Peptides- Charge and Isoelectric Points, Buffering Capacity

More chemistry revision! Unfortunately this topic can be a bit of a drag...

Understand the relationship between pH, pKa and charge on ionisable amino acid side chain chains and the concept of isoelectric point. Be able to estimate net average charges on amino acid side chains at specific pHs, given the pKas of the ionisable groups. 

This was already covered in my previous post about the acid base chemistry of amino acids.

Be able to to estimate the pH range in which the isoelectric point of a peptide occurs. 

To calculate the isoelectric point of a peptide, calculate the charge on the entire peptide at different pHs by adding together the charges on each amino acid (see my previous post for how to calculate the charges of amino acids). Not sure which pHs to test? Test the pKas of the various amino acids within the peptides. The isoelectric point is the point at which the charge on the entire peptide is 0. However, you might not be able to find this point via this method- you might instead find one point where the charge is +0.5 and another point where the charge is -0.5 (or +1/-1 etc.). The pH range of the isoelectric point will lie in between these two points, but for now we haven't learned how to calculate the exact isoelectric point (though there are programs that can do this).

Understand the idea of a buffer and that buffering occurs maximally around the pKa of the ionisable group(s) of the buffer. 

Understand the concept of how added –OH reacts directly with the protonated form of a buffer in the buffering range, thus reducing the the concentration of –OH that is left to directly react with protons in solution (H+, H+(aq), H3O+) to increase the pH. This type of argument also applies to the reverse titration, starting at high pH and adding acid, where added HA directly protonates the deprotonated form of the ionisable group of the buffer rather than directly generating H+ (H+(aq), H3O+). 

A buffer is, simply put, a substance that can absorb changes in pH. Weak acids and weak bases tend to make good buffers due to their reversible dissociation equations. Here's an example for a generic acid HA:

HA + H2O <--> H3O+ + A-

Here, any acidic H+ that is added to solution simply protonates A- and has little effect on the overall solution's pH. When more basic OH- is added to solution, however, it is protonated by HA and, once again, has little effect on the solution's pH. Overall, the effect is to absorb the effects of the H+ or OH- added into the solution.

The reason why buffering occurs maximally around the pKa of the buffer is that when pH = pKa, the amount of HA is equal to the amount of A-. Hence it is equally easy for the buffer to absorb an increase in H+ and OH-.

Understand why the buffering capacity of a buffer depends on its concentration. 

This is a relatively simple concept to explain. Essentially, the more buffer that you have, the more you have to "mop up" any extra acid or base that is added to the solution.

Understand the importance of charges on proteins. 

Charge is important as it provides another way in which proteins can interact with each other. For example, proteins that interact with DNA, such as histone proteins, often have positive charges in order to interact with negatively-charged phosphate groups on the DNA.

Understand that proteins can buffer pH changes, primarily owing to the side chains of histidine residues in the proteins.

As I've mentioned before, proteins can exist in protonated and deprotonated forms, just like acids and bases. Hence, proteins can also act as buffers. I think the reason why histidine is considered important here is because the imidazole ring of histidine has several ionisable groups.

Amino Acids- Acid Base Chemistry

Continuing on from yesterday's post about amino acids, today I'm going to go into a bit more detail about their properties.

Understand the pH scale and know the definition of of pH. Be able to calculate pH from [H+] and [H+] from pH. Understand that H+, H3O+ and H+(aq) are used interchangeably.

Know that high pH is alkaline and low pH is acidic, neutral pH is about pH 7.

Nice easy dot points to start off with :) The pH scale is, in a nutshell, a measure of how acidic or basic a substance is. It runs on a scale of 1-14, with lower pHs indicative of acidity and higher pHs indicative of basicity. A pH of 7 is neutral (neither acidic or basic).

pH is related to the concentration of hydrogen ions in solution. pH is equal to the negative log of the concentration of H+ ions in solution. Hence:

pH = -log [H+] and [H+] = 10^(-pH).

As for the last point, if I remember correctly from year 12 chemistry, hydrogen ions tend not to exist by themselves in solution, but rather associate with water molecules to form hydronium ions (H3O+). Hence H3O+ and H+ can be used interchangeably. (As for the H+(aq)... well that's just the same as H+ except it also has the state symbol.)

Know the difference between strong and weak acids and bases and understand the relationship between a conjugate acid and a conjugate base. 

The strength of an acid or base depends on how readily it dissociates into ions in solution. An acid or base that dissociates more fully in solution is considered to be a strong acid/base, whereas an acid or base that does not dissociate very much is considered to be a weak acid/base.

A conjugate acid is simply a base with a proton added, whereas a conjugate base is simply an acid with a proton removed.

Know the definition of pKa and be able to calculate it from Ka. Understand the relationship between pKa and acid strength. 

Ka is basically the equilibrium constant of the acid/base dissociation equation. If that doesn't make sense, let me try and explain with an equation. Let's say that you have an acid HA (where A is any anion). Its dissociation equation is this:

HA <--> H+ + A-

Equilibrium constants are basically the product of the concentration of the products divided by the product of the concentration of the reactants. Hence the Ka, or the equilibrium constant of this equation, is as follows:

([H+][A-])/[HA]

pKa is the negative log of Ka:

pKa = -log(([H+][A-])/[HA])

Lower pKas are indicative of stronger acids, whereas higher pKas are indicative of weaker acids. Let me try and explain why:

In a strong acid, there is a much higher concentration of products than reactants. Hence Ka is high. When taking a log, you'll also end up with a relatively high number, but this will then be reversed by the negative sign. (So for example if Ka was 10^3 you'd end up with a pKa of -3 by the end of it.) In a weak acid, on the other hand, the ratio between products and reactants is much smaller and hence Ka will be smaller. After taking a negative log you'll end up with a larger number. For example, if Ka was 10^(-6) the pKa would be 6.

That was a pretty terrible explanation, but moving on...

Know the derivation of the Henderson-Hasselbalch equation and the equation itself. Know that when the pH = pKa the concentrations of A- or conjugate base and HA or conjugate acid are equal. Be able to use the Henderson-Hasselbalch equation to calculate pH or pKa or concentrations of conjugate acid and conjugate base or net average charges. 

The Henderson-Hasselbalch equation is, simply put, a way of relating pH and pKa. This equation can be derived from the pKa equation (given above) and the knowledge that pH = -log[H+].

pKa = -log(([H+][A-])/[HA])
pKa = -(log([A-]/[HA]) + log[H+])
pKa = -log[H+]) - log([A-]/[HA])
pKa = pH - log([A-]/[HA])

This can be rearranged to give pH = pKa + log([A-]/[HA])

A neat feature of this equation is that it makes it easy to work out the isoelectric point- that is, the pH at which the concentration of A- is equal to the concentration of HA. You see, when pH is equal to pKa, they cancel out on both sides, giving log([A-]/[HA]) = 0. Now log 1 = 0, so ([A-]/[HA]) = 1. Hence A- must be equal to HA (since a number divided by itself gives 1).

Average net charges can be found by looking at the pKa of each ionisable group. Firstly you need to find the ratio of conjugate base (A-) to conjugate acid (HA). To do this simply plug in the pH and pKa given into the equation. If the ratio of conjugate base to conjugate acid is high, then assume that nearly all of the substance is in the base form, and calculate the charge accordingly. (This will depend on the substance- for example if you have NH3+/NH2, the deprotonated form will be neutral, whereas if you have COOH/COO-, the deprotonated form is negatively charged.) You can use similar logic if the ratio is very low, but in this case you would assume nearly all of the substance is in the acid form. If the substance is at its isoelectric point and half of it is protonated and the other half deprotonated, use the average of the charges of the two forms. For example if you had half NH3+ and half NH2, the average would be +0.5. From there, you can add up all of the charges to get an average net charge.

