[rss.914]

Can anyone help me on question 5... really dont get restriction mapping...

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

[Tamagotchi]

(Original post by rss.914)
Can anyone help me on question 5... really dont get restriction mapping...

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

a)i) the question says that the scientists integrated an unknown piece of DNA into a plasmid containing one restriction sequence for Kpn1 and BamH. When they added Kpn1 to the plasmid containing the unknown DNA, it gave two strands which means there must have been a restriction sequence on the unknown DNA strand otherwise you'd just get one long strand of DNA made up of the plasmid and the unknown DNA.

a)ii) the very large fragment is made up of a part of the plasmid and a part of the unknown DNA since both the plasmid and the unknown DNA contain restriction sites for Kpn1.

b) it the plasmid is broken into 3 fragments there must have been 3 restriction sites in total and since the plasmid contains 1, there must be 2 on the unknown DNA.

c)i) electrophoresis is concerned with how heavier structures move at a slower rate than lighter structures so shorter fragments will move faster and longer fragments will move more slowly.

c)ii)if you completely digest the plasmid and perform electrophoresis then you'll get a number of bands corresponding to different fragments with different number of bases. If you find the length of these bases and add them all up, if the plasmid has been completely digested then the number of base pairs should equal the original number of bases in the plasmid with the unknown DNA integrated. If the plasmid is not completely digested then you'll get larger pieces of DNA and when you add the number of bases from each strand, you'll get something larger than the original number of bases in the plasmid.

Hope it helps xD

[rss.914]

(Original post by Tamagotchi)
a)i) the question says that the scientists integrated an unknown piece of DNA into a plasmid containing one restriction sequence for Kpn1 and BamH. When they added Kpn1 to the plasmid containing the unknown DNA, it gave two strands which means there must have been a restriction sequence on the unknown DNA strand otherwise you'd just get one long strand of DNA made up of the plasmid and the unknown DNA.

a)ii) the very large fragment is made up of a part of the plasmid and a part of the unknown DNA since both the plasmid and the unknown DNA contain restriction sites for Kpn1.

b) it the plasmid is broken into 3 fragments there must have been 3 restriction sites in total and since the plasmid contains 1, there must be 2 on the unknown DNA.

c)i) electrophoresis is concerned with how heavier structures move at a slower rate than lighter structures so shorter fragments will move faster and longer fragments will move more slowly.

c)ii)if you completely digest the plasmid and perform electrophoresis then you'll get a number of bands corresponding to different fragments with different number of bases. If you find the length of these bases and add them all up, if the plasmid has been completely digested then the number of base pairs should equal the original number of bases in the plasmid with the unknown DNA integrated. If the plasmid is not completely digested then you'll get larger pieces of DNA and when you add the number of bases from each strand, you'll get something larger than the original number of bases in the plasmid.

Hope it helps xD

thankyou i get ci anc cii now but i dont understand part a or b? btw the answer to B should be 2 not 3

[Tamagotchi]

(Original post by rss.914)
thankyou i get ci anc cii now but i dont understand part a or b? btw the answer to B should be 2 not 3

For part a, if there wasn't a restriction site on the unknown DNA, Kpn1 would only cut at the restriction site on the plasmid and this would just cause the plasmid to open up and result in one fragment (ie, a single strand). But the question says the plasmid was broken down into two fragments meaning the enzyme Kpn1 must have cut twice to make two fragments. Imagine an elastic band, if you cut it once (ie, at the restriction site on the plasmid) you'll get one long strand. if you cut it once more, you'll get two strands and since there is only one restriction site for Kpn1 on the plasmid, the other one must be on the unknown DNA.

For part b, I should've made it clearer but the question asks for the number of restriction sites on the unknown DNA which I said is 2 at the end.

