BCMB 4010 Spring 2025 VSG #2 PDF

VSG #2 (All material is covered in lecture slides) For additional review: Chapter 2 I can't find these answers in my notes, does anyone else understand these topics: >Why is tryptophan used for UV absorbance? >Why are disulfide bonds important for protein stability? DrL>...

VSG #2 (All material is covered in lecture slides) For additional review: Chapter 2 I can't find these answers in my notes, does anyone else understand these topics: >Why is tryptophan used for UV absorbance? >Why are disulfide bonds important for protein stability? DrL> I’ll address this with a question; What are proteins made of? What makes a disulfide bond? Is this bond covalent or non-covalent? >proteins are made of covalently linked amino acids. The disulfide bond is a type of covalent bond? > Also for pKa of H, K, R, E, D, C, Y and Know the titratable protons for the residues in #2 on lecture four slides is the pKa for the overall structure and the titratable protons are for certain side chains? DrL> all of these were discussed in class. However, you don’t need to know the pKa’s for all the side chains until quiz 4. The exception of course is histidine, because we discussed how histidine could be used to make a “pH sensor”. 1)​ Of the two hydrogen bonds shown below, which one is the strongest? >I would say that the first one is the strongest. If we think about the charges, the second water molecule in the first diagram has two hydrogens which share the partial positive charge within water, while on the second diagram, the hydronium molecule has three hydrogens to share the charge across, resulting in the hydrogens having a lower partial positive charge compared to the first diagram, and thus the one on the left would have a higher separation of partial charges, resulting in a stronger attraction. DrL> I mentioned this in class, but we haven’t covered it in detail yet (we will) H-bonds are strongest when the pKa of the donor and the pKa of the protonated form of the acceptor are close (if they are the same, that’s a “low barrier hydrogen bond”, the strongest H-bond). Someone try again, this time consider the pKa’s. >The second one would be the strongest because hydrogen bonds are strongest when the donor and the protonated form of the acceptor have the same pka, which leads to more covalent character in the molecule. >The hydrogen bond shown on the right would be the strongest. H3O+ acts as the hydrogen bond donor and H2O acts as the hydrogen bond acceptor. When H2O accepts the hydrogen bond from hydronium it becomes protonated and becomes H3O+; since the donor and acceptor are both then H3O+ they have the same pKa and hydrogen bonds are strongest when the acceptor and donor have the same pKa. The picture on the left shows a hydrogen bond between 2 water molecules, when the hydrogen bond forms, one of those water molecules (the acceptor) becomes protonated, meaning the hydrogen bond is between one water a hydroxide molecule and one hydronium molecule, these have a very large difference in their pKa. >Water’s pKa is 15.4, while hydronium is -1.7. > Expanding on the pKa standpoint, the structure on the right is stronger because when the water is protonated its pKa will reach hydroniums which is -1.7 When it makes that jump and both molecules have a pKa of -1.7 it in turn is a stronger acid and a stronger bond. 2)​ Is it possible to increase the temperature high enough to make an endergonic and exothermic reaction spontaneous? > An endergonic reaction indicates a positive value for dG. Since the reaction is Exothermic, we know energy HEAT (not energy) is released, resulting in a negative dH. This tells us that the sign for dS must be negative, resulting in a positive value of “-TdS,” and that it must be greater in magnitude than dH, since the reaction itself is endergonic. Thus, if we increase the temperature, “-TdS” will become increasingly positive, increasing dG more. Thus this endergonic, exothermic reaction can not be made spontaneous by increasing temperature. DrL> good, what if the temperature is decreased, is it possible for the reaction to become spontaneous then? > yes, the reaction is exothermic (negative delta H) and a negative delta S. This means it can be spontaneous at low temperature because if the (T x delta S) term is smaller than the delta H value, the delta G will still be negative because the (T x delta S) term will not be large enough to make it positive. 