B/B UWorld missed q's 3/6/23

Q49. Which of the following conclusions about protein folding is best supported by the information in the passage? answer: higher levels of RNase A protein structure are encoded in the primary structure

for this I just had to rely on the general idea that higher levels of protein structure are encoded by lower levels of protein structure

Q23. Which of the following structures is the Fischer projection for beta-D-glucose?

got it right, but wasn't sure why../ didnt know clearly how to approach here's what I have to remember: -first identify the anomeric carbon (the one that connects to the cyclic O and not on the side of the CH2OH substituent) -if we have a beta anomer.. think of what that means in a hawthorne projection: beta means the substituent is pointing up and is cis to the CH2OH group instead of down and trans to the CH2OH group -since we know the substituent (OH), is pointing up we know that should be to the *left* on a fischer projection; "right down; so left up" thus, leaves us with option B. the only one that has the anomeric C1 carbon with OH on the left side

Q53. During muscle contraction, the power stroke is the pivoting motion of the myosin head that pulls the actin filament inward to shorten the sarcomere. Given this, which of the following events most immediately precedes this power stroke motion?

got this wrong bc honestly forgot how the power stroke cross bridge cycle worked but got it noww -so the cell is depolarized allowing Ca2+ to flow in from the sarcoplasmic reticulum into the cytosol -Ca2+ binds troponin which pulls on tropomyosin to expose the myosin-binding sites on the actin filaments -actin filaments now exposed! the myosin head is bound to ADP + Pi in its resting position but it now can bind strongly to actin forming a cross-bridge -the Pi then dissociates from the myosin/ADP+Pi complex, which is what initiates the power stroke! The ADP then also dissociates from myosin near the end of the power stroke -the sarcomere has been shortened; the contraction is complete -ATP then binds to myosin head and its now at its low-energy conformation -hydrolysis of the ATP is what allows it to then be broken into ADP+Pi (back to its high energy conformation), waiting for another exposure to actin filament myosin binding site

Q48. Based on the passage, the functional RNase A recovered after the removal of denaturants and subsequent treatment with oxidizing agent displays which levels of protein structure? I. primary II. secondary III. tertiary IV. quaternary

this is silly... just had to read that RNase A is a monomeric protein so it doesn't even *have* quaternary structure since the protein regains its normal fx, all other structure orders must be working just fine (primary secondary and tertiary)

Q 31. Which of the following conclusions is most strongly supported by the data in the passage? answer: The Anap residue moves from a surface-exposed to a buried position upon ligand binding.

All the info I needed was right there in the passage! makes total sense now. I think I got scared of the unfamiliar graph, general lack of detail, and like wavelength for some reason lol.... basically - we're told the wavelength of Anap in water and of ethyl acetate (hydrophobic). I had to connect this info given, to the graph which *coincidentally* had the same range of numbers... 490 and 420 as wavelengths..., but basically, they as ligand concentration increases, the wavelength drops from the one matching Anap in water to Anap in hydrophobic enviro; so this supports answer c) That as the ligand binds, Anap moves from aqueous to hydrophobic enviro

Q41. At physiological conditions, the reactions shown in Figures 1 and 2 are best described as having: answer: a small delta G near 0 and a large negative delta , respectively.

I got this right.. but I actually didn't know the rule that tells me why this is the answer. I'll remember from now on that: -reversible reactions have small positive deltaGs. -irreversible spontaneous reactions have large negative deltaGs (intuitive, makes sense)

Q51. The peptide bonds maintaining the amino acid sequence were left intact, despite the presence of denaturants. This is because the hydrolysis of peptide bonds: a) thermodynamically favorable but kinetically unfavorable b) kinetically favorable but thermodynamically favorable c) requires water, which was not supplied d) requires energy input from GTP, which was not supplied

I had a conceptual misunderstanding here.. I needed to know that the hydrolysis of peptide bonds is actually thermodynamically favorable! it has a negative deltaG and releases NRG. the reason it doesn't just happen in the body (and has a long-ish half life) is that it is kinetically unfavorable (has a big activation energy barrier to proceed). so option A. Its not D because hydrolysis of peptides does not require GTP, its the formation of peptide bonds that requires GTP input

Q3. A study indicates that endurance exercise performance is related to type 1 skeletal muscle fiber characteristics that allow for efficient performance of aerobic respiration to generate ATP for extended periods of time. Given this, which of the following characteristics is (are) most likely to be associated with type 1 skeletal muscle fibers? High mitochondrial density High numbers of capillaries around the muscle fiber High myoglobin concentration in the muscle fiber

