Now we've moved onto carbohydrates! (Actually, we covered nucleic acids after proteins, but there was more new material in the carbohydrates lectures so I'm going to talk about them first.)
Learn the terminology associated with carbohydrates.
Be able to classify the different
types of carbohydrates.
Wow, this dot point is broad. A carbohydrate is basically a sugar or a chain of sugar molecules, but the formal definition is something like "a polyhydroxyaldehyde or polyhydroxyketone, or a substance that gives these compounds on hydrolysis." Basically what this means is that a carbohydrate is either an aldehyde or a ketone with plenty of hydroxyl groups, or it's made up of several compounds like these.
A single polyhydroxyaldehyde or polyhydroxyketone is known more simply as a monosaccharide. Two monosaccharides can join together to make a disaccharide. Short chains of monosaccharides (generally around 2-10 monosaccharides) are known as oligosaccharides, whereas longer chains are known as polysaccharides.
Learn the general formula.
Simple. The general formula of a monosaccharide is CnH2nOn, where n is between 3 and 8.
Be able to give the number of
carbons for the category name.
Carbohydrates can also be categorised according to the number of carbons that they have. The number of carbons is given by the prefix tri-, tetra- etc., which is then attached to the suffix -ose to give triose (3-carbon sugar), tetrose (4-carbon sugar), pentose (5-carbon sugar) and so on.
Be able to identify common functional groups associated with carbohydrates.
I think this dot point here is referring to talking about aldehyde and ketone groups in sugars. When sugars are in their linear form, they are either aldehydes, with a C=O bond at the end of the molecule, or ketones, with C=O bonds somewhere in the middle of the molecule. Sugars that are aldehydes are known as aldoses, whereas sugars that are ketones are known as ketoses. You can combine the aldo- and keto- prefixes with the numbering prefixes to give ketohexose, aldetriose etc. However, normally the keto- or aldo- parts are omitted.
There are several other functional groups that may be attached to carbohydrates. One of these is the acetyl group, which is a C double bonded to an O and single bonded to a methyl group. There are also amino (-NH3+/-NH2) groups and amido groups (like acetyl groups but with an amine group instead of a methyl group).
Be able to describe chirality.
The definition of a chiral molecule is a molecule that cannot be superimposed on its mirror image. For example, your right hand is roughly a mirror image of your left hand, but you can't rotate it so that it looks the same as your left hand. Chirality generally occurs when a carbon atom is bonded to four different groups.
Be able explain the difference between D & L carbohydrates.
Know that the D form is the biological form.
D and L carbohydrates are named according to the orientation of the asymmetric carbon furthest from the carbonyl carbon in a Fischer projection (i.e. when the molecule is drawn in its linear form). In a D sugar, the hydroxyl group is on the right, whereas in an L sugar, the hydroxyl group is on the left. Most if not all sugars exist in the D form.
Understand the terminology associated with chirality.
Be able to distinguish between stereoisomers when presented with a figure.
A stereogenic centre is the centre around which chirality can take place. As I mentioned earlier, chirality generally occurs when a carbon atom is bonded to four different groups. In this case, the carbon atom is a stereogenic centre.
Enantiomers are pairs of stereoisomers that are mirror images of each other. Diastereomers, on the other hand, differ between one or more chiral centres in their molecules, but they are not complete mirror images of each other.
A meso compound is a molecule that has one or more chiral centres, but the molecule as a whole is symmetrical and therefore achiral.
Be able to number the carbons in a carbohydrate.
The carbons in a carbohydrate are simply numbered 1, 2, 3 etc., starting from the carbonyl carbon of an aldose, or starting from the end carbon nearest to the carbonyl carbon of a ketose.
Be able to describe an anomer.
Anomers are isomers that differ only in their orientation around the carbonyl carbon (a.k.a. the anomeric carbon) when the carbohydrate is in its ring form (I think).
Recognise the difference between the α and β anomers.
When the carbohydrate is in its ring form, there will usually be a terminal -CH2OH group. If the -OH on the carbonyl carbon is facing the side of the ring opposite to the -CH2OH group, it's in the alpha form, whereas if the -OH group is on the same side of the ring, it's in the beta form.
Know the difference between a Fischer and Haworth projection.
I've already alluded to this before, but a Fischer projection is basically a way of drawing the carbohydrate when it's in its linear form, whereas the Haworth projection is a way of drawing the carbohydrate when it's in its ring form.
Be able to recognise the difference between a pyranose and a furanose.
Simply put, a pyranose is a 6-membered ring whereas a furanose is a 5-membered ring. Generally one "member" of these rings is an O atom while all the other members are C atoms. Hexoses tend to be pyranoses and pentoses tend to be furanoses.
Be able to identify the different structural conformations of a pyranose sugar.
Pyranoses are not completely flat. If one carbon is above the plane of the ring, and another carbon on the opposite side is below the plane of the ring, the sugar is said to be in the "chair" conformation (if you search up some diagrams you'll see why). If one carbon is above the plane of the ring, and another carbon on the opposite side is also above the plane of the ring, the sugar is said to be in the "boat" conformation.
Be familiar with the structure of glucose, an important monosaccharide.
Glucose is a hexose- that is, it is a sugar with six carbons. It forms pyranose rings. Generally, in a Haworth projection, the hydroxyl groups on the 2nd, 3rd and 4th carbons will be in a "down up down" configuration, respectively.
Tuesday, November 3, 2015
Sunday, November 1, 2015
Proteins- Enzymes
Last lecture on proteins! w00t w00t! I can finally see the light!
You should know the difference between a 1st order and a 2nd order reaction and be able to write a rate equation for a simple reaction.
You should understand that in a reversible reaction at equilibrium there is no net reaction in either direction, even though the reaction is occurring in both directions, the rates of these reactions are equal. You should follow how this leads to the definition of the equilibrium constant for a reversible reaction and know what this is in terms of rate constants and equilibrium reactant and product concentrations.
Uhhhh I'm already stuck. (Clearly I stopped paying attention in these lectures after the first few. Admittedly, I had a very hard time staying awake during these lectures on proteins.) My notes weren't very helpful either, and neither were the slides, so a big thank you to the UC Davis ChemWiki, where I pinched the following from (paraphrased, of course- I'm not that dishonest).
A first-order reaction is a reaction in which the rate is directly proportional to the concentration of one of the reactants. Most of these take the form A -> B. In these reactions, the rate can be found by calculating -(dA/dt), or dB/dt, or k[A], where [A] is the concentration and k is some constant.
A second-order reaction is a reaction in which the rate is directly proportional to the product of the concentration of the reactants. These can take the form A + B --> C, or 2A --> B. In the first case, the rate is k[A][B], while in the second the rate is k[A][A], or simply k[A]^2.
Note that the above were all for irreversible reactions. In reversible reactions, you also need to take the products into account, as I'm going to show you now.
In a reversible reaction, you can calculate the forward and reverse reactions separately, as shown above for the first- and second-order reactions. The overall forward rate is equal to the forward rate minus the reverse rate. For example, in the reaction A + B <--> C + D, the overall forward rate is equal to kf[A][B] - kr [C][D] (f and r are meant to be subscripts).
Equilibrium is defined as the state in which there is no net reaction in either direction. In this case, kf[A][B] - kr [C][D] = 0. This can be rearranged to give kf[A][B] = kr[C][D] and then kf/kr = ([C][D])/([A][B]) (note that all of the concentrations in this case have to be the concentrations at equilibrium). kf/kr can also be called Keq (with "eq" as a subscript... when will Blogger put a subscript option? Maybe I should ask them to...), or the equilibrium constant.
You should understand the difference between ΔG and ΔG° and the relationship between ΔG° and the equilibrium constant for a reaction.
Delta G is the change in free energy for a reaction, while Delta G degrees is the change in free energy under specific conditions, namely the conversion of 1 mole of each of the products to 1 mole of each of the reactants. The relationship between these two can be summarised as follows (for a reaction A + B <--> C + D):
ΔG = ΔG° + RT ln(([C][D])/([A][B]))
where R is the gas constant (8.314J/mol/degrees Kelvin) and T is the absolute temperature in Kelvin.
When the reaction is at equilibrium, there is no net formation of the products (or the reactants, for that matter) and thus the free energy change is 0:
ΔG° + RT ln(([C][D])/([A][B])) = 0
This can be rearranged to:
ΔG° = -RT ln(([C][D])/([A][B]))
which, since the reaction is at equilibrium, can be simplified even further to
ΔG° = -RT ln K(eq)
You should know the difference between ΔG and ΔG‡ and know that it is ΔG‡ that determines the rate constant for the reaction and hence the rate of the reaction. You should understand that ΔG‡ is always positive, but ΔG can be positive or negative and that if it is negative the reaction will proceed spontaneously.
Delta G double dagger (sorry, not sure what to call it, but I think that's what the lecturer called it so I'm just going to copy him) is the activation energy of a reaction. I explained activation energy in one of my earlier Chemistry 3AB posts. Essentially, it's the energy required to reach the high-energy transition state. This is a considerable barrier to cross and thus the rate of reaction is limited by the proportion of molecules that have enough energy to cross this barrier. As the transition state is always of high energy, Delta G double dagger is always positive.
As I mentioned before, the free energy change is represented by Delta G. It can be positive, if the products are of a higher energy state than the reactants, or negative, if the inverse is true. If Delta G is positive, energy needs to be added to allow the reaction to occur; if Delta G is negative, the reaction will proceed spontaneously (though it may be very slow if the activation energy is high).
You should be able to sketch a free energy diagram illustrating the difference between a catalysed reaction and the same reaction that is uncatalysed. You should be able to estimate values of ΔG and ΔG‡ from a free energy diagram of a reaction where values are indicated on the free energy axis.
See my earlier post on reaction rates because I really cbf drawing diagrams right now.
You should know the definition of catalytic power.
Catalytic power is defined by the rate constant of the catalysed reaction divided by the rate constant of the uncatalysed reaction. If the catalysed reaction proceeds much faster, this ratio will be large, indicating a high catalytic power. If the catalysed reaction only proceeds a little bit faster, this ratio will be relatively small, indicating a low catalytic power.
You should understand that a catalytic site on a enzyme is part of its tertiary structure where substrates bind and catalysis occurs. You should also understand that the positioning of amino acid residues in the active site determines the position and orientation of bound substrates and also provides for the specificity of enzymes for particular substrates.