3)​ The standard free energy change for ATP hydrolysis at 37° C is -30.5 kJ mol-1 (T=310.15 K, R= 8.314× 10-3 kJ K-1 mol-1): ATP ADP + Pi (For hydrolysis reactions we ignore the contribution of water in the math). A red blood cell has the following concentrations of ATP (2.25 mM), ADP (0.25 mM) and Pi (1.65 mM). What is Q for this reaction? What is the ΔG of the reaction under these conditions? > See image below 4)​ Consider the two-state model for protein folding. Using the ‘rule of thumb’ What is the ΔGFolding, in kcal mol-1 for a protein that exists in a ratio of 1000:1 folded:unfolded states at room temperature (see ‘rule of thumb’ slide)? If a mutation destabilized the protein by 1.36 kcal mol-1, what would be the new ratio of folded:unfolded? > For 1000:1 the ratio is 10^3 which is 3 x -5.7 kj/mol which is -17.1 kj/mol which divided by 4.184 is -4.09 kcal/mol. The new ratio of folded to unfolded would be 100:1 since 1.36 kcal/mol is 5.7 kj/mol and the folded protein is becoming less stable. DrL> good, you even caught the kJ to kcal conversion. We have talked about mutations, specifically, the deletion of a charged residue and how that may influence the folding equilibrium. Can someone tell me if it would matter if the charged residue was on the surface of the folded protein, or buried in the core? Which case would have the greatest influence on the folding equilibrium and why? > Yes, it would matter if the deleted charged residue was on the surface or buried in the folded protein. If the deleted residue was on the surface of the protein, it doesn’t have as great of an effect on folding, there just wouldn’t be less ion-dipole interactions with the aqueous environment. However, if the deleted residue was buried in the protein core, this would more significantly influence the equilibrium as it takes energy to desolvate charged molecules, and no salt bridge will be formed to release energy to make up for the initial cost of desolvation. It is crucial that charged molecules in the hydrophobic core have a suitable pair, otherwise it can be very destabilizing to the core. DrL> I think I understand what you are saying, but it might be clearer if someone illustrated the two cases for us, and included both states (unfolded and folded) for each case (buried charge involved in a salt bridge and charge on the surface) > In the first case, the unfolded state must be desolvated before protein folding can occur which requires energy (to break the ion-dipole interactions between the ion and water molecules). If the charge is buried inside the core of the protein in the folded state, that input of energy is not compensated for by the formation of a salt bridge (due to the deleted residue) which decreases the favorability of the reaction. Attractions are broken in the unfolded state, but never replaced in the folded state. However, if the charge is on the surface of the protein in the folded state, then no desolvation is required for folding to take place; therefore, folding is more favorable because the input of desolvation energy is not required because the ion-dipole interactions are present in both states. 5)​ Know the 3 non-covalent interactions and rank them by their approximate strength. Identify the interactions that involve formal charges, dipole-dipole and induced dipole. >Ionic Bonds are the strongest, Dipole-dipole interactions are next strongest, and then VdW forces. DrL> okay, DRAW some examples below; DrL> Not sure I would use chlorine as an example of London dispersion forces. I would stay focused on the non-covalent interactions we see in enzyme/protein structure and function. 6)​ Know common H-bonding pairs. Why is ROH a good H-bond donor, but RCH3 a very poor (negligible) donor? There are two ways to answer this question, and I want both. The first involves the pKa, the second involves electronegativity & dipole strength. > ROH is a good H-bond donor as opposed to RCH3 because its pKa is much lower than RCH3. That is, ROH has a lower affinity for a proton than RCH3 and so it is going to be a better H-bond donor. In terms of electronegativity and dipole strength, oxygen is a very electronegative atom compared to H, so the O in ROH will have a partial charge, which creates a dipole. Carbon has about the same electronegativity as H, so there will be no partial charge/dipoles formed on RCH3. Since oxygen has a dipole, this allows ROH to be a good donor and form a strong H-bond with an H on other molecules. DrL> pretty good, pKa’s tell us many things, and for a biochemist, know the pKa’s of all the protonatable groups in the active site as well as on substrate/intermediate, helps