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Comments ...

 +0  (nbme18#21)

Stem should maybe say some lesions are lytic and some are sclerotic. Breast mets to bone is mixed type according to FA? If you go off the mets being purely lytic, one could think thyroid carcinoma is the correct primary tumor.

lae  thats also what I thought

 +7  (nbme24#5)

Platelet adherence and platelet aggregation are different things and this diferrence MATTERS A LOT. Fuck you, NBME. These differences supposedly matter on some questions and not on others. Where is the consistency? Hello?


 +1  (nbme23#18)

A possible integration to help with answering this question: Sketchymicro states one possible way to inactivate arenaviruses is heat. Arenaviruses are enveloped. If you can remember this, you can keep straight that enveloped viruses are heat-labile (sensitive)


 +3  (nbme23#28)

To be even clearer, this sounds like Fanconi syndrome, which has lead to Type II RTA


 +3  (nbme22#28)

Hemochromatosis associated with HLA-A3, but the appropriate screening test is serum transferrin saturation and ferritin levels


 +2  (nbme22#18)

It appears that although NRTIs are phosphorylated by thymidine kinase, resistance is actually due to mutations in the actual reverse transcriptase rather than thymidine kinase itself.

forerofore  mutations in thymine kinase are more frequent in herpes drugs

 +6  (nbme21#37)

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. Increased Km values result in 1/Km being smaller (closer to 0 on the x-axis), which is demonstrated in answer choice B.

A "normal" enzyme activity in the presence of higher B6 concentrations means that Vmax is not changing, but 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 ?

 +1  (nbme20#40)

Unusual that the patient is a young girl, OTCD is XLR inheritance.

eacv  yes !!!!! that make me doubt and choose the wrong one -.-




Subcomments ...

submitted by karljeon(23),

I don't know if there is an equation for this, but I basically pumped out every division across the table to get ~5% on average.

Here they are: 400 / 6,000 = 0.067 250 / 5,600 = 0.045 300 / 5,350 = 0.056 300 / 5,050 = 0.059 250 / 4,800 = 0.052

The average of these %s for all the years = 5.58%. So that's close enough to 5%.

seagull  good work. I found this question annoying and gave up doing those considering the amount of time we are given. +3  
vshummy  Well just don’t include the intake year... because that messed me up.. +4  
_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. +1  
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(303),

Keys were the:

-Glucosuria

-Phosphaturia

-Amino aciduria

Those should be re-absorbed by the PCT, so if they're not, Type 2 RTA.

lamhtu  To be even clearer, this sounds like **Fanconi syndrome, which has lead to Type II RTA** +3  
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(232),

Long answer ahead, but bear with me.

HINT: v looks kind of like y, whereas k looks more like x.

y-intercept = 1/Vmax

  • Vmax is the upper limit on how fast a reaction is catalyzed by enzymes.

x-intercept = 1/Km

  • Km is a ranking of how good an enzyme is at binding its substrate. An enzyme with a ranking of 1 is better at binding its substrate than an enzyme with a ranking of 5. (Lower Km = better enzyme)

Note that Vmax, as a measure of performance, can be altered through many things. Meanwhile, Km is a set characteristic of the enzyme, and cannot be altered.

In this example, the enzyme performance (Vmax) is increased by increasing the vitamin cofactor so that it reaches a "normal" activity. However, the enzyme is still inherently shitty due to a congenital defect, so the Km stays the same.

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 +1  
sbryant6  Yeah this explanation is wrong. +