As I've probably mentioned 50 billion trillion times before, structure is related to function. The catalytic site, or active site, requires that a substrate fit there before it can carry out its work. The ability of an enzyme to accommodate a substrate properly depends on the amino acid residues surrounding the active site, as well as the shape of the substrate itself.
You should be able to explain why the induced fit model of enzyme action better explains enzyme catalysis than the lock and key model and understand how an enzyme stabilises the transition state conformation(s) of a substrates(s) by forming extra bonds between this conformation and the active site of the enzyme.
The "lock and key" model is the one that you've probably heard all throughout high school (and which I've used a bit in this blog as well). Basically it goes like this: substrates have a particular shape, designed to fit in the binding sites of specific enzymes, just like how a key fits into a lock.
Even though the lock and key model is pretty elegant and easy to understand, there's an even better model available. You see, the lock and key explains how a substrate and an enzyme can fit together, but it doesn't explain why the substrate-enzyme complex remains stable even as the reaction proceeds. A better model is the induced fit model, which compares the enzyme as a glove and the substrate as a hand. Just as a glove may have to stretch to accommodate the hand, the enzyme may have to accommodate the substrate.
An important concept behind the induced fit model is that the enzyme can exist in a lower-energy state and a higher-energy state. When the enzyme is in a higher-energy state, it can accommodate the substrate. As the reaction proceeds, the enzyme can change shape slightly to accommodate the transition state, allowing for stability at this state as well. Eventually the products are released from the enzyme.
You should be aware of the requirement of some enzymes for cofactors.
Cofactors are other molecules etc. that some enzymes require for them to work. I've already mentioned one of these- pyridoxal phosphate (PLP). Aside from PLP, there are many other cofactors that can be found in enzymes, such as ions and vitamins. The activity of specific cofactors depends on the reaction in question.
You should understand why most enzymes lose activity above about 45°C and be aware of the existence of thermostable enzymes from thermophilic organisms.
At a high temperature, denaturation of enzymes may occur. Denaturation is the unfolding of a protein so that it no longer adopts its required configuration. This occurs at around 45 degrees Celsius due to the weak hydrogen bonding that is responsible for most of the folding of a protein. Some thermophilic organisms (organisms that are adapted to high temperatures) have more thermostable enzymes that allow them to survive in the heat. These may contain more covalent disulfide bonds which are more resistant to changes in temperature.
You should understand that in many cases the activity of an enzyme is dependent on pH because of the presence in the active site of amino acid side chains that undergo pH-dependent protonation or deprotonation, but which need to be in a particular protonation state in order to allow substrate binding or for catalysis to occur. You should also understand that a bell-shaped pH activity curve results from having 2 such ionisable residues in the active site, one of which needs to be protonated and one of which needs to be deprotonated for substrate binding/catalysis to occur.
As mentioned in my post about the acid base chemistry of amino acids, ionisable groups of amino acids can exist in a protonated or deprotonated form, depending on the pH. This affects charge, which in turn can affect substrate binding or catalysis. Sometimes there are multiple amino acids in the active site that are affected by pH. Hence, pH is often important for enzyme activity, with activity peaking at the optimal pH and decreasing as the pH increases or decreases.
YUSSSSSSSSSSSS NO MORE AMINO ACIDS OR PROTEIN LECTURES TO GO THROUGH FOR THIS UNIT!!!!!!!!!!!
Now it's time to go to sleep...
You should know the difference between a 1st order and a 2nd order reaction and be able to write a rate equation for a simple reaction.
You should understand that in a reversible reaction at equilibrium there is no net reaction in either direction, even though the reaction is occurring in both directions, the rates of these reactions are equal. You should follow how this leads to the definition of the equilibrium constant for a reversible reaction and know what this is in terms of rate constants and equilibrium reactant and product concentrations.
Uhhhh I'm already stuck. (Clearly I stopped paying attention in these lectures after the first few. Admittedly, I had a very hard time staying awake during these lectures on proteins.) My notes weren't very helpful either, and neither were the slides, so a big thank you to the UC Davis ChemWiki, where I pinched the following from (paraphrased, of course- I'm not that dishonest).
A first-order reaction is a reaction in which the rate is directly proportional to the concentration of one of the reactants. Most of these take the form A -> B. In these reactions, the rate can be found by calculating -(dA/dt), or dB/dt, or k[A], where [A] is the concentration and k is some constant.
A second-order reaction is a reaction in which the rate is directly proportional to the product of the concentration of the reactants. These can take the form A + B --> C, or 2A --> B. In the first case, the rate is k[A][B], while in the second the rate is k[A][A], or simply k[A]^2.
Note that the above were all for irreversible reactions. In reversible reactions, you also need to take the products into account, as I'm going to show you now.
In a reversible reaction, you can calculate the forward and reverse reactions separately, as shown above for the first- and second-order reactions. The overall forward rate is equal to the forward rate minus the reverse rate. For example, in the reaction A + B <--> C + D, the overall forward rate is equal to kf[A][B] - kr [C][D] (f and r are meant to be subscripts).
Equilibrium is defined as the state in which there is no net reaction in either direction. In this case, kf[A][B] - kr [C][D] = 0. This can be rearranged to give kf[A][B] = kr[C][D] and then kf/kr = ([C][D])/([A][B]) (note that all of the concentrations in this case have to be the concentrations at equilibrium). kf/kr can also be called Keq (with "eq" as a subscript... when will Blogger put a subscript option? Maybe I should ask them to...), or the equilibrium constant.
You should understand the difference between ΔG and ΔG° and the relationship between ΔG° and the equilibrium constant for a reaction.
Delta G is the change in free energy for a reaction, while Delta G degrees is the change in free energy under specific conditions, namely the conversion of 1 mole of each of the products to 1 mole of each of the reactants. The relationship between these two can be summarised as follows (for a reaction A + B <--> C + D):
ΔG = ΔG° + RT ln(([C][D])/([A][B]))
where R is the gas constant (8.314J/mol/degrees Kelvin) and T is the absolute temperature in Kelvin.
When the reaction is at equilibrium, there is no net formation of the products (or the reactants, for that matter) and thus the free energy change is 0:
ΔG° + RT ln(([C][D])/([A][B])) = 0
This can be rearranged to:
ΔG° = -RT ln(([C][D])/([A][B]))
which, since the reaction is at equilibrium, can be simplified even further to
ΔG° = -RT ln K(eq)
You should know the difference between ΔG and ΔG‡ and know that it is ΔG‡ that determines the rate constant for the reaction and hence the rate of the reaction. You should understand that ΔG‡ is always positive, but ΔG can be positive or negative and that if it is negative the reaction will proceed spontaneously.
Delta G double dagger (sorry, not sure what to call it, but I think that's what the lecturer called it so I'm just going to copy him) is the activation energy of a reaction. I explained activation energy in one of my earlier Chemistry 3AB posts. Essentially, it's the energy required to reach the high-energy transition state. This is a considerable barrier to cross and thus the rate of reaction is limited by the proportion of molecules that have enough energy to cross this barrier. As the transition state is always of high energy, Delta G double dagger is always positive.
As I mentioned before, the free energy change is represented by Delta G. It can be positive, if the products are of a higher energy state than the reactants, or negative, if the inverse is true. If Delta G is positive, energy needs to be added to allow the reaction to occur; if Delta G is negative, the reaction will proceed spontaneously (though it may be very slow if the activation energy is high).
You should be able to sketch a free energy diagram illustrating the difference between a catalysed reaction and the same reaction that is uncatalysed. You should be able to estimate values of ΔG and ΔG‡ from a free energy diagram of a reaction where values are indicated on the free energy axis.
See my earlier post on reaction rates because I really cbf drawing diagrams right now.
You should know the definition of catalytic power.
Catalytic power is defined by the rate constant of the catalysed reaction divided by the rate constant of the uncatalysed reaction. If the catalysed reaction proceeds much faster, this ratio will be large, indicating a high catalytic power. If the catalysed reaction only proceeds a little bit faster, this ratio will be relatively small, indicating a low catalytic power.
You should understand that a catalytic site on a enzyme is part of its tertiary structure where substrates bind and catalysis occurs. You should also understand that the positioning of amino acid residues in the active site determines the position and orientation of bound substrates and also provides for the specificity of enzymes for particular substrates.
As I've probably mentioned 50 billion trillion times before, structure is related to function. The catalytic site, or active site, requires that a substrate fit there before it can carry out its work. The ability of an enzyme to accommodate a substrate properly depends on the amino acid residues surrounding the active site, as well as the shape of the substrate itself.
You should be able to explain why the induced fit model of enzyme action better explains enzyme catalysis than the lock and key model and understand how an enzyme stabilises the transition state conformation(s) of a substrates(s) by forming extra bonds between this conformation and the active site of the enzyme.
The "lock and key" model is the one that you've probably heard all throughout high school (and which I've used a bit in this blog as well). Basically it goes like this: substrates have a particular shape, designed to fit in the binding sites of specific enzymes, just like how a key fits into a lock.
Even though the lock and key model is pretty elegant and easy to understand, there's an even better model available. You see, the lock and key explains how a substrate and an enzyme can fit together, but it doesn't explain why the substrate-enzyme complex remains stable even as the reaction proceeds. A better model is the induced fit model, which compares the enzyme as a glove and the substrate as a hand. Just as a glove may have to stretch to accommodate the hand, the enzyme may have to accommodate the substrate.
An important concept behind the induced fit model is that the enzyme can exist in a lower-energy state and a higher-energy state. When the enzyme is in a higher-energy state, it can accommodate the substrate. As the reaction proceeds, the enzyme can change shape slightly to accommodate the transition state, allowing for stability at this state as well. Eventually the products are released from the enzyme.
You should be aware of the requirement of some enzymes for cofactors.
Cofactors are other molecules etc. that some enzymes require for them to work. I've already mentioned one of these- pyridoxal phosphate (PLP). Aside from PLP, there are many other cofactors that can be found in enzymes, such as ions and vitamins. The activity of specific cofactors depends on the reaction in question.
You should understand why most enzymes lose activity above about 45°C and be aware of the existence of thermostable enzymes from thermophilic organisms.
