotic theory, organelles such as mitochondria and chloroplasts
are thought to have evolved from prokaryotes.
(i)
Explain what is meant by the term 'endosymbiosis'and how it relates to the origin
of these organelles in eukaryotes.
Endosymbiosis occurs when one organism live within another, and both
benefit from each other.
Endosymbiotic theory describes the origin of organelles such as
mitochondria and chloroplasts as very early prokaryotic organisms that
were engulfed by a larger host prokaryotic cell and enabled that host
to undergo respiration and photosynthesis.
[3 marks]
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02t07020lcAPE 2016
L
I
fl]il tilt ililt ililt ilil ll]l illl llll lilll lffi lil lil
0210702012
I
r
-I
-13-
(ii)
Discuss THREE lines of evidence in support of this theory.
1. Mitochondria and chloroplasts are similar in size to bacterial prokaryotes.
2. Mitochondria and chloroplasts have their own DNA, which is circular
3. Mitochondria and chloroplasts have their own ribosomes, which are 70S in size.
4. Mitochondria and chloroplasts divide by binary fission, just as bacteria do.
5. In the double membrane found in mitochondria and chloroplasts, while the
outer is similar to eukaryotic membranes, the inner is more similar to prokaryotic.
[6 marks]
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0210702010APE 20r 6
L
I
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0210702013
J
r
-l
-14(b)
Enzymes are globular proteins and are highly specific in the reactions catalysed. Explain
the basis of enzyme specificity and include in your explanation a brief description of the
globular nature of these molecules.
Enzymes have tertiary structures with 3D shapes. They consist of multiple polypeptide
chains joined to each other via covalent, ionic and hydrogen bonds.
Within the enzyme is a specific sequence of amino acids called an active site,
which forms bonds with complementary substrates like a lock and key. Sometimes,
the substrate fits into the active site but the enzyme must slightly change shape to
accommodate the substrate (induced fit).
[6 marksl
Total 15 marks
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02107020/CAPE 2016
L
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0210702014
J
ts.
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- 15 (a)
Evaluate the use of genetically modified organisms in agriculture. Include an explanation
of the term 'genetically modified organism'.
A genetically modified organism is one which contains genes transferred from
another species, or has had its gene altered.
GMO crops typically have increased crop yields and provide greater food security
since crops would be less susceptible to insects.
Some GMO crops can produce their own pesticides, which reduces the need for
more harmful pesticides, leading to less exposure to toxic chemicals.
Some GMO crops have enhanced nutrients to combat nutrient deficiencies in
humans living in impoverished regions, e.g. Golden rice containing beta-carotene
to mitigate night blindness.
GMO crops may be able to grow in areas where they usually won't since they can be
modified to withstand drought or frost.
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02t07020lcAPE 2016
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0210702015
[] ilr il
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-16-
-|
[7 marksl
(b)
Explain the steps involved in the use of recombinant DNA technology for the production
of human insulin from bacteria under the following headings:
Gene isolation
Gene insertion into a vector
GENE ISOLATION - The insulin-producing cells are obtained from the pancreas.
mRNA is extracted from the cells and incubated with reverse transcriptase, which
makes a complementary strand of DNA. The DNA strand is then incubated with
DNA polymerase and free nucleotides to form the DNA helix.
GENE INSERTION - Restriction enzymes are used to cut genes, leaving 'sticky ends'.
Small circular DNA plasmids from bacteria are also cut with the restriction enzymes
at identical points. The plasmids and cut DNA are joined using DNA ligase to form
recombinant plasmid DNA.
GO ON TO THE NEXT PAGE
02t07020lcAPE 20r6
L
I
rilil ull ll!] ll]t tilt tilt ilfl ilil ilil ll]l lil rill
0210702016
J
r
-17 -
-l
[8 marks]
Total 15 marks
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02107020/CAPE 20r6
L
I
tilil ffi ffilt tilt ilil flil ilil ilil lill rill lil lll
0210702017
I
r
6.
-l
-18-
(a)
Spore production is one type ofasexual reproduction seen in fungi
(i)
Using an example with which you are familiar, give an account of the process
involved in this type of asexual reproduction.
In penicillin:
Spores (conidia) occur externally on the cells of production. The hypha of the
fungus is a chain of haploid cells, where new cells are produced at the end
of the hyphae. Branches form at the tips of the hyphae and produce
strings of spores via mitosis.
The spores are protected by waterproof covering and are spread in the air.
[5 marks]
GO ON TO THE NEXT PAGE
02107020/CAPE 2016
L
I
tilil ffi ilil tilt uil tilil ilffi ilil ilu ilil til llll
0210702018
_t
t-
-l
-19-
(ii)
Fungi can also produce offspring by sexual reproduction. Discuss the
conditions under which asexual reproduction is more advantageous than sexual
reproduction.
- When sexual reproduction takes much longer to produce offspring or
more energy is needed to produce gametes.
- If a large number of offspring are needed in a short time.
- If individuals are already well-adapted to a stable environment.
- If food material is abundant and can support a rapidly growing population.
[3 marksl
GO ON TO THE NEXT PAGE
02t07020lcAPE 2016
L
I
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ffi ililt ilt llllt til tilt
0210702019
I
I-
-l
-20 (b)
Gonadal hormones coordinate activities of the ovary and uterus in human females. Give
a brief description of the THREE main phases in the uterine cycle and for EACH phase,
comment on the role of gonadal hormones. Include the names of the gonadal hormones.
FOLLICULAR PHASE - This is when thickening of endometrium tissue occurs,
as well as vascularization (development of blood vessels). Oestrogen signals
endometrium to thicken.
LUTEAL PHASE - Thickened endometrium must be maintained for implantation.
This occurs as progesterone is secreted by the corpus luteum.
MENSTRUAL PHASE - A decrease in progesterone levels leads to the
breakdown of the endometrium. This occurs if no fertilization of the secondary
oocyte has occurred.
[7 marks]
Total 15 marks
END OF TEST
IF YOU FINISH BEFORB TIME IS CALLED, CHECK YOUR WORI( ON THIS TEST.
021070201CAPE 2016
L
I
fl]il ilil ililt tilt ilil till ffi lill lil llill ill lil
0210702020
I
r
FORM TP 2017152
qfo
-l
TEST CODE O2IO7O2O
MAY/JLTNE 2017
CARIBBEAN EXAMINATIONS COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION@
BIOLOGY
UNITl-Paper02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULLY.
1
This paper consists of SIX questions in TWO sections. Answer ALL
questions.
2
Write your answers in the spaces provided in this booklet.
J
Do NOT write in the margins.
4
You may use a silent, non-programmable calculator to answer questions.
5
You are advised to take some time to read through the paper and plan your
answers.
6
If you need to rewrite any answer and there is not enough space to do so
on the original page, you must use the extra lined page(s) provided at the
back of this booklet. Remember to draw a line through your original
answer.
7
If you use the extra page(s) you MUST write the question number
clearly in the box provided at the top of the extra page(s) and, where
relevant, include the question part beside the answer.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright O 2015 Caribbean Examinations Council
All rights reserved.
L
ffi
02t07020lcAPE 2017
Ililililflililrililillllllllr l]lllfllllllrlllllllllllll
0210702003
_l
r
-l
-4SECTION A
AnswerALL questions.
Write your answers in the spaces provided in this booklet.
1.
(a)
Figure I is a stained animal cell as seen under a light microscope.
. , . t ...''
x 1000
Figure l. Animal cell
Source : http : //imgarcade. com/
(i)
In the box below, make a detailed drawing of the cell in Figure I and label FOUR
structures seen in the cell.
[6 marks]
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5
(ii)
Using the magnification provided, calculate the actual size of the cell, highlighted
by the bar line at the left of the image. State the calculated value to the nearest
micrometer. Show your working.
Length of cell (image) = 60 mm = 60,000 micrometers
Length of cell (actual) = Image / Magnification
= 60,000 / 1000
= 60 micrometers
Size
60 micrometers
[2 marks]
i-H:'
:1'Jf,,.11
:-a\f---
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(b)
An experiment is conducted to determine the amount of reducing sugar and starch in
bananas at different degrees of ripeness. Extracts of equal quantities of tissue samples of
bananas, at three different stages of ripeness, are prepared by mashing and grinding the
samples in water. For eaeh stage, sample extracts are placed in two sets of test tubes. A
fourth test tube is set up with only water and no banana extract. Tests for reducing sugar
and starch are conducted on all four test samples. The findings of the experiment are
summarized in Table 1.
