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KWAZULU-NATAL PROVINCE
EDUCATION
REPUBLIC OF SOUTH AFRICA
NATIONAL
SENIOR CERTIFICATE
GRADE 12
MARKS: 150
TIME: 2% hours
N.B. This question paper consists of 16 pages.
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Life Sciences Grade 12 Paper 2 KZN Prelim 2024 Memo
Grade 12 Life Sciences Paper 2 memorandum from the KZN Prelim Exam 2024, 27 pages. Read online or download the free PDF.
Life Sciences · Grade 12 · KZN Prelim Exam · 2024. Memorandum, 27 pages. Read online or download the PDF.
- Subject
- Life Sciences
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2024
- Exam period
- KZN Prelim Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 27
- File size
- 2.7 MB
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INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
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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Life SCS WA loa de d from Ss tanmorephysics.conv™"*™ Preparatory Examination
SECTION A
QUESTION 1
1.1. Various options are provided as possible answers to the following questions.
Choose the answer and write only the letter (A to D) next to the question
number (1.1.1 to 1.1.9) in the ANSWER BOOK, for example
1.1.10.D.
1.1.1. Which ONE of the following occurs in mitosis but NOT in meiosis?
Crossing over takes place
Two cells are formed at the end of the division
Homologous chromosomes arrange at the equator
Centrioles form at the poles of the cell
0QOWD>
1.1.2 The scientists who discovered that DNA is double helix.
Crick and Watson
Franklin and Wilkins
Watson and Wilkins
Wilkins and Crick
00ODND>
1.1.3. Which ONE of the following is an ac
Lamarck's theory of evolution? _»
eptable explanation based on
A __ An increasi at are resistant to insecticides
because they id reproduce
B__ Giraffes have lo ect d e have short necks
C Humans evolve 2
D- The ete ngers removed will give birth to
children withou P 5
1.1.4 Which ONE of the following is an acceptable explanation based on
Mendel’s law of segregation?
A Various ‘factors' controlling the different characteristics separate
out independently during gamete formation
B Two ‘factors' separate so that each gamete contains different
‘factors’
C Heterozygous individual for a particular characteristic will have the
dominant trait as the phenotype
D_ Two ‘factors' separate so that each gamete contains only one of
these ‘factors’
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1.1.5-..DNA was extracted from the cells of a cat and was analysed for its
nitrogenous base composition. It was found that 26% of the bases are
guanine.
What percentage of the bases are thymine?
A 26
B 52
Cc 24
D 48
1.1.6 A female that has received two identical alleles for haemophilia from
each parent is described as being ...
A dominant
B homozygous
C homologous
D heterozygous
1.1.7. The diagrams below represent the four phases of meiosis in no
sequence.
ACU)
K
XZ
Which ONE of the following is the correct sequence in which the phases
of meiosis occur?
PPAw
Oane
NNBPN
=W0-—
0C0NOWDDY
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1.1.8--A mother has blood type O and father blood type AB. They have two
biological children and two adopted. The genotypes of all children are
represented in the table below.
CHILDREN GENOTYPE
Peter a
Paul ii
Penny 18)
Patricia Ae
Children adopted are ...
won P -2
1.1.9 The diagram below shows the process fertilisation in humans.
Peter and Paul
Paul and Penny
Penny and Patricia
Patricia and Paul
Gamete A
Gamete B
Which ONE of the following correctly represents the chromosomes in
gamete A and B respectively?
0 QO BD PY
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GAMETE A GAMETE B
23 autosomes and Y 21 autosomes and X
gonosome gonosome
22 autosomes and Y 22 autosomes and X
gonosome gonosome
22 autosomes and X 22 autosomes and X
gonosome gonosome
22 autosomes and X 23 autosomes and X
gonosome gonosome
(9 x 2) (18)
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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. A disorder due to the presence of an extra chromosome on the
chromosome pair 21 in somatic cells of humans
1.2.2 Similar structures that are inherited from a common ancestor and
modified for different function
1.2.3. Asegment of a chromosome that codes for a particular characteristic
1.2.4 The stage in protein synthesis during which mRNA is coded from DNA
1.2.5 Undifferentiated cells that have the potential to form any body tissue
1.2.6 The study of the distribution of living organisms found in different
regions
1.2.7 The type of nucleic acid that carries a specific amino acid
1.2.8 The chromosome condition describing the presence of a single
set of chromosomes in a cell
1.2.9 An allele that is expressed phenotypically only in the homozygous
condition
(9x 1)
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 1.3.2) in the
ANSWER BOOK.
