Biochemistry Final

Question 1

9-17

where is glycine a good buffer?

Answer 1

-glycine has two spots where it can go from positive to negative cahrge and experience net charge of zero
- one region where it has a net charge of zero is between about ph 4 adn ph 8, that is where zwitterionic form will be
2 and 9 when moelcules become protonated and deprotonate dthey can act as a buffer they are the moelcules absorbing a hydrogen ion and as a result the ocnentration of hydronium in solution does not change and the pH remains relatively well buffered* around the pKa where these molecules can accept or donate a hydrogen ion to become protonated or deprotonated this is where these molecules will function as a buffer, because glycine has 2 pkas for aminoum and carboxylate 2 ph distint regionds where it will function as a buffer

Question 2

9-17 Do you think it is energenically favorable or unfavorable to form amide bonds?

Answer 2

so from the 3 aa do you think its exogonic or endergonic? for 3 aa to brign together to form polymer chain energentically unfavorable and will not happen sponateously in fact we find there is a favorable release of free energy when we hydroyzle the amide bond why we develop these tyrosine crystals as we age cheese overtime and individual protein chains begin to degrade into 3 amino acids..?

• Energetically nto favorable to form these amide bodns, take home message: polypeptides are thermodynamically unstable but we have been talking about the fact that thermodynamics is not kinetics, so even though there is a tendency towards hydrolysis of peptide chain but the half life of amide bodn is at ph 7 is 7 yeasr, so kinetically these things are stable and when our cells synthesis protein it is reasonably stable overtime even if present over aqueous conditions when can presumably be hydrolyzed, we will see this over and over again when a molecule or pathway is under kinetic control, that the cell is able to create enzymes that can conditionally speed up these reactions so that the cell is actually able to control the rate at which these reactions can happen
• protein may be kinetically stable, cell may want to degrade particular protein product, cells have mechanism to target proteins for destruction and degrade them at relatively fast rates*

Question 3

why trans amide bond?

9-17

Answer 3

The answer is that there are steric clashes =

There are steric clashes btw sidechain groups when we have the amide group in the cis confirmation, where we create that zero degrees for the amide bond we will see tht the side chain groups particularly at the c beta position, and create steric clash, that clash is alievaiated if we rotate bond in opposite orientation creating trans amide bond, now we position sidechain groups on opposite side of polypeptide chain, like aleiviating that steric clash just like elevating eclipsing interaction btw methyl and protane??

Question 4

9-17 why proline 10% cis?

Answer 4

• Polypeptides have a strong preference for this trans arrangement, except for one amino acid in which this preference is not so strong
• Its b/c it’s a cyclic side chain wraps around and forms a covalent bond back to the backbone nitogen
• side chain not only takes off from calpha position it also is attached onto the backbone nitrogen as well, cis-amide in proline is more prevalent than in amides from primary amines
• In the cis orientation this shows the steric clash we tend expect btw two substituents at backbone C alpha positions, but here have to understand that even in the trans confirmation of amide group we still have a steric clash this time its with the portion of proline side chain that is covalently attached to the backbone nitrogen
• So because we have some extent of steric clash in both cis and trans, for proline groups the energeic difference for porlene is less than other 18 amino acids and as a result the energy difference is much less, the prediction would be then if take a look at protein structures, you would expect to see proline residues would have a significant observable population in which the dihedral angle corresponds to cis amide bodn where omega dihedral angle would be zero, that’s exactly what we see, the majority of those amide bonds are trans and omega angle is 180, and we have a significant population of cis amide bonds and protein structures and vast majority of those are formed by proline amino acids* so now we understood omega bond

Question 5

9-17

phi and psi

Answer 5

• Need to think about types of steric clashes that may come about as we rotate around this phi and psi dihedral angle, if remember amide bond constitutes planes can think of polypeptide chains as a set of these planes connected through hinge elements, we will rotate these planes relative to one another as we rotate these phi and psi dihedral angles
• Rotations in psi and phi determine overall confirmation of backbone, these amide groups can see on plates/planes relatively rigid and we can then rotate these plates relative to one another by rotating around the phi and psi angles
• Again we should be able to identify which bonds are being rotated around as we rotate around phi and psi, rotation of psi is around c-c bond and rotation around phi bond is around the nitrogen c apha position

Question 6

9-17 Linus Pauiling exam 1 info

Answer 6

• he began to think about how can we put together polypeptide chain that would minimize steric clashes and maximize number of H bonds that can ne formed by the polypeptide amide oxygen NH groups and repeating amide groups in a polypeptide, he was in England caught a cold because of miserable weather lying in bed sickwith his cold began to sketch a polypeptide chain, and tried to fold that piece of paper in such a way to point carbonyl oxygen toward the amide NH group to form an amide bond and tried to see whether he could create a repeating pattern of h bonds in that origami folded piece of paper depening this poly peptide chain, allowed him to predict helical structure of protein
• also i and i+4 positioning!
  a. But which dihedral angles are favored in a protein chain? Minimize steric clahses, maximize H bonds!

Question 7

9-17 Ram plot what he was trying t do

Answer 7

what he did was he began to consider at what angles for psi and phi could he alleviate the steric clashes in a polypeptide chain, and what he saw was that there were actually very specific regions that correspond to low energy structures

Question 8

9-17

Ram 2

Answer 8

• plotd them out realized oculd make a 2d plot for each position in polypeptide chain that sepcizlies in A psi and A phi dihdeal angles and some of these pairs will alleviate the clashes**
• Plot dark green low energy states, light green more accessible and white denotes high energy confirmations
• Dark green are favorable regions and other regions give rise to clahses and these are disfavored or high energy states, and these plots witht eh dark green show low energy states, light green energetically accessible states and white is the disfavored states, so particular regions in this plot named after him associated with repeating arrangements of the polypeptide chain*
• thser egions, like with blue spots, are part of repeating units that are part of secondary structure elements*
• w e can assocaite them with different secondary strcuture elements, this combination for ex of phi and psi is associate withteh alpha helix* etc.

Question 9

9-17 alpha helix

Answer 9

-Repeating pair of phi and psi dihedral angles do it at every position create a helix, see one full turn of polypeptide chain every 3.6 residues within the protein
-We also see that as we create this particular arranagment we position backbone atoms to form a repeating network of h bonds btw amide carbonyl oxygens and amide h groups oriented in exact arrangement that can give rise to this network of hydrogen bonds
What defines an alpha helix=phi and psi dihedral angles about negative 60-50 in Rha’s plot allowed regions, we have all amide H and carboxy groups hydrogen bodned to one another except for those at the end of the helix for which there is no particular partner, as a result the termini of alpha helixes have exposed amide /NH and exposed carbonyl oxygens and these tend to be POLAR and located on the surface of proteins8 tned ot be locate don the surfaces of proteins
e. so our characteristic pattern of hydrogen bonds if we take the carbonyl oxygen of one amide group and create a hydrogen bond in this arrangement specified by the alpha helix we will make the h bond with the nh group of the residue at the 4th position down the chain, take that carbonyl walk 1,2,3,4 down chain form h bond to that Nh group creates regular pattern, do that at each and every position as go down the peptide chain, we make that network repeating hydrogen bonds and that is what defines an alpha helix, see this all the time  I to i+4

Question 10

9-17 left handed helix

Answer 10

• we see this rarely becuase it is such high energy state

Question 11

9-17

i to i+3/ or i+4

Answer 11

• i is at close proximity of side chains as positions i+3 to i+4 amide groups in backboen but also possibility to create interactions between chemical groups within th die chains, which amino acids at position i are going to interact with i+3 and i+4 becuase they are on teh same side of that cylinder formed by the helix SOO side chain groups can then pariticipate in electrostatic interactions, pie stacking and van der waals intearctions*

Side chains at i position can interact with side chains at i+3/i+4 positions using salt-bridges, hydrogen bonding, pi-stacking or van der Waals interactions

Question 12

9-17 alpha helix hydrophobic/hydrophilic details

Answer 12

1. Helices are often amphipathic
2. Helices often have a hydrophobic face, which points into the protein interior, and more charged/polar faces, which point into solution.
3. this turns out to be very nice when packing secondary elements into protein structure, in this situation take individual helix and barring it into the interior of folded proteins, we see a typical arrangement will be for all of the amino acids to be hydrophobic very foten we will see helix at the surface where one face of the helix is exposed to solvent other face of helix is pointing inward towards protein so tends to be amphiphilic
4. orient amide groups along alpha helix position to create H bonds to one another as we do that we align individual dipoles in each amide group and get them all pointing in the same direction, as a result overall dipole associated with alpha helix call that a helix dipole

Question 13

9-17

helical wheel

Answer 13

1. the charged and polar groups oritented on one side of the helical wheel and other side is hydrophobic residues other individual helices fully exposed to solvent very often all of those amino acids around the helix are fully charged and polar amino acids
2. 3.6 residues per turn
3. Two turns of the helix bring sidechains into the same projection from the helix axis
4. Residues at i, to i + 7 are in proximity

Question 14

9-17

Linus pauling i and i+4

Answer 14

• he though about how he can fold up this chain so that every oxygen is associated with a H bond with soem other NH group in the chain, realized forvable arrangement but also instead of bending paper aroudn could also bend it around to go from i+3 amide NH group results in different secondary structure will talk about in next lecture

Question 15

9-17

Some residues prefer to be in the helical conformation more than others

Answer 15

1. Helix propensity is favored by Ala, destabilized by Gly or Pro.
2. Leu stabilizes helix more than I or V. Crowding Cβ is less favorable.
3. Side chains interact at spacing of i,i+3 and i,i+4: acid base bridges (E,D to K,R) stabilize helix, hydrophobic groups too.

