BCMB 4010/6010 2025 Virtual Study Group #1 PDF

‘Study Guide 1 (Lectures 1 & 2) For chemistry review: Chapter 1.1, 1.2, 1.3 & 1.4 Implications of the peptide bond: Chapter 3.2 and 4.1 1) Proteins play the dominant catalytic role in cell function, it is therefore imperative that you know the chemical struc...

‘Study Guide 1 (Lectures 1 & 2) For chemistry review: Chapter 1.1, 1.2, 1.3 & 1.4 Implications of the peptide bond: Chapter 3.2 and 4.1 1) Proteins play the dominant catalytic role in cell function, it is therefore imperative that you know the chemical structure, as well as the three- and single-letter names for the twenty standard amino acids. You should have this committed to memory before quiz 4. To get the VSG started, DRAW each amino acid, with the appropriate charges, as it would appear in an aqueous solution at pH=2.0, pH=7.0, and pH=12.0. DrL> Don’t forget, we have agreed that, for the ease of memorization, the pKa for the main chain carboxy group and amino group will be 3.0 and 9.0, respectively. Refer to Lehinger for the pka of the R-groups. 2) Why is water a good solvent for life?  > Because of hydrogen bonding, it is one of the strongest dipole-dipole interactions >Because it is plentiful > pure water is 55M, super concentrated, entropy of water can drive protein folding > Water solubilizes ions DrL> Both H-bonding and the water dipole are important, but just to be clear, they are two different interactions. Can you draw water making a hydrogen bond and water involved in a dipole interaction? >Water’s polar nature is capable of dissolving a variety of substances such as sugars, amino acids, and salts. I agree that water does have strong hydrogen-bonding interactions (dipole-dipole), but it can also participate in ion-dipole interactions DrL> good, in the case of the ion-dipole (also called dipole-charge) interaction, you should highlight the dipole of water (partial positive and partial negative). As drawn, this is incorrect as the oxygen atom has two lone pairs, not one. The point is correct, however, the properties of water allow it to serve as a good solvent for many different types of molecules, allowing them to react with one another. This includes lipids, or hydrophobic molecules. Does anyone know how water solubilizes a hydrophobic molecule? > Hydrophobic molecules aggregate to more structures such as lipid bilayers when they are in water. When the two are introduced, water molecules rearrange to form a “shell” around the hydrophobic molecule. Water tends to cluster causing the hydrophobic molecule to aggregate together. Is this picture below the correct drawing of the ion-dipole interaction… DrL> Do water molecules “cluster” or do the hydrophobic molecules? Can’t be both, but is that what is really going on? Are the hydrophobic molecules really attracted to one another? If so, what is the non-covalent interaction that drives that? > It’s the hydrophobic molecules that cluster together, but it’s not due to an “attraction” between those hydrophobic molecules. Rather, it is because water bonds strongly with itself, which pushes the hydrophobic molecules together. When water molecules surround each hydrophobic molecule, they are in an ordered state and are not as free to bond with other water molecules. DrL> KNOW the scientific name of this water structure, what is it similar to? > I could be wrong, but is the water structure encapsulating the hydrophobic molecule called a clathrate cage? However, when hydrophobic molecules cluster together, some of the water molecules surrounding them are released, increasing entropy and driving the aggregation of more hydrophobic molecules. This non-covalent interaction is the hydrophobic effect. DrL> Just to be clear, the hydrophobic effect is NOT a non-covalent interaction. Hydrophobic molecules, especially aromatic rings etc, may interact via van der Waal’s (the third non-covalent interaction). This is the resonance created when electronic orbitals interact within a certain proximity (Think of base stacking in DNA, or hydrophobic side chains in the core of a protein). DrL> the ion is not a partial charge, but yes, water has partial negative and partial positive charges. DrL> A typical quiz question, related to water and non-covalent interactions, might ask you to identify, a salt bridge, ice-like structure, hydrogen bonds, dipole-charge, and van del Waals in the image below; DrL> Also, considering the same image above, what would happen to the Keq for the folding reaction if the negatively-charged amino acid was removed? (i.e. an Asp to Ala mutation? How would the relative energies of each state change?