# Life Sciences Grade 12 P2 North West Mock 2023 Paper & Memo

> Grade 12 Life Sciences Paper 2 question paper and memo from the North West Mock Exam 2023, 28 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-life-sciences-2023-north-west-mock-exam-question-paper-and-memorandum-paper-2/6ddde48d-aea2-4887-aab2-884a578af1a7
- Updated: 2026-09-28
- Subject: Life Sciences
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 28

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## About this document

Life Sciences · Grade 12 · North West Mock Exam · 2023. Question paper and memorandum, 28 pages. Read online or download the PDF.

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## Document text

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education

Department:

Education
North West Provincial Government

REPUBLIC OF SOUTH AFRICA

MARKS: 150
TIME: 2% hours

IMD

12832E

X10

CL

This question paper consists of 16 pages.

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. Answer ALL the questions.

2. LL the answers in the ANSWER BOOK.

°

he answers to EACH question at the top of a NEW page.

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

5. Present your answers according to the instructions for each question.
6. 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.

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

10. You must use a non-programmable calculator, protractor and compass
where necessary.

11. Write neatly and legibly.

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SECTION A

QUESTION 1

S options are provided as possible answers to the following questions.
se the answer and write only the letter (A-D) next to the question
bers (1.1.1 to 1.1.9) in the ANSWER BOOK, for e.g. 1.1.10 D.

Q : ‘

fica During which phase of meiosis do spindle fibres start to form?
A Prophase |
B- Metaphase |
C Anaphase |
D_ Telophase |

1.1.2. The following list includes some features that are studied in
evolution.
(i) Bipedalism
(ii) Dentition

(iii) Brow ridges
(iv) Stereoscopic vision

Which ONE the following key features separate the human line of
evolution from that of the African Apes?

A (i) only

B (i) and (iv) only

C (ii) and (iv) only

D (i), (ii) and (iii) only

1.1.3. Asmall section of mRNA has the following sequence of bases that
codes for different amino acids:

ccc CGU UAA

Which ONE of the following is the representation of the anticodons
and number of amino acids coded for by this section?

ANTICODONS NUMBER OF AMINO ACIDS

A GGG | GCA AUU 9
B GGG | GCA AUU 3
Cc GGG | GCA ATT 9
D GGG | GCA ATT 3:
Taar
fanah

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1.1.4. Ina family with four children, each child has a different blogg Brou:

The genotypes of the parents must be ...

A Wiand [Bi
B 1B and ii
C Biand Ie
D iandii

Which ONE of the following monohybrid crosses wh

; ere complete
dominance applies, will result in a phenotypic ratio of 3

21?

A Both parents are heterozygous

B One parent is heterozygous, and the other is homozygous
recessive

C Both parents are homozygous for the dominant characteristic

D

One parent is heterozygous, and the other is h

jomozygous
dominant

QUESTIONS 1.1.6 AND 1.1.7 ARE BASED ON
SHOWING THE POSSIBLE EVOLUTION OF S$:

THE TIMELINE BELow
OME HOMINIDS:

Australopithe cus
afarensis

Ss

Homo erectu

Len

3
i

1

0
it

(|

Homo
habilis

Ardipithecus
ramidus

Q

Homo
Sapiens

years ago did this species appear?

A Australopithecus afarensis; 4 mya
B Homo erectus; 2 mya

C Homo habilis; 2.2 mya

D

Ardipithecus ramidus; 5mya

o0QnaPY
Na0og

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Which species inhabited the Earth for the lon

gest time, and how many

How many genera are represented on the timeline above?

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NW/September 2023
NSC

1.1.8 The diagram below represents different phases of meiosis.

Identify the correct phase in diagram 2

A Anaphase II
B- Metaphase II
C_ Prophase |
D- Metaphase |

1.1.9 Ifa DNA molecule contains 20 000 nitrogenous bases, of which 20%

are cytosine, how many adenine molecules will be present?

A 1000

B 2000

Cc 6000

D 8000 (9x2) (18)

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132

1.3

Give the correct biological term fon each tof tthe following descriptions.
Write only the term next to the question numbers (1.2.1 to 1.2.9) in’the
ANSWER BOOK.

