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A-Level Physics Tuition Singapore/H2 Physics Tuition/JC Physics Tutor

Hi A-level/H2/JC Physics Tuition students

Topic 1 : Physical Quantities and Units – Part 1

1. All physical quantities consists of a numerical magnitude and a unit.

2. Base Quantities
A base quantity is defined in terms of a standard. It is not defined in terms of other physical quantities.

The are seven base units defined in the SI system.

Base Quantity Unit
Name Symbol
Mass kilogram kg
Length meter m
Time second s
Electric Current ampere A
Temperature kelvin K
Amount of Substance mole mol
Luminous Intensity candela cd

3. Derived Quantities
A derived quantity in Physics can be obtained from the multiplication or division of the base quantities; no numerical factors are involved.

4. Homogeneity of physical equations

A physically correct equation must be homogenous.

A physical equation is said to be homogeneous if each of the terms separated by plus, minus or equality signs on the left and right side of the equation has the same dimensions, i.e., have the same base units.

i.e If the equation A = B + C is homogenous, than the units of A = the units of B = the units of C
Note that you cannot add units together.

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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor

TOPIC 2: DNA AND GENOMICS – Part 1

LEARNING OUTCOME

(a)Describe the structure and roles of DNA and RNA (tRNA, rRNA and mRNA).(Mitochondrial DNA is not required.)

ESSAY ANSWER
STRUCTURE OF DNA

The DNA molecule consists of two polynucleotide chains twisted around each other to form a double helix. Each nucleotide is made up of deoxyribose sugar, phosphate group and the nitrogenous bases can be adenine, thymine, cytosine or guanine.

In each polynucleotide chain, the alternating pentose sugars and phosphate groups form the sugar phosphate backbone. The nitrogenous bases project into the helix. The bases of both chains are flat structures and can therefore be stacked. The bases lie perpendicular to the central axis. The nitrogenous bases are 0.34 nm apart.

The two polynucleotide chains (or strands) are antiparallel. The two polynucleotide strands are held together by hydrogen bonds that form between the nitrogenous bases of opposite strands. Complementary base-pairing rule govern the formation of hydrogen bonds. Adenine pairs with thymine and forms two hydrogen bonds, while cytosine pairs with guanine and forms three hydrogen bonds.

Hydrogen bonds between bases of two polynucleotides, together with hydrophobic interactions between the stacked bases, stabilises the structure of the double helix. The strong covalent phosphodiester bonds between adjacent nucleotides maintain the integrity of the DNA base sequence.

The width between the two sugar-phosphate backbones of a DNA molecule is constant at 2 nm. This is equal to the width of a purine and a pyrimidine. Each complete turn of the helix is 3.4 nm; there are 10 bases per turn in each chain.

ROLE OF DNA
DNA functions in DNA replication. DNA replication ensures that the amount of DNA in the daughter cells is the same as that in the parent cells at the end of mitosis. It also ensures that genetic information is passed down accurately from parent cell to daughter cells during mitosis.

STRUCTURE OF mRNA
mRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.
ROLE OF mRNA The role of mRNA is to serve as a template for protein synthesis in translation.

STRUCTURE OF tRNA
tRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.

tRNA has a clover leaf structure that results from the hydrogen bonds between complementary base pairs at certain regions. The 5’ end of the tRNA always ends in guanine. The 3’ end is the amino acid attachment site. It binds to an amino acid specific to the tRNA molecule. The anticodon is made up of a specific triplet base sequence. The triplet base sequence at the anticodon is complementary to the codon
found on the mRNA.

ROLE OF tRNA
The role of tRNA is in matching the correct amino acid to its codon in mRNA, in the process of translation.

STRUCTURE OF rRNA
rRNA consists of 1 polynucleotide chain. It is made up of nucleotide monomers. Each nucleotide monomer consists of a phosphate group, ribose sugar and the nitrogenous bases adenine, uracil, cytosine or guanine.

ROLE OF rRNA
The role of rRNA is in ribosome synthesis. rRNA is synthesized in the nucleolus. The nucleolus combines proteins imported from the cytoplasm with rRNA to form the large and small subunits of the ribosomes.

