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REPUBLIC OF SOUTH AFRICA
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Life Sciences Grade 12 Paper 2 North West Prelim 2024 Memo
Grade 12 Life Sciences Paper 2 memorandum from the North West Prelim Exam 2024, 29 pages. Read online or download the free PDF.
Life Sciences · Grade 12 · North West Prelim Exam · 2024. Memorandum, 29 pages. Read online or download the PDF.
- Subject
- Life Sciences
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2024
- Exam period
- North West Prelim Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 29
- File size
- 2.8 MB
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INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1.
PF ON
= © @Onoeaga
11.
Answer ALL the questions.
Write ALL the answers in the ANSWER BOOK.
Start the answer 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, tabels or flow charts only when asked to do so.
The diagrams in this question paper are NOT necessarily all drawn to scale.
Do NOT use graph paper.
You must use a non-programmable calculator, protractor and compass where
necessary.
Write neatly and legibly.
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SECTION A
QUESTION 4
1.1 Various options are provided as possible answers to the following questions.
Choose the answer and write only the letter (A-D) next to the question numbers
(1.1.1 to 1.1.9) in the ANSWER BOOK, e.g 1.1.10 D.
1.1.1 During which phase of meiosis do homologous chromosomes split and
move to the poles?
A Anaphase II
B Telophase |
(o} Anaphase |
D Prophase |
1.1.2 The list below describes features of evolutionary theories:
(i) Organisms show a great deal of variation
(ii) Long periods of time during which species do not change
(iii) Frequent use of organs results in them becoming stronger,
better and bigger
(iv) Favourable characteristics are passed on to the next
generation
Which ONE of the following combinations of features gives the
CORRECT characteristics of natural selection?
A (i) and (iv) only
B (iii) and (iv) only
Cc (ii) and (iii) only
D (i) and (iii) only
1.1.3. The table below shows the DNA triplets that code for some amino acids.
DNA triplet
a: ==
The anticodon for the amino acid leucine will be ...
A ACA.
B CUU.
Cc GAA.
D AGA.
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1.1.4— In humans, brown eye colour is dominant over blue eye colour.
When one parent has blue eyes and the other parent has brown
eyes and both are homozygous for the trait, what are the chances
that their children will have blue eyes?
A 100%
B 75%
Cc 50%
D 0%
Which ONE of the following monohybrid crosses will result in a
phenotypic ratio of 1: 1?
A Both parents are homozygous for the dominant
characteristic.
B Both parents are homozygous.
Cc a néte! ozygous and the other parent is
D jgous and the other parent is
The charact oth humans and African apes are ...
A al dimorphism and bipedalism.
B ual dimorphism and quadrupedalism.
Cc large cranium, small canines and prognathous jaw.
D small cranium, large canines and prognathous jaw.
The first primate to use tools constantly was ...
A Homo neanderthalensis.
B Homo sapiens.
Cc Homo floresiensis.
D Homo habilis.
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1.1.8 The diagram below represents different phases of meiosis.
Diagram 1 Diagram 2 Diagram 3
The sequence in which the phases of meiosis occur are diagrams ...
A 1,3 and 2.
B 3, 1 and 2.
Cc 2, 1 and 3.
D 1,2 and 3.
1.1.9 Asample of DNA has 80 guanine bases and 40 adenine bases.
How many phosphate molecules would you expect in ONE strand of
this sample of DNA?
A 40
B 60
Cc 120
D 240
(9x2) (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 numbers (1.2.1 to 1.2.9) in the
ANSWER BOOK.
