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submitted by sugaplum(324),
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hsTi ueqitsno si ianskg atbou VDJ aretmagennerr chwhi snphape ni het oneb wrma.or heT seegn ear lal edpphoc pu cuabsee teh B lcle is rtyign to egreneta a uienuq ibtnmoaoicn orf tis etopercr
mlspe i otcpn.s.ec. dod gnwoird

rahtCep 3 fo o"wh het nmeimu msseyt krow"s - emwoaes obko

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 oyu owkn taht 5%6 of eht atad lwli flal iwtihn D1S fo het m.aen oS if uyo ctbsratu 06150- you liwl tge .35 Wichh mneas taht boaut 1%6 ilwl lalf vbeao adn 1%6 lwli lalf owble 1 .SD Thye ear niksga ofr owh yman wlil llaf ovaeb 1 DS. Im' user three is a eertbt ayw fo ogndi iths, ubt sthta owh I tgo it .llo

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(125),
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oudt'Wln ceuat ohoclal icntopnusmo evne in eeadtmro monaut scaeu erlesierbv htiapce rlclealu yuinjr cerarzihteadc by aelurlcl bogloa?nlin tI should be eht itgrh ewansr elsnsu het uieqsont smte nemas sen"edWe"k

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. +