Note: For the Fehling's test the amount of the precipitate is measured as the height of
precipitate (cm) at the bottom of the test tube.
TABLE 1: OBSERVATIONS OF SAMPLE EXTRACTS AFTER TESTING
F'ood Test
Fehling's
Solution
Height of
precipitate
(cm)
Intensity of
colour
Iodine
(D
Observation
after heating
Green Raw
Banana
Half-ripe
Banana
Ripe
Banana
Greenish
Greenish
Greenish
solution
solution
solution
with red
precipitate
with red
precipitate
with red
precipitate
0.2
Intense blue-
black colour
0.4
Medium
blue-black
colour
1.0
Pale blue-
black colour
Waterwith
no Banana
Extract
Blue solution
0.0
Brown
colour
Suggest a hypothesis for this experiment.
Green bananas contain more starch and less reducing sugars compared
to ripe and half-ripe bananas
Starch is converted to reducing sugars during ripening of bananas. [1 markl
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(ii)
Compare the results obtained for the amount of reducing sugar and starch in the
banana samples. Include in your comparison reference to the observations recorded.
Shown by the presence of the red precipitate, reducing sugar is present at all stages
of ripeness. Shown by the presence of a blue-black colour after exposure to iodine,
starch is present at all stages as well.
As ripening occurs, the amount of starch decreases (blue-black colour intensity
decreases) and the amount of reducing sugar increases (height of precipitate
increases).
[3 marksl
(iii)
Comment on the significance of the findings of this experiment in relation to the
use of bananas as a food item.
Riper bananas contain more reducing sugar and would be sweeter and softer,
easier to consume and digest. Green bananas contain more starch and less
reducing sugars, so would be good for a diabetic to consume (after cooking).
[2 marks]
(iv)
State the purpose of testing a sample with no banana extract.
Acts as a control.
[1 markl
Total 15 marks
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0210702007
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2.
-8(a)
-|
Haemophilia is a recessive sex-linked condition in humans in which the blood does not
clot properly, leading to excessive bleeding. Use the symbols XH for the normal allele
and Xh for the haemophilia allele in your responses.
(i)
State the genotypes of the following
H
X Y
A normal clotting male ..............
A normal clotting carrier female
XHXh
[2 marksl
(iD Using the genotypes stated in (i) as parental genotypes, construct a Punnett square
diagram to show how haemophilia-affected offspring can result from normal
clotting parents. State the phenotype of all offspring.
[3 marksl
(iiD
Give an explanation as to why a man with haemophilia cannotpass on the condition
to his son.
The allele for haemophilia is sex-linked and is found on the X chromosome.
Sons of men would receive the Y chromosome from their fathers and thus not
the X chromosome with the faulty allele.
[1 markl
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0210702008
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-9 (b)
A geneticist crossed two types of pea and obtained 5474 plants with round seeds and
1850 plants with wrinkled seeds in the F, generation. You are required to use the Chisquare test to show that these results are consistent with the 3:1 ratio normally expected
from a monohybrid cross.
(D
State the null hypothesis for this test.
There is no significant difference between the observed and expected numbers.
[1 mark]
(ii)
Calculate Chi-square using the formula, Chi-square : I(O - E)'lE, where O is
the observed and E the expected number of plants. Show your working in tabular
form as follows.
Plant1}pe
Round
seeds
Wrinkled
seeds
Observed
Expected
o-E
5474
5493
1831
1850
(o - E)'
(o - E)r/E
-19
361
0.066
19
361
0.197
Chi-square
0.26
[2 marksl
(iii)
Use the data in Table 2 on page l0 to make an inference based on your calculated
Chi-square value.
The Chi-square value (0.26) is much smaller than the critical p-value (>5%)
of 3.84. Therefore, the null hypothesis will be accepted. (no sig. difference)
[2 marks]
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TABLE 2: STATISTICAL TABLE - CRITICAL VALUES OF CHI.SQUARE DISTRIBUTION
Degrees of
Number of
Freedom
Classes
I
2
0.46
1.64
2.71
3.84
6.64
10.83
2
)
r.39
3.22
4.61
5.99
9.21
13.82
J
4
2.37
4.64
6.25
7.82
1r.34
16.27
4
5
3.36
5.99
7.78
9.49
t3.28
18.47
Probability [p] that
chance alone could
produce the deviation
0.50
0.20
0.10
0.0s
(s0%)
(20%)
( r0% )
(s%)
0.0i
(r%)
(0.r%)
(c)
12 Yalues
0.001
The variation in size of an organism in a population is depicted in Figure 2.
a
6l
E
E
o
L
€)
!
z
Size
Figure 2.Slr,e distribution of an organism in a population
(i)
Using the axes provided in Figure 3, illustrate the effect of disruptive selection on
the population distribution in Figure
2.
I mark]
0
6l
E
"E,
o
L
()
E
E
z
Size
Figure 3. Effect of disruptive selection
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0210702010
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(iD Using Darwin's theory of natural selection, explain how disruptive selection can
lead to the formation of new species.
Using salmon as an example, there can be small, intermediate and large sizes.
This is due to natural variation in the population. Intermediate sizes are selected
against due to conditions favouring the other two extremes (small can stealthily
fertilize eggs, while large can physically fight better).
The successful small and large individuals reproduce and pass on their traits to their
offspring, while the intermediate individuals do not. New species may arise if the
genetic variants can no longer interbreed.
[3 marks]
Total 15 marks
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0210702011
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3.
-l
-t2(a)
An experiment is conducted to investigate the effects of type of culture medium and
sucrose concentration on the pollen germination rate of a species of palm. Pollen grains
are collected from newly opened flowers of the same bunch on the same day. The grains
are inoculated into two types of medium: a liquid culture medium and a solid culture
medium. The solid culture medium consists of different concentrations of sucrose in agar.
The liquid medium is similar to the solid medium with respect to the sucrose concentrations
but does not contain agar. The results of the investigation are given in Table 3.
TABLE 3: EF.FECTS OF CULTURE MEDIUM AND SUCROSE CONCENTRATION
ON POLLEN GERMINATION RATE IN A SPECIES OF'PALM
7o Pollen Germination
Sucrose Concentration
Liquid Culture Medium
Solid Culture Medium
0
5l
27
20
52
39
40
73
87
60
50
57
80
4I
66
100
0
4l
(el')
Source: AmericanJournal of Plant Sciences. 2013, 4, 1669*1674
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(i)
On the grid provided in Figure 4, plot line graphs for the data given in Table 3.
[4 marksl
Figure 4. Effects of culture medium and sucrose concentration on pollen germination
(iD
Briefly describe the overall trend for the effect of different sucrose concentrations,
in both culture media, on the germination rate of the palm pollen.
For both liquid and solid culture media, germination rate increased with
increasing the sucrose concentration, reaching a maximum rate at 40 g/l.
Beyond 40 g/l, the germination rates decreased as sucrose concentration
was increased.
[2 marksl
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0210702013
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(iii)
Based on the results shown in Table 3, what can be deduced about the effect of
the type of culture medium on the maximum rate of pollen germination?
The solid culture medium works faster for germination compared to the liquid.
The liquid culture medium has higher germination rates at sucrose concentrations
beyond 40 g/l than the solid culture medium.
[2 marks]
(b)
Figure 5 is a drawing of a section through a human placenta in situ.
I
Allantois
Fallopian
tII
tube
Developing
embryo
Cervix
Figure 5. Section through a human placenta
(D Identifu and describe the main function of EACH of the structures labelled I, II
and III.
I:
Chorionic villi - facilitates exchange of materials between the maternal
and foetal blood supply.
II:
Uterus lining (endometrium) - Supports attachment and development of
placenta and foetus.
III
Umbilical cord - Connects foetus to placenta to allow delivery (e.g. glucose) or
removal of materials (e.g. urea, carbon dioxide).
[3 marksl
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(ii)
-l
Comment on TWO roles of the structure labelled Y in the development of the
foetus.
(Y is the amniotic sac/amnion)
The amnion acts as a shock absorber to protect against mechanical damage.
It also helps to keep the temperature of the foetus stable and acts as a reservoir
of nutrients. It also helps support the developing foetus, holding it in a safe position.
[2 marksl
(iii)
Outline TWO ways in which the placenta acts as a protective barrier for the
developing foetus.
The placenta prevents large molecules from entering the foetus' bloodstream,
such as maternal hormones.
It also reduces the blood pressure of the maternal blood, as materials enter
the foetus.
The placenta also delivers antibodies to the foetus to facilitate natural passive
[2 marksl
immunity.
Total 15 marks
L-,
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0210702015
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_16_
SECTION B
AnswerALL questions.