COLUMN | COLUMN II
1.3.1. | The random arrangement of A: Metaphase |
chromosomes at the equator singly B: Metaphase II
1.3.2 | The natural shape of the DNA molecule A: Double helix
B: Single helix
1.3.3. | The site of meiosis in plants A: Stigma
B:_ Anther
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(3 x 2)
September 2024 Preparatory Examination
(9)
(6)
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1.4 The diagram below shows a process during meiosis.
The letters represent the alleles of genes in the chromosomes.
1.4.1
1.4.2
1.4.3
1.4.4
1.4.5
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Name the:
(a) Process shown in the diagram
(b) Phase of meiosis represented above
Identify:
(a) PointD
(b) PartB
Write down the LETTER and NAME of the part that join together two
chromatids.
Write down the genotype of the resulting chromatid A after the process
named in QUESTION 1.4.1 (a).
State how many:
(a) Characteristics are represented in the diagram above.
(b) Different genes are represented by the alleles in the diagram
above.
September 2024 Preparatory Examination
(1)
(1)
(1)
(1)
(2)
(1)
(1)
(1)
(9)
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1.5 Fossils are used as the evidence for human evolution.
The table below shows the period of existence of some hominids.
HOMINID PERIOD OF EXISTENCE
Australopithecus aferensis 4-3 mya
Australopithe: ricanus 3.2-2 mya
2.2-—1.4 mya
2-0.4 mya
0.4 - 0.05 mya
0.2—0 mya
1.5.1 With refere’ f above, name TWO:
(a) Genera
(b) Species that existed for the longest period of time.
1.5.2 State how long did Homo habilis live on earth.
1.5.3 Name:
(a) TWO fossils of Australopithecus africanus that were discovered in
Sterkfontein.
(b) The family to which Homo sapiens belong.
TOTAL SECTION A:
September 2024 Preparatory Examination
(2)
(2)
(1)
(2)
(1)
(8)
50
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SECTION B
QUESTION 2
2.1. Pepsin (protein digesting enzyme) is a protein secreted in the stomach as an
inactive pepsinogen.
The TABLE 1 below shows the sequence of bases on the mRNA coding for
some of the amino acids in a pepsin protein.
1 2 3 4 5 6 if
UUC | AAG | CUG | UGG | GAU | AUC | UGG
2.1.1. Give the number of:
(a) Different anticodons needed to form this part of a pepsin protein. (1)
(b) Phosphates on the mRNA base triplets in the table above. (1)
2.1.2 Name the organelle where pepsin is synthesised in a human cell. (1)
2.1.3. The TABLE 2 below shows the amino acids and DNA base triplets.
DNA BASE
AMIND, ed TRIPLET
ci [tac
With reference to the TABLE 1 in QUESTION 2.1 and TABLE 2 above:
(a) | Give the correct sequence of the tRNA base triplet that brings
tyrosine for the synthesis of pepsin. (1)
(b) | Give the amino acid coded for by the mRNA base triplet number 3. (1)
(c) A mutation caused the mRNA base triplet number 6 to change
from UUC to AUC.
Explain the effect of this mutation on the protein formed. (4)
(d) | Describe the type of mutation referred to in QUESTION 2.1.3 (c). (2)
(11)
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2.2 Read the information below.
DNA profiling is a forensic technique in criminal investigations. It is a tool for
determining the identity of a specific DNA sample to solve legal disputes and
notably help to determine paternity.
Faulty preparation procedures during DNA profiling may cause problems. Other
errors that may occur during DNA profiling include contamination of samples,
intentional planting of DNA samples, and mistakes during interpretation of
results. All this may lead to discrepancies in legal proof in court cases.
2.2.1. From the information above, state ONE:
(a) Use of DNA profiling other than determining paternity. (1)
(b) Effect of faulty preparations during DNA profiling. (1)
2.2.2 Explain how the DNA profiling is used in paternity testing. (6)
(8)
2.3 In cattle, long horns (L) and a twisted horns (T) are dominant over short horns
and straight horns.
The diagram below shows part of a genetic cross between a bull and cow. A
cow is heterozygous for the length of horns.
P1 Bull Xx Cow (Long and straight horn)
Genotype i Xx A
Phenotype 6 long and twisted horns _ 6 long and straight horns
offspring
2 short and twisted horns 2 short and straight horns
2.3.1 Name the type of dominance displayed by the horn length and shape. (1)
2.3.2 Give:
(a) The phenotype of a bull. (1)
(b) ALL the possible genotypes in the gametes of a bull. (3)
(c) The genotype at A above. (1)
(d) The phenotypic ratio for short and straight horns to long and twisted
horns offspring. (1)
(7)
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2.4 The diagram below shows the inheritance of fur colour in rabbits.
