Chemistry

Question 1

What is the difference between peptides and Proteins

Answer 1

Peptieds are usually less than 50 amino amino acid residues linked by peptide bonds whereas proteins are more than 50 amino acid residues

Question 2

What potential problems can come from dipeptide synthesis?

Answer 2

There are 3 main problems,
Firstly, Peptides are directional, so in for example the peptide we wanted to synthesise had alanine an the N-terminus and Phenylalanine at the C-terminus there is always the possibility of the peptides being swapped around so Alanine at the C-terminus as although this is not the peptide we were looking for there is nothing that stops it from ocurring

Secondly, In a reaction vessel, Peptides are able to self-condense. This simply means they are able to react with other peptides of themselves as so to form Ala-Ala or Phe-Phe instead of reacting with each other.

Finally, there is a possible problem in peptide synthesis which Depends on the side chains. Reactive side chains are able to participate in reactions. For example Lysine, containing a secondary amine group can participate in the reaction in the place of the other amine group.

Question 3

How do we use protecting groups?

Answer 3

Protecting groups can be used to protect amino acid residues. They are able to be easily removed under mild conditions when required, though will still fufill the role of preventing something like unwanted side reactions that might occur. It makes it so that only the groups we want to react can occur.

After that condensation reaction between the 2 peptide we remove the protecting group potentially for a new cycle of reactions

With amnio acids that have reactive R-groups, we also need to protect these so hence we need more Protecting groups but otherwise the cycle is the same

Question 4

Since peptide bonds do not just spontaneously form, how do we make it so that this bond formation does infact occur?

Answer 4

We need to chemically activate and then couple amino acid derivatives

We activate the carboxyl group by attatching a leaving group to the Acyl carbon of the carboxyl component to facilitate the attack of the amino acid component

This is necessay as otherwise the two amino acids at ambient room temperature will just form a salt

Question 5

How do we activate the carboxyl Group?

Answer 5

We can convert it into an acyl halide which provides a good leaving group to ensure that this happens rapidly under mild condition

Question 6

Describe how Resonance plays a role in the amide bond

Answer 6

The reso structure of the Amide bond has partial double bond character. This involves charge splitting with N being positive and O being negative

As a result the bond is rather chemically inert. The amide group is also planar

Question 7

What is a dihedral angle

Answer 7

A dihedral angle involves for angles and are measured clockwise

Angles of 0 or 120 are called eclipsed and angles of 60 or 180 are called staggered

Essentially, we are looking through the B and C angle

Question 8

What are the Dihedral angles in proteins

Answer 8

Phi, Psi and Omega

Phi is the angle between the NH and the R group The 4 atom which it involves are The C carbon on C=O, N, The C on the R group and The C on the Carbon on the C=O

Psi is from Angle between the C=O and R group. Its dihedral angle is One NH TO NH

Omega is the peptide bond

Question 9

Describe the influence of steric facors on condormaiton

Answer 9

Steric factors influence the conformation, when the R groups are on 2 different sides, and is staggered(180), instead of being eclipsed(0) the peptide bond is preferred as a Cis conormation of the peptide bond can result in steric clashes which are unfavourable

We say that the Phi and Psi angles aer co-dependant. When one is 0 the other cannot be 0 and this is simlplt to avoid steric clashes

Question 10

Describe the role in H bonding in protein structure

Answer 10

Back-bone hydrogen bonding is associated with Scondary structure, whereas side chain Hydrogen bonding is associated with tertiary structure

Question 11

What is sequence conservation and why is it useful

Answer 11

In general, evolution conseves amino acids which are important to structure and function across protein species, we are able to determine this by the alignment of multiple homologue sequences of a particular protein in different species which will reveal the residues which are conserved and those which are not.

Conserved residues tell us the important funcional parts of the protein sequence. Typically only a limited amount of a protein’s sequence is absolutely conserbed, Howeber insulin is a proein which is highly conseved which makes insulin therapy with different species a possiblity

Question 12

Describe protein structure stabilising forces?