Write your answers in the spaces provided in this booklet.
4.
(a)
(i)
Compare the cell size and THREE structural features ofprokaryotic and eukaryotic
cells
1. Eukaryotic cells contain double-membraned organelles such as mitochondria
and chloroplasts, while prokaryotic cells don't.
2. Eukaryotic cells contain a nucleus, while it is absent in prokaryotic cells
(typically storing DNA in a circular nucleoid region)
3. Eukaryotic ribosomes are larger (80S) than prokaryotic (70S).
4. Eukaryotic cells are much larger than prokaryotic cells.
[4 marks]
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0210702016
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(ii)
Outline the endosymbiont theory for the origin of eukaryotic cells. Explain how
characteristics of mitochondria and chloroplasts provide evidence in support of
this theory.
The endosymbiont theory states that mitochondria and chloroplasts were once
prokaryotic organisms (before they were organelles) that invaded larger host
cells and integrated within them symbiotically.
EVIDENCE
- Both organelles are similar in size to prokaryotic cells.
- Both organelles have circular DNA, similar to prokaryotes.
- Both organelles reproduce by binary fission, similar to prokaryotes.
- Ribosomes within mitochondria and chloroplasts are of a similar size (70S)
to prokaryotic ribosomes.
[5 marksl
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(b)
Discuss the importance of protein structure for determining enzyme specificity and mode
of action.
- Enzymes are globular proteins that have tertiary (3D) structures. They are catalysts
for biological reactions. They do this by lowering the activation energy for the reaction.
- Enzymes contain active sites, which are regions of specific size and amino acid
sequences, for the binding of substrates. Substrates bind to these active sites due to
their complementary shapes, similar to a lock and key mechanism.
- After binding, an enzyme-substrate complex is formed. The enzyme can slightly
change shape in response to binding (induced fit).
[6 marks]
Total 15 marks
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0210702018
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(a)
5.
-l
-19Describe the roles of DNA and RNA in protein synthesis.
- Sequences of DNA (genes) carry code for protein synthesis.
- There are several types of RNA involved in protein synthesis including:
1. mRNA (messenger RNA) - Used for transcription, which is when genetic code is
copied from DNA into codons. During this process, DNA is 'unzipped' to expose
bases, allowing RNA nucleotides to pair up to form complementary mRNA. The mRNA
takes the information away from the nucleus to the ribosomes.
2. tRNA (transfer RNA) - Used for translation, which is done to produce amino acids
from ribosomes reading the mRNA codons. A polypeptide chain is produced as
various tRNA molecules deposit respective amino acids to the ribosome.
3. rRNA (ribosomal RNA) - These form ribosomes, which move along mRNA
molecules and facilitate the assembly of polypeptide chains, which then bond and
fold into 3D proteins structures.
[5 marksl
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0210702019
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(b)
Genetic engineering and gene therapy are two closely related technologies with promising
potential for improving human health.
(i)
Outline the principle of genetic engineering, including a brief description of the
basic steps of this technique.
Genetic engineering involves the transfer of particular genes from one
organism and placing it into another, thus altering its genome. It can also
involve the alteration of an existing gene.
STEPS:
1. Identify and isolate the gene (using restriction enzymes to 'cut' DNA).
2. Join the selected gene with a vector (plasmid), using DNA ligase.
3. Introduce the vector to a host cell (e.g. E. coli bacteria)
4. Cloning/reproduction of cells.
5. Selection of transformed host cells with recombinant DNA to obtain product/
[5 marks]
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0210702020
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(ii)
Despite its promising potential, one of the arguments sometimes raised against
the use of gene therapy is that it is technically too dangerous. With reference to
cystic fibrosis, discuss reasons for this viewpoint.
1. Viral vectors may result in infection and inflammation of the lungs,
especially if the patients are immunocompromised. The viral vectors
may also mutate in patient. Death has been noted in a few CF treatment patients.
2. The therapy will only work if the gene is delivered to a large number of
tissues. Vectors may become trapped by cilia or mucus (a common symptom of CF)
The vectors may deliver DNA to non-target cells and thus may be inefficient.
[5 marksl
Total 15 marks
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0210702021
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6.
-l
aa
(a)
Giving examples, discuss TWO advantages of vegetative propagation of plants.
Vegetative propagation is a form of asexual reproduction, which allows new plants
to be produced from processes such as grafting, budding and cuttings.
It allows rapid multiplications of plants of identical genotype, so as to maintain
favourable characteristics and ensure that they are well-adapted to stable
environmental conditions. Tissue cultures ensure plants are also disease-free.
For some species, such as bananas, it is the only viable method of reproduction.
Bananas are reproduced via cuttings. Grafting can also be used to produce
hybrid versions of plants with mixed characteristics.
[5 marks]
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0210702022
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(b)
f..-1
:=.:
-l
-23 Discuss TWO genetic consequences of sexual reproduction.
--i]l-.
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j=+.
...\
.tar
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1. Sexual reproduction leads to increased variation in organism, ensuring high
-f5+
:riia
biodiversity in populations. This allows members of the species to overcome
,z
]!+'
ili{
.Ha
selective pressures such as pests and disease, and possibly facilitating speciation.
:!i+
.ff
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,*;
.-t--
2. The formation of seeds during sexual reproduction improves distribution of
,,f
.s
members away from parent plants and thus, reducing competition and increasing
:-tiJ.
..q
chances of colonizing new areas.
....-.-.:
.:-i::'.
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-:&
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.-*.
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:-_n"
,-.ti
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[4 marks]
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:.Gl
:-{ii
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0210702023
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(c)
Describe the development of fruit and seed structures from floral structures after fertilization.
- The zygote is formed and eventually becomes an embryo after mitosis.
- The endosperm is formed, supplying nutrients to the embryo.
- A large basal cell and suspensor cell column attaches embryo to ovule wall.
- The embryo develops a radicle, plumule and early leaves (cotyledons).
- The ovule develops into a seed and the ovary develops into the fruit.
- The ovary wall develops into pericarp (flesh) of the fruit.
[6 marksl
Total 15 marks
END OF TEST
IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK ON THIS TEST.
021070201cAP82017
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0210702024
J
The phospholipids form the bilayer by having their hydrophilic heads facing
outward and their hydrophobic tails facing inward, away from the cell exterior
and the cell cytosol.
Two layers of phospholipids form the bilayer.
Cholesterol
Intrinsic (or integral) proteins span the entire length of the phospholipid bilayer.
They can take the forms of channels, which allow passage of large, polar materials
in and out of the cell via facilitated diffusion.
They can also form carrier proteins, which are gated channels activated by certain factors
such as voltage or a chemical (ligand). These usually require ATP to activate
(active transport).
Triglyceride
They contain a higher amount of energy; three fatty acids store more energy
due to the long carbon chains with hydrogen bonds.
Lipid properties allow insulation/protection as adipose (fatty) tissue builds up
within animals.
Triglycerides are needed in the body to absorb vitamins A, D, E and K.
R
R
r
r
R
r
1:1
A gene is a sequence of DNA nucleotides that codes for a particular
polypeptide/protein.
- An allele is one or more alternative forms of a gene.
(e.g. R is a gene. R and r are alleles)
Prophase
Anaphase
Interphase
1. The concentration of mercury in foetal blood is higher than in the maternal blood
(or vice versa)
2. As the number of maternal amalgam fillings increase, the concentration of mercury
increases in both maternal and foetal blood.
It is an active process, as it would require ATP to do so.
This is because the foetal blood always has a higher concentration of mercury
than that of the mother, so the movement must occur against a concentration
gradient.
Intine
Exine
Tube nucleus
Generative nucleus
- Pollen grain germinates to form pollen tube. On descent, the generative nucleus
divides by mitosis producing two male haploid gametes.
- Once in the embryo sac of the ovule, one male haploid gamete fuses with ovum
nucleus (to form the diploid zygote).
- The other haploid male nucleus fuses with the diploid endosperm nucleus
in the embryo sac (to form the triploid endosperm).
In some crop seeds such as rice, wheat and maize, the endosperm remains
after the embryo has matured and provides the main source of nutrition for human
consumption.
In some crops the nutrients in endosperm is transferred to the maturing embryo
(cotyledons) and this is used for human consumption. Sugars are contained
in the pericarp, which was once the ovary wall.
- Enzymes are globular proteins with a 3-D shape and tertiary structure.
They have an active site on the 3-D structure to which a specific substrate attaches for
a reaction to occur.
- Competitive inhibitors have a similar shape to the substrate molecule.