1.2.1___ The mode of locomotion where an animal uses four limbs

1.2, The bond formed between amino acids

1.2180 1 The part of the skull that houses the brain

—

1.2.4 Achange in the sequence of nitrogenous bases

1.2.5 A disorder due to the presence of an extra chromosome on the
21°" pair of chromosomes in somatic cells of humans

1.2.6 Genetic evidence that is used to support the ‘Out of Africa’
hypothesis

1.2.7 The process by which genetically identical organisms are formed
using biotechnology

1.2.8 Type of dominance in which both alleles are expressed in the
phenotype

1.2.9 Similar structures that are inherited from a common ancestor and
modified for different functions (9x1) (9)

Indicate whether each of the statements 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 numbers
(1.3.1 to 1.3.3) in the ANSWER BOOK.

COLUMN | COLUMN II
1.3.1. The genotypeofaperson |A: XY
with haemophilia B; Xhxh
1.3.2 Variation in human height | A: Continuous variation
B: Discontinuous variation
1.3.3 Involved in the discovery of |A: Eldredge
DNA molecule B: Gould

(3x2) (6)

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1.4 Three ladies each gave birth to a baby boy in a hospital. The babies were
accidentally swopped.

The diagram below shows a technique used in paternity testing and the
results of each parent and their baby boys.

SMITH STEVENS JONES BABY
MR MRS MR MRS MR] MRS 41 2 3
aay

1.4.1 Identify the technique shown above. (1)
1.4.2. Which baby (1, 2 or 3) belongs to the Smiths? (2)

1.4.3 State TWO other uses for the technique named in
QUESTION 1.4.1. (2)

(5)

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41.5
on bacterial resistance.

 NWiSentermber, 202:

An investigation was carried out to determine the-éffect. oF ar’antibion<

The diagram below shows the effect of an antibiotic on a bacterial

population.

ulation before
tment with an

antibiotic Key
Key
Resistant Bacteria
Bacteria population (2) |

Bacteria

Non-Resistant

Cc)
G

Bacteria population
after treated with an
antibiotic

New bacterial

population
1.5.1. Give the:
(a) Dependent variable (1)
(b) Independent variable (1)
1.5.2 Name the type of evolution that occurs in this bacterial
population. (1)
1.5.3 What is the change in the environment that brings abut the
change in the bacteria population? donar (1)
1.5.4. Which variation can be found in greater proportion na
population? = (1)

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455° Whatpercentage of the bacieriain the original papulatio7 was
‘resistant to antibiotics” (1)

(6)

1.6 Agenetic disorder is caused by a recessive allele (a).

An ‘individual with the disorder is described as affected, and an individual
fit is described as unaffected. The affected individual is indicated by
circle or square.

‘pedigree diagram below shows the inheritance of this disorder in a

HO

1.6.1 How many of each of the following are present in the diagram?

(a) Males (1)

(b) Generations (1)

(c) Children of individuals A and B (1)
1.6.2 Give the:

(a) Possible genotype of individual D (2)

(b) Phenotype of individua! C
(6)

TOTAL SECTION A: 50

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NSC
SECTION B
QUESTION 2
2.1 “e below represents chromosomes in a cell that is undergoing
m .
fon)
=e ——>
B
Cc
(—__—___
A
2.1.1 Identify:
(a) Chromosome pair A (1)
(b) Structure B (1)
(c) Partc (1)
2.1.2 Which process is illustrated in the diagram? (1)
2.1.3 During which phase of meiosis does the process named in
QUESTION 2.1.2 take place? (1)
2.1.4 What is the importance of the process named in
QUESTION 2.1.2? (1)
2.1.5 Describe ONE other process in meiosis which will ensure the
same outcome as the one stated in QUESTION 2.1.4. (3)
(9)
2:2 Tabulate THREE differences between meiosis | and meiosis II. (7)
(7)

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2.2> “Lamarckisminefers tg
imvarck

Answer the following questions on Lamarck's theory

2.3.1_State the Two
28volution, Principles (laws) that Lamarck used to explain 6
Onn...
a8 pePlain ONE reason why most scientists reject this theory. (2)
dt (4)
2.4 Human blood types are determined b i
: : Y multiple alleles. Each pe ha:
only two alleles in their DNA that code for thelr blood type. es
Explain how an analysis of blood groups can be used to determine
paternity. (5)
(5)

2.5 Scientists wanted to determi

ne which type of inheritance accounted for
most of the selected genetic

disorders in dogs.