Question 16

rectiation 2pm 9-28

know diff btw chaperonins and chaperons, what are their mechanisms of action, what stage they act

Answer 16

1. know for CHAPEroNES* which are typically smaller proteins they actually do not REFOLD misfolded proteins, hwo they act during the process of protein folding, the chaperones bind to specific some patches of aa along backbone of polypeptide, they help prevent misfolding of proteins so chaperones do not refold misfolded proteins they help a polypeptide chain fold to its correct confirmation by preventing misfolded states from being populated, the ATP hydrolysis in this case is used for a slightly different purpose, chaperones actually use energy of ATP hydrolsys to undergo confirmational change to either bind to polypeptide or release polypeptde
2. but with chaperinons each 7 subunits within the chamber hydrolzyign atp to undergo also confirmational change but in that case each confirmational change created a new interaction with polypeptide chain so chain either extending out or getting a chance to refold back**

Question 17

9-15 I think?

addition of salt for ex can be used to alter protein structure and participate proteins as we purify them?

Answer 17

Yes because salts have positive and negative charges, if you out a bunch of negative charges by the protein cause poisitive chrges t be attract t them or opposite so that can change the confirmation of the protein that way

Question 18

Groel vs GRoES-

Answer 18

confirmation change that chamber undergoes, as ATP hydroylzed to ADP chamber expands and as it expands expsoes hydrophobic surface area on the interior, that is complementary to the hdyriphobic groups in the improperly folded protein now gives that improeply folded proteins ubstrate and opportunity to open up to detach hydrophobic interactions that are intermolcualr and incorrect and reasocate them with the envirornment* this in essence unfolds the protein from a properly folded sate, once it does that essentially whats going on it has been puled back up to top of that energy funnel** folding funnel he showed us before** so that hydrolysis of atp is paying energy to create a new environment inside that chamber and to pull that protein into an expanded form from which it can try again to fold properly*** partial polypeptide chain pulled apart, and let fo pulled apart and let fgo- while that happens in lower chamber ti chamber left open, not happening in top

Question 19

champeron versus champeronin

Answer 19

1. Any small protein, dimer- clamp like chameprone
2. Any big like structure whith chamber like structure inside it is champeronin
3. Clamplike chaperones

Question 20

heat shock proteins

Answer 20

1. Small proteins, ehat shock proteins because way discovered when ppl raised the growth temerpature of bacteria cultures to higher temperatures saw expression of proteins in bacterial cells, as if bacterial cells trying to do something to mitigate effect of inc temperature, so we know when temp of system high proteins start to be unstable or amy unfolded protein has harder time folding into its structure
2. So these proteins called heat shock proteins were making higher amounts inside bacterial cells to prevent misfolding of other proteins or to help other proteins folding into the correct structure** so first tthign to remember HSP heat shock protein DO NTO REFOLD MIS FODLED PROTEINS*****
3. They act DURING the folding reaction of a polyp chain to prevent chainf rom going to misfolded state, or they try to bias folding reaction ofpolypepyide chain toward folded structure, if piolyup misfolded cannto rescule polypepyides into correctly folded structure

Question 21

aggregate proteins why need chaperons

Answer 21

1. So say a polypeptide is folding into its 3 structure, generates intermediate structure say a bunch residues unstructured still say many are hydrophobic regions, these hydrophobic regions instead of forming hydrophobic interior cluster, hydrophobic regions btw diff polyp chains can nonspeiically interact with each other to give rise to a bunch of aggregates proteins, thse aggregates are not stabilized by intarmolecualr interactiosn they are stabilized by INTER moelcuarl lnteractions btw two polypeptide chains by hydrophobic residues by some other interaction generating aggregates
2. Polypeptides in misfolded state in aggregates have no use to us, in order to prevent the folding intermediates or actual initial protein starting protein reaction to form aggregates what we can do if have certain prptein grabs onto folding polypeptide chain covers up region that can potentially undergo intermolecular interactiosn what will happen is these clamp proteins prevent intermeolcaurl interactiosn from happening and gives polypepide chance to generate intramoelcualr itneractiosn for correct strucurre or drive protein to right pathway by forming right pathway itnermediates, grab on undergo confirmational change, clamp like champerones

Question 22

chaperones

Answer 22

are in more extended confirmation when bond to atp, when interact in open bdingin cliff interact with bidnign peptides can either bidn to completely unfolded peptides or trapped folded intermeidiates

-once interact with bidnign clamp, hydrolysis ATP undergoes and results in confirmational change in clamp like proteins that go from closed state to an open state, in that closed state it has a much higher affinity for plypeptide bidnign to, ithodsl onto the polypeptide, rest gets a chance ot generate folded structure or intetermeidate, and after a little bit a new protein come sin calld ncueltoide change factor kicks out ADP generated by hydrolysis of atp chamber like protein and replaces it with another atp molecule then clamp like goes back ot original confirmation elases paritally foled or almost totally foldd protein, then protein biased to form intermolecular contacts to give rise to native structure and clamp like champerone is recycled boudn to atp and in open cofniramtion when this another poilypeptide can hit another protein and go through same process
o Pulloing entire reaction towards favorign folded state of protein because its preventing surfaced exposed regins of folding intermediates ot fold itnermeocluarl interactions that can lead ot aggregations ormisfodld structures of intermediates*
o If already misfolded calmp cant do anything, but they can bind to it and deliver to chaperoneins

Question 23

Two conditions happen if misfolded:

Answer 23

If conditions of cell are not condusive to recovering misfolded protein then prteins can be sent to degregation pathway were certain small tags given to protein sent to proteosome death chamber, under certain conditions instead of degrading proteins given a second chance to correctly fold back into native structure
o A bunch of chaperones bidn to these misofodled structure and send them to chaperonins*

Question 24

How do you ensure only misfolded protein gets into GorEL complex??

Answer 24

So what happens is when proteins misfolding happens which is a huge area of study by itself, when proteins are NOT in their most stable confirmation in their cell, they have certain residues* that are exposed on tehri surface, in many cases many different types of residues exposed many cases hydrophobic residues typically should be buried inside protein, because of misfolded or stable misfolded  intermediate they are exposed on surface, so there are certain proteins which just like the chaperones we talked about before this, groel groes examples of chaperonin proteins, also something called chaperons that help proteins fold, smaller soluble proteins that help other proteins fold
o	class of chaperoines that bind to hydrophobic surface of misfolded proteins and delivery these misfolded proteins to these chambers in the absence of these additional chaperon proteins, the chances of the misfolded protein being delivered to the groel groes complex would be much lower as opposed to higher probability of being degraded, if proteins chaperons  not to bind and deliver to champeronins those misfoled proteins would actually be sent to different protein chamber inside the cell called proteosome where they are degraded, death chamber
o	there is a continuous partitioning of misfolded proteins between either going to degraded pathway or chaperon mediated refolding pathways**

Question 25

groel groes 2

Answer 25

- Small chamber like structure GroEL open confirmation or closed confirmation, GRoES assocaton of seven proteins, 7 subunit chamber like exactly identical arrangement in bottom half, so can think of it as two parts each made up of 14 proteins*/ 7 form big chamber and 7 form top cap part, chamber GRoEL and cap part GROES - Although symmetric structure at top and bottom look identical, they are structurally identical but not doing the same thing at the same time*** - top chamber open so a bunch of misfolded proteins and polypeptides can be delivered into chamber by other heat shock proteins, notice while this top chamber is open something going on in the bottom chamber, its not allowing anything new to come in cap closed on bottom chamber os bottom chamber at this point is busy trying to unfold the misfolded protein and give it a chance to refold - each 7 units in bottom chamber bound to atp contiuosuly burning through cycles of ATP hydrolyzed ot ADP* then the whole structure of the chamber changes and whithint eh chamber polypeptides are puled apart and release dback os itgets a chance ot refold into correct structure*

Question 26

groel vs groes 3

Answer 26

- Once polypeptide in chamber reaches stable folded structure becomes energetically ufnvaorable to unfold it keep in energentic state, at that point cap comes off polypepyide released, at that point groel closes on top chamber, so at sam time cant have both chambers closed** - When top closed bottom not right orientation to fit cap, once top opens up bottom goes through confirmational change so cap fits on, by that time top has a protein come in refolded once in stable confirmation allos folded polypeptide to go out and when that happens and top chamber opens up another rufnodled or misfolded polypeptide has entered the bottom caber and whole process keeps repeating, unfolding and refolding going on in both chambers but only at one hcamber at a time** atp alters itneractiosn of aa lining up inner wall of chamber with misfodld polypeptide inside chamber because that cell essentially triedto fold and rfold polypeptide*

Question 27

groel-groes 4 what is the function of symmetric architecture with two parts the top and bottom?

Answer 27

-we call the cap the groES and the chamber the groEL each made up of 7 subunits, • there are different things going on in top and bottom chamber at the same time point • as shown in top left hand corner, we have a top chamber where the chamber is open and the cap is not present, say partially folded protein entering at this point bottom chamber is actually closed, cap is closed what the bottom chamber is doing, the bottom chamber each subunit is continuously hydrolyzing atp to adp make that peptide fold and unfold, backbone of polypeptide side chains are interacting with aa that line interwall of side chamber, interactions of polypeptide those lining inner cavity of chamber continuously changing because of atp hydrolysis and associated conformational change* * visualize it in a way that partially peptide chain is being pulled apart and let go and pulled apart and let go, by doing that giving it a change to refold into proper structure • while that is happening in lower, same thing not happening in top part, protein has reached stable confirmation from which it becomes energetically unfavorable to fold and unfold as oppose to keeping it in folded state at that point cap of chamber opens up of lower chamber, lets go of folded protein stable intermediate structure, at that point top chamber undergoes conformational change, in its ATP bound state, and that conformational change allows it to close the cap on top of it, so whatever misfolded protein enters the top chamber gets trapped in, the bottom chamber is open, now top chain does ATP hydrolysis cycles to open up and try to fold it back, while that process is happening in top chamber the bottom chamber is still opening • so once it reaches stable structure on top, cap comes off lets that go and by that time another partially folded protein or misfolded protein has reached inside cavity of bottom chamber, so when the cap of top one comes off the cap on the bottom goes back on, process repeated, both chambers doing the same thing not at the same time, take turns in either remaining open or closed***

