1. Given the following short stretch of mRNA sequence: 5’ – GGG CCC UGA UGU AAC

Question

1. Given the following short stretch of mRNA sequence:
5’ – GGG CCC UGA UGU AAC AUA GAU AGG – 3’
A. What is the sequence of the DNA template from which it was transcribed? (make sure to indicate polarity).

--I got: 3' CCC GGG ACT ACA TTG TAT CTA TCC 5'


B. It is possible to translate RNA in vitro using cell-free preparations obtained from E. coli lysates. Under certain artificial experimental conditions, an initiator methionine codon is not required and translation can be initiated at any codon in the RNA molecule. If you were to use just this short RNA in such a system, how many different peptides would you obtain? Consider all reading frames and consider a peptide to consist of two or more amino acids. Show how you got your answer.

I'm assuming 3^8 because there are 8 possiblesets of 3 codons in the above sequence. Not sure..

C. Would the same peptides be generated by the RNA sequence that is complementary to the sequence given above?

D. Assume that the RNA sequence given above is actually a fragment of a larger mRNA that is synthesized in E. coli, and, that this mRNA is translated but you do not know which reading frame is used. Answer the questions below and BRIEFLY (10 words or less) explain your reasoning.
i) Could this RNA be from the beginning of a protein coding region?
ii) The middle?
iii) The end?

Don't really understand C and D.. please guide me in the right direction!
Thanks!

Explanation / Answer

1. A. 5’ – GGG CCC UGA UGU AAC AUA GAU AGG – 3’ 3' - CCC GGG ACT ACA TTG TAT CTA TCC - 5' Congrats. You were correct! B. Okay so here's what you have in RNA: GGG CCC UGA UGU AAC AUA GAU AGG (8 codons labelled 1 - 8 below) 1 2 3 4 5 6 7 8 Let's assume we start our reading frame at the first letter in a codon and are starting at codon one (or we are starting with codon GGG). From here we can make 7 different chains by adding more codons to the mix ( GGG CCC, GGG CCC UGA . . . 7) Now lets keep the same reading frame and start at codon two, we can now only make 6 different chains (CCC UGA, CCC UGA UGU . . . 6). everytime we begin at a later codon, we can make one less chain. So . . . 7 + 6 + 5 + 4 + 3 + 2 + 1 , or 7! is how many chains we can make from that reading frame. Now let's change our reading frame so that we start at the second codon: GGC CCU GAU GUA ACA UAG AUA notice that when we do this we have reduced our number of codons to seven. The same patter follows: starting at the first codon, 6 chains are possible starting at the second, 5 chains and so on . . . 6 + 5 + 4 + 3 + 2 + 1 or 6! The same pattern will occur at the last reading frame: GCC CUG AUG UAA CAU AGA UAG Notice that again there are only 7 codons, so: 6 + 5 + 4 + 3 + 2 + 1 or 6! Therefore your answer is 7! + 6! + 6!, assuming that each chain is unique (remember that when codons are transcribed to amino acids there is overlap - where more than one codon can code for the same amino acid) C.No, heres why: This is your original sequence: GGG CCC UGA UGU AAC AUA GAU AGG And this is the sequence complimentary to it: CCC GGG ACU ACA UUG UAU CUA UCC The complementary sequence has the exact opposite bases as the original - theres no way they could code for the same peptide chains! D. i) No, there is no start codon (UAC). ii) Yes it could, there is nothing preventing it from being there. iii) No, there is no stop codon (UTT, AUC, ACU). Answered to the best of my abillity - Good Luck :)

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