[rss.914]

(Original post by Tamagotchi)
For part a, if there wasn't a restriction site on the unknown DNA, Kpn1 would only cut at the restriction site on the plasmid and this would just cause the plasmid to open up and result in one fragment (ie, a single strand). But the question says the plasmid was broken down into two fragments meaning the enzyme Kpn1 must have cut twice to make two fragments. Imagine an elastic band, if you cut it once (ie, at the restriction site on the plasmid) you'll get one long strand. if you cut it once more, you'll get two strands and since there is only one restriction site for Kpn1 on the plasmid, the other one must be on the unknown DNA.

For part b, I should've made it clearer but the question asks for the number of restriction sites on the unknown DNA which I said is 2 at the end.

oh so for part b because the diagram only shows one recognition site for bamh1 there must be two on the unknown plasmid.

what about aii/?

[Tamagotchi]

(Original post by rss.914)
oh so for part b because the diagram only shows one recognition site for bamh1 there must be two on the unknown plasmid.

what about aii/?

The 'very large fragment' must be made up of a part of the plasmid and a part of the unknown DNA because we know that both of these contain a Kpn1 restriction site.

If you represent the plasmid containing the unknown DNA as a rubber band with a small strip in a different colour (to represent the unknown DNA) and if you cut once on each colour (since there are Kpn1 restriction sites on the plasmid and the unknown DNA) then youll end up with two strands that both contain bits of the two colours.

[asaaal]

(Original post by Tamagotchi)
a)i) the question says that the scientists integrated an unknown piece of DNA into a plasmid containing one restriction sequence for Kpn1 and BamH. When they added Kpn1 to the plasmid containing the unknown DNA, it gave two strands which means there must have been a restriction sequence on the unknown DNA strand otherwise you'd just get one long strand of DNA made up of the plasmid and the unknown DNA.

a)ii) the very large fragment is made up of a part of the plasmid and a part of the unknown DNA since both the plasmid and the unknown DNA contain restriction sites for Kpn1.

b) it the plasmid is broken into 3 fragments there must have been 3 restriction sites in total and since the plasmid contains 1, there must be 2 on the unknown DNA.

c)i) electrophoresis is concerned with how heavier structures move at a slower rate than lighter structures so shorter fragments will move faster and longer fragments will move more slowly.

c)ii)if you completely digest the plasmid and perform electrophoresis then you'll get a number of bands corresponding to different fragments with different number of bases. If you find the length of these bases and add them all up, if the plasmid has been completely digested then the number of base pairs should equal the original number of bases in the plasmid with the unknown DNA integrated. If the plasmid is not completely digested then you'll get larger pieces of DNA and when you add the number of bases from each strand, you'll get something larger than the original number of bases in the plasmid.

Hope it helps xD

c)ii)if you completely digest the plasmid and perform electrophoresis then you'll get a number of bands corresponding to different fragments with different number of bases. If you find the length of these bases and add them all up, if the plasmid has been completely digested then the number of base pairs should equal the original number of bases in the plasmid with the unknown DNA integrated. If the plasmid is not completely digested then you'll get larger pieces of DNA and when you add the number of bases from each strand, you'll get something larger than the original number of bases in the plasmid.
i dont understand this ^

Lets say it should add up to 20 and we get
5 + 2+ 6 +4 + 3 which would be if it is fully digested = 20
But if tits not fully digested we will get lets say:
(5+2) +6 +4 +3 = still equal 20

so what do you mean?

[Tamagotchi]

(Original post by asaaal)
c)ii)if you completely digest the plasmid and perform electrophoresis then you'll get a number of bands corresponding to different fragments with different number of bases. If you find the length of these bases and add them all up, if the plasmid has been completely digested then the number of base pairs should equal the original number of bases in the plasmid with the unknown DNA integrated. If the plasmid is not completely digested then you'll get larger pieces of DNA and when you add the number of bases from each strand, you'll get something larger than the original number of bases in the plasmid.
i dont understand this ^

Lets say it should add up to 20 and we get
5 + 2+ 6 +4 + 3 which would be if it is fully digested = 20
But if tits not fully digested we will get lets say:
(5+2) +6 +4 +3 = still equal 20

so what do you mean?