us to determine the mechanism of an enzyme. Organic chemists are designing molecules and want their reactions to go to completeness, so they look at pKa’s similarly. If they are doing a substitution reaction, they might use a molecule that has an Iodine atom incorporated. Why would they use Iodine over bromine or fluorine? > They might choose a structure with iodine because it is a great leaving group. The best leaving groups are those with a low pKa (for their conjugate acid). The pKa of HI is -10, so I- is a great leaving group. DrL> correct, In contrast to the synthetic chemist, why does a biochemist care about pKa’s? (Hint: We discussed this in the context of two reactions) > Is it because reactions we are interested in do not typically occur in water? For Like CO2 altering the pH of blood? DrL> I don’t understand this comment literally ALL of the reactions a biochemist, pharmacist, dentist, doctor, biotech engineer, etc. etc. cares about occur in water. >I do not agree with the above statement. The blood is composed of ~50% water. I think pkas are more important to a biochemist due to the buffer system of the body. It is more significant to understand how the pka’s play a role in this system, as it deals with life, but also there are a significant amount of organic molecules that can interact. Knowing the pkas and how they work allow us to understand the arterial blood gas system. In contrast to a synthetic chemist, they can often pick the buffers they want, and control the system heavily, using the solvents and solutes they often desire. [pt 2] I would also like to elaborate more on why Iodine is a better leaving group than bromine or fluorine. It has a low pka of the conjugate acid, due to its size and ability to stabilize the negative charge as an anion. DrL> good, why is it useful to identify good leaving groups, how does this help the biochemist (think about the goal of biochemistry, what are we trying to do/explain?) > It’s useful to identify good leaving groups so that we are able to better understand biochemical reactions (such as amino acid condensation), mechanisms of an enzyme, how substrates are catalyzed by enzymes, etc. All of this allows us to study how structure gives rise to function, how the function of an enzyme can change based on pH, and how biomolecules interact with each other in the cellular environment. >I agree. In addition, identifying what characteristics of an atom that constitute a good leaving group allow us to understand what leaves during reaction, and how the reaction proceeds. Furthermore, this is drastically important in biochemical fields such as pharmacy which Dr. L mentioned above. 7)​ Based on the answer to #5, what can you conclude about the relationship between pKa and H-Bond strength? >The closer the pkas are between the donor and acceptor of the hydrogen bond, the stronger the bond is. > More specifically I think you want the protonated pKa of the acceptor to be close to the pKa of the donor. > Agreed ^ the donor pKa should be equal to the protonated acceptor DrL> good. Does this make sense to everyone? In order for a H-bond to have the most covalent nature (strongest) then the proton must be “equally shared”, in order to be equally shared, the donor and acceptor must “equal affinity” for the proton, and pKa’s are a measure of the proton affinity! 8)​ What two noncovalent interactions does water make with proteins or unfolded peptides? > Ion- dipole interactions and Dipole-dipole interactions DrL> Dipole-charge covers this, what is the other interaction? >hydrogen bonds + ion-dipole interactions 9)​ Why are H-bonds between water molecules in ice stronger than those in the liquid state? Is ice melting at 25° C driven by enthalpy, entropy, or both? Is it exothermic? > It is driven by entropy. It is endothermic > why would it be endothermic as opposed to exothermic? The entropy is going up since there is more degrees of freedom resulting in a positive dS and negative dG, or am I missing something here? DrL> You don’t think melting ice is endothermic? Can someone help us here? Perhaps there is some confusion with the terms like endergonic and endothermic? > enderGonic (dG) and endertHermic (dH). Endothermic because it is requiring heat DrL> why are the H-bonds in ice stronger? >The H-bonds are stronger in ice because at the low temperature, more LESS “flickering” occurs between the H2O molecules in ice. In flickering, one molecule is the acceptor in an H-bond, making its pka close to that of H3O+, which makes it a stronger H