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submitted by sugaplum(323),
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siTh itnoseuq si kisnga tbauo VDJ rtrneearnmage whhci psehnpa in teh nebo .armwor Teh seneg era all opepdhc pu sceaueb eht B lcle si ytngir to eraneetg a euinuq tbainiocnom orf sti pcerorte
seimlp tncp..soce. odd rodniwg

Cerhpta 3 fo h"wo the mmnieu sytesm osw"rk - amoewes ookb

varunmehru  in the question stem, they are asking about a constant region. VDJ rearrangement is for the variable. It doesn't make sense :( +3  
sallz  Both the constant (heavy chains) and the light chains undergo gene rearrangement. The heavy chain undergoes V(D)J random recombinations, while the light chain undergo VJ random recombinations. So gene rearrangement could work for both regions. +6  
azibird  The constant region does not undergo recombination. That's why it's called constant. It's just right next to the variable region though, so they get expressed together as one protein. That's why the constant-labeled DNA region is variable length here. +1  


submitted by strugglebus(163),
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oS you wonk atht %65 fo teh tada lwli lafl nhiiwt SD1 of het .nmae So fi uoy abtsctur -10056 uyo lilw teg 53. hichW manse ahtt aobut %16 lwil allf beoav nad 6%1 lwli lfla wbole 1 DS. ehyT ear isagkn for who mnya ilwl allf veoba 1 DS. mI' eusr rehte si a tbetre awy fo gindo ,hsit ubt satht who I got it ol.l

sympathetikey  Same! +6  
sympathetikey  Except according to FA, it's 68% within 1 SD, so 34%, which split in half is 17%. +2  
amirmullick3  Sympathetikey check your math :D 100-68 is 32 not 34, and half of 32 is 16 :) +8  
lilyo  Can anyone explain why we subtract 68 from 100? This makes me think that we are saying its 35% of the data that falls within 1SD as opposed to 65. HELLLLLLP +  
sallz  @Lilyo If you consider 1 SD, that includes 68% of the population (in this case, you're saying that 68% of the people are between 296 and 196 (1SD above and 1 below). This leaves how many people? 32% outside of that range (100-68=32); half of those would be above 296 and the other half below 296, so 16% +5  


submitted by youssefa(124),
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l'tWdnuo autec aclohol mptsunnoico eenv in odemater otnamu ecasu ebreslierv aciehtp llauerlc inrjyu hracezeiactrd yb cellralu anoioblnl?g tI odlsuh be het ghtir wrnaes lsnues eth tsoqnuei tesm emnas de"sknWee"

hello  No. The order of liver damage due to alcohol is: fatty changes --> cellular swelling (cellular balooning) --> necrosis. This Q stem states to the patient consumed large amount of alcohol on a weekend -- he has acutely drank a large amount of alcohol on one weekend --> this corresponds with fatty changes +3  
et-tu-bromocriptine  It's not in pathoma, but I have it written in (so he or Dr. Ryan may have mentioned it) - Alcoholic hepatitis is generally seen in binge drinkers WITH A LONG HISTORY OF CONSUMPTION. +  
krisgsxr600  Its kind of in pathoma Chapter 1, "free radical Injury", Section 2 "examples of free radical injury" goes over how free radicals (caused by drinking) lead to fat accumulation +  
sallz  You can't get the steatohepatitis before getting the steatosis (fatty change). All the FAs caused by the alcohol consumption eventually lead to cytokine release, inflammation and finally the hepatitis seen in balloon swelling. +