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A Level GP/General Paper Tuition Singapore

The nutrition puzzle

THINK. As you read article 1, consider these questions:
o What are the effects of malnutrition?
o What are some measures to improve nutrition and what are the challenges faced?

In Eldorado, one of São Paulo’s poorest and most misleadingly named favelas, some eight-year-old boys are playing football on a patch of ground once better known for drug gangs and hunger. Although they look the picture of health, they are not. After the match they gather around a sack of bananas beside the pitch. “At school, the kids get a full meal every day,” explains Jonathan Hannay, the secretary-general of Children at Risk Foundation, a local charity. “But in the holidays they come to us without breakfast or lunch so we give them bananas. They are filling, cheap, and they stimulate the brain.” Malnutrition used to be pervasive and invisible in Eldorado. Now there is less of it and, equally important, it is no longer hidden. “It has become more visible—so people are doing something about it.”

If Eldorado’s slum children today eat better, it is partly thanks to José Graziano da Silva. He ran Brazil’s Fome Zero(zero hunger) campaign, a policy that has helped to cut hunger by more than a third in Latin America’s largest country. Now Mr Graziano wants to apply the lessons he has learned more widely: he recently took over as head of the United Nations’ Food and Agriculture Organisation (FAO). And he stands a better chance of success than his predecessors. His appointment coincides with a shift in the world’s approach to fighting hunger.

Governments around the world are paying increasing attention to nutrition. In 2010 donors, charities and companies drew up a how-to policy guide called SUN (which stands for scale up nutrition). Britain’s Department for International Development and other aid agencies are devoting more of their money to nutritional projects. The World Bank has nailed its colours to the mast with a book called “Repositioning Nutrition as Central to Development”. Save the Children, an international charity, talks about “galvanising political leadership” behind the effort. Underlying all this is a change in thinking about how best to improve nutrition, with less stress on providing extra calories and food and more on improving nutrition by supplying micro-nutrients such as iron and vitamins.

In many countries nutritional standards vary according to the season. Often both the amount and quality of food drop alarmingly in the months before the main harvest. Nutrition varies also within households. Mothers eat less in bad times to leave more for their older children, which harms the suckling child. Culture adds to the problem. In rural Bangladesh an attempt to improve nutrition by educating young mothers backfired, because the family diet turns out to be determined not by mothers, but by mothers-in-law. And nutrition can also be improved in all sorts of ways, including by better sanitation, which reduces intestinal diseases and enables people to absorb more nutrients; by investing in smallholder farming, to increase dietary variety; by vaccinating children against diseases; by educating women to breastfeed babies for longer, to improve immunity. Marie Ruel, of the International Food Policy Research Institute in Washington, DC, ticks off some of the tasks: focus on the first 1,000 days of life (including pregnancy); scale up maternal-health programmes and the teaching of good feeding practices; concentrate on the poor; measure and monitor the problem.

All this implies that a successful effort to improve nutrition has to push all the buttons at once. Brazil’s Fome Zero has 90 separate programmes run by 19 ministries. It embraces everything from a conditional cash-transfer scheme, called Bolsa Família, to irrigation projects and help for smallholders. Such an effort is hard to organise and cannot work unless politicians support it. “Malnutrition reduction needs powerful champions who know how to get things done across government, avoid gobbledygook and finish the story,” says Lawrence Haddad, director of Britain’s Institute of Development Studies.

Hence the importance of Mr Graziano, the FAO’s new boss. Interest in improving nutrition is growing; so is alarm at the failures of fighting malnutrition so far. He will not find it easy to cajole more countries into a large, broad-based effort. Governments are reluctant to change and want clear evidence. And just as the damage from malnutrition builds up over a lifetime, so better nutrition reveals its benefits only over many years, as well-fed mothers pass on good health to well-fed children. At a recent FAO conference someone was heard to remark that “at the moment nutritionists are in a position similar to environmentalists in the 1990s.” That is depressing, because it means progress will be slow; but it is encouraging, because progress will come eventually.

APPLY. Now, try answering this essay question: To what extent is the individual responsible for his own health?

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A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

IONISATION ENERGIES

• Ionisation involves the removal of electron(s), forming a cation.