donor. While it is donating an H to an H-bond acceptor, (another H2O molecule) its pka increases and it becomes a much better proton acceptor itself. This flickering leads to stronger H-bonds, and it is more prevalent in ice because the molecules are not moving around as much. > I am confused by your response. You state initially that there is less flickering in ice, then later on you state that flickering strengthens the H donor, and that it creates strengthened H-bonds. I think that there is less flickering between H2O molecules in ice, and they are more rigid, and thus because of this, Hydrogen bonds are not being flickered, or exchanged, as often. Thus they are able to stay in one orientation more comparatively, and will form 4 bonds with adjacent water molecules which will form the crystal lattice type structure. Someone correct me if i am incorrect, but flickering refers to the process of the breaking and reformation of hydrogen bonds between water molecules, but is this driven by temperature? Meaning that if temperature increases, the hydrogen bond breaks and then the water reforms with another or is this related more towards entropy? DrL> good, yes, as flickering increases, the H-bonding becomes weaker. 10)​When you shake a bottle of oil & water to disperse the neatly separated layers into a chaotic mess, explain what happens to the entropy of the system. Is this reaction enthalpically favorable? >The water molecules form clathrate cages around the hydrophobic molecules, making the system become more ordered, and there are less microstates for both the hydrophobic and water molecules, so the entropy decreases. It is enthalpically favored because new H-bonds are formed when the clathrate cage is made, which releases heat into the solution. 11)​Why is a ‘buried, unsatisfied’ charge or polar group (H-bond donor/acceptor) in the core of a protein unfavorable? HINT: to answer, you must consider the unfolded state of the peptide and the folded state. >A “buried, unsatisfied” charge or polar group in the core of a protein is unfavorable because in order for the protein to fold in the first place, desolvating must occur. Disrupting the ion-dipole or hydrogen bonds between the aqueous solvent and the charged or polar group costs energy. This energy cost must be balanced by some energy release which can occur if there is something, like a group of opposite charge, to stabilize the group. Otherwise, this would be unfavorable. DrL> very good. > I would state that “unsatisfied” is a vague term for a charge as it implied the charge has a state of autonomy, however I assume this means unpaired, or without its opposite charge. However I overall agree with person 1. In the unfolded state, the water molecules orient themselves around the ion, forming an ion dipole interaction. And without two oppositely charged components, the energy needed to break the ion-dipole will not be offset by the energy released during a potential salt bridge formation. DrL> very good. 12)​Why is the pKa of acetic acid in water different than in solvent DMSO? What does this tell you about the pKa of an acid buried in a protein core? > Firstly, acetic acid has a terminal carboxylic acid group which can form hydrogen bonds, while DMSO cannot. DrL> not sure why the acid is being compared to the solvent DMSO. What H-bonds are you specifically talking about, and in what context. For a question like this, start by DRAWING each reaction. Here there are two reactions, the deprotonation of acetic acid in water and the deprotonation of acetic acid in DMSO. Thus in the presence of water, the pKA of acetic acid will be different, as it can form hydrogen bonds and donate H+ to water molecules, while this cannot happen in DMSO. Acetic Acid is thus more ‘acidic’ in Water compared to DMSO, and thus will have a lower pKa with respect to water, and will have a higher pKa in DMSO. This is one reason why it is important to consider the solvent when looking at a system. DrL> this hasn’t explicitly explained why the acetic acid gives up the proton easier in water. > The acetic acid gives up the proton easier in water because the acetate ions can hydrogen bond with the water in solution. This stabilizes the negative charge more so than in DMSO. In DMSO, the methyl groups on the sulfur atom cannot hydrogen bond making it less favorable for the acetic acid to get rid of its proton and have a negative charge. > for the pKa of an acid buried in a protein core: the pKa would be higher since there is less water to solubilize to conjugate