1.2.1. The type of vision present in both apes and humans that allows for
depth perception
1.2.2 The monomers of nucleic acids
1.2.3 An explanation describing evolution as consisting of long phases of
little change alternating with short phases of rapid change
1.2.4 The phase in the cell cycle during which DNA replication occurs
1.2.5 The type of variation in a population in which there is no range of
intermediate phenotype
1.2.6 A sub-group of the hominids that includes only modern humans and
early human ancestors
1.2.7. Undifferentiated cells that can develop into any type of cell, tissue or
organ
1.2.8 The characteristics that are influenced by alleles carried on the
gonosomes
1.2.9 The biotec nological p in of genetically identical offspring
ree (9x1) (9)
1.3 Indicate whether nents in COLUMN | apply to A ONLY,
B or NONE of the items in COLUMN II. Write A only,
B ONLY, BOTH
B only, both Aa none next to the question numbers (1.3.1 to 1.3.3)
in the ANSWER |
COLUMN | COLUMN II
1.3.1 The genotype of a male x¢Y
with colour blindness x9x¢d
1.3.2 Evidence of evolution Modification by descent
Fossil record
Raymond Dart
Watson and Crick
1.3.3 Discovered the structure
of the DNA molecule
DPMP) OS
(3x2) (6)
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1.4 The diagram below shows the DNA profiles of a girl, her mother and four males.
1.4.1. Which male is the girl's biological father?’
r Males
Girl Mother 7 2 3 7
ees |
ey | es | es | ee
— ees | ee | ee
Se
— == ==
=o
==
fee | aoe
rss Ss
=
1.4.2 Explain your answer in QUESTION 1.4.1.
1.4.3 State TWO other functions of DNA profiling.
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(1)
(2)
(2)
(5)
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1.5 Some farmers add low doses of antibiotics to the cattle feed which could
result in the evolution of antibiotic resistant bacteria.
The graph below shows the effect of different doses of antibiotics on the
number of harmful bacteria.
Effect of different doses of antibiotics on the number of
harmful bacteria in live stock
1200
1000
Number of harmful bacteria
2 4 6 8 10
Antibiotic dose (mg)
1.5.1. Name the type of evolution that could result in the evolution of
antibiotic resistant bacteria. (1)
1.5.2 Identify the:
(a) Independent variable (1)
(b) Dependent variable (1)
1.5.3 Describe the trend observed in the graph. (2)
1.5.4 Use evidence from the graph above to explain why higher doses of
antibiotics will benefit the farmer economically. fA
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1.6 Albinismis caused by an autosomal recessive allele (a).
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The diagram below shows the inheritance of the albinism in a family.
oO
John Lizzy Mary
| Ha
oe) ae
Normal male C) Normal female
mi Affected * Affected
male female
KEY :
1.6.1 Give:
(a) Mary's phenotype (1)
(b) Lizzy’s genotype (1)
(c) John’s genotype (1)
1.6.2 Explain why Bob and Ann are both heterozygous for albinism. (2)
(5)
TOTAL SECTION A: 50
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SECTION B
QUESTION 2
2.1 The diagram below represents an animal cell in a phase of meiosis.
2.1.1 Name structure:
(a) A (1)
(b) B (1)
2.1.2 Identify the phase represented in the diagram above. (1)
2.1.3. Give a reason for your answer to QUESTION 2.1.2. (2)
2.1.4 | Which phase occurred before to the one represented in diagram? (1)
2.1.5 Explain the consequences if the process, causing variation, did not
occur between chromatids from the cell above. (3)
2.1.6 | How many cells will be formed at the end of this cell division? (1)
2.1.7. Tabulate TWO differences between Meiosis | and Meiosis II (5)
(15)
2.2 Lamarck’s theory of evolution is based on TWO Laws or Principles.
The Law of use and disuse and the Law of inheritance of acquired
characteristics
2.2.1 Explain the Law of inheritance of acquired characteristics
according to Lamarckism. (2)
2.2.2 Explain why Lamarck’s ideas are NOT accepted in the science
community today. (2)
(4)
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2.3 There are four blood groups in humans.
Blood group A, B, AB and O are regarded as phenotypes.
2.3.1. Explain why blood groups are controlled by multiple alleles. (2)
2.3.2. How many of these alleles can an individual inherit in the
genotype? (1)
2.3.3. Explain the role of blood grouping in paternity testing. (2)
(5)
2.4 Scientists wanted to determine what type of genetic disorder are found
mostly in cats.