Question 28

Limolene question from quiz

Answer 28

Q: dissolving nonpolar susbtances dissolving loimnune solution- when for clathrates UFNAVORABLE BECAUSE water molecule ordered becomes ordered when hydrpphobic effect energetically favorable because water moelcuels don’t have to order as much, so why is delta S is negative for limnuole because if visualize whats going on in terms of trying to dissolve nonpolar molecule in water, when trying to dissolve it in water, the limoluene moelcules themselves are not interacting with water, so you ar tryuing to fore them in water, so since wate rmoelcules are not interacting with limolene instead they are interacting btw themselves to form this network surrounding the limolene its trapped inside a cage the cage made up of water moelcuels and not interacting with limolene, if limolene trapped isndie cage itself cant be free to move around, watermolecuels hydrogen bodnign to eachother in that clathrate structure are also NOT free to move around, so that’s the reason if you start form situation where water moelcules randomly diffusing around and limolene just added to solution you end up in situation where entropy of water molecule as well as limolene IS LOWER WHY HAVE DELTA S NEGATIVE FOR LIMOLENEwhy have delta S negative for limolene**

Question 29

hydrophobic effect, so two diff situaton oen where limolene interacting with istelf the other interacting with water

Answer 29

• Now the other question hydrophobic effect is a slightly diff thing what we are talking about here happens ot nonpolar moelcules in solution, hydropbobic effect consequence of this, have high cocnentraiton of nonpoar moelcuels they tend ot cluster together and when they cluster together the water moelcuels that were otherwise at this interface of the clusters and forming cage like structure snow when two hdyrophobci groups come together the water moelcuels from that interface are released so after you have a hydrophobic interaction or have two hydrophobic groups packed against each other its at that point, have net number of wate rmoelcules released from this interface, so inc the overall entropy of wate rmoelcules, so the reduction in entriopy of wate rmoelcuels when they form a clathrate around the individual limoleuen moelcuels is what is entripoically unfavorable so tthats why delta S is negative, why if have high enough cocnetnration of limoelen moelcuels would rather cluster together, eliminat as much water from surface as possible instad of remaining indivdaully in solution, two diff sitations limolene by itself or interacting with another limolene molecule****

Question 30

• What stabilizies interact ion btw two limolene moelcules in an aq solutionc

Answer 30

answer to that would be that when two limolene come together the water moelculs releated by clathrates at interface increase the entropy of overall bulk molecule and that’s what makes the interaction btw limolene moelcules overall energically favorable because of inc in entropy of wate rmoelcuesl** by being released by clathrate structures at interface -but if question is just asking you whether just adding limolene to water is entroigcally fvaporable or not then the answr is NO B/C when add JUST limolene to water, not asking if limolene is forming hydrophobic interactions are not, we are just asking you if add limolene to water by itself without interacting ohter limolene moelcules is that entropically favorable? No its no other nopolar moelcyules is that an entropicly favorable process or not? In that case its entropically unfvaorable because of clathrate system**

Question 31

if there is a high concentration of limolene and let reaction proceed to equilibrium eventually will it reach some sort of a thermodyanmci favorable situation or not

Answer 31

if you do have a high enough concentration of hydrophobic moelcules so can interact with each other or IF just let enough time pass so at some point the hydrophobic molecules will interact with eachother, can only keep them stabilized in solution because of inc ientrpy of wate rmoelcules at the end of the day

Question 32

Ram plot 2

Answer 32

- The reason this is important is because we know that when a polypeptide chain folds** not all confirmations are equally and energetically fvaorble because in certain confriamtions can expect a lot of stric crowding, repulsion btw simairly charged side chains - If you have certain oreintations of side cahins/ oreintations of the groups are around alpha C or backbone certain orientations create a high energy sitiaton when collision btw side chains and groups on backbone and certain lower energy combination of torsion angles create lower energy situaiton which are thermodynamically more favorable**

Question 33

ram plot 3 why do we care?

Answer 33

you care about these- values of phi and psi is to come up with a way to predict with backbone confrimations would be more favorabel than others when a polypeptide chain folds* in modern days more importantly when working on solving structure of protein often from experimental data generate a model, when trying to piublsih that model you cant publish it if majority of torsion angles calcautled form that model are in violation of Ram plot*** so plot is a benchmark that tells you that there are certain combinations of torsion angles in backbone of polypeptide chain that will be energetically more favorable than others and certain combinations that are absolutely not allowed, so if your model shows there is a torsion angle combination btw angle phi and psi that is allowed and unless you have a really good reason why its true, model would not be accepted for publication it would eb flat out rejected, more practical use of it

Question 34

6 atoms

Answer 34

6 atoms can be thought of as relatively coplanar, distribtued around central amid ebond the CN bond in the C of carbonyl part of amide, and N being the nitrogen part of amide - CN bond becuase its paritally double bonded in nature other atoms coplanar with it C alpha attached to carbonyl carbon and C alpha also attached to N of amide bond so if thinkign of polypeptide backbone, these two C alphas represent the amino acid rpeciding and follwoign amide bond, all tehse six atoms can think of them as being more or less on the same plane*

Question 35

Ram plot 4

Answer 35

-Some phi and psi combiantions are v unfavorable: STERIC CROWDING of bkacbone atoms with other atoms in the backbone or side chains -Some phi and psi angle combinatns are more favorable: FORM FAVORABLE H BOND INTERACTIONS ALONG THE BACKBONE** o instead there is a limited region that is energentically favorable because we have to worry about steric clashes btw different groups in the peptide those two angles actually don’t get an opportunity to explore the different possibilities they essentially are binned, confirmatonal possibilities fall inyo particular low energy values in that landscale, those are the message get from rhamashamdra plot, then can use those to understand the features of the most common structures we observe in proteins that o fhte alpha helix and the beta strand*

Question 36

rotating bonds in peptide

Answer 36

only phi or psi bonds can rotate, capable fo free rotaiton connects C alpha to nitrogen of preceeding amide bond or the C alpha connects amino acid to next carbonul carbon of next amide bond, cant rptate around amide bond itself because partially doubel bonded in character so cannot rotate it itself, almost always stuck in transf confirmation, too C alphas in side of two planes are potinign in opposite directins trans, or sometimes cis when both C alphas pointing in same directin***

Question 37

ram plot 5 summary

Answer 37

ultimately 1. phi and psi combinations are very unfavorable: STERIC CROWDING of backbone atoms with other atoms in the backbone or side chains 2. some phi and psi combinations are more favorable: form favorable H bond interactions along the backbone 3. polypeptide cahins fold in ways so as to minmize collison between side chains hence only certain values of torision angles are energetically allowed**

Question 38

how to read ram plot from Mitra

Answer 38

- since two torsion angles go from 180-0-180 can plot 2 d plot, way to look at this looking at intersectin of two axis, vertical axis Y acis represents all possible values of torsion agnle psi, if go form center of this Y axis to the top go from o to +180, 0 to minus 180 values of psi, same as x axis which intersects y axis, it start at midpoint of that sectiona dn go to right you are talking about al the phi values that go frm zero to plus 180 and go from zero to negative side of x axis 0 to -180** of values of phi torsion angle - so since now defiend this interspace using a coordinate system of vertical axis and hroizotnal axis itnersectign each other, v precise values of axis can represent any combination of phi and psi in coordinate system where say phi is +150 and psi is say plus 120+ so should be able immediately to look at what quandrant* where is that combination such that that point represents psi +150 and phi 120 should obviously first look at positive places on this x axis, psi going from 0 to 180 reprsents positive avlues so the quantrand focusing on is top righthand coordinate becuae positive present values of pho and psi going from 0 t +180** so if find out where ont eh phi axis +150 is, and wherw on psi axis +120 is and then drop a lien from where they meet would represent coordination points that defines phi +150 and psi +120 that’s how youcome up with these figures, so ones that gve you more steric hindrance he indicated them by poijnt on diagram and essentially at the end of the day can see clear map that results from that, all regions shown in blue are regions represent energetically favorable confirmation phi and psi angles - darker blue color better in terms of energetic stability represent much lower energy combinations, lighter bleuu indicates still allowed but energies are still slightly higher* still negative energy but slightly higher, then any area blank white all represent phi/psi combos which would create significant steric collisions btw side chaisn or backbone and WOULD NOT be energetically allowed*

Question 39

Buffers * see sharp change in ph as function of adding base start devaitign and enter plateaued region flat, plato region in titration indictes buffer*8 whatever molecule is there in solution is resisting a change in pH why?

Answer 39

- Because in beginning when add in a little amount of base sharp rise in pH once get to flat area need a lot more base to achive the same change in ph because entered range where carbxolic acid is undergoing reversible dissocaiton asscoaiton - so everytime add base it is trying to before contributing abse to aq solution to inc ph a part of the base is forcing deprotonation of COOH so proton released by COOH is bidnign to that base to restore neutral water meocluel* so as long as you are in that range of the pka value of carboxylic acid, will need a lot more base in order to achieve the same amount of rise in pH if compared to basene of anything trying to slow down that rise* so that is why right around tha tpka value have a whole region of pH values or base concengratiosn wher eneed a lot fo abse to change the pH of the solution once pH of the solution ahs goen beyond tha point, COO fully deprotoatnd, ntohign preventing base from contributing more hydroxide ions to solution so even with small additions of OH get sharp inc in pH fo solution until hit second ionazbel group of gylcine the amino group - if keep adding base instead of contributing to raising pH of solution will try to deprotonate alpha amino gorp, whatever base adding deprotonates alpha amino group, OH bidns to H grenerates H20 instead of adding to basicity of solutont aht entire range exists over 1 unit ok pka of amino group once ph of solution goes beyond that nothing else to reduce to create that effect of absorbing hydroxide ions so now even with a little addition of hydroxide ions see an inc in pH - so why for glycine 2 regions in soltuin where glycine moelcuel tries to reseit change in pH of the soluton becuas ethsoe ranges correspond to pka of those ionazble groups*** so like hsitide does the same thing but now will do that at 3 different ranges of pH values* so if trying to generate a buffer where want buffer to acta s buffer not just at 1 ph range but at mutlipel* then moelcuels like amino acids amphiptes mutlipel ionizable groups can eb really good interms of buffering abilties*

Question 40

tyrosine and cysteine are special* why?

