AQA BIOL5 Biology Unit 5 Exam - 22nd June 2011

Reply 1640

mrt23498

14

Original post by Somali-Pir8

just did the specimen paper
does anyone have any clue as to what the grade boundaries are

i dont know but the specimen mark scheme is dreadful

Reply 1641

Flux_Pav

I have 278 UMS in AS - 80UMS in biol4 and 42UMS in biol6 - so i need 120 UMS in tomorow's exam, right?
according to june 2010, 120UMS was 66 marks

Reply 1642

NRican

Original post by Somali-Pir8

just did the specimen paper
does anyone have any clue as to what the grade boundaries are

I would guess a bit higher than the June 10 paper so around 65 for an A and -7/6 mark as you go down each grade

Reply 1643

bobsaeed

Original post by pip91uk

Quick question about 'sticky ends'. I know that every restriction enzyme will only recognise a specific sequence, but within that recognition sequence, will it always cut in the same place? Basically, I'm looking at page 248 in NT textbook and it states that an enzyme (HindlIII) has a 6bp recognition sequence, but in the diagram there are only 4 bases exposed on each sticky end. I always thought the enzyme cut at the start/end of the recognition site. :confused:

it always cuts in the same place, it recognised 6 but only cuts 4, that not uncommon i think its got something to do with the palindromic nature of the cut..... but that might be wrong. but it defonatly cuts in the same place every time

Reply 1644

Carpet21

12

Original post by Flux_Pav

I have 278 UMS in AS - 80UMS in biol4 and 42UMS in biol6 - so i need 120 UMS in tomorow's exam, right?
according to june 2010, 120UMS was 66 marks

No! You need 80UMS for an A!

Reply 1645

NRican

Original post by Flux_Pav

I have 278 UMS in AS - 80UMS in biol4 and 42UMS in biol6 - so i need 120 UMS in tomorow's exam, right?
according to june 2010, 120UMS was 66 marks

for an A u need 80 /140 UMS

Reply 1646

Carpet21

12

Original post by Abby :)

[QUOTE=Abby [excludedFace]smile[/excludedFace];32355599]Ditto. Hahahhaha so much to learn. :K:

Yet we're on here. :rofl:

All I can think of is how I'm glad I showered this morning so I have more time to cram tonight...

Reply 1647

Tericon

8

Original post by animal*nature

Could someone please explain why the answer is three on question 9bi on the spec paper thanks :smile:

xx

Four fragments made, so draw a line like this : ------------------- and then mark it until you get four fragments, like this:

----I-------I----------I------ = three

Reply 1648

Alison1992

13

Original post by Somali-Pir8

just did the specimen paper
does anyone have any clue as to what the grade boundaries are

the specimen paper is oddly hard in comparison with last years paper and the questions in the book don't you think?

the boundaries for June 2010 were

A*=69 A=62 B=55 C=49 D=43 E=37

because this paper was never actually set, the specimen never has grade boundaries because how well people do on it isn't monitored, you'll just have to take it at at standard percentages i think or base them on the above.

(edited 12 years ago)

Reply 1649

Zozza93

Original post by M | k e

DNA primers are just short strands of DNA that are complementary to the start of a target gene. They're used in Polymerase Chain Reaction (PCR) to bind to the start of the target gene, firstly to stop the 2 DNA strands of target gene just re-joining, and secondly because DNA polymerase needs a double strand to start from to work for the rest of PCR :tongue:

.

Well that's off the top of my head, pretty sure it's right though :smile:

.

And DNA fingerprinting is using the non-coding repeating sequences, in introns, to compare their lengths between different people. These repeating sequences will occur at different parts of the DNA, and chances of someone having same length as even one other person is extremely slim, but chances of them having same length at more than one part of the DNA is even slimmer. So these sequences are taken out from the DNA and replicated using PCR. All these fragments are labelled (e.g add fluorescence to them). They then undergo electrophoresis, where they're placed in a conductive gel at one end. Then an electrical current is passed through, and because DNA is negative they move towards the positive electrode at the far end of the gel. The smaller the repeating sequence length was, the further it will be able to travel. Then using UV light you can see bands of the DNA fragments, and compare how far they moved along the gel to the length of the repeating sequences.

Thankyou for taking the time to write this for me! I have one question however..
You say the repeating sequences, does that mean that introns all have the same sequence, but just their position in everyone's DNA differs? Thanks!

