Shannon G.

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Biomedical Science (Bachelors) - Oxford, Oriel College University

4.9

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

215 completed lessons

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#### Ratings & Reviews

4.9from 72 customer reviews

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

#### Qualifications

BiologyA-level (A2)A*
ChemistryA-level (A2)A*
MathsA-level (A2)A*
FrenchA-level (A2)A*

#### General Availability

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#### Subjects offered

SubjectQualificationPrices
BiologyA Level£24 /hr
ChemistryA Level£24 /hr
MathsA Level£24 /hr
BiologyGCSE£22 /hr
ChemistryGCSE£22 /hr
MathsGCSE£22 /hr
ScienceGCSE£22 /hr

### How do I find the turning points of a curve?

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=x2-5x+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).

Once turning point is identified, you can work out if it is a maximum or minimum by finding d2y/dx2.

d2y/dx2<0 - maximum

d2y/dx2.>0 - minimum

Thus for our example above

d2y/dx2=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=x2-5x+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).

Once turning point is identified, you can work out if it is a maximum or minimum by finding d2y/dx2.

d2y/dx2<0 - maximum

d2y/dx2.>0 - minimum

Thus for our example above

d2y/dx2=2 - minimum

3 years ago

18710 views

### How do I solve a quadratic equation by factorising?

A quadratic equation is one that includes xas the highest power of x. Factorising is achieved in 3 steps. Let’s consider the example x2-3x-3=1

1) Put the equation into the form ax2+bx+c=0

x2-3x-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

- This technique can be applied to finding the points of intersection on the x axis for a quadratic graph. For example, y=x2-3x-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.

2x2 + 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

2x2 + 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

x2 + 6x + 8=0

x2 – 8x + 16 = 0

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

y=x2 – 14x + 48

and sketch this function

A quadratic equation is one that includes xas the highest power of x. Factorising is achieved in 3 steps. Let’s consider the example x2-3x-3=1

1) Put the equation into the form ax2+bx+c=0

x2-3x-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

- This technique can be applied to finding the points of intersection on the x axis for a quadratic graph. For example, y=x2-3x-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.

2x2 + 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

2x2 + 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

x2 + 6x + 8=0

x2 – 8x + 16 = 0

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

y=x2 – 14x + 48

and sketch this function

3 years ago

1068 views

### How does meiosis achieve 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.

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.

3 years ago

1109 views

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