SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP)
GRADE 12
SUBJECT:
LIFE SCIENCES
TEACHER’S BOOKLET 2023
(With solutions)
SESSIONS 1-8
1
(Page 1 of 43)
TABLE OF CONTENT
A
Session 1: Skills needed in Life Sciences
3
B
Session 2: DNA: location, structure and
function
9
C
Session 3: DNA replication and DNA profiling
12
D
Session 4: RNA: location, structure and
function.
17
Protein synthesis
E
Session 5: Meiosis
22
F
Session 6: Diversity of reproductive
strategies & Human reproduction
29
G
Session 7: Gametogenesis & Menstrual cycle
34
H
Session 8: Fertilization and implantation
38
2
SESSION 1: SKILLS NEEDED IN LIFE SCIENCES
SECTION A
EXAM GUIDELINE
Orientation to Life Sciences:
 How science work.
Science is based on: • fundamental knowledge built on scientific evidence and
verified findings (articles that are published in journals or at conferences: peer
review); • observing; • investigating; • making measurements and understanding the
importance of scaling; • collecting and presenting data in the form of drawings,
written descriptions, tables and graphs; • understanding the limitations of scientific
evidence; • identifying patterns and relationships in data; • communicating findings;
and • taking societal aspects of scientific evidence into account.
Scientific skills involve: • importance of biological principles such as relationship
between surface area and volume/size, the relationship between structure and
function • biological drawings: principles that apply • translating 3 dimensional
objects or specimens into 2 dimensional drawings and photographs and interpreting
2 dimensional drawings and photographs: transverse and longitudinal sections •
introduction to graphs: different kinds of graphs and when to use them; interpreting
graphs. • calculating
(CAPS, GR.10)
NOTES & TIPS
It is important that Life Sciences learners have the following in their pencil cases:







Ruler
Blue ink pen
Pencil
Eraser
Protractor
Compass
Non-programmable calculator
Skills are tested in EVERY paper and the questions can be set on any topic. These
are easy marks to collect if you can use and apply the skills of Life Sciences.
3
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION: 1
STEP 1:
Variables:
Independent: Position of the lamp
Dependent: Diameter of pupil
STEP 2 & 3
Diameter of the pupil (mm)
4.8
4.0
3.2
2.4
1.6
0.8
1
2
3
4
5
6
7
Distance of the lamp (m)
STEP 4, 5 & 6
A line graph indicating the diameter of the pupil of a person if a lamp is
placed at different distances from the person.
Diameter of the pupil (mm)
4.8
4.0
3.2
2.4
1.6
0.8
4
1
2
3
4
5
6
Distance of the lamp (m)
7
EXERCISE: 2
QUESTION: 2
Calculations:
Transport
= 25/100 x 360° = 90°
Residential
= 27/100 x 360° = 97.2° ≈ 97o
Industrial
= 15/100 x 360°= 54°
Commercial = 28/100 x 360° = 100.8° ≈ 101o
Other
= 5/100 x 360° = 18°
Carbon dioxide emissions in different sectors of a city
Other
5%
Residential
27%
Transport
25%
Commercial
28%
Industrial
15%
Discuss these
assessment
criteria with the
learners
Mark allocation for the pie chart
Correct type of graph
1
Title of graph (CO2 emission + Sector)
1
5
Calculations:
1: 1-4 calculations correct
2: All 5 calculations correct
Correct proportion for each labelled slice
1:
1 to 4 slices correct
2:
All 5 slices correct
(6)
QUESTION: 3
Number of deaths (%)
1.
20
18
16
14
12
10
8
6
4
2
0
Bar graph showing four top causes of death of children
under the age of one year in 2008
Diarrhoea
HIV/AIDS
Pneumonia
TB
Causes of death
2. a. Number of deaths in percentage.
b. Causes of death
QUESTION: 4
Histogram showing the number of decomposing bacteria found in soil with different
pH levels
6
No. of decomposing bacteria
QUESTION: 5
1.
Percentage decrease: 32-45 x100
45
= 28.88%
2. a)
P=FXS
M
P = Estimated total number of individuals in the population.
F = Number caught and marked in the first catch.
S = Number caught in the second catch.
M = Number marked in the second catch
F= 32
S = 26
M=4
P=FXS
M
= 32 x 26
4
= 208
b) Percentage marked sea snails in Pool B in second sample:
6 x 100
21
=28.57%
c)
Average sea snails caught in both pools during first sample:
32+42
2
= 37
7
3.
1cm = 10000 µm. 12cm = 120000 µm 1.2 cm = 12000 µm.
Actual size of cell =measured length of cell x number on scale
measured length of scale
=
12 0000 x 2
12000
= 20 µm.
SECTION C
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION: 6
1.
1.1
1.2
2.
2.1
2.2.
More/fewer dark peppered moths/ pale peppered moths survive
in the polluted /unpolluted environment than in the unpolluted /
polluted environment
OR
No difference in the number of dark/ pale peppered moths that
survive in both environments
Max
(3)
- Was not a closed system so moths could fly in and out of the
environment/migration may have taken place
- The number of predators might have been different in both
polluted and unpolluted environment
- Both environments could have been different with regard to
vegetation found in them
- Both environments could have been different with regard to
climatic conditions 
- Human error in sampling/counting/recording/no repeats
Any 3
(3)
An increase/decrease in the concentration of sodium disulphate
will result in an increase/decrease in the percentage germination
of oats seeds 
OR
An increase/decrease in the concentration of sodium disulphate
will have no effect on the percentage germination of oats seeds
(2)
Concentration of sodium disulphate(IV) 
(1)
(1)
(2)
8
2.3
Temperature
Water
(2)
2.4
Average estimateincreases reliability 
(2)
2.5
Increasing concentrations of sodium disulphate decreased the
germination of oats seeds
(2)
(9)
Session 2: DNA: location, structure and function
SECTION A
EXAM GUIDELINE

