Pls follow the link below for tips:
http://berryberryeasy.com/2012/10/stpm-biology-2012-%E2%80%93-tips-and-predictions-for-papers-9641-9642-and-9643/
Friday, November 23, 2012
Wednesday, October 10, 2012
ANSWER - STPM TRIAL EXAM (IBM) - U6 2012 - MTC, P1
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1. C
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2. C
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3. D
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4. A
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5. B
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6. B
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7. C
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8. B
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9. A
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10. A
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11. E
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12. D
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13. C
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14. B
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15. B
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16. C
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17. A
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18. D
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19. D
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20. E
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21. D
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22. D
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23. E
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24. A
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25. D
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26. E
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27. A
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28. E
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29. D
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30. A
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31. B
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32. B
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33. B
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34. D
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35. C
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36. B
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37. C
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38. B
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39. B
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40. A
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41. C
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42. D
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43. C
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44. C
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45. A
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46. C
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47. C
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48. D
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49. C
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50. C
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Sunday, September 2, 2012
STPM Biology - chapter 3 (old syllabus); chapter 4 (new syllabus): Revision essay question 18
18. Define the term "enzyme" and relate its function and properties to its structure.
Answer:
Enzymes are substrate specific which means that it has a specific shape that allows only the wanted substrate to bond and react with it.Because of its structure and shape it can carry out its function in speeding up the reaction by lowering the activation energy needed for the reaction to take place.
by sharon teoh
STPM Biology - chapter 1 (old syllabus); chapter 3 (new syllabus): Revision essay question 36
36.
Give an illustrated account of the detailed structure and functions of the plasma membrane.
Answer:
(a) structure
Plasma membrane made up of two layers are composed of phospholipid .The hydrophilic heads of the phospholipids are in contact with water and extracellular fluid in the outer layer and the inner layer in in contact with the cytoplasm.The hydrophobic tails face each other in the middle of the bilayer.The protein molecules float about in phospholipid bilayer forming a mosaic pattern.There are two types of membrane protein which are globular protein.Intrinsic protein are the protein that are fully or partially embedded in the phospholipid bilayer. Extrinsic protein are the proteins that are found floating on the external or internal surface of the plasma membrane.Some of the membrane protein and phospholipid have short ,branched carbohydrate chains ,forming glycoprotein and glycolipid.Lipids such as cholesterol are inserted among phospholipid molecules.
(b) function
It is the protective membrane surrounds the outside of both Eukaryotic and Prokaryotic cells.It is made up of a double layer of phospholipid and controls the movement of various substances into and out of the cell.It also allows cell identification.It seperate interior of a cell from external environment.It is selectively permeable and amphipathic in nature.
Give an illustrated account of the detailed structure and functions of the plasma membrane.
Answer:
(a) structure
Plasma membrane made up of two layers are composed of phospholipid .The hydrophilic heads of the phospholipids are in contact with water and extracellular fluid in the outer layer and the inner layer in in contact with the cytoplasm.The hydrophobic tails face each other in the middle of the bilayer.The protein molecules float about in phospholipid bilayer forming a mosaic pattern.There are two types of membrane protein which are globular protein.Intrinsic protein are the protein that are fully or partially embedded in the phospholipid bilayer. Extrinsic protein are the proteins that are found floating on the external or internal surface of the plasma membrane.Some of the membrane protein and phospholipid have short ,branched carbohydrate chains ,forming glycoprotein and glycolipid.Lipids such as cholesterol are inserted among phospholipid molecules.
(b) function
It is the protective membrane surrounds the outside of both Eukaryotic and Prokaryotic cells.It is made up of a double layer of phospholipid and controls the movement of various substances into and out of the cell.It also allows cell identification.It seperate interior of a cell from external environment.It is selectively permeable and amphipathic in nature.
