# Biology maths question help :S

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Hi please can someone help me on these maths related biology questions. The mark scheme only tells you the answer which isnt very helpful .

(Ill upload the question in a minute )

(Ill upload the question in a minute )

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#2

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Hi please can someone help me on these maths related biology questions. The mark scheme only tells you the answer which isnt very helpful .

**kiiten**)Hi please can someone help me on these maths related biology questions. The mark scheme only tells you the answer which isnt very helpful .

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(Original post by

What questions?

**Kvothe the Arcane**)What questions?

Heres the first ques:

Spoiler:

Answer is 209

Show

Answer is 209

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#6

Depends on the number of volunteers.. there's not enough information there, unless there are 100,000 volunteers

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(Original post by

Is that the only information you're given?

**Matte0**)Is that the only information you're given?

(Original post by

Depends on the number of volunteers.. there's not enough information there, unless there are 100,000 volunteers

**AortaStudyMore**)Depends on the number of volunteers.. there's not enough information there, unless there are 100,000 volunteers

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#9

So the first answer is 4.1 x 6 x 8.5, because every year, 6 out of 100,000 people get the brain cancer, which means 6 x 410,000/100,000 get it every year in this sample, and then to find the amount in 8.5 years you just multiply it by 8.5. As for the second question, you're going to have to give us all the information :P no one can answer a question if not all the info is there. The reason we need to see the whole question is because we need to know which of the offspring are recombinant, and to determine which are recombinant, we need to know both of the parents' phenotypes, so you've got to tell me the F1 phenotype, we only have the phenotype for the pure-bred plants!

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#10

Actually hold that thought, I assume the parental phenotype for the F1 plants is green leaves and smooth fruit, so basically the % cross over value is 17+15/100 = 32%. Someone correct me if I'm wrong

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(Original post by

So the first answer is 4.1 x 6 x 8.5, because every year, 6 out of 100,000 people get the brain cancer, which means 6 x 410,000/100,000 get it every year in this sample, and then to find the amount in 8.5 years you just multiply it by 8.5. As for the second question, you're going to have to give us all the information :P no one can answer a question if not all the info is there. The reason we need to see the whole question is because we need to know which of the offspring are recombinant, and to determine which are recombinant, we need to know both of the parents' phenotypes, so you've got to tell me the F1 phenotype, we only have the phenotype for the pure-bred plants!

**AortaStudyMore**)So the first answer is 4.1 x 6 x 8.5, because every year, 6 out of 100,000 people get the brain cancer, which means 6 x 410,000/100,000 get it every year in this sample, and then to find the amount in 8.5 years you just multiply it by 8.5. As for the second question, you're going to have to give us all the information :P no one can answer a question if not all the info is there. The reason we need to see the whole question is because we need to know which of the offspring are recombinant, and to determine which are recombinant, we need to know both of the parents' phenotypes, so you've got to tell me the F1 phenotype, we only have the phenotype for the pure-bred plants!

(Original post by

Actually hold that thought, I assume the parental phenotype for the F1 plants is green leaves and smooth fruit, so basically the % cross over value is 17+15/100 = 32%. Someone correct me if I'm wrong

**AortaStudyMore**)Actually hold that thought, I assume the parental phenotype for the F1 plants is green leaves and smooth fruit, so basically the % cross over value is 17+15/100 = 32%. Someone correct me if I'm wrong

I dont really understand what youve done there. Also why is the denominator 100, wouldnt it be 82+17+15+86 = 200 ?

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#14

(Original post by

Thank you - yes that makes sense now

Yep sorry :3 i forget to include the whole question. See attached

I dont really understand what youve done there. Also why is the denominator 100, wouldnt it be 82+17+15+86 = 200 ?

**kiiten**)Thank you - yes that makes sense now

Yep sorry :3 i forget to include the whole question. See attached

I dont really understand what youve done there. Also why is the denominator 100, wouldnt it be 82+17+15+86 = 200 ?

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#15

(Original post by

The other maths question i dont know how to answer.

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**kiiten**)The other maths question i dont know how to answer.