Be aware of the approximate pKas of the α-amino and α-carboxyl groups of amino acids. 

This one'll be quick to answer. The approximate pKa of the alpha-amino group is 9.0 while the approximate pKa of the alpha-carboxyl group is 2.0.

Know the structures of the ionisable groups of amino acid side chains e.g. Asp/Glu: –COOH <--> COO- + H+, Lys: -NH3 + <--> - NH2 + H+, Cys: -SH <--> S- + H+ and hence know the charge on the conjugate acid and conjugate base forms of the ionisable groups. 

Generally, for polar acidic amino acids, the protonated form is uncharged while the deprotonated form is negatively charged. For polar basic amino acids, and other polar uncharged amino acids with -NH2 groups, the protonated form is positively charged while the deprotonated form is uncharged. Finally, for cysteine, the protonated form is uncharged while the deprotonated form is negatively charged.

Amino Acids

Time to get down to revising for my CHEM1004 exam!

I was originally going to do a series of posts going through textbook questions, but there were so many questions (plus so many that required diagrams) that after going through five or so chapters I realised that it wasn't an efficient use of my time. Besides, I'm sure my answers wouldn't have been that interesting to read either. Now I'm going to do the whole "expand on lecture key points" thing again because, let's face it, I have no imagination.

In this post, I'm going to talk about amino acids. To be honest, amino acids and proteins was probably my least favourite topic in this unit, but I guess it's pretty important because proteins have a wide variety of important functions in our bodies. So, without further ado, let's dig in:

Be able to draw the general structure of an amino acid in its cation, zwitterion and anion forms and know at which general pHs each form exists.

Amino acids can exist in a cation (positively charged), anion (negatively charged) or zwitterion (containing both positive and negative charges) forms. In the cation form, the amino group is positively charged (-NH3+) and the carboxyl group is neutral (-COOH). In the anion form, the amino group is neutral (-NH2) and the carboxyl group is negatively charged (-COO-). In the zwitterion form, the amino group is positively charged (-NH3+) and negatively charged (COO-).

Cation forms are more likely to exist at acidic pHs. This is due to the higher concentration of hydrogen ions which can protonate amino acids. The inverse is also true: anion forms are more likely to exist at basic pHs due to the lower concentration of hydrogen ions. Zwitterions are most commonly seen at around neutral pH.

Know that in all amino acids except glycine the α-carbon is a chiral centre and why this is the case.

In most amino acids, the amino carbon (i.e. the carbon next to the carboxyl group) is attached to four different groups: a hydrogen atom, the carboxyl group, the amino group and the amino acid side chain. Due to this, most amino acids are chiral. The exception is glycine as glycine's side chain is a single H. Thus the amino carbon of glycine has two identical -H side chains and is therefore achiral.

Understand the difference between R- and S- enantiomers and be able to identify the enantiomer given a structure of a chiral compound or be able to draw a structure given the enantiomer. This involves knowing the basis of priority for groups attached to the chiral carbon centre.

The first thing I'm going to attempt to quickly explain here is the concept of priorities (I'll probably cover this in further detail in a later post). Basically each chiral carbon atom is attached to four different groups. These groups can be given "priorities" for the purpose of determining whether the molecule is an R- or an S- enantiomer. Firstly, look at the four atoms directly attached to the chiral carbon and rank them in order of molecular weight (highest to lowest). So Cl would rank higher than O, O would rank higher than C, and so on. If you have a tie, take a look at the atoms attached to those atoms and go from there until you break the tie. (As I said, I'll probably explain this more clearly in a later post.)

Once you've assigned priorities to all four groups, mentally rotate the molecule so that the group with the lowest priority is facing away from you. If the other three groups are now going clockwise from highest to lowest priority, the molecule is in the R-enantiomer. If the three groups are going anticlockwise from highest to lowest priority, the molecule is in the S-enantiomer.

When it comes to drawing enantiomers, just be aware of your groups and their order, and make sure to place them accordingly. It makes it easier if you can draw them with the lowest priority atom already pointing away from you.

Understand that enantiomers rotate plane-polarised light either clockwise or anticlockwise by equal and opposite degrees and that this specific rotation, [α], is named dextrorotatory (D-, (+)) and laevoratatory (L-, (-)) respectively. Note that the direction of rotation cannot be predicted from the enantiomeric form of the chiral compound.

This dot-point kind of speaks for itself. Plane-polarised light, to my understanding, is basically like a long, flat beam of light. When it travels through optically active substances, the plane of light can rotate and this rotation can then be measured. Those that rotate the light clockwise are dextrorotatory and those that rotate the light anticlockwise are levorotatory.

Understand that most amino acids in biology are designated L-amino acids, but this does not refer to their specific rotation (some have (+) specific rotations some have (-) specific rotations), rather to their stereochemical relationship to L-glyceraldehyde. Know that most amino acids are thus S-enantiomers, except Cys that is R- and know why Cys is an exception. 

Once again, this dot point is kind of self-explanatory. After explaining what D- and L- means, I now have to take it back and say that that doesn't apply for amino acids. Instead, amino acids are designated as D- or L- depending on whether their structure is more similar to D-glyceraldehyde or L-glyceraldehyde. If you want to indicate specific rotation for amino acids, you'll have to stick to +/-.

Most amino acids are S-enantiomers (like L-glyceraldehyde), with -NH2 as their highest priority groups, followed by -COOH and then the amino acid side chain. Cysteine is an exception to this rule as its side chain is -CH2SH which ranks higher than -COOH due to the higher molecular weight of S compared to O.

Understand the importance of chirality in biology in relation to substrate binding in enzymes.

Chirality is important as different forms of the same molecule have different shapes, and enzymes will only bind substrates of the right shape. The wrong diastereomer, therefore, may not have any effect in a biological system.

Know which amino acids belong in each grouping and recognise the three-letter nomenclature for each amino acid. Know that Asp and Glu have carboxyl groups in their side chains, and the structures of the side chain groups of Lys, His and Arg. Also, know that Ser, Thr and Tyr all have –OH groups and that Cys has an –SH group.

I hate memorising stuff but apparently FAMILY VW is a good way to remember all of the hydrophobic amino acids. If only I could remember what each of the letters stood for...

Okay, well I remember that alanine is hydrophobic, so that must be the A. And then there's isoleucine, leucine and valine that are all also hydrophobic (the I, L and V). That's 50%! A pass (in Australia, at least)! w00t!

Hmm what else can I dig out of my brain... phenylalanine is hydrophobic, as it's basically just alanine attached to a benzyl (phenyl) group. (Just double-checked my book and it turns out that phenylalanine is F.) Also tryptophan is hydrophobic as that's a benzene ring attached to a five-carbon ring. Oh and then there's also proline, which is basically just a ring structure. Just one more...

Okay, stuff it, I'm just going to look in the textbook. Apparently the missing link is methionine, which I completely skipped over because I got fooled by the fact that it has an S in there (it's buried between two methyl groups, so it's probably not that reactive because of that).

Now for the other groups!

The polar acidic amino acids are aspartic acid and glutamic acid. Kind of easy to remember because there's only two of them and they both have "acid" in their names.

