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nbme21/Block 3/Question#37 (68.2 difficulty score)
A 5-year-old boy who has homocystinuria ...
The line on the graph labeled ‘B’🔍,📺
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 +12 
submitted by sahusema(145),
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GUYS! uoY lylrae heva to ookl ta hte xias sale.bl The noteqius sasy eth t.'sp toaicyhnetnis nytsshae vytticai )yai(-xs si ALRMNO twhi an EDARICENS munota fo oplxrdyai peasthpoh x(i)sa.x- Yi-axs si teh esam as ROML,NA axxi-s fthssi ot eht rhtgi whisgno NEAIDERCS pdioxayrl sophatphe




 +10 
submitted by amirmullick3(60),
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eTh eeervnwaiL Brku Ptol is yilamn ptenetrrdei yb tis X nda Y erntTeshcip.e t itensouq satest hatt ggnivi 6B secinraes the itytavci ot malnor ,lelvse whihc ensma eth iacyivtt ushlod be het emas sa teh , eeaonmHclr.n hirte Y ipcnertet odlshu eb het se,am os ew rea weebtne scoecih B adn .C A nad D od not vaeh het smea Y erecnttip sa gnia lrHr.meho ocnnnceoiarstt fo 6B deausc ivyticat to meeboc orm,nla adn so xVma lilw nto be ighagcnn dan we nac lcanec A nad AD . sha a rweol axVm nda D sah a hucm hiehrg Vxm .Teha feferiecdn weneebt B and C is in treih K.m Mingov to hte tlfe no eth X iasx esmka eth mK erlwo and estll ouy that infyatfi si giosher h yuo louwd ton eedn erom .B6 utB in uor acse fitinyfa of het emnyze fro B6 is layler wl,o hicwh is hyw ew ened a ont roem 6.B

In mamsu,ry we wtan eht mesa aVmx adn a hihrge Wm. Ke twna eht ra"lo"nm tctiviay (msea axvm as )lronma dna we deen giherh ouamsnt of 6B ofr ssuescc at siffoiyn fo teh enemzy fro 6B si yplrobba vrey o.wl

hoCcei B has a -1/K()m vleua clrose to 0 ihwhc enams Km si rwole dna nfyifiat fo teh mzynee fro sit ebsautrst si eprus how ls.iT seamk snees sa nggiiv hhgeri uotanms fo the t"impeivtoe"c ssuttabre 6B si ielphg.n

apurva  This explanation is wrong!!! X axis is not 1/km in the question, it is 1/pyridoxal phosphate. (1/km should be 1/homocysteine) Also the question is asking about allosteric activation of cystathione synthase due to addition of pyridoxal phosphate. Which means the Km should decrease (affinity increases after addition of pyridoxine). Considering the x axis to be 1/pyridoxal phosphate, now apply simple logic of maths, increase pyridoxal phosphate = bringing line more close to zero (because it is -1/pyridoxal phosphate). +11  
apurva  Had the X axis be 1/km, the answer should be C +  



 +9 
submitted by lamhtu(120),
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digeeNn herhi"g tecoannns"corti fo hte B6 orf enyzme yttiiacv si rheaotn awy of ignays Km si iehrhg icsne reom is ireqrdeu for 2/1 axvm ttyvii.ac esanIdcre Km eusalv rstuel ni /Km1 gnieb lremasl (soercl to 0 no eth -)s,xiax hwcih is rsteeoatmdnd ni rswnea eiccoh B.

A rao"nlm" nyzmee iiycvtat ni eth pcsenree fo irhegh B6 atoncroietnncs nmaes ttah Vmax is ont ch,aingng tbu mK s.i

d_holles  In other words, Normal enzyme activity = Vmax Adding more B6 to a mutated enzyme will improve it back to normal (Vmax stays constant, so Km will decreased due to ↑ affinity of the enzyme). +  



 +2 
submitted by navevan3(2),
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roF hist so,qniuet I swa nhkgntii atht yeth sdeka uoy camroep ot a la”rnom“ n,spreo so the amxV ln’uotdw hncaeg nsice htey staet htta shi aytivtic snretru ot a maornl lveel. ynlO eth x-isax uwdol ensic ti uwdol eb at iaendresc rntctaionnsoec ni om.pniacros




 +1 
submitted by trazobone(35),

Some of these explanations really doing the most. If the enzyme is able to increase to normal activity with a higher level of the substrate, that means we’re in a competitive inhibitor situation. Competitive always crosses on the graph, so we can eliminate A and D.

Now we analyze Km. We want small Km’s for potent enzymes. Size does not always matter 😉. Except in this case we don’t want that potency, we want to look for a 1/(really big Km), (value that is closer to zero), bc we want a weak big Km to be able to overpower the competitive inhibitor. Comparing the remaining 2 choices, B has a Km value closer to zero so there’s your answer.