• When the electrons in the atom occupy the lowest energy levels, the atom is said to be in its
ground state.

• When an electron is promoted to one of the higher energy levels, the atom is unstable and is said to be in an excited state.

(a) M (g)—> M+ (g)+e– H = 1st I.E.

First ionisation energy (1st I.E.) is the energy required to remove 1 mole of electrons from one mole of gaseous atoms in the ground state to form one mole of gaseous singly- charged cations.

(b) M+ (g) —> M2+ (g)+e– H = 2nd I.E

Second ionisation energy (2nd I.E.) is the energy required to remove 1 mole of electrons from one mole of gaseous singly-charged cations to form one mole of gaseous doubly- charged cations.

• Ionisation energies affect the type of bond formed by the atom with other atoms. Elements with low ionisation energies will find it easy to lose an electron to form a cation, resulting in ionic bonds being formed.

• Ionisation energies are positive values (i.e. endothermic) since energy is absorbed during ionisation to overcome the attraction between electron and nucleus.

• The 2nd I.E. > 1st I.E. because more energy is required to remove an electron from a positive ion (compared to a neutral atom) due to greater electrostatic attraction between the positive ion and the valence electron.

Factors influencing ionisation energies

Ionisation energy (I.E.) of an atom is influenced mainly by two factors:

Effective nuclear charge, Zeff
• An electron in the atom faces two main electrostatic forces, namely the attractive force by the nucleus (nuclear charge) and the repulsive force by electrons closer than itself to the
nucleus (shielding effect).

• Effective nuclear charge is the combined effect of nuclear charge, Z and shielding effect, S, caused by inner electrons:

Zeff = nuclear charge (Z)  shielding effect (S)

• Higher Zeff ⇒ stronger forces of attraction between nucleus and valence electron,
⇒ higher ionisation energy

(a) Size of (positive) nuclear charge, Z
• indicates the electrostatic forces of attraction between the protons in the nucleus and the valence electrons
• nuclear charge increases with an increase in proton number
• stronger attraction between the positive nucleus and valence electrons
• more energy is required to remove the valence electron

(b) Shielding (or screening) effect of the inner electrons, S
• shielding of the valence electrons from the electrostatic attraction of the positively charged nucleus by the inner electrons
• electrons in the same subshell offer poor shielding for one another
• shielding effect increases with an increase in the number of inner electrons
• weaker attraction between the positive nucleus and valence electrons
• less energy is required to remove the valence electrons

Distance of the valence electron from the nucleus (i.e. the size/radius of the atom)
• attraction of the positive nucleus for the valence electron decreases as distance of electron from nucleus increases
• ionisation energy decreases as n increases

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O-Level Additional Mathematics Tuition Singapore

The Factorial n!

The notation n! is read as ‘n factorial’. It is defined for a positive integer n as the product of the first n positive integers.

Examples

(a) 3! = 3 x 2 x 1 = 6

(b) 4!= 4 x 3 x 2 x 1 = 24

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O Level E Maths Tuition Singapore/Tuition O Level E Maths/Tutor

Simplification of Algebraic Expressions

1. In algebra, symbols like x, y, xy, z2 are used to represent numbers and variables.

2. We add or subtract the like terms by adding or subtracting the coefficients. For example:
3a + 5a – 7a = a

3. We do not add or subtract the coefficients of unlike terms. Hence, adding 3a and 5b gives 3a + 5b.

4. If an expression within brackets is multiplied by a number, each term within the
brackets must be multiplied by that number when the brackets are removed. For example:
8 (x + 2y) = 8x + 16y

Simplify the following expressions:
i. 6a + 2a – 4a
ii. 4b + 7b – a – 2b
iii. 7ab – 3ab + 3bc – 5/4 cb 2
iv. 4x – (2x – 3x)

Ans
i. 4a
ii.9b-a
111. 11/2ab – 1/2bc
iv. 5x

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O-Level Singapore/O-Level/Physics and Chemistry Tuition/Physics Tutor

EXPERIMENTAL CHEMISTRY – Do you Know?

1. Experimental Design

Burettes and pipettes are both used to measure accurate volumes of liquids because they have been accurately calibrated.