At a high temperature, denaturation of enzymes may occur. Denaturation is the unfolding of a protein so that it no longer adopts its required configuration. This occurs at around 45 degrees Celsius due to the weak hydrogen bonding that is responsible for most of the folding of a protein. Some thermophilic organisms (organisms that are adapted to high temperatures) have more thermostable enzymes that allow them to survive in the heat. These may contain more covalent disulfide bonds which are more resistant to changes in temperature.
You should understand that in many cases the activity of an enzyme is dependent on pH because of the presence in the active site of amino acid side chains that undergo pH-dependent protonation or deprotonation, but which need to be in a particular protonation state in order to allow substrate binding or for catalysis to occur. You should also understand that a bell-shaped pH activity curve results from having 2 such ionisable residues in the active site, one of which needs to be protonated and one of which needs to be deprotonated for substrate binding/catalysis to occur.
As mentioned in my post about the acid base chemistry of amino acids, ionisable groups of amino acids can exist in a protonated or deprotonated form, depending on the pH. This affects charge, which in turn can affect substrate binding or catalysis. Sometimes there are multiple amino acids in the active site that are affected by pH. Hence, pH is often important for enzyme activity, with activity peaking at the optimal pH and decreasing as the pH increases or decreases.
YUSSSSSSSSSSSS NO MORE AMINO ACIDS OR PROTEIN LECTURES TO GO THROUGH FOR THIS UNIT!!!!!!!!!!!
Now it's time to go to sleep...
Proteins- Modifications
Yeah, this protein stuff is becoming a bit of a drag (especially because I'm having to frantically look back through lecture slides and notes for this stuff)... at least this is the second last post :)
You should know the reaction for proline hydroxylation and the reason why ascorbate is required to maintain proline hydroxylase in an active state.
Well fffffffffffffffff-
Okay. Breathe. Look at notes. Look at lecture slides. You can do this.
Apparently the reaction is proline + alpha-ketoglutarate + O2 --> 4-hydroxyproline (a.k.a. 4-hyp) + succinate + CO2.
The enzyme that catalyses this reaction is proline hydroxylase. It associates with a cofactor, Fe2+, which oxidises to Fe4+ in the reaction. Ascorbate is required to reduce Fe4+ back to Fe2+ and therefore maintain proline hydroxylase in an active state.
You should understand why proline hydroxylation is essential for the correct folding and stability of collagen and that collagen that does not contain 4-hydroxyproline is unstable and this leads to the degeneration of connective tissue seen in Scurvy. You should know why ascorbate prevents Scurvy.
The ring structure of 4-hydroxyproline (4-hyp) has a slightly different conformation to that of proline (4-hyp is exo while normal proline is endo). The difference in structure makes the correct placement of Pro and 4-hyp essential for the correct folding of collagen, which is a fibrous protein found in many of the connective tissues in our body. As ascorbate is required to maintain proline hydroxylase and thus the production of 4-hyp, a lack of ascorbate (vitamin C), as seen in scurvy, can lead to the degeneration of connective tissue. This can be prevented by sufficient uptake of ascorbate.
You should be able to draw a protein phosphorylation reaction involving Ser/Thr or Tyr (you do not need to know the structure of ATP, other than the three phosphoryl groups). You should be aware that MgATP and MgADP are actually used as substrates by enzymes and be aware that phosphoryl group transfer from ATP is energetically favoured because of the charge separation that ensues.
Serine, threonine and tyrosine all have OH groups. The oxygen is nucleophilic and can attack the gamma phosphate (i.e. the phosphate furthest away) in order to become phosphorylated. Some other amino acids can also be phosphorylated, such as histidine, which can be phosphorylated on the N-H groups of the imidazole ring.
Now for a little bit more about ATP. ATP has three negatively-charged phosphate groups, all joined end to end. You might think that all of those negative charges would repel each other, and you're right. Part of the reason why phosphorylation is so energetically favourable is because of this repulsion between negative charges.
The negative charges between phosphate groups are somewhat "shielded" from other molecules by association with a cation, usually magnesium (Mg2+). This is why MgATP and MgADP, rather than pure ATP/ADP are usually used as substrates by enzymes. There are other reasons why Mg is useful here: it may help position phosphoryl groups, can facilitate binding by forming complexes with other amino acids in enzymes, and can withdraw electrons from the furthest phosphoryl group in order to make it more susceptible to nucleophilic attack.
You should know that protein phosphorylation is catalysed by protein kinases and that dephosphorylation (hydrolysis) is catalysed by protein phosphatases. You should know that there are Ser/Thr kinases and Tyr kinases and that each protein kinase phosphorylates only amino acids in particular amino acid sequences in a protein.
As stated above, kinases are a group of enzymes that add phosphate groups to proteins, whereas phosphatases remove them. Serine and threonine are reasonably similar in structure, which is probably why there are Ser/Thr kinases. Tyrosine, on the other hand, is quite different to the other two, so there are special kinases for tyrosine residues. The reason why protein kinases can be specific not only to Ser/Thr/Tyr but also to particular sequences is that the active site may bind to several amino acids at once.
You should be aware that phosphorylation/dephosphorylation can change the structure of a protein and thus change its biological activity e.g. it can activate or deactivate an enzyme. In addition, it can affect the way the protein interacts with other proteins leading to degradation or translocation within the cell for example.
As I've probably mentioned several times by now, structure is critical to function. Adding or removing a phosphate group changes the shape as well as introduces a negative charge. Hence phosphorylation and dephosphorylation can result in activation or deactivation of an enzyme.
One common misconception is that kinases always activate enzymes while phosphatases always deactivate them. This is not necessarily true. Some enzymes are active with a phosphate group added, while others are active once the phosphate group is removed. It really depends on the enzyme in question.
You should know the "in-line" nature of phosphoryl transfer reactions and be able to draw this.
Okay, this is something that I didn't know until I went back through the slides. Essentially, when a nucleophile attacks a phosphorus group, it attacks in line and opposite to the leaving group on the other side of the phosphorus. This causes the transition state to be planar bipyramidal- the "planar" part coming from the phosphorus and the oxygens being in a plane. The final product also has an inversion of configuration around the phosphorus group (i.e. from S-configuration to R-configuration or vice versa) though this is only really detectable if the phosphorus group is attached to different isotopes of oxygen. (I'll explain chirality and S/R-configurations in a later post.)
You should know the purpose of lipidation is to anchor proteins to membranes, as the lipid group inserts into the membrane bilayer.
Lipidation is the addition of a lipid group to a protein. As lipids are obviously lipid-soluble, lipidation provides a great "anchor" to attach the protein to the membrane.
You should know one example of a lipidation reaction in detail and be able to draw it, using the side chain or terminal group to which the lipid is attached (sketches of the protein structures are not needed) and you can represent the palmitoyl, myristoyl or farnesyl groups by their names, no need to learn their structures. You should know what the substrate is in each of the three lipidation reactions and the type of bond formed with the substrate protein.
Fggfdfaljksdgl I hate learning reactions in detail.
Palmitoylation is the addition of a palmitoyl group onto a protein. It appears to occur at -SH groups on proteins (probably at cysteine residues). Palmitoyl CoA (i.e. palmitoyl attached to coenzyme A) is attached here via palmitoyl acyl transferase, releasing CoA. A thioester bond is formed.
Myristoylation is the addition of a myristoyl group on a protein. It occurs at N-terminal glycine residues. Myristoyl CoA is attached here via myristoyl transferase, again releasing CoA. An amide bond is formed.
Farnesylation is the addition of a farnesyl group on a protein (yup, getting super creative here). It occurs at cysteine residues near the C-terminus of the protein. There may be several different recognisable motifs that follow the cysteine residue, such as AAX (alanine-alanine-any amino acid). Farnesyl transferase attaches farnesyl pyrophosphate onto the cysteine residue, releasing pyrophosphate in the process. Afterwards, a converting enzyme releases the last few residues of the protein.
You should be able to draw the sulfation reaction. You should know the general purpose of protein sulfation.
This is another one where I had to frantically look back through the lecture slides. Thankfully, this one isn't too difficult.
Sulfation is basically the addition of a sulfate group onto a protein- a bit like phosphorylation but with sulfate instead of phosphate. Instead of ATP, there's another molecule called PAPS: 3'-phosphoadenosine 5'-phosphosulfate. This molecule is a bit like ADP, but with a sulfate in place of the second phosphate, and with a phosphate group also attached to the 3' carbon of the ribose sugar. Tyrosyl protein sulfotransferase (TPST) catalyses the movement of the sulfate group from PAPS onto the protein.
Protein sulfation is thought to be involved in protein-protein and cell-cell interactions.
You should be able to draw the acetylation reactions. You should understand the different roles of N-terminus acetylation and the reversible acetylation of Lys residues in histones. You do not need to be able to reproduce the complex diagrams of nucleosome and chromatin structures.
Acetylation basically involves the movement of an acetyl group from acetyl CoA onto either the N-terminus of a protein or onto lysine residues, releasing CoA in the process. This is catalysed by N-acetyl transferase (in the case of addition onto the N-terminus) or histone acetyl transferase (in the case of addition to lysine residues). N-terminus acetylation is thought to be able to protect proteins from degradation, whereas acetylation of lysine residues in histones is thought to serve as "markers" that can be read by other proteins that regulate gene expression.
Proteins- Folding
Yup, yet another post about proteins...
You should understand Anfinsen's experiment and know the four major conclusions from it.
In Anfinsen's experiment, a protein was denatured by being placed in a reducing environment, breaking disulfide bonds. The protein was then placed back in an oxidising environment to see if the same disulfide bonds would form. Out of the 104 possible combinations of disulfide bonds that could have formed, only one formed- the one that allowed for correct functioning of the protein. From this experiment four major conclusions could be drawn:
1. Correct protein structure is necessary for correct function.
2. Disulfide bonds stabilise proteins- when broken, the protein is unable to carry out its proper function.
3. Correct disulfide bonds are formed as proteins fold.
4. All of the information for a protein to fold correctly is contained in the protein sequence, so even if a protein is denatured it may be able to renature again if placed in a favourable environment.
You should be able to describe the different types of van der Waals interactions.
Van der Waals interactions involve the formation of weak dipoles in molecules. A molecule with, say, a C=O bond may have a dipole, with the O being slightly negatively charged and the C being slightly positively charged. This slight dipole can allow the molecule to interact with other molecules with slight dipoles. Additionally, the dipole can induce a dipole in other molecules- the electronegative O may repel electrons at the end of a neighbouring molecule, for instance.
