NMBE 3 block 1...goforward, iqbalian, sammy, any1? - ia

[ArchivalUser]

Can anyone help explain for me please questions 4, 15 and 16?

Thanks

[ArchivalUser]

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[ArchivalUser]

4 :. From the graph it is apparent that the cell has been placed in a hypotonic solution. Remember that one millimole of NaCl has '2' osmotically active particles. So one millimole of NaCl = 2msom/L. There is only one value in the options that if doubled ( to get the value in mOsm/L) remains lesser than 300mSom/L ( the osmolarity of plasma). This value is 75 obviously. For the next small value of 150, the mOsm becomes 300..which is isotonic with plasma, not hypo.


15:. They say the locus is present on Y chromosomes. So it must be in the male mouse. The recepient offspring is A x B by genome. So he or she must have the Y chromosome and the mentioned allele. If he has to be genetically complimented by the donor, the donor has to carry the allele of Y chromosome. That is only possible if it receives the transplant from the "MALE" mouse product of A x B cross.

16. this is about arranging the values in the 2x2 table..
I did it like this

Positive PPD.............Negative PPD..

1...............................1___________> Tuberculus

99..............................9999________> Non tuberculous
_________________________
100.............................10,000

so by formula..RR = incidence in PPD positive patients/ Incidence in PPD negative patients

This becomes 1/100 divided by 1/10,000.
or 10,000/100= 100..

[ArchivalUser]

Sammy, thank you so much