Answer 12

These are essential to keep ptoreins stable.

They can be grouped into covalent interactions(peptide bonds and disulfide bridges)

Non-covalent bonds(Pi-Pi interactions, VDW forces, Hydrogen bonds, Electrostatic interactions, Hydrophobic interactions)

Question 13

What are the surface characteristics on a protein?

Answer 13

Hydrophillic residues are predominantly on the surface with hydrophobic regions predominantly being in the core, an exception of this may be membrane proteins which are in a lipohillic environment so will tend to appear more on the surface.

If soluble proteins have hydrophobic surface region, they have a particular function, for example for ligand binding in the acive site of an enzyme.

Question 14

Describe Folding and conformational changes in a protien

Answer 14

A chain of amino acids fold into their tertiary structure with its main driving force being that the non-polar groups want to clump into the middle to avoid contact with water. Drom this, it will maximise the other interactions.

Proteins fold to their native process through a directive pathway going through several intermediate states. A molten globule can for by hydrophobic collapse.

Chaperone proteins assist in the folding of proteins in the cell which are unable to do this by themselves

Question 15

Descibe me 3 differernt ways we can determine protein structure

Answer 15

X-ray crysollogaphy, where the proteins are growin in crystals and by exposing it to an X-ray source, we can detect it with a detector to see its electron density and get a model of the Structure

NMR spectroscopy

Cryo-electron microscopy

Question 16

Describe Covalent bonds in proteins

Answer 16

Peptide bonds are very stable and require harsh conditions with 6M concentration of acid or allkali to break them apart

Disulfide bridges can form between cysteine residues via thiol groups. They can be broken down into cysteines via B-mercaptoethanol to re-form thiols.

Question 17

Describe Non-covalent bonds in proteins

Answer 17

Hydrogen bonds vary in strength but contribute most to protein stability in the core as they are the firsthest away from water which would disrupt it. They can be disrupted by heat

VDW forces are generally weak, short range interactions due to dipole dipole interactions between close atoms

Pi-Pi oceralp occurs between Pi elecron clouds delocalised over rings and bonds. These can also be disrupted by heat.

Electrostatic bonds are sronger interactins which are stronger when two charged groups are closer. They can be broken down high ionic strengths which may compete with the existing salt

Hydrophobic interactions are interactions between hydrophoic sife chains which are generally forced to gether to minimise the exposure to water

Question 18

Describe The structure of insulin

Answer 18

Insulin is originally produced in the cells as preproinsulin. Once it enters the Endoplasmic reticulum it becomes Preonsulin as the signalling peptide is cleaved. It is then processed through the Golgi apparatus to become the mature form of insulin.

In its mature form,Insulin consists of 2 Chains. Its A chain is linked to the B chain with 2 disulfide bridges, with there being an additional bridge in the A chain on its own.

The A chain consists of 21 residies, with he B chain containing 30.

Insulin is a Globular hormone which normally exists as a Monomer. When it however is in higher concentrations insulin dimerizes and in the prescence of zinc further associates into hexamers. However only the monomer is the biologically active species

Under conditions where the native state is desyabilised, insulin aggregates to form amyloid fibres which can cause biomedical complications as it can interfere with the requirements of functional insulin in each dose. This mainly occurs at hugh insulin concentration

Question 19

Explain protein Engineering in Insulin

Answer 19

There are a number of insulin analogues which have been liscenced for treatment including

Rapid acting insulin analogues. After SC injection the proportion that is bound in the for of Hexamers and dimers is lower which means that the monomeric form of the molecule can be absorbed at the point of injection faster- insulin lispro, insulin aspart and insulin glulisine are examples

In the Insulin Aspart the proline is substituted with an aspartic acid residue which as increased charge repulsion and prevents the formation of hexamers

Question 20

Describe the ionisation states in the Histidine side-chain

Answer 20

Histidine contains an imidazole ring which hsa a pKa value of approximately 6. This means that in Physiological conditions, relatively small shifts in pH values below the pH of 6 will cause the imidazole ring to become protonated and carry charge equally between the two structures, This can be represented with two equally important resonance structures

