Febi S. GCSE Biology tutor, A Level Biology tutor, GCSE Chemistry tut...

Febi S.

Currently unavailable: until 17/10/2016

Degree: Medicine (Bachelors) - Imperial College London University

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About me

Hiya! I’m Febi, a first year medical student at Imperial College, offering to tutor Maths to GCSE and Chemistry and Biology to A level! I’m known to be friendly, encouraging and patient, qualities which I definitely bring to my tutoring. I had taught a wide range of ages at the tuition centre Explore Learning for over year, giving me invaluable tuition experience, and learning how to make learning so much more fun and enjoyable. Plus, I tutored GCSE and A-level biology and chemistry during sixth form at school, whether one-to-one or as a revision class, so I’m not short of experience! I really love tutoring, and I really love my subjects too (hence why I’m studying medicine!) and I really hope you might develop this same enthusiasm too!

Before we start the tuition sessions, we can chat about what you’re struggling with, and form an action plan about how to reach your goals. We will cover the basic understanding of the subjects, enough so that you can explain it to me, then we can work on the exam technique that goes with it. We can also work together to find out what type of learning or revision works for you, and will help you get the grades you want!

Honestly, I want you to be confident enough and able enough to not need me for too long! And we can have fun with it! There’s no pressure to ‘get it right’ straight away (you probably get enough of that from school anyway!).

If you have any questions about me (my degree course and application, any jobs I’ve done, my teaching style etc), feel free to contact me! I’m open to questions and happy to help. :)

Subjects offered

SubjectLevelMy prices
Biology A Level £20 /hr
Chemistry A Level £20 /hr
Biology GCSE £18 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr
Maths 13 Plus £18 /hr
-Medical School Preparation- Mentoring £20 /hr

Qualifications

QualificationLevelGrade
BiologyA-LevelA*
MathsA-LevelA*
ChemistryA-LevelA*
UKCATUni Admissions Test722.5 avg
BMATUni Admissions Test5.2,4.9,3.5A
EPQA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

05/05/2016

Currently unavailable: until

17/10/2016

Questions Febi has answered

How does a deletion of a base in the DNA base sequence lead to a mutated protein?

It is important to note that bases on the DNA (and mRNA) are read in separate sets of 3s along a DNA/(mRNA) stand (this is called a triplet code in DNA, or the codon in mRNA). So, if a base is deleted, the whole sequence undergoes a frame shift in which the base after the deleted one is read w...

It is important to note that bases on the DNA (and mRNA) are read in separate sets of 3s along a DNA/(mRNA) stand (this is called a triplet code in DNA, or the codon in mRNA). So, if a base is deleted, the whole sequence undergoes a frame shift in which the base after the deleted one is read with the original group (E.g. ATTGCAC --> ATGCAC). This means that the triplet code (the set of 3 bases) is different. Note above how both triplet codes have been altered by this. This will happen to ALL the sequences following the deletion, so all those triplet codes will be altered and incorrect.

As the codon on the mRNA is transcribed from this altered DNA sequence, it will also have a different codon. The codon on the mRNA is translated into a polypeptide, with each codon of 3 bases coding to 1 specific amino acid. Hence, if the codon is wrong, the amino acid is wrong, so the polypeptide is wrong.

Polypeptides are long chains of amino acids strung together, and are folded into specific 3D shapes (the tertiary structure) to form a protein. They do this by forming hydrogen, ionic and/or disulphate bonds with the specific amino acids in the chain. If these amino acids aren’t there because a different one has been coded for by the mRNA due to the deletion in the DNA, then these bonds cannot form. Hence, the 3D shape will not be formed so the protein is not formed correctly, hence mutating it. 

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3 months ago

86 views

What are surds and how does multiplying them work?

A surd is an irrational number which cannot be expressed as a fraction or a recurring number. E.g. root 2 = 1.4142135…. and it continues(but it does NOT repeat like a recurring fraction though). E.g. pi would actually be a surd too, because it cannot be accurately presented as a simple number ...

A surd is an irrational number which cannot be expressed as a fraction or a recurring number. E.g. root 2 = 1.4142135…. and it continues (but it does NOT repeat like a recurring fraction though). E.g. pi would actually be a surd too, because it cannot be accurately presented as a simple number like a fraction or recurring decimal. 1.4.... or 32/9 are not surds for that matter.

If you were to round 1.4142135… to 1.414, and then square it when trying to get back to 2, you will in fact get 1.999396. This is because rounding is a type of approximation, so you will not get the exact figure for 2 but for a number close to it. This would be the same with all surds. Root 2 would be the only fully accurate representation of the number.

Hence (root 2)2 would be 2. To help understand this, imagine it was root 16.
Root 16= 4.
42 = 16.
Hence, (root 16)2 = 16.
In the same way, any rooted surd multiplied by itself = the number inside the surd.

And in terms of multiplying different surds…

Root 9 x Root 4 = Root 36

We can prove this is because

Root (9x4) = Root 36   = 6 and root 9 x root 4 = 3 x 2 = 6. Therefore, this is correct!

Hence, with any surd, the inside number of the surds can be multiplied to make a new surd. (e.g. root 3 x root 5 = root 15 ) :)

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3 months ago

109 views

You added 75cm^3 of 0.5moldm^-3 HCl to impure MgCO2, and some was left unreacted. The unreacted HCl reacted completely with 21.6cm^m of 0.5moldm^-3 NaOH. So what is the percentage purity of the MgCO3 sample?

MgCO3 +2HCl --> MgCl2 + H2O +CO2 For this kind of question, I would always try to convert what I can into moles because that is the link between the solid (the grams of the MgCO3) and the liquid (the HCL and NaOH). Here is my process: Since mol=conc x vol, the mol of HCl initially added to th...

MgCO3 +2HCl --> MgCl2 + H2O +CO2

For this kind of question, I would always try to convert what I can into moles because that is the link between the solid (the grams of the MgCO3) and the liquid (the HCL and NaOH). Here is my process:

Since mol=conc x vol, the mol of HCl initially added to the impure MgCO3 is 0.5moldm-3 x 75x10-3 dm3 = 0.0375 moles.

The mol of NaOH that reacted with the excess HCl moldm3-3 is 0.5 x 21.6x10-3 dm3 = 0.0108 moles.

Because the NaOH and HCL are in a 1 to 1 molar ratio: (1NaOH + 1HCl --> NaCl +H2O), the mole of excess HCl is also 0.0108.

(All in moles) Total HCl initially added to MgCO3 sample – HCl that actually reacted with (the pure) MgCO3 = Excess HCl (which was reacted with the NaOH).
So, 0.0375 - HCl that actually reacted with (the pure) MgCO= 0.0108.
HENCE, HCl that actually reacted with (the pure) MgCO3 = 0.0375 – 0.0108 = 0.0267 moles.

As the (pure)MgCOand HCl reacted in a 1:2 ratio (see equation above) the mol of MgCO3 is 0.0267/2 = 0.01335

The molecular mass (Mr) of (pure) MgCO3 is 84.3 (worked out from the periodic table)

So as the mass(of pure MgCO3) = mol x Mr, the mass = 0.01335 mol x 84.3 = 1.125g

Since the mass of impure MgCOis 1.25g, the percentage mass of pure MgCOis 1.125/1.25 x100% = 90%

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3 months ago

90 views
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