As you said, if we simplify the situation and say the whole plasmid is made up of 20 bases then if it is completely digested, you would get those fragments which add up to 20. If some restriction sites are not cleaved then you'll get longer fragments. I think you're assuming that only one plasmid is being digested which isn't true (I could be wrong but I don't think you can isolate just one modified plasmid, digest it and identify the resulting fragments using electrophoresis). Instead of getting (5+2)+6+4+3 which is what you would get if you only had one plasmid, you actually get something more like (5+2)+5+2+6+4+3, with the (5+2) fragments coming from partially undigested plasmids and the 5 and 2 coming from those that have been fully digested.

Sorry if my explanation isn't clear :/

[asaaal]

(Original post by Tamagotchi)
As you said, if we simplify the situation and say the whole plasmid is made up of 20 bases then if it is completely digested, you would get those fragments which add up to 20. If some restriction sites are not cleaved then you'll get longer fragments. I think you're assuming that only one plasmid is being digested which isn't true (I could be wrong but I don't think you can isolate just one modified plasmid, digest it and identify the resulting fragments using electrophoresis). Instead of getting (5+2)+6+4+3 which is what you would get if you only had one plasmid, you actually get something more like (5+2)+5+2+6+4+3, with the (5+2) fragments coming from partially undigested plasmids and the 5 and 2 coming from those that have been fully digested.

Sorry if my explanation isn't clear :/

can you go over that part please?
so youre saying we have 2 plasmids in this example?
why is it just the (5+2) thats been added onto the original dna?
PLEASE PLEASE help me

[Tamagotchi]

(Original post by asaaal)
can you go over that part please?
so youre saying we have 2 plasmids in this example?
why is it just the (5+2) thats been added onto the original dna?
PLEASE PLEASE help me

The number of plasmids doesn't matter as long as there's more than one.
The (5+2) is just an example of what could happen. You could also get (2+6), (2+6+4) and so on. In the example, I've assumed that the plasmids have been cut in two ways: one that has been completely digested and another that has had everything digested (resulting in the same fragment as the first type) apart from the cut between the 5 and 2 fragment (resulting in the (5+2) fragment).

[rss.914]

(Original post by asaaal)
can you go over that part please?
so youre saying we have 2 plasmids in this example?
why is it just the (5+2) thats been added onto the original dna?
PLEASE PLEASE help me

(Original post by Tamagotchi)
The number of plasmids doesn't matter as long as there's more than one.
The (5+2) is just an example of what could happen. You could also get (2+6), (2+6+4) and so on. In the example, I've assumed that the plasmids have been cut in two ways: one that has been completely digested and another that has had everything digested (resulting in the same fragment as the first type) apart from the cut between the 5 and 2 fragment (resulting in the (5+2) fragment).

how does restriction mapping determine the base sequence of a gene

[Tamagotchi]

(Original post by rss.914)
how does restriction mapping determine the base sequence of a gene

Are you referring to the Sanger method?

[rss.914]

(Original post by Tamagotchi)
Are you referring to the Sanger method?

the specification says...

The use of labelled DNA probes and DNA hybridisation to locate specific genes.
Once located, the base sequence of a gene can be determined by
• restriction mapping
• DNA sequencing.

ii dont understand how you'd determine the sequence by restriction mapping or sequencing

[asaaal]

(Original post by rss.914)
the specification says...

The use of labelled DNA probes and DNA hybridisation to locate specific genes.
Once located, the base sequence of a gene can be determined by
• restriction mapping
• DNA sequencing.

ii dont understand how you'd determine the sequence by restriction mapping or sequencing

right well, first you have 4 test tubes. each with a single standed dna that you want to determin the dna sequence of.
You put the dna in 4 seperate test tubes bcos theres 4 nucleotides
In each test tube you have ALL the nucleotides (normal) and one terminator nuceleotide so in test tube one youd have a terminator G molecule, in the second A etc etc.
The point of terminator molecules is that they STOP any more of the dna from combining to the dna.
so what will happn in each test tube is that normal nucelotides and terminator nucleotides will bind to the single stranded dna and as its a random process you may sonetimes be left with a short dna molecule but sometimes a really long dna because you just dont know where the terminator nucleotide will bind!
Then we simply use electrophoresis to seperate out all the different strand sizes which will be determined by where the terminator molecule has binded.