Revision of the structure of the cell with an emphasis on the ribosome,
cytoplasm and the parts of the nucleus (Gr.10 content)

Location of DNA:

Makes up the genes on chromosomes (nuclear DNA)

Present in mitochondria (mitochondrial DNA)

Present in chloroplasts (plants)

Brief history of the discovery of the DNA molecule (Watson & Crick, Franklin &
Wilkins)

Structure of DNA
 The natural shape of the DNA molecule is a double helix
 Each strand of the helix is made up of a sequence of DNA nucleotides

Three components of a DNA nucleotide:

Nitrogenous bases linked by weak hydrogen bonds:
o Four nitrogenous bases of DNA: adenine (A), thymine (T), cytosine (C),
guanine (G)
9


o Pairing of bases in DNA occur as follows: A: T and G: C
Sugar portion (deoxyribose in DNA)
Phosphate portion

Stick diagram of DNA molecule to illustrate its structure

Functions of DNA:

Sections of DNA-forming genes carry hereditary information

DNA contains coded information for protein synthesis
NOTES & TIPS
This topic is an important building block for the understanding of genetics and
inheritance as well as evolution. It is important that learners understand this section
well so that they could link it with the other topics that will be dealt with in the rest of
the year.
The following points are important to remember:





The structures of DNA and RNA is NB
The differences between DNA, mRNA and tRNA are important.
DNA profiling is important for application questions.
DNA replication and transcription are confused - ensure differentiated
understanding.
Transcription and translation must be learnt by heart.
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
1.2
1.3
1.4
1.5
B
D
C
B
B





(10)
QUESTION 2
10
2.1
Gene 
2.2
Deoxyribose 
2.3
Hydrogen bond 
2.4
Hydrogen bond 
(4)
QUESTION 3
3.1
None 
3.2
B only 
3.3
Both A and B 
(6)
QUESTION 4
4.1
4.2
B – Nitrogenous base 
C – Phosphate 
(2)
(a) Nucleotide 
(b) (James) Watson 
(Francis) Crick 
(Maurice) Wilkins 
(Mark first TWO only)
(a) Nucleus /Mitochondrion/Chloroplast
(Mark first ONE only)
(1)
(2)
(1)
(6)
SECTION C
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION 5
Location
- The DNA is located in the nucleus
11
- and mitochondria and
- chloroplasts
Structure
- DNA is a double strandedmolecule that
- forms a helix
- It is made up of nucleotides
- Each nucleotide has a deoxyribose sugarmolecule
- a phosphate group and
- a nitrogenous base
- The bases are A, T, C and G
- which join to form complementary pairs/ (A to T and C to G)
- held by hydrogen bonds
(Any 2)
(2)
(Any 7)
(7)
(9)
Session 3: DNA replication and DNA profiling
SECTION A
EXAM GUIDELINE

Process of DNA replication:

When in the cell cycle it takes place

Where in the cell it takes place

How DNA replication takes place (names of enzymes not required)

The significance of DNA replication

Interpretation of DNA profiles

Uses of DNA profiles
NOTES & TIPS

The term ‘DNA fingerprinting’ is NO LONGER accepted. We only use the
term ‘DNA Profiling’.
12


In paternity testing by the means of a DNA profile, ALL the bands of the
DNA profile of the child must match some of the mother’s bands and
some of the father’s bands from their DNA profiles.
At a crime scene, the DNA profile from the DNA found, must completely
match that of the DNA profile taken from a DNA sample of a suspect to
make the suspect the perpetrator.
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
1.2
1.3
1.4
1.5
A
C
B
C
B





