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A level Biology: Post your doubts here!

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shazmina

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princess Anu said:
View attachment 51583
Can somebody help :( why is it D not A?
Click to expand...
We know that when there is more than 1 polypeptide chain the protein molecule is said to have Quaternary structure .. and also Quaternary structures do have a prosthetic groups.. take hemoglobin as an example .. it has the Quaternary structure as well as the prosthetic group Fe2+ ion .. therefore there is an ionic bond in Quaternary structure ..And also we know that tertiary structures do have ionic bond between them... but im not sure about ionic bonds in Secondary structure ...
Mahmood Magdy said:
Well the same question was driving me crazy. A is wrong as secondary structures do not have peptide bonds. Apparently, they do have ionic bonds(I'm not even sure how).
Click to expand...
So broh why do u say that there is ionic bonds in secondary structures???
 
princess Anu

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This is crazy! I don't get the logic behind this question :/. Haven't we just studied Secondary structure has Hydrogen bonds and nothing else!! :sleep:
And I don't think the prosthetic group is bonded by an ionic bond :s
 
Mahmood Magdy

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shazmina said:
We know that when there is more than 1 polypeptide chain the protein molecule is said to have Quaternary structure .. and also Quaternary structures do have a prosthetic groups.. take hemoglobin as an example .. it has the Quaternary structure as well as the prosthetic group Fe2+ ion .. therefore there is an ionic bond in Quaternary structure ..And also we know that tertiary structures do have ionic bond between them... but im not sure about ionic bonds in Secondary structure ...

So broh why do u say that there is ionic bonds in secondary structures???
Click to expand...
Naah I meant according to CIE secondary structures have ionic bonds. :O
shazmina said:
http://onlineexamhelp.com/wp-content/uploads/2014/08/9700_s14_qp_22.pdf
Can any1 please help me with Question 6 ( b) (i) & (ii) of this paper...
And here is the mark scheme ...
http://onlineexamhelp.com/wp-content/uploads/2014/08/9700_s14_ms_22.pdf
Click to expand...
Your question is pretty simple, think about it. The cardiac cycle starts with the Atrial systole. The Atria contract once and then relax for the rest of the cycle.
If we consider the ventricles, they are relaxed while the Atria contract, after that they contract then they relax. Here is a picture to make things easier.
 

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shazmina

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Mahmood Magdy said:
Naah I meant according to CIE secondary structures have ionic bonds. :O

Your question is pretty simple, think about it. The cardiac cycle starts with the Atrial systole. The Atria contract once and then relax for the rest of the cycle.
If we consider the ventricles, they are relaxed while the Atria contract, after that they contract then they relax. Here is a picture to make things easier.
Click to expand...
Ohkkkk :p

Thanks for the explanation broh :D
 
Mahmood Magdy

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Karim Nabil said:
what is a lac operon ? operon ?
#BIOA2
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If you are giving CIE A2, then congrats, you do not need to study it extensively(although it will be included in the new Syllabus 2016).
However, to understand the Lac operon you have to get what an operon is at the first place.
An operon is made up of a promoter, regulator and a gene. The promoter is a length of DNA that is situated next to genes to make sure that the gene is expressed and transcribed. The regulator is simply the start and stop codons etc and the gene would be coding for the polypeptide we want. The operon as a whole switches on and off according to the environment.
Now back to the lac-operon, it is present in some bacteria like E.coli and is switched on when lactose is present thus 'lac'-operon. So normally the bacteria will synthesize an enzyme to metabolize lactose, when exposed to lactose itself. However in Genetic Eng, we added the gene that codes for insulin to the bacteria. So when the bacteria was exposed to lactose, the lac-operon was switched on and the first part of the enzyme was synthesized along with insulin. Hope to have helped ;)
 