To do this they sequenced the genomes of many dogs of the same breed
that suffered from the genetic disorders.

The table below represents the type of inheritance for genetic disorders in

dogs.
TYPE OF PERCENTAGE OF GENETIC
INHERITANCE DISORDERS (%)

Autosomal recessive 71

Autosomal dominant 11

Sex-linked 10

Other 8
2.5.1 Define the term genome. (2)
2.5.2 State ONE factor that was kept constant in this investigation. (1)

2.5.3 Draw a pie graph to represent the information provided in the 6)
table.

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2.6 The diagram below shows parts of two different processes that take place in
the nucleus and a process taking place at the ribosome.

Bg

Ribosome
2.6.1 Identify:
(a) Molecule A (1)
(b) Organelle B (1)
(c) Molecule C (1)

(d) The number of codons shown in the diagram on the mRNA (1)

2.6.2 Give the anticodon for the second codon from left to right. (2)

2.6.3 Name AND describe the process that is taking place at X. (6)

264 Describe the role of mRNA in protein synthesis. (4)
20

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QUESTION &

3.1 The baobab tree belongs to the genus Adansonia. Different species are
found in Africa and Australia. Originally these two continents formed part
of Gondwanaland, as shown in the diagram below.

Gondwanaland

\

Australian baobab

Explain how the different species of baobab could have evolved on the
different continents. (6)

|

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3.2. Read the extract below.

Warblers/songbirds belong to a large group of birds grouped together in
the same bird family. In Siberia, Russia, two distinct species of greenish
blers coexist, one in the west and one in the east. A large region of
t, where they cannot survive, separates the two groups of warblers.
istributions only narrowly overlap in central Siberia, where they do
terbreed. The two species differ, in the songs that males sing to
t females. Also, females of each species do not recognise the song
other species but respond strongly to the song of the males from
their own species.

i

3.2.1. Name AND describe the reproductive isolation mechanism
described in the extract. (3)

3.2.2 Mention TWO other reproductive isolation mechanisms that keep
species separated. (2)

(5)

3.3. The diagrams below show the pelvic structure of two species not drawn

to scale.
A B
WIC

3.3.1 Which diagram represents the pelvis of a hominid? (1)

3.3.2 Explain ONE reason why Diagram B is better suited for a
bipedal mode of locomotion. (2)
3.3.3 Explain TWO advantages of bipedalism. (4)
(7)

aan]

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iloac estPy .
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The diagrams below r ,
epresert! the side and ventral views of the skulls

of three prmat

sapiens a re A gorilla. Australopthecus africanus and Homo

-_ he lagrams showing the site views are not necessarily in
e same order as the diagrams showing the ventral views

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3.4.1 Name the family that all these primates belong to. (1)

3.4.2 Which diagram (I, Il or Ill) represents the ventral view of
skull B? (1)

3.4.3 Give TWO visible reasons for your answer to
QUESTION 3.4.2. (2)

3.4.4. Describe THREE visible differences between the jaws in
diagrams II and Ill, which show trends in human evolution. (6)

3.4.5 Based on the differences in dentition, what conclusion.can be
made about the difference in the diet of the ogres in

diagrams II and III? (2)
He

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3.5 In guinea pigs, black fur (B), is due to a dominant gene and white fur (b)
is due to a recessive gene. Short fur (H) is due to a dominant gene and
long fur (h) is due to a recessive gene.
A bygeder made the following cross: BbHh x bbhh
3.940 State the phenotypes of the parents that the breeder crosse (2)

yi Give the genotypes of ALL the possible gametes of the parents. (2)

Give:
3.5:3
a) All the possible phenotypes of the offspring (2)
b) The dominant allele for fur colour (1)
3.5.4 Explain Mendel's law of independent assortment. (3)
(10)
3.6 Study the extract below.
Female fruit flies are approximately 2,5 mm long. Males are smaller and
possess a distinct black patch on their bodies. Females lay up to 400
eggs, which develop into adults in 7 to 14 days. Fruit flies can survive
and breed in small flasks containing a simple nutrient medium
consisting mainly of sugar. The gonosomes of fruit flies are similar to
that of humans (XX and XY).
3.6.1. Explain TWO reasons why the fruit fly is a useful organism for
studying genetic crosses. (4)
3.6.2 In fruit flies, the gene for eye colour is sex-linked and is carried
on the X chromosome.
The allele for red eyes R is dominant to the allele for white
eyes r.
A male with white eyes is crossed with a homozygous female
with red eyes.
Use a genetic cross to show the phenotypic ratio of the Fi
offspring. (8)
(10)
[50]

