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发表于 17-11-2009 01:26 AM
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Remember- do not confuse intercellular fluid and extracellular fluid.
Extracellular fluids are all fluids outside the cells including blood
plasma, tissue fluid, lymph, cerebro-spinal fluid, synovial (joint) fluid
and the aqueous humour of the eye. The term intercellular fluid refers
to tissue fluid only. Do not confuse either of these terms with intracelluar
fluid - the fluid inside cells. |
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发表于 17-11-2009 11:18 AM
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Typical Exam Question
Explain how the palisade mesophyl cells within a leaf are well
adapted for photosynthesis
closely packed to absorb more of incident light;
columnar shape / arranged at right angles to surface of leaf, to
reduce number of light absorbing cross walls;
large vacuole pushes chloroplasts to edge of cell;
chloroplasts on periphery of cell, short (diffusion) path for carbon
dioxide;
chloroplasts on periphery of cell to absorb light;
large number of chloroplasts / much chlorophyll, to absorb light ;
chloroplasts can move within cells to absorb as much light as
possible;
chloroplasts can move to prevent damage (in high light intensity) ;
cylindrical cells resulting in air spaces;
air spaces (between cells) to allow circulation of gases;
large surface area for, gas exchange/diffusion;
cell walls are thin, so short diffusion pathway/(greater) light
penetration;
air spaces act as reservoir of carbon dioxide;
AVP; e.g. non pigmented vacuole to allow light penetration |
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发表于 17-11-2009 11:26 AM
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Chloroplast - Structure and function
Chloroplasts are organelles bounded by a double membrane – this
separates the contents of the chloroplast from the rest of the cell,
allowing the photosynthetic reactions to take place without
interference from chemicals in the cytoplasm.
Plate-like membranes, called lamellae (thylakoids) run along the
length of the chloroplast from end to end – these are the site of the
light-dependent reactions.
Chlorophyll molecules (and other pigment molecules) that absorb
light energy are situated on the lamellae. At intervals, extra lamellae
are inserted to form structures called grana (singular:granum). A
chloroplast contains approximately 3000 lamellae -they increase the
surface area available for the attachment of extra pigment molecules
to trap more light energy. The pigment molecules are arranged in
such a way that they do not shade each other so that the maximum
amount of light energy can be absorbed.
The lamellae:
• Provide a large surface area for the attachment of the enzymes
and co-enzymes involved in the electron transfer chain of the
light dependent reactions.
• Partition the stroma so that different chemical reactions can
take place without interfering with each other.
The biconvex shape of the chloroplast is yet another way of
increasing surface area to maximise absorption of light energy
Sometimes in the exam you will be presented with an electron
micrograph of a chloroplast. Usually, the first question simply asks
you to label it.
Exam Hint:– Whenever a structure or cell possesses extra
membranes it increases the surface area available, in this
case for the attachment of extra pigment molecules, but in
other cells it can increase the rate of absorption (e.g. microvilli
of kidney tubule cells), cristae in mitochondria for attachment
of extra stalked particles).
Pigment molecules absorb light energy which ultimately becomes
converted to chemical energy incorporated into the glucose and
other compounds produced as a result of photosynthesis.
Chlorophyll a and chlorophyll b absorb only certain wavelengths
of light and therefore only a certain amount of the energy falling
on the chloroplast. Xanthophyll, carotene and phaeophytin are
accessory pigments that absorb light of other wavelengths,
ensuring more energy is trapped by the chloroplast.
The pigment molecules form clusters called photosystems. Each
photosystem consists of a central chlorophyll molecule called
the primary pigment molecule. Located around this are several
hundred accessory pigment molecules – light energy absorbed
by all these molecules is transferred to the central primary pigment
molecule.
*Note : accessory pigments absorb other wavelength of light to be excited ; consequently, they emit the electrons to the primary pigments , chlorophyll a (P680 in photosystem II and P700 in photosystem I). Primary pigments in photosystems are called the reaction centre. Their main task is to emit electrons to electron acceptor. Example of accesory pigments are other forms of chlorophyll a, chlorophyll b , xanthophylls, carotenoid , beta-carotene. |
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发表于 17-11-2009 01:18 PM
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原帖由 muse 于 2007/9/13 10:28 PM 发表 
1.Which of the following reproduction processes are not sexual reproduction?
I Paedogenesis
II Polyembryony
III Conjugation
IV Parthenogenesis
A.I and II
B.III and IV
C.I,II and IV
D.II, ...