I had the right idea here! skeletal type i muscle fibers rely on aerobic respiration (thus need oxygen) to function well; they are endurance muscle fibers I just got in my head about myoglobin vs hemoglobin and decided to omit option III - but now I know myoglobin is indeed what we'd find in muscle fibers! myoglobin is specifically responsible for supplying O2 to muscle fiber cells (made flashcard to review Type I (slow-oxidative), Type IIa (fast-oxidative-glycolytic) and Type IIx (fast-glycolytic) fiber types and their differences)

Q26. Glucose and BHB can each be converted to two acetyl-CoA molecules. How many NAD+ molecules (both cytosolic and mitochondrial) are consumed by each process, respectively? a) two, one b) two, two c) four, one d) four, two

Ok.. was on the right train of thought but lost track of the NAD+'s here. So *glycolysis* itself uses 2 NAD+s (1 NAD+ -> NADH) reaction for each G3P molecule that comes from 1 glucose molecule; that reaction is G3P to 1,3 biphosphoglycerate. Now to BHB. We're told and shown the reaction, that it forms 2 acetyl CoA molecules. my confusion was that the reaction looks like it just used 1 NAD+ but that produced acetoacetyl CoA.. so my assumption was that another NAD+ would be required to break acetoacetyl CoA into 2 acetyl Coas. but this is not the case! In fact - the acetoacetyl coa just has to go thru the final step of beta oxidation in order to form the two resulting acetyl CoAs. that final step is legit just a split.. (probs involves attack by a CoA molecule). but no NAD+ is necessary in this piece. so its just 1 NAD+ used to generate 2 acetyl CoAs - supporting option C.

Q43. Based on the pKa of the histidine side chain given in the passage, which of the following best describes the charges of the His-His dipeptide side chains at pH 7? given in the psg that pKa of His is 6.

The answer is: most of the side chains will have charge of 0, small percentage will be positive. I was using the trick to think of pka as the baseline, and pH as being an enviro that will be more or less protonated. this is good! *However* have to remember that when you have acidic or basic amino acids, this will impact what the overall charge of the AA will be. for some (basic) AAs, the deprotonated state when pH >pKa is a neutral charge (Arg, His, Lys). bc the protonated state is positively charged. for other AAs, the protonated state is when the charge is 0, and the deprotonated state makes the charge -1. (Glu, Asp, Tyr, Cys) steps to solve this: -identify whether the pH is higher or lower than pKa - will it be protonated or deprotonated? -what does that mean for this aa? if its basic, what charge results out of deprotonation/protonation?

Q6. The mutation G12C results in: a) achiral residue being replaced with S residue b) residue w/o RS designation being replaced by D residue c) D residue replaced by L residue d) residue without D/L designation replaced with an R residue

this was a straight discrete knowledge question. MUST know ins and outs of aa's. simply had to know that: -Glycine (achiral) does NOT have an R/S or D/L designation -all AA's have S absolute config (S = Status quo as far as abs config goes!) *except* for Cysteine which has an R config. (Really big Cyst on my amino acid tree...)

Q58. A researcher seeks to study proteins found in the mitochondrial matrix. Based on the information in the passage, which of the following procedures will be most successful at capturing and retaining mitochondrial matrix proteins on a chromatography column? Loading the protein mixture into: a) anion exchange column at pH7, high salt buffer b) cation exchange column at pH7, high salt buffer c) anion exchange column at pH7, high salt buffer d) anion exchange column at pH7, high salt buffer

was sort of a total guess here.. also probs low on time and randomly guessing bc of that now I know for this - just had to think of what the pH of the mito would be ( it would be acidic bc of all of those H+ in there) and so, in an acidic enviro a protein would become protonated and thus positively charged so im looking for the answer option that attracts positive charge remember: "you are what you attract" - so cation exchange column is what attracts cations next, salt level. Salts are used to elute chromatography columns.. if the q is asking what would help bind to the column, we *don't* want high salt levels.

Q24. Based on the passage, hepatocytes (ie, liver cells) that are producing BHB are also most likely to have: a) downregulate PFK2 b) downregulate phosphorylase kinase activity c) upregulate acetyl CoA carboxylase activity d) upregulated glycogen synthase activity

I knew I was between A and C here.. I incorrectly chose C. so c is wrong because glucose -> acetyl Coa, and then acetyl Coa becomes malonyl coA via acetyl Coa carboxylase, and then this malonyl Coa inhibits beta oxidation (which is like ultimately breakdown of stuff/glucose to ATP energy). we want to INCREASE glucose conc in this question; so we would want to INHIBIT the acetyl carboxylase enzyme to prevent malonyl coa formation (preserve our glucose source) A is right as I thought bc PFK2 activates PFK1.. it stimulates glycolysis to occur; so we'd want to inhibit this to maintain our stores of glucose

Q40. Hydrolysis of creatine phosphate can be coupled to phosphorylation of ADP because free creatine: I. is stabilized by resonance. II. tautomerizes to a lower-energy molecule. III. is relieved of charge-charge repulsion.

