some simple a-level questions....

[mak10]

I have come across certain simple questions in my MCQ past papers whose answers I am not really very sure of... so here it goes:

 

1) if there were 34 amino acids and DNA only contained two types of nitrogenous bases, what would be the minimum number of bases per codon that could code for proteins?

 

2) what single feature is correct for all enzymes?

 

3) A capillary tube is inserted into the phloem. what force/pressure causes the sap to flow out of the tube?

 

4) what proportion (in %) of the stored chemical energy in grass is converted into new tissue by a growing cow?

 

ask me for the options if you need'em.

 

-mak10

[VendingMenace]

Well, just a quick guess, let me know if you want explinations rather than verification

 

1)6 because 2^5 = 32 <36 so it must be 2^6, wich gives 64 possible combinations, enough to code for 34 ammino acids

2)They lower the activation energy of a reaction

3)Adhesive forces (ie. the molecules wish to stick to the tube quite abit) i could be wrong here, just a guess.

4) I have no idea.

[mak10]

could you explain the reasoning in q.1 ?? is there some kid of formula for these things?

 

-mak10

[VendingMenace]

could you explain the reasoning in q.1 ?? is there some kid of formula for these things?

 

sure thing....

 

Ok, so you have 32 amino acids and you need to code for each one of them. That means that you need to be able to make at least 32 disinct DNA sequences. Stipulated in the question was that there are only two different DNA bases (let's call then base A and base B). Thus, for each position on the DNA you have two choices (base A or base B). With me so far?

 

Ok then, if we were to consider a stretch of DNA 1 base pair long, we would have two choices, either A or B.

 

if we were to consider a stretch of DNA 2 base pairs long, we would have 4 choices,

AA

AB

BA

BB

 

if we were to consider a stretch of DNA 3 base pairs long, we would have 8 choices, namley;

AAA

AAB

ABA

BAA

ABB

BAB

BBA

BBB

 

ok, so, in general, if you take a sequence wich is N base pairs long and you have X number of base pairs to chose from and you are allowed to repeat your choices (ie. the string can have mulptiple of the same base) the formula for the number of possible permutations is;

 

X^N

 

in this case we see that if we have two base pairs, then a sequence of 5 bases will give us 2^5 or 32 different permutations. However, we require to be able to code for 34 ammino acids. Thus, we need at least 34 unique permutations. So, the sequence needs to be longer. If it is 6 base pairs, then it would have 2^6 = 64 possible permutations and would be able to code for 34 ammino acids.

 

Cool. I hope that makes sense. Let me know if it was unclear anywhere or if you need more information. ;D

[Mokele]

4) what proportion (in %) of the stored chemical energy in grass is converted into new tissue by a growing cow?

 

1-2%, 3% tops. Cows, as endotherms, are very wasteful, burning most of their food as heat.

 

Mokele

[mak10]

thx skye and mokele... i just had my exams today anyways. one of the questions was similiar to 1) and it was like a polypeptide is made of 'n' number of amnio acids and there are 'r' number of different amino acids available. so what is the expression that gives the minimum number of different polypeptides possible with these variables? i think i chose r^n.... unless i remember the question incorrectly. is it right?

 

-mak10

[VendingMenace]

r^n is correct -- good job!