To do this they took 200 cats of the same breed suffering from various
genetic disorders.
The table below represent the types of genetic disorders and percentage of
each:
a, Percentage of genetic
Type of genetic disorders disordars
Diabetes 45
Polycystic Kidney Disease 25
Deafness 20
Other 10
2.4.1 Explain the concept gene mutation. (2)
2.4.2 State ONE way in which the reliability of the investigation was
achieved. (1)
2.4.3 Draw apie chart to represent the data in the table above. (6)
(9)
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2.5 The diagram below shows the process of protein synthesis.
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aOoGOO—x
‘namo
z!
Process V Ss
2.5.1 Identify:
(a) Process V (1)
(b) Two places where molecule X is found (2)
(c) Nitrogenous base A (1)
2.5.2 Describe the process occurring at Y. (6)
2.5.3. Give the anticodon for the nitrogenous bases labelled Z. (1)
2.5.4 If the 3% base on the molecule S changed to C, explain the
consequences to the protein that will be formed. (3)
2.5.5 The table below shows the amino acids that correspond to
different DNA codes.
Amino acid DNA code
Glycine GGT
Arginine TCT
Methionine TAC
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Write down the sequence from left to right of amino acids coded by
molecule S.
(3)
(17)
[50]
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QUESTION 3
3.1 Shrubs of the family Proteaceae can be found in Australia, South America and
parts of Africa as shown on the map below.
South
America :
Africa Australia ?
3.1.1. Define the term species. (3)
3.1.2 Explain how the different species of shrubs could have evolved in
South America, Africa and Australia. (7)
3.1.3 Mention TWO reproductive isolation mechanisms that keep species
separated. (2)
(12)
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3.2 The diagrams below represent the pelvic structure and the ventral view of the
skulls of three organisms.
3.2.1 Write down the LETTER of the skull that shows a foramen magnum of
H. sapiens. (1)
3.2.2 Explain your answer to QUESTION 3.2.1. (2)
3.2.3 Describe how pelvis D is representative of a quadrupedal organism. (2)
3.2.4 Describe TWO visible differences between the jaws of skull K and L. (4)
(9)
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3.3 The diagram below shows a possible representation of human evolution.
aferensis
Ardipithec:
Australopithecus .”
omo sapiens
Homoerectus_ .
Homo habilis
ecus africanus
Paranthropus robustus
sis innit iad boisei
45 40 35 30 25 20 15 10 05 0
Million years ago
3.3.1. Name the type of diagram shown above.
3.3.2. How long ago did Australopethicus afarensis become extinct?
3.3.3 Identify TWO species that coexisted with Homo erectus.
3.3.4 State ONE advantage of a larger cranial capacity in H. sapiens.
3.3.5 Calculate the time difference between the evolution of Homo
erectus and Homo sapiens.
3.3.6 Describe how the fossils of the species in the diagram provide
evidence for the Out of Africa hypothesis.
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(1)
(1)
(2)
(2)
(2)
(2)
(10)
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3.4 In rabbits, brown fur (B) is dominant to white fur (b) and long ears (E) are
dominant to short ears (e).
A rabbit that is heterozygous for both characteristics is crossed with a
rabbit with white fur and short ears.
3.4.1. How many genes of the rabbits are considered in this cross? (1)
3.4.2 State the genotypes of the parents. (2)
3.4.3 Give:
(a) Phenotype of rabbit that is dominant for both characteristics (2)
(b) All the possible genotypes in the gametes of the
heterozygous parent (2)
3.4.4 State Mendel’s Law of independent assortment. (2)
(9)
3.5 Read the information below
Familial hypercholesterolemia, a disorder that is passed down
through families, causes LDL (bad) cholesterol level to be very high.
This disorder shows incomplete dominance.
Allele H“ codes for normal LDL receptor. Allele HS codes for the
inability to make a LDL receptor. Individuals with genotype HNH" are
normal and those with genotype HSHS have severe
hypercholesterolemia. Heterozygous (HHS) individuals have a mild
form of the disease.