STPM Biology - chapter 1 (old syllabus); chapter 3 (new syllabus): Revision essay question 36
The figure below shows a
section through part of the fluid mosaic model of the cell surface membrane
with a Na+/K+ pump protein.
a)
Explain why the cell surface membrane is
described as a fluid mosaic.
The cell surface membranes is
described as a fluid mosaic due to the scattered mosaic arrangement of proteins
that float in the phospholipid bilayer. The bilayer is fluid as the lipid and
proteins molecules can move laterally and rotate on their axis.
b)
Describe how the channel surface of the protein
differs from its surface next to the phospholipid tails.
The channel
surface is hydrophilic to allow the passage of ions. The surface next to
phospholipid tails is hydrophobic in order to interact with the hydrophobic
fatty acid tails so that the protein stays in the membrane.
c)
Explain why Na+ and K+
cannot pass freely across the phospholipid bilayer.
Na+ and K+
cannot pass freely across the phospholipid bilayer because the bilayer is
impermeable to ions. The hydrophobic tails in the phospholipid bilayer repel
the ions, hence transport of the ions requires the Na+/K+
pump.
d)
Cholesterol and glycolipids are associated with
the cell surface membranes.
Suggest one function of each compound in
membranes.
Cholesterol:
Functions as a plug to reduce the entry and exit of polar molecules across the
cell
surface membrane.
Glycolipid: Functions
in cell-cell recognition, acting as receptor sites for chemical signals.
by qian ying
STPM Biology - chapter 2:Revision essay question 10
10. Carefully study the
drawings in the figure below made from observations of electron micrographs.
a)
Cells are divided into two major groups
according to their structures. To which group do the above cells belong?
Bacterial cell is belonged to
prokaryotic cell while plant cell is belonged to eukaryotic cell.
b)
Name the structures indicated by the letters A,
B and C on the diagrams above.
A:
Mesosome
B:
Nucleoid
C:
Plasmodesma
c)
State one difference between the structures
labeled B and H.
B contains DNA that lies free in
the cytoplasm and is not associated with protein or RNA, but H is bounded by a
double membrane nuclear envelope and it contains DNA that is associated with
protein and RNA to form chromosome.
d)
State the main function of the following
structures:
D:
Contains chlorophyll to trap light energy for photosynthesis.
E:
Synthesis lipids and undergo detoxification.
F:
Synthesis protein.
G:
Site of aerobic respiration to synthesis ATP.
STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 10
10. The
results of an investigation of photosynthetic pigments in ivy
leaves (Hedera helix) is shown below.
by Wei kee
a)
Measure the distance
travelled by spot X.
The distance travelled by spot X is 65
mm.
b)
i.
The solvent front travelled 100mm.
Calculate Rf for spot X where
=
=
0.65
ii.
Identify the pigment in spot X using
the table of Rf values above.
The pigment in spot X is chlorophyll a.
iii.
State one way, apart from Rf
values, which could be used to identify the spot X.
Apart from Rf values, the
spot X can be identified by its colour.
c)
Deduce which of the
photosynthetic pigments listed in the table of Rf values was not
present in Hedera helix.
The photosynthetic pigment that not
present in Hedera helix is
phaeophytin.
Advantage for plants of having more
than one photosynthetic pigment is to cover more of the visible light spectrum
which enables them to capture more light energy.
by Wei kee
STPM Biology - chapter 1: Reision essay question 35
35.
By Wei Kee
a)
Define the term osmosis.
Osmosis is the diffusion of water
though a membrane from a region with high concentration to a region with a
lower concentration of water.
The word describes a plant cell which
has gained a maximum amount of water is turgid.
The word used to describe a cell where
the cytoplasm has pulled away from the cell wall is plasmolysis.
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Active
Transport
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Osmosis
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It requires energy.
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It does not require energy.
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It occurs against the concentration
gradient.
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It occurs down the concentration
gradient.
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It is the movement of molecules and
ions.
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It is the movement of water only.