Another type of van der Waals interaction is London dispersion forces. These occur between nonpolar molecules. Electrons in molecules do not remain in the same place all the time. At times, there will be more electrons in some parts of the molecule and fewer in other parts. This can create temporary dipoles which can then interact with each other. Larger molecules tend to display stronger London dispersion forces, probably due to their larger number of electrons and greater mass.
You should know how the placement of hydrophobic (nonpolar) and hydrophilic (polar) amino acid side chains in an alpha-helix determines the characteristics of the helix and hence how it packs into a protein structure. You should know the definition of an amphipathic alpha-helix.
As mentioned previously, all amino acid side chains face outwards from the helix. If these chains are hydrophilic they will interact better with water, whereas if the chains are hydrophobic they will interact better with lipids. If the protein is to exist in an aqueous environment such as in the cytosol, it is more energetically favourable if hydrophilic amino acids are facing towards the outside of the protein and hydrophobic amino acids are facing towards the inside. Alternatively, if the protein is to insert in the membrane, then it is more energetically favourable to have hydrophobic amino acids in the part of the chain that will be inserted into the membrane, and hydrophilic amino acids at the ends. Alpha-helices with hydrophobic parts and hydrophilic parts are known as amphipathic.
You should know how the placement of hydrophobic (nonpolar) and hydrophilic (polar) amino acid side chains in a beta-sheet determines the characteristics of the sheet and hence how it packs into a protein structure. You should know the definition of an amphipathic beta-sheet.
As also mentioned previously, every second amino acid side chain is above the beta sheet, while the others are below the sheet. Like alpha-helices, hydrophobic side chains tend to face into the protein while hydrophilic side chains tend to face outwards. Amphipathic beta-sheets are beta-sheets with hydrophobic parts and hydrophilic parts.
You should understand how hydrophobic regions of motifs and secondary structure determine how they pack in the tertiary structure of a protein.
As I just mentioned, hydrophobic regions tend to be hidden within proteins. Sometimes they are sandwiched between hydrophilic regions.
You should know the thermodynamic descriptions of the effects of protein folding on polar and nonpolar amino acid and water molecules. In particular you should know what the changes in enthalpy (Delta H) and entropy (-T Delta S) occur due to water molecules and hydrophobic (nonpolar) amino acids when a protein folds and why the entropic effect on water drives the folding.
A simplified explanation of enthalpy is that it is the change in heat. Generally heat is absorbed to break bonds and released when bonds are formed. A simplified explanation of entropy is that it is the change in "randomness" of the molecules. If I remember correctly, the second law of thermodynamics is that everything naturally wants to descend towards chaos, sorta like a class of schoolchildren who naturally get rowdier and rowdier the longer that they are left without a teacher.
One important concept is that the total free energy change, Delta G, is equal to the sum of Delta H for the chain and Delta H for the solvent minus T Delta S for the chain and T Delta S for the solvent. If Delta G is negative, then protein folding is favourable; if it is positive, then the folded conformation is not energetically favourable.
Let's have a look at what happens if you leave a protein to fold in a vacuum, where there is no solvent around. Delta H is favourable as there are plenty of hydrogen bonds and van der Waals interactions going on between different parts of the molecule, hence releasing energy as heat. Delta S, on the other hand, is unfavourable as this folding means that the protein is becoming more ordered and thus there is less entropy. Overall, however, Delta H is larger than Delta S, so the overall result is a negative Delta G and a shiny new folded protein.
Now let's have a look at what happens to hydrophobic groups in an aqueous solvent. Delta H for the chain this time is unfavourable. I think this is because the water gets in the way of some of the electrostatic reactions between parts of the molecule. Delta S for the chain is also unfavourable, again as the folding of an ordered chain means that there is a decrease of entropy. Delta H for the solvent is favourable, possibly because the "pushing aside" of water molecules by the protein as it folds means that the water molecules have more interactions with each other. Also, after folding, fewer water molecules are required to form a "cage" around the hydrophobic residues, resulting in greater "freedom" of the water molecules after folding than before folding. This means that entropy has increased. In fact, the entropy of water is one of the major driving forces behind the folding of proteins.
Hydrophilic groups in an aqueous solvent, however, do not have quite the same effect. Hydrophilic residues naturally want to interact with water, and so folding, which reduces these interactions, results in an unfavourable Delta H for the chain. Delta S for the chain remains unfavourable due to the formation of an ordered chain. The entropy of water is not decreased as much when hydrophilic groups are present as compared to hydrophobic groups. Overall, Delta G for this situation is relatively neutral.
Sorry if that was a terrible explanation- it's a topic I don't understand that well, partly because I don't have any kind of physics background.
You should understand the Leventhal paradox and the concept of an energy landscape funnel which enables a polypeptide chain to start from a large range of conformations but end up in a single native folded state. You should know that protein folding is an energetically favourable process and be aware of the possibility of relatively stable misfolded states.
One of the hypotheses around protein folding was that every combination is tested to find the lowest energy conformation. However, the Levanthal paradox states that there are so many folding possibilities for your average protein that it would take far too much time to test them all. Another model of protein folding involves an "energy landscape funnel" showing multiple different routes that a protein might undertake to reach its native "folded state." Along these routes, intermediates may be formed of progressively lower energy levels. There may, however, be some stable misfolded states mixed in among the intermediates.
You should know that some proteins require no assistance in folding (generally small simple proteins) whilst some require the assistance of heat shock proteins such as Hsp70 which bind to exposed hydrophobic regions in unfolded or partially folded proteins, protecting them from aggregating with other unfolded or partially folded proteins. You should know that the folding process can occur by association with Hsp70 but in larger more complex proteins the protein is delivered to the Hsp60/Hsp10 chaperones (GroEL/GroES in bacteria) which provide a hollow barrel shaped environment to allow folding to occur secluded from the rest of the cell. You should know why heat shock proteins are so called.
Whoa... that dot point was somewhat lengthy, and it covered quite a bit. The only thing that really needs additional explanation here is why heat shock proteins are called that. Heat shock proteins are called heat shock proteins as their synthesis tends to increase in situations of high heat. This may be because there is an additional risk of protein denaturation at higher temperatures, and thus more heat shock proteins are needed to counterbalance this.
You should understand Anfinsen's experiment and know the four major conclusions from it.
In Anfinsen's experiment, a protein was denatured by being placed in a reducing environment, breaking disulfide bonds. The protein was then placed back in an oxidising environment to see if the same disulfide bonds would form. Out of the 104 possible combinations of disulfide bonds that could have formed, only one formed- the one that allowed for correct functioning of the protein. From this experiment four major conclusions could be drawn:
1. Correct protein structure is necessary for correct function.
2. Disulfide bonds stabilise proteins- when broken, the protein is unable to carry out its proper function.
3. Correct disulfide bonds are formed as proteins fold.
4. All of the information for a protein to fold correctly is contained in the protein sequence, so even if a protein is denatured it may be able to renature again if placed in a favourable environment.
You should be able to describe the different types of van der Waals interactions.
Van der Waals interactions involve the formation of weak dipoles in molecules. A molecule with, say, a C=O bond may have a dipole, with the O being slightly negatively charged and the C being slightly positively charged. This slight dipole can allow the molecule to interact with other molecules with slight dipoles. Additionally, the dipole can induce a dipole in other molecules- the electronegative O may repel electrons at the end of a neighbouring molecule, for instance.
Another type of van der Waals interaction is London dispersion forces. These occur between nonpolar molecules. Electrons in molecules do not remain in the same place all the time. At times, there will be more electrons in some parts of the molecule and fewer in other parts. This can create temporary dipoles which can then interact with each other. Larger molecules tend to display stronger London dispersion forces, probably due to their larger number of electrons and greater mass.
You should know how the placement of hydrophobic (nonpolar) and hydrophilic (polar) amino acid side chains in an alpha-helix determines the characteristics of the helix and hence how it packs into a protein structure. You should know the definition of an amphipathic alpha-helix.
As mentioned previously, all amino acid side chains face outwards from the helix. If these chains are hydrophilic they will interact better with water, whereas if the chains are hydrophobic they will interact better with lipids. If the protein is to exist in an aqueous environment such as in the cytosol, it is more energetically favourable if hydrophilic amino acids are facing towards the outside of the protein and hydrophobic amino acids are facing towards the inside. Alternatively, if the protein is to insert in the membrane, then it is more energetically favourable to have hydrophobic amino acids in the part of the chain that will be inserted into the membrane, and hydrophilic amino acids at the ends. Alpha-helices with hydrophobic parts and hydrophilic parts are known as amphipathic.
You should know how the placement of hydrophobic (nonpolar) and hydrophilic (polar) amino acid side chains in a beta-sheet determines the characteristics of the sheet and hence how it packs into a protein structure. You should know the definition of an amphipathic beta-sheet.
As also mentioned previously, every second amino acid side chain is above the beta sheet, while the others are below the sheet. Like alpha-helices, hydrophobic side chains tend to face into the protein while hydrophilic side chains tend to face outwards. Amphipathic beta-sheets are beta-sheets with hydrophobic parts and hydrophilic parts.
You should understand how hydrophobic regions of motifs and secondary structure determine how they pack in the tertiary structure of a protein.
As I just mentioned, hydrophobic regions tend to be hidden within proteins. Sometimes they are sandwiched between hydrophilic regions.
You should know the thermodynamic descriptions of the effects of protein folding on polar and nonpolar amino acid and water molecules. In particular you should know what the changes in enthalpy (Delta H) and entropy (-T Delta S) occur due to water molecules and hydrophobic (nonpolar) amino acids when a protein folds and why the entropic effect on water drives the folding.
A simplified explanation of enthalpy is that it is the change in heat. Generally heat is absorbed to break bonds and released when bonds are formed. A simplified explanation of entropy is that it is the change in "randomness" of the molecules. If I remember correctly, the second law of thermodynamics is that everything naturally wants to descend towards chaos, sorta like a class of schoolchildren who naturally get rowdier and rowdier the longer that they are left without a teacher.
One important concept is that the total free energy change, Delta G, is equal to the sum of Delta H for the chain and Delta H for the solvent minus T Delta S for the chain and T Delta S for the solvent. If Delta G is negative, then protein folding is favourable; if it is positive, then the folded conformation is not energetically favourable.