Answer is IV
editted :不好误导大家。
[ 本帖最后由 darksider 于 17-11-2009 02:12 PM 编辑 ] |
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发表于 17-11-2009 01:55 PM
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发表于 17-11-2009 02:08 PM
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发表于 17-11-2009 06:58 PM
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发表于 18-11-2009 10:18 AM
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分享刚刚在recom的回复
just study them. Never enter the examination hall without knowing what they are 
Sliding filament hypothesis
-------------------------
You must learn these key points:
1 The presynaptic membrane is depolarized.
2 Ca2+ channels on the presynaptic membrane open.
3 There is an Influx of Ca2+2 ions from the synaptic cleft into the
synaptic bulb.
4 Vesicles move towards presynaptic membrane.
5 Vesicles fuse with presynaptic membrane.
6 Neurotransmitter e.g. acetylcholine is released from the vesicles.
7 The neurotransmitter diffuses across the gap or synaptic cleft.
8 The neurotransmitter binds to protein receptors on the
sarcolemma (post-synaptic membrane)which has a large surface
area for this purpose.
9 The binding of the neurotransmitter to the receptors on the
sarcolemma causes sodium channels to open causing an influx
of sodium ions.
10 An action potential is generated and there is depolarization
along the surface of the muscle/depolarisation spreads down
transverse tubules.
11 Calcium channels in the sarcoplasmic reticulum open.
12 Calcium ions diffuse out and bind to troponin.
13 Troponin moves tropomyosin.
• Many candidates believe that it is the calcium that moves
tropomyosin – it isn’t
14 This exposes myosin binding sites on actin molecules / thin
filaments
15 The calcium ions activate myosin which releases ATPase to
split ATP from mitochondria into ADP and Pi
16 This energy is used to move the heads of the myosin filaments
towards the now exposed binding sites on actin
17 The myosin head binds to the binding site on actin and cross
bridges are formed
18 The myosin heads tilt pulling the actin filaments past them
19 As they slide, the heads detach from one site when ATP molecule binds to the myosin head to break the cross bridge and they are now free to attach to another site in a ratchet mechanism.
20 As actin filaments from either end of the sarcomere move
towards each other, the muscle fibre contracts/ there is greater
overlap between thick and thin filaments
21 The sarcomere shortens as the I bands and H band shorten in each sarcomere.
Extract from Chief Examiner's Reports
What candidates get wrong on this first part
• Many candidates stated that vesicles move across the cleft
• Ion movement was often poorly described, with the direction
of movement often omitted. Many candidates incorrectly
referred to movement of chlorine or sodium ions at the
presynaptic knob
• Many candidates failed to mention binding to receptors on
the postsynaptic membrane
• Some made reference to depolarisation of the neurone rather
than the membrane
Oxygen dissociation curve
------------------------
Extract Chief Examiner’s Report
Many candidates had a poor understanding of the processes involved in
the transport of oxygen by haemoglobin and the dissociation curve. It
was often stated that haemoglobin has ‘ a high affinity for oxygen’, but
rarely was the answer qualified by explaining that it has a high affinity at
higher partial pressures, while oxyhaemoglobin readily dissociates at
lower ones. Only a small minority linked their explanation to the
dissociation curve.
A significant minority confused haemoglobin with red cells, and referred
to oxygen being transported in the ‘dip’ in the haemoglobin.
The commonest problem was the use of inappropriate terminology. For
example, descriptions of the haemoglobin ‘picking up’ or ‘dropping off’
oxygen were frequent. When describing the formation of oxyhaemoglobin,
candidates frequently made statements such as ‘haemoglobin joins with
4 oxygen molecules’.
Extract from Chief Examiner’s report
Many candidates explained that fetal haemoglobin has a high affinity
for oxygen, but only the better candidates went on to explain that this
means that fetal haemoglobin takes up oxygen when adult oxygen
haemoglobin dissociates.
Cell-mediated immunity
----------------------
-bacteria are ingested and digested by macrophages. The leftover antigen combines with the class II MHC protein and move to the surface of the macrophage to be present. The macrophage now becomes an antigen presenting cell(APC)
-helper T cell (T4 cell) with a matching receptor recognises class II MHC protein complex on the APC and its CD4 protein binds to the class II MHC protein.