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September 2024 Preparatory Examination
m
O-
Og |
© @
| Brown fur
* Brown fur OC) wnite fur
White fur
2.4.1 Name the phenotype that is controlled by the recessive allele.
2.4.2 Explain your answer in QUESTION 2.4.1.
24.3
2.4.4 Give a reason for your answer in QUESTION 2.4.3.
(1)
(4)
Identify the parents that support the conclusion that both rabbits with
brown and white fur coat belong to the same species.
(1)
(2)
2.4.5 State how many generations in the diagram have heterozygous rabbits? (1)
2.4.6 Rabbit 11 was crossed with a white fur coat colour rabbit.
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Using the letters B and b, do a genetic cross to show the phenotypic
ratio of their offspring.
(7)
(16)
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2.5 Read the passage below.
Farmers can kill weeds by spraying glyphosate which targets EPSPS
(enzyme) which is essential for plant life. Weeds are consistently developing
resistance to herbicides.
Soybean was genetically modified using glyphosate-resistant gene from the
cell_of a resistant weed. Genetically modified organism (GMO) such as
soybean has an impact on food security and resistance to diseases.
The environmental impact of genetically modified soybeans includes the
increased herbicide usage which pollutes air. The consumption of GMO foods
can cause allergic reactions and increased antibiotic resistance.
2.5.1. Define the term genetically modified organism. (2)
2.5.2 From the passage, give TWO:
(a) Negative effects of GMO in human health (2)
(b) Benefits of GMO in agriculture (2)
2.5.3 Explain why the gene of a weed cell was used in modifying soybean. (2)
(8)
[50]
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QUESTION 3
3.1. Read the information below.
Colour blindness (b) occurs when there is a problem with the pigments in
certain nerve cells of the photoreceptors in the retina. Parents pass down the
red-green colour vision deficiency to the offspring.
The diagram below represents the human somatic cell.
484.28 83 aK x3
“AR GH BH BS BS 4B
13 14 15 16 17 18
=
7
XR KR AA AA 09
19 20 21 22 23
3.1.1 Identify the diagram above. (1)
3.1.2 Name the type of photoreceptors referred to in the information above. (1)
3.1.3 Give the:
(a) Collective name for the type of chromosomes in pair numbered 23 (1)
(b) Genotype of the male who is colour blind. (1)
3.1.4 State THREE characteristics of the autosomal homologous chromosome
pairs. (3)
3.1.5 Give ONE reason why this diagram represents the human somatic cell. (1)
(8)
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3.2 There are insects called Fireflies, their blood contain a toxic chemical called
lucibufagin. They produce a glowing light to warn predators that they are toxic.
A scientist conducted an investigation to determine the relationship between the
toxic chemical and survival of the insects against predation.
Fireflies and non-toxic fruit flies were collected (200 each).
Both insects were released into the greenhouse structure in a controlled
environment.
e The lizards of the same species that are feeding on insects were released
into the greenhouse structure for one day.
e The number of survived insects was counted and recorded 3 times a day.
The percentage of insects survived are shown in the graph below.
100
90
80
70
60
50 =
40 - —
30 4— “all @Fruit flies
el OFireflies
20 7—
10 12 14
Time (Hours)
Percentage of insects survived
3.2.1. State the aim of the investigation. (2)
3.2.2 Identify the dependent variable of the investigation. (1)
3.2.3 State THREE planning steps to consider when conducting this
investigation. (3)
3.2.4 State ONE way in which the scientist ensured the:
(a) Validity of the results (1)
(b) Reliability of the results (1)
3.2.5 Calculate the number of fruit flies that have survived from predation in
10 hours. Show ALL workings. (3)
(11)
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3.3. Describe the process of speciation through geographic isolation. (6)
3.4 Different insecticides were used to control weeds in fields. The insecticides kill
insects, Random mutations within the insects over time, has caused the insects
to be resistant to insecticides. Colorado potato beetle has a remarkable ability to
develop resistance to a wide range of insecticides.
Colorado potato beetle has shown a continuous resistance to different
insecticides since 1960.
The table below shows the number of different insecticides that were resisted by
the Colorado potato beetle over years.