Let's have a look at what happens if you leave a protein to fold in a vacuum, where there is no solvent around. Delta H is favourable as there are plenty of hydrogen bonds and van der Waals interactions going on between different parts of the molecule, hence releasing energy as heat. Delta S, on the other hand, is unfavourable as this folding means that the protein is becoming more ordered and thus there is less entropy. Overall, however, Delta H is larger than Delta S, so the overall result is a negative Delta G and a shiny new folded protein.
Now let's have a look at what happens to hydrophobic groups in an aqueous solvent. Delta H for the chain this time is unfavourable. I think this is because the water gets in the way of some of the electrostatic reactions between parts of the molecule. Delta S for the chain is also unfavourable, again as the folding of an ordered chain means that there is a decrease of entropy. Delta H for the solvent is favourable, possibly because the "pushing aside" of water molecules by the protein as it folds means that the water molecules have more interactions with each other. Also, after folding, fewer water molecules are required to form a "cage" around the hydrophobic residues, resulting in greater "freedom" of the water molecules after folding than before folding. This means that entropy has increased. In fact, the entropy of water is one of the major driving forces behind the folding of proteins.
Hydrophilic groups in an aqueous solvent, however, do not have quite the same effect. Hydrophilic residues naturally want to interact with water, and so folding, which reduces these interactions, results in an unfavourable Delta H for the chain. Delta S for the chain remains unfavourable due to the formation of an ordered chain. The entropy of water is not decreased as much when hydrophilic groups are present as compared to hydrophobic groups. Overall, Delta G for this situation is relatively neutral.
Sorry if that was a terrible explanation- it's a topic I don't understand that well, partly because I don't have any kind of physics background.
You should understand the Leventhal paradox and the concept of an energy landscape funnel which enables a polypeptide chain to start from a large range of conformations but end up in a single native folded state. You should know that protein folding is an energetically favourable process and be aware of the possibility of relatively stable misfolded states.
One of the hypotheses around protein folding was that every combination is tested to find the lowest energy conformation. However, the Levanthal paradox states that there are so many folding possibilities for your average protein that it would take far too much time to test them all. Another model of protein folding involves an "energy landscape funnel" showing multiple different routes that a protein might undertake to reach its native "folded state." Along these routes, intermediates may be formed of progressively lower energy levels. There may, however, be some stable misfolded states mixed in among the intermediates.
You should know that some proteins require no assistance in folding (generally small simple proteins) whilst some require the assistance of heat shock proteins such as Hsp70 which bind to exposed hydrophobic regions in unfolded or partially folded proteins, protecting them from aggregating with other unfolded or partially folded proteins. You should know that the folding process can occur by association with Hsp70 but in larger more complex proteins the protein is delivered to the Hsp60/Hsp10 chaperones (GroEL/GroES in bacteria) which provide a hollow barrel shaped environment to allow folding to occur secluded from the rest of the cell. You should know why heat shock proteins are so called.
Whoa... that dot point was somewhat lengthy, and it covered quite a bit. The only thing that really needs additional explanation here is why heat shock proteins are called that. Heat shock proteins are called heat shock proteins as their synthesis tends to increase in situations of high heat. This may be because there is an additional risk of protein denaturation at higher temperatures, and thus more heat shock proteins are needed to counterbalance this.
Proteins- Levels of Structure
Now we've moved past from amino acids to proteins, which are larger structures made up of amino acids. Hopefully these next few posts should be at least marginally more interesting than those about amino acids.
Recognise that the linear sequence of bases in the coding strand of DNA code for the linear sequences of bases in mRNA which in turn codes for the linear sequence of amino acids in a protein polypeptide chain.
I thought that this is something that I've covered earlier, but apparently not. The bases in the coding strand can be copied to messenger RNA in a process known as transcription. This mRNA then leaves the nucleus for the cytosol, where it is transcribed on ribosomes into an amino acid sequence. This amino acid sequence can then fold to create a protein.
Know which is the amino (N)-terminus and which is the carboxyl (C)-terminus of a polypeptide chain and how amino acids in a polypeptide chain are numbered starting from the N-terminus.
The amino terminus is the end with the amine group while the carboxyl terminus is the end with carboxylic acid group. As for numbering, I'm sure they're just numbered 1, 2, 3 etc. from N-terminus to C-terminus.
Understand that the polypeptide chain has to fold into a defined 3-D structure in order for the protein to be biologically functional.
As I'm fairly sure I've mentioned before, structure is critical to function as enzymes and so on function through the way that proteins and other molecules fit together. Hence, for a protein to be biologically functional, it needs to fold into its correct structure first.
Know how H-bonding between peptide bonds stabilises the helical structure, with bonding formed between one amino acid carbonyl oxygen and the amide proton of an amino acid 4 residues along the polypeptide chain, leading to 3.6 amino acid residues per turn. Especially note that the side chains of the amino acids face outwards from the helix and are not inside the helical structure where there is no room.
Know why an α-helix has a dipole.
The carbonyl oxygen has a slightly negative charge, while the amide proton has a slightly positive charge. These conditions allow hydrogen bonds to be formed between the carboxyl groups of amino acids with the amide groups of amino acids four residues away. This forms a helical structure known as an alpha helix (a common secondary structure of proteins). In the alpha helix, all of the carboxyl groups face in one direction whereas all the amine groups face in the other, as this helps with hydrogen bonding between residues. As such, alpha-helices have dipoles: a slightly negative end that contains more carboxyl groups, and a more positive end that contains more amine groups.
Know that H-bonding between peptide bonds on two segments of β-sheet stabilise the structure and what the terms parallel and antiparallel mean. Also, it is important to know that the amino acid side chains lie above and below the plane of the sheet.
Beta-sheets are different secondary structures of proteins. In beta sheets, you essentially have chains of amino acids lying side by side. These chains of amino acids are held together by hydrogen bonds between the carboxyl groups of some amino acids with the amide groups of other amino acids, just like in alpha-helices. Chains can be parallel if they are all running in the same direction (N-terminus to C-terminus) or antiparallel if every second one is running in the opposite direction. As stated in the dot point, amino acid side chains lie above and below the plane of the sheet. Due to the zig-zag bending of the chain due to bond locations etc., half of the amino acids lie above the sheet and the other half lie below.
Know what a β-turn is and where it is found and the reasons why Gly and Pro are often found in them.
A beta-turn is basically a small loop that connects two antiparallel strands of a beta sheet (parallel strands have a beta-loop instead, which is much bigger). As this requires quite a tight turn, glycine, which only has a hydrogen atom in its side chain, is often used so as not to interfere with the other closely-packed amino acids. Proline is also often used because its tight ring-structure forces the chain to turn sharply, especially when proline is in the cis-form.
Be able to define the angles Φ and Ψ and explain why some combinations angles of Φ and Ψ do not occur and why some combinations are favoured. Be able to relate this to the Ramachandran diagram which shows that common secondary structures fall in areas of highly favourable combinations of Φ and Ψ.
Phi is the angle around the C-N part of the peptide bond, whereas psi is the angle around the C-C part of the peptide bond. Some angles are not allowed because they would bring the carboxyl groups or the amino groups too close to each other. A Ramachandran diagram is a diagram showing the sterically allowed values of phi and psi. As expected, commonly found structures contain favourable combinations of phi and psi.
Understand that pieces of secondary structure often combine into motifs that are components of the final tertiary structure of a protein, know a couple of examples that between them have α-helices and β-sheets as components.
As mentioned before, alpha-helices and beta-sheets are common secondary structures. These secondary structures can combine to form motifs, which can then combine further to form the overall tertiary structure of the protein. One of these motifs is the beta-alpha-beta motif, which is essentially two parallel strands of a beta-sheet, but the beta-loop is made up of an alpha helix. The beta-hairpin motif is essentially several antiparallel beta strands next to each other with beta-turns between them. There is also an alpha-alpha motif which consists of several alpha helices side by side. Alpha-helices and beta-sheets can also combine to form "barrels" through which other substances may be able to pass through, as in the case of pores and channels in the cell membrane. Common structures here are beta-barrels and alpha/beta-barrels. (Not sure if there are alpha-barrels.)
Understand the different graphical representations of models of protein structures.
Unfortunately I don't have any non-copyrighted pictures to show you, but proteins as displayed in textbooks etc. may be displayed in several different ways. Sometimes a ball-and-stick model is shown with every atom and every connection between them. Other times you might see alpha-helices displayed as cylinders or spirals and beta-sheets displayed as arrows, with thin tubes connecting them. There are several other ways that proteins might be displayed which I'm not going to go into here because it's kind of hard to explain without pictures.
Understand how the different levels of protein structure relate to each other.
Know the nomenclature of quaternary structures (subunits,α,β etc.) and how subunits bind to each other.
The primary level of protein structure is the amino acid sequence. The amino acid sequence, and the order of the side chains, affects the way that the protein will fold.
The secondary level of protein structure is the formation of alpha-helices and beta-sheets. This is facilitated by hydrogen bonds between the carboxyl and amino groups of the amino acids.
The tertiary level of protein structure is the combination of alpha-helices and beta-sheets into motifs and ultimately a completely folded polypeptide chain.
The quaternary level of protein structure is the association of several different polypeptide chains to form a protein. This doesn't happen for every protein- some proteins consist of only a single polypeptide chain. In proteins that do have multiple polypeptide chains, each polypeptide chain is called a subunit and is given a name such as alpha or beta. For example, haemoglobin has two alpha and two beta subunits. Subunits bind mostly through noncovalent interactions, such as hydrogen bonds, but sometimes interchain disulfide bonds do occur.
Recognise that the linear sequence of bases in the coding strand of DNA code for the linear sequences of bases in mRNA which in turn codes for the linear sequence of amino acids in a protein polypeptide chain.
I thought that this is something that I've covered earlier, but apparently not. The bases in the coding strand can be copied to messenger RNA in a process known as transcription. This mRNA then leaves the nucleus for the cytosol, where it is transcribed on ribosomes into an amino acid sequence. This amino acid sequence can then fold to create a protein.
Know which is the amino (N)-terminus and which is the carboxyl (C)-terminus of a polypeptide chain and how amino acids in a polypeptide chain are numbered starting from the N-terminus.
The amino terminus is the end with the amine group while the carboxyl terminus is the end with carboxylic acid group. As for numbering, I'm sure they're just numbered 1, 2, 3 etc. from N-terminus to C-terminus.
