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Comments ...

 +1  (nbme24#19)

In India, our nurses are smart enough to perform PV and even to deliver the neonate!!!

I am crying due to my 60usd loss!!!!


 +0  (nbme24#28)

Lol i thought trauma would hit more superficial structure than deep.. haha


 +0  (nbme24#42)

I think for nbme 24, i wasted my hard earned 60 usd


 +0  (nbme24#40)

Pharyngeal arch is not same as pharyngeal floor, pharyngeal arch is mesoderm from which nerve artery muscle cartilage develop. Pharyngeal floor is from where tongue develops which is endoderm


 +0  (nbme22#32)

Chronic diarrhoea == Vit D malabsorption = Hypocalcemia (say in crohns)

ACute diarrhoea = Hypernatremia, Hypokalemia, hyperphosphatemia

Dehydration can also cause hyperuricemia and ppt gout attack, but for young pt i think this will be irrelevant


 +0  (nbme21#45)

Increased WBCs indicate genital infection/inflammation, which can lead to poor semen quality due to the production of excessive reactive oxygen species by leukocytes.

Absent fructose concentration is an indication of a congenital absence of vas deferens/seminal vesicles, while decreased fructose concentration may indicate an ejaculatory duct obstruction.

pH > 8.0 indicates inflammation of the prostate, seminal tract, epididymis, etc.; pH < 7.2 indicates seminal vesicle dysfunction or obstruction of the ejaculatory ducts.


 +0  (nbme21#50)

I wonder i chose “sexual transmission” because same background of question stem was in AMBOSS stating the acute serum sickness like presentation for Hep B.

None the way, NBME wins.



 +1  (nbme21#14)

The real thing is TSH, T3, T4 and thyroglobulin cannot cross placenta. TRH, Iodine, TSI can cross. If mother has high TSH (considering primary hypothyroidism) —> Poor brain development If mother has TSI (Hashimoto) —> cross the placental barrier —Increase in thyroid gland (May present with stridor at birth.

demihesmisome  Ahhhhh Thank you. This was what I was trying to understand.
apurva  please note that free t4 can cross the placenta




Subcomments ...

submitted by apurva(19),

The real thing is TSH, T3, T4 and thyroglobulin cannot cross placenta. TRH, Iodine, TSI can cross. If mother has high TSH (considering primary hypothyroidism) —> Poor brain development If mother has TSI (Hashimoto) —> cross the placental barrier —Increase in thyroid gland (May present with stridor at birth.

demihesmisome  Ahhhhh Thank you. This was what I was trying to understand. +  
apurva  please note that free t4 can cross the placenta +1  


Any good material to prepare for this kind of stuffs?

apurva  Lord Jesus +1  


submitted by mcl(454),

This image is useful. Note that the stain used makes myelin appear dark.

Vignette is typical for Parkinson's disease. Area D is the substantia nigra.

oznefu  Oh nice! Thanks! +  
bend_nbme_over  Great image thanks! Even though it was an MSU link :P Go Blue! +  
apurva  Saved My life +  


submitted by niboonsh(219),

This is a case of acute transplant rejection. weeks to months after the transplant, recipient cd8 and/or cd4 t cells are activated against the donor (a type 4 HSR) and the donor starts making antibodies against the transplant. This presents as a vasculitis with dense interstitial lymphocytic infiltrates. (FA2018 pg 119)

ls3076  Actually was confused about this due to a UW explanation. UW said acute txp rejection has two types - humoral and humoral and cellular. Humoral has Neutrophilic infiltrate + necrotizing vasculitis while cellular has lymphocytosis. Can anyone simplify/explain this please? +1  
apurva  We usually look for c4d complement for humoral response in acute graft rejection. Because c4d makes covalent bond with the endothelium can can be found on staining because it is long lasting. +  


the infarct occurred 16 hrs ago, from 12-24hrs after the infarct there will be Red Neurons.

tsl19  FA 2018 - p. 496: ischemia -> pyknosis within 12-24 hours. +  
d_holles  yeah the infarct occurring 16 hr ago is key. i zoomed in only on the died 1 hr later +  
apurva  Me too zoomed on “1 hour later” and marked no change +  


The Lineweaver Burk Plot is mainly interpreted by its X and Y intercepts. The question states that giving B6 increases the activity to normal levels, which means the activity should be the same as the normal. Hence, their Y intercept should be the same, so we are between choices B and C. A and D do not have the same Y intercept as normal. Higher concentrations of B6 caused activity to become normal, and so Vmax will not be changing and we can cancel A and D. A has a lower Vmax and D has a much higher Vmax. The difference between B and C is in their Km. Moving to the left on the X axis makes the Km lower and tells you that affinity is higher so you would not need more B6. But in our case affinity of the enzyme for B6 is really low, which is why we need a ton more B6.

In summary, we want the same Vmax and a higher Km. We want the "normal" activity (same vmax as normal) and we need higher amounts of B6 for success so affinity of the enzyme for B6 is probably very low.

Choice B has a (-1/Km) value closer to 0 which means Km is lower and affinity of the enzyme for its substrate is super low. This makes sense as giving higher amounts of the "competitive" substrate B6 is helping.

maleehaak  amazing explanation +1  
bswsu  thank you very much. Very thorough explanation. +  
apurva  This explanation is wrong!!! X axis is not 1/km in the question, it is 1/pyridoxal phosphate. (1/km should be 1/homocysteine) Also the question is asking about allosteric activation of cystathione synthase due to addition of pyridoxal phosphate. Which means the Km should decrease (affinity increases after addition of pyridoxine). Considering the x axis to be 1/pyridoxal phosphate, now apply simple logic of maths, increase pyridoxal phosphate = bringing line more close to zero (because it is -1/pyridoxal phosphate). +2  
apurva  Had the X axis be 1/km, the answer should be C +  


The Lineweaver Burk Plot is mainly interpreted by its X and Y intercepts. The question states that giving B6 increases the activity to normal levels, which means the activity should be the same as the normal. Hence, their Y intercept should be the same, so we are between choices B and C. A and D do not have the same Y intercept as normal. Higher concentrations of B6 caused activity to become normal, and so Vmax will not be changing and we can cancel A and D. A has a lower Vmax and D has a much higher Vmax. The difference between B and C is in their Km. Moving to the left on the X axis makes the Km lower and tells you that affinity is higher so you would not need more B6. But in our case affinity of the enzyme for B6 is really low, which is why we need a ton more B6.

In summary, we want the same Vmax and a higher Km. We want the "normal" activity (same vmax as normal) and we need higher amounts of B6 for success so affinity of the enzyme for B6 is probably very low.

Choice B has a (-1/Km) value closer to 0 which means Km is lower and affinity of the enzyme for its substrate is super low. This makes sense as giving higher amounts of the "competitive" substrate B6 is helping.

maleehaak  amazing explanation +1  
bswsu  thank you very much. Very thorough explanation. +  
apurva  This explanation is wrong!!! X axis is not 1/km in the question, it is 1/pyridoxal phosphate. (1/km should be 1/homocysteine) Also the question is asking about allosteric activation of cystathione synthase due to addition of pyridoxal phosphate. Which means the Km should decrease (affinity increases after addition of pyridoxine). Considering the x axis to be 1/pyridoxal phosphate, now apply simple logic of maths, increase pyridoxal phosphate = bringing line more close to zero (because it is -1/pyridoxal phosphate). +2  
apurva  Had the X axis be 1/km, the answer should be C +  


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