so in each test tube you have:
loads of single stranded dna that youve seperated using dna helicase - this gives us a single stranded template strand
a mixture of nucleotides - A G T C
1 type of nucleotide in each test tube to stop the binding because what they so is not form bonds with nucleotides either side and dont allow DNA polymerase to complement the template strand
DNA polymerase which im sure you know what itll be used for - just to catalyse dna synthesis

I hope thats helped you a bit

[rss.914]

(Original post by asaaal)
right well, first you have 4 test tubes. each with a single standed dna that you want to determin the dna sequence of.
You put the dna in 4 seperate test tubes bcos theres 4 nucleotides
In each test tube you have ALL the nucleotides (normal) and one terminator nuceleotide so in test tube one youd have a terminator G molecule, in the second A etc etc.
The point of terminator molecules is that they STOP any more of the dna from combining to the dna.
so what will happn in each test tube is that normal nucelotides and terminator nucleotides will bind to the single stranded dna and as its a random process you may sonetimes be left with a short dna molecule but sometimes a really long dna because you just dont know where the terminator nucleotide will bind!
Then we simply use electrophoresis to seperate out all the different strand sizes which will be determined by where the terminator molecule has binded.


so in each test tube you have:
loads of single stranded dna that youve seperated using dna helicase - this gives us a single stranded template strand
a mixture of nucleotides - A G T C
1 type of nucleotide in each test tube to stop the binding because what they so is not form bonds with nucleotides either side and dont allow DNA polymerase to complement the template strand
DNA polymerase which im sure you know what itll be used for - just to catalyse dna synthesis

I hope thats helped you a bit

thanks, how do you use restriction mapping to determine a sequence though by counting the base with the longest kb until the shortest?

[asaaal]

(Original post by rss.914)
thanks, how do you use restriction mapping to determine a sequence though by counting the base with the longest kb until the shortest?

well if you have a piece of long DNA .. lets say 10kb
and you know that 100% it has 2 recognition sequence of AGGA and enzyme A will recognise this then if you add an unknown gene to this dna if you add enzyme A you will still have only 3 pieces of fragments during electrophoresis but if the unknown dna also has AGGA then we will get e.g
4 pieces of dna if it has one recognition sequences of AGGA
5 pieces of dna if it has two recognition sequences of AGGA etc etc

its hard to explain without you showing me an example, post a question youre stuck on and ill talk you though it

[rss.914]

(Original post by asaaal)
well if you have a piece of long DNA .. lets say 10kb
and you know that 100% it has 2 recognition sequence of AGGA and enzyme A will recognise this then if you add an unknown gene to this dna if you add enzyme A you will still have only 3 pieces of fragments during electrophoresis but if the unknown dna also has AGGA then we will get e.g
4 pieces of dna if it has one recognition sequences of AGGA
5 pieces of dna if it has two recognition sequences of AGGA etc etc

its hard to explain without you showing me an example, post a question youre stuck on and ill talk you though it

haha im confused ... and the only restriction mapping questions i know about are in the specimen and june 11 paper and iv done them :/

[asaaal]

(Original post by rss.914)
haha im confused ... and the only restriction mapping questions i know about are in the specimen and june 11 paper and iv done them :/

well if youve got them then you obviously know it!
btw do you know where i can get the mark scheme for the spec?

p.s sorry for confusing you more

[rss.914]

lol i dont get it man and yeah here it is...

http://www.school-portal.co.uk/Group...urceId=3087251

[Tamagotchi]

(Original post by rss.914)
the specification says...

The use of labelled DNA probes and DNA hybridisation to locate specific genes.
Once located, the base sequence of a gene can be determined by
• restriction mapping
• DNA sequencing.

ii dont understand how you'd determine the sequence by restriction mapping or sequencing

To find the sequence of bases in a gene, you don't use restriction mapping - the Sanger method is sufficient.