(i.e. Energy of the unfolded state compared to the energy of the folded state) 3) Draw the resonance forms of a peptide bond. DrL> almost, but not correct, what’s missing, can someone fix this? >For the resonance structure on the right, the central nitrogen should have a positive charge and the left side amine group should also have a positive charge. The left side alpha-carbon should also have an R group attached, but I think everything else is correct. DrL> not quite, still missing something. >I believe the central nitrogen has the wrong charge. N has 5 valence electrons and has no charge when it has 3 bonds. The top left image should have a negative charge on central nitrogen, while the peptide on the right should have no charge… > i see it as the lone pair that it originally had it leaving and nitrogen doesn’t like to make 4 bonds thus leaving a positive charge on the image on the right >This is how I drew it below > why did your R groups disappear on the right structure? DrL> good catch! Where did the R-groups go? They should still be there, otherwise the drawing is incorrect! > oh i missed the hydrogen bond hydrogen atom bonded to on the nitrogen when i drew it my mistake! DrL> EXCELLENT! The above discussion is a great example of how you can use the VSG to correct, refine, and perfect your solutions!  4) Draw the hydrolysis reaction of a dipeptide with side chains labeled ‘R” and circle the nucleophile (atom) and draw an arrow to the electrophile. Do not attempt a detailed mechanism or the transition state, simply products and reactants. DrL> is this the “hydrolysis” reaction? >I think the one shown above is showing the condensation reaction, what I drew below is what I think is the hydrolysis reaction >I have added my drawing down below, I’ve drawn the same reaction, just with bond lines instead of the condensed form. 5) How many amino acids are in a tripeptide? Be able to draw a tripeptide with sidechains labeled “R”. Based on your answer to #3, identify the ‘peptide plane’, charged groups, and H-bond donors/acceptors. Which bonds in the main chain can ‘freely’ rotate? Why  can’t the peptide bond ‘freely’ rotate? For an additional challenge, draw the tripeptide HKL. > There are three amino acids in a tripeptide. The phi (α-C to nitrogen) and psi (α-C to carbonyl C) bonds can freely rotate. The peptide bond cannot freely rotate because it is sp2 hybridized (trigonal planar), resulting in resonance. DrL> good, clearly this is a tripeptide at a neutral pH. The peptide plane boxes are pretty clear, but can you simply name all six atoms that are in the peptide plane? >I think that the carbonyl carbon, the carbonyl oxygen, the nitrogen and its attached hydrogen, the carbon alpha to the carbonyl, and the carbon alpha to the nitrogen are the six atoms in the peptide plane. DrL> good, a more accurate way to state that would be; (from left to right) the alpha carbon, carbonyl carbon, carbonyl oxygen atoms of the first amino acid, and then the nitrogen, amide proton, and the alpha carbon of the second amino acid. Amino acid sequences are written from the N-terminus to the C-terminus. How about this… I propose that everyone knows how to draw the following tripeptide; A-G-A for the first quiz. I will make it worth five points. If so, someone please draw this tripeptide below; >This is my attempt below, I feel like I’m on the right track, but it doesn’t feel right. Can someone help explain defining which carbon is the alpha carbon? > The alpha carbon is the carbon which is directly adjacent to the carbonyl. DrL> yes, adjacent to the carbonyl on each amino acid. Someone remembers their organic nomenclature! So on the above figure, it is the carbon next to the carbonyl which has an ‘R’ and ‘H’ on it. There can be multiple alpha carbons on a chain because it is a way of saying where a carbon is relative to the carbonyl (and the above structure has multiple carbonyls that are part of amino acids). >Thank you! In the drawing down below in the A-G-A drawing, are there 3 alpha carbons then? > That’s correct >I don’t think the above structure for AGA is right, the carbon that is connected to the Nitrogen of Alanine should be the carbonyl carbon and there should be a carbon to the right of the carbonyl which connects to Nitrogen and the R group DrL> I don’t have a problem with the drawing of AGA per se, but I would prefer if we draw our amino acids in a cis conformation (all main chain atoms in the plane of the paper/page. One other thing, draw a dashed box around all the atoms in each peptide plane. Failure to identify all atoms in each peptide plane will not get full credit. > for AGA and the planes highlighted, i could be very wrong, so feel free to correct my drawing. > And just for clarification, the general rule for peptide planes is sort of like n-1. Suppose there were a tripeptide, and since there are 3 2 peptide bonds, you would have 2 or n-1 (n standing for peptide bonds in this scenario) of peptide planes? DrL> does a tripeptide have 3 peptide bonds? >No only 2 peptide bonds and 3 amino acids > So for every amino acid in a tripeptide there are 2 peptide bonds and 2 planes: The nomenclature of dipeptide had me confused. 