Degree: Molecular & Cellular Biochemistry (Masters) - Oxford, Oriel College University
Hello! A warm welcome to all…
My name is Adam and I am a biochemistry student at Oxford University. In brief, Biochemistry encompasses all things science but particularly requires a deep and thorough understanding of maths, organic chemistry and cellular biology, subjects I adore! Throughout my course I will apply advanced mathematical techniques learnt at A-level and beyond to unravel the hidden world of chemistry that governs life itself.
As an ex-elite gymnast myself, I am no stranger to hard-work, strategic planning and dedication. Over the past 6 years, coaching the sport has also shaped me into a patient and friendly character who has a strong set of problem-solving and communication skills which will aid me as a tutor. To add to my teaching experience, I initiated and lead many extracurricular clubs at my old sixth form for younger students to take part in, the most successful being an early morning “Fit4Figures” club. We offered support to year 7 students who struggled a lot with maths by tutoring individuals on a one-to-one basis. As well as tutoring myself, I was also responsible for planning a years-worth of sessions, providing relevant resources and most importantly motivating the students enough to get out of bed and into school for 7am every week! Finally, in 2014 I travelled to Chennai with the school’s India Direct team where we visited a couple of children’s homes and were fortunate enough to teach some basic English. My teaching experience, as well as being tutored before when I was younger, means I am very familiar with the roles and responsibilities of a tutor.
Free trail session
During our free trial session I will create a student profile. This will include: what level of study you are in (GCSE, AS or A2); your exam boards; your preferred style of learning; current/target grades (not essential to me); what your future plans are e.g. what you hope to study at A-level or university. This profile will help me to tailor the sessions to your specific needs and will also help me to identify strengths and weaknesses.
Right from the get go, we will get to work right away. Notice the word ‘we’. My tutorials will not involve me lecturing you for the whole time because at the heart of true learning is cooperation between the teacher and student. Prior to our first session, I will have made a list of specific and measurable learning outcomes and prepared practice questions to apply what you have learnt, all corresponding to your current level of understanding and student profile.
I appreciate that students learn in different ways so I will adjust to the most suited style of teaching for you. I will go at your pace and help to break topics down into more manageable chunks.
My final, but most important, point about my tutorial sessions is that you can literally ask me any question. Through high school and sixth form, teachers always commented at parents evening about how I ask a lot of questions, sometimes too many! In my opinion, there is no such thing as being "too inquisitive", or a silly question, so fire away!
EXAM TECHNIQUE: The most important skill you need as a student, especially in light of the current syllabi for the sciences, is exam technique. I often found that simply understanding a topic was never enough. Once you understand the topic (which is the hardest bit), it is then about trying to “read the examiners mind” in your exams, as my teachers used to say, and to do this requires a strong exam technique e.g. identifying what exactly it is that the question is asking of you, using the correct terminology, knowing when to suggest ideas and think outside the box etc. To train you up and develop these skills, I will regularly set you past paper questions matched to the covered topic. After working through it together and then marking your work, we will assess how well you are understanding the topic and what needs further improvement.
I am happy to help with science/maths UCAS applications including med. school preparation (I did A LOT of work experience during Yr12 as I was originally applying for med. school) as well as Oxbridge applications and interview tips. After all, I was once in your shoes and had to write that dreaded personal statement, so I know how daunting it all is!
I am also happy to help with any E.P.Q. issues. Whatever the project please don't be afraid to ask for help. My project, which scored 50/50, was about life on Mars and so definitely stretched me academically and taught me a lot about time management, organisation of work etc.
So, what next?
If you have any questions, send me a ‘WebMail’ or book a ‘Meet the Tutor Session’! I am generally available during the evenings of weekdays but I am definitely willing to arrange tutorials around what is convenient for you-contact me for more details
I can’t wait to work with you!
Adam Waldron :P
|Biology||A Level||£20 /hr|
|Chemistry||A Level||£20 /hr|
|Extended Project Qualification||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|-Personal Statements-||Mentoring||£20 /hr|
|Physics (AS-LEVEL)||A-Level||A (AS-LEVEL)|
|Extended Project Qualification||A-Level||A*|
A key point to remember here is that a^x could mean the base (a) is not “e”-that special number which has a gradient function, dy/dx (the differential), EQUAL to the function itself, y=e^x. When "a" is any real number, you must treat it differently to functions with the base "e".