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By Wei Kee
STPM Biology - chapter 5 (old syllabus) chapter 5 (new syllabus): Revision essay question 10
10. The
diagram represents some of the biochemical reactions of aerobic respirations.
Study the diagram, and then answer the question which follows.
a)
Although oxygen is not mentioned, state how the diagram indicates that aerobic
respiration is occurring, rather than anaerobic respiration.
b)
Which reaction, represented by a numbered arrow, indicates that a molecule of
carbon dioxide is released at that point?
Oxidation occurred when
Isocitrate loses a Carbon Dioxide. Number 6.
c)
i) At what numbered stage is ATP used?
3
ii)
Why is ATP need at this stage?
To breakdown glucose into
pyruvate, as pyruvate is use in Krebs Cycle.
d)
Which numbered stage represents a coenzyme being released for recycling?
7
e)
Which numbered stage represents the point at which a respiratory substrate
enters the mitochondrion?
5
f)
What type of enzyme would be necessary at stage 6?
Aconitase
g)
What happens immediately to the removed 2H?
Dehydrogenation occurred.
NAD will convert into NADH+H.
h)
Where precisely would you expect to find acetyl CoA in a cell?
Matrix of Mitochondria
STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 9
9.
a)
Write a chemical equation to summarise how two molecules of glucose are joined
to form maltose. Structural formulae are not expected.
α Glucose + α Glucose Maltose + Water
b)
State two ways by which the reaction you have shown in (a) could be reversed.
Maltose could be broken
down into 2 glucose molecule by Maltase through hydrolysis.
ci)
Maltose and sucrose both have the same empirical formula, but different
structural formulae. What biochemical term is used to describe this?
Maltose is from aldehyde
group and sucrose is from keto group.
cii)
Maltose is a reducing sugar but sucrose is a non- reducing sugar. If you were
given a solution suspected to contain a mixture of these two sugars and asked
to prove their presence, describe the procedure you would use.
To test for the presence
of monosaccharides in Maltose and Sucrose, Maltose and Sucrose are dissolved in
water, and a small amount of Benedict’s reagent is added. During a water bath,
which is usually 4-10 minutes. A colour change will be observed. Brick red
precipitate will be obtain in Maltose because of the presence of glucose. On
the other hand, the sucrose solution will remains blue because it does not
react with Benedict’s reagent. If dilute hydrochloric acid is added into the
test, positive result will be obtain because acidic condition will break the
glycosidic bond in sucrose through hydrolysis. The products of sucrose
decomposition are glucose and fructose.
Friday, August 31, 2012
STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 9
9. Describe the mechanism of action in synthesis of lipid by photosynthesis.
Answer
Answer
- Lipid Synthesis
- Fatty acid synthesis is catalyzed by fatty acid synthetase using the substrates acetyl-CoA and malonyl-CoA, the reductant NADPH, and a small protein called acyl carrier protein, which carries the growing fatty acid chain; the fatty acid is lengthened by adding two carbons at a time to its carboxyl end
- Triacylglycerols are formed from the reduction of dihydroxyacetone phosphate (a glycolytic pathway intermediate) to glycerol 3-phosphate, which then undergoes esterification with two fatty acids to form phosphatidic acid; this can then be used to produce triacylglycerol
- Phospholipids are also produced from phosphatidic acid using a cytidine diphosphate (CDP) carrier
•Lipid Anabolism - lipid synthesis is known as lipogenesis
•
Øglycerol is
synthesized from dihydroxyacetone phosphate (an intermediate product of glycolysis)
Ø
Ømost lipids, including nonessential fatty
acid chains and steroids, begin with acetyl-CoA
Ø
Ølipogenesis can use almost any organic
substrate because lipids, amino acids, and carbohydrates can be converted to
acetyl-CoA
Ø
Ønot every fatty acid chain that can be broken
down can be built; thus, fatty acids that cannot be built must be obtained in
the diet and are called essential fatty
acids (e.g. linoleic acid and linolenic acid
are essential fatty acids)
animation
STPM Biology - chapter 4(old syllabus) chapter 6 (new syllabus): Revision essay question 10
10. Briefly describe the structure of chloroplast
Answer
Answer
Monday, August 20, 2012
STPM Biology - chapter 18: Revision essay question 4
4. Given a population of 35% of white mice. Knowing
that the white colour skin is caused by the double recessive genotype, “aa”.
