Could someone please explain how B is the correct answer here
http://i.imgur.com/j9EJx.png.
From what I understand of the question. It seems to me that F should be the correct answer since having one copy of Beta-A-Globin (wild type allele) should give you 2 fragments (0.2 and 1.15kb) and having one copy of Beta-S-Globin (sickle cell allele) should give you 1 fragment (1.35kb). So shouldn't a carrier have all three fragments?
Haha, nice. Good timing, Beardsley. I actually just encountered this same question on USMLE Rx THIS MORNING.
Since the pt has sickle cell TRAIT, he/she's got one normal allele and one faulty one. The bottom one shows that the mutant allele lacks the restriction site, so the probe binds to and interprets the fragment as one continuous piece, which is 1.35Kb total. That's just the bottom one though. But at least you know that 1.35Kb will show up on the gel.
Then for the top one (the normal allele), the probe binds to JUST the 1.15Kb fragment because the LOCATION of the probe indicates homology with only that portion of the DNA. Since the MstII restriction site is intact, the probe will remove just the 1.15Kb fragment. The probe doesn't demonstrate homology with the .2Kb fragment, so .2Kb will not show up on the gel.
I got this question wrong too btw. The concept is easy in the end, but the drawing of the probe looked a bit awkward and it was easy to get thrown off. It would have been helpful to have had the probe shown either attached to the strand itself or with the probe in its current position with arrows indicating the homology.
This nevertheless reinforces the value of the QBank. There's nothing hard about this question, but it's easy to get tripped up, especially when you're rushing.
Hope that helps,
~Phloston