I only I wasn't super sure here.. I saw there was resonance so def I, I thought there could be tautomerization, but that will not really happen here (no ketone/enol like situation going on; its just movement of the positive charge shifting to different N's) - so not II and regarding charge-charge repulsion, the negative phosphate is actually next to a positive nitrogen... so there isn't really a charge charge repulsion, if anything there is somewhat of attraction

Q57. Each of the following may explain why the proteome pI distribution is bimodal EXCEPT: a) proteins are typically most active when pH = pI, so a bimodal distribution allows a different population of proteins to activate upon pH disturbance b) proteins are typically least stable and more prone to aggregation when pH = pI, resulting in an underrepresentation of PI values of 7.4 c) most proteins are more soluble if their pI is either above or below but not equal to the surrounding pH d) the pKa distribution of the AA side chains is bimodal, leading to a bimodal distribution of pI values in the proteins they compose

I really didn't know how pH affects protein activity so rly just guessed. now I know that when the pH = pI of the protein, the protein is NOT stable; its electrically neutral.. it can't interact with the polar substances around it.. so it won't really dissolve; instead it will form aggregates.. so overall it is not active if pH is higher or lower than pI - this is good for protein activity! the protein will either be positively or negatively charged based on whether its basic or acidic; so it will be able to dissolve in solution and interact with substrates thus, most proteins have a pI either greater than 7.4 (basic), or less than 7.4 (acidic) to keep in sln and prevent aggregation

Q5. RASAL is a protein that regulates Ras activity. When RASAL binds Ras, it lowers the energy of the transition state of the GTP hydrolysis reaction. RASAL is best described as: b) inhibitor of Ras's enzymatic fx and of cell growth and division c) activator of Ras's enzymatic fx but an inhibitor of cell growth and division

I was so close here, I had a really solid conceptual understanding here! was just a matter of realizing what exactly is meant by the "enzymatic fx of Ras" the Ras's enzymatic function consists of its ability to perform the hydrolysis reaction. so even tho catalyzing this hydrolysis reaction renders Ras inactive, this still counts as "activating Ras's enzymatic function" so option C.

Q29. If the molecular weight of the Protein X polypeptide is 50.0 kDa, what is the theoretical maximum amount of full-length Protein X that could be produced by adding 50.0 μmol of the chemically synthesized Anap mixture to a culture of E. coli? (Note: Assume all other amber stop codons of the E. coli genome have been removed via mutation.) a) 0.5 g b) 1.25 g c) 2.5 g d) 5g

Lol so I'm well aware that I read this math, skimmed the passage - saw nothing that I could immediately recognize to solve this.. guessed+moved on. maybe a good method? one to come back to. but NOW I know - a little more skimming of last para would tell us that 1 molecule of Protein X incorporates 1 Anap molecule! so the 50 umol of Anap used means we are working with 50 umol of Protein X. The trick here is that we are also told that the Anap they produced was optically inactive (which I know means there are equal amounts of the L and D configurations present). *So* this actually means that only half of it is usable to form Protein X , because only the L amino acids are used to make proteins /are recognized by enzymes. Thus, 0.25 umol of Anap used = 0.25 umol of Protein X * (1 mol / 1000 umol Protein X) * (50,000 g/ 1 mol of Protein X) = 1.25 g we know 50,000 g is the MW of the protein x bc we're given MW of 50kDa = 50,000 Da = 50,000 g/mol 1 Da = 1 g/mol so we land on B) 1.25g

Q9. The two time constants measured in Experiment 2 were τD and τI. These time constants were pH dependent and pH independent, respectively. Given this, the most likely interpretation is that: a) tD represents the ligand binding reaction and Ti represents the covalent attack b) tD represents the covalent attack and Ti represents the ligand binding reaction c) lifetimes represent ligand binding and covalent attack but cannot be assigned w/o further info d) lifetimes cannot be correlated to ligand binding or covalent attack bc the fluorescent AA is far away from the binding and attack site

answer is B- got it correct but had literally completely guessed. I understood what the q was asking: basically, which is pH dependent and independent for both the ligand binding part of the rxn, and the covalent attack? had to just realize that per the reaction 1: it looks like for this nucleophilic rxn we need the cysteine S to have that negative charge (so it must be in pH dependent, namely basic/deprotonated conditions) so that it can attack and form the product. and ligand binding here does *not* depend on pH! we actually are told this in the psg... had to pick up the detail here that "the Kd for ARS-853 (i.e. the ligand) interactions with RasG12C (the enzyme)... was not significantly pH dependent" - bam they sneakily told us right in the psg (so must be option B)