(10)
QUESTION 2
2.1
Replication 
2.2
Template 
2.3
Complementary strand 
2.4
DNA profile 
2.5
Interphase 
(5)
QUESTION 3
3.1
A only 
3.2
B only 
3.3
Both A and B 
(6)
QUESTION 4
4.1
DNA profiling
(1)
13
4.2
Male 3
4.3
-
The bands of the child’s DNA are a combination of the DNA from
each parent 
Three bands are identical to that of the mother
- The remaining three bands correspond with that of male 3
(3)
To investigate crimes/resolve disputes
To identify organisms from their remains
To identify family relationships other than paternity e.g. siblings
or cousins
- To test for the presence of specific alleles/genes that cause a
genetic disorder
- To establish matching tissues for organ transplants
Any
(Mark first TWO only)
(2)
(7)
-
4.4
(1)
-
QUESTION 5
5.1
DNA replication 
(1)
5.2
Nucleotide
(1)
5.3
1 – guanine
2 – cytosine
(2)
5.4
Double helix
(1)
5.5
Hydrogen bond
(1)
(6)
QUESTION 6
6.1
Weak hydrogen bond
6.2
Adenine with thymine/ A-T 
(1)
Guanine with Cytosine/ C-G 
6.3
(2)
(a)





DNA unwinds 
and unzips /weak hydrogen bonds break
separating into two strands 
each strand receives
free-floating DNA nucleotides 
14
 from the nucleoplasm 
 forming
complementary
strands
thymine/guanine with cytosine
 this process is enzyme controlled
/adenine
with
(Any 4)
(4)
(b) interphase 
(1)
(c)
-
doubles the genetic material so it can be shared between the
resulting daughter cells during division 
- Results in the formation of identical daughter cells during
mitosis 
(Mark first ONE only)
(1)
(9)
QUESTION 7
7.1
DNA profiling 
(1)
7.2
Jennie 
(1)
7.3
- All of the bands on her DNA profile 
- match the bands from the DNA sample  from the crime scene
(2)
7.4
- Proof of paternity 
- Tracing missing persons 
- Identification of genetic disorders 
- Establishing family relations 
- Matching tissues for organ transplants 
- Identifying dead persons 
Any
(1)
(5)
SECTION C
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION 8
15
- A child received DNA from both parents 
- The DNA profiles of the mother, child and the possible father are
determined 
- A comparison of the DNA bands of the mother and the child is made 
- The remaining DNA bands are compared to the possible father’s DNA
bands 
- If all the remaining DNA bands in the child’s profile match the possible
father’s DNA bands 
- Then the possible father is the biological father 
- If all the remaining DNA bands in the child’s profile does not match the
possible father’s DNA bands 
- Then the possible father is not the biological father 
Any
(6)
QUESTION 9
9.1
9.2
Heila and Leo
(Mark first TWO only)
(2)
- All of the DNA bands from Heila and Leo
- match with the DNA bands of the mother and the father
OR
- None of the DNA bands from Priya
- match with the DNA bands of the mother and the father
9.3
- Tracing missing persons
- Identification of genetic disorders
- Identification of suspects in a crime
- Matching tissues for organ transplants
- Identifying dead persons
(Mark first THREE only)
(2)
Any (3)
(7)
16
Session 4: RNA: location, structure and function.
Protein synthesis
SECTION A
EXAM GUIDELINE
RNA: location, structure and function

Location of RNA:
 mRNA is formed in the nucleus and functions on the ribosome
 tRNA is located in the cytoplasm

Structure of RNA
 A single-stranded molecule consisting of nucleotides
17

Three components of an RNA nucleotide:
 Nitrogenous bases
- Four nitrogenous bases of RNA:
- adenine (A), uracil (U), cytosine (C), guanine (G)
 Sugar portion (ribose in RNA)
 Phosphate portion

Stick diagram of mRNA and tRNA molecules to illustrate their structure
Function of RNA:
 RNA plays a role in protein synthesis

Protein synthesis

The involvement of DNA and RNA in protein synthesis:

Transcription
o The double helix DNA unwinds.
o The double-stranded DNA unzips/weak hydrogen bonds break
o to form two separate strands.
o One strand is used as a template
o to form mRNA
o using free RNA nucleotides from the nucleoplasm.
o The mRNA is complementary to the DNA.
o mRNA now has the coded message for protein synthesis.

mRNA moves from the nucleus to the cytoplasm and attaches to the
ribosome.

Translation
o Each tRNA carries a specific amino acid.
o When the anticodon on the tRNA
o matches the codon on the mRNA
o then tRNA brings the required amino acid to the ribosome.
(Names of specific codons, anticodons and their amino acids are not to be
memorised.)
o Amino acids become attached to each other by peptide bonds
o to form the required protein.

