# Life Sciences Grade 12 Paper 2 KZN Mock 2023 Paper & Memo

> Grade 12 Life Sciences Paper 2 question paper and memo from the KZN Mock Exam 2023, 27 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-life-sciences-2023-kzn-mock-exam-question-paper-and-memorandum-paper-2/62b7f642-b151-4226-90bf-8b5a72bf18bb
- Updated: 2026-09-28
- Subject: Life Sciences
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 27

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## About this document

Life Sciences · Grade 12 · KZN Mock Exam · 2023. Question paper and memorandum, 27 pages. Read online or download the PDF.

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### Page 1

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xt F
anes } KWAZULU-NATAL PROVINCE
yy

/ EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

Sees ee ee ee esses

MARKS: 150

TIME: 2% hours

This question paper consists of 17 pages.

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1.

2.

10.

11.

Answer ALL the questions.
Write/ALL the answers in the ANSWER BOOK.
Start-the answers to each question at the top of a NEW page.

Number the answers correctly according to the numbering system used in this
question paper.

Present your answers according to the instructions of each question.

Do ALL drawings in pencil and label them in blue or black ink.

Draw diagrams, tables or flow charts only when asked to do so.

The diagrams in this question paper are NOT necessarily drawn to scale.
Do NOT use graph paper.

You may use a non-programmable calculator, protractor and a compass.

Write neatly and legibly.

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SECTION A

QUESTION 1

NSC

Chooseithé answer and write only the letter (A to D) next tothe ques'

1.1. Various options are provided as possible answers to they ickewing ay ul

(1.1.1 t07/1.10) in the ANSWER BOOK, for example 1.1.14 all i

1.1.1

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An anticodon is a sequence of three nucleotides on wlll n

A DNA.
B_ tRNA.
C rRNA.
D mRNA.

An individual that has received a different allele from each parent for a
particular characteristic is described as being ...

A homologous,
B homozygous.
C_ recessive.

D heterozygous.

Which ONE of the following occurs in meiosis, but NOT in mitosis?

A Centrosomes form spindle fibres

B_ Reduce chromosome number by half

C Chromosomes are pulled towards opposite poles
D Chromosomes arrange at the equator of the cell

During which ONE of the following phases of meiosis does the random
arrangement of chromosomes occur?

A Prophase |
B_ Prophase II
C Metaphase II
D. Anaphase |

A horse has 64 chromosomes in a somatic cell. The gonosomes are the
same as those in humans.

A normal horse sperm cell will have ...
A 31 pairs of autosomes and an X chromosome.
32 autosomes and a Y chromosome.

B
C 32 autosomes and an X chromosome.
D_ 31 autosomes and a Y chromosome

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1.1.6

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NSC
The insulin molecule contains 51 amino acids.

How many mRNA nucleotides code for this insulin molecule?

A 51
B, 17
@ 153
D 152

Which ONE of the following statements regarding chromosomes is
correct?

22 pairs of chromosomes in humans are autosomes and 2 pairs of
chromosomes are gonosomes

Chromosomes are made up of DNA molecules and proteins
Chromosomes are found in somatic cells only

Chromosomes are thread-like structures located inside the nucleus
of animal cells only

0uOWD >

There is uncertainty about who the biological father of the child is. To
establish paternity, DNA profiling was conducted.

The diagram below shows the DNA profiles of a child, her mother and
four men.

Mother Man 1 Man 2 Man 3 Man 4

Q
Es

Which ONE of the following men is most likely to have been the father
of the child?

Man 1
Man 2
Man 3
Man 4

0OoODU>Y

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1.1.9 The phylogenetic tree below shows the period of existence of some

hominids.
Homo sapiens

0 5 —

05 Homo erectus

104 Homo habilis
a 1.5 a Australopithecus africanus
os
4 2.0 4 Australopithecus
2 ened 4 4 aethiopicus
2 2.5 = =
2 ‘ee 7
= 3.0 a
= went

3.6 — Australopithecus

4.0 4 SS afarensis

I Ardipithecus ramidus
45

Which ONE of the following tables best represents the information in the
phylogenetic tree about the Homo species?