TOTAL SECTION B: 100
GRAND a 150
Uf

i

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education

Department:

Education

North West Provincial Government
REPUBLIC OF SOUTH AFRICA

NATIONAL SENIOR CERTIFICATE

a” © i © i if ie ie jennie
uu ty
" LIFE SCIENCES P2 "
" SEPTEMBER 2023 1"
" MARKING GUIDELINES :
“Sc cccccesdccccccccsesceccc we eccccceccceeeeese’

MARKS: 150

These marking guidelines consists of 12 pages.

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NSC — Marking Guidelines

PRINCIPLES RELATED TO MARKING LIFE SCIENCES

1.

10.

11.

12.

13.

14.

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 theYfirst three irrespective of whether all or some are correct/incorrect.

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

If comparisons are asked for, but descriptions are given
Accept if the 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 the sequence is muddled and links do not make sense

Where sequence and links are correct, credit. Where sequence and links are
incorrect, do not credit. If the sequence and links become correct again, resume
credit.

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

Wrong numbering
If the answer fits into the correct sequence of questions, but the wrong number
is given, it is acceptable.

If the language used changes the intended meaning
Do not accept.

Spelling errors
If recognisable, accept the answer, provided it does not mean-something else
in Life Sciences or if it is out of context.

If common names are given in terminology
Accept, provided it was accepted at the national memo discussion meeting.

If only the letter is asked for, but only the name is given (and vice versa)
Do not credit.

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15.

16.

17.

18.

19.

NSC — Marking Guidelines

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

Be sensitive to the sense of an answer, which may be stated differently.

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

Code-switching of official languages (terms and concepts)

A single word or two that appear(s) in any official language other than the
learner's assessment language 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.

Changes to the memorandum
No changes must be made to the memorandum. The provincial internal
moderator must be consulted.

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SECTION A
QUESTION 1
1.10 1.1.1
1.1.2
1.1.3
1.1.4
1.1.5
1.1.6
1.1.7
1.1.8
1.1.9
1.2 1.2.1
1.2.2
1.2.3
1.2.4
1.2.5
1.2.6
1.2.7
1.2.8
1.2.9
1.3 1.3.1
1.3.2
1.3.3
14 1.4.1
1.4.2
1.4.3
1.5. 1.5.1
1.5.2
1.5.3
1.5.4
1.5.5

NSC — Marking Guidelines

AVY
Biv v
BY v
AvY
Avy
BY
Byv
Dvv
Cvv

QuadrupedalismY

Peptidev

Craniumy

Gene mutation”

Trisomy 21v/Down Syndrome
Mitochondrial DNAY /mt-DNA
CloningY

Co-Dominancev

Homologous structures”

Both A and B¥v
A onlyv¥v
Nonevv

DNA profiling’
Baby 3vv

— Identifying suspects in a crimev
— Tracing missing personsY
— Identifying genetic disordersv

— Matching tissues for organ transplantsv

— Identifying dead personsY
(Mark first TWO only)

(a) Bacterial resistance”
(b) Effect of antibioticsv

Natural Selection’ /Punctuated Equilibrium
AntibioticsY
Resistant Bacteriav

25%

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NW/September 2023
(9x2) (18)
(9x1) (9)
(3x2) (6)

(1)
(2)
Any (2)
(5)
(1)
(1)
(1)
(1)
(1)
(1)
(6)

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NSC — Marking Guidelines

16 161 (a) 5Y (1)
(b) 3v (1)
(c) 3v (1)
1.6.2 Jtfa) Aav/AAY (2)
(b) Affected femalev (1)
(6)
TOTAL SECTION A: 50
SECTION B
QUESTION 2
2.1. 2.1.1. (a) HomologousY chromosomes/bivalent (1)
(b) Centromere” (1)
(c) Chromatidy (1)
2.1.2 Crossing overY (1)
2.1.3. Prophase Iv (1)
2.1.4 Ensures genetic variation’ (1)
2.1.5 — Chromosomes/chromatids arrange themselves randomly”
— oneither side of the equatory”
— during metaphase lv/II (3)
(9)