Simple diagram to illustrate transcription and translation in protein synthesis
NOTES & TIPS
18
This topic is an important building block for the understanding of genetics and inheritance as
well as evolution. It is important that learners understand this section well so that they could
link it with the other topics that will be dealt with later.
When we talk about DNA profiling, we no longer refer to the pattern of bars as a DNA
fingerprint.
When working out the possible father in paternity testing, you MUST compare DNA profiles
of the mother, child and possible father using the following steps:
1. Compare the mother’s profile with the child’s profile.
2. Cancel out all the bands in the child’s profile that match the mother’s profile.
3. Look at the bands that are left and compare these to the possible father’s profile.
4. If these bands ALL match bands in the possible father’s profile, then he is likely the
father of the child.
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
C

19
1.2
1.3
1.4
1.5
C
C
D
C




(10)
QUESTION 2
2.1
Transcription 
2.2
Uracil
2.3
Peptide bond 
2.4
Translation 
2.5
Amino acid 
2.6
Ribosome 
(6)
QUESTION 3
3.1
B only 
3.2
A only 
3.3
Both A and B 
(6)
QUESTION 4
4.1
4.2
(a) tRNA 
(1)
(b) Anticodon 
(1)
(a) UGG  (in correct order)
(2)
(b)TGG  (in correct order)
(2)
(6)
QUESTION 5
- The double helix DNA unwinds  and
- unzips /weak hydrogen bonds break
- to form two separate strands 
- One starnd is used as atemplate 
20
-
using free RNA nucleotides from the nucleoplasm 
The mRNA is complementary to the DNA 
The coded message for protein synthesis is thus copied onto mRNA 
Any 6
(6)
QUESTION 6
Structure
 RNA is a single stranded molecule
 made up of nucleotides
 Each nucleotide is made up of a ribose sugar
 a phosphate group and
 four different nitrogenous bases
 adenine, uracil, guanine and cytosine/ (A, U, G and C)
 The phosphate group alternates with the ribose sugar
 and the nitrogenous base is attached to the ribose sugar
 mRNA is made up of many bases
 arranged in triplets called codons
 Bases in tRNA are arranged in triplets called anticodons
 tRNA has a place of attachment for an amino acid
Any 10
(10)
SECTION C
HOMEWORK QUESTIONS
QUESTION 7
7.1 Nucleus
(1)
7.2 (a) Deoxyribose
(b) Uracil/U 
(1)
(1)
7.3
(Mark first TWO only)
1 mark for table + (Any 2 x 2)
(5)
(8)
21
QUESTION 8
8.1
Gene mutation
8.2
8.3
(1)
There is a change in the (sequence of) nitrogenous bases /
nucleotides
(a) 5/Five
(b) UAU
(c) - The codon CCG in the original sequence
- changed to CUGin the mutated sequence
- The amino acid proline was replaced by leucine
- This resulted in a different protein/no protein being
formed
(1)
(1)
(1)
(4)
(8)
SESSION 5: MEIOSIS
SECTION A
EXAM GUIDELINE
Meiosis introduction
Revision of the structure of a cell, with an emphasis on the parts of the
nucleus, the centrosome and the cytoplasm
Structure of chromosomes:
•
Chromosomes consist of DNA (which makes up genes) and protein
•
The number of chromosomes in a cell is a characteristic of an organism (e.g.,
humans have 46 chromosomes)
22
•
Chromosomes which are single threads become double (two chromatids
joined by a centromere) as a result of DNA replication
Differentiate between:
•
Haploid (n) and diploid (2n) cells in terms of chromosome number
•
Sex cells (gametes) and somatic cells (body cells)
•
Sex chromosomes (gonosomes) and autosomes
Revision of the process of mitosis
Meiosis the process
Definition of meiosis
Site of meiosis in plants and in animals
Meiosis is a continuous process, but the events are divided into different
phases for convenience
Events of interphase:
•
DNA replication takes place
o
Chromosomes which are single threads, become double
o
Each chromosome will now consist of two chromatids joined by a centromere
o
DNA replication helps to double the genetic material so that it can be shared
by the new cells arising from cell division
The events of the following phases of Meiosis I, using diagrams:
•
Prophase I - Including a description of crossing over
•
Metaphase I – including the random arrangement of chromosomes
•
Anaphase I
•
Telophase I
The events of each phase of Meiosis II, using diagrams:
•
Prophase II
•
Metaphase II – including the random arrangement of chromosomes
•
Anaphase II
•
Telophase II
23
The importance of meiosis:
•
Production of haploid gametes
•
The halving effect of meiosis overcomes the doubling effect of fertilisation,
thus maintaining a constant chromosome number from one generation to the
next
•
Mechanism to introduce genetic variation through:
o
Crossing over
o
The random arrangement of chromosomes at the equator
Abnormal meiosis
Non-disjunction and its consequences
Non-disjunction of chromosomes at position 21 during Anaphase in humans to
form abnormal gametes with an extra copy of chromosome 21
gamete (23 chromosomes) may lead to Down syndrome
Comparison of meiosis and mitosis
Similarities of mitosis and meiosis
Differences between mitosis and meiosis
NOTES & EXAM TIPS