base, thus leading to the acid having a higher affinity for the proton. It takes less energy to to bury a protonated acidic groups. DrL> very good extension of the DMSO example and application to a protein core! 13)​Charles Tanford viewed protein H-bonds act as a folding constraint that selects against misfolded intermediates. What did he mean? HINT: why is it that H-bonds do not directly stabilize a folded protein? >H-bonds do not directly stabilize a folded protein, but they serve as a limiting factor for the number of configurations a folded protein can have. When an unfolded protein is desolvated and the interactions with water are broken, it becomes very high in energy. In order to form a low energy viable folded protein, the folding must occur so that all of the H-bonds that occurred between the water molecules and the protein must be reformed within the protein. Generally, only the native folded structure of the protein satisfies this requirement. DrL> that’s a pretty good explanation, this person needs to help with the next question below… 14)​What are the most important contributions to protein stability? What role do salt bridges and H-bonds play in protein structure and stability? > Some of the most important Contributions to protein stability are the interactions in between the interior portions of the polypeptide chain such as salt bridges and H-bonds. These salt bridges are ionic attractions between positive and negative species of a polypeptide and contribute heavily to the structure of a folded protein, and contribute a large portion towards stability. H-bonds have this same effect. DrL> not quite, “stability” refers to dG, do H-bonds and salt bridges contribute to the dG of the protein folding reaction? > The formation of salt bridges and hydrogen bonds in a folded protein give off energy when these bonds form. Thus dG will become more negative, as energy is released to the surroundings, resulting in a more favorable instance of protein folding. With a lower dG, the folded protein produced is more ‘stable,’ as it is more thermodynamically favorable for the reaction to occur. [My Revision to my above statement]: I know the entropy of water drives  proteins the fold, however I am not quite sure what results in a more stable vs less stable protein. I would presume that if the interior structures of the protein interact with each other well, such as hydrogen bonding and nonpolar VDW interactions, although weak, would help hold the protein together. I am confident that salt bridges and hydrogen bonds themselves contribute to the structure of the folded protein, and help it form its natural state. Then, if stability refers to dG, or the free energy of folding, it would be more favorable, and thus more “stable” for dG to be lower. dG is reliant upon the H-bonds and salt bridges forming within the protein to release the energy that was absorbed when the H-bonds in the clathrate cages were broken. [I am slightly confused overall] DrL> For simplicity, I introduced a “two-state” model for protein folding. The unfolded state is at a much higher free energy than the native folded state. As we all know, the free energy for folding to the native state is driven entirely by the entropy of water. To resolve some of that confusion, ask yourself the following; do you really think that protein folding only involves two states? If there are other folding intermediates (intermediate states), where would the free energy of those states be relative to the unfolded state and the native state (functional state)? How might H-bonding and salt bridges get you from one state to another? >There are lots of intermediate states that can occur during protein folding. However, only the native (correct) structure of a protein will reform ALL the hydrogen bonds, salt bridges/charges that were desolvated. This means that the native structure will be the lowest in energy, which makes it harder for proteins to misfold. Other intermediates will not reform all the hydrogen bonds and salt bridges, making them higher energy and less favorable. > when we look at this graph the energy differences are equal, but it does not automatically mean one state is preferred over another. I have to disagree with the statement that H-bonds and salt-bridges stabilize protein folding, and agree with the person above that H-bonds and salt bridges influence folding of the protein rather than stabilizing it. DrL> yes, I guess another way of saying it is that H-bonds and salt bridges select for the correct fold. They help guide the folding protein to the correct structure.  