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NBME 21 Answers

nbme21/Block 3/Question#37 (65.0 difficulty score)
A 5-year-old boy who has homocystinuria ...
The line on the graph labeled ‘B’🔍
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 +11 
submitted by sahusema(138),
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GU!SY ouY elyral ehav to olok at hte sxia esalbl. Teh uesnqoit ssya hte .tps' ihycntstnoeai nsseytha tacitviy y)i(axs- is LMARNO with an ESIDAENRC utoman of idrolaxyp sahphpote ax(-x.si) axYs-i si hte eams sa OLR,MAN -xisax sshtif to the rhgti oniswgh RNCAEDISE darpoxliy psoahehtp




 +10 
submitted by amirmullick3(59),
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hTe eervinLawe urkB loPt si aiymln ettndpreeri yb ist X dna Y t.seiereT ncthp onesuiqt ssetta atth inggvi B6 rnesaesic the cyttavii to lronma se,vlel hiwhc nmesa teh civatiyt sdulho eb eht meas sa eht nrelcema,n o.H irteh Y citntrepe hsoudl be the s,aem os ew are ebeentw hcceosi B nda .C A nad D do ont eavh teh easm Y tectierpn as .igr noeHmlarh orinntccseaotn of B6 asucde iivacytt ot mcbeoe on,lram adn so Vmax lliw nto be nnacgigh dan ew acn nacecl A nad .D A ash a woler Vmxa and D has a umhc iehhgr hVa T.mex cfefidnere eetbwen B nad C si in htire m.K goMvin to eth ltef no het X sxai mekas the Km oelrw nda sltle yuo ttah aftfniiy is hhsgeo ir uoy wulod nto dnee emro .B6 Btu ni uro ecas nftiiyfa of eht ymneez fro 6B si ryleal lw,o ihchw is yhw we eden a ont omre .B6

nI ,mmarysu we want teh seam xmaV nad a hehirg . WmKe wtna the mln"ora" ctatyiiv mse(a mvax sa ra)nlom dan ew eedn rihegh atosnum of B6 rof seuccss nf fiiyatso fo teh zemyen orf 6B si pblraoby vrye .wlo

Cheico B ash a /)(m1-K elvau crosel to 0 ihhwc sanme Km si olrew adn ainifytf of eht zeynem fro tsi raesbutst is perus h Twios.l msake snees sa vggnii hgrieh atonsmu of teh m"pvote"ctiie ettsuasbr 6B si l.npihge

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). +10  
apurva  Had the X axis be 1/km, the answer should be C +  



 +9 
submitted by lamhtu(113),
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ingNdee ehhrgi" "noresncanttoci fo the B6 for yemenz aitticyv si ornheat ayw fo isynag Km si hirehg nseci erom is eqrdiuer ofr 21/ vxam iyatic.vt rendesIca Km usavel rlutse in /m1K igenb srellma erclos( to 0 on het )xs-,xai chwhi si daotsemtednr ni wsanre echcoi .B

A nro"aml" zyenme titaivcy in het pesecren of eirhhg 6B stnanicrcntoeo sanme thta maVx is ton anicng,gh tub Km i.s

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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For sith ot,qensui I aws gniktnih atth yhet deksa you earpmoc to a lo“m”rna osen,rp so hte mVxa wdntol’u hgenac inecs heyt setta taht ihs ticytavi nrtreus ot a mranlo elvle. yOnl teh xxias- ldwou nesic ti olwdu be at raeniescd iecncntoonarst in nos.riamcop




 +0 
submitted by usmleuser007(377),
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ermbmeeR 1) itvetpcimoe triinoihsb ~ enils srsCo ta s-)yeitne2rtc oceipNiv-nttoem obrishiint ~ on lnei rssoc




 +0 
submitted by minhphuongpnt07(6),
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eb earwa atht hte -asxix for sith qnesiuot si ont Km"" (vi6Bt is tno eht etbattashruts ' [/BV1]i6t. to ,pyfisim ubeeacs het Vmxa is aonng ysta eth ms,ae uoy juts eedn to enceirsa B[]tv6i ot tge het asme aVxm (aka stfih ot rithg no hte x)xsa.i- n'Dto eb eoodfl auonrd yb the qesuniot eitr.rw

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! +  



 -1 
submitted by stapes2big(12),
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The ueinqtso is aiaclslyb sigtten up a iaiuttson ilarmis ot that of a vcopietitem niihrbto.i yTeh asy htat rghieh rcntansiotonec of prdlaxioy hsepotaph resiscnea hte atciyvti of teh yneezm to oanmrl svlle.e mlialSriy, newh ouy hvae a oetiictvmpe hiniiortb and ncerisea eht nronitcntoeac fo atsrtue,sb eth menyze nca vchiaee the amse axVm sa if thiuwto the .irinohbit




 -5 
submitted by hungrybox(977),
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gnLo nswrae ,haeda btu reba htiw me.

T:HNI v kosol kdni of klie y, erhswae k olsko rome kile .x

tt-iyercnpe = V1xa/m

  • Vamx is eth perpu mliit on woh tsfa a crieotna si aldzaeyct by zmenys.e

rp-ttxnieec = Km/1

  • mK is a innakgr of woh ogdo na emzyen is at iibndng tsi erbtsatu.s nA zeymne tiwh a akngrin of 1 is betrte at inbindg tsi tsbrueast ahtn na eyzemn wtih a kringna fo 5. wr(eLo Km = erbtte eme)yzn

Noet hatt ,Vaxm as a ereausm fo e,nremcaprfo anc be erltade ghthruo namy signt.h ,hwaiMleen mK is a ets ictteaascihcrr fo eht ,ynzeem nda notanc be ld.etare

nI sith am,xlpee eht eyenmz feoamprcern V()xma is aesrdniec yb sinerincga the mivntai fraotcoc so htat it aerhcse a ro""almn .ititcayv ,Hrvoewe het eeyzmn si ilstl nynitrheel tyhist ude ot a icaeolntgn etefcd, os hte Km syats het a.ems

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 +11  
sbryant6  Yeah this explanation is wrong. +