Remember IPMAT to describe the phases of meiosis
They usually give diagrams in the question papers and ask questions.
Learners must identify the correct event in a phase.
Learners must know the differences between meiosis I and meiosis II
For example: Look at metaphase 1 and 2. What are the differences?
Metaphase 1 has a pair of homologous chromosomes at the equator of the cell
while Metaphase 2 have single chromosomes at the equator. Anaphase 1 has
chromosomes moving to the poles while anaphase 2 has chromatids moving to
the poles.
If they ask you to identify the phase you must give the full description for
example anaphase 1 not only anaphase, you must write the phase as well.
Learners must know the difference between a centrosome, centromere and a
centriole
Centrosome: Structure that is responsible for the formation of spindle fibres
during cell division in animal cells. Composed of centrioles.
24



Centromere: Structure that holds two chromatids together in a replicated
chromosome and which also attaches the chromosome to the spindle thread
during cell division.
Avoid using the term bivalent for homologous chromosomes.
The definition of cytokinesis is a popular question.
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
1.2
1.3
B
D
B
(6)
QUESTION 2
2.1
Spindle fibres
2.2
Cytokinesis
2.3
Interphase
2.4
Down syndrome
2.5
Chiasma
2.6
Spindle fibre
2.7
Non- dysjunction
(7)
QUESTION 3
25
3.1
B
3.2
A
3.3
A
3.4
Both A and B
3.5
B
(10)
QUESTION 4
4.1
Prophase 1
(1)
Anaphase 1
(1)
4.2
(5)
(7)
Criteria for marking
Only two cells have been drawn
Each cell contains only two unreplicated
chromosomes
Each chromosome is the corret size and
correctly shaded
Any TWO correct labels
1 mark
1mark
1 mark
2 marks
QUESTION 5
26
5.1
D- Chromatid
E- Centromere
5.2
23
5.3
(a) E
(b) C/B
(5)
QUESTION 6
6.1
6.2
6.3
6.4
6.5
23
a) Centromeer
b) Chiasma/chiasmata
Ovary
a) Crossing over
b) Prophase I 
c) ova/gamete/sex cells
C→B→ A (correct sequence)
(1)
(2)
(1)
(3)
(1)
(8)
QUESTION 7
7.1
7.2
7.3
7.4
7.5
7.6
7.7
Centriole /centrosome
Anaphase 1
- the spindle fibres contract
- the centromeres split
- each chromatid is pulled to the opposite poles
Crossing over
It leads to genetic variation
46 
- Structure B consists of two DNA molecules / contains a
double thread /Is made up of two chromatids
- because of DNA replication
- Structure C consists of one DNA molecule / contains a single
thread/chromatid
- because it is unreplicated / as a result of the splitting of the
chromosome during anaphase 2
(1)
(1)
(2)
(1)
(1)
(1)
(3)
(10)
SECTION C
27
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION 8
8.1
8.2
8.3
8.4
a) Meiosis 1
b) Prophase 1
Ovary
C-centromere
3
(1)
(1)
(1)
(2)
(1)
(6)
QUESTION 9
9.1 5/Five
(1)
9.2. Gonosomes/Sex chromosomes
(1)
9.3
(a) Down syndrome/Trisomy 21
(b) Non-disjunction
(1)
(1)
9.4
Male
(1)
(5)
QUESTION 10
10.1
(a) 20
(1)
(b) 50
(1)
- A sperm cell is a gamete
- formed by meiosis
- and must be haploid
- to overcome the doubling effect of fertilisation
(4)
10.2.
10.3
10.4
(1)
Anaphase I
- Spindle fibres shorten/contract
- Chromosome pairs separate and
- move to the opposite poles
(3)
(10)
28


SESSION 6:
Diversity of reproductive strategies
Human reproduction: Structure of the male and female reproductive system
& Puberty
SECTION A
EXAM GUIDELINE
29
Diversity of reproductive strategies