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NSC — Marking Guidelines

2.2 Meiosis | Meiosis Il

Crossing over takes placev | No crossing over takes
placev

In metaphase, the (In metaphase)
chromosomes align on the chromosomes align singly~
equator in homologous on the equator
pairsv
During anaphase, During anaphase,
chromosomesvY move chromatids” move towards
towards the poles the poles

Homologous chromosomes | Chromosomes occur singly

in prophase | ” in prophase Il v7
The chromosome number is_ | The chromosome number
halved” /changes from remains the samev
diploid to haploid
Results in two daughter Results in four daughter
cellsv cellsv
(Mark first THREE only) Any 3 x 2 +1 for table (7)
2.3 2.3.1 'Law' of use and disusev
‘Law' of the inheritance of acquired characteristics” (2)
2.3.2 — Acquired characteristics”
— are not inheritedv/do not cause any change to the DNA of
an organism's gametes (2)
(4)
2.4 — The blood the of mother, possible father and the child must be
comparedv

—_ If this shows that it is not possible that these parents can produce a
child with his/her blood group”

— then this man is not the father”

— If this shows that it is possible that these parents can produce a
child with his/her blood groupY

— then he may/ may not be the fatherY

— because other males have the same blood group” (5)
Any (5)

2.5 2.5.1 The mapping of the exact position of all the genes in allthe
chromosomes of an organism/humanv ~ (2)

2.5.2 The breed of the dogsv
(Mark first ONE only) (1)

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NSC — Marking Guidelines

2.5.3 Calculations

Total: 71 + 11+ 10+ 8 = 100

Autosomal recessive: 71/100 x 360 = 255,6°
Jautésomal dominant: 11/100 x 360 = 39,6°

dOn,
an 10/100 x 360 = 36°
er: 8/100 x 360 = 28,8°

PERCENTAGE GENETIC DISORDERS CAUSED BY

DIFFERENT TYPES OF IN
Other

8%

Sex -linked
10%

Autosomal
dominant
11%

Criteria for marking graph

HERITANCE IN DOGS

CRITERIA

Pie chart drawn (T)

Caption showing the relationship between
the two variables(H)

Correct calculation (C) to determine the
proportions for the labelled sectors

Plotting of Sectors (P)
Correct proportions for labelled sectors

1-3 sectors drawn and labelled correctly

All 4 sectors of the graph are drawn and
labelled correctly

Copyright reserved

Autosomal
recessive
71%
MARK
1
1
2
1 loon
—
2
(6)
(9)

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NSC — Marking Guidelines

2.6 2.6.1 (a) DNAY (1)
(b) NucleolusY¥ (1)

(© tRNAY (1)

(a) 3v (1)
2.6.2—UCCvv (2)

2.6.3 DNA Replicationv*
— The DNA (double helix) unwinds” and
—  _unzipsv/hydrogen bonds break
— to form two separate strandsY
— Both DNA strands serve as the templatesY
— to build a complementary DNAv’/(A to T and C to G)
— using free (DNA) nucleotides” from the nucleoplasm
— This results in two identical (DNA) moleculesY
- Each molecule consists of one original strand and one
new strandv *4 Compulsory + Any 5 (6)

2.6.4 -— mRNA forms during transcription”

— by copying the coded message from DNAY

— and moves out of the nucleusv

— and attaches to the ribosomev

— the anticodons on the tRNA match the codons on the

mRNAY
Any (4)

(16)
[50]

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NSC — Marking Guidelines

QUESTION 3

3.1 — The common ancestor of the original baobab
population/the baobab population of Gondwanaland was
separated into two populationsY

2) by the ocean/seav*

“| There was no gene flow between the African baobab and
Australian baobabv /two baobab populations.

— Each baobab population were exposed to different
environmental conditions ~/other selection pressures.

— Natural selections occur independentlyv in each of the
two baobab populations.

— Such that the individuals of the two baobab populations
become very different from each other.