15)​Maintaining proper blood pH is critical, please describe the reaction that is responsible for buffering your blood. Does this reaction have to be catalyzed? ​ > I think it is the carbonate-bicarbonate buffer which keeps the blood at proper pH. It responds to change in blood pH by shifting using Le Chatlier’s principle. One side of the reaction carbonate and the other side is dissociated into a proton and bicarbonate. The equation is driven by the desire to be at equilibrium and resist changes in pH. carbonic anhydrase is used for the carbonic acid ​ > I agree with person 1. This reaction occurs frequently in the Kidneys, within the arterial blood gas system, where the pH in the blood is altered through Ventilation and via the urinary system. DrL> First, carbonic anhydrase is a ubiquitous enzyme, it is expressed in every single cell in the body. Start with the reaction catalyzed by carbonic anhydrase. In order to “describe” a reaction, you must write out the reaction! DrL> okay, so based on your understanding of human physiology, what will happen to the pH of my blood if I start to hyperventilate? > When we hyperventilate, we are expelling more CO2 than we are producing. I think this would cause our blood pH to go up and become more basic. The equilibrium of the bicarbonate buffer reaction would shift from the bicarbonate toward the carbonic acid. 16)​When we consider enzyme mechanisms and aqueous solutions, pKa tells us three very important things, what are they? ​ > pKa gives us an indicator of acidity, where we can tell what components of a reaction or solution are likely to behave as an acid. This is also useful in determining whether a group will be protonated or deprotonated in a certain medium. I am not sure as to the third. >Affinity for a proton, what is a good leaving group, buffering range > It tells us affinity because the lower the pKa the less they want the proton and the higher the more they want it. It tells us a good leaving group because the lower the pKa of a leaving group (the protonated version) the better it is at leaving. It tells us the buffering range which is pKa ∓ 1 pH unit. 17)​Is the formation of clathrate structures an endothermic or exothermic reaction? Explain your answer. ​ > The formation of clathrate structure involves the formation of hydrogen bonds between water molecules around a nonpolar structure or substance. This formation of bonds will decrease entropy since water is becoming more ordered. In addition, since hydrogen bonds are forming, it will release energy in the form of heat to the solution, thus this is exothermic. DrL> forming ice-like structures (H-bonds) is definitely exothermic (enthalpically favored). 18)​How does solvent influence the pKa of a protonatable group? How does this apply to protein folding, structure, and function? > The solvent influences the pka of a protonatable group by influencing its state of protonation. DrL> that statement makes no sense at all. A good habit to get into when writing, is to read what you have written, slowly and out loud. Becoming a good writer takes an incredible amount of time and effort. Writing and rewriting over and over and over, but this is something you should practice as writing well is important to all professionals. For example if the solvent in the system has a significantly low pka, it will have a high tendency to donate H+ atoms, and thus protonation of a protonatable group will likely be easier compared to a solvent with a higher pka. DrL> okay, now you are starting to think about both sides of the reaction, that’s good. However, what if the solvent does not have a pKa? (cannot donate or accept a proton) Moreover, the ONLY solvent we care about in biochemistry is water, so the solvent is NOT changing. You have the right idea, if you stabilize one side of the reaction, then the pKa will change. As you state above, if it becomes “easier to protonate” a functional group, what has happened to the pKa of the group? Has it gone up or down? >If it is “easier to protonate” a functional group, its proton affinity (pKa) has increased. DrL> easier to protonate is the same thing as saying harder to deprotonate, so does everyone agree? As far as the second component, the structure of the protein will determine its function, thus it is desired for the protein to successfully fold into its native folded state. As pkas shift, it can alter the ability for side chains to become protonated, or deprotonated, thus this can interfere with the