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NBME 21 Answers

nbme21/Block 3/Question#37 (58.9 difficulty score)
A 5-year-old boy who has homocystinuria ...
The line on the graph labeled ‘B’🔍
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The Lineweaver Burk Plot is mainly interpreted by its X and Y intercepts. The question states that giving B6 increases the activity to normal levels, which means the activity should be the same as the normal. Hence, their Y intercept should be the same, so we are between choices B and C. A and D do not have the same Y intercept as normal. Higher concentrations of B6 caused activity to become normal, and so Vmax will not be changing and we can cancel A and D. A has a lower Vmax and D has a much higher Vmax. The difference between B and C is in their Km. Moving to the left on the X axis makes the Km lower and tells you that affinity is higher so you would not need more B6. But in our case affinity of the enzyme for B6 is really low, which is why we need a ton more B6.

In summary, we want the same Vmax and a higher Km. We want the "normal" activity (same vmax as normal) and we need higher amounts of B6 for success so affinity of the enzyme for B6 is probably very low.

Choice B has a (-1/Km) value closer to 0 which means Km is lower and affinity of the enzyme for its substrate is super low. This makes sense as giving higher amounts of the "competitive" substrate B6 is helping.

maleehaak  amazing explanation +1  
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). +8  
apurva  Had the X axis be 1/km, the answer should be C +  



 +9  upvote downvote
submitted by lamhtu(86),
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ingeNde "ihhreg oaccetnnsortin" of teh B6 ofr emnezy itaticvy si roatnhe ayw fo gniays mK si rghihe iscen reom si dieerruq orf /21 xmav aycitt.vi rdsIeneca Km aevusl rsletu in /m1K bnige mellrsa o(rlecs ot 0 no teh )xxs-ia, chwih is enstradmteod ni wreasn ceocih B.

A n"a"orml ezeymn ctyiavti ni eth esnpeecr of gherhi 6B sncootintanrce mesna tath Vxam si ont ,ngainchg btu mK 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). +  
wolvarien  why the answer is no C ? +  



 +8  upvote downvote
submitted by sahusema(117),
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SYUG! oYu yllare ehav to olok ta het xias blae.sl hTe usieqnto sysa teh .spt' sciynteiohtna enhtassy tcitvyia -(iyx)sa is NMLRAO with na SEAINECRD otmuan fo adxylroip tshaeophp s(xx)i-.a i-Yxas is eht asem as O,NMALR xsixa- sithsf ot eth griht siogwhn IASEECDNR adlyxprio sophhpate




 +2  upvote downvote
submitted by navevan3(2),
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rFo ihst s,itneuqo I swa nhiiktng htat ehyt asedk uyo rcmapoe to a lam”orn“ sonepr, so hte xmVa n’twlduo enaghc cnsie ythe aestt htta ish ctviaiyt rerunst ot a mlaron lele.v Olyn teh x-saix ulowd enics it uolwd be ta inedseacr nntstreccooina ni omsnropaic.




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reemmeRb 1 ) vctoiieeptm onirisbith ~ neisl Csosr ta ce-)yeinr 2stt em-pintNvetcooi rsiohbinti ~ no lnie srcos




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eb weraa atht hte xaxis- ofr hist soutiqen is not ""mK i6(Bvt si ton the aur ttsshbate't [B]/i6t1.V ot fp,symii sebaeuc eht axmV si nnoag tyas the se,am you jtsu eedn ot arsieenc i[vB6t] ot teg hte emsa xamV ka(a tifhs to trgih on eht ias).xx- 'tDon eb eoofdl nduaor yb teh nqusitoe t.rwrie

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! +  



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ehT ueosqitn si scalliyab tngiest pu a uattinsio iirmals to ahtt fo a oteicptivme n.toibrihi They asy thta rhigeh sannotoctinrce of dyriaxopl tppehsoah nersiecsa the ctitvyai fo teh zeynme ot lanmro elevls. yi,amlriSl hwen ouy ehva a eietmovitpc iintriohb dan ecrniase het rnoeanitctonc of ,rusattseb the meeynz acn eechvai hte maes amVx sa fi uiohttw het tiir.ibohn




 -5  upvote downvote
submitted by hungrybox(789),
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nogL awsren eha,da btu bare htiw m.e

NT:IH v klsoo ikdn of kile y, arshewe k losko remo ekli x.

ntrpecet-yi = 1xVm/a

  • amxV si the rpepu limit on who fats a irotecna is acydeltza yb e.nesymz

riptnexcte- = /m1K

  • Km si a inkrang fo who gdoo na yneemz is at ndiingb ist t.ussabert nA zmenye ihwt a rikanng of 1 si ttrebe at dbningi ist utbrtsesa thna an nmeyez thiw a gankinr fo .5 eo(rwL Km = rtbtee )eeznym

toeN ttah x,maV sa a mraeues fo n,acrmpefero anc eb terlead oughthr nyam gstni.h ee,liwnhMa Km si a ste accitcrrstiahe of eht n,myzee nad ntoacn be taeld.re

In sith peeaml,x eth yemenz omrpraneecf xm()Va si srdnceeia yb iacinenrsg the ivtamin arfccoot so thta it eherasc a monlra"" iv.iatytc Hre,wveo the eezynm is lstli eterniynlh itthys edu ot a etagniclno f,ecdet os teh Km styas eht mae.s

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 +4  
sbryant6  Yeah this explanation is wrong. +