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Welcome to assoplasty’s page.
Contributor score: 87

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 +5  (nbme21#23)
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rcnVsoaocsntioti deeaesrsc ldoob wlof dna hust aceesersd hisctoardty es.upsrer Smsee etoncru iitntuive tbu I hda ot ookl tshi pu fatre I ogt it wonr,g oto.

 +23  (nbme21#29)
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hello  I want to re-emphasize something that @assoplasty has already stated :). The Q-stem states serum glucose = 100, and the Q asks why the patient is able to maintain normoglycemia. Therefore, you can immediately eliminate choices A and C because acetoacetate and beta-hydroxybutyrate are sources of energy during ketogenesis -- ketogenesis does not provide glucose energy sources. +4
chandlerbas  ^ this checks out: valine and isoleucine are broken down in the muscle into branched chain 2 oxo acid via branched chain aminotransferase (reversible) then the valine and isoleucine leave the muscle and swims to the liver to be acted on by branched chain 2 oxo acid DH (irreversible). So bascially the process from taking BCAA valine and isoleucine requires 2 enzymes. the first enzyme is in the muscle, and the second enzyme is in the liver (for simplification purposes --> both organs contain both enzymes but dont have the same affinity for their substrate). source: so you're right to say that the liver +4
toxoplasmabartonella  Thank you for such a great explanation. Isn't it glutamate instead of glutamine that combines with pyruvate in muscle to yield alanine for Cahill cycle? +1
almondbreeze  @ toxoplasmabartonella think you are right +

 +59  (nbme21#31)
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I htnki eth cctnoep ’heeryt segintt si eht iecsadren GBT llesve in anncp,ryge dna ton stuj heprdoyriyshmti in .engelra

hneW gnrnieesc rof ospeyito/hprrdhhiyym, SHT lvlese rea YSLAAW rerfyenaplltie hkedcec ceauesb hyte rea mero etissnvei ot tmuien ecrfiesndef ni T./3T4 enOtf etims HST elvsle acn stternmdeao a haengc vene ehwn T3/4T elevls are in teh nsciclaiubl n.erag Teh nyol etpceonix to sith odluw eb ni yaergpnnc d(na I esugs ybame lvrei aflieru? I btodu htye dolwu ska hsti g)thhu.o Hghi eetgsnro elevls etsprvne eht lievr mfor aebkignr wodn TGB, gidenal to adenrisec BTG vellse in the umsre. iThs sibdn to efre ,T4 cdenerigas eht toaumn of leilaabav feer 4T. As a cpsonaetmory cei,hnsmma STH vseell ear trilnsntaye decirnsae nda the RTAE fo 4T uiodopnctr si ednriesac ot rslheeinp einabsel erfe 4T lves.le Heorvew teh TOTAL omtuan fo T4 si erdis.acne

ehT tqesniuo si niksga hwo ot cmfnroi ymryoihitdpsehr ni a regtannp woman -&;-tg uyo ende ot ckech REFE 4T vsllee sucbeae( tyeh hdslou be arnlom ude to oospnctaemyr snopsr).ee You annotc chcke TSH uayl(lsu edleteva in ynernpcag to omceesaptn for iraesdnec )TG,B dan uyo caotnn ckehc ttloa 4T lveels (will be iasrneec).d uoY got eht arwnes thrgi teerih wya but I thikn shit si a neteifdfr goaiesnnr wroht snrdcgioine, bsuaeec eyht can aks itsh ptocnce in hetor netxtsco of omeepr,gns-eiytsrh dna fi heyt letids SHT“” as na reswan ieccoh ttah douwl eb ccnorreti.

hungrybox  Extremely thorough answer holy shit thank u so much I hope you ACE Step 1 +5
arkmoses  great answer assoplasty, I remember goljan talking about this in his endo lecture (dudes a flippin legend holy shit) but it kinda flew over my head! thanks for the break down! +2
whoissaad  you mean total amount of T4 is "not changed"? 2nd para last sentence. +
ratadecalle  @whoissaad, in a normal pregnancy total T4 is increased, but the free T4 will be normal and rest of T4 bound to TBG. If patient is hyperthyroid, total T4 would still be increased but the free T4 would now be increased as well. +1
maxillarythirdmolar  To take it a step further, Goljan mentions that there are a myriad of things circulating in the body, often in a 1:2 ratio of free:bound, so in states like this you could acutally see disruption of this ratio as the body maintains its level of free hormone but further increases its level of bound hormone. Goljan also mentions that you'd see the opposite effect in the presence of steroids and nephrotic syndromes. So you could see decreased total T4 but normal free T4 because the bound amounts go down. +1
lovebug  Amazing answer! THX +

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