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BIOL4 Biology Unit 4 Exam - 13th June 2011 watch

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    (Original post by Tericon)
    So I've always had trouble with ratios, and I'm terrfied they're going to come up, like they did in this paper:

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF

    Question 7, fig. 3

    I know the answers, but I don't understand ratios, i thought ratios were 3:1 not a decimal number.

    Can someone please just talk me through what on earth they mean and how to deal with them? Not just in the context of this question, because I really don't get them at all.

    + rep and a plate of cookies for the person kind enough to help a maths phobic
    This is the bit where I struggled too. If you remember the sa:volume ratio from unit 2.
    The higher sa:volume ratio meant the surface area is greater than volume. Followed by greater loss of heat via radiation and all those things.
    The lower sa:vol meant the volume is greater than surface area, therefore more control in heat.

    The question 7 fig 3(d) asks us why the ratio of animal plant is always lot less than "1." Now if you think about it, when a greater no. divides smaller no. the answer is always less than 1.
    In this case plant(seaweed) being the producer has "the greater no." the animal gets energy from seaweed, thus (small no). I hope it makes sense.
    Plus I am not 100% sure of my reasonings.
    Please do correct me.
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    (Original post by SK-mar)
    the ratio has always got to be something : 1 ........ so for example in one of the past papers it said give the ratio of respiratory losses in cows to the amount of food absorbed in their gut. and these values were given in the table above. for respiratory losses the value was 12.25, and for absorption it was 12.50 ...... so the ratio of R:A (respiration : absorption) would be

    12.25:12.50 ..... however to make it something:1 ... you divide the first value by the 2nd .... so 12.25/12.50 ... = 0.98:1 .....

    in your case the values are ratios to 1 .... so for example if they gave a value of 1.25 ... it would mean 1.25:1 .... hope this helps...
    Ok, thank you, but in the same paper, question 7c. the ratios in the table are numbers such as '0.20' (ratio of dry biomass of animals to abundance of animals)what does this mean? 0 biomass for 20 animals?!

    Also question 7ei. , i know the answer, but don't understand how the ratios show this.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF

    Sorry, its just these ratios tend to turn up a lot with AQA
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    (Original post by Tericon)
    Ok, thank you, but in the same paper, question 7c. the ratios in the table are numbers such as '0.20' (ratio of dry biomass of animals to abundance of animals)what does this mean? 0 biomass for 20 animals?!

    Also question 7ei. , i know the answer, but don't understand how the ratios show this.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF

    Sorry, its just these ratios tend to turn up a lot with AQA
    yep so just do the same.... im thinking 0.2 means 0.2:1 ....
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    (Original post by SK-mar)
    yep so just do the same.... im thinking 0.2 means 0.2:1 ....
    This is when we need mathematician anyway..
    I think( not entirely sure) the mystery behind 0.20:1 ratio is:
    Lets suppose, there were 12 animals and its biomass was like 60 (if we include the unit the no. wont look suspicious)

    So,
    60:12
    then,
    12/12=1--->>> 60/12=5 ( Ratio 1:5)
    So what does it tell us ( according to the empirical formula rule from chemistry)
    this means for every 1 animal the biomass will be approximately 5.

    The ratio 0.20:1 (maybe)means that for every weed the bioass will be 0.20(some unit)

    this should be right in chemistry term not sure about biology..
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    Any hot topics people think is going to come up!!??
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    is anyone able to help me with the hardy weinberg equation on past paper June 10 question 3d. I have tried every angle but can't get 44% as answer?? thanks!
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    I'll help you!

    You know that the frequency of the dominant allele (long haired/H) is 0.33 in London. Therefore you know the frequency of p. As p+q=1, you know q, because it is 1-0.33=0.67.

    Cause you're trying to find heterozygous genotypes in the population you have to us 2*p*q. This gives you 0.44. And because you need to find the PERCENTAGE of the whole population, times that by 100. This will give you 44%.
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    (Original post by Nat1986)
    is anyone able to help me with the hardy weinberg equation on past paper June 10 question 3d. I have tried every angle but can't get 44% as answer?? thanks!
    q = 0.33
    p+q = 1

    therefore p = 1-0.33 = 0.67

    then think p^2 +2pq + q^2 but only use the 2pq bit as that refers to those that are heterozygous ...... so 2 x 0.33 x 0.67 = 0.4422 ... as a percentage thats 44.22 % or 44% rounded.
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    (Original post by Nat1986)
    is anyone able to help me with the hardy weinberg equation on past paper June 10 question 3d. I have tried every angle but can't get 44% as answer?? thanks!
    OK I've always had trouble with Hardy-Weinberg, but it gets better with practice! SO do loads of these questions until you get the hang of it.
    I'll copy and paste the question here so we can see what's going on.

    Hair length in cats is determined by a single gene with two alleles. The allele for long hair (h) is recessive. The allele for short hair (H) is dominant.
    Use the information in the table and the Hardy–Weinberg equation to estimate the percentage of cats in London that are heterozygous for hair length. Show your
    working.

    First things first, you need to use data they've given you. In the beginning of the question, there's a table giving you frequencies of different phenotypes. We're obviously interested in long-haired cats from London. If we look at the table, we can see that the frequency of the allele for long-hair in London is 0.33.

    The wording of the data is important. If it says allele, you consider only the allele, NOT the genotype. So in this instance, in London, the frequency of the allele for long-hair in cats is 0.33.

    Now, we let p = frequency of "H", q = frequency of "h"
    We're told long-hair is coded for by a recessive allele, so now we know that
    q = 0.33.