-the APC then releases interleukin-1. Interleukin-1 stimulates the helper T lymphocyte to secrete interleukin-2 which has the folloing effect:
*stimulate T cells to multiply , forming many forms of T lymphocytes(cytotoxic T cells, suppressor T lymphocyte, trigger T lymphocyte, memory T lymphocyte for example)
*stimulate B cells to produce antibodies
*promotes inflammation
-Cytotoxic cell defends the body by actively destroying virus-infected nucleated cell or malignant cancerous cell.
-The viral or tumour antigens are presented on the surface of the nucleated cell by class I MHC protein.
-This protein is recognised by protein receptor and surface protein CD8 of cytotoxic T cell.
-Once in contact with these protein, the cytotoxic T cell releases perforin and granzyme in the presence of interleukin-2 (from helper T cell).
-the perforin perforates the nucleated cells and this allows the granzyme to enter the nucleated cell and then causes apoptosis, resulting in the death.
-the activation and killing action of cytotoxic T cell is known as cell-mediated immune response. |
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发表于 18-11-2009 10:19 AM
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mechanism of opening and closing of stomata
--------------------------------------------
1)starch-sugar hypothesis
-during the day, photosynthesis occurs
-concentration of CO2 drops and pH increase
-starch is converted to sugar(sucrose, for eg)
-this lowers the water potential of guard cells
-water enters from subsidiary cells which have higher water potential by osmosis
-guard cells become turgid
-stomata open
-during the night, photosynthesis does not occur
-concentration of CO2 increaes and pH decreases
-sugar is converted to starch
-water potential increases as glucose is converted to starch (starch is osmotically inactive)
-water leaves guard cells to adjacent cells by osmosis.
-guard cells become flaccid
-stomata close
K+ ions accumulation hypothesis (better hypothesis)
-------------------------------
-during the day, blue light activates the proton pump in guard cells which pumps H+ ions out of the guard cells. This creates a proton gradient in which the H+ concentration is higher outside the cell.
-The loss of H+ ions from the guard cells generate a negative membrane potential
-The negative membrane potential causes influx of K+ ions into the guard cells
-This lowers the water potential of guard cells
-water enters from subsidiary cells by osmosis.
-guard cell becomes turgid
-stomata open
During water shortage
---------------------
-During water shortage, abscisic acid is produced.
-abscisic acid binds to the receptor on membrane of guard cells, which inhibits the proton pump and causes the Ca2+ channels in the plasma membrane to open.
-influx of Ca2+ ions into the cytosol of guard cells causes depolarisation. This , in turn, lead to opening of anionic channel which facilliates the efflux of chloride ions and malate ions.
-Loss of these anions would further depolarise the membrane , open the K+ ions channel and trigger the passive extrusion of K+ ions into the adjacent cells.
-water potential of subsidiary cells decreases, so water enters from guard cells into them by osmosis
-guard cells become flaccid
-stomata close.
Quote:
Originally Posted by sugarspice 
^Wah, where you got those from?
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The chief examiner's report and the sliding filament theory(modified;I include :ATP are used to break cross bridge, myosin ATPase) are copied from the bio fact sheets.
Formation of resting membrane potential
---------------------------------------
-the resting neurone membrane is impermeable to sodium ions
(a common misconception usually made by student by stating that the neurone membrane is permeable to sodium ions and of course I'm not talking about the sodium ion voltage-gated channel)
-three sodium ions are actively transported to the outside by sodium pumps for every two potassium ions transported into the neurone by potassium pump. (potassium pump is a weak pump and its effect is minimal. It is therefore too weak to account for the high concentration of potassium ions in the neurone). This causes the concentration of the sodium ion to be higher on the outside relative to the inside.
This also creates a positive ions deficit within the neurones.
-At the same time, the resting neurone membrane is freely permeable to potassium ions and these ions diffuse down the concentration gradient to balance the positive ion deficit within the neurone. However,the influx of potassium ions never quite catches up with the outflow of sodium ions (by sodium pump) and therefore , a positive ion deficit within the resting neurone is always maintained.
-Within the neurone are many large negatively charged proteins which cannot pass through the resting neurone membrane.
-Due to all the factors above, a resting membrane potential of approx -70mV is produced.
