Biology P5

[musa]

yar it is also stated in mark scheme except 114 as well
reAD THE MARK SCHEME AGAIN NOTHING WEIRD IN IT

THE EXPECTED VALUE OF 80 IS OBTAINED BY ADDING ALL OBSERVED POPULATION LIKE 114+36+90 =240
AND DIVIDING IT INTO ALL CASES EQUALLY LIKE 240/3=80 FOR EACH

 

[zeebujha]

For gel electrophoresis:
http://learn.genetics.utah.edu/content/labs/gel/

 

[MukeshG93]

Can somebody please help me with this particular question.

October / November 2007, Question 3 A.
I have no idea how to do entire 3a and b i and ii.

well null hypothesis is that there is no effect of grazing on moth population
expected value should be same for all according to our null hypothesis so consider it 114 for all

well for bi degree of fredom is total-1 =2 in this case
bii simply process and check table value as in done in every chi square question

But the mark scheme says that the Expected result is 80 for all. :s
And then they do some weird stuff to the chi square result after that. Divide it by some number :s

So, we expect no change at all (as stated by the null hypothesis), so the results should be the same, regardless of how many years the grassland is left grazed or ungrazed (which is not even a word in the dictionary, CIE really needs to check what they write). So, whatever results we get, we would expect those results to be evenly distributed between the three sets of data (according to the null hypothesis). The total is 36 + 90 +114 = 240, and when distributing between the three of them, you get 80 for each, justifiable, no? Then, I'm sure you can help yourself with the mathematics.

 

[MukeshG93]

yar it is also stated in mark scheme except 114 as well
reAD THE MARK SCHEME AGAIN NOTHING WEIRD IN IT

THE EXPECTED VALUE OF 80 IS OBTAINED BY ADDING ALL OBSERVED POPULATION LIKE 114+36+90 =240
AND DIVIDING IT INTO ALL CASES EQUALLY LIKE 240/3=80 FOR EACH

Dude, calm down. Turn off the caps lock and please be friendly. Some of us require explanations and we are not ready to feed off the marks scheme so easily. Sorry if this is rude, but please, calm down. Thank you.

 

[musa]

dear nothing wrong my caps was switched on by mistake and well if u not satisfied with m y answer no problem
what i mean we in null hypothesis stae that the variation in the data is only due to chance we expect that the results should not varry due to other factors
i am again sorry if any body is didturbed from my comments

 

[zeebujha]

So, we expect no change at all (as stated by the null hypothesis), so the results should be the same, regardless of how many years the grassland is left grazed or ungrazed (which is not even a word in the dictionary, CIE really needs to check what they write). So, whatever results we get, we would expect those results to be evenly distributed between the three sets of data (according to the null hypothesis). The total is 36 + 90 +114 = 240, and when distributing between the three of them, you get 80 for each, justifiable, no? Then, I'm sure you can help yourself with the mathematics.[/quote]
There is another way to approach this problem as well. We assume that the grazing doesn't affect the moth population. So, the population determined in a field that has never been grazed should be the same as that has been grazed. The longest ungrazed period mentioned in 30 years. So, it is the most accurate data as factors that would affect the population in the scenario of not grazing have had their maximum effect in 30 years as compared to 10 years. Thus, we can presume that the ungrazed population after 2 years of grazing or 10 years of not grazing has to be 114 and thus we can take that to be the expected value.

 

[zeebujha]

yar it is also stated in mark scheme except 114 as well
reAD THE MARK SCHEME AGAIN NOTHING WEIRD IN IT

THE EXPECTED VALUE OF 80 IS OBTAINED BY ADDING ALL OBSERVED POPULATION LIKE 114+36+90 =240
AND DIVIDING IT INTO ALL CASES EQUALLY LIKE 240/3=80 FOR EACH

Dude, calm down. Turn off the caps lock and please be friendly. Some of us require explanations and we are not ready to feed off the marks scheme so easily. Sorry if this is rude, but please, calm down. Thank you.

yeps buddy, keep it cool!

 

[zeebujha]

The method for averaging works as well because we expect all the three values to be the same (null hypothesis again). But the data aren't same. So, the only reason could be experimental error. The three data are so different that we can not exclude any one of them as being anamolous. So left with three very different data which are actually supposed to be the same, we have no option but to average the three and find an average expected value!

 

[Xam]

HELP!!!! 2010 Onv variant 53 question 1 part C (i) plz how will we find "n"????
in mark scheme its 12.... BUT HOW!!!!

 

[angel eyez]

u ppl r doing a gr8 job...helping out ppl...
just 1 thing i need to ask..
is there any difference b/w t-test n chi-test...???
if yes den how will v get to noe...
n will the qs of both the tests r xpected to b different...??
plz do help..m w8ing..