A B
HOMINIDS PERIOD OF HOMINIDS PERIOD OF
EXISTENCE EXISTENCE
(mya) (mya)
Homo erectus 2-0.5 Homo erectus 2-05
Homo sapiens |_2- present Homo sapiens 0.3 - 0.2
Homo habilis 2.33 - 1.4 Homo habilis 2.6-1.4
Cc D
HOMINIDS PERIOD OF HOMINIDS PERIOD OF
EXISTENCE EXISTENCE
(mya) (mya)
Homo erectus 2-1.5 Homo erectus 2-05
Homo sapiens | 0.2 - present Homo sapiens | 0.2 - present
Homo habilis 2.33 - 1.4 Homo habilis 2.33 -1.4

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1.1.10 The effect of speciation through geographic isolation on a population of

a single species includes ...

natural selection occurs dependently in each of the two populations
gene flow between the two populations

the two populations become different from each other

the decrease to the variety of species that live on earth

ROD >

(10x2) (20)

1.2 Give the correct biological term for each of the following descriptions. Write only
the term next to the question number (1.2.1 to 1.2.9) in the ANSWER BOOK.

1.2.1

1.2.2
1.2.3

1.2.4
1.2.5
1.2.6
1.2.7

1.2.8

Undifferentiated cells that may be stimulated to develop into any type of
body cell

The point of crossing over between two adjacent chromosomes

The process by which genetically identical organisms are formed using
biotechnology

The nitrogenous base found in DNA but not in RNA.

A sex-linked disorder that affects the photoreceptors in the eye

A sudden change in the sequence of nitrogenous bases of a nucleic acid
The breeding of organisms by humans to achieve a desirable phenotype

The type of evidence for human evolution that includes tool-making
(8x1) (8)

1.3 Indicate whether each of the descriptions in COLUMN | applies to A ONLY, B
ONLY, BOTH A AND B or NONE of the items in COLUMN II. Write A only, B
only, both A and B, or none next to the question number (1.3.1 to 1.3.3) in the
ANSWER BOOK.

COLUMN | COLUMN II
1.3.1 | Division of the cytoplasm A: Meiosis |
B: Meiosis II
1.3.2 | Crossing over A: Telophase |
B: _Prophase II
1.3.3 | Principle of Segregation A: Watson
B: _ Gregor Mende}

(3x2) (6)

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NSC

1.4 The diagram below shows part of a process in a nucleic acid.

1.41

1.4.2

1.4.3

I= _
‘mt |

Name the:
(a) Process shown in the diagram
(b) Nucleic acid shown in the diagram

(c) Natural shape of the nucleic acid named in QUESTION 1.4.1 (b).

Identify:
(a) Bond 1
(b) Part2

(c) Structure 3

Name the organelle that is found in plant cells only, where the nucleic acid
named in QUESTION 1.4.1 (b) is located.

4000]

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(1)
(7)

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1.5. The diagrams below show the pelvis of chimpanzee and Homo sapiens.(not
necessarily in that order)

1.5.1

1.5.2

1.5.3

1.5.4

Give the LETTER of the pelvis ecies that has the:

a i

(a) Foramen magnum in a more
(b) S-shaped Spine ,
Name the species that is represented by the pelvis shown in diagram:
(a) A
(b) B

State TWO visible structural differences between the pelvis in diagram A
and B.

Give ONE significance of the shape of the pelvis shown in diagram A.

TOTAL SECTION A:

Lon

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(1)
(1)

(1)

(4)

(1)
(9)

50

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SECTION B
QUESTION 2
2.1 The diagrams below show part of a phase in meiosis.

2.1.1

Diagram | Diagram II

Identify:

(a) The phase represented in diagram Il.

(b) PartA

Name the type of cell that results from meiosis.

How many chromosomes were present in the cell at the beginning of
meiosis?

The cell in diagram I is undergoing abnormal meiosis:
(a) | Give the term for abnormal meiosis.

(b) Name the genetic disorder that would result in humans if this abnormal
meiosis occurs in chromosome pair number 21.

Explain why the cell in diagram Il does not belong to a human.

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2.2 The diagram below represents part of the process of protein synthesis.

Ww 1
—=>——_——_ tRNA
pv
yy
© UAA
G i X
2
Tk
‘SY Y A UU
AN)
ery z
2.2.1 Identify:
(a) Organelle X
(b) Monomer 1
2.2.2 Write down the DNA sequence for the base triplet 2.
2.2.3 Name and explain the importance of process W in protein synthesis.
2.2.4 Describe the role of tRNA in the process of protein synthesis.
2.2.5 The table below shows the codons that code for different amino acids.

CODON AUC UAC UCG AUU UGC

AMINO ACID | Isoleucine | Tyrosine | Serine | Isoleucine | Cysteine

With reference to the diagram in QUESTION 2.2 and the table above:
(a) State the amino acid that is coded by the DNA base triplet AGC.

(b) If the codon AUU on molecule Z change to AUC, explain the effect it
would have on this particular protein molecule.

2.3. Duchenne muscular dystrophy (DMD) is an inherited conditionle€aused by a sex-
linked recessive allele (X4).

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(1)
(1)

(1)
(3)
(3)

(1)

(3)
(13)

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NSC

The diagram below shows the inheritance of DMD in a family.

2.3.1
2.3.2
2.3.3

2.3.4

2.3.5

2.3.6

2.3.7

Key
© B Unaffected
Cd CQ C) | DMD male
c D Unaffected
female
Og @ ow female

E F

Name this type of a diagram.
How many males are unaffected in the second generation?
State the phenotype of individual C.

Give all the possible genotypes of individual:

Explain why individual A can only be heterozygous for this condition.

Individual E marries a heterozygous female. Calculate the percentage
chance of their daughters having an allele for DMD. Show all working.

Explain why DMD cannot be passed from father to son.

2.4 Tay-sachs disease is an autosomal recessive (r) genetic disorder.

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(1)

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NSC

KZN Prep exam/2023

A boy has Tay-sachs disease and his sister is homozygous and unaffected.

Use a genetic cross to determine the genotypes and phenotypes of their parents. (6)

2.5 Three «men were seriously injured in a motor car accident and could not be
identified2One of them is suspected to be Sipho’s father. The blood types of the
three men was established and compared to the blood groups of Sipho and his
mother-This evidence was not conclusive and DNA profiling had to be conducted.

The table below shows the blood types of Sipho, his mother and the three men.

2.5.1

2.5.2

2.5.3
2.5.4

QUESTION 3
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INDIVIDUAL BLOOD TYPE
Sipho {e}
Sipho’s mother B
Man 1 A
Man 2 AB
Man 3 {e)
Give the:
(a) Genotype of Sipho’s mother (1)
(b) Blood type that is controlled by the recessive allele (1)
Name the type of dominance that is shown by the genotype of Sipho's
mother in QUESTION 2.5.1 (a). (1)
Which man could NOT be Sipho’s father? (1)
Explain why Sipho’s father could not be identified by using blood groups
only. (5)
(9)
[50]

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3.1. In dogs, black fur (B) is dominant over white fur (b) and prick ear (E) is dominant
over drop ear (e). Drop ear is caused by a recessive allele
A dog breeder crossed a black, drop-eared dog with a white, prick-eared dog.

The diagram below shows the parent dogs that were crossed and the genotypes
of their-three offspring.

