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DELETED ACCOUNT
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#241
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#241
Are the June 10 boundaries for real? :lolwut: Doubt it will be anywhere near that.
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emah123
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#242
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#242
2 days enough to revise for this retake?? got a B in jan 74 UMS hoping to get it up to an A....
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TH4RG3
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#243
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#243
(Original post by frogs r everywhere)
Are the June 10 boundaries for real? :lolwut: Doubt it will be anywhere near that.
Yeah thats what I thought I got 43/75 and thought id done rubbish but 39 was an A*??

I think some of the questions were awful such as the shrews and stuff.
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saaaarn
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#244
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#244
Could someone please explain Jan10 3bi) please? :confused:
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laser174572
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#245
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#245
(Original post by saaaarn)
Could someone please explain Jan10 3bi) please? :confused:
I can give it a go
The first thing to do is identify which curve represents the survival curve of a developed country - C - as it has low infant mortality and high life expectancy. Next, recall the definition of "average life expectancy", it is the age at which half of the population have died; therefore find the percentage of the maximum age for curve C which corresponds to 5000 people remaining from the original population of 10,000, by eye on my screen it looks to be about 90%. You would then multiply the maximum age, given to you in the intro, 95, by 90% to find the mean life expectancy giving, 85.5 years.
Hope that helps
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s24a
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#246
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#246
does anyone know where i could find loads of how science works questions? its the one thing that i find the hardest to get my head around
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saaaarn
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#247
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#247
(Original post by s24a)
does anyone know where i could find loads of how science works questions? its the one thing that i find the hardest to get my head around
Got these from a thread a while back, theyre not mine
hope they help
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sals1234
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#248
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#248
(Original post by Hakh)
You do birth rate - death rate
20-5 = 15 this is per 1000 so you do 15/1000
and then you times the value by the total population of 107,000,000 and you get 1605000
I hope that helped!
Thank you very much!
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saaaarn
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#249
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#249
(Original post by laser174572)
I can give it a go
The first thing to do is identify which curve represents the survival curve of a developed country - C - as it has low infant mortality and high life expectancy. Next, recall the definition of "average life expectancy", it is the age at which half of the population have died; therefore find the percentage of the maximum age for curve C which corresponds to 5000 people remaining from the original population of 10,000, by eye on my screen it looks to be about 90%. You would then multiply the maximum age, given to you in the intro, 95, by 90% to find the mean life expectancy giving, 85.5 years.
Hope that helps
Thank you!!!
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nathaan5
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#250
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#250
(Original post by saaaarn)
Got these from a thread a while back, theyre not mine
hope they help
Where did you get them from? Link won't open for me
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hollywils
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#251
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#251
Can someone please help me with a question from a previous past paper.
June 2010 question 3d. I don't understand how to get the correct answer. The value 0.33 in the table is applied to recessive allele frequency (q) (as seen in mark scheme) and not the genotype frequency (q squared). I'm confused because it clearly says '(h) is recessive' so surely that would means the long haired cats in London would have two of the recessive alleles (q squared) to have long hair and the 0.33 value should apply the genotype frequency rather than the allele frequency part of the hardy Weinberg equation?! No idea if the above actually will make sense to anyone but worth a try!


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TH4RG3
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#252
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#252
(Original post by hollywils)
Can someone please help me with these a question from a previous past paper.
June 2010 question 3d. I don't understand how to get the correct answer. The value 0.33 in the table is applied to recessive allele frequency (q) (as seen in mark scheme) and not the genotype frequency (q squared). I'm confused because it clearly says '(h) is recessive' so surely that would means the long haired cats in London would have two of the recessive alleles (q squared) to have long hair and the 0.33 value should apply the genotype frequency rather than the allele frequency part of the hardy Weinberg equation?! No idea if the above actually will make sense to anyone but worth a try!


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This is what I got...