The three basic amino acids are lysine, histidine and arginine. Histidine has a five-membered imidazole ring in it (it's basically a ring with two Ns in it), arginine has two NH2 groups at the end, and lysine has a five-carbon chain (including the alpha carbon, not including the carbonyl carbon) with an -NH2 group at the end.

All other amino acids are polar uncharged. These are asparagine, cysteine, glutamine, glycine, serine, threonine and tyrosine.

Understand how polar and nonpolar amino acid side chains interact with water and how this influences their placement in a protein. Understand how hydrogen bonds are formed and be able to recognise potential H-bonding between amino acid side chains.

Water is relatively organised, with each water molecule hydrogen-bonded to other water molecules. When a hydrophobic nonpolar side chain enters the water, the water molecules have to rearrange to form a "cage" around it. This is energetically unfavourable, especially if there are a lot of hydrophobic chains exposed to the water. If all of the hydrophobic amino acids cluster together, however, with hydrophilic amino acids on the outside, this disruption of water molecules is reduced. Hence the outside of proteins tend to be hydrophilic with hydrophobic centres.

Hydrogen bonds are generally formed between relatively positive hydrogens and relatively negative oxygens or nitrogens. As there are plenty of =O and -NH groups in amino acids due to the peptide linkages, peptide chains can form intrachain hydrogen bonds. As well as in peptide linkages, certain amino acids (see my answer to the previous question) have groups such as -OH, -NH2 etc. which are also capable of forming hydrogen bonds. Hence hydrogen bonds can also form between amino acid side chains.

Sensitive and Self-Sensitive Cells

Last one! Then I need to stop procrastinating and do the online test... *shudders* (Even though it's only worth 2%, I am a massive perfectionist so it still makes me nervous.)

1. Explain the ionic basis of membrane potentials.

A membrane potential is essentially the difference in charge between the two sides of the membrane. Generally, the inside of cells is more electrically negative than the outside. Membrane potentials are generated by the selective permeability of cell membranes to certain ions. For example, K+ ions can diffuse out of the cell, leaving negatively-charged electrons behind. Not too many K+ ions can diffuse out, however, as eventually the cell becomes negatively charged enough for the K+ to be drawn back in. This eventually creates an equilibrium which becomes the membrane's resting potential.

2. Understand the principle of the Nernst equation and electrochemical equilibrium.

As I mentioned in the previous paragraph, eventually an equilibrium is created where there is no net movement of charge or ions across the membrane. The value that the membrane voltage would have to be for the K+ ions to be in equilibrium can be determined by the Nernst equation:

Vm = -60log(([K+]i)/([K+]o)) millivolts

where Vm is the value of the membrane voltage (that m is meant to be a subscript, by the way), [K+]i is the concentration of K+ inside the cell, and [K+]o is the concentration of K+ outside of the cell (i and o are also meant to be subscripts).

3. Understand the principle of the Goldman equation and how it relates to the steady state membrane potential.

Electrochemical equilibrium does not happen inside of cells, as there are many different ions that can move around and create multiple ion gradients. Hence the Goldman equation, instead of the Nernst equation, can be used for cells. The Goldman equation is similar to the Nernst equation, except the ([K+]i)/([K+]o) part is different. It's going to be hard to explain, so let me give you an example.

For example, let's say that you're looking at K+, Na+ and Cl-.

The numerator would be (PK)(Kin) + (PNa)(Nain) + (PCl)(Clout)
The denominator would be (PK)(Kout) + (PNa)(Naout) + (PCl)(Clin)
(Stuff in italics is meant to be subscripted. The P__ is the relative membrane permeability of the ion in question, whereas the _in and _out is the concentration of the ion inside and outside of the membrane.)

I really hope that you can see the pattern here, because I'm not sure how to explain it. Note that the Cl is swapped around- its extracellular concentration is on top, and its intracellular concentration is on the bottom. This is because Cl- is a negative ion whereas the other two are positive. If you have other ions that you need to look at, you can add them in as well.

4. Describe & explain the ionic basis of electrical signalling in excitable (nerve and muscle) cells.

The main basis of signalling in nerve and muscle cells is the action potential. Nerve and muscle cells have voltage-gated ion channels in their cell membranes. These are channels that open when they are stimulated by a high enough voltage. At low levels of stimulation, the channels do not open, but when there is enough stimulation (enough to raise the membrane potential to -55mV), Na+ channels open, allowing Na+ to rush in and depolarise the membrane. Na+ channels begin to close as K+ channels open, allowing K+ to leave and restore the charge. (Na+/K+ pumps restore the original concentrations of both ions, but this takes a small amount of time, meaning that following an action potential there is a "refractory period" during which the cell cannot fire.) The influx of Na+ ions can also stimulate nearby channels, leading to a rapid wave of depolarisation down the axon. This causes synaptic vesicles at the end to be released into the synapse, where they can then stimulate other neurons or muscle cells.

Membrane Transport- Pumps and Carriers

Second last! w00t!

This lecture doesn't have a "lecture objectives" slide, so I'll just list the dot points in the summary and expand on each one as much as I can.

1. Pores, channels and transporters increase permeability.

As I mentioned in my previous post, some substances do not diffuse through the membrane readily and thus require pores/channels or transporters to carry them through.

2. Transporters can be passive carriers or active pumps. 4. Primary active transport use an ATPase. 5. Secondary active transport uses a co-transporter to move one thing down its concentration gradient and another up its gradient.

Transporters, or carrier proteins, work by binding a substance to one side, then changing conformation to release the substance on the other side. If the substance is being carried down its concentration gradient, this process requires no energy (as the energy is provided by the concentration gradient).

Some carriers carry multiple substances. These are called symports, if they carry substances in the same direction, or antiports, if they carry substances in opposite directions. Carriers can carry substances against their concentration gradient if they are also carrying another substance down its concentration gradient. In this way the energetically unfavourable movement of a molecule up its concentration gradient is coupled to the "favourable" movement of a molecule down its concentration gradient. This is called secondary active transport.

Active pumps carry substances up their concentration gradients, but unlike the other passive carriers mentioned above, they require hydrolysis of ATP to do so. This is called primary active transport.

3. Transporters show saturation.

Transporters, unlike channels, can only transport a certain number of molecules at a time (as they must bind and change shape). There are also only a certain number of transporters in a cell's membrane. Hence, once all of the transporters are occupied, no more molecules can be transported across at the same time. When all of the transporters are occupied and the rate of transport ceases to increase, the transporters are said to be saturated.

6. Active transport keeps the contents of cells far from equilibrium.

As active transport can transport molecules up their concentration gradients, they can create a situation in which there is a large concentration of a molecule on one side of the membrane and a low concentration on the other. This is the case in the Na+/K+ pump: active transport keeps most of the Na+ outside the cell and most of the K+ inside the cell. This creates concentration gradients which can be used to drive other processes in the cell, such as the transport of glucose into the cell via the Na+/Glucose symport.

Membrane Diffusion and Osmosis

Down to the last three lectures that need revising for now! Yay :) (Also there are only 3 more lectures left for the semester! Almost there!)

1. Understand the random movement of molecules down a concentration gradient.

As you should know by now, molecules diffuse from an area of high concentration to an area of low concentration. (To find out more, review my earlier post on diffusion and osmosis.)

One common misconception with diffusion is that particles move to an area of low concentration. This isn't strictly true. Molecules don't wake up in the morning and think "hmm, looks like there's less of us over there, better get moving then." They move around randomly, dispersing themselves over the available area. Throughout this process of dispersion, you get roughly even amounts of the molecule throughout the area. Hence the net movement of molecules is from a high concentration to a low concentration, but the motions of individual molecules is still random.