Pipettes are calibrated to measure fixed volumes such as 10.0 cm3, 25.0 cm3. Burettes measure up to an accuracy of 0.1 cm3.

2. Uses of filtration

Insoluble salts such as copper(II) oxide, lead(II) iodide are separated from water by filtration.
The filter paper has tiny holes (called pores) that enable the particles of liquid (e.g. water, ink dyes, dissolved sodium chloride) to pass through, retaining behind the larger solid particles (e.g. sand, copper(II) oxide).

Solutions such as aqueous sodium chloride can be collected as a filtrate as the sodium and chloride ions are small enough to pass through the pores of the filter paper.

3. Crystallisation

A saturated solution is one that contains the maximum amount of solute that can possibly dissolve in it at a given temperature. A hot solution can dissolve more solute than a cold one. Hence, on cooling, the bulk of the solute is obtained as crystals.

A hot saturated solution gives large crystals when cooled slowly e.g. cooling at room temperature gives larger crystals compared to freezing. This is because in freezing, particles in a saturated solution have a shorter period of time to pack closer together to form a larger crystal.

Crystals can be re-crystallised (that is dissolved again in the same solvent, then repeating the entire crystallisation process) in order to obtain purer crystals.

After crystallisation, the crystals can be weighed and the percentage purity of the impure salt can be calculated using:

Purity = (Mass of salt obtained/ Initial mass of impure salt )x 100%

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Energy,Work and Power

1. Energy is the ability to do work.

2. Energy can exist in many forms. It can be converted from one form to another.

3. Gravitational potential energy Ep is the energy possessed by an object due to its position in a gravitational field.

Ep = mgh

4. Kinetic energy Ek is the energy possessed by an object due to its motion.

Ek = 1/2 m v^2

5. The Principle of Conservation of Energy states that the total amount of energy in an isolated system is constant.

Total initial energy = total final energy

6. Work done on an object W is the product of force F acting on the object and distance s travelled by the object in the direction of the force.

W = Fs

7. The SI unit of energy and work done is the joule (J).

8. Power P is the rate of work done W. Power can also be defined as the rate of energy conversion E.

P = E/t

9. The SI unit of power is the watt (W). It can also be expressed in joule per second (J s-’).

10. Efficiency is the ratio between useful energy output Eoutput. and energy input Einput.

Efficiency = Eoutput/Einput

If you need help in the above topics, please contact Angie @96790479 or Mr Ong @98639633

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O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

EXPERIMENTAL CHEMISTRY – Do you Know?

1. Experimental Design

Burettes and pipettes are both used to measure accurate volumes of liquids because they have been accurately calibrated.

Pipettes are calibrated to measure fixed volumes such as 10.0 cm3, 25.0 cm3. Burettes measure up to an accuracy of 0.1 cm3.

2. Uses of filtration

Insoluble salts such as copper(II) oxide, lead(II) iodide are separated from water by filtration.
The filter paper has tiny holes (called pores) that enable the particles of liquid (e.g. water, ink dyes, dissolved sodium chloride) to pass through, retaining behind the larger solid particles (e.g. sand, copper(II) oxide).

Solutions such as aqueous sodium chloride can be collected as a filtrate as the sodium and chloride ions are small enough to pass through the pores of the filter paper.

3. Crystallisation

A saturated solution is one that contains the maximum amount of solute that can possibly dissolve in it at a given temperature. A hot solution can dissolve more solute than a cold one. Hence, on cooling, the bulk of the solute is obtained as crystals.

A hot saturated solution gives large crystals when cooled slowly e.g. cooling at room temperature gives larger crystals compared to freezing. This is because in freezing, particles in a saturated solution have a shorter period of time to pack closer together to form a larger crystal.

Crystals can be re-crystallised (that is dissolved again in the same solvent, then repeating the entire crystallisation process) in order to obtain purer crystals.