Name:  1370703410100.jpg
Views: 197
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But apparently it is meant to be 44?? I think the markscheme is wrong, and why an A* was only 39/75!

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TH4RG3
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#253
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#253
(Original post by hollywils)
Can someone please help me with a question from a previous past paper.
June 2010 question 3d. I don't understand how to get the correct answer. The value 0.33 in the table is applied to recessive allele frequency (q) (as seen in mark scheme) and not the genotype frequency (q squared). I'm confused because it clearly says '(h) is recessive' so surely that would means the long haired cats in London would have two of the recessive alleles (q squared) to have long hair and the 0.33 value should apply the genotype frequency rather than the allele frequency part of the hardy Weinberg equation?! No idea if the above actually will make sense to anyone but worth a try!


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Found why it is the allele not genotype...

Name:  1370703503679.jpg
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Im guessing most people got that wrong...

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nrp_95
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#254
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#254
(Original post by hollywils)
Can someone please help me with a question from a previous past paper.
June 2010 question 3d. I don't understand how to get the correct answer. The value 0.33 in the table is applied to recessive allele frequency (q) (as seen in mark scheme) and not the genotype frequency (q squared). I'm confused because it clearly says '(h) is recessive' so surely that would means the long haired cats in London would have two of the recessive alleles (q squared) to have long hair and the 0.33 value should apply the genotype frequency rather than the allele frequency part of the hardy Weinberg equation?! No idea if the above actually will make sense to anyone but worth a try!


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From what I've gathered, 0.33 applies to q because the table is showing allele frequency not phenotype frequency. So it's showing the number of ALLELES for long hair in the population, not the number of cats with long hair. So these alleles could be in homozygous or heterozygous cats, not just those showing the long hair phenotype as the allele is recessive.


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#255
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#255
(Original post by TH4RG3)
This is what I got...

Name:  1370703410100.jpg
Views: 197
Size:  39.9 KB

But apparently it is meant to be 44?? I think the markscheme is wrong, and why an A* was only 39/75!

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(Original post by hollywils)
Can someone please help me with a question from a previous past paper.
June 2010 question 3d. I don't understand how to get the correct answer. The value 0.33 in the table is applied to recessive allele frequency (q) (as seen in mark scheme) and not the genotype frequency (q squared). I'm confused because it clearly says '(h) is recessive' so surely that would means the long haired cats in London would have two of the recessive alleles (q squared) to have long hair and the 0.33 value should apply the genotype frequency rather than the allele frequency part of the hardy Weinberg equation?! No idea if the above actually will make sense to anyone but worth a try!


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The mark scheme is correct

The frequency of the allele "h" is 0.33.
The frequency of the allele "H" must therefore be 0.67 (by using p+q=1)


To find out the amount of heterozygous carriers, all we do is:

use 2pq

2 x 0.33 x 0.67= 0.4422
Multiply answer by 100 to get 44.22%.
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hollywils
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#256
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#256
(Original post by nrp_95)
From what I've gathered, 0.33 applies to q because the table is showing allele frequency not phenotype frequency. So it's showing the number of ALLELES for long hair in the population, not the number of cats with long hair. So these alleles could be in homozygous or heterozygous cats, not just those showing the long hair phenotype as the allele is recessive.


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Ah thanks!!!


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s24a
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#257
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#257
(Original post by saaaarn)
Got these from a thread a while back, theyre not mine
hope they help
thanks!
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#258
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#258
(Original post by TH4RG3)
Found why it is the allele not genotype...

Im guessing most people got that wrong...
This is what the examiner's report said concerning that question

Many candidates failed to complete the calculation successfully, either because they assumed the information in the table referred to genotype rather than allele frequency or neglected to produce the answer as a percentage, as required.
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hollywils
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#259
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#259
(Original post by frogs r everywhere)
This is what the examiner's report said concerning that question
oh hah need to make sure I read it properly!


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hollywils
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#260
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#260
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