Reply 1650

mrt23498

14

Original post by Flux_Pav

I have 278 UMS in AS - 80UMS in biol4 and 42UMS in biol6 - so i need 120 UMS in tomorow's exam, right?
according to june 2010, 120UMS was 66 marks

So you have so far 400 UMS, you need 480 for A, hence you need 80/140 UMS marks,
For A* you need 148/140 UMS hence impossible

Reply 1651

M | k e

Original post by pip91uk

Quick question about 'sticky ends'. I know that every restriction enzyme will only recognise a specific sequence, but within that recognition sequence, will it always cut in the same place? Basically, I'm looking at page 248 in NT textbook and it states that an enzyme (HindlIII) has a 6bp recognition sequence, but in the diagram there are only 4 bases exposed on each sticky end. I always thought the enzyme cut at the start/end of the recognition site. :confused:

I thought the cut itself didn't leave a specific sticky end, sometimes it may not even leave a sticky end.

However thinking of vectors, surely it would have to be the same each time or using the same restriction enzyme to cut both the vector and the target gene wouldn't work if they didn't leave complementary sticky ends.

So as a guess I'll say the sticky end is always the same, but just doesn't have to be at the start and end of the recognition sequence because also in a different book I have the EcoRI restriction enzyme only cuts the middle 4 out of its 6 bp recognition sequence :tongue:

.

Reply 1652

Flux_Pav

Original post by Carpet21

No! You need 80UMS for an A!

yeah!
for A* 120 right?

Reply 1653

Nat1986

Original post by Sparkly-Star

I have the answer but I don't get it, I hate homeostasis questions. :mad2:

me too! I've read over this question loads and still struggle with all the info. hope nothing like this comes up tomorrow! :frown:

Reply 1654

NRican

How many DNA molecules made from 1 molecule after 6 cycles in PCR.

Is it something to do with square numbers??

Reply 1655

Tericon

8

Original post by NRican

for an A u need 80 /140 UMS

Can you work out my UMS needed please too? :smile:

I got 240 at AS (but have resat Unit 1...) 72/100 Unit 1 116/140 Unit 2 52/60 Unit 3

A2: Resat unit four again, but first time round I got 74/100

I reckon I'll have gained at least 10 UMS for unit 1, so that'll make it 82/100 ums for unit 1, and possibly 5 on unit 4 = 79/100.

So forgetting Unit 6, whats the max I need in this exam for an A overall?

(Could you calculate pre resits too as well as my resit estimates, just so i know the best and worst case scenairo)

Thank you :smile:

Reply 1656

Flux_Pav

Original post by mrt23498

So you have so far 400 UMS, you need 480 for A, hence you need 80/140 UMS marks,
For A* you need 148/140 UMS hence impossible

oh thats a shame

Reply 1657

Sparkly-Star

OP

13

Original post by mrt23498

So you have so far 400 UMS, you need 480 for A, hence you need 80/140 UMS marks,
For A* you need 148/140 UMS hence impossible

An overall A - 480 UMS + a combined 90% from A2 modules is an A*. Dunno about her other modules but it could be possible.

Reply 1658

pip91uk

Original post by M | k e

I thought the cut itself didn't leave a specific sticky end, sometimes it may not even leave a sticky end.

However thinking of vectors, surely it would have to be the same each time or using the same restriction enzyme to cut both the vector and the target gene wouldn't work if they didn't leave complementary sticky ends.

So as a guess I'll say the sticky end is always the same, but just doesn't have to be at the start and end of the recognition sequence because also in a different book I have the EcoRI restriction enzyme only cuts the middle 4 out of its 6 bp recognition sequence :tongue:

.

Thanks

That makes sense and I don't have time to contemplate it any more so that'll do! Haha. I hate how unpredictable AQA papers are, you never know which information to be fussy about understanding!

Reply 1659

Tericon

8

Original post by NRican

How many DNA molecules made from 1 molecule after 6 cycles in PCR.

Is it something to do with square numbers??

Its always double the number of strands, so theres two strands in 1, so cycle 1 = 4, cycle 2 = 8, cycle 3 = 16, cycle 4 = 32, cycle 5 = 64, cycle 6 = 128.

My maths is a bit rusty, but I think thats square numbers.... :smile:

AQA BIOL5 Biology Unit 5 Exam - 22nd June 2011

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