The role of the following reproductive strategies in animals in maximising
reproductive success in different environments (using relevant examples):
•
External fertilisation and internal fertilisation
•
Ovipary, ovovivipary and vivipary
•
Amniotic egg
•
Precocial and altricial development
•
Parental care
Introduction: Human reproduction
Revision of the schematic outline of the human life cycle to show the role of
meiosis, mitosis and fertilisation
Structure of the male reproductive system
Structure of the male reproductive system, using a diagram
Functions of the testis, epididymis, vas deferens, seminal vesicle, prostate
gland, Cowper's gland, penis and the urethra
Structure of the female reproductive system
Structure of the female reproductive system, using a diagram
Functions of the ovary, Fallopian tubes, uterus lined by endometrium, cervix,
vagina with its external opening and the vulva
Structure of the ovary, using a diagram, showing the primary follicles, the
Graafian follicle and the corpus luteum
Puberty
Main changes that occur in male characteristics during puberty under the
influence of testosterone
Main changes that occur in female characteristics during puberty under the
influence of oestrogen
NOTES & EXAM TIPS
REPRODUCTIVE STRATEGIES IN VERTREBRATES


The topic accounts for 8 marks out of 150 in paper 1. Terminology is very
important.
HUMAN REPRODUCTION
Learners have to know the front and side view of the male and female
reproductive systems.
30
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
B
C
D
B
C
D
D
C
D









(18)
QUESTION 2
2.1
Vagina
2.2
Puberty
2.3
Prostate
2.4
External
2.5
Amnion
2.6
Amniotic 
2.7
Precocial
2.8
External fertilisation
2.9
Altricial development
2.10
Internal fertilisation
2.11
Parental care
2.12
External fertilisation
2.13
Internal fertilisation
(13)
QUESTION 3
31
3.1 B only 
3.2 B only 
3.3 B only
3.4 None 
(8)
QUESTION 4
4.1
Male fertility 
(1)
4.2
Measuring the sperm count 
(1)
4.3
- Age
- Diet
-Exercise
- Activity level
- Lifestyle
-Occupation etc
( Accept factors that are not related to health and race)
(2)
4.4
4.5
4.6
4.7
TY inhibits the secretion of testosterone
Spermatogenesis cannot take place/no sperms will be produced
The higher temperature/pressure on the testes due to the tight
underwear
Could decrease the sperm count/sperm production lead to the
production of abnormal sperm
To determine if TU is still effective after 12 months
To see if the sperm count returns to normalwhen the treatment stops
No sperm will be transported
From the epididymis to the urethra
Semen without sperm will be released
(2)
(2)
(1)
(2)
QUESTION 5
5.1
Eggs are retained/hatch in the female body and the young are born
live
(2)
32
5.2
5.3
5.4
2
-The egg has the highest yolks/energy content
-that will allow maximum development before hatching
1
(1)
(2)
(1)
(6)
QUESTION 6
6.1
6.2
6.3
(a) Internal fertilisation
(b) Viviparous
Foetus develops in the mother’s uterus
- Protected as it develops in the mother’s pouch
- Nourished with mother’s milk
(Mark fist TWO only)
(1)
(1)
(1)
parental
care

(2)
(5)
SECTION C
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION 7
7.1
7.2
7.3
7.4
7.5
Seminal vesicle
Transports semen out of the body
(Mark first ONE only)
- Transports its secretions in ducts/ secretion not directly in blood
- Does not produce a hormone
(Mark first TWO only)
Spermatogenesis
- The fluid is alkaline 
- to neutralise the acidity of the vagina/ urethra
- The fluid contains nutrients
- for the sperm to generate energy for movement
(1)
(1)
(2)
(1)
(4)
(9)
QUESTION 8
8.1
8.2
External fertilisation
- Their embryos develop inside eggs
- which are laid/ hatch outside the body of the female
(1)
(2)
33
8.3
8.4
8.5
- The males release semen all around the female when she releases the eggs 
- A large number of gametes/ eggs are produced
(2)
Graph X
(1)
T- hey will have a higher number of surviving embryos/eggs
- Because their fertilised eggs are attached to the vegetation 
- where they are protected from predators/washing away
(3)
(9)
SESSION 7:
Gametogenesis & Menstrual cycle
SECTION A
EXAM GUIDELINE
Gametogenesis
34











Formation of gametes (gametogenesis) by meiosis

Male gametes formed by spermatogenesis

Female gametes formed by oogenesis
Spermatogenesis:

Under the influence of testosterone

diploid cells in the seminiferous tubules of the testes undergo meiosis

to form haploid sperm cells
Structure of a sperm, using a diagram
Functions of the parts of a sperm cell (acrosome, head with haploid nucleus, middle
portion/neck with mitochondria and a tail)
Oogenesis:

Diploid cells in the ovary undergo mitosis

to form numerous follicles.

At the onset of puberty

and under the influence of FSH,

one cell inside a follicle enlarges and undergoes meiosis.

Of the four cells that are produced, only one survives to form a mature,
haploid ovum.

