Asked by Anonymous Anonymous
Can i use your diagrams in a first year university essay?
Hi, which diagrams are you speaking of?
- Also known as the beta pleated sheet due to the pleated appearance of the protein structure from a side view.
- No strict rules to how they are formed because the hydrogen bonds can be formed between distant amide hydrogen and carbonyl oxygen.
- They are two or more strands distant from each other in the primary structure that form hydrogen bonds with each other side by side.
- There are two types of beta sheet structures; parallel and anti-parallel. Parallel beta sheets have strands that run in the same direction as each other and anti-parallel beta sheets have strands that run in opposite direction to each other. The hydrogen bonding in anti-parallel beta sheets are usually more linear.
- The N-H and C=O groups on the outer edge of the beta sheet structure are not hydrogen bonded to other strands of the primary sequence.
- If the R-groups along the outer edge of the beta sheets are polar, it can interact with solvents such as water. If they are non-polar, they can interact with hydrophobic structures such as lipids.
- They can also pack closely against side chains of nearby alpha helix structures.
- Almost all of the polar amide groups are hydrogen bonded to each other in the beta sheet structure.
- Parallel sheets are almost always buried in the inside of the structure of a protein and anti-parallel sheets are mostly exposed to solvent due to the amino acids that make up that part of the structure. Therefore, anti-parallel sheets are seen as being more stable structures than parallel sheets.
- Parallel sheets usually have other structures, such as helices, separating them from other parallel sheets.
- They have a right handed twist to the beta strands due to the steric factors of the L-amino acid configuration.
- Isoleucine and valine are often found in these beta sheets because they are hydrophobic.
- Beta strands can be amphipathic because of the alternating side chains of amino acids next to each other. These amphipathic strands are found on the surface of proteins.
- A large anti-parallel beta sheet can also form a barrel structures (such as retinol binding protein). The last strand of the beta sheet is hydrogen bonded to the first strand so it forms a closed barrel shape.
- The exterior of the structure is usually surrounded by solvent as it is hydrophilic and the interior is where the hydrophobic residues are found so non-polar species can be found here (e.g retinol).
The only bonds in the secondary structure of a protein are regular repeated hydrogen bonds from the peptide bonds of the amide groups and carbonyl groups. You can also classify protein families by their secondary structures.
- This is the most common secondary structure of proteins.
- The carbonyl oxygen atom act as a hydrogen bond acceptor and the hydrogen attached to the nitrogen atom of the amide group act as the hydrogen bond donor.
- The carbonyl oxygen atom (n) bonds to the hydrogen of the amide group four residues along (n+4) by hydrogen bonding.
- Can be formed when the R-groups in the primary structure is not ‘bulky’.
- If proline is present, the hydrogen bond formation would stop because the amine group is bonded to the R-group in proline. This is why proline is known as the ‘helix breaker’.
- The first amine group and the last carbonyl group of the helix are not involved in the hydrogen bonding of the secondary structure.
- The helix forms a cylinder shape, with the hydrogen bonds forming the walls of the cylinder and the R-groups pointing outwards.
- The properties of the R-groups that make up the primary structure dictates the interactions the helix has with other parts of the protein chain and with other molecules. It also means that the helix can be amphipathic (e.g. one side of the helix is hydrophobic and the other side is hydrophilic).
- They can be right-handed (going clockwise) or left-handed (anti-clockwise) but left-handed helices are very rare.
- The helices can be any length because they have no limit.
- Other variations of the alpha helix exist but are also very rare.
Primary structure of proteins
The primary structure of a protein is extremely important in governing the structure and interaction of the protein. It is made up of a chain of amino acids that are coded for by DNA. Amino acids are always quoted/drawn from the N group to the C group. The amino acids involved also give the protein various chemical properties to allow it to arrange into the different levels of protein structure.
- The amino acid sequence: determines everything about the protein structure.
- Peptide bonds: formed between the amino group of one amino acid to the carboxyl group of another amino acid. It can form hydrogen bonds and is involved in the secondary structure.
- R-group: extremely important in determining the tertiary structure of the protein.
The peptide bond
Polypeptides have a trans arrangement most of the time (R-groups above and below the plane of the polypeptide) because this makes the polypeptide more stable due to less obstruction from neighbouring R-groups. The peptide bonds are planar (can’t rotate) but the covalent bonds either side of it can rotate depending on the R-groups of the amino acids present, which gives it the trans arrangement. This also restricts the number of arrangements the polypeptide can have. The rotation between C-C is called the psi (ψ) angle and the rotation between the N-C bond is called the phi (φ) angle.
They are formed when a hydrogen atom attached to a very electronegative atom is bonded to a very electronegative atom with a lone pair of electrons. The atom that is attached to the hydrogen atom is called the hydrogen bond donor and the non-bonded atom is called the hydrogen bond (H-bond) acceptor. Single H-bonds are relatively weak but many combined together can make the overall H-bond binding strength very strong.
Side chains (R-groups)
- Hydrophobic: these R-groups interact with each other by van der Waals and tend to pack together to avoid the water.
