Genetic methylated fragment interpretation

[shigella123]

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How do we interpret this? Can anyone please help? I'm so lost.

 

[witzelsucht]

I believe the answer is D because she has prader wili, so the endonuclease will not cut the 12kb fragment since it is methylated, so she will have one 12kb fragment, since she is not getting the mother's good copy

 

[notbobtrustme]

should be A, D has a thicker band, indicating more DNA in that lane. Assuming everything has been standardized and normalized, a person with two copies will have a band that is greater in intensity. The kid has only one allele at that locus, so the band should be half as intense.

Basically, this question is asking you whether or not this kid's DNA is methylated. If it's not methylated, the restriction enzyme will cut the DNA fragment into two pieces giving answer choice C or the parental lanes. Half of all the DNA should be methylated, but the other half shouldn't be, as that would be expressed in the wild type. That's what gives the two bands in the parental lanes. Since this kid has only the methylated copy, the restriction enzyme won't recognize the DNA sequence and the DNA fragment will remain intact.

 

[dyspareunia]

should be A, D has a thicker band, indicating more DNA in that lane. Assuming everything has been standardized and normalized, a person with two copies will have a band that is greater in intensity. The kid has only one allele at that locus, so the band should be half as intense.

Basically, this question is asking you whether or not this kid's DNA is methylated. If it's not methylated, the restriction enzyme will cut the DNA fragment into two pieces giving answer choice C or the parental lanes. Half of all the DNA should be methylated, but the other half shouldn't be, as that would be expressed in the wild type. That's what gives the two bands in the parental lanes. Since this kid has only the methylated copy, the restriction enzyme won't recognize the DNA sequence and the DNA fragment will remain intact.

I thought PWS/Angelman are due to having two methylated (silenced) copies?

 

[notbobtrustme]

I thought PWS/Angelman are due to having two methylated (silenced) copies?

isn't it because of a deletion in either maternal or paternal chromosome that leads to abnormal methylation in the remaining chromosome? There aren't two alleles in either Angelmann or PW.

 

[dyspareunia]

isn't it because of a deletion in either maternal or paternal chromosome that leads to abnormal methylation in the remaining chromosome? There aren't two alleles in either Angelmann or PW.

Right, I was thinking about uniparental disomy (I think?)... which is much less common.

I agree that the answer is A. PWS/Angelman are most commonly a deletion from either paternal or maternal DNA. The presentation of PWS/Angelman are due to lacking that particular part of chromosome 15.

So the way I understand it (I hope this is right) is that paternal chromosome 15 has all the things that prevent PWS sxs and maternal chromosome 15 has all the things that prevent Angelman sxs.. In other words, if you have a maternal deletion and only the paternal 15, you have Angelman.

 

[shigella123]

The answer is A.

notbobtrustme, you are awesome at this stuff. It made sense after you explained it. I think this is the 2nd question you helped me with.. You are a great teacher....

In Prader Willi we have paternal allele deleted and maternal's methylated= inactivated. So, is that the reason you said, "The kid has only one allele at that locus, so the band should be half as intense."

So, this is what I understood so far. Since the pt has only 1 allele we should be looking at only 1 band answers. And since it's methylated it should not be thick band, but the other. Right?
I'm so happy I finally understand it .... You are awesome! Thank you very much..

 

[notbobtrustme]

The answer is A.

notbobtrustme, you are awesome at this stuff. It made sense after you explained it. I think this is the 2nd question you helped me with.. You are a great teacher....

In Prader Willi we have paternal allele deleted and maternal's methylated= inactivated. So, is that the reason you said, "The kid has only one allele at that locus, so the band should be half as intense."

So, this is what I understood so far. Since the pt has only 1 allele we should be looking at only 1 band answers. And since it's methylated it should not be thick band, but the other. Right?
I'm so happy I finally understand it .... You are awesome! Thank you very much..

hey, thanks. I was pointing out that A and D are both potentially right answers, but that D has a thick band, which should correspond to twice as much DNA. Since the patient only has half the number of genes at that locus (since the other chromosome has nothing at that part) the band is too thick to be right.

Methylation is going to determine whether or not the second restriction enzyme will cut the DNA fragment. Since this person has one locus and that locus is methylated, only A and D can be right answers.

Glad I could help. I was doing research for the past 10 years, so I should know this stuff pretty well.

 

[shigella123]

"but that D has a thick band, which should correspond to twice as much DNA. Since the patient only has half the number of genes at that locus (since the other chromosome has nothing at that part) the band is too thick to be right."

So if methylated it should be thick or thin band? If its' methylate it should be the thin band since it's inactive, right? My only other question is why not B?

Thanks for helping...

 

[notbobtrustme]

Thickness of the band has to do with how much DNA is in that particular spot. PW syndrome is a deletion, so at that particular locus, the patient has less DNA, which shows up as less DNA in that lane, which shows as a thinner/lighter band.

The sizes of said bands have to do with methylation. If the DNA is methylated, then the restriction enzyme will not recognize the DNA sequence and end up not cutting that DNA fragment. B is excluded because the size of the DNA fragment does not correspond to anything possible given the experimental setup. C is essentially wild-type, so this is excluded as well.

So knowing that, we can narrow our choices down to either A or D. Since the patient has a deletion, there should be less DNA in that lane, which corresponds to A, the lane with a thinner band.

 

[dyspareunia]

"but that D has a thick band, which should correspond to twice as much DNA. Since the patient only has half the number of genes at that locus (since the other chromosome has nothing at that part) the band is too thick to be right."

So if methylated it should be thick or thin band? If its' methylate it should be the thin band since it's inactive, right? My only other question is why not B?

Thanks for helping...

Ok I finally figured this out completely.

The top line is the 12kb fragment, the bottom band is the methylated fragment. Wildtype has both a methylated fragment and a 12kb fragment, which is the basis of the entire PWS/Angelman thing. Wildtype has one methylated (silenced) sequence and one active sequence.. hence if you're missing the active sequence then you have only the silenced sequence and end up having PWS or Angelman, depending on which silenced sequence you are missing (paternal or maternal).

The answer can't be "B" because "B" is the non-methylated sequence and if the patient had that sequence then they wouldn't have PWS.


So, bottom line is this. Wildtype has two anti-PWS sequences, but one is silenced. PWS patients have only the silenced anti-PWS sequence, therefore they have PWS. In this question, the active anti-PWS sequence is represented by the bottom band. Thus, the patient must have only the top band, or they wouldn't have PWS.

 

[shigella123]

Thanks a lot dyspareunia and notbobtrustme for breaking this down.