ability for the polypeptide to undergo hydrogen bonding. If this impact is negative, hydrogen bonds may not be able to form, and thus the protein could result in a non-native folded state, and this can hinder its function. DrL> don’t overthink the question. In the case of a native protein structure, everything has folded correctly. So, the question is essentially asking you about a native structure and what is happening to the pKa’s of the protonatable side chains as the protein transitions from unfolded to folded. >If I wanted to simplify this, I would say that this is why sometimes actual pkas in proteins differ from the standard ones in the textbook. When a protein folds, interactions are forming which can influence how strongly a group wants to hold onto its proton. For example, if there was an -OH group in the unfolded structure, we would expect its pka to be around 15. However, if the protein folded and this OH was close to a NH group, the OH pka would go down. This is because the NH group could shift the reaction towards the deprotonated state by stabilizing the negatively charged oxygen with hydrogen bonds. DrL> that’s good, it’s always important to think of all the states in a given reaction and how both sides of a reaction (in this case, loss/gain of a proton) can be influenced. Stabilizing the negative charge on the conjugate base of any acid will definitely shift the reaction in favor of deprotonation, lowering the pKa. 19)​How is the blood pH “sensed” by hemoglobin? > Blood pH is sensed by hemoglobin as an increase in CO2 decreases the pH making it more acidic, hemoglobin recognizes that and in turn binds to the excess H+ ions. Conversely hemoglobin can release H+ ions when the blood is too basic. DrL> How does hemoglobin bind/release protons? > Protons bind to hemoglobin at histidine residues (pKa = 6) when blood pH becomes low and can be released when hemoglobin undergoes a conformational change from a taut state to a relaxed state when blood pH becomes high. DrL> If your blood pH was below 6 (required to protonate if pKa=6.0) then you would be dead. Moreover, are you saying there are multiple histidines that get protonated? I don’t think that is correct. >Since histidines pka is close to physiological pH, it is easily reversible between its protonated and deprotonated states, making it a pH sensor. (this next part could be wrong), Since its pka is so close to pH, DrL> close to what pH? You didn’t give an explicit value. typically, there is about a 50:50 concentration of deprotonated:protonated histidine. DrL> You MUST be explicit, where does this ratio come from if you haven’t stated the pKa or the pH? If the blood becomes too acidic or too basic, these relative concentrations will shift, which is how the body monitors blood pH. DrL> concentrations of what? Histidine? I am confused… >since histidines pka of 6 is close to physiological ph of 7.45. it is easily reversible between its protonated and deprotonated states, making it a pH sensor. (this next part could be wrong), Since its pka is so close to pH, there is typically a 50:50 ratio of protonated to deprotonated histidine. 20)​Which is the better leaving group (make sure you know the rule)? 1) COO- vs NH3, 2) OH- vs H2O, 3) COO- vs H2O, 4) NH3 vs OH-? ​ > 1) NH3 2) H2O 3) H2O 4) NH3 DrL> no. ​ >1) COO- 2) H2O 3) H2O 4) NH3 DrL> TOPICS THAT WE DISCUSSED IN CLASS ARE BELOW; DrL> You have used ProtParam and your protein sequence to calculate the extinction coefficient (at 280 nm) of your enzyme (4790 M-1 cm-1). After isolating and concentrating your protein sample you measure the absorbance at 280 nm in a 1 cm cuvette and see a reading of 0.85. What is the concentration of your protein? >Using Lambert Beer Law (I absolutely ) butchered the name, we have A=eLC where absorbance is equal to the product of the extinction coefficient, pathlength, and the concentration. Thus we get:​ 0.85 = (4790 M-1cm-1)(280 nm)(c)​ Here we want to solve for C, but its is important to check units to confirm the final unit will be moles. We then need to convert nm to cm, or vice versa. 