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Yh sorry it should be 200! So basically, recombination is where 2 alleles of the same gene swap over between the 2 parental chromosomes during meiosis. It's actually whole chunks of DNA that swap but for simplicity let's say it's just one gene. Now, if the offspring inherited non-recombinant DNA, then they would have exactly the same phenotype as the parents, ie either green leaves + smooth fruit or mottled leaves + peach fruit. This is because, if you work it out, you'll find that 50% of the F2 offspring will be homozygous recessive and the other 50% will be heterozygous (just take my word for it, I did the punnet squares :P), this means that statistically 50% should be green leaves + smooth fruit and 50% should be mottled leaves + peach fruit. The reason that the F2 generation plants can't inherit a mixture of the traits (if we assume there is no recombination) is because we know that green + smooth fruit are both dominant traits (because the F1 plants were green + smooth fruit even though half of the parents had mottled + peach fruit), so the F2 generation are either going to be green + smooth (ie heterozygous) or mottled + peach (ie homozygous recessive), the offspring can't be green + peach for example because they've inherited a whole chromosome from each parent, e.g. with the F1 plants as a parent, the F2 plants have either inherited the chromosome with both dominant alleles from the pure-bred green+smooth plants or they've inherited the chromosome with both recessive alleles from the original mottled and peach plants. This has probably made no sense, but it's 3am so it's hard to explain even though it's an easy concept, but the take-home message here is that without recombination, the F2 offspring must inherit 1 or the other of the parents' phenotypes. So, this means that if they inherit a mixture of the phenotypes, then they are the recombinants, and as the equation neatly states, the % cross over value is number of recombinants (ie the number of F2 plants with a MIXTURE of traits) divided by the total number of plants (which you rightly corrected to 200), so the answer is 17+15/200 (x100), which will just be half of the number I wrote earlier.

**AortaStudyMore**)Yh sorry it should be 200! So basically, recombination is where 2 alleles of the same gene swap over between the 2 parental chromosomes during meiosis. It's actually whole chunks of DNA that swap but for simplicity let's say it's just one gene. Now, if the offspring inherited non-recombinant DNA, then they would have exactly the same phenotype as the parents, ie either green leaves + smooth fruit or mottled leaves + peach fruit. This is because, if you work it out, you'll find that 50% of the F2 offspring will be homozygous recessive and the other 50% will be heterozygous (just take my word for it, I did the punnet squares :P), this means that statistically 50% should be green leaves + smooth fruit and 50% should be mottled leaves + peach fruit. The reason that the F2 generation plants can't inherit a mixture of the traits (if we assume there is no recombination) is because we know that green + smooth fruit are both dominant traits (because the F1 plants were green + smooth fruit even though half of the parents had mottled + peach fruit), so the F2 generation are either going to be green + smooth (ie heterozygous) or mottled + peach (ie homozygous recessive), the offspring can't be green + peach for example because they've inherited a whole chromosome from each parent, e.g. with the F1 plants as a parent, the F2 plants have either inherited the chromosome with both dominant alleles from the pure-bred green+smooth plants or they've inherited the chromosome with both recessive alleles from the original mottled and peach plants. This has probably made no sense, but it's 3am so it's hard to explain even though it's an easy concept, but the take-home message here is that without recombination, the F2 offspring must inherit 1 or the other of the parents' phenotypes. So, this means that if they inherit a mixture of the phenotypes, then they are the recombinants, and as the equation neatly states, the % cross over value is number of recombinants (ie the number of F2 plants with a MIXTURE of traits) divided by the total number of plants (which you rightly corrected to 200), so the answer is 17+15/200 (x100), which will just be half of the number I wrote earlier.

(Original post by

For this one you need to do a chi squared test, do you know how to do one of those? Basically, for those values, you get a chi squared of about 277, which equates to a tiny p value (p<0.00001), this means that there is an absolutely ridiculously small probability that the difference in the results in that table are due to chance (there is a less than 0.001% probability that the results are due to chance), which means the scientists can safely conclude that taxol is effective because there a very very small likelihood that the differences in the results between the controls and people on taxol were just due to chance. If you don't know how to do chi squared, then let me know, but you should be expected to know it at your level I think. I hope I've been of help anyway

**AortaStudyMore**)For this one you need to do a chi squared test, do you know how to do one of those? Basically, for those values, you get a chi squared of about 277, which equates to a tiny p value (p<0.00001), this means that there is an absolutely ridiculously small probability that the difference in the results in that table are due to chance (there is a less than 0.001% probability that the results are due to chance), which means the scientists can safely conclude that taxol is effective because there a very very small likelihood that the differences in the results between the controls and people on taxol were just due to chance. If you don't know how to do chi squared, then let me know, but you should be expected to know it at your level I think. I hope I've been of help anyway