They fit temporarily onto the active site and compete with substrate for entry to the active site.
The effect of competitive inhibitors decrease as the substrate concentration is increased.
- Non-competitive inhibitors bind to another part of the enzyme molecule (allosteric site) or binds
permanently to active site. Distortion of the (3-D) structure of enzyme can occur.
Distortion of the enzyme structure can prevents the substrate from binding.
Effects of non-competitive inhibitors are not reversed if substrate concentration is increased.
1. Epidermis – outermost layer of root can offer some protection as a physical barrier / Some
cells have root hair extensions. Provide larger surface area for the uptake of water and nutrients
2. Parenchyma – beneath the epidermis (Cortex). These cells are relatively unspecialised.
They have thin cell walls and facilitate the movement of water. They provide support when cells are
turgid. They may also serve for storage of starch (energy reserve)
3. Endodermis – single layer of cells surrounding the stele / vascular tissue. Cell walls are
waterproofed to form the Casparian strip. Controls the movement of water and mineral salts
into the vascular tissue.
4. Xylem tissue – vascular tissue - contains dead, empty cells with lignified side walls and no end
walls (vessel element). Lignin gives strength to the organ for anchorage. Vessel elements arranged
end to end to form continuous tube for water transport throughout the plant
5. Phloem tissue – vascular tissue – contains phloem sieve elements. Living cells with perforated end
walls arranged end to end. Allows sucrose solution to flow to root and other parts of plants to supply
OTHERS:
energy and building materials.
- Pericycle
- Meristematic tissue
- The information stored in the gene is in the form of a sequence of bases which is
used to sequentially arrange amino acids to synthesize a protein.. mRNA is the
molecule which is used as the template to organize the primary structure of proteins.
- The process by which DNA is used to synthesize RNA is called transcription /
producing mRNA with a complementary base sequence to one strand of DNA.
The DNA strand unzips so that both strands separate (catalyzed by DNA helicase)
- One strand is used as the template strand. Complementary bases are added to
strand being copied (in 3’to 5’ direction, catalyzed by RNA polymerase). A pairs with
U and C pairs with G.
- Condensation reactions occur (phosphodiester bonds form) between bases (in a
5’to 3’ direction)to form mRNA.
- Three bases (triplet codon) on the strand code for 1 amino acid.
- Phenotype is the external observable characteristics of an organism. It is determined by the kind of
proteins produced in the body.
- Proteins are expressions of genes. A particular gene codes for a particular protein.
- Cells of a similar phenotype form tissues (different tissues form organs, organs form organ
systems), which together form the organism that expresses the trait.
For example, the protein haemoglobin is packed into an erythrocyte giving it its characteristic red
colour and biconcave shape. With melanin, when there are larger amounts in the basal epidermal
layer of the skin, it would appear darker in colour.
1. Removal of small group of cells from plant (called explant)
2. Disinfection of the explant immediately after removal from plant.
3. Immerse explant in sterile aerated solution/culture medium containing hormones
and nutrients
4. Undifferentiated cells of explant divide repeatedly to form callus – can be
maintained indefinitely in culture
5. Callus cells sub-cultured on sterile medium and induced to form shoots and roots
by varying plant growth substances
6. Plantlets transplanted to sterile soil once they are large enough
7. Plant tissues are totipotent – Each cell contains all the information required to
produce the entire plant (totipotency)
Differentiated into
a head, neck,
middle piece and
flagellum (tail).
Spherical cell surrounded
by follicle cells and
zona pellucida.
The tail provides
motility (allows
swimming).
Much larger than sperm
cell. Contains more
cytoplasm and stores
more nutrients.
Haploid.
Delivers male
chromosomes.
Haploid.
Delivers female
chromosomes.
Contains glycoproteins that
facilitate union
with oocyte.
Contains microvilli
that absorbs nutrients
and has proteins that
bind to proteins on
sperm cells.
Numerous
mitochondria.
Numerous
mitochondria as well.
Provides energy
for motility.
Provides energy
for metabolic
processes and
development.
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SUBJECT
BIOLOGY – UNIT 1 – Paper 02
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FORM TP 2019156
CARIBBEAN
02107020
MAY/JUNE 2019
E XAM I NAT I O N S
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®
BIOLOGY
UNIT 1 – Paper 02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULLY.
1.
This paper consists of THREE questions. Answer ALL questions.
2.
Write your answers in the spaces provided in this booklet.
3.
Do NOT write in the margins.
4.
You may use a silent, non-programmable calculator to answer questions.
5.
You are advised to take some time to read through the paper and plan your
answers.
6.
If you need to rewrite any answer and there is not enough space to do so
on the original page, you must use the extra lined page(s) provided at the
back of this booklet. Remember to draw a line through your original
answer.
7.
If you use the extra page(s), you MUST write the question number
clearly in the box provided at the top of the extra page(s) and, where
relevant, include the question part beside the answer.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright © 2018 Caribbean Examinations Council
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Module 1 — Cell and Molecular Biology
1.
(a)
Figure 1 shows a light micrograph of the transverse section through the primary stem of
a Dahlia (Dahlia sp.) plant, a typical dicotyledon.
Figure 1. A light micrograph of a transverse section of a Dahlia plant stem
Source: https://www.sciencephoto.com/image/572998/large/C0197068
-Dahlia_stem,_light_micrograph-SPL.jpg
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In the space below, construct a tissue diagram from the light micrograph of the
plant section in Figure 1.
[3 marks]
(ii)
Provide annotations to describe EACH labelled structure, A–E.
is the outermost layer of (dicot) stem that is covered by thick cuticle
A Epidermis
............................................................................................................................
/ has multicellular root hairs / Protects the underlying tissue
.................................................................................................................................
is present below the epidermis in several layers. Can be differentiated into
B Cortex
............................................................................................................................
endodermis, parenchyma, collenchyma. Used for mechanical support
.................................................................................................................................
Pith/medulla occupies the central portion of the ground tissue. Stores
C ...........................................................................................................................
materials/water/nutrients. Allows gaeous exchange through air spaces.
.................................................................................................................................
Xylem, part of the vascular bundle. Includes dead empty cells with lignified side walls
D ............................................................................................................................
and no end walls. Transports water and dissolved ions (from root to leaves).
.................................................................................................................................
Phloem lies outside of the vascular bundle / is composed of (living) sieve tubes/sieve
E ............................................................................................................................
elements/companion cells. Transports synthesized organic food (from source to sink).
.................................................................................................................................
[5 marks]
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(b)
The graph in Figure 2 shows the change in the rate of uptake of two substances,
Substance A and Substance B, as the concentration of each substance increases across a
lipid bilayer.
Figure 2. Graph of change in the rate of substance uptake
(i)
Describe the changes in the rate of uptake of Substance A and Substance B as the
concentration of solute increases as depicted in Figure 2.
A – Linear increase in the rate of uptake of substance as concentration of solute
.................................................................................................................................
increases. / Rate of uptake is directly proportional to solute concentration.
.................................................................................................................................
.................................................................................................................................
B - Initial increase in the rate of uptake as concentration of solute increases
.................................................................................................................................
followed by a plateau / no further increase beyond a certain concentration.
.................................................................................................................................
.................................................................................................................................
[2 marks]
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Identify the types of diffusion illustrated for Substance A and Substance B in Figure 2.
A - Simple/normal/regular diffusion
.................................................................................................................................
B - Facilitated diffusion
.................................................................................................................................
.................................................................................................................................
[2 marks]
(iii)
Compare and contrast the two types of diffusion identified in (b) (ii).
1. For both simple and facilitated diffusion, molecules and ions move down the
.................................................................................................................................
concentration gradient (From high concentration to low concentration)
.................................................................................................................................
2. For both simple and facilitated diffusion, no energy (as ATP) is required/passive
.................................................................................................................................
Any two -->
process / For both, movement is due to the random movement of molecules and ions
.................................................................................................................................
.
3. Simple diffusion occurs through phospholipid bilayer whereas facilitated diffusion
.................................................................................................................................
occurs through transport/transmembrane proteins.
.................................................................................................................................
4. Small and nonpolar molecules move by simple diffusion whereas large or polar
.................................................................................................................................
molecules move by facilitated diffusion.
.................................................................................................................................
5. Facilitated diffusion (may be initially faster than simple diffusion but) plateaus/becomes
.................................................................................................................................
saturated as solute concentration increases, whereas simple diffusion is linear
.................................................................................................................................
[4 marks]
(iv)
Arrange the following molecules according to their ability to diffuse through the
lipid bilayer: CO2, glucose, water, RNA. Begin with the molecule that diffuses
the easiest. Explain the order of your arrangement.