Answer 40

- Tyrosine net charge 0 but because goes from an OH to O-, two grousp ionize at higher pH and give moelcuel negative charge going frm 0 to -1 - But only two aa that take you from 0 to -1 which are cystine and tyrosine** - Cystine we know deprotonates around 8.3 or so so when cystine exists in protonated form is SH uncharged, if deprotonate is S- thiolate form gives you negative charge, provided Cystine not in a disfulded bond state if by itself as thiol group - For tyrosine in order to figure out what would be charge at ph 10 so if carbxolic acid deproronate at 3, so keep raising pH of solution that carboxilic acid gets deprotoatned at ph of 3 get negative charge on carboxyclic acid at 4, ph 8 enter deprotonate amino acid, by time at 8.5 or higher will be deprotonated, so by time get pH of 9 alpha amino completely deprotonated, carbox acid still protaonte so have net minus charge from just those two groups - the phenolate group, as start to get to ph of ten protonate hydroxyl group goes from 0 to -1** since pka of group is at 10 and ph of solution at as 10, IT WILL BE HALF DEPROTONATED SO will be negative half charge, so net charge on tyrosine at ph of 10 is -1 frm COO- then -0.5 from thiolate so -1.5, already lost positive charge by deprotonating alpha amino * *Only two groups cystine and phenol of tyrpsone** uncharged state has a zero charge if deprotonated go to phenolate anion -1 charge, or cystine thiol group if not disfuilde group if free thiol group has zero charge** if deprotoatned* because thiol group of cysteine has pka of8, then thiolate anion negatively charged****

Question 41

Rules for titration!

Answer 41

-two aa side cahins deprotonate at lower ph those are yoru acids* cabrxolic acids which will deprotonate at say pkas or ph around 4 if protonated form neutral when deprotonated they will make group negatively charged go from zero to minus 1 both happen*** o Lysine or arg go from +1 to 0** acids= 0 to -1 tyrosine and cystine go from 0 to -1*

Question 42

coiled coiled

Answer 42

keratin fibrous protein made uo of coiled coil structures, have typical right handed helix but right handed helices wrapped around each other that is going in left handed, each alpha helix is right handed but two wrapped around each other then that helical wrapping would be left handed* so actually have opposite coordination compared original handedness of individual helices o Typically speaking if helices have specific right handed ior left hadnedness, the triple helix will be the opposite the overall quarntary assocaiton of coiled coiled will hve the oppsite, of he indiivudla helix* same for collagen, so collage individual helices are left handed overall structure right handed triple helix

Question 43

collagen helix

Answer 43

triple helical structure, individual helices are left hadned, but when they are helically intertwined with each other that overall helical strucutre is right handed so called right handed triple helix* o Typically speaking if helices have specific right handed ior left hadnedness, the triple helix will be the opposite the overall quarntary assocaiton of coiled coiled will hve the oppsite, of he indiivudla helix* - when make coil coiled we take two alpha helixces adn twist around each other, see quite a lot of examples in which there are hydrophobic interactions present at the interface* and can interdigitate* with respect to one another*

Question 44

collage 2

Answer 44

h bodnignw ithin each chain but so within typical right handed helix have I to I+4 rule, collagen doesn’t follow that rule much more extended structure means have to go over more aa to complete one full circle longer H bodnign interaction but also weaker btw individual chains - what is actually holding them together is when, if just take a single strand of three strand structure v hard ot get into left handed helical confirmation, that left handed helical confiramtion is uniquely formed when all three wrapping each other in righted handed confirmation b/c special groups like hdyroporline. and amid ebonds btw backboens optimal h bond pattern stabilzign triple helical structure within context of that structure, within that strucutre they are left handed helixes, unless part of right handed overlal structure the individual left handed structure wouldn’t be stable - H bodnign btw backbones of the individual helices to each other as well as hydrogen bond of hydroxprolines btw the chains* - Think of triple helix as more like a beta strand interaction with h bonding btw the strands in this case not strands they are forming helices, but h bonds more btw strands as opposed to within the same strand** **hydroxyprolines have hydroxyl group which can hydrogen bond as well which has lone pairs on the oxygen*

Question 45

quiz question urea 2

Answer 45

h bonds with wate rmolecules, so dec entropuy makes it rigid - there is a net entrooy and entrpy change with respect to urea and indivdiaul water molecules, so if take solid urea and dissolve in water actually breaking lattic of crystal structure of urea and dissolving in water to so they become more mobile in water, but in this case we were saying when urea is in water what happens? not solid state of urea - urea very polar molecule, has to interact with water so no way it cannot effect entropy of urea as long as its in water - delta s engative if consdie rtwo sititosn wate rwithotu urea and water with urea, no urea nothing holding wate rmoelcuesl to each other othe than hydrogen bonds extremely transient, as soon as add urea bunch of urea tighly holding on to water moelcules tightly holding onto it, each urea can form bonds, amide hydrogens with another two so like 6 water moelcuels surround urea separate from clathrate structure hydration layer, so all wate rmoelcuesl sticking around urea have lowered entropy of water moelcule*

Question 46

quiz question urea versus limolene urea quiz question cnt.

Answer 46

different from limonene b/c limolene is NOT interacting with water molecules but entropy of water moelcuesl still reduced water molecules were forming cage like structures by formign H bonds btw themselves, in this case watermoelcueslf ormign hydrogen bodns with solute still results in dec of those wate rmolecules adding to solvation layer around urea delta s positve ? NO not possible, if lots of wate rmolecuels forming tight interactiosn with polar surrounding it cant inc ability of water molecules random diffusion not possible so correct answer is B, so we know urea interacts with water so d wrong, and e is wrong becuase random motion of urea signicantly inc in water, because when urea is in solid state doesn’t have high entropy but moment dissolve it in water then its entropy is inc -solid urea versus urea in water dissolved, entropy will inc when add to solution, but once equilbiruatedf rom poitn of addition at that poitn entropy would not be high, each urea moelcule surrounded by water moelcule, although more mobile than they would be in solid state where urea molecuyles cant move at all, then in that perspective yes can see where confusion is, more mobile than solid satte a cant move at all, liquid can still move around, then their entropy would inc compared to solid state, but inside the water if you compare solvated molecule vs unsolvated moelcuel then solvated entropy is lower than nonsovlated*

Question 47

2 reasons why dissolving urea in water is energetically favorable??

Answer 47

* **Addition of polar molecule to wate would have signicaint contribution** so say urea add urea to water forms a lot of H bondsiwth wate rmoelcuesl so two reasons why dissovlign urea is thermodynamically favorable** 1. its forming a lot of hydrogen bodnign interactions with water molecules creating a NEGATIVE enthalpy change, 2. other reason being that as we were just talking about if consider urea going from solid state to solvated ionic state, that is actually energetically favorable for urea molecule itself* but one major reason for entropy dissolving urea in water is entropically favorable all hydrogen bonds forming contribute to enthalpy of system stronger H bonds think of it as H bonds btw water molecules being replace with H bonds in urea , on 1;1 basis nothing changing but H bonds to urea n water are stronger than H bonds btw two water moelcules becusae urea is much mroe polar than water molecules, enthalpically favorable*, compared to solid urea entropy of urea solvated form goes up? But free urea vs solvated urea then that goes DOWN entropy of water moelcules def goes down becusa entropy of water meolcuesl without urea is just water molecules diffusing around in solution, added urea water molecules sticking around urea will have lower entropy*

Question 48

9-22 | alpha helix details with aa 2

Answer 48

proline alpha helix breaker becasue it does not hydrogen bond when form amide linkage for proline, no NH group so can't participate in the H bond network its looking for; no amide H to donate - glycine too fleixble, often times find this at the end of the alpha helix when teh chain wants to bend around, flip flops so also a helix breaker not favorable; would lead to large entropy PENTALTY for restraining where aa needs to be in teh alpha helical confirmation*** - interesting note Leu s fairly compatible with alpha helix, but not Ile which is almost the same thing does not have this great propensity turns out steric crowding close to backbone b/c branch at C beta position, not just a matter of size its because its branched and bulky right next to hte backbone that disfavors the alph helix, Leu branched at gamma C, leu branched further down chain for some reason its that bullky branched structure right next to the backbone that seems to be unfavporable for alpha helix

Question 49

9-22 | alpha helix stabilizing interactions

Answer 49

- want stabilizing interactons 1. great way is to be AT that interface where we ar eburrying surface area, lets put hydrohobic aa there bcuase will give us favorable noncovalent interactions and effect, create a pattern so those aa are at that interface by putting them at repeating i adn i+3 or i and i+4 that put sthose aa ON THE SAME FACE OF HT EHELIX to contact a neighborign helix - so can engage in favorable noncovalent interactions like plus or minus charge interactions these can also stabilize an alpha helix when have hydrophobic aa side chaisn that will also favor formation of alpha helix why paritcular spacing is so improtant i and i+3 or 4

Question 50

9-22 | why branched, i and i+4 for alpha helix?

Answer 50

you get i adn i+3 spacing becuase branched and they have good hydrophobic surface area btu also as we put two leucines together they can udnergo van der waals contact adn they can actually branched side chains snuggle up next to eahc other, sometimes methyl groups will fall into place next to each other call knobs into holes packing, leucine groups hve nice hydrophobi c contacts nice vanderwaals interactions, create a nice geasy jigsaw puzzle to create noncovlaent interactions - also alternating hydrophobic and polar** amino acids are great stabilizing pattern interactions, give you diff aa around helix commonf eature on oen sdie of secondary structure to optimize packing when forming proteins tructure beta sheets is when thats good gets you that stripe***

Question 51

9-22 thr vs ser

Answer 51

threonine is branched at c btea position versus seroine, both have Oh side chain, threonine less likely to form alpha helix than seroine becuase branched at beta C***

Question 52

9-22 | helix dipole for apha helix, what is the best aa to put there?