3.5.1 Describe what is meant by incomplete dominance. (2)
3.5.2 Give evidence from the information above that shows that
inheritance of familial hypercholesterolemia is an example of
incomplete dominance. (2)
3.5.3 A man, heterozygous for familial hypercholesterolemia marries a
woman who has normal LDL receptors.
In a genetic cross show the percentage chance of them having
children with severe hypercholesterolemia. (6)
(10)
[50]
TOTAL SECTION B: 100
GRAND TOTAL: 150
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education
Department:
Education
North West Provincial Government
REPUBLIC OF SOUTH AFRICA
SERB BEER BEB REEE BEEBE BEEBE
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SERB ERE RR RRR RRR RRR ERR
MARKS: 148
These marking guidelines consist of 13 pages
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NSC — Marking Guidelines
PRINCIPLES RELATED TO MARKING LIFE SCIENCES
1s
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 but 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 are
incorrect, do not credit. If sequence and links become 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 recognisable, accept the answer, 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 discussion meeting.
If only letter is asked for and only name is given (and vice versa) Do not credit.
If units are not given in measurements
Candidates will lose marks. Memorandum will allocate marks for units separately.
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NSC — Marking Guidelines
16. Be sensitive 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’ 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.
19. No changes must be made to the marking guidelines without consulting the
Provincial Internal Moderator.
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NSC — Marking Guidelines
SECTIONA
QUESTION 1
1.1 wih Cyvv
5 Se 9 AYvY
177.3 Cvv
1.1.4 Dvv
121:5, Dvv
4; 6 i AJ
14
1
A. (9x2) (18)
unctuated equilibriumy
1.2.4 Interphasev
125 Discontinuous” (variation)
1.2.6 HomininsY¥
1.2.7 Stem cellsY¥
1.2.8 Sex-linkedv (characteristics)
1.2.9 CloningY (9x1) (9)
1.3 133.1 Aonly¥¥
1:3,2 Both A and BY ¥
13.3 B only¥v (3x2) (6)
1.4 1.41 3v (1)
1.4.2 The 2/3 DNA bands/black bars on the DNA profile of male 2/3Y
match with the 3 bands on the profile of the girlY (2)
1.4.3 - Identify suspects of crimev
- Determine the probability or causes of genetic defects /identify
the cures for genetic disorders
- Establish the compatibility of tissue types for organ transplants”
- Identify missing relatives
- To identify dead bodiesY Any (2)
(Mark the first TWO only) (5)
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NSC — Marking Guidelines
1.5 45.4 Natural selection’ (1)
1.5.2 (a) (Different) doses of antibioticv (1)
(b) Number of harmful bacteriaY (1)
L543 Higher doses of antibioticY results in lower number of harmful
bacteriav
OR
Lower doses of antibiotic Y result in higher number of numbers of
bacteriaY (2)
1.5.4 It decreases the number of harmful bacteriaY the most thereby
preventing diseases in the cattlev/resulting in less medical
expenses.
OR
Decreasing mortality of cattlkev by maintaining the number of cattle
to sell” /breed/make profit
(Mark first ONE only) Any (1 x 2) (2)
(7)
1.6 1.6.1 (a) Affectedv female (1)
(b) Aav (1)
(c) aay (1)
1.6.2 They have an affected son/a son with albinism/son with
aa genotype who had inherited one recessive allele from each of the
parentsY
OR
Both Bob and Ann each have parents who are affectedvY and
therefore inherited recessive alleles from each parent”. (2)
(5)
TOTAL SECTION A: 50
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SECTION B
QUESTION 2
2.1 2.1.1
2.1.2
2.1.3
2.1.4
2.1.5
2.1.6
2.1.5
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(a)
hysics.com Nw/September 2024
NSC — Marking Guidelines
Spindle fibrev/spindle thread (1)
Cell membraneY (1)
(b)
Anaphase I|¥
The centromere splits/spindle fibres contractY and the (two)
chromatids are pulled to opposite poles ¥
Metaphase Il “
(1)
(2)
(1)
- Result in similar/same gametesY
- offspring of the same genotype and phenotype” being formed
- reduced chance of survival/increased chance of extinction and
- loss of biodiversityY
(Mark first THREE only)
Four” /4
Any
Tv
(3)
(1)
Meiosis |
Meiosis Il
Crossing over takes placeY
Crossing over does not takeY
place
Homologous chromosomes
arrange at the equator in
pairs¥
Chromosomes arrange at the
equator singlyY
Whole chromosomes move to
opposite poles of the cell¥”
The chrom umber
halved from di oh
idv
atids move to opposite
of the cell ¥
chromosome number!