Year Number of insecticides
that were resisted
1960 2
1970 10
1980 28
1990 38
2000 40
2010 40
3.4.1. State ONE way in which the resistance of Colorado potato beetle was
determined in the investigation over years. (1)
3.4.2 Describe the change in insecticide resistance over time. (3)
3.4.3. Explain the economic impact caused by the Colorado potato beetle
resistance to insecticides. (3)
3.4.4 Use Darwin's theory of natural selection to explain the evolution of
insecticide-resistant Colorado potato beetle. (6)
[13]
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3.5 The dia
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grams below show the pelvises of the African ape and modern human.
The diagrams are not drawn to scale.
3.5.1
3.5.2
3.5.3
3.5.4
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Identify a diagram of the pelvis above that belongs to an African ape?
Tabulate TWO visible differences between pelvis in diagrams A and B.
Explain the significance of the position of the foramen magnum that is
associated with the pelvis in diagram B.
The ‘Out of Africa’ hypothesis is one explanation of the evolution of
modern humans.
Describe how genetic evidence is used to support the ‘Out of Africa’
hypothesis.
TOTAL SECTION B:
GRAND TOTAL:
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(1)
(2)
(4)
(12)
[50]
100
150
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KWAZULU-NATAL PROVINCE
EDUCATION
REPUBLIC OF SOUTH AFRICA
NATIONAL
SENIOR CERTIFICATE
GRADE 12
~ =i § ,
gamorephwies.com
MARKS: 150
This marking guideline consists of 11 pages.
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ices P2 September 2024 Preparatory Examination
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PRINCIPLES.RELATED TO MARKING LIFE SCIENCES SEPTEMBER 2024
10.
11.
12.
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.
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13.
14.
15.
16.
17.
18.
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ices P2 September 2024 Preparatory Examination
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If common names given in terminology
Accept provided it was accepted at the National memo discussion 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
Be sensitive to the sense of an answer, which may be stated in a different way.
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 appears in any official language other than the learners’
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.
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SECTION A
QUESTION 1
1.1 fie AyByvY
1.7.2- Avv
1-1-3- 9D
11.9 Cvv
(9x2) (18)
1.2 1.2.1 Down Syndrome ¥
1.2.2 Homologous structure Vv
1.2.3 Genev
1.2.4 Transcription Vv
1.2.5 Stem cells v
1.2.6 Biogeography Vv
1.2.7 tRNAY
1.2.8 Haploid v
1.2.9 Recessive Vv
(9x1) (9)
1.3. 1.3.1 Bonlyvv
1.3.2 Aonlyvv
1.3.3 Bonly vv
(3x2) (6)
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14 1.44
1.4.2
1.4.3
1.4.4
1.4.5
1.5 1.5.1
1.5.2
1.5.3
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(a) Crossing over v
(b) Prophase | ¥
(a) Chiasma v
(b) Chromosome ¥
CY Centromere v
rTnD V (any order)
(a) 4v
(b) 4v
(a) - Australopithecus “
- Homo ¥
(b) - Australopithecus africanus
- Homo erectus ¥
0.8 million years V / 800 thousand years
(a) - Mrs Ples v
- Little Foot v
(b) Hominidae v
September 2024 Preparatory Examination
(2)
(2)
(1)
(2)
(1)
(8)
TOTAL SECTION A: 50
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SECTION B
QUESTION 2
21 244 (a) 6v (1)
(b) 21” (1)
2.1.2. Ribosome ¥ (1)
2.1.3. (a) AUGY (1)
(b) Leucine ¥ (1)
(c) iffe in willbe formed v
leucine Y in a protein Any (4)
(d) -C e of bases v / nucleotides
-a gene mutation will occurY Any (2)
(11)
2.2 2.2.1 (a) Determining identity of criminals ¥ (1)
(Mark the first ONE only)
(b) Discrepancies in legal proof in court cases v (1)
(Mark the first ONE only)
2.2.2 -Achild received DNA from both parents v
- DNA profiles of the mother, child and the possible father are
determined v
- DNA bars of the mother and the child are compared v
- Remaining DNA bars are compared to the possible father's DNA barsv
- If all the remaining DNA bars in the child’s profile match the possible
father’s DNA bars¥
- then the possible father is the biological fatherv
- If all the remaining DNA bars in the child’s profile does not match the
possible father DNA barsv
- then the possible father is not the biological fathery’ Any (6)
(8)
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2.3 2.3-4
2.3.2
24 2.4.1
2.4.2
2.4.3
2.4.4
24.5
2.4.6
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Complete dominanceY
(a) Long and twisted hornsv
(b) -LT
-Ulyyy
-IT
-It
(c) Littv
(da) 1:3¥
White fury”
- Both individual 3 and 4 have brown fur”
- The offspring 7 has white furw”
- which is expressed in the phenotype of homozygous condition’
- Each parent must be heterozygousv /carrying recessive allele
- The offspring inherited recessive allele from both parentsY
5 and 6 v
- They interbreed v
- and produce a fertile offspring v
4v
P, Phenotype Brown fur x White fury”
Genotype Bb x bby
Meiosis
Gametes
Fertilisation
Fy Genotype
Phenotype
Phenotypic ratio:
Pi and Fiv
September 2024 Preparatory Examination
(1)
(1)
(3)
(1)
(1)
(7)
(1)
(4)
(1)
(2)
(1)
Meiosis and fertilisation’ Compulsory mark 1* + Any 6
OR
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P, Phenotype Brown fur x white fury’
Genotype Bb. x bbv
Meiosis .
