AQA BIOL4 ~ 11th June 2013 ~ A2 Biology

Are the June 10 boundaries for real? Doubt it will be anywhere near that.

Could someone please explain Jan10 3bi) please?

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

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!

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

Got these from a thread a while back, theyre not mine
hope they help

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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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!


Posted from TSR Mobile

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!


Posted from TSR Mobile

This is what I got...



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

Posted from TSR Mobile

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!


Posted from TSR Mobile

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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Got these from a thread a while back, theyre not mine
hope they help

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