Answer 52

- so electrostatic interactions at helix termini intearting with helix dipole** is good 1. because amide groups all aligned individual dipoles associated withe each amide group all positioned they can in effect all sum up to create realtive region of positive charged density at N terminus, adn negative charged desnity at C terminus of helix, if have charged density dipole can position side chain groups to have favorable interactions with that*** 2. another way to think abotu helix dipole, all H bonds up adn down are accoutned for except towards N temrinus, lots of NH groups that dont have partners*** Carbonyl O not partnered upa dn if can swign a side cahin around and form a sid chain to backboen H bond that will also be favorable, these side cahin interactions btw amino acids at positions 1 to i+3 or +4 deicde those positions are important becuase they end up ont eh same side of the alpha helix after made one full turn***like asparate at i and lysine at i+3 positive and ngatie charges close enoguh to get favorable electrostatic interactions to stabilize the ehlix*** so put positive charges near C- terminal, and for N teriman with N+ put acids with negative charges***

Question 53

9-22 destabilizing for alpha helix

Answer 53

trypotphan and trptophan next to each other too big, destabilizing becuase alpha helix is realtively compact*

Question 54

9-22 cation pi interactions

Answer 54

aromatic aromatic interactions when aromatic groups sometimes have favorable interactiosn with cationic side chains**

Question 55

9-22 Helix core/ helix bundle

Answer 55

pack multiple helices together to make a bundle, see thsi quite a bit in folded protiens as well, a lot of the same type of noncovalent interactions we are talking about come together, fi we were to assemble a bunch of alpha helices, may fund you would want to bury/sequester aa side chaisn that were hydrophobic** provide you same type of driving force for describing hydrophobic interactions* - at core of two coiled coiled, have a lot of leucien side cahins present at th einterface, hydrophobic aa at a/d positions which are great, leu great because provides a nic eamoutn of hydrophobic surface area, have side chains that can interdigitate adn create a lot of stabilizing contacts

Question 56

9-22 | beta strands

Answer 56

beta strands make nonlocal H bond contancts, not possible to make local H bond contacts like ahelix antiparallel is more stable because we like LINEAR arrangmenet of hydrogen bonds, those h bonds will be a lhave a little bit more of an energetic payoff than parallel - amphiphatic - see pattern polar-hydrohphobic-polar-hydrophobic alternating pattern

Question 57

9-22 | why do beta strands have alternating amino acids polar/hydrophobic?

Answer 57

b/c it gives you the opporutnity to have side chains presented up adn down, ex on slide ile -argaine-val-etc -beta sheets like ahelix pitch of 2 residues per turn, one residue points down adn one resdiue point sup so if want to have hdyrophoboc usrface 1,3,5 then you want those to be hydrophobioc thats why get alternatign pattern of hydrophobic and polar aa - H bond network can positon sidechain groups so there are favorable interactions not only across strands btu also stacked one on top of another can put those sheets together adn take hydrophobic side chains liek ala and get them to nestle within each other and get them to create favorable contacts across 3dimension and really stabilize this think, like silk includes repeating secondary structure element these beta sheet structures, which self assemble to create a very tough mateiral, extrmeely strong**

Question 58

9-22 | entropy

Answer 58

if you take a polymer chain that is intrinsically flexible and you out it in a situation where it undergoes folding, compaction and transformation to one confirmation of that polypeptide chain and all the compoennts are precisely positioned in 3d space, the entropy of that protein chain will be diminished FOR SURE*** 43. The overall entropy of the system is a different matter because you need to include a consdieriation of water in which tha tpreotien chain may be folding, protein chain iself by definition if you fix it into one state, that’s ONLY ONE state low entropy condition for that polymer chain** within the random coil tehn that unfolded state populates a whole bunch of confirmatiosn that are energetically very vye rsimilar, when we take a look at the experiemtanlyl determine structure of proteins, we don’t see all that often a protein is completely helical or completely composed of strands v often find both secondary structure types are included

Question 59

9-22 | Anfinsen 1

Answer 59

Anfinsen postulated that the native structure of a protein is the thermodynamically stable structure; it depends only on the amino acid sequence and on the conditions of solution, and not on the kinetic folding route. - from chemical standpoint only one convenient way to syntehsize it

Question 60

9-22 fiburous proteins

Answer 60

-So for these structural fibrous proteins tend to see small set of aa repeated over and over again, in contrast to globular porteins they are composed of a small set of aa repeated look on slide for what goes in silk, collagen etc - keratin forms coiled coiled creat dimers when ptu two proteins tgoether, when assmble mutlipel timers tgoether great protofilaments look on ppt; remember ala and glyc are small so allwoed to pack tightly - the interdigitation gives it such amazing strucutral integrity pure unit weight actually stronger than steel

Question 61

9-22 | collagen proline details

Answer 61

-individual left handed helices wrap around each other in right handed sense to make triple helical structure, H bonds are established btw the chains, why between the chains? We don’t see H bodn network within each individual molecule so much, because of the fact collagen sequencs are proline rich and realize previously prolien groups in backbone cant find hydrogen bonds, we do see repeating motif three aa repeated over and over again, often hydroxproline at third position, can see its v similar to proline itself - Another image showing three diff peptide chains coming together each oen colored differently shows us the way that glycine aa is forming h bonds between individual strands this shows us helical arrangements of each individual protein chain - Here is our 4 hydroxy proline, wait that wasn’t one of the 20 aa natural that nature uses to synthesize proteins on ribosome** in fact ribosome does not accept hydroxyproline as makes structure1 where do you think hydroxproline comes from where is it generated - not just porline at every pitch, need glycine at every 3rd amino acid, but very proline rich sequence

Question 62

9-22 note about polyproline

Answer 62

porline doesnot want to form alpha helices, which are secondary structures with a particular pattern of dihedral angles and H bonding, but you can pack proline them up and form polyproline helices, characterized by set of phi adn psi dihedral angles NOT BY H BOND interactions beccaseu no H bond network of backbone**** so individual chains are going to fit into one of those bins from teh RAm plot realtively low energy region avoids steric clashes, types of noncovalent interactions are imperative in definiting the arrangement of polypeptide backbone - so no H when wrap them around create very favorable vanderwaals contacts adn also some very interesting stereoelctric effects where there is an interaction tbw poliarzed electron dnesities between funcitonal groups on differnet peptide chains - noncovalent interactions that stabilize proline helical arranagements paricualrly in collagen type structures greatly insipred by his work at mit... so can have stable helix without H bonds but if have proper other noncovalent interactions to specify a prepeating set of dihedral angles we will get a helix.. his pHD dissertation, helical strucure turns otu no hydrogen bondign within backboen of these molecules - so h bond is great but not hte only way to stabilize a helix*** * if want to design two units joined together by fairly rigid spacer may use poly proline sequence*

Question 63

protein folding energy, funnel model= golf ball model ex | ****

Answer 63

at some point get lucky find energy well drops down that is obviously not what is going on for protein folding, not a good description, instead what we realize is that the whole energy landscape is kind of directed downhill toward the global free energy minimum that means as we knock that golfball around it has an intrinsic propensity to end up going towards flobal free energy inimum, as protein chain is collapsing and beocmign more compact becomes restricted in a way that drives it to the proper folded form • Another aspect can consider in funnel like landscape there is an entropy can associate with the width of the funnel, at the top can see boundary is very broad, means protein can wander around in many many arrangements, as move down the width of that landscape becomes constricted , protein chain becomes more compact, and it has less entropy and less states are available to it, entropy is dec, free energy is dec protein chain is more compact and its being driven toward the proper folded state • Big conclusion we have from appreciating this funnel model for protein folding is that it is not a random search that the protein chain is limited in the number of different confirmations it needs to explore, and will end up getting into its final folded state fairly efficiently as the chain becomes compacted*

Question 64

how to draw cis proline / cis polyproline info

Answer 64

* Is C alpha pointing in same direction of C alpha, or opposite direction of other c alpha that’s how you draw a cis proline*** always keep C alaphas on same side so cis! like field goal in football * What happens with polyproline 2 have repeated prolines in the trans form and that will kind of create a more extended version of prolien strand* whereas if you have Polyproline1 because cis confirmation of proline, which actually creates bent confirmation of baclkbone naturally so like every residue one of the other in bent confirmation then it wouldn’t not be an extended out confirmation it would be more like a compact confirmation where polyproline tract would be bent on itself* so that’s the major diff btw trans and cis polyproline, trans more extended confirmation and the cis more compact because polyproline track bends on itself***

Question 65

9-28 recitation 3:30 pm | i adn i+4 more details

Answer 65

o This structure is stabilized predominantly by hydrogen bodning nteractions btw polypeptide backbone itself*** so carbonyl groups and h of amide nitrogen forms h bond interactions with a very specific pattern such that if you start with a residue, say I residue it’s the i+4 resideu that it forms a H bond on backbone

Question 66

9-24 | tertariy structure

Answer 66

allow secondary elemnts to assocaite with each othe arranging hydrophobic surface area towards interior adn sequewsting hydrophobic surface area providing a driving force for foldign** hydrohobic heme packed away in itnerior like for hb or myoglobin - beta sheets most stable when algined with right handed helix/right handed twist, for sture stems from stereochemsitry f each individual amino acid, if use D aa compared to L, then D may have all these come together with left hadned twist who knows

Question 67

9-24 protein folding

Answer 67

- proteins can fold sponateously under right conditions, but if alter the conditions, make harsh can froce proteins tolose it sdisticnt functional 3d arrangment of aa in molecules, called protein denaturation or unfolding - folding and unfolding is a highly cooperative process, this means that essentially an all or non phenomenal as we undergo changes in einvriomehter with temp or pH we radpily progress from conditions in which protein is all fodled through a v narrow transition region to one in which the protein is completely unfolded** - look on slide of graph shows unfolding RNase A which is 100% folded by below 20 degrees C but by time reach 35 degrees C entire protein chain is UNFOLDED**

Question 68

9-24 | Anfinsen's experiment 1

Answer 68

Whats required to get it to fold? Molecule substrate needs to be manipulated by some machinery in cell to be converted from folded state to unfolded state, other possibility is that proteins are just capable of spontaneously self assembling into correct 3D arrangement, so famous protein scientist Chris Anfinsen- postulated nature structure most thermodynamically stable structure, essentially all information required encoded within sequence of AA, should be able to unfold protein and isolate it and on its own under proper conditions should be able on its own to refold without assistance of any other entity, so he did this experiment with the protein ribonuclease -Presence of cystine disulfide is crucial for stabilizing ribonuclease fold* Does narrow transition window mean all or nonhalf unfolded? Cooperativity usually means protein itself is either present completely folded or completely unfolded, sometimes as go through these transition regions one portion of protein change will undergo excursions or become partially unfolded*

Question 69

9-24 | Anfinsen's experiment 2

Answer 69

- Not only small subfraction of proteins can spontaneously unfold, not ALL PROTEINS CAN SPONATENOUSLY unfold but many can, ribonuclease is one ex of a protein that can on its own spontaneously unfold* - he knew that ribonuclease can be denatured and unfolded in presence of concentration of urea -If take native ribonuclease in presence of 8 M urea, can get many secondary structure elements to unfold, also incubate protein in presence of beta-mercaptoethanol REDUCE disulfide linkages and convert to sulfhydryl groups, i.e. SH, so did two things= 1. unfolded protein chain and 2. broken disulfide linkages by adding both these chemical influences, two different chemical influences and 2 different effects on protein structure***

Question 70

9-24 Anfinsen's experiment 3 why urea?

