e-n-i-g-m-a
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#1
Report Thread starter 13 years ago
#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?
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Comp_Genius
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#2
Report 13 years ago
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well there are two bases, and you want 34 codons:
2^5 = 32
2^6 = 64, so you want 6 bases per codon.
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e-n-i-g-m-a
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#3
Report Thread starter 13 years ago
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heres another one ...
ONe complete turn of the double helix of DNA contains 10 pairs of bases and is 3.4 nm long. What is the approximate length of the DNA coding sequence of lysozome, a protein of 129 amino acids ?
A. 132 nm
B. 113 nm
C. 66nm
D. 44nm
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Comp_Genius
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#4
Report 13 years ago
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well each turn contains 10 bases (on one chain), you have 129 amino acids, so you have 387 codons. This means 38.7 turns are required. Since each turn is 3.4nm long, the length is approximately 131.58nm. Thence, A is the answer.
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e-n-i-g-m-a
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#5
Report Thread starter 13 years ago
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i dont understand that how 38.7 turns are required ??
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oxymoron
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#6
Report 13 years ago
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(Original post by e-n-i-g-m-a)
i dont understand that how 38.7 turns are required ??
Each amino acid is coded for by 3 bases - so 129 amino acids requires 129 x 3 bases of DNA to code for it = 387 bases.

As you are told that one turn of the helix is 10 bases, that means 387 bases will be 387/10 turns = 38.7 turns.
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