This occurs in a monthly cycle.
Structure of an ovum, using a diagram
Functions of different parts of an ovum (jelly layer, haploid nucleus,
cytoplasm)
The menstrual cycle includes the uterine and ovarian cycles
Events in the ovarian cycle:

Development of the Graafian follicle

Ovulation

Formation of the corpus luteum
Events in the uterine cycle:

Changes that take place in the thickness of the endometrium

Menstruation
Hormonal control of the menstrual cycle (ovarian and uterine cycles) with
reference to the action of FSH, oestrogen, LH and progesterone
Negative feedback mechanism involving FSH and progesterone in controlling the
production of ova
NOTES & EXAM TIPS

The diagrams of the ovum and sperm cell are VERY important.

Learners must be exposed to graphs based on the menstrual cycle, ovarian
cycle and uterine cycle.
35
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
QUESTION 1
1.1
1.2
1.3
B
C
B



(6)
QUESTION 2
2.1
LH
2.2
Corpus Luteum
2.3
Menstruation
2.4
Graafian follicle
2.5
Middle piece
2.6
Acrosome 
(6)
QUESTION 3
3.1
B only 
3.2
B only 
3.3
None 
(6)
QUESTION 4
4.1
4.2
4.3
4.4
a) Jelly layer/zona pellucida
b) Cytoplasm
c) Acrosome
Oogenesis
D
E ; F
36
(6)
QUESTION 5
5.1
5.2
5.3
-The high levels of progesterone in the pills
-will inhibit the secretion of FSH from the pituitary gland
-No follicle will develop
-and hence no oestrogen will be secreted
The increase in the progesterone IeveI
indicates that corpus luteum has been formed
The increase in the progesterone IeveI
indicates that corpus luteum has been formed
(4)
(2)
(1)
(7)
QUESTION 6
6.1
6.2
6.3
a)
b)
FSH
Ovulation
- It causes the endometrium
- To become thicker
a)
The levels of progesterone drop
Therefore FSH secretion is no longer inhibited
A new follicle starts to develop
b)
The levels of progesterone drop
Therefore the endometrium is no longer maintained
and menstruation takes place
(2)
(2)
(3)
(3)
(10)
QUESTION 7
7.1
7.2
7.3
7.4
7.5
Cervix
The site of fertilisation
The site of zygote division
The transfer of the ovum/embryo to the uterus (any one)
Diploid cells in the ovary undergo mitosis
To form numerous follicles
Under the influence of FSH
One cell undergoes meiosis
To form an ovum
It us a hollow organ
It has a muscular wall
It has a blood-rich lining/endometrium
No follicle will develop
No oestrogen produced
(1)
(1)
(4)
(1)
37
And no progesterone produced
Therefore, the endometrium will not develop* to be shed during
menstruation
* Compulsory mark + Any 2 (Explain to learners why)
(3)
(10)
SECTION C
ANSWERS FOR HOMEWORK QUESTIONS
QUESTION 8
8.1 - Stimulates ovulation
- Stimulates the development of the corpus luteum
(Mark the first TWO only)
(2)
8.2 (a) - A high concentration of FSH
- will stimulate more follicles to develop
- Therefore, more ova will be produced increasing the chances to
fall pregnant
(3)
(b) - A peak in hormone B
- will indicate that ovulation is about to happen
- therefore, an ovum will be available for fertilisation
Any (2)
8.3 - The levels will remain low because
- the high progesterone levels during pregnancy
- will inhibit the pituitary gland from secreting FSH/Hormone A
(3)
(10)
SESSION 8:
FERTILIZATION AND IMPLANTATION
SECTION A
EXAM GUIDELINE
38
Fertilisation and development of zygote to blastocyst

Definition of copulation and fertilisation

Process of fertilization

Development of zygote  embryo (morula and blastula/blastocyst)  foetus
Implantation, gestation and the role of the placenta

Definition of implantation

The role of oestrogen and progesterone in maintaining pregnancy

Structure of the developing foetus in the uterus, using a diagram

Functions of the following parts:

Chorion and chorionic villi

Amnion, amniotic cavity and amniotic fluid

Umbilical cord (including umbilical artery and umbilical vein)

Placenta
NOTES & TIPS


In this section diagrams are very important – know the structure and
function
Know the sequence of development:
Zygote  Morula  Blastocyst (not blastocyte)  Foetus


Know the functions of the amniotic fluid, placenta and umbilical cord
The period of development of the foetus in the uterus is ‘gestation’ which
spans from conception to birth as opposed to ‘pregnancy’ which refers to all
changes (hormonal, physical, emotional) that take place in the body of a
female as a result of the developing foetus
SECTION B
ANSWERS FOR TYPICAL EXAM QUESTIONS
39
QUESTION 1
1.1
1.2
B
D
(2)
(2)
(4)
QUESTION 2