2. Understand the importance of molecular weight, lipid solubility and charge on membrane permeability.

Molecules with a lower molecular weight tend to diffuse more rapidly, probably because they don't have to "push aside" other molecules as much.

Lipid solubility is another important factor influencing membrane permeability. Molecules that are more lipid soluble are more permeable as they are soluble in the cell membrane (which is a lipid bilayer).

Charged particles, as they tend to be hydrophilic and lipid insoluble, have a very low permeability. They can exert their effects by binding to receptors on the cell surface (see my post on biochemical messengers) or by diffusing through ion channels (I'll talk about these later).

3. Define the concept of flux and its relationship to membrane permeability.

Flux is the movement of molecules across a given area. If you are looking at the flux across the membrane, you are essentially also looking at the membrane permeability (i.e. the ability of molecules to move across the membrane). Hence flux is directly proportional to membrane permeability.

4. Contrast the diffusion of solutes through a membrane and through a pore/channel.

I'm not really sure what there is to contrast here. While lipid-soluble substances diffuse through the membrane, ions and other charged substances often have to travel through channels. The general principles still apply, but sometimes the channels can open and close.

5. Understand the concepts of osmolarity/osmolality and tonicity.

Osmolarity is the number of particles dissolved in a litre of solution, while osmolality is the number of particles dissolved in a kilogram of solution. These terms are pretty much interchangeable on Earth.

Tonicity, on the other hand, is the ability of a solution to affect a cell. A hypotonic solution will cause a cell to burst while a hypertonic solution will cause a cell to lyse. Sometimes a hypotonic solution will also be hypoosmotic, but not always. Sometimes a solution might start out with the same osmolarity as the cell, but it contains certain particles that can diffuse into the cell, thus changing the osmolarities of the cell and the solution.

6. Describe the distribution of water and solute concentrations in the body compartments.

I'm not really sure what is required here. There's a diagram in the slides that shows that most water (roughly 25L) is in the cells, followed by extracellular fluid (roughly 13L), followed by blood plasma (roughly 3L). There's also another 1L marked "transcellular fluid," which, according to Google, is basically fluid that is between cells but separated off from the main interstitial fluid. All but the transcellular fluid has an osmolarity of around 290 mOsm (i.e. roughly the same amount of particles dissolved in all of the fluids in the body).

7. Understand the importance of osmolality and hydrostatic pressure in the regulation of fluid balance.

As blood travels to the tissues, water diffuses into the interstitial fluid via osmosis (a process that is partially counteracted by hydrostatic forces). In the capillaries in the arterial side, the blood pressure is much greater, causing more water to diffuse into the interstitial fluid than out of it (or at least that's my understanding). The opposite occurs in the venous capillaries. This process mostly "balances out" the movement of fluid into and out of the interstitial space, with any extra fluid picked up by the lymph system. If not enough water diffuses back into the blood on the venous side, however, water can build up in the interstitial space, resulting in swelling, or oedema.

Cellular Protein Trafficking and Secretion

Yes, yet another revision post :( Still a few more to go...

1. Outline how proteins with distinct localisation signals are delivered to the nucleus and mitochondrion.

Proteins have localisation signals in their amino acid sequence that indicate where they will be needed in the cell.

Proteins that are needed in the nucleus have a nuclear localisation signal (NLS) in their amino acid sequence. This sequence can be located anywhere in the protein and is not removed during translation. To be imported into the nucleus, the NLS binds to a protein called importin, or a nuclear import receptor. Once the bound complex enters the nucleus, Ran-GTP binds to importin, releasing the protein. Importin with Ran-GTP bound can then exit the nucleus, where a GAP (GTP activating protein) hydrolyses GTP, resulting in Ran-GDP, which dissociates from importin, leaving it free to import another molecule.

Export of proteins from the nucleus can also occur in a similar way. Some proteins have nuclear export signals. They bind to a special nuclear export receptor along with Ran-GTP. The entire complex travels out of the nucleus, where Ran-GTP has its GTP hydrolysed to GDP. Ran-GDP then dissociates along with the protein being carried out.

Proteins that are needed in the mitochondria travel there a little differently. Once again, they require a localisation signal, but in this case the localisation signal must be at the end that is translated first. They are translated in the cytoplasm, and then their folding is blocked by hsp70 as a folded protein would be too bulky to travel into the mitochondria. The mitochondrial membrane is somewhat harder to cross than the nuclear membrane and as such there are TOM and TIM protein translocators in the mitochondrial membrane that make this job easier. The localisation signal first binds to the TOM (transporter outer membrane complex) where it is fed through to the TIM (transporter inner membrane complex). From there, it is dragged into the mitochondrion. Energy for this process is provided by the electron transport chain, which is provided through oxidation of food molecules (if I understand correctly). Once safely inside the mitochondrion, the localisation signal is cleaved off.

This isn't super relevant, but it was in the lecture slides, so I'll include it anyway. Aside from the TOM and TIM complexes, there are also complexes that serve to help anchor proteins into the membrane. The SAM complex helps anchor proteins in the outer membrane, while the OXA does the same for the inner membrane.

2. 3. Be able to list the steps involved producing a secreted protein. Outline the functions of vesicles.

Like proteins that are needed in the mitochondria, proteins that will eventually be secreted have a special signal at their N-terminus. These proteins are first translated on free ribosomes in the cytosol. Once the signal has been translated, however, signal recognition particles (SRPs) recognise the signal, pause translation, and take the protein to SRP receptors on the endoplasmic reticulum. The ribosome and peptide are inserted onto a translocator, where translation continues.

Translated proteins are released into the inside of the endoplasmic reticulum. The endoplasmic reticulum can then "bud off" into vesicles which can carry proteins around to different areas of the cell. For example, proteins can go to the Golgi apparatus, where post-translational modifications are made, or to the cell membrane, where they can be excreted. I'll talk a bit more about vesicles in the next part...

4. Identify the roles of coat proteins, Rabs, and SNAREs.

Coat proteins are required for formation of vesicles. Cargo receptors in the endoplasmic reticulum bind to their cargo on the inside of the cell, and to coat proteins on the outside of the cell (at least, that's my understanding of it). Coat proteins form a sort of "case" around the budding vesicle. They are discarded before the vesicle fuses to another membrane. There are several types of coat proteins: clathrin assembles on vesicles moving between the Golgi and plasma membrane, COPI forms around vesicles moving through the Golgi or moving from the Golgi to the ER, and COPII forms around vesicles moving from the ER to the Golgi.

Rabs are proteins which act as a sort of address tag on the protein. They bind to Rab effectors, which are motor or tether proteins located on the outside of other membranes in the cell.

SNARE proteins mediate fusion of vesicles to other membranes. SNAREs located on the vesicle are known as vSNAREs, while those on the target membrane are known as tSNAREs.

5. Describe the difference between regulated and unregulated exocytosis in secretion. 

In a nutshell, vesicles in regulated exocytosis require a signal before their contents are secreted, while vesicles in unregulated, or constitutive, exocytosis do not. As far as I know, only the latter type have special proteins on the outside of their vesicles that probably act as some kind of signal, telling the cell to target them and move them to the membrane for secretion right away.

Biochemical Messengers

This revision post will be derived from the lecture objectives. This is partly because I haven't been bothered enough had the time to read the book, plus the book covers a lot more than I need to know for first year anyway.

