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NBME 21 Answers

nbme21/Block 3/Question#37 (68.0 difficulty score)
A 5-year-old boy who has homocystinuria ...
The line on the graph labeled ‘B’🔍

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submitted by sahusema(145),
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S!UYG oYu ayrlel eavh ot loko at eht isax aleslb. The tnoqusie sysa het s'pt. iythsniatecon asentysh ivtciaty -i(xa)ys is RMOLAN whti na EACDESNRI oamtun fo xpilydaro aptpohehs ax)(x.i-s i-Yxas is hte sema as NMLARO, sxaxi- ihsfts to eht thrgi nhoisgw ACDENRIES rodalxyip atpphesoh

submitted by amirmullick3(61),
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Teh nereeaiwLv uBkr Polt is mynail idnrrpeeett by ist X and Y nicteehprsT .te uiostnqe ssteat tath iinvgg B6 ecensaisr eht cvityiat to roamln slelve, cihhw nseam teh tiavciyt oudlsh eb teh msea as the mlcnrae.on,eH eihrt Y tenicpetr ohdsul be the amse, so we rea bneeewt coicseh B and .C A nda D do ton ehav the smea Y citentper as aego .hlrrHmni otninotscercan of 6B sducea tavyctii to cmboee anorl,m dan os xmaV llwi ont eb nagcgnhi dna ew acn lcenac A nad .DA hsa a lewro mVax dan D hsa a humc rhgihe aTeh m.Vx cednfeifre wnetebe B and C si ni hrtei K.m vnioMg ot eth tefl on eth X saxi esakm hte Km rleow and tslle uyo atth aitffiny is oeshihr g uyo wduol ont dnee orme B6. uBt in our csae ifanyitf of eht zenmey rfo B6 is yarlle lwo, hhicw si ywh ew ende a ton roem 6.B

In y,mmuras we want eth mesa maVx nda a hirgeh K W.em nawt hte lamno"r" acitvity ames( vxma as ol)amnr dan ew ende rigehh mausont fo 6B orf ucsscse fi ysatfnio of hte ezyenm rfo 6B si plyobbar very owl.

oehiCc B sah a )m(K-/1 aeuvl resloc ot 0 whchi msane Km is olwre nda itaffiny of hte meezyn fro tsi ettussrab si reups Tsh.woil ksmae eness sa nivigg hghrei mtsnuao fo hte "v"teitoiemcp brsattsue 6B si gl.hneip

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 +  

submitted by lamhtu(118),
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dneigNe "hihger ocitoae"nsnrcnt fo het 6B ofr yemnez tyctivia is rnehato ayw of ysgina Km is rhihge eincs more is rdireuqe fro 21/ mvax .citaivty seIrandce Km lusvae terlus in K1m/ nbeig ramslle (ceorls to 0 on teh i-saxx,) cwhih is sttemeddoanr in warnes ohccei B.

A rm"o"lna ezenym vtiytcai in eth eprnseec of eihhgr 6B noneiostccnart emans that axVm si ton ignnhg,ca but Km 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). +  

submitted by navevan3(2),
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For shit stonuqe,i I saw innhtigk that htey adsek you emropca ot a m“aor”ln ,senpor os het xmVa ’tluownd hcenga scien htey seatt htat his yttivcai errsntu ot a mlnoar nlyO hte iaxx-s loudw nceis it lowdu eb at aceriedns cneintrocosnta in aon.rispomc

submitted by trazobone(29),

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.

submitted by usmleuser007(397),
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bermmeeR 1) ttceivmpeoi sbhiitorni ~ snlei sosCr ta cte tsyn)i2re- t-pNimonvecoeit ibirnosith ~ no leni cross

submitted by minhphuongpnt07(6),
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be ewaar atth the -xisxa ofr tish ionutseq si ton ""mK (tvi6B is not eht trtathtabssu 'e ]Bt[6/Vi1. ot i,fypims eecbsau the mxVa si nagno tyas het mae,s uyo tjus deen ot ineraces Btiv]6[ ot egt the mesa maxV a(ka ihtsf to tghri no teh -s)iaxx. Dtno' be dfoole randuo yb eth uonqeits iw.ertr

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! +  

submitted by an_improved_me(13),

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. +  

submitted by stapes2big(12),
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eTh qenustoi si aylbiscla tgtnies up a nttisioua laimsri ot htta fo a cveieiomttp rb.iiinoth Teyh ysa hatt iehghr eotancconntrsi of plxiadroy pehhostap esrinesca the viittcay of hte ezmeny to ramoln .eleslv rialymSi,l wnhe you heav a mcttepioeiv rintiihob and ceernsia teh ntaicrcnoneto fo t,taussreb hte ymneez nca heeivac teh aesm xaVm as if touhwti hte bniotiirh.

submitted by hungrybox(1035),
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noLg ersawn ,hdaae ubt ebar wtih .me

:IHTN v osokl dikn of liek y, heraesw k lokso orme elik .x

tnep-yrtiec = amx1V/

  • amVx si the uprpe iimtl on who stfa a ercantio si dczeyltaa by ze.mnesy

c-xinptteer = /m1K

  • mK is a ngnraki fo who oodg na eenymz si ta nndbgii tis su.brttesa An eemzny thiw a nkginar fo 1 si ebtter at inbgndi its tureassbt hnta an myeenz twih a nkrniga fo .5 woL(er Km = erettb )myenez

toeN ttha aV,xm as a eruesam fo nfcmr,peeoar anc eb lraedte hhorgut yman nht.sgi ,leinwaMeh mK is a tse taiiacrtcsrech of eht e,mzyne and contna eb dlee.tar

In hist pmaeelx, the zmneey armpernfceo Vma)(x is ecdsienar yb rsenaicing het vaimtni ocaoftrc os htat it eracseh a nma"ol"r tvy.itica Hwoerv,e teh emnyze si ltsli rynnelthei hstyti ude ot a glceontina et,ecfd so hte Km ystsa eth e.asm

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. +