.................................................................................................................................
Molecules from easiest to hardest: CO2, water, glucose, RNA
.................................................................................................................................
.................................................................................................................................
1. CO2 is small/non-polar/lipid soluble (and diffuses the easiest).
.................................................................................................................................
2. Water is small and polar. Diffuses through aquaporins in bilayer.
.................................................................................................................................
3. Glucose is large/polar and requires transmembrane proteins.
.................................................................................................................................
4. RNA is very large/largest/highly charged/very polar.
.................................................................................................................................
.................................................................................................................................
[6 marks]
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(c)
Phospholipids are major components of all cell membranes.
With the aid of an annotated diagram, explain the organization of membrane lipids.
Discuss the structure of a cell membrane specifying the organization of membrane lipids
and the various ways in which proteins can associate with the membrane.
Draw diagram here.
.............................................................................................................................................
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1. In aqueous environments, he hydrophilic (water-loving) or polar end (‘head’) will
..............................................................................................................................................
interact with the water and face outwards into the aqueous environment inside and
..............................................................................................................................................
outside the cell.
..............................................................................................................................................
2. The hydrophobic (water-hating) or non-polar end (‘tail’) end will stay shielded from the
..............................................................................................................................................
water on the inside and face inwards creating a hydrophobic interior.
.............................................................................................................................................
3. Glycolipids (and glycoproteins) help stabilize the membrane structure by forming
.............................................................................................................................................
(hydrogen) bonds with water molecules outside the membrane (also important for cell
.............................................................................................................................................
recognition; receptors for hormones etc.)
Any ..............................................................................................................................................
3 points -->
4. Most proteins float about in the phospholipid bilayer forming a fluid mosaic pattern.
..............................................................................................................................................
5. Some proteins penetrate only part of the way into the membrane while others
..............................................................................................................................................
penetrate all the way through (integral/intrinsic).
..............................................................................................................................................
6. Proteins can be found at the outer and inner surface of the phospholipid bilayer
.............................................................................................................................................
(peripheral/extrinsic) loosely bound to the hydrophilic (polar) surfaces.
.............................................................................................................................................
7. Cholesterol helps to maintain the fluidity of the membrane as temperatures change /
.............................................................................................................................................
prevents membrane from becoming too rigid/firm/stiff at low temperature and too fluid at
..............................................................................................................................................
high temperature
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
.............................................................................................................................................
.............................................................................................................................................
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[8 marks]
Total 30 marks
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Module 2 — Genetics, Variation and Natural Selection
2.
(a)
Table 1 shows the relative growth rate data for sensitive strains and resistant strains of S.
typhimurium bacteria, exposed to varying concentrations of tetracycline antibiotic.
TABLE 1: RELATIVE GROWTH RATE DATA FOR SENSITIVE STRAINS
OF BACTERIA
Tetracycline
Conc. (μg/ml)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
(i)
Relative
Growth Rate of
Sensitive Strain
1.00
0.67
0.47
0.31
0.23
0.16
0.13
0.08
0.08
Relative Growth
Rate of Resistant
Strain
1.00
1.02
1.04
1.01
0.99
1.01
1.01
1.01
1.01
On the grid provided on page 11, plot a line graph showing the relative growth
rate (y-axis) vs tetracycline concentration (x-axis) for the sensitive and resistant
strain of bacteria.
[6 marks]
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(ii)
Outline the effect of EACH of the following antibiotic concentrations on BOTH
bacterial populations at 0.1 ug/ml and 0.7 ug/ml antibiotic concentrations.
- The resistant strain is (apparently) unaffected/slight increase by this
0.1 ug/ml ................................................................................................................
concentration of antibiotic. / Relative growth rate is close to 1.
.................................................................................................................................
- The sensitive strain growth rate decreases/is reduced. Relative
.................................................................................................................................
growth rate reduced to 0.67.
.................................................................................................................................
- The growth rate of resistant strain is still unaffected at this concentration.
0.7 ug/ml .................................................................................................................
Relative growth rate is close to 1.
.................................................................................................................................
- At this concentration of tetracycline, the growth rate of the sensitive
.................................................................................................................................
strain is greatly/considerably/severely reduced/decreased / Relative
.................................................................................................................................
growth rate reduced to 0.08.
[4 marks]
(iii)
Explain how the differences between the growth rates of the bacteria could lead
to natural selection.
1. Bacterial strains differ in susceptibility to the antibiotic. / Lower growth rates in
.................................................................................................................................
sensitive bacteria strain on exposure to the antibiotic. / Higher growth rates in
.................................................................................................................................
resistant strain.
.................................................................................................................................
2. When exposed to the antibiotic, higher growth rate promotes survival in resistant
.................................................................................................................................
strain. / Lower growth rate reduces survival in sensitive strain.
.................................................................................................................................
Any 4 points -->
3. Growth and reproduction continues in the resistant strain on exposure to the
.................................................................................................................................
antibiotic. / Growth and reproduction reduced in the sensitive strain.
.................................................................................................................................
4. The antibiotic acts as an environmental/natural selective force/pressure.
.................................................................................................................................
Selection for the resistant strain / Selection against the sensitive strain.
.................................................................................................................................
5. The surviving members of the sensitive strain exposed to high levels of tetracycline
.................................................................................................................................
[4 marks]
can eventually proliferate or reproduce.
6. The surviving members of the sensitive strain can thus become a population
of a new resistant strain.
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Figure 3 below illustrates how mutation brings about genetic variation.
Figure 3. Mutation and genetic variation
(i)
Complete the missing codons in the boxes labelled A and B in Figure 3.
CAC
A ............................................................................................................................
GUG
B ............................................................................................................................
[2 marks]
(ii)
With reference to the mutation in Figure 3, explain how mutation brings about
genetic variation at the DNA, protein and cellular level.
Effects at the DNA level: A single point mutation/substitution resulting in a
.................................................................................................................................
nucleotide being replaced by another / during transcription, the mRNA
.................................................................................................................................
molecule would have differed as well
.................................................................................................................................
Effects at the protein level:
.................................................................................................................................
- Coding for the amino acid VAL instead of GLU. Protein primary structure is
.................................................................................................................................
affected / Polypeptide chains affected. Protein 3D/folding/globular structure is
.................................................................................................................................
distorted/impaired. Protein function is adversely affected.
.................................................................................................................................
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Effects at the cellular level:
.................................................................................................................................
- Cell structures can be adversely affected / e.g. abnormal haemoglobin
.................................................................................................................................
causes red blood cells to have a sickle shape)
.................................................................................................................................
- Cell function can be adversely affected / e.g. Sickle-shaped red blood
.................................................................................................................................
cells die prematurely, which can lead to anaemia. / inflexible, sickle.................................................................................................................................
shaped cells get stuck in small blood vessels and can cause serious
.................................................................................................................................
[6 marks]
medical complications / lower affinity to oxygen.
(c)
Natural selection was proposed by Darwin to be the mechanism of evolutionary change.
Discuss the process of natural selection.
1. Variation occurs between individuals in a population / variation due to
..............................................................................................................................................
mutation/crossing/sexual reproduction.
..............................................................................................................................................
2. Each generation produces more offspring than the environment can support. /
..............................................................................................................................................
Tendency for overpopulation.
..............................................................................................................................................
3. Population increases but resources are finite so there is competition among
..............................................................................................................................................
individuals.
..............................................................................................................................................
4. The environment applies selection pressure to select the fittest.
..............................................................................................................................................
5. Individuals that are better-adapted/have advantageous genes/alleles/traits
8 points..............................................................................................................................................
--->
survive. / Survival of the fittest.
..............................................................................................................................................
6. Individuals that survive reproduce / transmit advantageous genes/alleles/traits
..............................................................................................................................................
to their offspring. / beneficial alleles passed on to the next generation/inherited
..............................................................................................................................................
7. Individuals with the non-advantageous genes/alleles do not get to pass them on
..............................................................................................................................................
to offspring as they do not survive.
..............................................................................................................................................
8. Frequency of the favourable/beneficial genes/alleles/traits increases in the
..............................................................................................................................................
population over time
..............................................................................................................................................
9. After many generations, the population evolves / population is mainly comprised
..............................................................................................................................................
of individuals with the advantageous genes / Population becomes better adapted
..............................................................................................................................................
to its environment
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[8 marks]
Total 30 marks
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Module 3 — Reproductive Biology
3.
(a)
Gametogenesis is the development and production of the male and female germ cells
required to form a new individual.
