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# BIOL4 Biology Unit 4 Exam - 13th June 2011 watch

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1. biologist i have Question for you..

Q. A species of insect, only found on remote island, has a characteristic controlled by a pair of codominant alleles, CM and CN

(a) There were 500 insects in the total population.In this population, 300 insects had the genotype CMCM, 150 had the genotype CN CN. calculate the actual frequecy of the allele CN by using this figure.

(b) use your answer to part (a) and the hardy weinberg equation to calculate the no. of insects that would be expected to have the genotype CN CN

NB: M and N are the powers of C.
2. (Original post by NRican)
It's not working for me?
Yeah it didn't work for me to here's a better link ;http://www.dabbleboard.com/draw?b=Gu...14afdd9d8312aa

(Original post by al_habib)
biologist i have got Question for you

There were 500 insects in the total population.In this population, 300 insects had the genotype CMCM, 150 had the genotype CN CN. calculate the actual frequecy of the allele CN by using this figure.

NB: M and N are the powers of C.
325 CM 175CN so the frequency is 35%
3. (Original post by al_habib)
biologist i have got Question for you

There were 500 insects in the total population.In this population, 300 insects had the genotype CMCM, 150 had the genotype CN CN. calculate the actual frequecy of the allele CN by using this figure.

NB: M and N are the powers of C.
can u tell us the answer please if you have it so we can see if we are correct ourselves.

I got 0.35 (also 35%).
4. (Original post by al_habib)
biologist i have Question for you..

Q. A species of insect, only found on remote island, has a characteristic controlled by a pair of codominant alleles, CM and CN

(a) There were 500 insects in the total population.In this population, 300 insects had the genotype CMCM, 150 had the genotype CN CN. calculate the actual frequecy of the allele CN by using this figure.

(b) use your answer to part (a) and the hardy weinberg equation to calculate the no. of insects that would be expected to have the genotype CN CN

NB: M and N are the powers of C.

i got 61 for part b.) is this correct?
5. (Original post by al_habib)
biologist i have Question for you..

Q. A species of insect, only found on remote island, has a characteristic controlled by a pair of codominant alleles, CM and CN

(a) There were 500 insects in the total population.In this population, 300 insects had the genotype CMCM, 150 had the genotype CN CN. calculate the actual frequecy of the allele CN by using this figure.

(b) use your answer to part (a) and the hardy weinberg equation to calculate the no. of insects that would be expected to have the genotype CN CN

NB: M and N are the powers of C.
Hmmm....I don't get b) what does it mean by expected?
6. (Original post by User12399)
can u tell us the answer please if you have it so we can see if we are correct ourselves.

I got 0.35 (also 35%).
EXPLAIN PLEASE? thanks... did you just do 500- 450 ... = 50 .... and divide by 2 cos the codominant heterozygous alleles are in same frequencies... and add 25 to each one? so 325 cm and 175 cn?
7. I got 57% for a)
8. (Original post by SK-mar)
EXPLAIN PLEASE? thanks... did you just do 500- 450 ... = 50 .... and divide by 2 cos the codominant heterozygous alleles are in same frequencies... and add 25 to each one? so 325 cm and 175 cn?
That's what I did, it feels like you gave too much information normally I would have to work from CMCM only.
9. It would really help of the OP gave us the answer so we know we were correct. :/
10. (Original post by SK-mar)
EXPLAIN PLEASE? thanks... did you just do 500- 450 ... = 50 .... and divide by 2 cos the codominant heterozygous alleles are in same frequencies... and add 25 to each one? so 325 cm and 175 cn?
Genotype CMCM= 300 and CNCN=150
Because theres two "CM" allele in genotype CMCM, the total no. of allele would be 625.. As you mentioned above for heterozygous theres one of each Cm and Cn.
So the no of Cn allele = 325...
11. (Original post by Rose1990)
Genotype CMCM= 300 and CNCN=150
Because theres two "CM" allele in genotype CMCM, the total no. of allele would be 625.. As you mentioned above for heterozygous theres one of each Cm and Cn.
So the no of Cn allele = 325...
so if thats the case then surely with 50 more insects that are CMCN there would be 50 more CM alleles and 50 more CN alleles ? so there would be 650 CM, and 350 CN ....

so the percentage frequency is the same at 35% .....0.35
12. (Original post by SK-mar)
so if thats the case then surely with 50 more insects that are CMCN there would be 50 more CM alleles and 50 more CN alleles ? so there would be 650 CM, and 350 CN ....

so the percentage frequency is the same at 35% .....0.35
I suck at maths. Thnxs for pointing out my mistake...
13. (Original post by User12399)
It would really help of the OP gave us the answer so we know we were correct. :/
how did u get 0.35 ? :s
14. Could someone be really helpful and list the sort of mathematical equations/skills we will need for this paper
because these questions always seem to stomp me! and make me panic for about 20 minutes:|

Obviously we need hardy weinberg and mark-release-recapture
but things like
percentage increase/decrease etc?
does anybody know for sure/ could give quick formulas!
thank youu!
15. (Original post by Rose1990)
I suck at maths. Thnxs for pointing out my mistake...
no problem at alllll.....
16. so for jan 11 guys Q2C

why cant u do do 1- 0.16 first?

why square root

many thanks!

i also dont get Q 2bi so confusing????
17. (Original post by User12399)
can u tell us the answer please if you have it so we can see if we are correct ourselves.

I got 0.35 (also 35%).
MS goes as 0.25/25% = 2mks

Cn = 250 / 1000 = 1mk
18. (Original post by SK-mar)
i got 61 for part b.) is this correct?
hhm am afraid no.

p2 = 0.0625 (ans from part a)

writing down herdyweinberg equation earns you a mark

the final ans = 31.25 / 31
19. (Original post by kingsmod1)
so for jan 11 guys Q2C

why cant u do do 1- 0.16 first?

why square root

many thanks!

i also dont get Q 2bi so confusing????
The questions gives, resus negative( phenotype/ genotype) does not give us allele frequency.

So resus negative= 16% =0.16
Q^2=0.16 ( Q^2 represent homozygous recessive genotype)
q=0.4 , p+q=1.
P=0.6. 2pq= 48%

This is how I make sense of this kind of questions.
20. (Original post by al_habib)
MS goes as 0.25/25% = 2mks

Cn = 250 / 1000 = 1mk
ok can you explain exactly how that answer is achieved as everyone who answered it got it wrong... btw where is this question from and are you doing AQA? cheers...

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