 +0 
submitted by usmleuser007(397),
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ereRbemm )1 tcipoeievmt ibitiroshn ~ lesni orssC at eritt-eyns c)2 ticovmoNpe-tine nstrbiiioh ~ on elni sscro




 +0 
submitted by minhphuongpnt07(6),
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be raeaw htta het xiaxs- orf htis iotsuneq si not K""m tiB(6v si otn eht stttseth'uabar i]6B.Vt/1[ to i,mfpyis useebca the xVma is ngano tsya hte asm,e ouy utsj dnee to irecsane ][i6Btv ot get teh asem xVma aka( shfti ot irght no teh )aisx-x. t'onD eb dlofeo uodanr yb teh unqtseoi wrre.ti

poisonivy  this makes sense to me, X cannot be Km!! because if it was Km then line B would be showing a "competitive inhibitor" effect, which cannot be the case since they state that the enzyme activity increases to normal after B6 administration so it favors the reaction instead of decreases it. I got this one wrong of course, too tricky for me! +  



 +0 
submitted by an_improved_me(16),

So i found this question ridiculous... the explanations provided did not really clear it up for me, but i think i understand it now (please correct me if i'm wrong).

First and foremost, I'm not even sure if this is truly a lineweaver-burke plot, since the X-axis is not 1/Km.

But, more generally, I tried to understand what happened at both extremes of [PLP] and Velocity.

PLP: Ask yourself: where on the graph is [PLP] highest? Well, if the X-axis is equal to 1/PLP, PLP would be at its highest at the value closest to 0 on the x-axis. (think: what is 1 divided by a billion? VERY CLOSE TO ZERO). This would correspond to a point along the y-axis for the enzymes activity. In other words, PLP is at its highest concentration at the graphs Y-intercept.

Remember what we also said: when the mutated enzyme is around lots of PLP, it basically functions like a nml enzyme. This means, that the Vmax of the mutated enzyme is equivalent to the Vmax of the normal enzyme. Therefore, The 1/Vmax of both enzymes are equivalent... at the y-intercept! This is big! If you get this far, then you can eliminate (A) and (D), since these graphs have different Y-intercepts (different V-maxes) for their respective enzymes.

Moving on...

Velocity (V): When V is at its highest (i.e. Vmax), the value of 1/V would be at its lowest. (Again: think of it practically: what is 1 divided by 1 billion? ~0!) Therefore, the value of 1/V is at its lowest when the y-axis is equal to 0. Therefore, The Vmax of this occurs when the y-axis is close to zero... Heres the thing tho.. these enzymes require different levels of PLP to operate at V-max. Remember, the V-max of the MUTATED enzyme will happen at a much HIGHER [PLP]. Or conversely, the V-max of the NORMAL enzyme will happen at a much LOWER [PLP]...

an_improved_me  Meaning, when we look at the X-intercept (when V is highest), the mutated enzyme will require more PLP. More PLP means a "lower" value on the x-axis (closer to zero) relative to the normal enzyme. +  
an_improved_me  Putting it ALL together: y-intercept will be the same for both mutated/normal enzme x-intercept will be closer to 0 for the mutated enzyme, relative to the normal enzyme. The only answer that has both correct relationships is (B). I think i spent 5 minutes on this question trying to figure it out. I probably would have needed 10 minutes to actually get there. Hopefully now I can get it quicker if something like this shows up on the real thing. +  



 -1 
submitted by stapes2big(12),
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ehT iuenstqo is cblasyila steting up a atisoiunt isralim to hatt of a tteimocpive hitobnir.i hTye yas thta hhrige ttnioscrncnoea of ordlipyxa easphohtp rsncsaiee hte iatciytv fo eth eemzny ot mloanr .svelle malS,liiyr newh uoy aveh a iotitpemcev nthioirbi and ciaesrne the otcocrannniet of tbes,tsuar the emyenz acn hcevaei the seam Vmxa as fi oihuttw hte trbiihino.




 -5 
submitted by hungrybox(1044),
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oLng anserw aea,hd utb aber thwi m.e

H:TIN v osokl indk fo leik y, whseaer k ksool remo liek x.

tcneterpiy- = 1/Vxam

  • mVxa is the rppeu tiilm no owh saft a neitroca si tlazecady yb .esmnzey

t-reicntpex = K1/m

  • Km si a knragni of hwo ogdo na zmeney si ta ibinngd sit ebtrtss.ua An yenemz htwi a nngiakr fo 1 is terebt at nngbiid ist asbsteurt nath an ezenym hwit a nakring fo 5. wrL(eo Km = etrbte zmyne)e

tNoe taht x,Vma sa a emerusa of rpemrfneo,ca nac be dalreet ghhurto myna nhist.g en,hlwMiea mK si a ets hcicttraceiars of eth mee,zny nda ncnaot be atrde.el

nI iths pm,eealx eth eenzym omracepfren (xaVm) si draeeinsc by nincrsagei eth vtiiamn orcctfoa os thta it eachesr a nlr""moa tt.yivcia wov,erHe het zeynme si itlls teynhrinle tiyths ued ot a naielonctg detfec, os het mK sstay eth aes.m

mnemonia  Awesome. +  
ht3  wait line B shows the vmax doesn't change and that the km is getting larger (enzyme is still shitty so larger km) so -1/km would be a smaller number and would approach 0 +1  
lamhtu  You say Km cannot be altered and its staying the same, but the answer of the graph demonstrates a higher Km value. Needing "higher concentrations" of the B6 for enzyme activity is another way of saying Km is higher since more is required for 1/2 vmax activity +13  
sbryant6  Yeah this explanation is wrong. +