1: Have an understanding of the various mechanisms cells use for communication.

Cells have several mechanisms for communicating.

Firstly, they can release hormones and other signalling molecules. These cells may either diffuse through interstitial fluid to nearby target tissues (in the case of paracrine signalling) or they may travel through the blood to their target organ (in the case of endocrine signalling). Neurons interact through a similar mechanism, but instead neurotransmitters are released into a synapse- a small space between neurons.

Secondly, cells may carry signalling proteins on their cell surfaces. These "kick in" when the cell makes contact with a receptor cell. If I remember correctly, some cells in the immune system work by utilising this system.

2: Be able to list components of cellular communication: receptors and signal molecules, intracellular signalling pathways and messengers, effectors.

Signal molecules are, as their name implies, molecules that send out a signal of some kind. Some are hydrophilic and can travel through the blood easily, but cannot cross the plasma membrane due to the high lipid concentration. Instead, they signal by docking to receptor molecules on the cell surface. Other signal molecules are hydrophobic and require binding proteins to carry them through the blood. They can cross the plasma membrane and dock to receptor molecules within the cell.

After docking onto a receptor, intracellular signalling pathways are often activated. This involves the release of other molecules (also known as second messengers) that can carry out other functions in the cell. A common second messenger is cAMP, or cyclic AMP, which can activate protein kinases, which in turn can activate other molecules in the cell. Second messengers are often small and as such are able to diffuse through the cytoplasm quickly.

Effector proteins are simply the "end target," i.e. the protein whose function needs to be activated in order for a desired cell function to be carried out.

3: Describe the functions of molecular switches (kinases/GTPases) in signalling.

The shape of proteins is critical in order for the protein to be able to interact with its desired targets and carry out its functions in the cell. Kinases can modify the shape of proteins by adding phosphate groups. This often activates the protein, but can also deactivate it. (Phosphatases, which remove phosphate groups, have the opposite effects.)

Certain types of proteins, known as GTPases, have GTP (guanosine triphosphate) bound to them. Most of these proteins are active while carrying GTP. However, once the GTP is hydrolysed, by a GAP (GTP activating protein) or otherwise, only GDP (guanosine diphosphate) remains, hence affecting the function of the protein in the cell. This usually has a deactivation effect on the protein. This effect can be reversed by the effects of GEF (guanosine exchange factor) proteins, which remove GDP, allowing GTP to bind. (GTP is present in much higher concentrations in the cell, so it is far more likely that GTP will bind, rather than GDP.)

4: Be able to describe an example of common cell signalling mechanisms in detail.

(The lecturer did say that this was just extra and not something examinable, which is good, because I can never remember specifics.)

The example given in the lecture is that of the EGF-RTK-Ras-MAPK signalling pathway.

EGF is a signalling molecule that binds to EGF receptors. These EGF receptors are examples of enzyme-coupled receptors. Enzyme-coupled receptors are receptors that either act as enzymes once they have a substrate bound, or can directly activate enzymes. In this case, when EGF binds, two EGF receptors dimerise to become a kinase that can phosphorylate itself. These phosphate groups then interact with a protein called GRB2, which interacts with another one called SOS (like SOS I'm dying under the weight of all of these protein names). SOS is a GEF (see question 3), which means that it is able to activate GTPases. The GTPase that it interacts with is called Ras.

Ras can then bind to and activate Raf, a kinase which phosphorylates another protein called MEK. MEK is also a kinase which phosphorylates and activates MAPK (mitogen-activated protein kinase).

MAPK has two functions, both of which aid in transcription of the gene c-Fos. Firstly, MAPK directly phosphorylates and activates a transcription factor called TCF. MAPK also phosphorylates and activates a protein called p90RSK, which in turn can phosphorylate and activate other transcription factors (at least, I think that's what they are...) called SCF. Once all of these are in place, the gene can be transcribed.

Okay, that was quite complicated. Let's see if I can find a simpler one...

This next example revolves around the activation of a protein kinase that is imaginatively called protein kinase C.

Activation in this case starts with a G-protein coupled receptor. G-protein coupled receptors essentially act as GEFs (proteins that can activate GTPases, as covered above). Actually, the G-proteins that interact with the G-protein coupled receptors are pretty special in that they have three subunits, and when activated, they break off into two parts: an alpha subunit and a beta/gamma subunit. Both of these subunits can interact with other molecules within the cell.

In this case, the G-protein, Gq, interacts with a phospholipid called PI 4,5-biphosphate, breaking it down into inositol 1,4,5-triphosphate (IP3) and diacylglycerol. IP3 can open Ca2+ channels in the endoplasmic reticulum, releasing Ca2+ into the cytosol. Once that has occurred, both Ca2+ and diacylglycerol can interact with protein kinase C, activating it and allowing it to phosphorylate other proteins in the cell.

The Cell Cycle

Yet another revision post!

Q1: Since there are about 10^13 cells in an adult human, and about 10^10 cells die and are replaced each day, we become new people every three years. (True/False, explain why)

This almost seems more like a philosophical question than a biological question...

Anyway my answer is false, since not all of the cells in our body die and are replaced. For example, skeletal muscle cells and neurons do not get replaced. The latter is particularly significant because we often think of our brains as making us who we are.

Q2: In order for proliferating cells to maintain a relatively constant size, the length of the cell cycle must match the time it takes for the cell to double in size. (True/False, explain why)

Yes, it is true that everything within the cell has to be doubled before cell division so that all daughter cells will be of a "normal" size.

Q3: While other proteins come and go during the cell cycle, the proteins of the origin recognition complex remain bound to the DNA throughout. (True/False, explain why)

I think this is true. To my understanding, origin recognition complexes remain bound to the DNA throughout, but they require the proteins of the pre-replicative complex (preRC) to allow initiation of replication to take place.

Q4: Chromosomes are positioned on the metaphase plate by equal and opposite forces that pull them toward the two poles of the spindle. (True/False, explain why)

Yes, that is indeed how they are aligned at the spindle. Once at the metaphase plate, chromosomes often oscillate gently as they are tugged back and forth by the two spindle poles.

Q5: Meiosis segregates the paternal homologs into sperm and the maternal homologs into eggs. (True/False, explain why)

This statement is false. During the first meiotic division, the paternal and maternal homologs cross over before being pulled apart into different cells, and hence chromosomes contain a mixture of paternal and maternal DNA. In males, all of the daughter cells in meiosis become spermatids that can undergo spermiogenesis to become sperm, while in females, one daughter cell becomes an egg while the others degenerate. This process takes place regardless of whether the chromosomes within the sperm or eggs originated from the father or the mother.

Q6: If we could turn on telomerase activity in all our cells, we could prevent ageing. (True/False, explain why)

Probably false. In any case, turning on telomerase may instead cause cancer- many cancer cells have the ability to produce telomerase and therefore stop cells from undergoing replicative cell senescence (i.e. not replicating after a certain number of cycles).

Q7: Many cell-cycle genes from human cells function perfectly well when expressed in yeast cells. Why do you suppose that is considered remarkable? After all, many human genes encoding enzymes for metabolic reactions also function in yeast, and no one thinks that is remarkable.

I'm just going to take a wild stab in the dark here. I think that this is possibly due to humans being more complex organisms than yeast. Humans have a variety of different cells that have different requirements with regards to the cell cycle, whereas yeast does not. Hence perhaps it is remarkable that so many human cell-cycle genes are still able to function in yeast.