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shazmina

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Mahmood Magdy said:
If you are giving CIE A2, then congrats, you do not need to study it extensively(although it will be included in the new Syllabus 2016).
However, to understand the Lac operon you have to get what an operon is at the first place.
An operon is made up of a promoter, regulator and a gene. The promoter is a length of DNA that is situated next to genes to make sure that the gene is expressed and transcribed. The regulator is simply the start and stop codons etc and the gene would be coding for the polypeptide we want. The operon as a whole switches on and off according to the environment.
Now back to the lac-operon, it is present in some bacteria like E.coli and is switched on when lactose is present thus 'lac'-operon. So normally the bacteria will synthesize an enzyme to metabolize lactose, when exposed to lactose itself. However in Genetic Eng, we added the gene that codes for insulin to the bacteria. So when the bacteria was exposed to lactose, the lac-operon was switched on and the first part of the enzyme was synthesized along with insulin. Hope to have helped ;)
Click to expand...
Amazing ............................. :p :D
 
Karim Nabil

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Mahmood Magdy said:
If you are giving CIE A2, then congrats, you do not need to study it extensively(although it will be included in the new Syllabus 2016).
However, to understand the Lac operon you have to get what an operon is at the first place.
An operon is made up of a promoter, regulator and a gene. The promoter is a length of DNA that is situated next to genes to make sure that the gene is expressed and transcribed. The regulator is simply the start and stop codons etc and the gene would be coding for the polypeptide we want. The operon as a whole switches on and off according to the environment.
Now back to the lac-operon, it is present in some bacteria like E.coli and is switched on when lactose is present thus 'lac'-operon. So normally the bacteria will synthesize an enzyme to metabolize lactose, when exposed to lactose itself. However in Genetic Eng, we added the gene that codes for insulin to the bacteria. So when the bacteria was exposed to lactose, the lac-operon was switched on and the first part of the enzyme was synthesized along with insulin. Hope to have helped ;)
Click to expand...
Thanks
 
Sadi66

Sadi66

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Mahmood Magdy said:
Dude Dude you are confusing yourself, your graph has 2 atrial systoles which is incorrect, you have only one atrial systole per cardiac cycle. The answer C is the correct one. The picture has all the explanation you need to know, May/June 2014 paper 22 last question has the timing thing. With those, I guess you will understand it. The question says AV valves and SL valves have to be closed, and the total time, which will be from 2-1(arrows on the graph) and from 4-3(on the graph too).
Click to expand...
Mahmood Magdy said:
Dude Dude you are confusing yourself, your graph has 2 atrial systoles which is incorrect, you have only one atrial systole per cardiac cycle. The answer C is the correct one. The picture has all the explanation you need to know, May/June 2014 paper 22 last question has the timing thing. With those, I guess you will understand it. The question says AV valves and SL valves have to be closed, and the total time, which will be from 2-1(arrows on the graph) and from 4-3(on the graph too).
Click to expand...
 
princessnoor

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Plz anyone help me solving these parts i cant figure out
Bio p52 nov10 Q1 b(i) and nov 10 p53 Q1c (i)
Nov11 p53 Q2b(ii)
 
Mahmood Magdy

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princessnoor said:
Plz anyone help me solving these parts i cant figure out
Bio p52 nov10 Q1 b(i) and nov 10 p53 Q1c (i)
Nov10 p52 Q2b(ii)
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Well P52 Q1b(i), they are asking for the value of t for INTERNODE LENGTH , X1 and X2 are the mean values of both the types of leaves. In the table 1.1, the values are written in the form of Mean +- Standard deviation, therefore 23-19, S2 and N2 are the standard deviation and the number of samples and thus 3^2 / 30, and therefore the final answer 8.9.

Nov 2010 P53 Q1c(i), the formula is for the Standard error, simply you will have to divide by the square root of the number of samples, therefore, you will divide by square root of 12, and then round the answer of 1.15 to 1 :D

Nov 2011 P53 Q2b(ii) , Chi-square test, you need the total number of samples first, which will simply be 112 (40+72) or from table 2.1 add all the values of Cross1, as offspring with tail is recessive lets say, 0.25 * 112 = 28, and therefore 112-28= 84. Just put these values in the formula given and add both these results at the end to get 6.86(after rounding to 2dp)
 
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