SULIT
4551/2
BIOLOGI
Kertas 2
Ogos
2015
2 ½ jam
JABATAN PENDIDIKAN NEGERI PULAU PINANG
MODUL
PENGAJARAN BERFOKUS
BAGI
PENINGKATAN PRESTASI SPM 2015
SKEMA
PERMARKAHAN
BIOLOGI
KERTAS 2
SET 3
Modul 3 Biologi SPM 2015 Kertas 2 : Skema Jawapan
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Soalan
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Skema Pemarkahan
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Sub Markah
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Jumlah Markah
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|
1(a)
(b)
(c)
(d)
(e)(i)
(e)(ii)
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Able to
name the phase U.
Sample
answer:
U :
Interphase
Able to
describe the processes at sub phases X, Y and Z during
phase U
Sample
answer :
X : Cell
synthesises protein / new orgenelles formed
Y : DNA
is synthesized / is replicated / 2 sister chromatids
formed
Z : Cell
accumulates energy / synthesise energy / prepare for
cell
division
Able to
draw a daughter cell based on the following criteria:
No. of
chromosomes are haploid / 3 chromosomes
Types of
chromosomes/ non homologous
New
genetic combination
Able to
explain how radiotherapy can treat cancer.
Sample
answer :
F :
Radiotherapy uses radiation / high energy rays
E1 :
destroy the nucleus of cancerous cells
E2 :
cancerous cells die / cannot divide mitotically
E3 : cell
cycle stops
Able to
name the method and explain the advantages of the
method in
increasing crop yield.
Sample
answer :
T :
Tissue culture / Cloning
E1 :
Large numbers of clones can be produced
E2 :
Within a short period of time / any time
E3 :
Clones inherited good characteristics/ resistance to diseases
/ fast
growth rate / large fruit / good genetic traits
Able to
state one problem :
Clones
can be destroyed completely if they do not have the
resistance
to new diseases / pest.// No variation
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1
1
1
1
1
1
1
(Any 2)
1
1
1
1
Any 1E=1
1
1
1
1
T=1m
Any 2E
=2
Any 1
|
1
3
2
2
3
1
12
|
|
Soalan
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Skema
Pemarkahan
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Sub
Markah
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Jumlah
Markah
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2 (a)(i)
(a)(ii)
(a)(iii)
(b)(i)
(b)(ii)
(c)
(d)
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Able to
name the type of fingerprints of students X and Y
Answer:
X - Loop
; Y- Composite
Able to
state one factor that causes variation in the fingerprints
of
students X and Y.
Answer:
Genetic
factor
Able to
state how the factor in (a) (ii) causes variation
Answer:
Genetic
recombination during crossing over
results
in the formation of different
Able to
state the type of variation
Answer:
Continuous
variation
Able to
state two traits, other than fingerprint, which show the
same type
of variation as in (b)(i)
Answer:
The
ability to roll tongue
Types of
hair
Able to
explain the differences between the type of variation
shown by
fingerprints and height.
Sample
answer:
Height
Types of fingerprint
- Shows
normal distribution Shows discrete distribution
-
Affected by environmental Not
affected by environmental
factors factor
Able to
explain how variation can ensure the survival of a species
Sample
answer:
- Can
differentiate from one individual to another / no one is the
same
- Able to
adapt to a new environment
- Able to camourflage to run away from any predators
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1
1
1
1
1
1
1
1
1
1
1
1
|
2
1
1
1
2
2
3
12
|
|
Soalan
|
Skema
Pemarkahan
|
Sub
Markah
|
Jumlah
Markah
|
|
3(a)(i)
(ii)
(b)
(c)
(d)(i)
(ii)
(e)
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Able to label the alleles for F1 genotype.
Able to
state the phenotype for F1 generation
Answer : Round
, yellow (colour)
Able name
the process that occurred during meiosis which produced
different
gametes in second possibilities.
Answer
: Crossing-over // cross-over
Able to
draw gametes J and gamete K which are produced in second
possibility.
Gamete J / K
Able to
state which possibilities will cause more variation to the
offsprings
Answer : Second
possibility
Able to
explain one reason for your answer in (d)(i).
Sample
answer
F :
crossing over occurred between (chromatids from a pair of )
homologous chromosomes
E1 : in
prophase 1 /meiosis 1 / meosis
E2 : (the
exchange of parts between chromatids) results in new
Genetic Combinations // a different
genetic composition.
E3 :
(four) different gametes produced.
E4 :
(thus, each time) gametes from two individuals fertilize
randomly,it produced large number of
variations between
offspring Any three
Able to complete
Diagram 3.3 by filling in F1 generation gametes
drawn in
(c), genotype of F2 generation and phenotypeof F2 generation which will be
produced.
Criteria
:
All Gametes from F1
generation correct = 1mark
All Genotype of F2
generation = 1 mark
All Phenotype of F2
generation = 1mark
Sample
answer
 |
1
1
1
1+1
1
1+1+1
1+1+1
|
1
1
1
2
1
3
3
12
|
|
Soalan
|
Skema
Pemarkahan
|
Sub
Markah
|
Jumlah
Markah
|
|
4(a)
(b)(i)
(b)(ii)
(c)
(d)
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Able
to name.