Q8: Hoechst 33342 is a membrane-permeant dye that fluoresces when it binds to DNA. When a population of cells is incubated briefly with Hoechst dye and then sorted in a flow cytometer, which measures the fluorescence of each cell, the cells display various levels of fluorescence as shown in the figure.

(Description of figure: relative fluorescence per cell on x-axis (0, 1, 2), number of cells on y-axis (no scale). There is a large peak at relative fluorescence of 1 and a smaller peak at 2. There are some cells in between the two as well.)

A. Which cells in the figure are in the G1, S, G2 and M phases of the cell cycle? Explain the basis for your answer.

At a relative fluorescence of 1, there is relatively little DNA in the cell (i.e. the DNA has not replicated yet). Hence I would say that cells with a relative fluorescence of 1 are in the G1 stage.

Between relative fluorescence levels of 1 and 2, there is between the normal amount and twice the amount of DNA in each cell. The cells here are probably in S phase, where the DNA is beginning to replicate but the replication process is not quite complete.

At a relative fluorescence of 2, the DNA has already doubled. Hence, I would say that cells with this fluorescence are in the G2 and M stages.

B. Sketch the sorting distributions you would expect for cells that were treated with inhibitors that block the cell cycle in the G1, S or M phase. Explain your reasoning.

I can't be bothered drawing anything, so I'll just describe here.

If the cell cycle was blocked in the G1 phase, the cell would be stuck with the base amount of DNA. Hence all of the cells would have a relative fluorescence of 1.

If the cell cycle was blocked in S phase, the cell would either be stuck with the base amount of DNA (if it was stuck in the beginning of S phase), twice the amount of DNA (if it was stuck at the end of S phase), or somewhere in between. I'm not entirely sure what the relative fluorescence of the cells would be- maybe most cells would be in between.

If the cell cycle was blocked in M phase, the cell would be stuck with twice as much DNA. Hence all of the cells would have a relative fluorescence of 2.

Q9: The yeast cohesin subunit Scc1, which is essential for sister-chromatid cohesion, can be artificially regulated for expression at any point in the cell cycle. If expression is turned on at the beginning of S phase, all the cells divide satisfactorily and survive. By contrast, if Scc1 expression is turned on only after S phase is completed, the cells fail to divide and they die, even though Scc1 accumulates in the nucleus and interacts efficiently with chromosomes. Why do you suppose that cohesin must be present during S phase for cells to divide normally?

I think that this is because cohesin is required to make sure that the two sister chromatids are closely joined as they are generated. If the chromatids are generated first, the lack of cohesin may cause them to break apart, and adding cohesin to them at this point may result in chromatids from one chromosome joining to chromatids from other chromosomes. Haphazard joining of chromatids would cause cells to be unable to divide normally.

Q10: High doses of caffeine interfere with the DNA damage response in mammalian cells. Why then do you suppose the Surgeon General has not yet issued an appropriate warning to heavy coffee and cola drinkers? A typical cup of coffee (150mL) contains 100mg of caffeine (196g/mole). How many cups of coffee would you have to drink to reach the dose (10mM) required to interfere with the DNA damage response? (A typical adult contains about 40 litres of water.)

40 litres x 10mM = 400 millimoles of caffeine are required to interfere with the DNA damage response.

0.1g caffeine in a cup of coffee / 0.196 g/mmol = 0.51 millimoles of caffeine in a cup of coffee (2 d.p.)

400/0.51 = 784 cups of coffee (nearest whole number).

That's a lot of coffee.

Q11: How many kinetochores are there in a human cell at mitosis?

There are 46 chromosomes in a human cell. At mitosis, each of these is composed of two sister chromatids, each with its own kinetochore. Hence there are 46 x 2 = 92 kinetochores in a human cell at mitosis.

Q12 is just a series of pictures to put in order. It's not something I can type up here so I'll just skip this one.

Q13: Down syndrome (trisomy 21) and Edwards syndrome (trisomy 18) are the most common autosomal trisomies seen in human infants. Does this fact mean that these chromosomes are the most difficult to segregate properly during meiosis?

No, it does not. The fact that Down syndrome and Edwards syndrome are the most common autosomal trisomies seen in human infants may simply mean that these are the autosomal trisomies most compatible with life. Other autosomal trisomies may occur at the same rate, but these foetuses do not make it to full term.

Q14: The human genome consists of 23 pairs of chromosomes (22 pairs of autosomes and one pair of sex chromosomes). During meiosis, the maternal and paternal sets of homologs pair, and then are separated into gametes, so that each contains 23 chromosomes. If you assume that the chromosomes in the paired homologs are randomly assorted to daughter cells, how many potential combinations of paternal and maternal homologs can be generated during meiosis? (For the purpose of this calculation, assume no recombination occurs.)

Well, there are two options for chromosome 1 (paternal or maternal), and then two for chromosome 2, etc. This gives a whopping 2^23 combinations before considering recombination.

The Cytoskeleton

Part two of my panicky revision series!

This is probably one of my weaker topics, so if you're going to use my answers here as your "study guide," be warned that you might be hindered more than helped unless you cross-check everything as well. (And if you do cross-check and find discrepancies, please let me know so that I don't screw up anyone else :) )

Q1: The role of ATP hydrolysis in actin polymerisation is similar to the role of GTP hydrolysis in tubulin polymerisation: both serve to weaken the bonds in the polymer and thereby promote depolymerisation. (True/False, explain why)

I think that this is true. The hydrolysed forms of ATP and GTP (ADP and GDP, respectively) do indeed promote depolymerisation as the free-energy change for a subunit with ADP/GDP bound is less than that for ATP/GTP.

Q2: Motor neurons trigger action potentials in muscle cell membranes that open voltage-sensitive Ca2+ channels in T tubules, allowing extracellular Ca2+ to enter the cytosol, bind to troponin C, and initiate rapid muscle contraction.

True. This is indeed how muscle neurons work to "fire" a muscle cell, provided that I comprehended the textbook correctly ;)

Q3: In most animal cells, minus-end directed microtubule motors deliver their cargo to the periphery of the cell, whereas plus-end directed microtubule motors deliver their cargo to the interior of the cell.

False. Microtubules grow out from a microtubule organising centre (MTOC), with minus ends at the middle and rapidly-growing plus ends radiating outwards. Hence minus-end directed motors take their cargo towards the MTOC (i.e. towards the interior of the cell), whereas plus-end directed microtubule motors travel away from the MTOC (i.e. towards the periphery of the cell).

Q4: The concentration of actin in cells is 50-100 times greater than the critical concentration observed for pure actin in a test tube. How is this possible? What prevents the actin subunits in cells from polymerising into filaments? Why is it advantageous to the cell to maintain such a large pool of actin subunits?

It is possible for the concentration of actin to be much greater than the critical concentration due to the presence of certain proteins that make polymerisation less favourable. One of these enzymes is thymosin. The large pool of actin filaments may serve as reserve "building blocks" when actin filaments are required for cell surface protrusions etc.

Q5: Detailed measurements of sarcomere length and tension during isometric contraction in striated muscle provided crucial early support for the sliding-filament model of muscle contraction. Based on your understanding of the sliding-filament model and the structure of a sarcomere, propose a molecular explanation for the relationship of tension to sarcomere length in the portions of figure Q16-1 marked I, II, III and IV. (In this muscle, the length of the myosin filament is 1.6 micrometres, and the lengths of the actin thin filaments that project from the Z discs are 1.0 micrometres.)