6) The polarity of chemical bonds is essential to macromolecular structure and function. Draw the following molecules and label the partial charge(s) where appropriate; water, carbon dioxide, ammonia, and hydrogen sulfide. DrL> good, just note that CO2 is not a polar molecule due to the negative charge on either side of the carbon atom.  7) Draw the reaction coordinate (free energy diagram) for a reaction that has a ∆G= -10 kJ mole-1, a reaction that has ∆G= -5 kJ mole-1, and a reaction that has a ∆G= 5 kJ mole-1. Of the three reactions, which reaction is the most spontaneous? Which reaction occurs the fastest? Explain. None of these are the most fastest per se, we need to first see the activation energy. However, the positive delta G reaction coordinate graph will never be spontaneous while the other two are spontaneous, but “most spontaneous” still can’t be decided as activation energy is unknown. DrL> The Y-axis is correct, but the X-axis should be “reaction coordinate” or “reaction progress”. Also the first two reaction coordinates look the same yet have been clearly labelled with -5 versus -10 kJ/mol. When asked a question like this be sure you label each plot so there is no confusion and emphasize the differences in the data. Reactants and products should be labeled as well. Can someone redraw this with complete labels and emphasize the difference between the plots? This is important because part of critical thinking involves the ability to recognize differences in data that provide clues to what is happening. Does this make sense? Yes sorry I was just trying to save time with the original graphs I apologize. DrL> great! You really want to emphasize differences like this, it really demonstrates that you understand the concepts being tested. 8) Describe four types of non-covalent bonds that facilitate macromolecular structure and function. Which two interactions are more closely related? Why do you think it is important that macromolecules are influenced by many, relatively weak, bonds instead of fewer stronger bonds? What interactions do you think are most prevalent in the core of a protein? Justify your answer. > The four types of non-covalent bonds are hydrogen bonds, ionic bonds (electrostatic interactions), hydrophobic interactions, and Van der Waals forces. Hydrogen bonds and ionic bonds are more closely related because hydrogen bonds are interactions formed between an H and O, N, F (which are electronegative) making it partially electrostatic in nature, and ionic bonds are also formed through electrostatic forces between a positively and negatively charged atom. I think it is important that macromolecules are influenced by many relatively weak bonds instead of fewer stronger bonds because it allows for the complex interactions between molecules. For example, the active site of an enzyme needs to be able to form weaker bonds instead of stronger ones so that it is able to release a substrate once it has done its job. The presence of many weaker bonds allows for it to hold onto that substrate in the first place, but not so strongly that the substrate can’t dissociate from the enzyme. I think hydrophobic interactions are the most prevalent in the core of a protein because similar to a phospholipid bilayer, amino acids with hydrophilic side chains will experience more electrostatic interactions with an aqueous environment, leading them to be more present on the outside of a protein. Meanwhile, the amino acids with hydrophobic side chains will be sort of pushed towards the inside of a protein, minimizing its contact with the aqueous environment. DrL> very good, regarding the question of many weak versus fewer stronger interactions, your point about dissociation is spot on (for both substrates and products), but there is more to this, and it has to do with structural dynamics and enzyme function. > Another reason why weak bonds are important for enzymes is that many enzymes need to do an induced fit which may require flexibility which is easier to achieve if the substrate bonds with the enzyme through weak bonds versus a few strong bonds. 