Q56. A typical plasma membrane protein with a pI value of 8.0 is trafficked to an endolysosome prior to its degradation. How will the membrane protein's net charge change as the endolysosome acidifies? The protein's charge changes from: a) positive to negative b) positive to more positive c) negative to positive d) negative to more negative

answer is B. again it was a question with pKas so that triggered my flight or flight lol but basically - if pKa is 8, and its placed in a plasma membrane normal enviro, it will become protonated (its in a solution that is more acidic than its pKa) then if you put it in an acidic endolysosome, it'll only get more protonated (more positive) I think my thinking was way too simplistic in picking A thinking that "oh the pKa is 8 its a basic aa and basic aa's are positive; so it'll go from basic to acidic i.e. positive to negative"; instead next time I will use this technique of thinking about how the environment will affect the protonation of the molecule, and go from there); will it be getting more protonated/positive? or less

Q36. Based on the information given in the passage, the radius of the P13 channel pore can be best described as being: a) between 300-400 pm b) between 400-500 pm c) between 600-700 pm d) greater than 1000 pm

answer is C. again totally had a great conceptual understanding... I just missed it bc of a silly oversight/not thinking thru.. so it was good that I recognized that at 700-900 pm radius, we're back to no obstruction and baseline conductance (i.e. NEs are prob not getting thru to the porins bc they are now too big) I had to realize that this then means that the radius of this P13 pore must be 600-700 pm. this allows the smaller ones in, but won't allow the big guys. I think I knew this, but accidentally said 300-400 looking at the MW (like the actual names of these NEs which are called like PEG-300 and PEG-400) but the q asked about the radius

Researchers purified protein X, experimentally measured its pI value through isoelectric focusing (IEF), and found that it differed significantly from the theoretical pI calculated as described in the passage. The theoretical pI value may have differed from the experimental pI if the theoretical calculation did not consider: posttranslational modifications on protein X. the tertiary and quaternary structure of protein X. the pH in the subcellular localization of protein X.

answer is I or II only. didn't know this /rushed/ felt uncomfortable with the pIs and pKas and pH again lmao but feeling a little better now I needed to know that protein structure and posttranslational modification can both affect the pI of the protein ex: if the protein has a postttrnaslational modification like a negative phosphate, this will impact the charge of the overall molecule causing it to have a different pI experimentally and then also: the structure of the protein can impact pI. what if in a the actual protein (as opposed to individual isolated amino acids) there are 2 amino acids with opposite charges near each other that form electrical connections and mitigate overall charge? what if there is destabilization of some part of the protein due to charge-charge repulsion? these things can impact the equilibrium of proton dissociation, and thus impact Ka, or pKa (resulting in altered pI for the molecule experimentally, that may not have been considered theoretically)

Q32. Based on the data shown in Figure 2, Protein X can best be described as: a) cooperative protein, Hills coefficient < 1 b) cooperative protein, Hills coefficient = 1 c) cooperative protein, Hills coefficient > 1 d) noncooperative

so from this question I learned lol that *only* positively cooperative curves will be sigmoidal negative cooperativity will show up more hyperbolic- NOT sigmoidal. so despite the trend looking like it is negative; it is demonstrating positive cooperativity: as the concentration of ligand increases the wavelength decreases as the fraction of Protein X increases (shallow, steep, shallow) = S curve = positive cooperativity (option c).

Q7. At a pH of 7.0, the G12C cysteine residues of mutant Ras proteins are, on average: a) more deprotonated than free Cys side chains b) less deprotonated than free Cys side chains c) protonated to the same extent as free Cys side chains d) full 100% protonated

the answer is A. I wasn't really sure how to approach this also because pKas and q's about acid-base protonation/deprotonation scared me. BUT! got it now. its easiest to think first about how protonated is the solution? at low pH you have plenty of H+s available second thing I had to remember is that the pka is the pH at which you have an equal amount of protonated and deprotonated versions of the molecule; *so* if you then make the solution more acidic (increase the amt of H+s) you will increase the amount of the protonated version of the molecule that exist. the bigger the change from the molecules pKa to the pH you're changing to, the greater the difference you'll see - so like here, since there is a full 1.0 pH drop from 8 to 7 for the free Cys residues, there will be a lot more protonated Cys in this pH7 sln than the G12C mutant proteins that only have a 0.6 pH drop and will thus have less protonated and *more deprotonated* molecules in sln (option A)