; a
4 €
Parent 1 Parent 2
bbee Je BbEe
Offspring 1 Offspring 2 Offspring 3
3.1.1 Name this type of genetic cross. (1)
3.1.2 Describe what a recessive allele is. (2)
3.1.3 How many of the three offspring are white? (1)
3.1.4 Give the:
(a) Phenotype of offspring 3 (1)
(b). Genotype of parent 4 (2)
(c) Possible genotypes of the gametes of parent 2 (2)
3.1.5 Identify the offspring that:
(a) Has the same phenotype and genotype to that of parent 2 (1)
(b) Is homozygous for drop ear (1)

(11)

3.2. Read the extract below:
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Salmonellosis is a bacterial disease that affects the human intestinal tract. This
disease is caused by salmonella bacteria. Most infected people have diarrhoea,
stomach cramps and fever. The antibiotic, Ciprofloxacin, is used to control
salmonella in humans. This antibiotic kills the salmonella, but it can also cause
nausea“and eye discomfort. It has been found that salmonella developed
resistance to Ciprofloxacin treatment.

Salmonellosis was treated with Ciprofloxacin from 2005 to 2011 and the
percentage of Ciprofloxacin resistant bacteria was recorded.

The results are shown in the graph below.

35
=
gs 3
g
& 2,5
2
s 2
2
wn
215
£
S 1
2
05
2
° 9
2005 2006 2007 2008 2009 2010 2011
Year
3.2.1 From the extract, identify TWO:
(a) Negative side effects of ciprofloxacin (2)
(b) | Symptoms of salmonellosis (2)
3.2.2 Explain how the use of the antibiotic resulted in the evolution of
ciprofloxacin resistant bacteria. (6)
(10)

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3.3

3.4
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KZN Prep exam/2023
NSC

Dark brown mice migrated to the Sand Hills of Nebraska. A mutation in one gene
for fur colour resulted in light brown mice. Light brown mice camouflage well
against the sand. Predators feed on the mice in the sand.

The colour of the mice depends on the mutation in a gene for fur colour.
Scientists/¢arried out an investigation to determine the effect of fur colour on the
percentage of mice killed by predators. The mice were not allowed to migrate from
1970 t04.974 and the percentage of mice killed by predators was recorded every
year:

The table below shows the results of their investigation.

MICE KILLED BY PREDATORS (%)
YEAR LIGHT BROWN MICE DARK BROWN MICE
1970 2 4
1971 1.5 6
1972 2 6.5
1973 1 7
1974 0.5 10
https://flexbooks.ck12.org>lesson
3.3.1 Identify the:
(a) Dependent variable (1)
(b) Independent variable (1)
3.3.2 Name ONE advantage of the mutation to the mice. (1)
3.3.3 Give ONE way in which the reliability of the results was ensured. (1)
3.3.4 State ONE conclusion that can be drawn from the results of the investigation. (2)
3.3.5 Draw two bar graphs for the first three years on the same set of axes to
represent the information in the table. (6)
(12)

Read the passage below on human evolution.

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NSC

KZN Prep exam/2023

Fossil evidence indicate that modern humans originated in Africa and then
migrated to other continents. The following fossils serve as evidence for this
migration: Ardipithecus, Australopithecus and Homo (Homo habilis, Homo
erectus.and Homo sapiens).

3.4.1 iiName the hypothesis on human evolution that is described in the

| “passage.

3.4.2

3.4.3

3.4.4

From the passage, identify TWO of the:

(a) Oldest fossils that were found in Africa

(b) Fossils that were found in Africa only
Name the family to which Homo sapiens belong.

Give ONE evidence other than the fossil indicating that modern humans
originated in Africa.

3.5 The diagrams of giraffes below show evolution.

3.5.1

3.5.2

3.6 Read the passage below:

Give ONE reason why Giraffe B may be described as a transitional
species?

Describe how Lamarck had explained the evolution of the long neck of the
giraffe.

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(1)

(2)
(2)
(1)

(1)
(7)

(1)

(4)
(5)

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Squirrels are believed to have been evolved from one common ancestor
species.

Scientists conducted an investigation to find out whether the grey and brown
squirrels’belong to the same species. They allowed grey and brown squirrels to
interaétlin the same habitat for three years. They observed that the population
of brown squirrels bred from April to June only, whereas grey squirrels bred from
August-to October only.