9) If we know the change in enthalpy (∆H) and the change in entropy (∆S) for a reaction, then the Gibbs Free energy equation can be used to predict whether or not a reaction  may be spontaneous. Since both ∆H and ∆S can be positive or negative, describe the four possible reaction conditions and whether or not each condition is indicative of a favorable reaction. We know dG = dH - TdS (little d = delta), thus if both dH and dS are positive, to get a favorable reaction, defined with a negative dG, then the temperature must be high. Thus with a low temperature, and dH and dS being positive, then the reaction would not be favorable since dG would be positive. Similarly, if dH and dS are both negative, then for low temperatures, dG will be negative and the reaction then favorable. However for high temperatures, dG is positive, and thus the reaction is not favorable. If dH is positive, and dS is negative, then the reaction will never be favorable regardless of temperature. If dH is negative, and dS is positive, the reaction will always be favorable regardless of temperature. DrL> To help drive home the possible conditions, could someone make a table showing the four possibilities ( both ∆H and ∆S positive, both ∆H and ∆S negative, +∆H and -∆S, as well as both -∆H and +∆S > DrL> well that was quick! > I had it pulled up! 10) Reconsider question #7 and describe what limits the rate of a reaction. Identify the rate limiting step on a free energy plot. The rate limit of a reaction is the highest peak of the reaction coordinate. This highest peak is just the transition state that requires the most amount of input energy in order for that state to transition. DrL> Okay, but can someone rephrase this, using language that is a bit more scientific? Think in terms of the energy of each of the states (reactants, products, and transition state). >The rate of the reaction is determined by favorable molecule interactions that result in product formation, so anything that would limit that including low temperatures which is low kinetic energy or restriction on how the molecules can interact such as physical barriers impeding the reactant molecules contacting each other in the correct orientation. DrL> not sure where you are going with this. Don’t overthink things. Describe the free energy associated with each state, keeping in mind that there are millions of molecules in solution. What limits the rate of conversion from reactants to products for the reaction? > The rate limiting step is the slowest step of a reaction, which can be identified by the transition state with the highest activation energy on the free energy plot. There are several factors that can limit the rate of conversion from reactants to products, including the concentration of reactants and temperature of a solution. Initially, the reactant starts out at a certain level of energy. Then, when it gains enough energy to overcome the activation energy barrier, it will convert into the transition state, which has the highest energy of the reaction. As the reaction proceeds, the molecule will decrease in energy as it converts into the product, either at a higher or lower energy level than the initial reactant. DrL> This can be a lot more succinct. Yes, the transition state is the highest energy state of the reaction. For the peptide hydrolysis reaction, this would be a tetragonal carbon atom with an oxyanion AND a hydroxyl group, not a very long-lived state because the hydroxyl could easily take electrons from the oxyanion, pick up a protein and leave. The higher the energy of the transition state, the less likely that state is to be populated. Reactions can only proceed if the transition state is populated. FOR TWO AND A HALF POINTS on QUIZ #1, I want you to propose a reasonable hypothesis connecting the rate of chemical reactions with the energy of the transition state and the probability of that state being populated. Please put your attempts below; > The probability that molecules will reach the transition state decreases with increasing transition state energy/activation energy. The rate is proportional to the population of the transition state, which is determined by the Boltzmann distribution based on the energy barrier at the transition state. 11) If a reaction is exothermic, is it always favorable? Explain. >No, exothermic just refers to the enthalpy of the reaction, where a negative enthalpy (exothermic) is where the energy in a reaction of products being formed of bonds being made is greater than the energy required to break bonds. If entropy of the reaction decreases in the reaction enough so at a high enough temperature, then the Gibbs free energy will be positive and not spontaneous. This is summarized in the equation DeltaG= DeltaH-TDeltaS  12) Why is cis-azobenzene a higher energy molecule than trans-azobenzene? The electron clouds are overlapping in cis-azobenzene, resulting in a higher electron-electron repulsion compared to trans-azobenzene, where the clouds are far enough apart that they don’t interact. 