(Figure Q16-1 is a line graph with sarcomere length in micrometres along the x-axis and tension in % along the y-axis. At roughly 1.3 micrometres, there is 0% tension, at 1.6 micrometres, there is over 75% tension, at 2 and ~2.2 micrometres, there is 100% tension, and at 3.6 micrometres there is 0% tension.)

I'm going to go from longest to shortest, simply because that's easier for me. This is a question that I'm pretty shaky about, so please do correct me if there are errors in my logic.

At 3.6 micrometres, there is 0% tension because the filaments are simply laid out end-to-end. You see, there is a 1.0 micrometre actin filament, plus a 1.6 micrometre myosin filament, then another 1.0 micrometre actin filament, leading to a total of 3.6 micrometres at full length.

At 2 and ~2.2 micrometres, there is 100% tension because the sarcomere has contracted enough for the two actin filaments on either side to meet.

Since I've just noted that 2 micrometres is the length of the filament when actin filaments on either side have met, the question remains of how sarcomeres can shorten further. Perhaps myosin can continue to "crawl down" the actin filaments, with the actin filaments in the middle either overlapping or peeling off. This would probably contribute to the decrease in tension at 1.6 micrometres. However, if this were to continue, eventually the Z-discs would get in the way. At the moment I can't imagine how the sarcomere would be able to get down to 1.3 micrometres (less than the length of the myosin filament) without it "popping out" which may be why the tension has dropped down to 0%.

As I've said, I'm really shaky about this, so don't take my word for it.

Q6: At 1.4mg/mL pure tubulin, microtubules grow at a rate of about 2 micrometres/minute. At this growth rate, how many alpha/beta-tubulin dimers (8nm in length) are added to the ends of a microtubule each second?

Maths question... at least this one looks reasonably easy. 2 micrometres is equivalent to 2000 nanometres. 2000 nanometres divided by 60 seconds gives 33.3 nanometres per second (1 d.p.). 33.3/8 gives 4 alpha/beta tubulin dimers per second (nearest whole number).

Q7: A solution of pure alpha/beta-tubulin dimers is thought to nucleate microtubules by forming a linear protofilament about seven dimers in length. At that point, the probabilities that the next alpha-beta dimer will bind laterally or to the end of the protofilament are about equal. The critical event for microtubule formation is thought to be the first lateral association. How does lateral association promote the subsequent rapid formation of a microtubule?

I have no idea. My best guess is that it provides a larger surface area and more places where new molecules can make contact.

Q8: How does a centrosome "know" when it has found the centre of the cell?

If I remember correctly, the microtubules that radiate out from the centrosome will all exert roughly the same amount of tension when they are all roughly the same length. This helps the centrosome "know" when it has found the centre of the cell.

Q9: (warning this is long) The movements of single motor-protein molecules can be analysed directly. Using polarised laser light, it is possible to create interference patterns that exert a centrally directed force, ranging from zero at the centre to a few piconewtons at the periphery (about 200nm from the centre). Individual molecules that enter the interference pattern are rapidly pushed to the centre, allowing them to be captured and moved at the experimenter's discretion.

Using such "optical tweezers," single kinesin molecules can be positioned on a microtubule that is fixed to a coverslip. Although a single kinesin molecule cannot be seen optically, it can be tagged with a silica bead and tracked indirectly by following the bead. In the absence of ATP, the kinesin molecule remains at the centre of the interference pattern, but with ATP it moves toward the plus end of the microtubule. As kinesin moves along the microtubule, it encounters the force of the interference pattern, which stimulates the load kinesin carries during its actual function in the cell. Moreover, the pressure against the silica bead counters the effects of Brownian (thermal) motion, so that the position of the bead more accurately reflects the position of the kinesin molecule on the microtubule.

A. As shown in the figure (in the textbook), all movement of kinesin is one direction (toward the plus end of the microtubule). What supplies the free energy needed to ensure a unidirectional movement along the microtubule?

As movement only occurred in the presence of ATP, I think that it is safe to assume that ATP supplies the energy required for movement. I'm not sure if this alone actually guarantees unidirectional movement, however, or if it's related to some other part of kinesin's structure.

B. What is the average rate of movement of kinesin along the microtubule?

From the diagram in the book, it appears to be a bit less than 10nm/s- maybe around 9nm/s but hard to tell from the scale.

C. What is the length of each step that a kinesin takes as it moves along a microtubule?

Again from the size of the "steps" in the diagram in the book, it appears to be around 9-10nm/s, so it appears that kinesin takes one "step" per second.

D. From other studies it is known that kinesin has two globular domains that can each bind to beta-tubulin, and that kinesin moves along a single protofilament in a microtubule. In each protofilament, the beta-tubulin subunit repeats at 8-nm intervals. Given the step length and the interval between beta-tubulin subunits, how do you suppose a kinesin molecule moves along a microtubule?

Since kinesin has two globular domains, it wouldn't surprise me if kinesin walked in a sort of "hand-over-hand" motion, with the two domains acting as the "hands." These "hands" would grip onto beta-tubulin subunits.

E. Is there anything in the data that tells you how many ATP molecules are hydrolysed per step?

I don't think that there is, though I could be wrong.

Q10: A mitochondrion 1 micrometre long can travel the 1 metre length of the axon from the spinal cord to the big toe in a day. The Olympic men's freestyle swimming record for 200 metres is 1.75 minutes. In terms of body lengths per day, who is moving faster: the mitochondrion or the Olympic record holder? (Assume that the swimmer is 2 metres tall.)

1 metre = 1000 millimetres = 1 x 10^6 micrometres. The mitochondrion thus travels 1 x 10^6 body lengths in one day.

If the swimmer is 2 metres tall, then he travels 100 body lengths in 1.75 minutes. 100/1.75 = 57.14 body lengths a minute (2 d.p.). 57.14*60*24 = 82 286 body lengths in one day (nearest whole number).

Hence, in terms of body lengths per day, the mitochondrion is travelling faster.

Q11: Cofilin preferentially binds to older actin filaments and promotes their disassembly. How does cofilin distinguish old filaments from new ones?

New actin filaments have ATP bound to them. Over time, this ATP hydrolyses to form ADP. Cofilin can distinguish old filaments from new ones by recognising the presence of ADP or ATP.

Q12: Why is it that intermediate filaments have identical ends and lack polarity, whereas actin filaments and microtubules have two distinct ends with a defined polarity?

While the monomers that make up intermediate filaments have distinct ends (an amino and a carboxyl terminus), they line up anti-parallel with each other, so that half of the monomers are pointing in one direction and the other half are pointing in the other direction. This gives a net polarity of zero.

Actin filaments and microtubules, on the other hand, have distinct ends and do not line up anti-parallel with other monomers. This gives these filaments negative and positive ends.

Q13: How is the unidirectional motion of a lamellipodium maintained?

The unidirectional motion of a lamellipodium is maintained by the assembly of actin filaments at the leading edge, with depolymerisation, aided by cofilin, occurring away from the leading edge. This gives a net effect of the filament network appearing to "move forward."

Cell Membrane Structure

So I have another online quiz coming up which I'm nervous about even though it's only worth like 2%... I was about to keep frantically reading through all of the textbook chapters that are related to what we've covered in the lectures, then I realised that that really isn't an efficient use of my time. Instead I'm going to do a mixture of answering textbook questions and making questions out of the "Lecture Objectives" dot points.