Understand that the polypeptide chain has to fold into a defined 3-D structure in order for the protein to be biologically functional.
As I'm fairly sure I've mentioned before, structure is critical to function as enzymes and so on function through the way that proteins and other molecules fit together. Hence, for a protein to be biologically functional, it needs to fold into its correct structure first.
Know how H-bonding between peptide bonds stabilises the helical structure, with bonding formed between one amino acid carbonyl oxygen and the amide proton of an amino acid 4 residues along the polypeptide chain, leading to 3.6 amino acid residues per turn. Especially note that the side chains of the amino acids face outwards from the helix and are not inside the helical structure where there is no room.
Know why an α-helix has a dipole.
The carbonyl oxygen has a slightly negative charge, while the amide proton has a slightly positive charge. These conditions allow hydrogen bonds to be formed between the carboxyl groups of amino acids with the amide groups of amino acids four residues away. This forms a helical structure known as an alpha helix (a common secondary structure of proteins). In the alpha helix, all of the carboxyl groups face in one direction whereas all the amine groups face in the other, as this helps with hydrogen bonding between residues. As such, alpha-helices have dipoles: a slightly negative end that contains more carboxyl groups, and a more positive end that contains more amine groups.
Know that H-bonding between peptide bonds on two segments of β-sheet stabilise the structure and what the terms parallel and antiparallel mean. Also, it is important to know that the amino acid side chains lie above and below the plane of the sheet.
Beta-sheets are different secondary structures of proteins. In beta sheets, you essentially have chains of amino acids lying side by side. These chains of amino acids are held together by hydrogen bonds between the carboxyl groups of some amino acids with the amide groups of other amino acids, just like in alpha-helices. Chains can be parallel if they are all running in the same direction (N-terminus to C-terminus) or antiparallel if every second one is running in the opposite direction. As stated in the dot point, amino acid side chains lie above and below the plane of the sheet. Due to the zig-zag bending of the chain due to bond locations etc., half of the amino acids lie above the sheet and the other half lie below.
Know what a β-turn is and where it is found and the reasons why Gly and Pro are often found in them.
A beta-turn is basically a small loop that connects two antiparallel strands of a beta sheet (parallel strands have a beta-loop instead, which is much bigger). As this requires quite a tight turn, glycine, which only has a hydrogen atom in its side chain, is often used so as not to interfere with the other closely-packed amino acids. Proline is also often used because its tight ring-structure forces the chain to turn sharply, especially when proline is in the cis-form.
Be able to define the angles Φ and Ψ and explain why some combinations angles of Φ and Ψ do not occur and why some combinations are favoured. Be able to relate this to the Ramachandran diagram which shows that common secondary structures fall in areas of highly favourable combinations of Φ and Ψ.
Phi is the angle around the C-N part of the peptide bond, whereas psi is the angle around the C-C part of the peptide bond. Some angles are not allowed because they would bring the carboxyl groups or the amino groups too close to each other. A Ramachandran diagram is a diagram showing the sterically allowed values of phi and psi. As expected, commonly found structures contain favourable combinations of phi and psi.
Understand that pieces of secondary structure often combine into motifs that are components of the final tertiary structure of a protein, know a couple of examples that between them have α-helices and β-sheets as components.
As mentioned before, alpha-helices and beta-sheets are common secondary structures. These secondary structures can combine to form motifs, which can then combine further to form the overall tertiary structure of the protein. One of these motifs is the beta-alpha-beta motif, which is essentially two parallel strands of a beta-sheet, but the beta-loop is made up of an alpha helix. The beta-hairpin motif is essentially several antiparallel beta strands next to each other with beta-turns between them. There is also an alpha-alpha motif which consists of several alpha helices side by side. Alpha-helices and beta-sheets can also combine to form "barrels" through which other substances may be able to pass through, as in the case of pores and channels in the cell membrane. Common structures here are beta-barrels and alpha/beta-barrels. (Not sure if there are alpha-barrels.)
Understand the different graphical representations of models of protein structures.
Unfortunately I don't have any non-copyrighted pictures to show you, but proteins as displayed in textbooks etc. may be displayed in several different ways. Sometimes a ball-and-stick model is shown with every atom and every connection between them. Other times you might see alpha-helices displayed as cylinders or spirals and beta-sheets displayed as arrows, with thin tubes connecting them. There are several other ways that proteins might be displayed which I'm not going to go into here because it's kind of hard to explain without pictures.
Understand how the different levels of protein structure relate to each other.
Know the nomenclature of quaternary structures (subunits,α,β etc.) and how subunits bind to each other.
The primary level of protein structure is the amino acid sequence. The amino acid sequence, and the order of the side chains, affects the way that the protein will fold.
The secondary level of protein structure is the formation of alpha-helices and beta-sheets. This is facilitated by hydrogen bonds between the carboxyl and amino groups of the amino acids.
The tertiary level of protein structure is the combination of alpha-helices and beta-sheets into motifs and ultimately a completely folded polypeptide chain.
The quaternary level of protein structure is the association of several different polypeptide chains to form a protein. This doesn't happen for every protein- some proteins consist of only a single polypeptide chain. In proteins that do have multiple polypeptide chains, each polypeptide chain is called a subunit and is given a name such as alpha or beta. For example, haemoglobin has two alpha and two beta subunits. Subunits bind mostly through noncovalent interactions, such as hydrogen bonds, but sometimes interchain disulfide bonds do occur.
Amino Acids- Metabolism
Just had a look at the dot points for this. Urggghhhh, I think this is the point where I looked at the lecture schedule to see how many more lectures we had on proteins and amino acids and groaned when I saw how many more there were.
Know the nitrogen cycle and and that the two major routes for the assimilation of nitrogen involve reduction of NO2 or N2, using complexes of enzymes in which complex electron transfer pathways occur, that both result in the formation of NH4 +.
The nitrogen cycle is, simply put, a series of reactions in which nitrogen can be converted into different forms, including NO2, NO3-, N2 and NH4+. NH4+ is the form in which we take nitrogen into our bodies, and it can be formed through the aerobic process of nitrate assimilation from NO3- to NO2 by nitrate reductase and then to NH4+ via nitrite reductase, or through the anaerobic processes of denitrification of NO3- to N2 and then nitrogen fixation from N2 to NH4+. (These processes are carried out by other organisms, not by us.) Of these routes, nitrogen assimilation is the most common.
Know that NADH and NADPH are reducing agents that supply electrons in a number of reactions.
I *think* NADH and NADPH help supply some of the electrons for the assimilation of nitrogen into the body. NADH also helps with ATP synthesis, while NADPH helps with synthesis reactions in the body. I'll probably go into these in more detail in a later post.
Know details of the three major reactions used in biology for NH3/NH4 + assimilation.
Urgh I hate having to learn details as my memory is not that great- I'd much rather just learn general principles and be able to apply them to any situation. Nevertheless, here are the three major reactions:
1. Carbamoyl-phosphate synthetase
In this reaction, NH4+, bicarbonate and ATP react to form carbamoyl phosphate and ADP. It starts with bicarbonate becoming phosphorylated by ATP. Ammonia then nucleophilically attacks the carbon atom in bicarbonate, displacing the phosphate group to form carbamate. Carbamate can then be phosphorylated by another molecule of ATP to form carbamoyl-phosphate, which can then enter the urea cycle to produce urea.
2. Glutamate dehydrogenase
In this reaction, NH4+ and alpha-ketoglutarate are reduced by NADPH to form glutamate. NH4+ first replaces the ketone group of alpha-ketoglutarate to form an intermediate with an H2N+ group, which is then reduced by NADPH to form the H3N+ group of glutamate.
3. Glutamine synthetase
In this reaction, NH4+, glutamate and ATP form glutamine and ADP. First, glutamate is phosphorylated by ATP to form a gamma-glutamyl phosphate intermediate. NH4+ then nucleophilically attacks gamma-glutamyl phosphate, releasing the phosphate group and forming glutamine.
Know the difference between an essential and non-essential amino acid (you do not have to learn which amino acids fall into each category).
In a nutshell, non-essential amino acids can be synthesised by us, whereas essential amino acids must be taken up in the diet.
Understand that amino acids are synthesised from metabolic precursors which are components of glycolysis, the TCA cycle and pentose phosphate pathway, that have other major roles in cellular metabolism (no need to learn details of the intermediates and which intermediates give rise to which amino acids).
Hehe, this is something that I think that I shut out of my mind because I got overwhelmed by the diagram. The dot point is pretty self-explanatory, but I'll just give a quick recap over what the processes mentioned are. Glycolysis is the process in which glucose is broken down to produce some ATP as well as pyruvate. Pyruvate then enters the citric acid cycle (a.k.a. the TCA cycle) in which electrons are released and picked up by NADH to power the synthesis of ATP (that's not exactly how it works- as I mentioned earlier, that's a topic for another post). The pentose phosphate pathway is a sort of "side pathway" of glycolysis in which 5-carbon sugars (pentoses) and NADPH are produced. I reckon that it's pretty neat that intermediates in these pathways can also be used to produce amino acids.
Know details of the transamination reactions.
Urgh this is another "know details of these reactions" dot points again. Transamination reactions are reactions in which amino groups are transferred from one molecule to another. They are used in the synthesis of many amino acids, in which an amino group is transferred from glutamate to the ketone group of an alpha-keto acid (basically an amino acid with a ketone group instead of an amino group and hydrogen).
Transamination reactions are catalysed by the pyridoxal coenzyme complex. The attached coenzyme, pyridoxal phosphate, essentially picks up the amino group from glutamate and transfers it on to the alpha-keto acid. I think there are more steps involved in this, but this is the extent of what appears on the slides so I hope that this is all that we have to know.
Understand that amino acids are categorised as glucogenic or ketogenic and know the definition of these terms (no need to learn the specifics of which amino acids fall into which category nor the specifics of metabolites that they give rise to).
The terms glucogenic and ketogenic are related to the catabolism (breakdown) of amino acids. Glucogenic amino acids will break down to form intermediates in the synthesis of glucose, whereas ketogenic amino acids will break down to form intermediates (generally acetoacetate or acetyl CoA) in the synthesis of fatty acids or ketone bodies.
Understand the reasons for the different modes of excretion of nitrogen derived from amino acid catabolism.