    We want the percentage of cats in London that are heterozygous for hair length. So that means we want the genotype "Hh".

    p + q = 1, so p + 0.33 = 1, so p = 1 - 0.33 = 0.67
    (p + q)^2 = 1, so p^2 + 2pq + q^2 = 1
    We're only interested in 2pq because we want the heterozygous genotype "Hh". Remember to use the "2" as well
    So: 2 x 0.67 x 0.33 = 0.4422, the question asks for a percentage, so multiply by 100

    0.4422 x 100 = 44.22% = 44% (2sf)
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    (Original post by dominicdiep)
    I'll help you!

    You know that the frequency of the dominant allele (long haired/H) is 0.33 in London. Therefore you know the frequency of p. As p+q=1, you know q, because it is 1-0.33=0.67.

    Cause you're trying to find heterozygous genotypes in the population you have to us 2*p*q. This gives you 0.44. And because you need to find the PERCENTAGE of the whole population, times that by 100. This will give you 44%.
    Long-haired is recessive?
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    lol i just realised that, but in this case it doesn't matter in terms of working out the equation, because knowing 0.33 (even if it was recessive or dominant) would still tell you the frequency of the other allele, so you would still be able to use the equation. But yeah, you're right
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    Hello, can anyone help me with this?

    Its question 5b on the Jan 11 paper (attached at start of this thread) Its about a graph, the graph axis says 'net' yet we are asked to calculate the gross?

    I got 2.4 just from reading the graph, but according to MS that's wrong, so how is this question worked out?

    Thanks
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    I got stuck on this one too - go to page 11 and the answer is there
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    (Original post by Tericon)
    Hello, can anyone help me with this?

    Its question 5b on the Jan 11 paper (attached at start of this thread) Its about a graph, the graph axis says 'net' yet we are asked to calculate the gross?

    I got 2.4 just from reading the graph, but according to MS that's wrong, so how is this question worked out?

    Thanks
    use the equation : net = gross - respiratory losses.....

    so gross = net + respiratory losses.

    reading the value for net of the side gets around 2.4 for medium light at 20 degrees. then you read off the respiration line below (but remember to use the graduations on the right hand side of the graph for the CO2 intake bit) ... and thats about 0.3-4 or something ....answer is around 2.75-2.81
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    (Original post by SK-mar)
    use the equation : net = gross - respiratory losses.....

    so gross = net + respiratory losses.

    reading the value for net of the side gets around 2.4 for medium light at 20 degrees. then you read off the respiration line below (but remember to use the graduations on the right hand side of the graph for the CO2 intake bit) ... and thats about 0.3-4 or something ....answer is around 2.75-2.81
    I completely forgot about that equation, thanks
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    Can someone please help me with this question on energy transfer. Im finding it really hard to understand this....

    Its question 4 (a) June 2007 - the old papers .... heres the link

    and its BYA5 question paper

    http://www.freeexampapers.com/past_p...+A%2F2007+Jun/
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    (Original post by uasif93)
    OK I've always had trouble with Hardy-Weinberg, but it gets better with practice! SO do loads of these questions until you get the hang of it.
    I'll copy and paste the question here so we can see what's going on.

    Hair length in cats is determined by a single gene with two alleles. The allele for long hair (h) is recessive. The allele for short hair (H) is dominant.
    Use the information in the table and the Hardy–Weinberg equation to estimate the percentage of cats in London that are heterozygous for hair length. Show your
    working.

    First things first, you need to use data they've given you. In the beginning of the question, there's a table giving you frequencies of different phenotypes. We're obviously interested in long-haired cats from London. If we look at the table, we can see that the frequency of the allele for long-hair in London is 0.33.

    The wording of the data is important. If it says allele, you consider only the allele, NOT the genotype. So in this instance, in London, the frequency of the allele for long-hair in cats is 0.33.

    Now, we let p = frequency of "H", q = frequency of "h"
    We're told long-hair is coded for by a recessive allele, so now we know that
    q = 0.33.

    We want the percentage of cats in London that are heterozygous for hair length. So that means we want the genotype "Hh".

    p + q = 1, so p + 0.33 = 1, so p = 1 - 0.33 = 0.67
    (p + q)^2 = 1, so p^2 + 2pq + q^2 = 1
    We're only interested in 2pq because we want the heterozygous genotype "Hh". Remember to use the "2" as well
    So: 2 x 0.67 x 0.33 = 0.4422, the question asks for a percentage, so multiply by 100

    0.4422 x 100 = 44.22% = 44% (2sf)
    Thank you for that! I was going wrong by thinking 0.33 was the q^2 so kept trying to square root it. makes much more sense now and I need to remember to read what info the question gives!
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    can anyone help with question 2b on past paper Jan 10 with regards to rate of growth?
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    (Original post by SK-mar)
    ey? we need to know forensic science? also what did your teacher say is likely to come up?
    hey forget about what i said ok our syllabus is different i am edexcel candidate the date and unit is same so i got confused
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    (Original post by Nat1986)
    can anyone help with question 2b on past paper Jan 10 with regards to rate of growth?
    The maximum rate of growth is where the graph is at its steepest
    Divide the change in mass (Y axis) by the time taken (X axis) to find the rate (but make sure the points you pick for the change in mass are in the steepest part of the graph otherwise it isn't maximum rate growth!)
    So say between 100mg and 200mg, the time taken for this growth is 2.5 days
    So the change in mass (100) divided by time taken (2.5) = 40
    Hope that makes sense!
 
 
 
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