After crystallisation, the crystals can be weighed and the percentage purity of the impure salt can be calculated using:

Purity = (Mass of salt obtained/ Initial mass of impure salt )x 100%

If you need help in the O level Chemistry, please contact Angie @96790479 or Mr Ong 98639633

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O-Level Singapore/O-Level/Pure Physics Tuition/Physics Tutor

Energy,Work and Power

1. Energy is the ability to do work.

2. Energy can exist in many forms. It can be converted from one form to another.

3. Gravitational potential energy Ep is the energy possessed by an object due to its position in a gravitational field.

Ep = mgh

4. Kinetic energy Ek is the energy possessed by an object due to its motion.

Ek = 1/2 m v^2

5. The Principle of Conservation of Energy states that the total amount of energy in an isolated system is constant.

Total initial energy = total final energy

6. Work done on an object W is the product of force F acting on the object and distance s travelled by the object in the direction of the force.

W = Fs

7. The SI unit of energy and work done is the joule (J).

8. Power P is the rate of work done W. Power can also be defined as the rate of energy conversion E.

P = E/t

9. The SI unit of power is the watt (W). It can also be expressed in joule per second (J s-‘).

10. Efficiency is the ratio between useful energy output Eoutput. and energy input Einput.

Efficiency = Eoutput/Einput

If you need help in the above topics, please contact Angie @96790479 or Mr Ong @98639633

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A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor

Hi A-Level/H2 Math Students

Normal Approximation to the Binomial Distribution

X ~ B (n, p) can be approximated by a normal distribution with μ = np and σ2 = np(1 – p)
i.e. X ~ N (np,np(1- p))approximately if
(i) n is large, such that n > 30 ;
(ii) p is large enough, such that np > 5 and n(1- p) > 5 .

Note that X ~ B(n, p) has a discrete distribution
whereas X ~ N(np,np(1-p)has a continuous distribution.

Therefore, we need to adjust the values of X by using a continuity correction

Normal Approximation to the Poisson Distribution

X ~ Po(λ) can be approximated by a normal distribution with μ = λ and σ2 = λ
i.e. X ~ N(λ,λ) approximately if λ is large (λ > 10).

Note that Po(λ) has a discrete distribution whereas N(λ,λ) has a continuous distribution.
Therefore, we need to adjust the values of X by using a continuity correction

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A-Level Economics Tuition Singapore/H2/H1 Economics Tuition

Topic 2.2 Macroeconomic Aims, Problems/Issues, Consequences and Policies

Syllabus
Supply Side Policy and its Effectiveness

Outcome
Explain supply side policies.
Analyse the effects of supply side policies on the economy.
Evaluate the effectiveness of supply side policies in achieving macroeconomic aims.

Macroeconomics Lecture 17 : Supply Side Policies

17.1 Definition

Supply-side policies are government policies designed to affect Aggregate Supply (AS) directly, specifically policies to reduce costs or raise productivity. They are also to minimize friction in markets and to improve information flow. It seeks to achieve economic gains by long-term measures designed to raise the rate of economic growth by shifting the long run AS curve to the right rather than manipulating AD.

(a) Reducing Government Expenditure – Release More Resources to the Private Sector

A more efficient use of resources within the public sector and a reduction in the size of the sector can allow private investment to increase with no overall rise in AD. Thus the supply-side benefits of higher investment could be achieved without the demand-side costs of higher inflation.

Grants and subsidies are reduced. When subsidies are reduced, many nationalized industries will increase their prices and reduce their costs since their objective is to maximize profits. However, these policies are not without problems. In some sectors, the effect may be to cut services rather than increase efficiency. Since it may be easier to reduce long term capital expenditure (e.g. for developmental purposes) than recurrent expenditure (e.g. wages), there may be a sharp decline in new roads, schools etc.

(b) Reducing Taxes to raise work incentive

One way to encourage savings is to reduce personal income tax rates and shift the emphasis to taxes on consumption so that any income saved is subject to minimal income taxes.

Supply-side advocates also claim that high taxes discourage work. Hence, to encourage people to work (thereby shifting the AS to the right), income taxes should be lowered.

Business income is taxed firstly as the income of firms when it is earned and secondly as income of households when it is paid out as dividends. Income taxes reduce business profits and lower the return to investing in company shares. Thus, households will be discouraged from investing in businesses thus limiting the growth of firms which in turn limits economic growth.

