Greetings all~~ I have copied a problem below (with the answers included). I amt
ID: 5413 • Letter: G
Question
Greetings all~~I have copied a problem below (with the answers included). I amtotally confused on parts a-c to why certain numbers were used as abase (like 3 ^ 3 veses 2^3). Can someone simplify what this problemis trying to show?
Other features of the reading of mRNA into proteins being the sameas they are now (i.e., codons must exist for 20 different aminoacids), what would the minimum codon length be if the number ofdifferent bases in the mRNA were each of the numbers below, insteadof four?
a. two
Answer: Codon length would be 5 bases. If only twobases existed in mRNA and codons were only four bases long, therewould only be 24 = 16 possible codons, not enough tocode for 20 amino acids. So, in order to encode 20 different aminoacids, one would need to read codons 5 bases long since there wouldbe 25 = 32 possible codons.
b. three
Answer: Codon length would be 3 bases: With threebases in mRNA, 33 = 27 possible combinations.
c. five
Answer: Codon length would be 2 bases: With fivebases in mRNA, 52 = 25 possiblecombinations.
Explanation / Answer
The problem is asking you that IF there weren't four types ofbases, but instead the number that they gave you, what would be theminimum codon length... so for a., for example, you need to find the minimum number ofbases that you need in order to reach 20. (since there are 20 aminoacids). They're telling you that you only have 2 bases. So, askyourself, 2 times what equals something 20 or greater? The answeris 5. Therefore, the codon length is 5 bases. Apply this to therest.Hope this helps!
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