Answers:
Phase S : Prophase 1
Membrane U : Nuclear membrane
Able
to explain the chromosomal behavior during phase S.
Sample
answer :
·
Exchange of genetic material between
homologous chromosome
·
Crossing over
Able
to explain how the chromosomal behaviour contribute to the survival of an
animal species.
Sample
answer :
Able to
explain the chromosomal behaviour during Metaphase I
Sample
answer :
Able to complete
the diagram of the daughter cells and explain the occurrence.
Sample
answer :
V W
|
1
1
1
1
1
1
1
1
1+1
1
1
1
Any 2
|
2
2
2
2
4
12
|
|
Soalan
|
Skema
Pemarkahan
|
Sub
Markah
|
Jumlah
Markah
|
|
5(a)
(b)
(c)
(d)(i)
(ii)
(e)
|
Able
to explain the effect of FSH on structure P.
Sample
answer :
·
(FSH) stimulates the growth / development
of primary follicle
·
Structure P will grow / develop / becomes
secondary follicle / Grafiaan follicle.
Able
to explain the effect if the level of hormone Y is low.
Sample
answer :
·
Ovulation will not occur
·
Grafiaan follicle will not release the
secondary oocyte (into the Fallopian tube).
Able
to explain one difference between the primary oocyte and the secondary
oocyte.
Sample
answer :
·
Difference : Primary oocyte is diploid
while secondary oocyte is haploid.
·
Explanation : Primary oocyte has undergo
meiosis (I).
Able
to complete the graph that shows the thickness of endometrium wall is
decreasing / thinning.
Able
to explain the graph drawn.
Sample
answer :
·
Corpus luteum degenerate
·
No progesterone is secreted (to thicken the
endometrium wall).
Able
to explain the importance of structure Q during foetal development.
Sample
answer :
·
Q secrete progesterone
·
(progesterone will thickened and)
maintained the endometrium wall
·
Endometrium wall ready for implantation of
embryo // prevent miscarriage / abortion.
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1
1
1
1
1
1
1
1
1
1
1
1
|
2
2
2
1
2
3
12
|
|
No
Soalan
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Skema Pemarkahan
|
Markah
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6.(a)
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P 1 The first phase of mitosis is prophase
P2 During
prophase, chromatids becomes condensed and visible. Nucleolus disappears
P3 Centriols move to the opposite poles of the cell.
P4 Nuclear membrane disintegrates, spindle fibres
are formed.
P5 The second phase is metaphase.
P6 During metaphase, the sister chromatids are
arranged at the cell equater.
P7 Each centromere attaches itself to the spindle
fibre.
P8 Centromere splits at the end of metaphase.
P9 The third phase of mitosis is anaphase.
P10 During anaphase, chromatids are pulled towards
the opposite poles.
P11 The chromatids reach the opposite poles of the
cell at the end of anaphase.
P12 The fourth phase of mitosis is telophase.
P13 During telophase, spindle fibres disintegrate.
P14Nuclear membrane reforms around the chromosomes.
P15 Nucleolus reappears.
P16 At the end of telophase, cytokinesis occurs.
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1
1
1
1
Any 3
1
1
1
1
Any 3
1
1
1
1
1
1
1
Any 3
1
Total
10 marks
|
|
(b)
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F Process of Cloning in an animal such as the sheep.
P1 Somatic cells (from the mammary gland cells) are
removed and grown in a low culture medium.
P2 The starved cells stop dividing and enter a non-dividing phase.
P3 An unferterlised egg cell is obtained. The
nucleus is sucked out, leaving the cytoplasm and organelles without any
chromosomes.
P4 An electric pulse stimulates the fusion between
the somatic cell and the egg cell without nucleus.
P5 The cell divides repeatedly, forming an embryo.
P6 The embryo is then implanted into a surrogate
mother (the same breed of sheep as the ovum donor sheep)
P7 The cloned sheep of the somatic cell donor, is
born.
Advantages
P8 To enable individuals with desired
characteristics to be produced in short time.
P9 Prevent extinction of species.
Disadvantages
P10 Costly
and requires professional skill
P11 Cloning reduces genetic variation and limits the biodiversity, as
cloning produces genetically identical
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1
1
1
1
1
1
1
1
Any 6
1
1
1
1
Total
10
|
|
TOTAL
|
20
|
|
|
7 (a)
|
P1 Q is
corpus luteum
P2 A Graafian follicle that releases its secondary
oocyte will develop into corpus luteum.
P3 Corpus luteum is a big and yellowish mass of
cells.
P4 The corpus luteum acts as a secreation gland.
P5 If pregnancy occurs, the corpus luteum
secretes oestrogen and progesterone.
P6 These hormones ensure the endometrium becomes
thick and vascularised.