Question 21

Describe how Post-translational modifications to proteins can alter them

Answer 21

Post-translational modification is the process of modifying a protein after translation

An example of modification is the Phosphorylation of Hydroxyl groups of Ser, Thr or Tyr with the result being an added charge. This is believed to be key to some signalling pathways in the cell

Glycosylation is additional Sugar moieties to Ser, Thr or Asn resideus which can alter the solubility of the protein

Hydroxylation is the addition of OH groups to Pro it Lys residues and can alter Hydrogen bonding

Disulfide bridge formation. The joining of two cysteine residues to produce a more stable protein with additional covalent linkage

Methylation, is the addition of a methyl group to an Oxygen or Nitrogen. This is the addition of a hydrophobic group

Question 22

What is a possible consequences of too many groups

Answer 22

A protein with too many basic groups, will have a net positive charge in physiological conditions as the groups will all be protonated

A protein with too many acidic chains will have a net negative charge.

Question 23

What is the Isoelectric point

Answer 23

The isoelectric point is where the overall protein carries no net charge.

The individual amino acids may be charged but the Overall protein is not.

It depends on the amino acids present(How many basic or acidic groups the protein has) and the pH of the solution environment

Question 24

Describe the role of the isoelectric point in Long-acting insulin

Answer 24

Insulin glargine is a long-acting analogue which contains 2 extra arginine residues at the end of the B chain to alter the Isoelectric point. This increases the IEP of insulin and alters the solubility. By increasing the IEP it is more soluble in lower acidic pH conditions which are used in formulations but less soluble upon injections

This works as the inulin is soluble in solution but when injection ist becomes less soluble and begins to form dimers and hexamers which form by precipitation. This then means when they enter the body they will take a longer time to monomeric forms of insulin which are active thus producing a longer acting effect.

Question 25

Describe Gel electrophoresis

Answer 25

This is a method of separation and analysis of macromolucules like proteins

The principle is that molecules can be separated according to their size and charge.

Proteins are electrophoreised within a matrix or gel and then in an electric field, charged molecule migrate either towards an anode or athode dependent on their charge

The force of attraction to the electrode depends on two different factors. The size of the charge on the molecule, and the size of the electic field

We also need to take into account forces which prevent movement of molecules such as Friction and/or repulsion in medium

So overall Ion velocity =electrometric mobility x applied electric field

Question 26

What is absorbtion spectrophometry

Answer 26

In absortion spectrophotometry, a dissolved substance will absorb light of a specific wavelength characterises by that substance.
We can via the extinction coefficient determine the Concentration

Question 27

What is the use of UV absorbtion in measuring protein concentration

Answer 27

UV absorbtion is a good way in detecting protein concentrations as no additional reagents or incubates are required. Furthermore, No protein standard is needed to be prepared. The assay also does not consume protein. The relationship between the absorbance to protein Conc is linear.

However, any non-protein component in the solution that absorbs UV light will interfere with the assay

Question 28

What protein detection methods can we use post Gel-electrophoresis

Answer 28

• Typical detection methods include
  
  • Coomassie Brilliant Blue dye staining
  • Western blot(Protein immunoblot)
    
    • Transfer of gel contents onto a membrane and detection via labelled antibodies

Question 29

Describe SDS-PAGE

Answer 29

The native structure of proteins is maintained during electrophoresis. Separation is done according to size and charge. The Acrylamide gels serve as a size-selective sieve during separation. As proteins move through a gel in response to an electric field, the smaller molecules travel more rapidly in comparison to the larger proteins. In addition to this, Highly negative molecules will migrate faster than less negatively charged molecules towards the anode. These 2 effects in combination mean that a highly negatively charged larger molecule will migrate faster than a less negatively charged smaller molecule. We might do this to assess the quaternary structure of the protein to see if it is a hexamer, a dimer, or a monomer.