Probe question kaplan bs section test 3

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SaintJude

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Passage excerpt:
Topic: Sickle cell anemia is caused by two copies of the allele for abnormal hemoglobin, HbS

...In addition, when the gene for the β chain of HbA (normal hemoglobin) is digested by the restriction enzyme MstII, a 1.1-kb fragment is generated. When the gene for the β chain of HbS, which lacks a MstII site, is digested by MstII, a 1.3-kb fragment is generated
Q: A DNA sample is digested by MstII, yielding a 1.3-kb fragment that is then successfully hybridized with a probe specific for the 1.1-kb fragment of the HbA gene. Does this prove that the DNA sample contains the HbS gene?

A. Yes, because the mutation causing sickle-cell anemia in the HbS gene eliminated the MstII site.
B. Yes, because the probe specific for the 1.1-kb fragment successfully hybridized with the 1.3-kb fragment.
C. No, because the 1.3-kb fragment from the HbS gene would not normally hybridize with the probe specific for the 1.1-kb fragment of the HbA gene.
D. No, because mutations other than the one responsible for HbS production could have eliminated the MstII site.

Can someone explain why each is wrong/right? Thank you much in advance!

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i'd go with B?

My reasoning goes, A and D are opposites and B and C are opposites
B because since the enzyme digested a 1.3 AND it bound to a 1.1 probe
C doesn't work because the fragment WOULD bind to the probe from HbA fragment, since the MstII was missing, it means the HbS is almost the same as HbA, just longer (1.3 compared to 1.1)

This is a piss poor attempt
 
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I want to go with B too because the difference in the DNA sequence that causes sickle cell is only 1 base. So the probe would still stick to the mutant gene, but it's fragment would be longer since that 1 base change eliminated the RE recognition site.
 
Well crap, I guess now that I think about it, that makes sense. If the 1 base change were 2 bases further down it would be in the wobble position and not matter, but it could still eliminate the cut site. Damn, got me! Haha
 
I've taken a cell bio class and I have no idea...

edit: The two strands successfully hybridized so there's obviously a large amount of complementarity between the two single strands of DNA. However, this doesn't mean absolute complementarity. I guess this eliminates B&C because successful hybridization is not a absolutely conclusive in either case? A&D, I have no idea. Nothing in the passage suggests that the elimination of MstII is/isn't caused by the mutation. It only says that the MstII site isn't found in the mutant strand.
 
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Passage excerpt:
Topic: Sickle cell anemia is caused by two copies of the allele for abnormal hemoglobin, HbS

Q: A DNA sample is digested by MstII, yielding a 1.3-kb fragment that is then successfully hybridized with a probe specific for the 1.1-kb fragment of the HbA gene. Does this prove that the DNA sample contains the HbS gene?

A. Yes, because the mutation causing sickle-cell anemia in the HbS gene eliminated the MstII site.
B. Yes, because the probe specific for the 1.1-kb fragment successfully hybridized with the 1.3-kb fragment.
C. No, because the 1.3-kb fragment from the HbS gene would not normally hybridize with the probe specific for the 1.1-kb fragment of the HbA gene.
D. No, because mutations other than the one responsible for HbS production could have eliminated the MstII site.

Can someone explain why each is wrong/right? Thank you much in advance!

I don't have much to add content wise, but the way I would approach this would be through process of elimination. This kind of question seems more like a VR question rather than BS.
A. This is too much of an assumption. The questions stems has clearly given you a hint that there is something odd for the fact that 1.1-kb fragment successfully hybridized with 1.3-kb fragment.
B. This does not explain an HbS. The passage does not say that in order to be a HbS gene it must meet this requirement - binding between 1.1 and 1.3 kb fragments.
C. This would also be an assumption (assuming you've given us all the relevant information from the passage) so it is not safe to choose it, since we have no background knowledge in this specific topic.
D is the best choice out of all. It doesn't make any statement that disagrees with the passage or goes beyond it. It just gives a possibility of another cause of the shortening of the fragment reminding us that is not safe to assume a HbS gene is present.

I know its not founded in BS knowledge, but I hope it helps.
 
Thank you. That was perfect explanation actually. Also, allow me to add Kaplan's explanation to for B.

Changing any one of the amino acid (not just the one that is changed in sickle cell anemia) could result in successful hybridization, because the vast majority of the base would still be complementary to the probe

Hybridization doesn't mean that every single base is exactly complementary. It just means that, generally the degree of "stickiness" is high enough for majority of sample to hybridize.

And also:
There is no reason why the only mutation that results in the elimination of this site has to be the mutation that produces the HbS gene.

This was an important lesson for me to understand. Generally, I should be careful about making assumptions. Any mutation that occurred within the MstII site of the HbaA gene would produce a gene without the MstII site, resulting in the 1.3 kb fragment.
 
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