Answer 70

Why it assists in unfolding proteins urea assists because has two NH2 groups and carbonyl oxygen, very small molecule very dense number of H bond donors and H bond acceptors, and as a result can compete with internal H bonds and proteins structures and form alternative h bond connections that assist in breaking apart native structure of protein, in addition urea can form H bonds with water, recognize water in a way that influences solvation arrangement of water molecules around protein and also affects free energy folding of protein chain* - H bond contacts important for maintain 3 structure of protein, backbone to backbone H bonding seen in alpha helices, side chain to backbone and side chain to side chain all H bond type interactions can be critical for defining protein structure and urea competes with INTRA molecular H bonds and break apart folded structure

Question 71

9-24 Anfinsen's experiment 4 why betamercapethanol?

Answer 71

-Whats going on with mercaptoethanol- it does have an SH group itself on its structure it can play a catalytic role in breaking the disulfide bond that is present in protein structure** promotes disulfide exchange*** and assists in converting disulfide to the free thiols**

Question 72

9-24 Anfinsen's experiment 5 what he did

Answer 72

1. He began with ribonuclease, an enzyme catalyzes the hydrolysis of RNA molecules**** and it is an active enzyme when it is properly folded under native conditions and low temperatures, and proper aqueous environments 2. Then added urea and beta-mercapethanol then this allowed him to completely unfold the protein chain and break apart the protein, so no longer forming proper secondary or tertiary structure, now inactive says on slide 3. Then he asks the question can I get that protein chain to refold? In some ways this was a simple matter in some ways of removing components from solution responsible for denaturation, can dilute urea out, and remove beta-mer. Recreate folding conditions that are proper and sure enough that ribonuclease spontaneously on its own with no other molecules around was able to refold and create the correct disulfide bonds and regain enzymatic activity*** 4. interesting but crucially he introduced a wrinkle* which is that he tried to do this experiment in which he first removed beta-merc. And then he removed the urea and what he found was yes he could recreate the disulfide linkages by removing beta-merc. But then when he tried to allow protein chain to refold he got almost no activity back, and why was that, because when he recreated folding conditions the protein chain had already been locked into the incorrect arrangement of disulfide bonds between 8 different cysteine aa in protein chain*** **** e5. so that in the absence of urea* those covalent linkages were stapled in place and those covalent crosslinks prohibited the protein chain from regaining its proper three dimensional structure 5. he never compltely removed BME

Question 73

9-24 Anfinsen's experiment 6 how to overcome the issue then with protein being locked with wrong C-SH bonds? Q on slide: Why does it impair protein folding to reform disulfides first, then remove denaturant? How does beta-mercapoetanol help reform native structure step 3?

Answer 73

- if he then added back trace beta-mercapethanol just a little bit just enough to allow disulfide linkages to reform, to exchange with one another then the protein chain was able to wonder around with those covalent cross links transiently opening and closing in order for it to explore conformational space and explore all different possibilities until it ended up in proper 3d structure *****regaining its enzymatic activity - yes its just a simple matter of some chemical species are denaturing and when remove them can get a protein to refold, but its important to think about why it is that removing beta-mercapethanol and then urea leads to inactive protein and how beta-mercapethanol can in v small amounts assist in recovering improperly folded protein and allow it to refold?!** =Overall conclusions from his work- proteins can adopt native confirmation SPONTANEOUSLY, that is all that is needed, the individual molecule on its own has/contains all information required to specify its 3d arrangement =That is sequence can uniquely determine structure

Question 74

9-24 | Anfinsen's experiment 7

Answer 74

-Its been pointed out not all proteins are capable of spontaneously folding, many small globular proteins can, many complex structures or proteins that need to end up in membrane have difficulty spontaneously refolding and talk about machinery in cell to assist proteins in refolding or to prevent proteins from improperly folding - If remove urea and beta-m simultaneously the protein chain was able to refold, the time that he had real difficulty was when he first removed beta-merca, formed disulfide links and then tried to get the protein chain to refold - We are asking shouldn’t the reaction be reversible? Yes forming disulfide links in presence of beta-mercaptoethanol is reversible, can form and exchange them in presence of trace amount of beta-mer, will swap with one another and beta-merca. Will assist with opening and closing, that’s what its doing/provides a catalytic function in facilitating the overall rearrangement of protein chained typology

Question 75

9-24 | energy assocaited with protein folding

Answer 75

Well first of all for proteins that don’t have disulfides don’t need any beta-merc, can unfold the protein in urea and then the protein will spontaneously refold in the absence of urea when the urea is removed, and same thing for thermal denaturation, heat up protein and unfold it, often but not always when reduce temperature the protein can reversibly refold again and that unfolding is a spontaneous process* so in this case it is spontaneous, the molecule is refolding in essence on its own all information required for folding coming from protein chain itself why call it spontaneous, its spontaneous and going into favorable delta g to form the native protein chain arrangement***

Question 76

9-24 How do we understand thermodynamics of protein folding- We know Gibbs free energy term is an outcome of both entropic and enthalpic contributions, what are those contributions when think about thermodynamics of protein folding?

Answer 76

- he just tried to convince us change in free energy for folding should be NEGATIVE AND FAVORABLE- certainly conclusion from Ainsefiend experiment - What will provide a favorable entropy change, HYDROPHOBIC EFFECT it’s the burial of hydrophobic surface area when we bring those nonpolar aa into the interior of the protein chain and sequester them away from solvent, that is a favorable negative -T delta S term, that is opposed by unfavorable change in the confirmational entropy (? make sure entropy) of protein chain itself, taken something random, flexible multiple states from random, multiple protein chain and said no not giving you any wiggle room told you exactly which one confirmational state you need to be I, locking you down that unfavorable entropy term -T delta S will be unfavorable!!! - overall net entropy term, net negative t delta S for folding is often unfavorable but not always, btu we see that when that net entropy term is unfavorable it can be overcome by a favorable enthalpy term for folding where will that come from? All those noncovalent interactions talking about, van der waals with packing, electrostatic terms/turns**** hydrogen bonds, all provide favorable enthalpy term for folding!!! – encounter usually these individual terms are quite large, and net G is small but quite favorable and sufficient enough to be folded under physiological conditions - ->so overall we consider folding of protein begins with protein in NONATIVE STATE_ random coil arrangement, high energy state under folding condition--> protein chain itself is in a high entropy state, undergo folding and favorable delta G to reach this overall low energy state and low entropy state of protein folded chain, folded state is low entropy

Question 77

9-24 | thermodynamics- creation of hydrophobic core is key force driving folding? Why does hit provide favorable entropy term?

Answer 77

a. Two main forces stabilize folding- hydrophobic effect and hydrogen bonding plays major role in providing favorable enthalpy*** term both of these arrows to right on slide are STABILZIING proteins, and countervailing that is unfavorable energy term associated with confirmational ENTROPY of protein chain, when see H bonds forming in protein structure, keep in mind those H bonds are very often are present in the denatured state, made with solvent and then they are just swapped with intramolecular H bonds with protein chain is folding, nevertheless H bonding does provide favorable enthalpy term so countervailing influences, conformational entropy opposing, HH bonding and hydrophobic effect favoring folding overall net free energy change is favorable but can see summing HUGE terms here and this is just a handwaving example these are reasonable numbers but again many protein molecules will see real differences from these particular values, btu we are summing v large individual energy changes to get a net favorable free energy change that is quite small* so what is the feature we can see here or the outcome? predicting protein folding is difficult and predicting protein structures in previous lecture is very difficult just like this

Question 78

9-24 How do proteins fold? look on slides

Answer 78

Proteins fold by progressive stabilization of Beginning of helix formation and collapse surface for the overall process of protein fol ized as a funnel (Figure 2.60). -percent of residues in proper native confirmation inc as the protein forms a more and more compact structure, finally end up as fold down funnel go down in free energy creating more and more compact states and end up in overall free energy minimum the native structure, so its not a global free energy random search** -the compacts states are favored and lend themselves to creating more compact structures that provide for a greater contract for proper native structure

Question 79

9-24 hydrophobic collapse model of protein folding or nucealation model of protein folding

Answer 79

hydrophobic collapse says unfolded protein chain has a tendency to form a compact species and within that compact species the protein chain forms secondary structures in the proper three dimensional arrangements, individual secondary structures form and then reassociate with one another pack against each other in the right way to form the properly folded protein - nucleation model says individual local region of protein structure form first with particular secondary structure elements associating with one another in correct orientation forms like foundation or scaffold upon which the rest of the structure can fold - Disulfide bonds there to correctly lock in place secondary structures, often find in places where exported from cell, because those disulfide bonds need to be present in oxidizing environment can help stabilize them, in cell conditions more reducing so often we don’t have disulfide bonds present***

Question 80

9-24 | chaperones- heat shock proteins

Answer 80

= heat shock protiens, cell makes these things under high temp conditions precisely when unfolded proteins created in cell** we will take particular look at chaperonins really large assemblies that refold proteins within a cavity*

Question 81

9-24 groel groes complex summary

Answer 81

- refolding chamber ATP driven hydrolysis of atp to form adp provides energy to unfold proteins within interior and allow them to refold, massive about million molecular weight entire piece of machinery and these two chambers operate in tandem** - each one of the two chambers can individually refold improperly folded proteins there is a cycle of unfolding protein binding along with ATP binding, once unfolded protein ends up inside chamber, cap can come down and close over the protein chain - ***ATP hydrolyzed to ADP provides energy to pull unproperly folded protein apart and give it opportunity to reassemble into properly folded species, then at that point chamber opens up and properly folded protein is popped out - In other chamber same thing happens, but these 4 steps are 180 degrees out of sink so coordinated but each chamber at a different point in the cycle as this continues

Question 82

9-24 | Is Protein folding is reversible?