13) Can bond breaking release energy to do work? How do molecules like peptides, anhydrides, phosphoanhydride, mixed anhydrides, & thioesters release energy during hydrolysis? No, it isn’t the bond breaking that releases energy. It’s the bonds that FORMED after the fact that release energy. All those molecules release energy because when one bond is broken, two more are formed with water and so the math shows that 2 bonds formed minus 2 bond broken still leads to a net decrease of potential energy as the products are more stable, thus energy is released overall. DrL> almost, but missing one important detail. Can someone comment on the energy of the bonds being broken relative to the bonds that are being made? (bond “swapping” is constantly happening in your cells) > As in a peptide bond hydrolysis reaction, energy is required to break bonds. When the bond breaks, the components can form lower energy bonds that release energy, and that energy can be used to do work. > simple answer is NO bond breaking does not release energy but it COSTS energy. Above answer does a good job of in depth understanding i would say. 14) Consider the following dissociation reaction: H3O+ + -OH 2H2O. What does equilibrium mean (exactly what about this reaction is equal)? Does the reaction stop once equilibrium is reached? Equilibrium can be thought of as the point in which the rates of the forward reaction and reverse reaction are equal. In other words the reactants are being converted to products at the same rate that the products are being converted to reactants. At equilibrium, the concentrations of the reactants and products should remain relatively constant with potential slight deviations. The reaction does not stop at equilibrium, as reactions are still undergoing, but the relative concentrations should not change.  15) Draw a free energy plot (i.e. reaction coordinate) for a reaction with a Keq < 1. Repeat for Keq >1 and Keq =1. If Keq is < 1, is the reaction favorable? Can Keq be negative? > If Keq < 1, the reactants are lower in energy than the products, so at equilibrium, the reaction doesn’t favor the formation of products but rather the reactants; therefore, the reaction is unfavorable. Keq, given that it is a ratio of concentrations, cannot be negative as concentrations are always measured as positive values. DrL> good. 16) What is the thermodynamic ‘rule of thumb’? How is this useful when considering protein stability? The thermodynamic rule of thumb can be referred to as the following equation, Delta G = G* - RTln[P]/[R] or Delta G = -5.7kJmol-1([P]/[R]) This is useful when considering protein stability as delta G looks at the relationship between folded and unfolded proteins. A negative delta G would represent more [P] or folded proteins, hence, indicating that folded proteins are more favored as compared to positive delta G where [R] is more in abundance, representing unfolded proteins, hence means that unfolded proteins are more favored. DrL> okay, thanks for posting this. I clearly need to go back and go through this again. The rule of thumb is simply a useful thermodynamic relationship that applies to ANY chemical reaction. It allows us to easily estimate how much energy is required to change the ratio of products/reactants relative to the ratio at equilibrium (Keq). > question: for example if you had the equation and the Keq ended up being say 10^2, would the delta G be -5.7 times 2. Is that the general correlation? The negative signs tend to confuse me a little, what does it mean/stand for conceptually? DrL> depending on the circumstance of details of the question, you may have to put energy INTO the system to affect Q in some cases, and in other cases you may be getting some energy out of the system. Adding energy would mean a positive value, whereas energy released would be negative.// > that boiled it down perfectly, thank you! > Small correction to the equation above, you just missed the log. ΔG = -5.7 kJmol-1 log([P]/[R]) DrL> good catch, log base 10, not the natural log. Log base 10 is much easier to use. DrL> If the Keq for a chemical reaction is 0.1, but when the metabolites (substrate and products) are measured in the cell Q=10, how much energy was required to increase the product/substrate ratio? > approximately 11.4 kj/mol. To do this you plug Q and K into the original problem of 𝑄 ∆𝐺 =− 𝑅𝑇𝑙𝑛( 𝐾 ) and using the constants. 17) For a reaction at equilibrium, how would you calculate the standard free energy change? > To calculate ΔG°, you can use the equation ΔG° = -RTlnKeq. Standard free energy change is at STP so we know that R is 0.008314 kJ/mol*K and T is 25 degrees Celsius (298 K when in the equation. If we have the Keq already, we can use that to plug into the equation. If we do not, we can use the concentrations of the reactants and products at equilibrium to find the Keq (by dividing concentration of products by concentration of reactants), which can then be plugged into the equation. Keq can be found this way because at equilibrium Keq is equal to Q. 18) Why are near-equilibrium reactions easily reversible? Why are some reactions considered ‘irreversible’? For example, ATP hydrolysis and Aldolase have similar free energy changes, but aldolase is reversible. >Near-equilibrium reactions are easily reversible because the products to reactants ratio is able to be manipulated to have the reaction go in the opposite direction. The activation energy of the reverse Aldolase reaction is so large that manipulation Q is not sufficient to drive the reaction in reverse. >ATP hydrolysis > did you mean the activation energy of the reverse ATP hydrolysis is so large that manipulating Q wouldn’t be enough to reverse it, making it irreversible? >Yes 19) Using the picture below, answer the questions that follow; DrL> NOTE: the Y-axis below should simply be Gibb’s Free Energy (G), note dG. a.) Which of the reactions above is spontaneous? > Reaction A is spontaneous as the Keq is greater than 1 in reaction A, whereas reaction B has a Keq of less than 1. > Could you say that it is relaxing back to the Keq point. (rubber band analogy)? DrL> yes. Basically all of the chemical reactions in your metabolic pathways are displaced from equilibrium in one way or another, but this depends on your metabolic needs, which change over time. For example, if you are well fed, your body will divert glucose into making glycogen. As the day goes on and you don’t eat, all those reactions have to be reversed! >Can someone explain why for B the reaction can’t go back to equilibrium with more reactants being formed overtime than products? > I believe it has something to do with le chateliers principle. In B, it's below a Keq of 1, making it to where you have to add more product to reach equilibrium, so you can’t really force it to go over in reactants as there aren’t enough to begin with. DrL> Not sure what is being asked here. ALL chemical reactions, left to themselves, will go to equilibrium. The question is asking you about favorability (exergonic versus endergonic). All chemical reactions can also be displaced from equilibrium, but this requires an input of energy. b.) DRAW the progress curves for each of the reactions above, assuming a single reactant is being converted to a single product. DrL> good, the color key is great, but just remember on the quiz to delineate products from reactants (most people probably don’t have colored pencils) by directly labeling each line. Bonus Question; (Not required but fun to think about) One of the greatest illustrations of how temperature influences equilibrium can be seen when studying a protein folding reaction. Based on the data below, clearly and concisely answer the following questions; T(ºC) T(K) [proteinunfolded] (M) [proteinfolded] (M) 50 323 5.1 x 10-6 2.0 x 10-3 100 373 2.8 x 10-4 1.7 x 10-3 a.) How does the equilibrium change as the temperature is raised? >As the temperature is raised, we can see that the relative concentration of unfolded protein increases, while the concentration of folded protein decreases. Thus we can  conclude that as temperature increases, equilibrium will shift to the reactants in the reaction: Unfolded Protein Folded Protein. DrL> good. You could use the rule of thumb to estimate how much energy is required to shift that ratio! > ΔG at 50 ℃ would be -15 kJ*mol-1 , at 100 ℃ it would be -4.5 kJ*mol-1 DrL> in reality more information is needed (enthalpy and entropy are both temperature dependent), however, this simple calculation does illustrate the point, specifically, a relatively small increase in temperature can change the folding equilibrium by several orders of magnitude. b.) What are the equilibrium constants for the folding reaction at each temperature? >50C: (2*10^-3)/(5.1*10^-6) = 392, 100C: (1.7*10^-3)/(2.8*10^-4) = 6 c.) What thermodynamic parameters are needed in order to calculate the free energy (∆G) of the folding reaction at the two temperatures >To calculate dG, we need the following values: Enthalpy change, Entropy Change, Temperature. >At standard state to calculate dGo which then can be used to calculate dG by dG= dGo+ RTlnQ >I agree, although I would go further to breakdown dG standard, Thus bringing the total list to: Temperature(given), Keq(can be calculated), R(a constant), and values of Q, or initial concentrations given above. I state that we would need Q along with the person above me, because i presume the concentrations given in the question are the equilibrium concentrations. Because of this, we cannot produce dG, since the reaction is at equilibrium already. There must be a shift from a non-equilibrium state towards equilibrium, or vice versa, in order to see a change in free energy. DrL> the problem is that you also need to know the ∆H and ∆S for the reaction, at the different temperatures. Enthalpic and entropic contributions are temperature-dependent variables. This is an advanced question, but solvable.(hint; Van’t Hoff)