(i)
Draw a well-labelled flow diagram of the process of oogenesis starting from the
primary oocyte.
[5 marks]
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State the role of ONE named hormone in oogenesis.
.................................................................................................................................
1. GnRH secreted by the hypothalamus stimulates the pituitary gland to
secrete LH and FSH.
.................................................................................................................................
2. FSH stimulates the growth of Graafian follicles in ovary / stimulates the
.................................................................................................................................
development of egg/oocyte within the follicle to form secondary oocyte.
.................................................................................................................................
3. LH induces the rupture of the mature Graafian follicle and the release of
.................................................................................................................................
any
1 --->
secondary oocyte / LH causes ovulation
.................................................................................................................................
4. LH promotes the development of corpus luteum (from ruptured graafian
[2 marks]
follicle)
5. Oestrogen at low levels inhibits FSH and LH (which stops development of
more follicles)/negative feedback
6. Oestrogen at high levels stimulates a surge in secretion of LH (which
causes ovulation)/positive feedback
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(iii)
Figure 4 shows a diagram of a sperm cell.
Figure 4. Diagram of a sperm cell
Provide annotations on ONE function of EACH of the structures labelled A–D in
the diagram of a sperm cell shown in Figure 4.
(Head) Contains nucleus, which has haploid set of chromosomes
A ............................................................................................................................
Contains acrosome and hydrolytic enzymes for penetration of egg.
.................................................................................................................................
(Neck) Connects head to tail/middle piece (since centriole attaches to filament
B ............................................................................................................................
of tail. Centrioles present form the sperm flagellum.
.................................................................................................................................
(Middle piece) The middle piece of human sperm contains the mitochondria
C ............................................................................................................................
used for ATP production.Provides energy for the propulsion of the sperm
.................................................................................................................................
(Tail) It helps in propelling the sperm cell forward to meet the egg / allows motility.
D .............................................................................................................................
- The tail moves in a wave-like/lashing/whipping motion / allows the sperm to swim
................................................................................................................................
[4 marks]
(in a fluid medium).
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Relate any ONE structural feature of a sperm cell to its functions.
Tadpole shape/streamlined/flattened head/cylindrical body/ elongated tail:
.................................................................................................................................
increases mobility/swimming
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
[2 marks]
(b)
(i)
Self-incompatibility is determined by multiple alleles at the S locus. Complete the
following table, using the key given below, to show the extent to which fertilization
is possible between the genotypes of the parents.
Key: (−) no fertilization possible
(+) fertilization possible by all pollen grains
(− +) fertilization only possible by half pollen grains
Genotype of Male Parent
S1S2
Genotype of Female Parent
S 1S 2
S1S3
S 3S 4
-
-+
+
[3 marks]
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(ii)
Differentiate between EACH of the following pairs of terms:
Self-incompatibility and male sterility
Self-incompatibility: Inability of a plant with functional pollen to set seeds
.................................................................................................................................
when self-pollinated. / Flowers will produce pollen grains which fail to fertilize
.................................................................................................................................
the same flower (or any other flower of the same plant).
.................................................................................................................................
Male sterility: Pollen is absent or non-functional/inactive. Failure of plant to
.................................................................................................................................
produce viable pollen grains / incapable of producing male gametes
.................................................................................................................................
.................................................................................................................................
[2 marks]
Monoecy and dioecy
- Plants that exhibit monoecy have unisexual reproductive organs or flowers,
.................................................................................................................................
with the organs or flowers of both sexes carried on a single plant. / separate
.................................................................................................................................
male and female flowers on a single/same plant
.................................................................................................................................
- A dioecious plant has separate sexes with individuals producing either
.................................................................................................................................
male
or female flowers. / male flowers on one plant and female flowers on another
.................................................................................................................................
separate plant
.................................................................................................................................
[2 marks]
Protandry and protogyny
- In protandry, the stamens/anthers/androecium develop ahead of
.................................................................................................................................
pistil/stigma/gynoecium. Pollen grains mature/are shed before the
.................................................................................................................................
pistils are ready to accept them.
.................................................................................................................................
- In protogyny, the pistils/stigma/gynoecium mature and accept pollen
.................................................................................................................................
ahead of the stamens/anthers/androecium.
.................................................................................................................................
.................................................................................................................................
[2 marks]
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Explain TWO major genetic consequences of EACH of the following methods of
pollination:
(i)
Self-pollination
1. Promotes inbreeding / less genetic diversity than cross pollination / Greater
.................................................................................................................................
chance of loss of alleles (through inbreeding) compared to cross pollination
.................................................................................................................................
2. Variation and hence adaptability to changed environment are reduced. /
.................................................................................................................................
Restriction of gene pool / Restriction of evolution of species.
.................................................................................................................................
3. Vigour decreases with prolonged self-pollination which increases
.................................................................................................................................
susceptibility to pests and diseases/decrease yields
.................................................................................................................................
Any two -->
4. Preserves genetic makeup of well adapted plants in a stable/unchanging
.................................................................................................................................
environment / If a given genotype is well-suited for an environment, self.................................................................................................................................
pollination helps to keep the trait stable in the species
.................................................................................................................................
.................................................................................................................................
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(ii)
Cross pollination
.................................................................................................................................
1. If a given genotype is well-suited for an environment, cross pollination
.................................................................................................................................
prevents this trait from being stable in the species. / may lose advantageous
.................................................................................................................................
phenotypes in a well-adapted stable environment
.................................................................................................................................
2. Cross pollination promotes outbreeding/prevents inbreeding which
.................................................................................................................................
increases the amount of variation in a species
.................................................................................................................................
3. Increases vigour/resistance to pests/diseases/less homozygous disease
.................................................................................................................................
Any two ---->
genes expressed
.................................................................................................................................
4. Cross pollination may introduce some undesirable characters but new
.................................................................................................................................
useful characters can also be introduced
.................................................................................................................................
5. Increases variation enhances adaptability to changing environments /
.................................................................................................................................
Utilises more of the gene pool – more variants for natural selection/more
.................................................................................................................................
evolutionary potential/development of a new species possible over time.
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
[8 marks]
Total 30 marks
END OF TEST
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TH
IS
T
NO
ON
E
PA
G
TE
RI
W
DO
CANDIDATE’S RECEIPT
INSTRUCTIONS TO CANDIDATE:
1.
Fill in all the information requested clearly in capital letters.
TEST CODE:
0
2
1
0
7
0
2
0
SUBJECT: BIOLOGY − UNIT 1 − Paper 02
PROFICIENCY:
ADVANCED
REGISTRATION NUMBER:
FULL NAME: ________________________________________________________________
(BLOCK LETTERS)
Signature: ____________________________________________________________________
Date: ________________________________________________________________________
2.
Ensure that this slip is detached by the Supervisor or Invigilator and given to you when you
hand in this booklet.
3.
Keep it in a safe place until you have received your results.
INSTRUCTION TO SUPERVISOR/INVIGILATOR:
Sign the declaration below, detach this slip and hand it to the candidate as his/her receipt for this booklet
collected by you.
I hereby acknowledge receipt of the candidate’s booklet for the examination stated above.
Signature: _____________________________
Supervisor/Invigilator
Date: _________________________________
BIOLOGY CAPE UNIT 1 JULY 2021 SOLUTIONS
QUESTION ONE
8 marks –





Any FOUR annotations above (other possible annotations: mesosome, slime capsule,
plasmid, cilia)
Drawing occupies at least ½ of box
Drawing has clean continuous lines, no shading.
Title is included, appropriate.
Magnification is included, accurately calculated.
TEST CODE 02107020
2022
MAY/JUNE
FORM TP 2022171
CARIBBEAN
EXAMINATIONS
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION@
BIOLOGY
UNIT 1 - Paper 02
2 hours 30 minutes
READ THE FOLLOWING INSTRUCTIONS CAREFULLY.
1.
This paper consists of THREE questions. Answer ALL questions.
2.
Write your answers in the spaces provided in this booklet.
3.
Do NOT write in the margins.
4.
You may use a silent, non-programmable calculator to answer questions.
5.
You are advised to take some time to read through the paper and plan your answers.
6.
If you need to rewrite any answer and there is not enough space to do so on the
original page, you must use the extra lined page(s) provided at the back of this
booklet. Remember to draw a line through your original answer.
7.
If you use the extra page(s) you MUST write the question number clearly in
the box provided at the top of the extra page(s) and, where relevant, include
the question part beside the answer.
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
Copyright 0 2020 Caribbean Examinations Council
All rights reserved.