- Hydrophilic: these R-groups can interact, by hydrogen bonding, to each other, peptide bonds, organic polar molecules and water.
- Amphiphatic: these R-groups can interact with both water (by hydrogen bonding) and away from water (by van der Waals interactions).
I just wanted to let you all know that I have now created a new site that will have my new science posts (and older ones too!) in a version that I believe will be easier to follow. I will be adding the same posts to both sites so don’t worry about missing anything if you don’t want to use the given website.
Asked by myscientificsummer myscientificsummer
This looks really helpful! I am doing my A2s this year! I also want to study biochemistry:) are you at university now? which did you get into? I also did the Nuffield Bursary and it was such a good experience!
I’m glad that you’re finding it useful! I am planning on creating a new website soon that will categorise these notes (and future ones) into different sections so it will be easier to read and find.
I’m currently studying biochemistry in UCL (University College London) and I highly recommend this course (and uni!) if you really love these two subjects at A-level.
The Nuffield Bursary will definitely help you hugely when you are applying for university. Make sure you mention it in your personal statement because it gives you a lot to talk about when you go for interviews etc.
Jenny (aka chubbylemonscience) :)
Asked by Anonymous Anonymous
Could you possibly do a chemistry problem for me? I think I know it but I need some sort of help to make sure it's correct? the question is: What is the empirical formula for the following comps: 2.1 percent H, 65.3 percent O, and 32.6 percent S? is it h2so4? i'm confused when i get to the number of moles of H because i get 2.06779. does that round to 2?
I have just worked the question out and I have also got H2SO4. I would suggest that you round the ratios you get at the end to the nearest whole number unless the ratio ends with .5. For example, if you ended with 2.5 for H, you would have to times the answer by 2 to get a whole number (to get 3). You would also times the answers for the other elements by 2 as well to get your final empirical formula.
I hoped that helped. :)
Asked by Anonymous Anonymous
do you do biology mind maps?
Yes, however, I have only done one mind-map for A2 biology. In some of my earlier posts, I have uploaded some coloured notes, which include the main points of certain topics of biology. I also have a few large mind-maps for AS biology in my older posts that you can check out if you are doing AS biology.
Asked by fv-k fv-k
hey i do chemistry at school and your probably gonna think im really dumb because im struggling with learning about limiting reactants and i really need help with it. is there any chance you could give me some tips or explain to me how to do it?? could you reply back to my ask also please :-) thanks
Hey, I don’t think you’re dumb for not getting this at all! I promise that it gets easier with every practice.
Example question: When 15g of Copper(II) Chloride reacts with 20g of sodium nitrate, copper(II) nitrate and sodium chloride are formed. What is the limiting reagent and how much sodium chloride is formed in grams?
Here are my steps for tackling these types of questions:
- Write down the triangles you need for the question. For this question, you only need this triangle.
2. Write down the equation of the reaction and balance it.
3. Work out the ratio of moles for the reactants.
4. Work out how many moles of reactant (CuCl2 and NaNO3) is used in the reaction by using the triangle.
- For CuCl2 : 15g/Mr of CuCl2 = 15g/[63.5+(35.5x2)]=0.112mol
- For NaNO3 : 20g/Mr of NaNO3= 20g/[23+14+(16x3)] = 0.235mol
5. From the ratio I worked out from step 3, I know that for 1 mole of CuCl2, 2 moles of NaNO3 is needed. So in theory, 0.112mol of CuCl2 should react with 0.224mol of NaNO3 (0.112x2), but I have worked out from step 4 that 0.235mol is used in the reaction so NaNO3 is in excess and CuCl2 is the limiting reagent.
6. To work out the grams of sodium chloride formed, I need the balanced equation and the triangle.
- Now that you know the limiting reagent is CuCl2, I will use the number of moles worked out from step 4 (0.112mol) to work out the number of moles for NaCl.
- From the balanced equation, I know the ratio of moles between CuCl2 and NaCl is 1:2.
- No. of moles for NaCl = 0.112x2 = 0.224mol
- Using the triangle, 0.224mol x Mr of NaCl = 0.224 x (23+35.5) = 13.104g
- Round 13.104g to 13g so that the decimal place is the same as those given in the question.
- Therefore, 13g of NaCl is produced.
I hoped that helped. If you follow similar steps when you answer these types of questions, you will get it right. My biggest advice is to remember to balance the equation and remember the triangles!!!!!!
Asked by Anonymous Anonymous
Are your notes edible? ^____^ Btw can I ask what did you get for F325 and do you have any more notes for it? You are an inspiration <3 xxx
Hehe, I wish they were! I actually haven’t taken my F325 exam yet, 3 more weeks to go before I take it…
Thanks for the message and I’m glad that it has helped :)
Asked by Anonymous Anonymous
which page does the AS chem start?
Page 13 :)
Asked by beautyhasmanymoods beautyhasmanymoods
i adore you right now.
Well, thank you! lol :)
Chemistry OCR A2 mind map 16
Chemistry OCR A2 mind map 15
Chemistry OCR A2 mind map 14