P7 Oestrogen stimulates the formation of the milk
glands.
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1
1
1
1
1
1
1
Any 7
Total
7
|
|
7 (b)
|
P1 Oogenesis refers to the process which forms the
oocytes.
P2 The formation of the ovum starts with mitotic divisions of the
germinal epithelium several times to
produce oogonia.
P3 The oogonia will grows to form primary follicle.
P4 The primary oocytes undergoes meiotic division to
produce secondary oocyte.
P5 The secondary oocytes becomes secondary oocytes
through meiosis I.
P6 When a female reaches purberty the process of
ovulation happens.
P7 During ovulation, the secondary oocytes is
released from the ovary.
P8 It moves along the Fallopian tube and waiting for
fertilisation.
|
1
1
1
1
1
1
1
1
Any 8
Total
8
|
|
7 (c)
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P1 Sperm is produced through spermatogenesis while
the ova are produced through oogenesis
P2 The sperms are formed in testes while the ova are
formed in the ovaries
P3 There are four sperms formed at the end of
meiotic division while there is only the ovum is formed at the end of meiotic
division.
P4 The size of the sperm is small while the size of
the ovum is big.
P5 The sperm has a head, middle piece and tail while
the ovule is round in shape.
P6 In meiosis I , two spermatocytes are produced
while in meiosis I, the secondary oocyte and polar body are formed.
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1
1
1
1
1
1
Any 5
Total
5
|
|
TOTAL
|
20
|
|
|
8 (a)(i)
|
P1 Haemophilia is a sex-linked diseased which is
carried by recessive gene on chromosome X
P2Haemophilia can occur in men because men only need
one recessive gene on chromosome X to cause the haemophilia character.
P3 A woman who has family pedigree of haemophilia
does not necessary has it because
P4 she can still exist in heterozygous condition
which means that the female is normal but is a carrier for the next
generation.
P5 In female, two recessive genes are needed to show
the haemophilia characteristic
P6 Male have haemophilia when the mother carried
recessive gene for haemophilia.
P7 Females only have haemophilia when both parents
carried recessive gene for haemophilia.
|
1
1
1
1
1
1
1
Any 7
Total
7
|
|
(a)(ii)
|
P1 The
inheritance of haemophilia can be avoided by making sure that every marriage
of the grand children does not involve carriers of haemophiliacs.
P2 Eventually, the recessive genes which cause
haemophilia will disappear after a few generation.
|
1
1
|
|
(b)(i)
|
  |
1
1
1
1
1
Label 1
Total
6
|
|
(b)(ii)
|
P1 During meiosis , Ali produces sperm with allele Io
.
P2 Mina produce ovum
with allele I o.
P3 During fertilization,
the sperm with allele I o and ovum with
allele Io
P4 combine and
P5 form zygote.
P6 The homozygous recessive IoIo
P7 cause the son Zaki inherited blood group O.
|
1
1
1
1
1
1
1
Any 5
Total
5
|
|
TOTAL
|
20
|
|
Soalan
|
Skema Pemarkahan
|
Sub Markah
|
Jumlah Markah
|
|
9 (a)
(b)
(c)
|
Able to
identify type of varation of the flies.
Answer:
Discontinuous
variation
Able to
give three reason.
Answer:
Able to
draw the graph correctly.
Answer:
7-9 correct
4-6 correct
1-3 correct
Able to explain
what happened to the fly that caused by variation.
Answer:
P1 1. Crossing-over
P2
Reciprocal crossing-over of genes between chromatids of homologous
chromosomes may occur during prophase I of meiosis. This produces new linkage
groups and so provides a major source of genetic recombination alleles.
P3 2. Independent assortment
P4 The orientation of the chromatids of
homologous chromosomes (bivalents) on the equatorial spindle during metaphase
l of meiosis determines the direction in which the pairs of chromatids move
during anaphase I. This orientation of the chromatids is random. During
metaphase II ,the orientation of pairs of chromatids once more is random and
determines which chromosomes migrate to opposite poles of the cell during
anaphase II. These random orientations and the subsequent independent
assortment of the chromosomes give rise to a large calculable number of
different chromosome combinations in the genetics.
P5 3.Random
fusion of gametes
P6 A third source
of variation occurs during sexual reproduction as a result of the fact that
the fusion of male and female gametes is completely random (at least in
theory). Thus, any male gamete is potentially capable of fusing with any female gamete.
P7 .Mutation
P8
(a) Deletion
P9 - Involves the loss of chromosomal
P10 (b) Insertion
P11 -Involves
a region of a chromosome breaking off and rejoining either the other end of
the same chromosome or another non-homologous chromosome
P12 (c) Duplication
P13
- Chromosome becomes duplicated so that an additional set of genes
exist for the region of duplication
P14 (d)Inversion
or substitution
P15
- Occurs when a region of a chromosome breaks off and rotates through
180° before rejoining the chromosome
|
1
1
1
1
1
1
1
(3)
(2)
(1)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Any 10
|
4
6
10
20
|
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