Fertilisation
Fy
Phenotype 2 brown fura
Phenotypic ratio: 1:1v*
Pi and Fiv
Meiosis and fertilisation’ Compulsory mark 1* + Any 6 (7)
(16)
2.5 2.5.1. Anorganism in which the genetic material is manipulated for the benefit of
a human desirable characteristicv Vv (2)
2.5.2 (a) - Allergic reactionsY
- Antibiotic resistancev (2)
(Mark the first TWO only)
(b) - Food security”
- Resistance to diseasesY (2)
(Mark the first TWO only)
2.5.3. - To produce a glyphosate-resistant soybeanv
- since weeds are resistant to glyphosateY (2)
(8)
[50]
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QUESTION 3
3.1 3.4.1-— KaryotypeY (1)
3.1.2 - Conesv¥ (1)
3:1.3.. (a) Gonosomesv/sex chromosomes (1)
(b) xeyv (1)
3.1.4 - Similar sizev
- Similar shapev
- Same location of centromereY
- Position of genev’/ alleles
- Genes coding for the same characteristicsY Any (3)
3.1.5 - 46 chromosomesv / 23 pairs of chromosome (1)
(8)
3.2 3.2.1. To determine the relationship between the toxic chemicals and survival
of insects Vv (2)
3.2.2 Survival of insects” (1)
3.2.3. - Obtain permission from the greenhouse owner to conduct investigation’
- Decide on the sample sizev
- Decide on the recording toolY
- Decide on time/datev
- Decide on the insect types”
- Decide on the predator typev Any (3)
(Mark the first THREE only)
3.2.4 (a) - Lizards of the same speciesY were used
- The ratio of fire-flies to fruit-flies”
- Same environmental conditions” in greenhouse structure Any (1)
(Mark the first ONE only)
(b) 200 insects of each were usedY (1)
(Mark the first ONE only)
3.2.5 rd v x 200¥
= 120 v (3)
(11)
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3.3. - Ifa population of a single species becomes separated by a geographical
barrierY (sea, river, mountain, lake)
- then the population splits into two”
- There is now no gene flow between the two populations”
- Since each population may be exposed to different environmental
conditions~/the selection pressure may be different
- natural selection occurs independentlyv in each of the two populations
- such that the individuals of the two populations become very differentv from
each other
- genotypically and phenotypically”
- Even if the two populations were to mix againv
- they will not be able to interbreedv
- The two populations are now different species” Any (6)
3.4 3.4.1 By counting the number of different insecticides that Colorado potato
beetle resistedv (1)
3.4.2 - Insecticide resistance increasedv
- from 1960 to 2000Y
- and remained constant in 2010v (3)
3.4.3 - Buying different types of insecticidesv
- will be very expensivev
- leading to a greater loss of money ¥ (3)
3.4.4 - Colorado potato beetle show variation’
- Some have insecticide-resistant and others do notY
- If there is an application of insecticidesY
- Colorado potato beetle with resistance surviveY
- whilst those without resistance died¥
- those that survive, reproducev
- and pass the allele for resistance to their offspringv
- The next generation will therefore have a higher proportion of Colorado
potato beetle with resistancev (6)
(13)
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35 35.1 AY
Rese F 4 ed
A B
It is longv It is shortv”
Lis, Itis widev
(1 mark for table + 4)
only)
35:35 ar jon’
rt th
= for bip v
Od : turev Any
3.5.4 - Every cell has mitochondrion with DNAv
- Mitochondrial DNA is only inherited through the maternal line
- The analysis of mutations on the mitochondrial DNAY
- shows that the oldest female ancestor was located in AfricaY
- and that all humans descended from herv /mitochondrial Eve Any
TOTAL SECTION B:
GRAND TOTAL:
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(5)
(2)
(4)
(13)
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