02107020/MJ/CAPE 2022
0210702003
I Ill IIIIl Ill IlIll IllIl Ill Il Ill Il Il Ill IllIl Ill Il Il Ill III Ill I
Module 1
(a)
Cell and Molecular Biology
Figure I shows a light Illicrograph of a cross section through a dicot
root.
geo-e.
Figure 1. Diagram showing vascular arrangement of dicot root
Source: https://www.sciencephoto.com/media/78322()/view
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02107020/MJ/CAPE 2022
0210702004
I I I I I Ill II Il I
I IlI Ill II IlI IIlIlII I Il IIlI I IIII
-5In the space provided, construct a plan diagram (o show at least the FOUR major
tissues labelled W, X, Y onclZ in the light nnicrogt•aphin Figure l,
Similar to the above, but the stele would be smaller
and the following labels would be included:
W - Cortex, X - Phloem, Y - Xylem, Z - Epidermis
Include the title: Plan diagram of a TS of dicot root
Include a magnification of the drawing compared
to the image micrograph.
[5 marksl
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02107020/MJ/CAPE 2022
0210702005
o.
(ii)
state the function of
the structures
Y and
labelled W, X,
Z in Figure I on
Cortex - Moves water to the centre of the root.
Phloem - Transport (translocation) of sucrose
Xylem - Transport of water and dissolved mineral salts to the leaves
Either: (Intracellular) Air space - Contains oxygen for aerobic respiration.
Epidermis - Protects the root; contains root hairs for water absorption
[4 marks]
(iii)
root of a plant is
Explain why the dicotyledenous
considered an organ.
An organ consists of numerous tissues, which are specialized to play
certain functional roles. The dicot root of a plant contains several tissues
such as phloem sieve tube elements, epidermal cells, endodermal cells,
cambium, xylem vessels and root hair cells. Together, these allow the
root to perform its role of absorption and transport of water, mineral ions
and sugars.
[3 marks]
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02107020/MJ/CAPE 2022
0210? 02006
(b)
(i)
Outline the basis of the enclosyjnbiont theory.
The theory states that mitochondria and chloroplasts arose as early
prokaryotic organisms, which were absorbed into a larger host prokaryote,
thus giving the host the ability to absorb oxygen for respiration and absorb
light energy for photosynthesis/sugar production.
12marks]
(ii)
Complete the table below by indicating whether the following features are
present, absent or sometimes present in a eukaryotic cell and a prokaryotic cell.
Eukaryotic Cell
Prokaryotic Cell
Plasmid
Sometimes present*
Present
Mitochondria
Sometimes present**
Absent
Nuclear envelope
Sometimes present**
Absent
Ribosome
Present
Present
Feature
[4 marks]
*All mitochondria and chloroplasts have 'plasmid-like' structures
so perhaps 'Present' can work for that category under Eukaryotic?
** Red blood cells don't have mitochondria and nuclei.
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02107020/MJ/CAPE 2022
IllIl
IlIll IllI
0210702007
IIllIll
(e)
he
in
protein
the
out
that
pH,
such
by
protein's structure. Enzymes
on the
dcpcndcnt
body is
Figure 2 shows how enzyme actionis
within cell,
substrate concentration.
concentration and
Tcmpcrnture
8
50
10 20 30 40
Temperature CC)
10 12 14
60
70
Substrate concentration
Enzyme concentration
3 x substrate
2x enzyme
2 x substrate
Ix enzyme
1 x substrate
No enzyme
8
10
10 12 14
12 14
Time
Time
Figure 2. Effectof varying factors on enzyme activity
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02107020/MJ/CAPE 2022
IIlI I II II IlII I I
0 210? oeooe
I IlIIIlI I IlI IlI I I III IlI I IlI IlII III I Il Il II
Explain the innpact ol' the 10110wingfactors on enzyme action using the information
provided in Figure 2.
pH
In this particular enzyme, the rate of enzyme action is the highest around a pH of 12,
which is alkaline. The enzyme is completely denatured at a pH of 8, and the rate of
enzyme action declines as pH is increase from 12 to 14.
This is due to pH affecting the formation and strength of ionic bonds holding the
tertiary 3D structure of the enzyme together. If these ionic bonds were to break
during denaturation, the active site would change shape and enzyme function would
be affected.
Temperature
The optimum temperature of the enzyme is approximately 27 degrees Celsius.
The rate decreases as temperature is increased and total denaturation occurs at
40 degrees Celsius. At this point, bonds such as hydrogen and ionic bonds would
have been broken between the secondary and tertiary structures of the protein. This
would alter to shape of the active site and make it lose its specific shape to allow
binding to the substrate.
At temperatures below 27 degrees Celsius, there is a decreased amount of kinetic
energy in both the enzyme and substrate molecules, decreasing the rate of reaction
between substrate and enzyme.
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02107020/MJ/CAPE 2022
o z 10702009
Enzynne concentration
From the graphs, it can be observed that the rate of product formation is doubled
as enzyme concentration is also doubled, observed by the steeper gradient of
the graph. This is due to an increased likelihood of enzyme availability for substrate
binding, thus more substrate can be converted to product if more enzymes are
present.
It can also be seen that product would be formed if no enzyme is present, but at
highly decreased rates. This is due to the activation energy for the reaction being
too high, thus increasing the times of conversions of substrate to product.
Substrate concentration
As substrate concentration increases, so does enzyme activity for the first 6 units
of time. This is because substrates are highly available for product conversion as
they interact with the enzymes. However, the rate of reaction decreases as time passes
due to enzymes becoming occupied with substrate and less being available.
As time passes also, there would be less and less substrate as more and more
product is formed. The plateau for each curve indicates that all of the substrate has
been converted to product, with 3x substrate having approximately 3x more product
than 1x substrate.
112marks]
Total 30 marks
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02107020/MJ/CAPE 2022
0210?02010
-12Module 2 — Genetics, Variation and Natural Selection
2.
(a)
The cell cycle consists of two phases, interphase and mitosis. Figure 3 shows two
of the phases in mitosis. In the appropriate spaces provided in Figure 3, make
detailed sketches to illustrate the TWO key stages of mitosis in the cell that are
NOT shown.
Stage I (Prophase)
stage 11
stage 111
stage IV
Figure 3. Diagram showing
(ii)
State the name of the missing
Stage Il
the stages of mitosis
[2 marks]
stages.
Metaphase
Stage Ill . Anaphase
Stage IV
Telophase
13marks)
02 J07020/MJ/CAPE 2022
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0210? 02012
NEXT PAGE
-13(iii)
Explain how the process of' l)iitosis ensures that cach daughter nucleus receives a
Cullset ol' clu•otnosonocs.
During interphase, which occurs before mitosis, growth of the cell occurs
as well as DNA replication, causing the amount of genetic material to be
doubled. Sister chromatids are formed during prophase, which are then
aligned in the equator during metaphase and separated during anaphase.
After telophase and cytokinesis occurs, each daughter cell has its own
sister chromatids before DNA replication and mitosis can occur once again.
[3 marks]
(b)
(i)
Define the term 'natural selection'.
The adaptation of organisms to their environments that allow them to overcome
selective pressures, survive and reproduce, passing on their advantageous
traits.
[2 marks]
List THREE of Darwin's observations concerning natural selection.
1. Members of a species vary from each other, and these varied traits can
be inherited.
2. All organisms produce a excess offspring (there is a struggle for existence).
3. Population numbers remain fairly constant over long periods of time.
[3 marksl
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02107020/MJ/CAPE2022
0210702013
(c)
In 1970, biologist John Encllet•did
the hypothesis that male
expcrinnentsto test
different Streams
from
selection. Guppies
natuctll
through
spots per male
evolves
coloured
colourntion in guppies
number of
The
pond.
at•tificinl
placed together in one
population in the pond was able to
The
1(),
was
recorded at the start of the experitnent
reproductive cycles.
reproduce several tinnesas guppies have nujny
divided
Aller six tnonths, the original population was
and placed into THREE separate
the bodies of a sample ofmales
of coloured spots on
ponds, A, B and C, The mean
predator that docs not prey on guppies
a
of
individuals
Six
pond.
was recorded for each
was added to Pond C'.
guppies
of
predator
wete added to Pond B. A voracious
recorded
of a sample of males was
The nnean nutnber of coloured spots on the bodies after 2() months (14 months after
and
after Il months (five months after the separation)
number of spots per male recorded
mean
the separation). Table I shows the results of the
for the three ponds during the experiments.
