As a biomedical science student at Oxford University, I am offer strong and knowledgeable support for biology, chemistry and maths from GCSE upwards. What makes me the right tutor for you? Here are a few of my personal qualities that I believe demonstrate my tutoring skills: 1) I am passionate about my subject and about education: I love learning. I get butterflies in my stomach sitting in lectures and tutorials discussing the topics I love. In the same way, I love teaching and grasp the opportunity to pass what I know to others with similar interests. 2) I have experience I have one to one tutoring experience with younger students in maths and English. Additionally, I can adapt my teaching skills as required to suit the student as shown presenting my research project at the Big Bang fair in London having gotten through to the final. I was praised for my ability to interact with audiences from primary school students to university professors. 3) I have the ability and the knowledge It's clearly essential to know what you are talking about when tutoring. My high grades in A level maths, biology and chemistry (A*) and the awards for highest achiever in biology in chemistry recognise that this is the case. Additionally, grade A in French A level and in Ethics AS level show that I am able to apply myself to other areas and manage a high work load. What's on offer? I offer high quality online tutoring in biology, chemistry and maths at GCSE and A level. Lessons will aim to engage students including video and audio where possible to aid interaction. I can adapt to student requirements by altering the length and format of tutorials to suit. Some examples of lesson format include: - Interactive white board work - Presentation style teaching - Working through past papers and sheets - Question and answer sessions - Revision sessions I believe that to make lessons as effective as possible, it is important to outline the aims and key points at the beginning and provide a written summary at the end to aid students when looking over notes. Additional support I am also keen to help with university applications and personal statements having had experience of the application process at Oxford and other universities.

As a biomedical science student at Oxford University, I am offer strong and knowledgeable support for biology, chemistry and maths from GCSE upwards. What makes me the right tutor for you? Here are a few of my personal qualities that I believe demonstrate my tutoring skills: 1) I am passionate about my subject and about education: I love learning. I get butterflies in my stomach sitting in lectures and tutorials discussing the topics I love. In the same way, I love teaching and grasp the opportunity to pass what I know to others with similar interests. 2) I have experience I have one to one tutoring experience with younger students in maths and English. Additionally, I can adapt my teaching skills as required to suit the student as shown presenting my research project at the Big Bang fair in London having gotten through to the final. I was praised for my ability to interact with audiences from primary school students to university professors. 3) I have the ability and the knowledge It's clearly essential to know what you are talking about when tutoring. My high grades in A level maths, biology and chemistry (A*) and the awards for highest achiever in biology in chemistry recognise that this is the case. Additionally, grade A in French A level and in Ethics AS level show that I am able to apply myself to other areas and manage a high work load. What's on offer? I offer high quality online tutoring in biology, chemistry and maths at GCSE and A level. Lessons will aim to engage students including video and audio where possible to aid interaction. I can adapt to student requirements by altering the length and format of tutorials to suit. Some examples of lesson format include: - Interactive white board work - Presentation style teaching - Working through past papers and sheets - Question and answer sessions - Revision sessions I believe that to make lessons as effective as possible, it is important to outline the aims and key points at the beginning and provide a written summary at the end to aid students when looking over notes. Additional support I am also keen to help with university applications and personal statements having had experience of the application process at Oxford and other universities.

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Archibald (Parent from Newport)

Very pleasant and patient to work with.

Veronica (Parent from Watford)

Dear Shannon, Many thanks for the tutorial provision that you offered. I very much appreciated the high-quality and detailed lessons that you provided. Best Wishes, Simeon

Gina (Parent from Slough)

June 8 2016

Yasmin got B's in her GCSE Science Double Award. Thanks Shannon.

Yasmin (Student)

May 16 2016

Amazing! Was dreading my Biology exam a few months ago and now I feel 100 times more confident about it! :D

At turning points, the gradient is 0. Differentiating an equation gives the gradient at a certain point with a given value of x. To find turning points, find values of x where the derivative is 0.

**Example:**

y=x^{2}-5x+6

dy/dx=2x-5

2x-5=0

x=5/2

Thus, there is on turning point when x=5/2. To find y, substitute the x value into the original formula.

y=(5/2)^{2}-5x(5/2)+6

y=99/4

Thus, turning point at (5/2,99/4).

**Additional points**

Once turning point is identified, you can work out if it is a maximum or minimum by finding d^{2}y/dx^{2. }

d^{2}y/dx^{2}<0 - maximum

d^{2}y/dx^{2.}>0 - minimum

Thus for our example above

d^{2}y/dx^{2}=2 - minimum

At turning points, the gradient is 0. Differentiating an equation gives the gradient at a certain point with a given value of x. To find turning points, find values of x where the derivative is 0.

**Example:**

y=x^{2}-5x+6

dy/dx=2x-5

2x-5=0

x=5/2

Thus, there is on turning point when x=5/2. To find y, substitute the x value into the original formula.

y=(5/2)^{2}-5x(5/2)+6

y=99/4

Thus, turning point at (5/2,99/4).