---------------------------------------------------------------------------------------
Advantage of saltatory conduction
-It speeds up the transmission of impulse along the axon, causing low response time / fast response. (this low response time is needed by usually voluntary action such as muscle contraction and also reflex action)
-It restricts ion movement to only the nodes of Ranvier. This causes less ATP is expended on the sodium potassium pump and also it makes repolarisation faster (for shorter refractory period , ie 1/2500, so the volley of impulse in myelinated neurones can reach 2500 impulses per second ; if the neurone is unmyelinated, the refractory period would only be 1/50 so the volley of impulse can only reach a maximum of 50 impulse per second which could be too slow for muscle contracton and other voluntary actions).
Electro-osmosis theory
----------------------
Companion cells actively absorbs K+ ions from one end of a sieve plate and actively secretes them to the adjacent end of another sieve plate. This creates a potential difference and this cause the polar water to moves down the potential gradient by osmosis. As water moves across by osmosis, solutes are carried along.
Advantage
-----------
-support bidirectional flow of solutes within sieve plates.
-requires energy in the form ATP ; uses the companion cell (edge all other theory such as Munch's mass flow hypothesis , cytoplasmic streaming , peristaltic wave hypothesis because according to these theories, companion cells are not involved in transport in phloem and if this is the case, companion cells would have disappeared due to evolution)
Disadvantage
-------------
-never explains how solutes acutally move across the small pores between sieve plates
Quote:
Originally Posted by stevew08
for bio how bout opening of stoma, plant hormone, aids, natural selection, speciation? ans also
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For plant hormones, please refer to your notes as they are relatively easy.
Remember 5 types of hormones : auxin , gibberellic acids, cytokinin, ether and abscisic acid.
remember which hormones work synergetically (synergism)
eg
-Auxin and Gibberellic acids which promote apical dominance ;
-GA and cytokinin promotes seed germination
and which hormones work antagonistically
eg
-Auxin promotes apical dominance and inhibits bud dominance while cytokinin promotes bud dominance but inhibit apical dominance.
-GA and cytokinin promotes seed germination while abscisic acid promotes seed dormancy |
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发表于 18-11-2009 10:19 AM
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Aids
----
Causes
--------
HIV (Human Immunodeficiency Virus) which attacks the T4 cells in the body. It does so by recognising and binding to the specific receptor called CD4 protein on the surface membrane of T4 cells.
-Its lipoprotein then fuses with the plasma membrane and release its RNA and enzymes to the interior of T4 cells.
-reverse transcriptase catalyses the formation of complementary/copy DNA (cDNA) from the viral RNA.
-The complementary DNA then pairs up using free DNA nucleotides in the T4 cells.
-The DNA then take controls of the T4 cells. It allows transcription to occur in the nucleus , producing many mRNA and at the same time, synthesis of protein coats in the cytoplasm.
-These protein coats(lipoprotein) and viral RNA(mRNA) then assemble to form new HIV virus and the host cell is lysed, and these new HIV virus are free to infect other T4 cells.
Treatment
----------
-Use anti revese transcriptase enzyme such as Azidothymidine (AZT)
-Use protease inhibitor such as indinavir and ritonavir so that these protease cannot cleave the host's proteins to amino acids for synthesis of protein coats.
Prevention
-------------
-avoid sharing needles
-avoid having multiple sex partners
-check for HIV viruses before doing any blood transfusion
Symptoms
----------
-early symptoms are fever, loss of appetite, nausea and severe migraine
-as time passes AIDS patient will be suffering with candidiasis , kaposi's sarcoma several other infections.
-most AIDS patients die of kaposi's sarcoma
Natural Selection
-----------------
This is a process by which organisms who can better adapt to the environment will be able to surivive and thus reproduce and pass on their alleles to the subsequent generation.
Species with favourable phenotypes will have the selection advantage while species who cannot adapt to the environment will have selection disadvantage. There are always selection pressures being exerted on a population. Example of selection pressures are predation by eagles, intraspecific competitions, fighting for foods , space availability and so on.
There are many types:
-Stabilising selection (stabilises towards one phenotype ; babies with higher body weight tend to have better immunity than those that have light weight so babies with higher weight are likely to survive but those with light weight die ... so as time passes on , only the former is seen)
-Disruptive selection (favours both extreme phenotypes ; eg, large beak size finches and small beak size finches )
-polymorphism (transient polymorphism (black peppered moth and white botched moth ) and balance polymorphism (btw alleles where only heterozygote individuals survive ; homozygous recessive and homozygous dominant patients will die ; in sickle cell anaemia)
-Directional selection (favours one extreme phenotype)
Speciation
----------
-speciation refers to the formation of new speies
-There are 2 types : allopatric speciation and sympatric speciaiton
-sympatric speciation refers to the formation of new species in the same area
-allopatric speciation refers to the formation of new species in different areas, ie , due to geographical barriers (separated by seas, moutains etc)
-sympatric speciation can be due to behavioural differences(different courtship behaviour, different diets, niche such as the case of finches with large beak and finches with small beak), mechanical incompatibility (different reproductive structure due to mutation) etc.