There are three ways that nitrogen can be excreted: as ammonia (NH3), as urea or as uric acid. The method used depends on the organism: fish and some aquatic animals excrete nitrogen as ammonia, birds and some reptiles excrete it as uric acid, whereas we (and many other terrestrial animals) excrete it as urea. Ammonia is very toxic and requires lots of water for it to be removed quickly and safely, which is probably why this mode of excretion is only seen in aquatic animals. Urea is not as toxic, but is very soluble in water so it takes plenty of water with it when it is excreted. Uric acid, on the other hand, can be excreted as a paste (bird poo...) with minimal water loss. (I wonder why we don't use uric acid then? Seems like we could save water that way, but maybe there's more to the story.)
Have a basic understanding of how the urea cycle operates, especially the importance of the transamination reactions, glutamate dehydrogenase and carbamoyl phosphate synthetase in feeding nitrogen into the cycle. You should know the overall reaction for the urea cycle.
The urea cycle is the cycle in which nitrogen from NH4+ and aspartate forms urea, which is later excreted. NH4+ is first converted into carbamoyl phosphate through the carbamoyl phosphate synthetase reaction, covered earlier in this post. Carbamoyl phosphate then reacts with ornithine to form citrulline, which then reacts with arginine to form arginosuccinate. Arginosuccinate can then be broken down into arginine and fumarate. Arginine can be hydrolysed to form urea and ornithine (the latter of which can then react with carbamoyl phosphate to start the cycle over). Fumarate, on the other hand, can be hydrolysed to form malate, which can then be converted to oxaloacetate via malate dehydrogenase, releasing electrons which can be picked up by NAD+ to form NADH. Oxaloacetate can then undergo transamination to reform aspartate.
The overall reaction is as follows:
NH4+ + HCO3- + aspartate --> urea + fumarate
Know the nitrogen cycle and and that the two major routes for the assimilation of nitrogen involve reduction of NO2 or N2, using complexes of enzymes in which complex electron transfer pathways occur, that both result in the formation of NH4 +.
The nitrogen cycle is, simply put, a series of reactions in which nitrogen can be converted into different forms, including NO2, NO3-, N2 and NH4+. NH4+ is the form in which we take nitrogen into our bodies, and it can be formed through the aerobic process of nitrate assimilation from NO3- to NO2 by nitrate reductase and then to NH4+ via nitrite reductase, or through the anaerobic processes of denitrification of NO3- to N2 and then nitrogen fixation from N2 to NH4+. (These processes are carried out by other organisms, not by us.) Of these routes, nitrogen assimilation is the most common.
Know that NADH and NADPH are reducing agents that supply electrons in a number of reactions.
I *think* NADH and NADPH help supply some of the electrons for the assimilation of nitrogen into the body. NADH also helps with ATP synthesis, while NADPH helps with synthesis reactions in the body. I'll probably go into these in more detail in a later post.
Know details of the three major reactions used in biology for NH3/NH4 + assimilation.
Urgh I hate having to learn details as my memory is not that great- I'd much rather just learn general principles and be able to apply them to any situation. Nevertheless, here are the three major reactions:
1. Carbamoyl-phosphate synthetase
In this reaction, NH4+, bicarbonate and ATP react to form carbamoyl phosphate and ADP. It starts with bicarbonate becoming phosphorylated by ATP. Ammonia then nucleophilically attacks the carbon atom in bicarbonate, displacing the phosphate group to form carbamate. Carbamate can then be phosphorylated by another molecule of ATP to form carbamoyl-phosphate, which can then enter the urea cycle to produce urea.
2. Glutamate dehydrogenase
In this reaction, NH4+ and alpha-ketoglutarate are reduced by NADPH to form glutamate. NH4+ first replaces the ketone group of alpha-ketoglutarate to form an intermediate with an H2N+ group, which is then reduced by NADPH to form the H3N+ group of glutamate.
3. Glutamine synthetase
In this reaction, NH4+, glutamate and ATP form glutamine and ADP. First, glutamate is phosphorylated by ATP to form a gamma-glutamyl phosphate intermediate. NH4+ then nucleophilically attacks gamma-glutamyl phosphate, releasing the phosphate group and forming glutamine.
Know the difference between an essential and non-essential amino acid (you do not have to learn which amino acids fall into each category).
In a nutshell, non-essential amino acids can be synthesised by us, whereas essential amino acids must be taken up in the diet.
Understand that amino acids are synthesised from metabolic precursors which are components of glycolysis, the TCA cycle and pentose phosphate pathway, that have other major roles in cellular metabolism (no need to learn details of the intermediates and which intermediates give rise to which amino acids).
Hehe, this is something that I think that I shut out of my mind because I got overwhelmed by the diagram. The dot point is pretty self-explanatory, but I'll just give a quick recap over what the processes mentioned are. Glycolysis is the process in which glucose is broken down to produce some ATP as well as pyruvate. Pyruvate then enters the citric acid cycle (a.k.a. the TCA cycle) in which electrons are released and picked up by NADH to power the synthesis of ATP (that's not exactly how it works- as I mentioned earlier, that's a topic for another post). The pentose phosphate pathway is a sort of "side pathway" of glycolysis in which 5-carbon sugars (pentoses) and NADPH are produced. I reckon that it's pretty neat that intermediates in these pathways can also be used to produce amino acids.
Know details of the transamination reactions.
Urgh this is another "know details of these reactions" dot points again. Transamination reactions are reactions in which amino groups are transferred from one molecule to another. They are used in the synthesis of many amino acids, in which an amino group is transferred from glutamate to the ketone group of an alpha-keto acid (basically an amino acid with a ketone group instead of an amino group and hydrogen).
Transamination reactions are catalysed by the pyridoxal coenzyme complex. The attached coenzyme, pyridoxal phosphate, essentially picks up the amino group from glutamate and transfers it on to the alpha-keto acid. I think there are more steps involved in this, but this is the extent of what appears on the slides so I hope that this is all that we have to know.
Understand that amino acids are categorised as glucogenic or ketogenic and know the definition of these terms (no need to learn the specifics of which amino acids fall into which category nor the specifics of metabolites that they give rise to).
The terms glucogenic and ketogenic are related to the catabolism (breakdown) of amino acids. Glucogenic amino acids will break down to form intermediates in the synthesis of glucose, whereas ketogenic amino acids will break down to form intermediates (generally acetoacetate or acetyl CoA) in the synthesis of fatty acids or ketone bodies.
Understand the reasons for the different modes of excretion of nitrogen derived from amino acid catabolism.
There are three ways that nitrogen can be excreted: as ammonia (NH3), as urea or as uric acid. The method used depends on the organism: fish and some aquatic animals excrete nitrogen as ammonia, birds and some reptiles excrete it as uric acid, whereas we (and many other terrestrial animals) excrete it as urea. Ammonia is very toxic and requires lots of water for it to be removed quickly and safely, which is probably why this mode of excretion is only seen in aquatic animals. Urea is not as toxic, but is very soluble in water so it takes plenty of water with it when it is excreted. Uric acid, on the other hand, can be excreted as a paste (bird poo...) with minimal water loss. (I wonder why we don't use uric acid then? Seems like we could save water that way, but maybe there's more to the story.)
Have a basic understanding of how the urea cycle operates, especially the importance of the transamination reactions, glutamate dehydrogenase and carbamoyl phosphate synthetase in feeding nitrogen into the cycle. You should know the overall reaction for the urea cycle.
The urea cycle is the cycle in which nitrogen from NH4+ and aspartate forms urea, which is later excreted. NH4+ is first converted into carbamoyl phosphate through the carbamoyl phosphate synthetase reaction, covered earlier in this post. Carbamoyl phosphate then reacts with ornithine to form citrulline, which then reacts with arginine to form arginosuccinate. Arginosuccinate can then be broken down into arginine and fumarate. Arginine can be hydrolysed to form urea and ornithine (the latter of which can then react with carbamoyl phosphate to start the cycle over). Fumarate, on the other hand, can be hydrolysed to form malate, which can then be converted to oxaloacetate via malate dehydrogenase, releasing electrons which can be picked up by NAD+ to form NADH. Oxaloacetate can then undergo transamination to reform aspartate.
The overall reaction is as follows:
NH4+ + HCO3- + aspartate --> urea + fumarate
Amino Acids- Peptide Bonds
Yup, yet another post about amino acids :( Looking at the lecture notes, this isn't going to be easy either...)
You should know the net reaction for peptide bond synthesis and the net hydrolysis reaction.
Okay, this bit I'm pretty sure I know. The net synthesis reaction is just a condensation reaction, where two amino acids bond together and water is released. The net hydrolysis reaction is the opposite: water is added in order to break the bond. (It's not that simple when broken down into individual steps, but that's the net reaction.)
You should know the principles of synthetic peptide synthesis but not detailed structures of the protecting groups and solid phase coupling group. You should understand that synthetic peptide synthesis proceeds from C-terminus to N-terminus.
One of the challenges of synthesising peptides is that you need to ensure that the correct amino acids bind to each other in the correct order. One way to achieve this is to use "protecting groups"- that is, to keep the ends covered until you want them to bind to something. Manipulating both ends can be a bit clunky, so the terminal amino acid is often bound to a polystyrene bead (the "solid phase) on the carboxyl terminus. The amino termini of all of the amino acids is then bound to an Fmoc group which protects amino acids from reaction until required.
When it's time for another amino acid to be added on, the N-terminus of the peptide already on the chain has its Fmoc group removed, and the amino acid to be added on is activated at its carboxyl group by another molecule called DCC. The alpha-amino group of the peptide then "attacks" the carboxyl group of the amino acid to be added. These steps are repeated as often as required, and then HF is used to wash the peptide off the bead.
You should know the amino acid activation reaction with ATP and the nucleophilic substitution reaction coupling the amino acid to tRNA. You should know how the peptide bond is formed between the alpha-amino group of the aminoacyl-tRNA and the carbonyl carbon of the ester link in the peptidyl-tRNA. You should understand that protein synthesis proceeds from N-terminus to C-terminus.
Aminoacyl-tRNA, simply put, is a tRNA with an amino acid attached to the 3' end.