! Stop and Think : How has Singapore adjusted its corporate and income tax rates over the last ten years?
Corporate tax in Singapore has reduced from 25.5% in 2001 to 17% in 2010.
Personal Income tax rate for those in the highest income bracket (taxable income exceeding S$320k has reduced from 22% in 2003 to 20% currently (since 2007). Even those in the lower taxable income bracket i.e. more than S$20,000, the tax rate has reduced from 4% to 2% in the same period.

Please contact Angie Hp 96790479 or Mr Ong 98639633 if you need help in Economics and complete Notes

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A-Level Physics Tuition Singapore/H2 Physics Tuition/JC Physics Tutor

Hi A-level/H2/JC Physics Tuition students

Mastering Qualitative Questions

Chapter 20 Nuclear Physics

1. Describe how two samples, one emitting alpha particles, and the other emitting beta particles can be distinguished through a simple school laboratory experiment, using a Geiger-Muller (GM) tube connected to a ratemeter. [3]

2. State one similarity and one difference between radioactive decay and nuclear fission.

3. Distinguish between the radioactive decay and the fission of a nucleus

4. Explain why tracks due to gamma radiation in a cloud chamber are hardly visible.

5. Why energy released during Nuclear Fusion is more than Nuclear Fission?

6. What are the Quantities that must be conserved in any radioactive decay process

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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor

TOPIC 1: CELLULAR FUNCTIONS – Part 19

LEARNING OUTCOME

(s)Describe, with the aid of diagrams, the behaviour of chromosomes during meiosis, and the associated behaviour of the nuclear envelope, cell membrane and centrioles. (Names of the main stages are expected, but not the subdivisions of prophase)

ESSAY ANSWER

Prophase I
Condensation of the chromatin occurs until the chromosomes become visible. Synapsis occurs, where homologous chromosomes pair
up, forming bivalents. Crossing-over occurs between non-sister chromatids of homologous chromosomes, at chiasmata. Crossing over enables exchange of genetic material between homologous chromosomes
and results in new combination of alleles on chromosomes of gametes. Nucleoli disintegrate and nuclear envelope disappears. Centrioles
moves to opp. poles

Metaphase I
2 rows of chromosomes are now aligned along the equator of the spindles, each chromosome is still paired with its homologous partner. The two hromosomes of each homologous pair faces opposite poles. If crossing over and exchange of genetic material occurred in Prophase I, non-sister chromatids will carry genetic material from the other chromosome. Independent assortment of chromosomes also occurs at this stage.
Kinetochore microtubules attach to the kinetochores of chromosomes

Metaphase I
2 rows of chromosomes are now aligned along the equator of the spindles, each chromosome is still paired with its homologous partner. The two chromosomes of each homologous pair faces opposite poles. If crossing over and exchange of genetic material occurred in Prophase I, non-sister chromatids will carry genetic material from the other chromosome. Independent assortment of chromosomes also occurs at this stage.
Kinetochore microtubules attach to the kinetochores of chromosomes

Anaphase I
The homologous chromosomes are separated. One homologous chromosome
is pulled to one pole, and the other to the opposite pole. Centromeres remained intact and sister chromatids are not separated at this stage. Kinetochore microtubules shorten to pull the chromosomes to their respective poles.

Telophase I & Cytokinesis I
Chromosomes reach opposite spindle poles. Separation of homologous chromosomes implies that chromosome number is halved after cytokinesis. Each nucleus has a haploid (n) chromosome set. Each chromosome exist
as two sister chromatids.

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A Level GP/General Paper Tuition Singapore

Unit Practice: Author’s Intention

Writers always aim to achieve a certain specific effect with each sentence or phrase. They carefully choose words, phrases, images, analogies, examples and even punctuation in order to achieve this. This not only adds layers of meaning to the text, but also lends precision—the reader is guided along the author’s line of thought with the thoughtful and strategic placement of language features.

The Author’s Intention question type is designed to test your sensitivity to language and to reward your ability to accurately distill

(1) the reason for the author’s approach or the effect the author aims to achieve, and often you are also required to

(2) explain the context or meaning of the language feature used to achieve it. Accurately tracking the writer’s ideas as well as his (positive or negative) attitude to the subject of the sentence can help you deduce his intentions in using these language features to present information.

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