The zygote divides by mitosis
to form a ball of cells
called the morula
which further divides to form a hollow ball of cells
called the blastula/blastocyst
Any
(4)
QUESTION 3
3.1
3.2
3.3
3.4
3.5
Acrosome
Mitochondria 
(a) 3 
(b) 1 
(c) 1 
B- Nucleus
Mitosis
(1)
(1)
(1)
(1)
(1)
(2)
(1)
(8)
QUESTION 4
4.1
(a) Ovum
(1)
(b) Luteinising hormone/ LH
(1)
(c) Implantation
(1)
4.3
46 
(1)
4.4
14 
(1)
4.5
- Zygote 
- divides by mitosis 
- to form a mass ball of cells 
- called a morula 
- Which further divides by mitosis to form a hollow ball of cells
- called blastocyst/blastula.
Any
(4)
40
4.6





FSH will not be produced 
No ovum will develop 
LH will not be produced 
therefore, no ovulation will occur .
No pregnancy will occur 
Any
(3)
(13)
QUESTION 5
5.1



5.2


The scar tissue 
may partially block the fallopian tube 
preventing the embryo from reaching the uterus /resulting in
implantation in the Fallopian tube
(3)
the other Fallopian tube is still present /not blocked
fertilisation may still take place in this Fallopian tube / the
developing embryo can move along this Fallopian tube
OR


During invitro fertilization /IVF
The resulting embryo is inserted into the uterus 
OR


5.3




The ovum can be placed after the blockage 
Allowing fertilization 
Insufficient space 
Poor/no placental development 
Decreased blood supply 
Insufficient nutrients /oxygen
Mark First TWO Only
(2)
Any
(2)
(7)
(Any order)
(2)
QUESTION 6
6.1


6.2
Umbilical artery 
Umbilical vein 
Blood to the foetus
Contains relatively high
Blood away from the foetus
Contains relatively low
concentration of nutrients 
41
concentration of nutrients 
(Any examples)
Contains no (metabolic) waste
products 
Contains relatively high
concentration of oxygen
Contains relatively
low concentration of carbon
dioxide 
6.3
6.4
Contains high concentration
(metabolic) waste products
Contains relatively low
concentration of oxygen
Contains relatively high
concentration of carbon
dioxide
Mark the first ONE only
1 x 2 + 1 mark for the table
When answering a table question the statements on a row
should be covering the same aspect. For example first row is
about nutrients, second row waste products etc. you cant
answer nutrients in first column and then talk about waste
products in the second column in row 1.
Diffusion

The placenta will not be fully functional / there is less surface

area for diffusion and therefore,

less oxygen/ nutrients will be supplied to foetus

leading to under development 

causes the accumulation of metabolic waste products in the

foetus/less diffusion of wastes to the mother

causing blood to be highly toxicand this

results in the miscarriage / death of the foetus
Any
(3)
(1)
(3)
(9)
SECTION C
HOMEWORK QUESTIONS
QUESTION 7
Acrosome
7.1

7.2
- Fuses with the nucleus of the ovum
- Carries genetic material
- Produce energy/ site for cellular respiration
- which is needed for movement of the sperm
- The oval/torpedo-shaped head
- will facilitate faster movement
7.3
7.4
(1)
Any
(1)
(2)
- The presence of an acrosome
- enables the nucleus/head to enter the ovum
42
- A longer tail
- ensures faster movement
(Mark first TWO only)
Any (2 x 2)
(4)
(8)
QUESTION 8
8.1
(Mark the first TWO only)
8.2
(Any 2 x 2 + 1) (5)
- Ectopic pregnancy ✓
- Intra-uterine foetal growth restriction ✓
- Abnormal placentation ✓
- Foetal malposition ✓
(Mark first TWO only)
8.3
(Any 2 x 1) (2)
- The high levels of progesterone ✓
- inhibit the pituitary gland ✓
- from releasing the FSH ✓
- Therefore, no new follicle will develop ✓
- and no ovum will be released ✓/ ovulation takes place
- for another fertilisation to occur ✓
8.4
(Any 4 x 1)
(4)
(Any 8 x 1)
(8)
(19)
- The embryo develops an outer membrane, the chorion ✓
- and an inner membrane, the amnion ✓
- The amnion forms a cavity ✓
- which encloses the amniotic fluid ✓
- The chorionic villi ✓that develops from the chorion
- together with the endometrium ✓
- forms the placenta ✓
- A hollow tube called the umbilical cord ✓ attaches
- the embryo to the placenta ✓
- The umbilical cord consists of an umbilical artery ✓
- and an umbilical vein ✓
43
44