ON A MALE GUPPY
TABLE 1: MEAN NUMBER OF SPOTS
IN THREE DIFFERENT PONDS
Mean Number of Colored Spots
Time (Months)
per Male
Pond A
Pond B
Pond C
6
11.8
11.8
11.8
11
12
13.8
10.5
20
13
13
9.5
mod/oucontent/
Source: Data adaptedfrom hitps://www.open.edu/openlearn/ocw
view.p?printable=l&id=2128
(i)
On the grid provided on page 15, plot a line graph to represent the data in Tablel
[5 marks]
for Pond A, Pond B and Pond C.
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02107020/MJ/CAPE2022
0210702014
-15-
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02107020/MJ/CAPE 2022
0210702015
-16(ii)
Using the infornmtion given on page 14, discuss how the results of John Endler's
experitnent, as it relates to variation in the number of spots in male guppies, may
support TI-IREEof Darwin's deductions regarding natural selection.
In Ponds A and B, where there were no predators that posed any threats to
guppies, the mean number of coloured spots on the male guppies increased.
This showed that in this environment, perhaps the males with coloured spots
were more successful at attracting females. As a result, these males would
have reproduced and produced offspring that would have inherited the trait
for a high number of coloured spots. Due to genetic variation, some of these
individuals would have more coloured spots than their fathers, thus making
them more attractive to females as well.
However, in pond C, where there was a voracious predator, a high number of
coloured spots were seen as a disadvantage. This could have been due to
increased visibility to the predator, causing them to be more easily spotted
and predated upon. This would have reduced the population numbers of
male guppies with a high number of coloured spots, thus favouring guppies
with a lower number of spots. These guppies with a low amount of spots
would have been best adapted in this particular environment. They would have
survived then reproduced and passed on this advantageous trait to
their offspring, with the number of coloured spots being reduced each
generation due to overproduction of offspring and genetic variation.
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02107020/MJ/CAPE2022
o z 10 toe 0 16
-18Reproductive Biology
Module 3
3.
(a)
the ideal glucose concentration for
A student sets up an expet•ilnentto determine
germination of pollen grains of impatiens.
percentage
the
tneasut•ing
by
gernlination
set-up. Each petri dish contains
Figure 4 shows an illustration of the experimental
of a glucose solution.
different numbers of pollen grains with varying
0.002%
0% glucose
concentrations
0.2% glucose
0.02% glucose
glucose
Figure 4. Illustration of experiment set-up
State a suitable hypothesis for this experiment.
Higher concentrations of glucose result in higher rates of germination
for pollen grains of impatiens.
12marks]
(ii)
State ONE possible source of error in the experiment.
The spacing between the pollen grains in the solution may be uneven,
thus may affect available glucose for uptake.
[1 mark]
(iii)
State ONE change in the experiment that could have overcome the error stated in
(a) (ii).
Each petri-dish can have their pollen grains positioned in the same manner,
evenly spaced between each other.
[1 mark]
(iv)
Identify ONE responding variable in the experiment.
Number of germinated pollen grains / pollen grains that spouted pollen tubes.
Il markl
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02107020/MJ/CAPE 2022
o z 10?02010
-19 _
(b)
State the function of the following structures in the anther of a stamen.
(i)
Epidermis
Protects the internal components, such as the pollen sacs.
Il mark]
Tapetum
Provides nutrition for developing pollen grains.
Il mark]
(iii)
Pollen mother cells
Divides by meiosis to form haploid gamete nuclei.
[1 mark]
(iv)
Starting from the pollen mother cell, outline the steps in pollen grain formation.
The pollen mother cell undergoes meiosis to form four haploid microspores.
These then separate and undergo mitosis to form a pollen tube nucleus and
generative nucleus. The pollen grain also has two layers, an external exine
and an internal intine.
[3 marksl
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02107020/MJ/CAPE 2022
0210? 02019
-20(v)
Explain the significance of double fertilizationin the
enlbryo sac of a flower.
The haploid egg cell is fertilized to become a zygote and the diploid polar nuclei
fuse with the male gamete to form a triploid (3n) endosperm nucleus. The
endosperm is used as a food storage for the growth and development of
the plant embryo and seed post-fertilization.
[3 marks]
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02107020/MJ/CAPE2022
o z 10 roe ozo
.1,
-21(c)
Figure 5 is a now chart
which shows the relationship among the hormones GnRH, LH,
FSI-I,testosterone and
inhibin in the process of spermatogenesis in the human male.
Ilypothnlamus
GnR11
Anterior lobc
of pituitary gland
FSH
Sertoli cells of
testis
Inhibin
Leydig cells of
testis
%%••Testosterone
Spermatogenesis
Figure 5. Flow chart of hormonal control of spermatogenesis
Using the information given in the flow chart, outline how the body controls
spermatogenesis in the human male.
GnRH from the hypothalamus stimulates production of LH and FSH from the anterior
pituitary gland. LH binds to receptors on the LEYDIG cells in the TESTES to secrete
TESTOSTERONE and begin spermatogenesis. FSH binds to the SERTOLI cells, making them more
receptive to testosterone. There is a NEGATIVE FEEDBACK system involved to regulate
testosterone concentration in the blood. Sertoli cells also release another hormone known as
INHIBIN that inhibits release of FSH from the pituitary gland. Testosterone also limits the release of
LH and GnRH.
Therefore, as testosterone and inhibin levels increase, FSH, GnRH and LH levels decrease,
thus also decreasing testosterone and inhibin levels. However, when testosterone levels
and inhibin levels have decreased, FSH, GnRH and LH levels once again increase, thus
raising testosterone and inhibin levels.
14marksl
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02107020/MJ/CAPE2022
0210702021
-22 (d)
There is increasing evidence that n mother's behaviour during pregnancy affects the
developtnent oc the Coetusand can cause long-term health consequences for the child.
Figure 6 shows the predicted effects of Inaternal smoking on birth weight based on a
population study in British Colonlbia between 2001 and 2006.
3500
3450
3400
3350
3300
3250
.E 3200
3150
3100
o
2
4
6
8 10 12
Cigarettes per day
14
16
Figure 6. Graph showing predicted effects of maternal smoking during
pregnancy on foetal birth weight
Adaptedfrom Anders Erickson, Aleck Ostry, Laurie Chan and Laura Arbour. (2016).
"Air Pollution, Neighbourhood and Maternal-level Factors Modify
the Effect of Smoking on Birth Weight:A MultilevelAnalysis in British Columbia, Canada.
BMCPublic Health. 16. 10.1186/s12889-016-3273-9.
(i)
Using the information from Figure 6, discuss the effects of maternal smoking on
foetal development. You must refer to Figure 6 in your discussion.
The overall trend observed is that birth weight decreases as the number of
cigarettes per day increases. From 0-2 cigarettes/day, there is a decrease from
3450g to 3490g (-60g). From 2-4, the decrease is 3390g to 3355g (-45g).
The rate of reduction in birth weight further decreases as number of cigarettes
increase past this point, with the graph almost reaching a plateau between
14 and 16 cigarettes/day. This is due to constriction of the placental arteries by
the action of nicotine, limiting the nutrient uptake such as amino acids
and calcium by the foetus, thus limiting growth and bone formation. The carbon
monoxide and tar from the cigarettes also limit oxygen intake by the mother
and thus, reduces rate of respiration in the foetus, delaying tissue
[4 marks]
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growth.
02107020/MJ/CAPE2022
0210702022
-23(ii)
A pregnant wotnan visits an antenatal
clinic and is advised by the physician that
as a result of her lifestyle the baby is
at high risk of having a low birth weight.
The physician strongly recommends that she makes lifestyle changes that can
prevent this outcome.
Discuss how TWO maternal behaviours, other than smoking may contribute to
low birth weight. You must include ONE lifestyle change that the physician may
recommend for EACH maternal behaviour discussed.
1. Consumption of alcohol may contribute to underdeveloped limbs, reduced
muscle tone and foetal alcohol syndrome. This is because alcohol is a depressant,
which reduces nervous transmission. These alcohol molecules can enter the foetus'
bloodstream and slow tissue growth and development. A physician would suggest
that the mother completely abstain from alcohol.
2. If a mother is malnourished, the foetus would not obtain the balance of
nutrients required for bone and organ formation and growth. A physician would
recommend that a mother consume a diet rich in calcium, iron, lipids and folic
acid. Respectively, these contribute to the development of bones, haemoglobin,
phospholipid bilayers (cell membranes) and the foetus' neural tube.
Other answers may include effects of exercise (to increase blood circulation to foetus),
and addictive drug intake (which impact foetus' tissue development).
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02107020/MJ/CAPE2022
0210?02023
Answers by Punished Sperwin
Document compiled by Orlando Ramkissoon