Calculate the allele and genotype frequency for this population.
Answer
White mice = 35
Frequency of aa genotype, q² = 0.35
Frequency of allele, q = 0.59
Frequency of allele, p = 1 – q
= 1 – 0.59
= 0.41
Frequency of genotype, AA = p²
= (0.41)²
= 0.17
Frequency of genotype, Aa = 2pq
= 2 (0.41) (0.59)
= 0.48
CHECKING:
p² + 2pq + q²
= 0.17 + 0.48 + 0.35
= 1
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
Answer
White mice = 35
Frequency of aa genotype, q² = 0.35
Frequency of allele, q = 0.59
Frequency of allele, p = 1 – q
= 1 – 0.59
= 0.41
Frequency of genotype, AA = p²
= (0.41)²
= 0.17
Frequency of genotype, Aa = 2pq
= 2 (0.41) (0.59)
= 0.48
CHECKING:
p² + 2pq + q²
= 0.17 + 0.48 + 0.35
= 1
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
STPM Biology - chapter 18: Revision essay question 3
3. In a given population, only the "A" and
"B" alleles are present in the ABO system. There are no individuals with type
"O" blood or with O alleles in this particular population. If 200
people have type A blood, 75 have type AB blood, and 25 have type B blood, what
are the alleleic frequencies of this population (i.e., what are p and q)?
Answer
Type A blood = homozygous AA = 200
Type B blood = homozygous BB = 25
Type AB blood = heterozygous AB = 75
The frequency of A = 2 (200) + 75
2 (300)
= 0.792
= p
Since p + q = 1,
then q = 1 – p
= 1 – 0.792
= 0.208
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
Answer
Type A blood = homozygous AA = 200
Type B blood = homozygous BB = 25
Type AB blood = heterozygous AB = 75
The frequency of A = 2 (200) + 75
2 (300)
= 0.792
= p
Since p + q = 1,
then q = 1 – p
= 1 – 0.792
= 0.208
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
STPM Biology - chapter 18: Revision essay question 2
2, Define the Hardy-Weinberg Law
Answer
A principle that in an infinitely large, randomly mating population in which selection, migration and mutation do not occur, the frequencies of alleles and genotypes do no change from generation to generation.
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
Answer
A principle that in an infinitely large, randomly mating population in which selection, migration and mutation do not occur, the frequencies of alleles and genotypes do no change from generation to generation.
It can be represented by the Hardy-Weinberg
equation.
p² + 2pq +q² = 1
where, p = dominant allele frequency
q = recessive allele frequency
p² + 2pq +q² = 1
where, p = dominant allele frequency
q = recessive allele frequency
p² = frequency of
homozygous dominant in a population
q² = frequency of homozygous recessive in a population
2pq = frequency of heterozygous in a population
q² = frequency of homozygous recessive in a population
2pq = frequency of heterozygous in a population
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
STPM Biology - chapter 18: Revision essay question 1
1. What is meant by:
Answer
i.
Gene pool
ii.
Allele frequency
iii.
Genotype frequency in a population
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
Answer
i.
Gene pool
The sum total of all the genes and
alleles that are present in a sexually reproducing population of a particular
species pf organism at a given time.
ii.
Allele frequency
The relative number of gene compared to the total
number of all alleles of the gene in a population.
iii.
Genotype frequency in a population
The
percentage of individuals with a specific genotype in a population
Assignment submitted by L.Y., Kong, S. H., Lim, S. C., Soo, C. W.
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