— genotypically and phenotypically”

— even if the two baobab populations were to mix again”

— they will not be able to interbreed. v
The two baobab populations are now different baobab
species (in the question) (No marks for this statement)

*41 Compulsory + Any 5 (6)

3.2 3.2.1 -— They use species-specific courtship behaviour”*
- Males sing songs to attract female matesY
— and females do not respond/recognise songs from other
males¥ *1 Compulsory +2 (3)

3.2.2 — Breeding at different times of the yearvY
— Plant adaptations to different pollinators~
— Producing infertile offspringy’
— Prevention of fertilisation’
(Mark first TWO only) Any (2)
(5)

3.3 3.3.1 BY (1)
3.3.2 — The pelvis is short and wideY

— to support the body weight of the upper body”
(Mark first ONE only) (2)

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NSC — Marking Guidelines

3.3.3 — Frees the armsv

so that they could carry offspringv /tools/food/manipulate
things

— Allows the ability to see further”
to spot dangerv /food

4! Reduces the surface area exposed to the sun,”
so, less heat is absorbedv /less heat lost/improved
thermoregulation

— Expose the genitalsv
to attract the opposite sexv

— Efficient locomotion’
allows travelling longer distancesY

(Mark first TWO only) Any2x2 (4)
(7)
3.4 3.4.1. Hominidaey (1)
3.4.2 Diagram Illv (1)
3.4.3. — The foramen magnum is in a more forward position’
— The palate is more roundedv’
(Mark first TWO only) (2)
3.4.4 — The jaw/palate is rectangular in Diagram Il” and rounded in
Diagram IIlv

— Large spaces between the teeth in Diagram IIV and small/
no spaces between the teeth in Diagram III

— Large canines/teeth in Diagram II Y and small canines/
teeth in Diagram III v

(Mark first THREE only) Any3x2 (6)
3.4.5  — This organism in Diagram II had a diet of hard, raw foodv
- and the organism in Diagram III has a diet of softerv/
cooked food (2)
(12)
3.5 3.5.1 Phenotype Black short fur” x White long fur” (2)
3.5.2 Gametes BH, Bh, bH, bhvY (2)
3.5.3 (a) Black Short fur; Black Long fur; White Short fur; White
Long fury ¥ (2)
(b) Black fur (1)

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NSC — Marking Guidelines

3.5.4 — Because of the random arrangement of chromosomes at the
equator during meiosis“~/any one of two alleles of a
characteristic can sort with any two of another characteristic

— the alleles of different genes move independently of each
other into the gametes”
“! They can therefore appear in the gametes in different

combinations” (3)
(10)

3.6 3.6.1 = Females lay many eggs” which will ensure that the ratios
obtained will be reliablev”
— The eggs develop very quicklyv/7 to 14 days
therefore, results will be obtained fast”
— Can clearly differentiate between male and femalev
making it easy to breedv
— Fruit flies survive and breed in small flasks~
making it easy to observe and work withv
— The food medium required is simplev/cheap
making it easy/cheap to conduct the investigationv
(Mark first TWO only) (Any 2 x 2) (4)

3.6.2

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NSC — Marking Guidelines

Pi Phenotype: White-eyed male x red-eyed femalev

Genotype: x'Y x XRXR Vv
Meiosis
G/gametes x, Y XR XR v
Fertilisation [ = S ]
F1 Genotype XRXr XRXT XRY  XRY v

Phenotype: 2 red-eyed males : 2 red-eyed females
1 red eyed male : 1 red-eyed femalev*

P; and Fiv
Meiosis and fertilisation v
(*1 Compulsory + Any 5)

OR
P1 Phenotype: White-eyed male x Red-eyed femaleY
Genotype: x'Y x XRXR v
Meiosis
Gametes XR XR
x XR xr XR xr

Fertilisation

sain Y XRY XRY

1 mark for correct gametes”
1 mark for correct genotypesY

Fi Phenotype: 2 red-eyed males: 2 red-eyed females
1 red-eyed male: 1 red-eyed femalev*

Px and Fi ¥
Meiosis and fertilisation Vv (6)
(*1 Compulsory + Any 5) (10)
[50]

TOTAL SECTION B: 100
GRAND-TOTAL: 150

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