3.6.1 Based on the passage, name the type of reproductive isolation
mechanism displayed by the grey and brown squirrels. (1)

3.6.2 State TWO reproductive isolation mechanisms in animals other than the

one named in QUESTION 3.6.1. (2)
3.6.3. Explain how would scientists confirm if the two squirrel types belong to
different species? (2)
(5)
[50]

TOTAL SECTION B: 100
GRAND TOTAL: 150

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KWAZULU-NATAL PROVINCE

‘oa? EDUCATION
ec REPUBLIC OF SOUTH AFRICA

FINAL

NATIONAL
SENIOR CERTIFICATE

MARKS: 150

This marking guidelines consists of 10 pages.

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Grade 12 — FINAL Marking guideline

PRINCIPLES RELATED TO MARKING LIFE SCIENCES SEPTEMBER 2022

1.

10.

11.

12.

13.

14.

15.

If more information than marks allocated is given
Stop marking when maximum marks are reached and put a wavy line and ‘max’ in the
right-hand margin.

If, for example, three reasons are required and five are given
Mark the first-three irrespective of whether all or some are correct/incorrect.

If whole process is given when only part of it is required
Read all and credit relevant part.

If comparisons are asked for and descriptions are given
Accept if differences / similarities are clear.

If tabulation is required but paragraphs are given
Candidates will lose marks for not tabulating.

If diagrams are given with annotations when descriptions are required
Candidates will lose marks

If flow charts are given instead of descriptions
Candidates will lose marks.

If sequence is muddled and links do not make sense
Where sequence and links are correct, credit. Where sequence and links is incorrect, do
not credit. If sequence and links becomes correct again, resume credit.

Non-recognised abbreviations
Accept if first defined in answer. If not defined, do not credit the unrecognized
abbreviation but credit the rest of answer if correct.

Wrong numbering

If answer fits into the correct sequence of questions but the wrong number is given, it is
acceptable.)
If language.used changes the intended meaning
Do not accept.

Spelling errors
If recognizable accept provided it does not mean something else in Life Sciences or if it
is out of context.

If common names given in terminology
Accept provided it was accepted at the National memo discussiah-meeting.

If only letter is asked for and only name is given (and vice versa)
No credit

If units are not given in measurements
Candidates will lose marks. Memorandum will allocate marks for units separately

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Grade 12 — FINAL Marking guideline

16. Besensitive to the sense of an answer, which may be stated in a different way.

17. Caption
All illustrations (diagrams, graphs, tables, etc.) must have a caption

18. Code-switching of official languages (terms and concepts)
A single word-or two that appears in any official language other than the learners’
assessmentianguage used to the greatest extent in his/her answers should be credited,
if it is correct>A marker that is proficient in the relevant official language should be
consulted. This is applicable to all official languages.

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Life Scieewwaloaded from Stanmorephysics.com
Grade 12 — FINAL Marking guideline

SECTION A
QUESTION 1

1.1 BY <
pvt
BY
cut
Dv
CVV
ByYY
BYY
Dvv
Cvv

Fa CO
Sa OO
SPOWNDOBRWN =

oO

1.2 Stemv cells
Chiasmatav
CloningY
Thyminev
Colour blindness”
Mutationw’
Artificial selection /Selective breeding

Culturalv evidence

Aaa aa
NNNNNNPND
ANOAnNBRWH>

1.3 1.3.1 BothA and BYY
1.3.2 Nonevv
1.3.3 Bonlyvv

14 1.4.1 (a) DNA replication’
(b) DNAv molecule
(c) Double helixv

1.4.2 (a) Hydrogen’ bond
(b) Deoxyribose sugar”
(c) Nucleotidev
1.4.3. Chloroplastv
15 15.1 (a) BY
(b) Av

1.5.2 (a) Homo sapiens
(b) Chimpanzee

1.5.3. - Pelvis Ais shortv whereas B is long”

