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chiddler

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5-bromouracil resembles thymine enough to become incorporated into DNA during replication. Once incorporated, it rearranges to resemble cytosine. If 5-bromouracil was present during the replication of the sense strand shown below, which of the following might be the sense strand formed in the following replication?

5' -GGCGTACG -3'

A. 5'- GGCGCACG -3'
B. 5'- GGCGATCG -3'
C. 3'- CCGCAGGC -5'
D. 5'- GGCGTGCG -3'

answer is D.

I've been trying to figure this out for way too long. I would write my logical progression, but it's gone all over the place and i'm extremely frustrated :<

If it helps, the book gives the following:

---------------------------------------------------------Original is 5'- GGCGTACG -3'
new anti sense strand in presence of bromouracil: 3'- CCGCABGC -5'
---------------------------------------------new sense strand: 5'- GGCTGGCG -3'

thanks.
 
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drdrdrdrdrdrdr

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Hi!

Those sense/antisense can be tough. I am not sure if I fully understood the question but this is what I did:

+ is sense
- is antisense

5' GGCGTACG 3' (+)
3' CCGCABGC 5' (-) <--Here problem says B becomes Cytosine upon incorporated. HENCE:
3' CCGCACGC 5' (still -)
5' GGCGTGCG 3' (+) is the answer

Hope that made sense!
 

MedPR

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The strand they give you -- 5' -GGCGTACG -3' -- gets replicated and instead of being
3'- CCGCATGC -5', you get 3'- CCGCABGC -5' because, as the question says, Bromomouracil can replace thymine, then convert to cytoseine.

So, the question is:

If 5-bromouracil was present during the replication of the sense strand shown below, which of the following might be the sense strand formed in the following replication?

It tells you to replicate the given sequence. You should know from the question that the first replication will produce a B (or whatever letter you want to call it, BU, whatever) instead of a T. The question also states that it will transform into a Cytoseine, so now you must be replicating something with a C where you ordinarily would expect a T if there hadn't been a mutation.
 

chiddler

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Hi!

Those sense/antisense can be tough. I am not sure if I fully understood the question but this is what I did:

+ is sense
- is antisense

5' GGCGTACG 3' (+)
3' CCGCABGC 5' (-) <--Here problem says B becomes Cytosine upon incorporated. HENCE:
3' CCGCACGC 5' (still -)
5' GGCGTGCG 3' (+) is the answer

Hope that made sense!

hi yourself!

The question says that the strand in question was replicated with bromouracil. This means that bromouracil is within the strand already shown. I don't understand which basepair corresponds to the bromouracil, though. Given that, I don't follow what you've done.
 

SLC

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This one's pretty simple.

Normally the sense strand will be identical replication to replication right?

But we've got 5-bromouracil that can be substituted for thymine.

So in the anti-sense strand when we base pair with the adenine of the original sense strand we use the 5-bromouracil.

Then we know that once in the anti-sense strand the 5-bromouracil rearranges to resemble cytosine so when we do our next base pairing to produce a new sense strand the guanine base pairs with the rearranged 5-bromouracil.

It's just a question that's trying to test your knowledge of replication while remembering a twist that changes one base pair. Tough to wrap your head around alone, so on the test you should be writing this process out on your scratch paper rather than using B in the anti-sense strand I'd just use a C for simplicity since you know that the end result is that the incorporated 5-bromouracil for all intents and purposes acts exactly like a cytosine.

so you have your original: 5'- GGCGTACG -3'
Then you have your anti-sense: 3'- CCGCACGC -5'
And your new sense: 5'- GGCGTGCG -3'

Try and make things as simple as possible, the 5-bromouracil is there to distract you and throw you off your game.

In these types of questions, just jump straight to the end result of the substitution (that it ends up acting like cytosine) and just substitute cytosine wherever adenine would be. Much simpler that way.
 

chiddler

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ugh. So bromouracil is present in the replication of this sequence, not the synthesis of this sequence.

that clears up very much.....
 

MedPR

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ugh. So bromouracil is present in the replication of this sequence, not the synthesis of this sequence.

that clears up very much.....


Yea, reread the question carefully and you'll see that they are telling you to do two replications, first one with bromouracil, then second one after bromouracil has changed into cytoseine. This tricked me the first time too since none of the answers made sense, but I reread it (very slowly) and it worked out.
 

SLC

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hi yourself!

The question says that the strand in question was replicated with bromouracil. This means that bromouracil is within the strand already shown. I don't understand which basepair corresponds to the bromouracil, though. Given that, I don't follow what you've done.

No it says that when you went to replicate the sense strand they've given you, there was bromouracil present. So you proceed as normal, just substituting C's for A's
 

jsp132

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hi yourself!

The question says that the strand in question was replicated with bromouracil. This means that bromouracil is within the strand already shown. I don't understand which basepair corresponds to the bromouracil, though. Given that, I don't follow what you've done.

Original Template Strand or sense strand is:

5' -GGCGTACG -3'

then

During replication of the template strand would become:

3' -CCGCATGC- 5' with Thymine representing the bolded letter

BUT since the question stated "bromouracil resembles thymine enough to become incorporated into DNA during replication"

it becomes 3' -CCGCABGC- 5' See how I replaced the Thymine with the bromouracil? that's what it's saying to do....

THEN it says "it rearranges to resemble cytosine"

So what do you think we do? We replace the Bromouracil with the Cytosine

it becomes 3' -CCGCACBC -5' See how there I replaced the Bromouracil with cytosine

So, since that is the antisense strand: the new sense strand would be:

5'- GGCGTGCG -3'
 
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