ains the same (haploid) “
Results in two daughte v
ults in four daughter
ee orern
2x2 +1 for table
(5)
(15)
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NSC — Marking Guidelines
2.2 2.2.1 The changes/modifications brought about by the frequency of
use or disuseY were transmitted/passed on to the
offspringY (2)
2.2.2 Organisms do not evolve because they were
determined/wanted to change/” but changes took place
randomly due to mutationsY (2)
OR
The characteristics acquired by an organism through use
or disuse do not affect the organism's genetic makeupY
therefore cannot be inheritedY
OR
There is no evidence V to show that acquired characteristics
are inherited ¥
OR
The characteristics acquired by an organism in its lifetime“
cannot be inheritedY
(4)
2.3 2.3.1 There are THREE different allele that control inheritance of
blood groups Vv ¥ (2)
23.2 2v (1)
2.3.3. It is used to exclude a man from being a child’s fathery v (2)
OR
To narrow down the possible fathersv ¥
(5)
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NSC — Marking Guidelines
24 24.1 The changes to the sequence” of nitrogenous bases in a DNA
moleculeY (2)
2.4.2 Used 200V cats (1)
2.4.3
Percentage of genetic disorders in cats
Other
10%
Diabetes
meee 45%
PKD ~—
25%
Calculations:
Diabetes: 45/100 x 360° = 162°
PKD 25/100 x 360° = 90°
Deafness: 20/100 x 360° = 72°
Other: 10/100 x 360° = 36°
vv
Rubric for marking the graph:
Criteria Mark
Correct type of graph (T) dAIO |
Caption/Heading with 2. (H) 1
variables
Correct calculations (C) A2u it |
Plotting (P) 1-3correct slices: 1
All four correct slices: 2
(6)
(9)
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NSC — Marking Guidelines
25 2.5.1_ (a) Transcription’ (1)
(b) Nucleus¥
MitochondrionY
ChloroplastY Any (2)
(Mark the first TWO only)
(c) AdenineY (1)
2.5.2 -EachtRNAcarries a specific amino acidY
- When the anticodon on the tRNAY
- matches the codon on the mMRNAY
- then tRNA brings the required amino acid” to the ribosomeY
- Amino acids become attached to each other by peptide bondsY
- to form the required protein’ Any (6)
2.5.3 GGUY (1)
2.5.4 -the anticodon will change to UAGY instead of UAC
- that will code for a different amino acidY
- resulting into a change in the (sequence)of amino acidsY
- leading to a slightly different structure of protein v Any (3)
2.5.5 Methioninev - glycineY - argininev (in correct sequence) (3)
(17)
[50]
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NSC — Marking Guidelines
QUESTION 3
3.1 3.1.1 Agroup of organisms sharing the similar/ same characteristicsY
that are able to interbreedv to produce fertile” (3)
3.1.2 -The original population of shrubs became separated into three sub-
populations” by continental drift/ocean/waterY
- no gene flow amongst the (three) sub populations”
- each population experienced different environmental
conditionsY
- and underwent natural selection independentlyY
- the shrubs in each population became different”
- genotypically and phenotypicallyY
- Even if they were to mix again/cross pollinate v’, they will not be able
interbreedv /to produce fertile population of shrubs
- because they are now three different speciesv” in South America,
Africa and Australia 2 compulsory “+ Any 5 (7)
3.1.3 - Species-specific courtship behavior”
- Breeding at different times of the yearY
- Infertile offspringv
- Adaptation to different pollinators”
- Prevention of fertilization’ Any (2)