Answer 82

-can argue in many cases it is, saw that within experience can unfold protein thermally decrease temperature and protein chains will refold so can be able to go up unfold proteins and decrease the temperature and get the proteins to fold, this is true but it does depend on environment and in particular if you are present in a very crowded environment with high concentration of proteins as individual proteins unfold and expose hydrophobic surface area, associate intermolecularly in such a way cant become disentangled and may not be able to refold as individual molecules, at low concentrations no problems, hydrophobic surface area created but under refolded conditions it can spontaneously refold**

Question 83

In animals, triacylglycerols are stored predominantly in what type of cells?

Answer 83

adipocytes

Question 84

The most abundant fatty acids in nature consist of unbranched chains of how many carbons?

Answer 84

12-20

Question 85

What type of fatty acid is most commonly used for biological energy storage?

Answer 85

long-chain saturated

Question 86

What type of fatty acid is most abundantly found in membranes?

Answer 86

saturated fatty acids

Question 87

Glycogen is a polysaccharide and has many water molecules associated with it due to its polar nature. Triacylglycerols are nonpolar. What is the relative energy storage density (kJ/g) of triacylglycerols to glycogen?

Answer 87

significantly greater, 6:1

Question 88

Glycogen branching occurs by the formation of what type of glycosidic bond?