Typical Exam Question
On Albemarle, the beak depths of Geospiza fortis ranged
between 11 and 16mm. Outline how meiosis and polygenic
inheritance may have contributed to this variation.
Markscheme
-beak size due to effect of more than one gene;
-different alleles of each gene are present in the population;
-meiosis gives new combinations of alleles / of genes / of DNA;
-via crossing-over;
-via independent assortment of chromosomes / in meiosis I;
-via independent assortment of chromatids / in meiosis II;
-‘meiosis gives new combinations of chromosomes;
-many combinations of alleles / chromosomes) are possible;
Typical Exam Question
Suggest how genetic isolation could have occurred if speciation
of the Galapagos finches had been:
(i) allopatric;
(ii) sympatric.
Markscheme
(i) Allopatric speciation could have occurred if finches were
separated by the sea i.e they were on separate islands or
they could have been separated by mountains
(ii) Sympatric speciation could have occurred if the finches were
separated:
Temporally i.e they had different breeding seasons or feeding
times
Behaviourally e.g. if they had different courtship displays /
different niches /habitats / feeding areas /
Mechanically – by a mismatch in reproductive parts
Reproductively e.g. by gamete incompatibility if the sperm
were killed in the female’s reproductive tract or via hybrid
inviability/infertility |
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发表于 19-11-2009 10:57 AM
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发表于 24-11-2009 04:17 PM
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我已經過了STPM好久了，看到在這里大家開心的討論，又有朋友指導，替大家開心  以前家里有Internet都還沒有盛行去Forum，全部都是靠自己 
大家加油！ |
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发表于 5-12-2009 05:02 PM
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想要请问大家，你们的insect specimens project真的都是自己去抓的吗？
有人买昆虫的标本吗？ |
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发表于 6-12-2009 12:37 AM
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本帖最后由 生物难题 于 6-12-2009 12:47 AM 编辑
这是其中一题pass year。
individual I-2 is blood group O, individual II-1 blood group B, and individual II-2 blood group AB.
The probability of individual of generations I, II and III with the blood group B is …
为什么答案是50%？
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发表于 6-12-2009 05:45 PM
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回复 414# 生物难题
possible I-1 = AA,BB,OO,AB,BO,AO....Since the I-2 IS OO,in order to have progeny II-1 is B blood group,I-2 must at least have 1 B allele,thus ,in six possible arrangment,AB,BB,BO are possibles.thus 3/6=1/2.
GENERATION II i dunno wat it is stated.
possible II-1 are BB or BO,so,genetic cross with AB will produce progeny either(in case of BB) AB OR BB (Show probability of 1/2),(in case of BO) producing AO,BO,BB,or AB (which is also probability of 1/2) |
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发表于 6-12-2009 10:02 PM
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想要请问大家，你们的insect specimens project真的都是自己去抓的吗？
有人买昆虫的标本吗？
south 发表于 5-12-2009 05:02 PM 
我是自己抓的
也有學校是買標本的
自己抓有它的樂趣所在 |
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发表于 6-12-2009 10:09 PM
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回复 生物难题
possible I-1 = AA,BB,OO,AB,BO,AO....Since the I-2 IS OO,in order to have pr ...
四月一日的小皮 发表于 6-12-2009 05:45 PM 
还不是很明白，不过还是谢谢你 |
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发表于 7-12-2009 04:24 PM
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还不是很明白，不过还是谢谢你
生物难题 发表于 2009/12/6 10:09 PM 
In generation I... individual I is B type and another type is O type. (so probability of B in gen I - 50%)
In generation II ... individual I is B type and another is AB type (so probability of B in gen II - 50%)
In generation III .. individual I can be A/B type and individual II can also be A/B type (so probability of B in gen II - 50%)
Therefore... the answer. |
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发表于 17-12-2009 04:37 PM
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学长学姐，你们有什么方法可以找到植物标本的名字吗？？ |
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发表于 22-1-2010 07:57 PM
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BIOLOGY STPM..你们都用什么品牌的参考书。？LONGMAN? |
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1954 
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