For this particular post, I'm going to try my hand at answering some chapter questions. Please tell me if I've gotten something wrong so that a) I can learn from it and b) I don't confuse other people :)

Q1: "Although lipid molecules are free to diffuse in the plane of the bilayer, they cannot flip-flop across the bilayer unless enzyme catalysts called phospholipid translocators are present in the membrane." (True/False, explain why)

This statement is true. The "flip-flop" of lipids from one side of the bilayer to the other is energetically unfavourable, and thus enzymes (in this case phospholipid translocators, a.k.a. flippases, are required).

Q2: "Whereas all the carbohydrate in the plasma membrane faces outward on the external surface of the cell, all the carbohydrate on internal membranes faces toward the cytosol." (True/False, explain why)

This statement is false. The carbohydrates always face away from the cytosol, no matter whether the plasma membrane is an internal or external one. This is due to their addition in the lumen of the Golgi apparatus.

Q3: "Although membrane domains with different protein compositions are well known, there are at present no examples of membrane domains that differ in lipid concentration." (True/False, explain why)

False. Areas with high protein concentrations would have to have lower lipid concentrations, as there is less room for the lipids there. Also, the different kinds of lipids can be found in different concentrations in the two different layers of the membrane.

Q4: When a lipid bilayer is torn, why does it not seal itself by forming a "hemi-micelle" cap at the edges?

I would assume that this would have to do with the shape of phospholipids. The two tails of phospholipids give them a more cylindrical structure, as opposed to the more "conical" structure of single-tailed lipids. Hence, while the latter can form micelles, phospholipids used in membranes cannot.

Q5: Margarine is made from vegetable oil by a chemical process. Do you suppose this process converts saturated fatty acids to unsaturated ones, or vice versa? Explain your answer.

I think that this process converts unsaturated fatty acids to saturated ones. Saturated fatty acids, due to lacking "kinks" in their chains that cis-unsaturated acids have, can pack together more closely. This property may account for the greater density of margarine as compared to vegetable oil.

Q6: If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? (Neglect the loss of lipid from the raft that would be required to accommodate the protein.)

Note: I have been informed that my original answer to this question was incorrect- I think I overcomplicated the question the first time. This is my re-attempted version- hopefully it's correct this time! (Also, to the people who are using this site to help with homework, please have a go at doing it yourself before looking on here! You'll learn more by actually doing the problems, and you might find other mistakes that I've made- I am only human, after all ;))

First things first, we need to calculate the area of the lipid raft and the area of the lipid molecules. To do this, we can use the formula for area of a circle, which is πr². Remember, the radius is half the diameter; hence the radius of the lipid raft is 35nm and the radius of a lipid molecule is 0.25nm. This gives us the following equations:

Area of the lipid raft = π * 35²
Area of a single lipid molecule = π * 0.25²

To get the number of lipid molecules, you need to divide the area of the lipid raft by the area of a single molecule:

No. of lipid molecules = (π * 35²) / (π * 0.25²)
No. of lipid molecules = (35²) / (0.25²)
No. of lipid molecules = 19 600

Hence, there are 19600 molecules... in a monolayer of the lipid raft. Remember, lipid rafts come in bilayers, so we need to double our answer to get the number of lipid molecules altogether:

No. of lipid molecules in the bilayer = 19600 * 2 = 39 200 lipid molecules.

Next, we need to find out how many protein molecules there are. If there are 50 lipid molecules per protein molecule, then we simply need to divide the number of lipid molecules by 50 to get the number of proteins.

No. of proteins = 39 200 / 50 = 784 protein molecules.

Hopefully this answer is correct this time...!

Q7: Monomeric single-pass transmembrane proteins span a membrane with a single alpha helix that has characteristic chemical properties in the region of the bilayer. Which of the three 20-amino-acid sequences listed below is the most likely candidate for such a transmembrane segment? Explain the reasons for your choice.
A- ITLIYFGVMAGVIGTILLIS
B- ITPIYFGPMAGVIGTPLLIS
C- ITEIYFGRMAGVIGTDLLIS

For a transmembrane protein, you would expect to see some hydrophilic amino acids at either end, since they are in contact with aqueous solution, and some hydrophobic amino acids in the middle, where they are surrounded by hydrophobic lipids.

The textbook has helpfully given me "FAMILY VW" as a convenient mnemonic for the hydrophobic amino acids (phenylalanine, alanine, methionine, isoleucine, leucine, tyrosine, valine and tryptophan). Hence I need to choose the protein that has more of these letters in the middle and fewer of these letters at the end.

Here they are again, with the "FAMILY VW" letters highlighted in blue. As these are hydrophobic amino acids, I would expect to see more of them in the middle of the chains, and fewer of them at the end.

A- ITLIYFGVMAGVIGTILLIS
B- ITPIYFGPMAGVIGTPLLIS
C- ITEIYFGRMAGVIGTDLLIS

There probably isn't the clearest distinction between these three proteins, but the first one does have a few more hydrophobic amino acids in the middle (besides, the second and third both have the same number of hydrophobic and hydrophilic amino acids in the same places, and given that this is a multiple choice question, process of elimination leaves A). Hence I would choose A for my answer.

Q8: You are studying the binding of proteins to the cytoplasmic face of cultured neuroblastoma cells and have found a method that gives a good yield of inside-out vesicles from the plasma membrane. Unfortunately, your preparations are contaminated with variable amounts of right-side-out vesicles. Nothing you have tried avoids this problem. A friend suggests that you pass your vesicles over an affinity column made of lectin coupled to solid beads. What is the point of your friend's suggestion?

Lectin binds to carbohydrates, such as those that bind to proteins and lipids on the outer membrane of the cell (see my answer for Q2). Hence, an affinity column containing lectin could help to "weed out" the right-side-out vesicles by binding to carbohydrates on their cell surfaces.

Q9: Glycophorin, a protein in the plasma membrane of the red blood cell, normally exists as a homodimer that is held together entirely by interactions between its transmembrane domains. Since transmembrane domains are hydrophobic, how is it that they can associate with one another so specifically?

Transmembrane domains can associate with each other through their specific shapes, which result from their sequences of amino acids.

(I feel kind of bad for giving answers that are shorter than the questions, but moving on...)

Q10 (reworded due to lack of diagrams): There are several mechanisms by which membrane-binding proteins bend a membrane. Some of these cytosolic membrane-bending proteins would induce an invagination of the plasma membrane. Could similar kinds of cytosolic proteins induce a protrusion of the plasma membrane? Which ones? Explain how they might work.

I'm going to go through the different mechanisms that I've learned about, and then try and discuss whether or not I think they could cause protrusions.

Mechanism #1: Insertion of proteins or lipid anchors to increase the area of one side of the plasma membrane. Increasing the area of one side- but not the other- causes the membrane to curve around. This could work if there were proteins or lipid anchors in the extracellular matrix (actually I could probably use this as a cop-out answer for all of the mechanisms here). I'm not sure how this could work from the inside, however. One idea I have is that enzymes on the inside could remove proteins or lipid anchors that are on the inside of the membrane, but this would only work if there were enough proteins or lipid anchors on the outside of the membrane to cause a bending effect afterwards.

Mechanism #2: Forming a rigid scaffold that deforms the membrane or stabilises an already bent membrane. This could probably work from the inside, if you have a protein that was of a different shape.

Mechanism #2: Causing particular membrane lipids to cluster together so that their differently-sized head groups can induce curvature. If there were enough "large-headed" lipids on the extracellular layer, clustering a lot of "small-headed" lipids in the intracellular layer may cause a protrusion.