0.85 = (4790 M-1cm-1)(280 * 10-7 cm)(c) Then solve for C. C = 6.33 M DrL> Incorrect, and you should probably have guessed it was incorrect based on the value. Think about this for a second. An average “small” protein has a molecular weight of 25,000-50,000 daltons. If the concentration was “6.33 M”, what would the density of that solution be? >would the pathlength be the length of the cuvette? Like in this picture below? DrL> Correct, this is an easy one. DrL> incorrect. M=moles/L >For the second density question, this is a guess based on a formula I found online. I don’t really know much about daltons DrL> In the early 1900’s scientists could not agree on the relative “hydrophobicity” for the 20 standard amino acids. Why didn’t the scientists agree? Describe the experiment and identify what the problem was. >The scientists could not agree on the relative hydrophobicity of the amino acids due to the manner in which it was being measured. The experiments consisted of starting with the amino acid in a water solution, then mixing it with an aprotic solvent, and seeing how much relative concentrations were in the aprotic solvent and water. The issue arises when scientists used different aprotic solvents, which resulted in changes in the hydrophobicity of the amino acids since amino acids can stay in solution better or worse depending upon the aprotic solvent used. DrL> good, DrL> In the space below, draw a titration plot for the amino acid K and another titration plot for E. Be sure to label the axis and all pKa’s ​ > I know the graph I drew below is incorrect, as it is a rough idea of how it should be done. I know the pka’s of the three protonatable sites are the points of rough horizontal components, and the buffer region is +- 1, however I am not quite sure how to know at which equivalent of OH- would result in which pH. I know they are sequential in order of pka, however I am not sure how to determine whether it is 1 eq. Or 0.5 eq etc. The purpose of the graph below, although incorrect, is to provide a basic form and for others to comment on what is incorrect for the purpose of my learning. In addition, I know we treat the -COOH group as a pka of 3 and the Nh3 group adjacent as 9 but do we just memorize the pkas for the side chains on each amino acid? DrL> Yes, and the graph below looks pretty good, with the exception of the first pKa, which should occur at 0.5 equivalents (halfway through the deprotonation of the -COOH group) >would this be right? DrL> looking at the titration on the left, why is the first pKa at 5.0? BOTH amino acids have a the same carboxy group, and we agreed that that pKa would be 3.0. DrL> Consider a serum protein that contains a salt bridge between an aspartic acid and histidine. The pKa for the aspartic acid is 4.5 and the pKa for the histidine is 6.5. Based on this knowledge, answer the following; What percentage of the serum protein will contain a salt bridge if the blood pH is 7.5? > [HA]/[A-] = 10^(pka-pH). For the salt bridge to form, the aspartic acid needs to be deprotonated and the histidine needs to be protonated. Aspartic acid: 10^(4.5-7.5) =.001P :1D = 1000:1 deprotonated (D) to protonated (P) Histidine: 10^(6.5-7.5) =.1P:1D = 1:10 protonated (P) to deprotonated (D) Since only one in every ten histidines will actually be protonated at this pH, it is the limiting factor, and there will be 10% of the protein that contains the salt bridge. DrL> nicely done, does everyone understand the logic laid out above? > could you explain how the ratio for histidine is the limiting factor, it’s been a while since gen chem for me and i’ve forgotten the rules. I wouldve thought that 1 in 1000 is the limiting factor, or is it because histidine would run out faster because theres only 10 to work with? >the ratio for histidine is the limiting factor because we need it to be protonated, and only 1 in every 10 molecules are PROTONATED. On the other hand, we need aspartic acid to be deprotonated for the salt bridge, and the ratio for DEPROTONATED to protonated is 1000:1, so there is ALOT more deprotonated acid than there is protonated base. So the protonated base would be the limiting factor, because it will run out and there will be excess deprotonated acid. DrL> Do you ever think the blood pH would ever get close to the pKa of the acidic amino acid? If the pH never gets within 2 pH units of a pKa, does that pKa really matter? Anyone else help me here? DrL> What percentage of the serum protein will contain a salt bridge if the blood pH is 6.5? > [HA]/[A-] = 10^(pka-pH). For the salt bridge to form, the aspartic acid needs to be deprotonated and the histidine needs to be protonated. Aspartic acid: 10^(4.5-6.5) =.01P:1D = 100:1 deprotonated (D) to protonated (P) Histidine: 10^(6.5-6.5) = 1:1 deprotonated (D) to protonated (P) Histidine has the smaller ratio, so it is the limiting factor and there will by 50% salt bridge formation at this pH DrL> Again, nicely done, does everyone understand the logic laid out above?