- Pelvis A is widev whereas B is narrow”
(Mark the first TWO only)

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September 2023

(10 x 2) (20)

(8 x 1) (8)

(3 x 2) (6)

(4)

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1.5.4

SECTION B
QUESTION 2

2.1 2.1.1

2.1.2
2.1.3

2.1.4

2.2 2.2.1

2.2.2

2.2.3

2.2.4

2.2.5

2.3 2.3.1

2.3.2

2.3.3

September 2023
Grade 12 — FINAL Marking guideline
Supports upper body weight” (1)
(Mark the first ONE only) (9)
TOTAL SECTION A: 50
(a) Metaphase Ilv (1)
(b) ChromosomeY (1)
Gamete~/sex cell (1)
6v/six (1)
(a) Non-disjunctionY (1)
(b) Down syndromev (1)
- Cell in diagram II has three chromosomes”
- Cell would have 23 chromosomesY (2)
(8)
(a) Nucleus~ (1)
(b) Amino acidv (1)
TAAY (1)
- Transcriptiony*
- mRNA copies coded message from DNAY
- which moves to the ribosomes”
- so that amino acids can be arranged in a specific sequencev
- for the formation of a specific protein’
Compulsory 1*+any 2 (3)
- Picks up specific amino acids”
- and bring them to the ribosomes”
- has specific anticodonsv that are
- complementary to the (mRNA) codonsvY Any (3)
(a) Serinev (1)
(b) - The anticodon will be UAGY
- instead of UAAY
- Amino acid isoleucine will not change”
- resulting in the same proteinY Any (3)
(13)
Pedigreey diagram (1)
av (1)
DMD malev (1)

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Grade 12 — FINAL Marking guideline

2.3.4 (a) XOxdv (1)
(b) - X2xPv
- XDxdy (2)

2.3.5 - Individual A is unaffectedv
- caused by dominant allele on X chromosomev/X°
- Offspring C inherits Y chromosome from individual Bv/father
- and X¢ from individual Av /mother
OR
- Individual A has an offspring C who is affected
- Therefore, receiving a recessive allele/X¢ from individual A
- individual A must have a dominant allele as well

- hence she is unaffected Any (3)
2.3.6 1 |” x100Y =50%Y (3)
2
2.3.7 - DMD is caused by a recessive allele on the X chromosomev /X?
- Sons inherit the Y chromosome from their fatherv’ (2)
(14)
2.4 Pa Phenotype Unaffected x Unaffected*
Genotype Rr x Rrv*
Meiosis "8
Gametes R, r x R, rv
Fertilisation — <<
Fi Genotype _RR\=* Rr Rr, try
Phenotype 3 unaffected and affected’
Pi and Fiv

Meiosis and fertilisation’
Compulsory mark 2* + Any 4

OR
Pu Phenotype Unaffected x Unaffectedv*
Genotype Rr x Rrv*

Meiosis Gametes R r

an R RR Rr
Fertilisation r Rr ir
Fs 1 mark for correct gametes

1 mark for correct genotypes
Phenotype 3 unaffected and affectedv

Pi and Fiv
Meiosis and fertilisation’ Compulsory mark 2* + Any 4 (6)

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Grade 12 — FINAL Marking guideline

25 2.5.1 (a) [iv (1)
(b) Ov (1)
2.5.2. Completew dominance (1)
2.5.3 Man2y (1)
2.5.4 - Mahe or man 3 could be the father of Siphow

since both man may have recessive allelev/i / I*i or ii”
The mother must have genotype I®iv since she is blood type B
Sipho would have inherited the recessive allele/i from both parents

- and he would have the genotype iiv’ (5)
(9)
[50]
QUESTION 3
3.1 3.1.1. Dihybrid’cross (1)

3.1.2 An allele that is masked/not shown in the phenotype when found in the
heterozygous conditionv ~

OR

An allele that is expressed in the phenotype when found in the
homozygous conditiony Vv (2)
3.1.3. Twov/2 (1)
3.1.4 (a) Black fur, prick-earedY (1)
(b) Bbeevv (2)
(c) - bEY (1)
- bev (1)
3.1.5. (a) Offspring 2” (1)
(b) Offspring 1” (1)