**Additional points**

Once turning point is identified, you can work out if it is a maximum or minimum by finding d^{2}y/dx^{2. }

d^{2}y/dx^{2}<0 - maximum

d^{2}y/dx^{2.}>0 - minimum

Thus for our example above

d^{2}y/dx^{2}=2 - minimum

A quadratic equation is one that includes x^{2 }as the highest power of x. Factorising is achieved in 3 steps. Let’s consider the example x^{2}-3x-3=1

*1) Put the equation into the form ax ^{2}+bx+c=0*

x^{2}-3x-4=0

2*) Factorise*

We need two numbers that

- add together to get -3

- Multiply together to get -4

-4x1=-4 and -4+1=-3

Thus, factorising gives (x-4)(x+1)=0

*3) Solve the equation!*

If two numbers are multiplied together to give 0, one of them must be 0. Thus:

x-4=0 and x=4

x+1=0 and x=-1

The equation has been solved

**Additional points:**

- This technique can be applied to finding the points of intersection on the x axis for a quadratic graph. For example, y=x^{2}-3x-4. At the x axis, y=0 so you can work out x as above.

- Harder quadratic equations can also be solved by factorising. For example when a isn't 1.

2x^{2} + 7x + 3=0

Find two numbers that multiply to give 2x3 (6) and add to give 7. In this case, 6 and 1.

Split 7x into 6x +x

2x^{2} + 6x+x + 3=0

Factorise each part by taking out a common factor.

2x(x+3)+1(x + 3)=0

The sames as

(2x+1)(x+3)=0

thus x = -1/2 or x=-3

__Practice questions__

1. Solve by factorising

*x*^{2} + 6*x* + 8=0

*x*^{2} – 8*x* + 16 = 0

2. Find the points of intersection with the x axis for

y=*x*^{2} – 14*x* + 48

and sketch this function

A quadratic equation is one that includes x^{2 }as the highest power of x. Factorising is achieved in 3 steps. Let’s consider the example x^{2}-3x-3=1

*1) Put the equation into the form ax ^{2}+bx+c=0*

x^{2}-3x-4=0

2*) Factorise*

We need two numbers that

- add together to get -3

- Multiply together to get -4

-4x1=-4 and -4+1=-3

Thus, factorising gives (x-4)(x+1)=0

*3) Solve the equation!*

If two numbers are multiplied together to give 0, one of them must be 0. Thus:

x-4=0 and x=4

x+1=0 and x=-1

The equation has been solved

**Additional points:**

- This technique can be applied to finding the points of intersection on the x axis for a quadratic graph. For example, y=x^{2}-3x-4. At the x axis, y=0 so you can work out x as above.

- Harder quadratic equations can also be solved by factorising. For example when a isn't 1.

2x^{2} + 7x + 3=0

Find two numbers that multiply to give 2x3 (6) and add to give 7. In this case, 6 and 1.

Split 7x into 6x +x

2x^{2} + 6x+x + 3=0

Factorise each part by taking out a common factor.

2x(x+3)+1(x + 3)=0

The sames as

(2x+1)(x+3)=0

thus x = -1/2 or x=-3

__Practice questions__

1. Solve by factorising

*x*^{2} + 6*x* + 8=0

*x*^{2} – 8*x* + 16 = 0

2. Find the points of intersection with the x axis for

y=*x*^{2} – 14*x* + 48

and sketch this function

Meiosis produces 4 haploid daughter cells and is essential for forming gametes that contain one copy of each chromosome. Meiosis allows genetic diversity which gives rise to new combinations of genetic alleles so that offspring have varied phenotypes and natural selection occurs.

Genetic diversity is achieved by 3 main mechanisms:

1) Random chromatid assortment

When paired homologous chromosomes line up on the equator in metaphase 1, paternal and maternal chromatids are assigned to either cell randomly. Thus, each cell receives a random combination of maternal or paternal chromosomes.

2) Crossing over

In order to keep homologous chromosomes paired, chiasmata form in which complementary sections of homologous chromosome cross over thus exchanging genetic material. This creates new allele combinations on the same chromosome. The further the genes are from the centre of the chromosome, the easier crossing over is and the more frequent recombination.

3) Random fertilisation

Any sperm can fertilise any ovum and thus again, a random combination of homologous chromosomes is produced in the zygote adding to genetic diversity.

Meiosis produces 4 haploid daughter cells and is essential for forming gametes that contain one copy of each chromosome. Meiosis allows genetic diversity which gives rise to new combinations of genetic alleles so that offspring have varied phenotypes and natural selection occurs.

Genetic diversity is achieved by 3 main mechanisms:

1) Random chromatid assortment

When paired homologous chromosomes line up on the equator in metaphase 1, paternal and maternal chromatids are assigned to either cell randomly. Thus, each cell receives a random combination of maternal or paternal chromosomes.

2) Crossing over

In order to keep homologous chromosomes paired, chiasmata form in which complementary sections of homologous chromosome cross over thus exchanging genetic material. This creates new allele combinations on the same chromosome. The further the genes are from the centre of the chromosome, the easier crossing over is and the more frequent recombination.

3) Random fertilisation

Any sperm can fertilise any ovum and thus again, a random combination of homologous chromosomes is produced in the zygote adding to genetic diversity.