(Mark the first TWO only)
(12)
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NSC — Marking Guidelines
32 3.2.4 MY (1)
S22 Foramen magnum is in a (more) forward positionY ¥ (2)
3.2.3 It is longY and narrow” (2)
3.2.4 Jaw of skull K is less prognathous/less protruding/non-
prognathousv, that of skull L is more prognathous/more
protruding/prognathousY
The palate of skull K is short and semi-circular/more roundedv ,
that of skull L Is long and rectangular/U-shapedY
The canines of skull K is short~/small, those of skull L is
longY/ large and pointed
There are no spaces between the teeth/diastema in skull Kv, in
skull L there are spaces between the teethv/diastema
(Mark the first TWO only) Any (2 x 2) (4)
(9)
3.3. 3.3.1 Phylogenetic tree/cladogramv (1)
3.3.2 3,1 /3,0 myav (1)
3.3.3. Homo habilis~”
Homo sapiens¥
Paranthropus robustusY
Paranthropus boiseiv Any (2)
3.3.4 Increase in brain size/bigger brain sizev v to allow for:
Increased intelligence VV
storing and
cessi
ssing of large informationy ¥
information quicklyY ~ Any1x2 (2)
3.3.6 The fossils of Ardipithecus, Australopethicus and Homo habilis are
found in Africa onlyY
The oldest fossils of Homo erectus and Homo sapiens are found in
AfricaY and
the youngest fossils are found in other continents (Asia, Europe)”
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Any (2)
(10)
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NSC — Marking Guidelines
3.4 3.4.1 Twov/2 (1)
3.4.2 BbEev, bbee Vv (2)
3.4.3 (a) Brown fur with long earsv ¥ (2)
(b) BE, Be, bE and bevY (2)
3.4.4 Alleles of different genes segregate/move independently of each
othervinto the gamete so that they appear on the gametes in different
combinationsY (2)
OR
The various ‘factors’ controlling the different characteristics
are separate entitiesv, not influencing each other in any way, and sorting
themselves out independently during gamete formation’
OR
A pair of alleles(factors) controlling a particular characteristic separate
independentlyY and then recombine differently to form gametesv.
(9)
3.5 3.5.1 Neither of the alleles are dominant over each other.
An intermediate phenotype is obtained when both alleles are
presentY (2)
OR
Neither one of the two alleles of a gene is dominant over the otherv,
resulting in an intermediate phenotype in the heterozygous conditionv .
3.5.2 Individuals with genotype HNHS/heterozygous genotype have a mild
form of the disease. Vv’. (2)
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NSC — Marking Guidelines
P1 Phenotype Man with x Normal Womanv/
Mild Awoman without
hypercholesterolemia hypercholesterolemia
Genotype HNHS x HNHN v
Meiosis
Gametes
Fertilization
Fi: Genotype nN 4 i 3 HNHSY
Phenotype
with mild
hypercholester:
0%" chance for children with severe
hypercholesterolemia
P1 and Fiv
Meiosis and fertilization V
“4 compulsory mark + Any 5
OR
P1 Phenotype Man with x Normal Woman v/
Mild Awoman without
hypercholesterolemia hypercholesterolemia
Genotype HNHS x HNHN vw
Meiosis
age Gametes HN Hs
Fertilization nN a TS
HN HNHN | HNHS
1 mark for correct gametes
1 mark for correct genotypes
Fi Phenotype 2 Normal individuals/without hypercholesterolemia:
2 Individuals with mild hypercholesterolemiaY
O%¥ “chance for children with severe
hypercholesterolemia
P1 and Fiv¥
Meiosis and fertilization ¥
“4 compulsory mark +Any5 = (6)
[48]
TOTAL SECTION B: 98
GRAND TOTAL: 148
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