 

[honeycoveredcookie]

u ppl r doing a gr8 job...helping out ppl...
just 1 thing i need to ask..
is there any difference b/w t-test n chi-test...???
if yes den how will v get to noe...
n will the qs of both the tests r xpected to b different...??
plz do help..m w8ing..

You use chi - test when you have discrete data.
T test is when you have continuous - non discrete data .

 

[Xam]

u ppl r doing a gr8 job...helping out ppl...
just 1 thing i need to ask..
is there any difference b/w t-test n chi-test...???
if yes den how will v get to noe...
n will the qs of both the tests r xpected to b different...??
plz do help..m w8ing..

n T-test gives SIGNIFICANT RESULT

 

[honeycoveredcookie]

So, we expect no change at all (as stated by the null hypothesis), so the results should be the same, regardless of how many years the grassland is left grazed or ungrazed (which is not even a word in the dictionary, CIE really needs to check what they write). So, whatever results we get, we would expect those results to be evenly distributed between the three sets of data (according to the null hypothesis). The total is 36 + 90 +114 = 240, and when distributing between the three of them, you get 80 for each, justifiable, no? Then, I'm sure you can help yourself with the mathematics.

There is another way to approach this problem as well. We assume that the grazing doesn't affect the moth population. So, the population determined in a field that has never been grazed should be the same as that has been grazed. The longest ungrazed period mentioned in 30 years. So, it is the most accurate data as factors that would affect the population in the scenario of not grazing have had their maximum effect in 30 years as compared to 10 years. Thus, we can presume that the ungrazed population after 2 years of grazing or 10 years of not grazing has to be 114 and thus we can take that to be the expected value.[/quote]

Okay, thanks a ton.

 

[MukeshG93]

u ppl r doing a gr8 job...helping out ppl...
just 1 thing i need to ask..
is there any difference b/w t-test n chi-test...???
if yes den how will v get to noe...
n will the qs of both the tests r xpected to b different...??
plz do help..m w8ing..

Look back in the thread. But if you are feeling too lazy, I can explain anyways. T-test is for continuous data using a normal distribution so it is a comparison using the means and standard deviation of data. Chi-squared tests use the actual data itself that has not been processed into a mean and standard deviation, so chi-squared is for discrete data and t-test is for continuous data (kinda like digital and analogue, or this might just be me being a physics guy) Well, the exam provides the formulae and the instructions for use and which test to carry out so you won't have a problem with getting to know or not. You probably won't be asked the difference between them, but, well, who knows? You might, lol. The questions won't be too different, comparing and relating and making references to the table is all we have to do. Cheers

 

[MukeshG93]

Alright, respirometer people! I'm gonna be back with some good notes, expect a lot of good stuff!

 

[zeebujha]

Alright, respirometer people! I'm gonna be back with some good notes, expect a lot of good stuff!

I am waiting :beer:

 

[angel eyez]

thanku so much... :good:
i get it now...
1 more qs i need to ask i saw an experiment regarding alginate beads in the thread m not sure if v really have to learn the whole of the exp b/c i never saw such kind of qs in the pastpapers...
r v really suppose to know n will v b xpected to write the exp....???
ANYONE??

 

[zeebujha]

thanku so much... :good:
i get it now...
1 more qs i need to ask i saw an experiment regarding alginate beads in the thread m not sure if v really have to learn the whole of the exp b/c i never saw such kind of qs in the pastpapers...
r v really suppose to know n will v b xpected to write the exp....???
ANYONE??

Yeps. Questions have been asked both pre-2007 and post 2007 regarding immobilisation and making alignate beads in the lab.

 

[honeycoveredcookie]

thanku so much... :good:
i get it now...
1 more qs i need to ask i saw an experiment regarding alginate beads in the thread m not sure if v really have to learn the whole of the exp b/c i never saw such kind of qs in the pastpapers...
r v really suppose to know n will v b xpected to write the exp....???
ANYONE??

Its very possible that they may ask you to write out an experiment for alginate beads.
Go through alginate beads experiment, respirometer, potometers, how to calibrate the microscope, photosynthesis experiments, possibly something about speciation since it hasn't come yet.

 

[angel eyez]

THANKU....
I REALLY APPRECIATE...

HAS ANY1 DONE N/09/51??
THERE'S A QS ASKING TO MAKE A GRAPH OF TEMP V/S LIGHT DEPENDENT STAGE...
I CAN NOT UNDERSTAND WHY IS THERE A HORIZONTAL LINE N THEN DECREASING....???
ISNT IT SUPPOSE TO B INCREASING INITIALLY...???

M REALLY CONFUSED...

ANYONE WHO HAS THE QS TO THIS QS??? :Search: