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Welcome to lamhtu’s page.
Contributor score: 87


Comments ...

 +2  (nbme18#21)
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meSt luhdos aymeb asy mose oessnli are tcliy nad meos are teiscc.olr Bseatr mest ot beon si xdmei ypet iarcocndg ot ?FA If you go fof teh emst nbegi leyurp ycl,it eno colud khint rhitdyo acmaornci si eht coetcrr iarrpym orm.ut

lae  thats also what I thought +1
pg32  Yeah I didn't pick breast for the same reason. Then I didn't pick thyroid because I doubt serologic studies would be normal in thyroid cancer (if you check T3/T4 and TSH). So I went with avascular necrosis -_- +
lynn  in the FA index, the only things listed under lytic bone lesions are adult T cell lymphoma, langerhan histiocytosis, and multiple myeloma. Obviously there's more than that but those might be the main ones we need to know. You could also say that a giant cell tumor is also technically lytic, considering they describe it as "osteoclastoma." Idk. I thyroid but looking at FA, none of the thyroid carcinomas describe metastatic lytic lesions. Medullary carcinoma might be the one to confuse you, but it secretes calcitonin which inhibits osteoclasts, so it shouldn't cause resorp or lytic lesions. Right?? +
prp5c  just a different view - I was between avascular necrosis and metastatic carcinoma, but ended up going with metastatic carcinoma because I figured avascular necrosis of the lumbar and thoracic region would be hard since you'd have the artery of Adamkiewicz as a dual supply to the vertebrae? +

 +46  (nbme24#5)
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Patlleet echreenad dna pltletae gngreiotaga rea ierfetdnf itngsh dan tihs edrfeicenr TETARMS A T.OL kFuc o,yu BN.EM sTehe ffeceseindr spseolupyd reattm on omes noesisuqt dna ton no t.rehos hWree si the nicyneost?sc e?lloH

hungrybox  Agreed. This is so fucking stupid. +
hungrybox  "Aspirin inhibits platelet aggregation and produces a mild bleeding defect by inhibiting cyclooxygenase, a platelet enzyme that is required for TXA2 synthesis." literally straight from Big Robbins +1
susyars  Im gonna upvote this bc i love to be right +2
regularstudent  It's always a horrible, horrible feeling to pick the wrong answer that you know they think is right. Amazing job NBME... +1

 +3  (nbme23#18)
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A lepibsso ngortiatine to pelh ithw iwersanng shti ntiquseo: kthrceymoicS sasett one epsislob ayw ot ttnvieaiac asrrusneeavi is etha. nAsseuarrive aer eoevpe.ndl If oyu cna bemeemrr shti, ouy acn ekep sihtratg ahtt dplneveeo urevsis era ihle-beltaa ievsnsite()


 +5  (nbme23#28)
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oT be eevn ,rerleac hits sudnos keli oncaFni n,smrodey wihch sah alde to eyTp II ATR


 +9  (nbme22#28)
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ocarmstHieoomhs soisetaacd tiwh -A,AL3H but eth ppaptaroeir ingnsrece test is mseru atsnrerfnri toiuanrtsa and nrriieft ellves


 +1  (nbme22#18)
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tI preaasp atth hthuaolg NTsIR era alrhtdppeoshyo by meidynthi eisna,k iteassernc si alauytlc due ot tnstauimo in eth taualc reveers cnreaspatsrit haertr htan yiitmhned ieaksn steilf.

forerofore  mutations in thymine kinase are more frequent in herpes drugs +3
mtfp  NRTI need to be phosphorylated by HOST CELL thymidine kinase, mutation in viral kinase has no role in NRTI resistance +10

 +9  (nbme21#37)
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iNenedg egihrh" n"asetnrtinococ of eth B6 rfo znemey ivctayit si eahtrno wya of iysagn mK si grehih seinc ermo is rieedruq orf 1/2 vxam ty.atiicv aenrecsdI Km ualsve sutler in /1Km being emlsral ecls(ro to 0 no hte xs,a)-xi hhwic si andestedmort ni nawsre ceohci B.

A "rmal"on nmyzee tviyatci in hte reenecsp of hgirhe 6B arenontnicscto nsame htta maVx si ton hnig,ancg btu Km .is

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 ? +

 +2  (nbme20#40)
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lsaUunu thta hte anitpte is a gnoyu ligr, OTDC is XRL rehcnnieai.t

eacv  yes !!!!! that make me doubt and choose the wrong one -.- +
ally123  FA p. 83 +




Subcomments ...

submitted by karljeon(89),
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I 'ndto ownk fi rhete is an uineaqot rfo sih,t btu I liaalsbcy upepdm uto eyerv vdoiisni scaors het ltaeb ot teg 5~% on .eeaavrg

eHre tyhe ae:00 4r / 06,00 = 0.2760 50 / 0506, = 453.0000 / 55,30 = 00560. 03 / 55,00 = 20550.09 / 0,084 = .5020

ehT eevgaar fo heest %s ofr lal het aerys = So.85.% 5 thtsa' socel ohegun to .%5

seagull  good work. I found this question annoying and gave up doing those considering the amount of time we are given. +4  
vshummy  Well just don’t include the intake year... because that messed me up.. +11  
_yeetmasterflex  How would we have known not to include the intake year? From average **annual** incidence? +  
lamhtu  Do not include intake year because the question stem is asking average annual incidence. The 4000 positives at intake could have acquired HIV whenever, not just in the last year. +5  
neels11  literally didn't think there was an actual way to figure this out. but my thought process was: okay incidence means NEW cases. so the annual average at the end of 5 years would be: (# of NEW people that tested positive at the end of year 5) / (# of people at that were at risk at the beginning of year 5) <--- aka at the end of year 4 250/5050 = 4.95% also if you look at year 5: you'll see that the at risk population is 4800 when 300 new cases were found the year before. 5050 at the end of year 4 MINUS the 300 new cases at the end of year 4 should give you 4750 as the new population at risk. but notice that end of year 5 we have 4800. idk if that means 50 people were false positives before or 50 people were added but in incidence births/death/etc don't matter it's kind of like UWORLD ID 1270. assuming average annual incidence is the same as cumulative incidence this was just a bunch of word vomit. sorry if it was unbearable to follow +  


submitted by sympathetikey(980),
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Keys ewre het:

oG-rciulsua

rauaiPthsp-oh

nm-iAo iarauicd

sThoe ohusld eb edbrobrse-a by hte ,PCT so if thy'ere ,ont ypTe 2 A.RT

lamhtu  To be even clearer, this sounds like **Fanconi syndrome, which has lead to Type II RTA** +6  
yb_26  To be even clearer: Wilson disease => Fanconi syndrome => type II (proximal) RTA +  
charcot_bouchard  To be even clearer, you all have been pretty clear +  
charcot_bouchard  To be even clearer, you all have been pretty clear +  


submitted by hungrybox(791),
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Long nesawr ,aeahd tbu reba with em.

NH:TI v lsook iknd of lkei ,y ehrewas k ooslk emro ikle x.

ntirpetcey- = xm1/Va

  • xamV is eth reupp milit no owh asft a eortianc is aytlcezad yb myn.eezs

cprietxent- = mK/1

  • Km is a igrnkna fo how dogo an neyzme is at giinnbd sti ssrubtaet. An emynez htwi a nkrngai of 1 si ebttre at iinbdng its butsartes naht na enzeym twih a aikrgnn fo .5 wLroe( Km = brteet ezn)yem

tNeo atth ma,Vx as a maruees fo rcf,pnearmeo nac eb laeertd hroutgh naym .gthsin lnwaihe,Me mK is a tse craticrtehacis of the z,meney nda nnoact be ee.tarld

In htis ae,xlepm teh myeenz areenmpocrf (Va)xm is driesnaec yb nrgncaiies hte naiivtm cotfcrao so taht it hsceaer a l"mnr"ao .tyaitvci wreo,eHv eht nmeeyz is lslti yerhnltein hityst ude to a nantogeilc dtee,cf so het mK sasyt teh sem.a

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. +