To attach an amino acid to tRNA, the amino acid must first be activated. The alpha-carboxyl of the amino acid attacks the alpha-phosphate of ATP (i.e. the phosphate closest to the ribose sugar), forming 5' aminoacyl adenylate, or aminoacyl-AMP. From there, there are two slightly different paths that the amino acid can take that both lead to the formation of aminoacyl-tRNA. Class I aminoacyl-tRNA synthases will attach the aminoacyl-AMP to the 2' end of the tRNA, releasing AMP. Transesterification then moves the amino acid on to the 3' end. Class II aminoacyl-tRNA synthases, on the other hand, attach the aminoacyl-AMP directly to the 3' end of the tRNA.
Peptidyl-tRNA is tRNA with a polypeptide chain attached to the 3' end. When an aminoacyl-tRNA comes in to the ribosome, the alpha-amino terminus of the new amino acid nucleophilically attacks the carboxyl terminus of the polypeptide chain on the peptidyl-tRNA. This creates a new peptide bond between the amino acid and the polypeptide chain, causing the polypeptide chain to move onto what was originally the aminoacyl-tRNA.The original peptidyl-tRNA, now deprived of its polypeptide, is soon released from the ribosome.
You should know what exopeptidases/exoproteases and endopeptidases/endoproteases are and the differences between them. You should also understand that proteins are broken down, via peptide hydrolysis reactions ultimately into their constituent amino acids, which are most commonly re-used to synthesise proteins.
Peptidases/proteases are enzymes that break down polypeptides. Exopeptidases/exoproteases break off amino acids one by one from the end of the chain, whereas endopeptidases/exoproteases break down bonds in the middle of the chain. Exopeptidases are not specific to any particular amino acids, but endopeptidases are. For example, trypsin will only break bonds at the C-terminal of arginine or lysine.
You should be able to draw the resonance and resonance hybrid forms of the peptide bond and understand that the partial double bond character of the peptide bond prevents rotation about it in proteins and peptides.
The two resonance forms of the peptide bond are one in which the double bond is between the C and the O, and one in which the double bond is between the C and the N with a negative charge on O and a positive charge on N. Resonance leads to stability, and stability in this case means that there is less rotation allowed around the peptide bond.
You should be able to draw H-bonding interactions between the peptide bond and other peptide bonds and with amino acid side chains.
I wrote a bit about peptide bonds in my first post about amino acids.
You should know how disulfide bonds form and that only some proteins have disulfide bonds that stabilise their structures, mainly extracellular proteins in an oxidising environment.
Disulfide bonds are S-S bonds formed between two Cys residues in an oxidising environment. As these are covalent bonds, they play important roles in stabilising the structure. They can be broken in a reducing environment.
You should know what isopeptide bonds are and that they occur between the side chain amino group of Lys and most commonly the side chain carbonyl carbon of Asp/Asn/Gln, but sometimes with a terminal carboxyl carbon. You should understand that the role of intracellular isopeptide bonds in providing extra stability to proteins. You should be aware that ubiquitination tags proteins targeted for degradation with the protein ubiquitin.
Isopeptide bonds are, simply put, peptide bonds that aren't between the alpha-amino and alpha-carboxyl groups but rather between the side chains of amino acids. As these bonds are covalent and have resonance, they are quite stable.
I'm not really sure how ubiquitination is relevant here, but yes, ubiquitin is sometimes added onto proteins as a way of marking them for degradation in proteases etc.
You should be aware that there are other amino acids other than the common 20. Two of these, selenomethionine and selenocysteine, where selenium replaces sulfur in the amino acids, are incorporated during protein synthesis via aminoacyl-tRNAs. Selenomethionine appears to be randomly incorporated instead of methionine whilst selenocysteine is incorporated via a special aminoacyl-tRNA at specific points in protein synthesis. Details of selenocysteyl-tRNA synthesis need not be learnt.
There's not much I can elaborate on here, other than that selenomethionine does not appear to have an effect on protein structure or function, whereas selenocysteine may be important for the activity of some enzymes.
You should be aware that other amino acids are derived from normal amino acids that are chemically modified (through the actions of enzymes) in the synthesised proteins in a general process called post-translational modification.
Amino acids can be modified by adding other groups, such as hydroxyl groups, onto them. For example, a hydroxyl group can be added to proline to form 4-hydroxyproline. These modified amino acids can be important- for example, 4-hydroxyproline is often found in collagen fibres (if I remember correctly).
You should know the net reaction for peptide bond synthesis and the net hydrolysis reaction.
Okay, this bit I'm pretty sure I know. The net synthesis reaction is just a condensation reaction, where two amino acids bond together and water is released. The net hydrolysis reaction is the opposite: water is added in order to break the bond. (It's not that simple when broken down into individual steps, but that's the net reaction.)
You should know the principles of synthetic peptide synthesis but not detailed structures of the protecting groups and solid phase coupling group. You should understand that synthetic peptide synthesis proceeds from C-terminus to N-terminus.
One of the challenges of synthesising peptides is that you need to ensure that the correct amino acids bind to each other in the correct order. One way to achieve this is to use "protecting groups"- that is, to keep the ends covered until you want them to bind to something. Manipulating both ends can be a bit clunky, so the terminal amino acid is often bound to a polystyrene bead (the "solid phase) on the carboxyl terminus. The amino termini of all of the amino acids is then bound to an Fmoc group which protects amino acids from reaction until required.
When it's time for another amino acid to be added on, the N-terminus of the peptide already on the chain has its Fmoc group removed, and the amino acid to be added on is activated at its carboxyl group by another molecule called DCC. The alpha-amino group of the peptide then "attacks" the carboxyl group of the amino acid to be added. These steps are repeated as often as required, and then HF is used to wash the peptide off the bead.
You should know the amino acid activation reaction with ATP and the nucleophilic substitution reaction coupling the amino acid to tRNA. You should know how the peptide bond is formed between the alpha-amino group of the aminoacyl-tRNA and the carbonyl carbon of the ester link in the peptidyl-tRNA. You should understand that protein synthesis proceeds from N-terminus to C-terminus.
Aminoacyl-tRNA, simply put, is a tRNA with an amino acid attached to the 3' end.
To attach an amino acid to tRNA, the amino acid must first be activated. The alpha-carboxyl of the amino acid attacks the alpha-phosphate of ATP (i.e. the phosphate closest to the ribose sugar), forming 5' aminoacyl adenylate, or aminoacyl-AMP. From there, there are two slightly different paths that the amino acid can take that both lead to the formation of aminoacyl-tRNA. Class I aminoacyl-tRNA synthases will attach the aminoacyl-AMP to the 2' end of the tRNA, releasing AMP. Transesterification then moves the amino acid on to the 3' end. Class II aminoacyl-tRNA synthases, on the other hand, attach the aminoacyl-AMP directly to the 3' end of the tRNA.
Peptidyl-tRNA is tRNA with a polypeptide chain attached to the 3' end. When an aminoacyl-tRNA comes in to the ribosome, the alpha-amino terminus of the new amino acid nucleophilically attacks the carboxyl terminus of the polypeptide chain on the peptidyl-tRNA. This creates a new peptide bond between the amino acid and the polypeptide chain, causing the polypeptide chain to move onto what was originally the aminoacyl-tRNA.The original peptidyl-tRNA, now deprived of its polypeptide, is soon released from the ribosome.
You should know what exopeptidases/exoproteases and endopeptidases/endoproteases are and the differences between them. You should also understand that proteins are broken down, via peptide hydrolysis reactions ultimately into their constituent amino acids, which are most commonly re-used to synthesise proteins.
Peptidases/proteases are enzymes that break down polypeptides. Exopeptidases/exoproteases break off amino acids one by one from the end of the chain, whereas endopeptidases/exoproteases break down bonds in the middle of the chain. Exopeptidases are not specific to any particular amino acids, but endopeptidases are. For example, trypsin will only break bonds at the C-terminal of arginine or lysine.
You should be able to draw the resonance and resonance hybrid forms of the peptide bond and understand that the partial double bond character of the peptide bond prevents rotation about it in proteins and peptides.
The two resonance forms of the peptide bond are one in which the double bond is between the C and the O, and one in which the double bond is between the C and the N with a negative charge on O and a positive charge on N. Resonance leads to stability, and stability in this case means that there is less rotation allowed around the peptide bond.
You should be able to draw H-bonding interactions between the peptide bond and other peptide bonds and with amino acid side chains.
I wrote a bit about peptide bonds in my first post about amino acids.
You should know how disulfide bonds form and that only some proteins have disulfide bonds that stabilise their structures, mainly extracellular proteins in an oxidising environment.
Disulfide bonds are S-S bonds formed between two Cys residues in an oxidising environment. As these are covalent bonds, they play important roles in stabilising the structure. They can be broken in a reducing environment.
You should know what isopeptide bonds are and that they occur between the side chain amino group of Lys and most commonly the side chain carbonyl carbon of Asp/Asn/Gln, but sometimes with a terminal carboxyl carbon. You should understand that the role of intracellular isopeptide bonds in providing extra stability to proteins. You should be aware that ubiquitination tags proteins targeted for degradation with the protein ubiquitin.
Isopeptide bonds are, simply put, peptide bonds that aren't between the alpha-amino and alpha-carboxyl groups but rather between the side chains of amino acids. As these bonds are covalent and have resonance, they are quite stable.
I'm not really sure how ubiquitination is relevant here, but yes, ubiquitin is sometimes added onto proteins as a way of marking them for degradation in proteases etc.
You should be aware that there are other amino acids other than the common 20. Two of these, selenomethionine and selenocysteine, where selenium replaces sulfur in the amino acids, are incorporated during protein synthesis via aminoacyl-tRNAs. Selenomethionine appears to be randomly incorporated instead of methionine whilst selenocysteine is incorporated via a special aminoacyl-tRNA at specific points in protein synthesis. Details of selenocysteyl-tRNA synthesis need not be learnt.
There's not much I can elaborate on here, other than that selenomethionine does not appear to have an effect on protein structure or function, whereas selenocysteine may be important for the activity of some enzymes.
You should be aware that other amino acids are derived from normal amino acids that are chemically modified (through the actions of enzymes) in the synthesised proteins in a general process called post-translational modification.
Amino acids can be modified by adding other groups, such as hydroxyl groups, onto them. For example, a hydroxyl group can be added to proline to form 4-hydroxyproline. These modified amino acids can be important- for example, 4-hydroxyproline is often found in collagen fibres (if I remember correctly).
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