3.2 3.2.1 (a) - Nauseav
- Eye discomfortv (2)
(Mark the first TWO only)

(b) - Diarrhoeav
- Stomach crampsv
- Fever’ Any (2)
(Mark the first TWO only)

Copyright reserved Please turn over

### Page 25

Lite scieewwaloaded from Stanmorephysics.com

Grade 12 — FINAL Marking guideline

3.2.3. - Natural selection’ occurs
- There is variation” /mutation in the population of bacteria
- Some are resistant to the antibiotic ciprofloxacin, some are non-
resistant’
- When ciprofloxacin is used”
- Thebacteria that are non-resistant are killedY by the ciprofloxacin
- Those that are resistant survive and reproducev
- The=characteristic for resistance to ciprofloxacin is passed on to the
offspringv’
- The next generation will have a higher proportion of ciprofloxacin
resistant bacteriav Any
3.3. 3.3.1 (a) Percentage of mice killed by predators”
(b) Fur colour’
3.3.2 It results in light brown fur that camouflage well against the sandv
Mark the first ONE only
3.3.3. Repeated the investigation for five years”
Mark the first ONE only
3.3.4 The light brown mice have lesser percentage killed by predators than dark
brown miceY v
OR
The dark brown mice have higher percentage killed by predators than
light brown micev ¥
Mark the first ONE only
3.3.5
Percentage of mice killed by predators over three years
7 As ‘ ——
J
= § morepH| =&com|
= 1
g 5
o
34 Key
Qa
B 3 [ight brown mice
3 Mibark brown mice
= 2
o
&
21
0
1970 1971 1972
Year
Copyright reserved

September 2023

Please turn over

### Page 26

Life Scieewwaloaded from Stanmorephysics.com

3.4 3.4.1

3.4.2

3.4.3

3.4.4

3.5 3.5.1

3.5.2

3.6 3.6.1

3.6.2

Grade 12 — FINAL

arking guideline

Guideline for assessing the graph

September 2023

CRITERIA

ELABORATION

MARK

Correct type of graph (T)

Bar graph drawn

Caption of graph (C)

Both variables included

Axes-labels (L)

X- and Y-axes correctly
labelled

Sealejfor X- and Y-axes (S)

- Equal space and width of
bars for X-axes and
- Correct scale for Y-axes

Plotting of co-ordinates (P)

1 to 5 co-ordinates plotted
correctly

Only first 3 years co-
ordinates plotted correctly

‘Out of Africa’v hypothesis
(a) - Homo erectusY
- Homo sapiens”
Mark the first TWO only
(b) - ArdipithecusY
- AustralopithecusY
- Homo habilis”
Mark the first TWO only
Hominidaev
Geneticv evidence
It has characteristics common to both giraffes A and BY
OR
It has intermediate characteristics between giraffe A and BY
Mark the first ONE only
- Aneestor.of giraffes was having short neck”
- Environment change to have long treesY”
- It stretehed the neck to feed on top branchesv
- The néck developed and became longY
- to feed on top branches”
- Giraffe passed on long neck to the offspring”
Breeding at different times”

- Species-specific courtship behavior”

- Infertile offspringy
- Prevention of fertilisation”
Mark the first TWO only

Copyright reserved

Any

Any

Any

(6)
(12)

(1)

(2)

(2)

(1)

(1)
(7)

(4)
(5)

(1)

(2)

Please turn over

### Page 27

Lite Scieewwaloaded from Stanmonephysics.com September 2023
Grade 12 — FINAL Marking guideline

3.6.3 - If they were allowed to interbreedv
- and cannot produce fertile offspringY

OR
- Analysis of DNAv
- to check the matching sequence” Any (2)
(5)
[50]

TOTAL SECTION B: 100

GRAND TOTAL: 150

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