For the function y=a^x, the rule is simply dy/dx=a^x*lna
(^ means “to the power of”. * means “multiply”)
The proof for this requires an understanding of implicit differentiation (differentiating both x and y terms within an implicit relation).
Start with the function
ln (natural log) both sides of the equation
(the implicit part means you differentiate with respect to y but multiply by dy/dx after...)
1/y*dy/dx=lna (<----"xlna" differentiates to simply lna because a is just a constant and so lna is also just a constant)
Multiply across by y
N.B. differentiating a^(f(x)) where f(x) is a function of x requires use of the chain rule too (first set the function f(x) to another letter, say u, so you have a^u. Differentiate a^u with the above rule. Differentiate u=f(x) separately, then use chain rule).
The easiest way to visualise this is to imagine a line of four dinner ladies serving you different parts of a school meal e.g. the first peas, the second mash potatoes, the third chicken and the fourth gravy. To move to the next station you must have the preceding station’s food on your plate (e.g. you must have mash to be served chicken).You're walking down the line of dinner ladies, you have your peas and mash, but then the lady serving chicken is reallyyyyy slow-she is slowing the entire process down! However fast you collect your peas and mash, it is impossible to reduce the time it takes to reach the gravy dinner lady because it’s the chicken collection that is the slowest stage, relative to the other workers. Similarly, in a multi-stage reaction, the stages usually follow on one from the other, the finishing materials of one stage acting as the starting materials of the next. Therefore, the RATE OF THE SLOWEST STEP WILL GOVERN THE RATE OF THE ENTIRE REACTION. This is the rate determining step.
N.B. any step that occurs AFTER the rate-determining step will not affect rate. Also, by studying the order of a reaction from the rate equation, you are able to learn more about the rate-determining step (what is involved, what isn’t).see more
Firstly, there are chemical differences between the two nucleic acids. DNA has the following structural properties:
-double stranded and anti-parallel polymer held together by (relatively) weak hydrogen bonds
-strong deoxyribose sugar backbone
-consists of the four nucleobases adenine, cytosine, guanine and thymine
-many C-H bonds that improve stability
-helical ('B' form)
On the other hand, RNA (mRNA, tRNA, rRNA, snRNA etc.) has a different structure:
-single stranded polymer that is more flexible
-ribose sugar backbone
-consists of the four nucleobases adenine, guanine, cytosine and URACIL (replacing thymine)
-C-OH groups make it less stable
-helical (but 'a' form)
The structure of the two polymers determines their functions. For example, the double stranded DNA is important for the replication of genetic information and conserving it for generations to come so the process can be repeated again and again (similar to photocopying a recipe book and then storing the photocopies in a safe for future use). This is known as semi-conservative replication. Each strand (after "unzipping" the DNA by breaking the hydrogen bonds) is used as a template to synthesise a new complementary strand. The strong backbone of deoxyribose sugar protects the base sequence, preventing random mutations changing the precious genetic code of a cell. This is reinforced by strong hydrocarbon bonds which increase its stability.
In comparison, RNA is single stranded meaning it can fold in on itself and form intrastrand complementary bonds (as we see in tRNA) and hence perform catalytic functions similar to the way a tertiary protein would. This catalytic ability plays an integral role in processes such as translation, where RNA is responsible for piecing together amino acids into a polypeptide chain (this occurs at the ribosome which is made of rRNA). Similarly, siRNA molecules in the cytoplasm utilise their catalytic ability to "chop" mRNA transcribed from a gene into tiny pieces. This prevents translation taking place and hence a protein is not manufactured, silencing the gene originally transcribed. Therefore, RNA has the ability to control gene expression. Finally, in contrast to DNA, the lack of stability means that RNA serves only as an intermediate rather than a "store" of genetic information.
In conclusion, structure and function go hand in hand. The differences in structure between DNA and RNA determines the differences in their biochemical functions in the cell.see more