Answer 88

α-1,6

Question 89

``` Which of the following is NOT found at elevated levels in lipid rafts? Choose one: A. Transmembrane proteins B. Glycerophospholipid C. Cholesterol D. Sphingolipid ```

Answer 89

Glycerophospholipid

Question 90

What does it mean that glycogen phosphorylase is a processive enzyme? A. Being regulated by phosphorylation makes glycogen phosphorylase processive. B. "Processive" refers to processing energy; therefore all enzymes in glycolysis are processive enzymes. C. Glycogen phosphorylase stays attached to glycogen substrates and continues to release glucose-1-P. D. There is a process by which glycogen phosphorylase matures; thus it is processive.

Answer 90

Glycogen phosphorylase stays attached to glycogen substrates and continues to release glucose-1-P.

Question 91

Which of the two figures above depicts an oil and why? (image A depicts one or two trans phospholipid, image B shows a bunch of trans and cis-phospholipids)

Answer 91

Figure B depicts an oil since there are numerous unsaturated lipids with cis double bonds

Question 92

drawing the helical wheel

Answer 92

• 7 atoms completes 2 full turns, 5 goes between 1 and 2, line conenctign 4 and 5 1 btw, 6 btw 2 and 3 7 would become three or 4 so once come to 8 it would be right under 1** very hard ot draw anything longer than 8 aa in this** if aa are all polar hydrophic fully exposed can create H bodns* can predict where helix will be in context of rest of protein* by time go form I to i+7 come back to same exact position technically 8 is right behind 1 here*

Question 93

9-28 recitation 3:30 | if helical wheel shows hydrophobic side chains...

Answer 93

- if helix surroudned by hydrohobic residues shown in yellow on slide, know this entire helix is surrounded on all side by hydrophobic side chains, extremely energetically unfavorable to put that in solution exposed to solvent SO ALWAYS BURY HELIX IN PROTEIN in hydrophobic core/ environment, if helix surrounded on al sides by hydrophobic side chains know for sure burried in sme hydrohobic regons, transmembrane protein lipid bilayer, always in hydrophobic envionrment can figure out if entire circumference is surrounded by similar amino acids, or certain side chains prefer specific environments

Question 94

surface exposed helix

Answer 94

- hydrophobic interacts with hydrpphobic - hydrophilic residues forms hydroden bond intearctionswith sovlent - called amphipathic helixces, can separate intot wo surfaces one predominatnly hydrophobic in nature the ohter side predominantly hydrophilic in nature

Question 95

fully exposed helix, can predict whta location..... in context of rest of the protein

Answer 95

exact opposite of first scenario, can have helix amino acids where uniform distributon of polar, charged or uncharged amino acids around helix, sovlent exposed helix entire helix expsoed to sovlent cna forn favorable hydrogen bondign interactions*

Question 96

collage i to i plus...

Answer 96

-Collagen left handed helix more extened don’t follow i+i+3 or itoI+4 because more xtend follow itoi+6 more stretchd out need smany more resideus to complete one full circle*

Question 97

ainsfield experiment 8 Mitra

Answer 97

o This protein Ribonuclease A extremely stable, in addition to a bunch o fnoncovelent interaction there are disfulide bonds, covalent bonds stabilizing regions of protein together which makes it extrmeley stable question is what would happen if somehow get rid of entire structure of protein -needs two types of reagents to do so, also there makes extremely stable, what would happen if you somehow get ird of its entire structure of the protein? You need two typesof reagents in order to do so, so if you use something like urea** urea can only break noncovalent interactions it cant break disulfide bonds**, complets with intra molecular H bonds so need coimbo of UREA AND REDUCIGNA GENT** btea-m reduces dsifuldie bonds to individual thiols*** so if you have taken a ribnucelase artiel add high concenreation of urea unfolds 3 structure to randomly structured polypeptide; say now if just remove BME from solution while keeping high concentration of urea in solution what will happen is BME removal creates an oxidizing environment where all these individual thiols in protein snap back and get disulfied bonded form*** -Urea is present in solution so no chance of noncovalent intearctiosn forming in protein so overall architecture of protein stabilized by both covalent and nocovalent bonds, noncovalent bonds lacking disulfide bonds/ surfur thiol groups of cystine* will most likely not be in the right orientations to form the exact same pattern of disulfide bonds as seen in native protein when remove BME and keep urea in solution**

Question 98

if keep BME in solution | A's experiment 9 mira

Answer 98

-if keep BME same high concentration remove urea completely don’t get folded protein back because in the absence of urea noncovlent intearctions of urea start forming bringin thiol groups close to each other to form disulfide bonds but every time form a disuflide bond will break open in high cocnengtration of BME and structure fall apart so to get back to native strucutre- have to remove urea BUT keep small amount of BME in solution, remove urea generate overall scaffold of protein to fold b/c only noncovlent strucures start forming, thiol groups start coming together in the right orientation to form disuflide bonds and lets say if you remove majority of bme now these thiol groups snap back into disfulide bonds in addition to right combination of disfulde bonds get a bunch of scrambled up disfuide bodns btw wrong pairs od side cahins!!! So in asbsence of BME once form those wrong disufilde bodns cant open up to form right structure of protein, ****if have small amount of BME in solution correctly keeps opening up disofulde bond and allows them to reform with correct partners so each cystine will finds its correct partner*** and form right pair of disulfide bonds

Question 99

so you may ask so once it has formed the right disufilde bond pairs, why don’t they open up in presence of BME to go back to misfolded structure? mitra A's experiment 10

Answer 99

Logic remove urea, kept small amount of BMR in solution because want incorrectly formed disfuilde bonds to be reduced back to thiol and reform correct pair, BME doesn't have a brain/ cant look at a protein structure and say whats right vs. whats wrong, will open up correctly paired cysteine residues and incorrectly, paired so what keeps correctly folded structure and rearranges incorrect structure? ***That is the whole pint of anisnfeld experiment by the way** and answer is NOT because oxidizing neviroment** with a little BME in solution still in mildly reducing evnrionment not oxidizing, so in mildly reducing environment all disulfide bonds will keep opening up and reforming, so bonds open up and form correct pair of disulfide bonds, when correct pair forms it DOESN easily open up and form a bad bond, under mildly reducing conditions ***because thermodynamically most stable** so the whole idea of this experiment is that the proposed that the thermodynamic hypothesis- out of all the possible structures a polypeptide chain can fold into its biological functional structure one in which has adapted to exist in its natural surrounding is its lowest energy state, what we call the native state** so that means any other structure that can form will be at a slightly higher energy level o So say all incorrectly formed disulfide pairs are at a higher energy, once form the right bonded combo of all noncovalent interactions and ds bonds energy at lower energy

Question 100

A's experiment 11 Mitra | Thermodynamic hypotehsis

Answer 100

SOO Anytime incorrectly formed ds bond in mildly reducing environment gets opportunity to form correct ds bond and goes down to lower energy level which in terms of thermodynamic energy is favored, once reach low energy sate even once open ds bonds would never jump back to the higher energy state** unless you provide additional energy to the system so in lowest energy state where correct ds bonds have formed even if BME transiently opens up disulfide bonds they immediately snap back and remain at that state because at that state sum of all interactions and both from noncovalent interactiosn taken together with ds bonds will put the molecule at the lowest energy state, can go from higher energy state to lowest energy state by forming right interactions when form the correct interactiosn even when open up transiently you will not jump back to the higher energy state, so that is why gradually* all the molecules over a period of time start rearrange incorrectly formed disulfide bonds and come to lower energy state** -****In presence of urea never able to form noncovalent interactions, which are necessary to generate overall scaffold of protein** if caked urea in solution break up ds bonds and then there is nothing there to hold the rest of the protein together, because if open up the rest of ds bodsn what holds protein together is all the other noncovalent interactiosn which will only form in absence of urea***, so need to remove urea and keep small amount of BME, so like hold a hgue part of structure together with noncovalent interactions allow disuflide bonds to rearrange within that scaffold***

Question 101

hydrophobic collapse model Mitar

Answer 101

, in this collapsed structure nothing 2 or 3 that is detectable all you see is big extendd moleculeor extended elongated moelcuel rapidly colapsign into compact form without any signfiican tstructure in that compact form, which purely happens because of the hydrophobic side chains of aa** trying to avoid contact with water moleculs* all try to interact with ach other and aovid contact with water moelcuels, once have compact structure more or less a simultaneously formation of 2 and 3 strucutre gives rise to final structure of protein, so main focus on this model si hydrophobic collapse of initial unfolded ensemble of protein structures has been observed in many rptoein models, start with unfolded state remove evyerhtign froms olutin that keeps it unfolded will first rapidly collapse into hydrophobic structure than rearrange within that structure give rise to tertiary structures folded protein o Framework model- when tart with fully unfolded polypeptide for these proteins what has been observed not rapid compaction with burred hydrophobic core***

Question 102

10-6 binding curve

Answer 102

- if protein more to the left curve, tighter binding, hitting that value of 1 at smaller concentraiton og lidangd indicative of tighter binding - easiest way to do it, go to 0.5 fractional sautration correspond to kd value of 10 nm for potein A and 60 fr prteoin B so smaller number of Kd*** value the tighter the binding***

Question 103

10-6 myoglobin has evolved to.. (color of meats at grocery store hint)

Answer 103

- to preferentially bind Oxygen over COs2, if inc ligand concentration of ligand Co2, even though doesn't bind well can push CO onto iron center and if we do that changes photophysical properties of that porphoringroup adn it will abosrb light thta gives it the beautiful red color, so if push CO into it it will look super fresh* does not taste better nasty packed with CO gas, like carbon monoxide treated tuna to retain color

Question 104

10-6 does R / T impact ability of oxygn to bind?

Answer 104

absolutely yes! as you bind oxygen, and you undergo these structural rearrangements the question is does it influence the ability of oxygen to bind? it abolutely alters the ability /strength of the binding -R state has a much higher affinity binding for O than T state, can see how slight rotation ofthe alpha beta subunits right interpahse brings these two things closer proxmity adn fills in this void in protein stucture

Question 105

10-6 | BPG

Answer 105

inlfuenes the binding curve of the hb pushes it ot the right, and can see this rightnad cure is in the presence of 8 nm PBG one way get adapted to high alttude hve in BPG concentration in uour circualtory system takes a while for egulation to be regulated dowanrds when climb to high elevations - in bindig site because of discomplementary presence of BPG stabilizes and favors the T stte reduces hte affinity of hemo for oxygen - it enhances ability of hemoglobin to deliver oxygen to teh ittuse

Question 106

10-8 | hemoglobin mutations

Answer 106

on slide see listen a number of mutations in beta and alpha subunit, these can result in a variety of different problems in hemoglobin function -down at the bottom can see some of these site mutations cause quaternary structure tetermatic assembly to dissociate - other cause an inability to bidn heme cofactor - others will crate changes in ability of hemoglobin to bind oxygen, one particular bad is sickle cell anemia

Question 107

10-8 | sickle cell anemia mutation

Answer 107

What goes on when mutate glu-val- go from polar negatively charged amino acid, to hydrophobic valine amino acid, this creates a site of addtionla hydrophobic surface area on the surface of the hemoglobin protein, as a result it distorts the quaternary structure** of these molecules in such a way that an individual tetramer now has a site of hydrophobic associton with another tetramatic assembly of hemolgobin f. Two tetramers come together in such a way that these interactions can be propagated that creates another site for interaction and as a result these chains of hemoglobin tetamers beign to line up and these polymrs of hemoglobin beign to grow, can see the site for this interacton btw two tetrameric stuctures is manifested at contact between ebta subuntis** phenalalnine and leucine on one tetramer interact with valine mutant on another beta subunit and we can see if this were a glutamic acid residue this hydrophobic assocaton would not take place*** 5. Further insight into sicke cell disease consequnces of this assocaiton a. Not only do individual tetramers begin to associate long polymer chains form and create strands and fibers withinr ed blood cell, alters affinity of hemioglobin for oxygen and because of prescen of fibers within red blood cells it alters the shape of the red blood cell itself, so the presence of this new hydrophobic amino acid creates a new patch on the surface that polyermizes the hemoglobin tetramers***

Question 108

10-8 Why create a concavity? When we want to create a hydrophobic binding site?

Answer 108

a. Its true that it could be concealed from solvent b. We want to hide hydrophobic – this is important because of aggregation! Why we want to tuck it away and make it not accessible! b/c= Aggregation like sickle cell will happen if put hydrophobic surface are binding sites in concavities precisely to avoid having them projected in a way that would lead to nonspecific assocation or undesirable quatanery interactions or sites of protein /protein aggregation*** c. So don’t want hydrophobic spot on surface, aggregates protein d. Sometimes we see binding site present at mobile region in protein, bidnign of ligand actually leads to a cleft closure as he two portions of protein structure swings down it will expel water from that site, why might that be energetically favorable to expel water from the surface of protein upon cleft closure? Studied this for a while leads to inc in total volume of system*** which may give us a hint as far as that goes e. Yes we do see structural rearrangement at ligand binding sites often associated with cleft closures, for protein/protein interactiosn to get good specifity see large surface areas for tight protein binding, if small interact surfaces typically sufficnet to bidn to small molecule but typically lead ot weak intearctiosn if trying to create protein /protein binding sites** f. think about creating complimentary of binding site of a protein, few amino acids at the bidnign site for protin concanavilina A protein bindinga sugar alpha mannoside, g. See hydrogen bonds from the amide NH group** and H bonds from side chains* and we see hydrogen bonds from water molecules entrained and captured within that protein bidnign site that play a specific role in mediating* protein ligand interactions* these are crystalgraphically resolved water molecules that play a role in allowing tight bidnign interactions to this saccharide

Question 109

10-8 How can the same enzyme function so differently relative to temerpature? What is difent between them to support the idffernt functions?

Answer 109

We can actually see remarkably similar organization of enzyme active site, can also see remarkably similar overall 3d structure for protien, same overall fold for example, but even some small number of alterations for amino acid sequence can give rise to huge changes in stability and thermal motion** a. so if we design in a few more disuflfide bridges or prolines at particular positions to breaka helix, to make it rigid where want to break ahelix, can have those sorts of mutations lead to a lot more thermal stability b. its very interesting to study the biophysical consequences of mutaitons that give rise to thermostability in these extremophiles** c. he asked us a questions imilar ot this on midterm- how modificaitons in protein give rise to less thermal motions and enhanced stability, same analysis give rise to less thermal motion and henhace stability- same analysis even smaller changes, a handful of alterations in sequence can play a big role in altering dynamics

Question 110

10-8 cofactors

Answer 110

A cofactor is an exogeneous** molecule it's not part of the protein itself, it is bound by the protein that allows it to carry out its function right, with regard to myoglobin and hemoglobin the ligand is oxygen that is what the protein is evolve to bind and transport the cofactor is the heme group that's the prosthetic group, that's the cofactor it is associated with the protein often times over a long time course to allow it to carry out its function, ike self splicing mrna** -some enzymes require co-factors is might be a metal might be a coenzyme in order to carry out the function, Jessie asked how do I know what a cofactor is rather than just a ligand?

Question 111

10-8 | Enzymes are catalysts

Answer 111

* Chemical reactions in cells require specific catalysts. * Enzymes are biomolecules (mostly proteins, but also RNA) which perform these functions. * Some enzymes require cofactors (metals, vitamins, coenzymes) for function. * Enzymes are classified by function they perform.

Question 112

10-8 kinases

Answer 112

transfer of a chemical group from one molecule to another I've already mentioned kinase has which take a phosphate group from ATP transfer it to another molecule that's phspho transfer reaction is carried out by transferases called kinases

Question 113

10-8 | equilbirum constants can be expressed as...

Answer 113

remember these equilbirum cosntants can be expressed as differneces btw on rate to off rate, tight bdinging and very engative delta Gs implies v small values of Kd*** which implies very small values of off rate, ligand goes onto protein and takes a LONG TIME for it to come off***

Question 114

10-8 why is it energetically favorable to displace water with a cleft closure?

Answer 114

Is it because the after cleft closure once ligand bidns would be hydrophobic on interior, more hydrogen bonding intearctions when water displaced? So this is another manifestation of oranizaiton of water at protein surface highly organized within these clefts* and as the ligand comes in and binds it displaces that water the cleft can close and push the wate rout and as it does so it undergoes transfer from beign at the protein interfafce to return to bulk solvent, where it has a lot more entropy* so this is an entropic driving force for bidning* when can create these cleft closures* so when open and no bidning yet the surface of those clefts would eb hydrophobic still, often times ther eis still hydrophobic surface area within those clefts** often combination ofa bunch of different amino acid side chains to provide complemtanry to whatever side chain/ substrate you are drealing with

Question 115

10-20 Typically found in fats and oils from both plants and animals. They are effective reservoirs of energy Why they do not store as fatty acids?

Answer 115

Fatty acids at higher reduction state relative to glycogen – more energy so as a result can be used to store more energy than if we were just storing away glcose, sucrose or sugars • Hydrophobic–do not need to be solvated when stored because removed charged sites on these triglycercols, removign charged sites creates entities that are more hydrophobic than free sugars and less charged than individual fatty acids so they do not need to be solvated hen they are stored glycerol three acyl groups created by esterification of fatty acids onto glyercol skeleton fantastic storage of fats if storing a bunch of glucose we would need to sequester no tonly sugar molecules but also wate rin order to hydrate them properly, but these hydrophobic molecules can just back into adipose tissue