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(Original post by

Yh sorry it should be 200! So basically, recombination is where 2 alleles of the same gene swap over between the 2 parental chromosomes during meiosis. It's actually whole chunks of DNA that swap but for simplicity let's say it's just one gene. Now, if the offspring inherited non-recombinant DNA, then they would have exactly the same phenotype as the parents, ie either green leaves + smooth fruit or mottled leaves + peach fruit. This is because, if you work it out, you'll find that 50% of the F2 offspring will be homozygous recessive and the other 50% will be heterozygous (just take my word for it, I did the punnet squares :P), this means that statistically 50% should be green leaves + smooth fruit and 50% should be mottled leaves + peach fruit. The reason that the F2 generation plants can't inherit a mixture of the traits (if we assume there is no recombination) is because we know that green + smooth fruit are both dominant traits (because the F1 plants were green + smooth fruit even though half of the parents had mottled + peach fruit), so the F2 generation are either going to be green + smooth (ie heterozygous) or mottled + peach (ie homozygous recessive), the offspring can't be green + peach for example because they've inherited a whole chromosome from each parent, e.g. with the F1 plants as a parent, the F2 plants have either inherited the chromosome with both dominant alleles from the pure-bred green+smooth plants or they've inherited the chromosome with both recessive alleles from the original mottled and peach plants. This has probably made no sense, but it's 3am so it's hard to explain even though it's an easy concept, but the take-home message here is that

**AortaStudyMore**)Yh sorry it should be 200! So basically, recombination is where 2 alleles of the same gene swap over between the 2 parental chromosomes during meiosis. It's actually whole chunks of DNA that swap but for simplicity let's say it's just one gene. Now, if the offspring inherited non-recombinant DNA, then they would have exactly the same phenotype as the parents, ie either green leaves + smooth fruit or mottled leaves + peach fruit. This is because, if you work it out, you'll find that 50% of the F2 offspring will be homozygous recessive and the other 50% will be heterozygous (just take my word for it, I did the punnet squares :P), this means that statistically 50% should be green leaves + smooth fruit and 50% should be mottled leaves + peach fruit. The reason that the F2 generation plants can't inherit a mixture of the traits (if we assume there is no recombination) is because we know that green + smooth fruit are both dominant traits (because the F1 plants were green + smooth fruit even though half of the parents had mottled + peach fruit), so the F2 generation are either going to be green + smooth (ie heterozygous) or mottled + peach (ie homozygous recessive), the offspring can't be green + peach for example because they've inherited a whole chromosome from each parent, e.g. with the F1 plants as a parent, the F2 plants have either inherited the chromosome with both dominant alleles from the pure-bred green+smooth plants or they've inherited the chromosome with both recessive alleles from the original mottled and peach plants. This has probably made no sense, but it's 3am so it's hard to explain even though it's an easy concept, but the take-home message here is that

**without recombination, the F2 offspring must inherit 1 or the other of the parents' phenotypes**. So, this means that**if they inherit a mixture of the phenotypes, then they are the recombinants, and as the equation neatly states, the % cross over value is number of recombinants**(ie the number of F2 plants with a MIXTURE of traits) divided by the total number of plants (which you rightly corrected to 200), so the answer is 17+15/200 (x100), which will just be half of the number I wrote earlier.So, basically the number of recombinant offspring are the ones who have different phenotypes to the parents (green + peach and mottled + smooth). Then as you said its 17+15/200 (x100)

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Maybe you misunderstood something but the markscheme says you should have

7.44 and 1.74

ratio: 4.28:1

% decrease = 76.6%

But i dont understand what theyve done (this is for the ques below - so you know what im talking about )

(Original post by

The other maths question i dont know how to answer.

Posted from TSR Mobile

7.44 and 1.74

ratio: 4.28:1

% decrease = 76.6%

But i dont understand what theyve done (this is for the ques below - so you know what im talking about )

**kiiten**)

The other maths question i dont know how to answer.

Posted from TSR Mobile

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#19

(Original post by

Ok i dont understand what you said at all (maybe because its one paragraph?) apart from the parts in bold - i think i understand from those few sentences Thank youuu

So, basically the number of recombinant offspring are the ones who have different phenotypes to the parents (green + peach and mottled + smooth). Then as you said its 17+15/200 (x100)

**kiiten**)Ok i dont understand what you said at all (maybe because its one paragraph?) apart from the parts in bold - i think i understand from those few sentences Thank youuu

So, basically the number of recombinant offspring are the ones who have different phenotypes to the parents (green + peach and mottled + smooth). Then as you said its 17+15/200 (x100)

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