Help please (6) Suppose that life is discovered in other parts of our solar syst
ID: 211210 • Letter: H
Question
Help please
(6) Suppose that life is discovered in other parts of our solar system, and the same ami he as are coded for by the mRNA found in these extraterrestrial forms of life. Surprisingly,t number of nucleotides in these alien cells differs from what is observed on our planet Cells from the noon of Saturn have only two types of nucleotides. Does this have implications for the process of translation? What would be the minimum size for a codon? Explain the reasoning for your answer. 1 point) b) Cells from Mars have only three types of nucleotides in their mRNA? Again, what would be the minimum size for a codon? Why? (1 point) c) Cells from a moon around Jupiter have five different nucleotides in their mRNA. What would be the minimum size for a codon? Explain your reasoning. (1 point)Explanation / Answer
Answer:
Based on the given information, the extraterrestial forms of life discovered in other parts of the solar systems have the same amino acids. That means, these life forms also have 20 amino acids.
Codon size should be adequate to cover 20 different amino acids.
Number of amino acids that can be encoded by a codon of size (C) with (N) different nucleotides is NC. For example for a triplet codon with 4 different nucleotides C = 3 and N = 4. Total number of possible combinations = 43 = 64. Due to degeneracy of code, these 64 different triplet codons codes for 20 amino acids.
Case 1) Only two different nucleotides:
In this case, the minimum codon size to cover 20 amino acids is 5. Number of possible combinations with a codon size of C = 5 and number of nucleotides N = 2 is NC = 25 = 32. It can't be less than 5 (for example 4) because in that case the number of unique combinations would not suffice to code for 20 different amino acids (24 = 16).
Case 2) Three different type of nucleotides:
In this case, the minimum codon size to cover 20 amino acids is 3. Number of possible combinations with a codon size of C = 3 and number of nucleotides N = 3 is NC = 33 = 27. It can't be less than 3 (for example 2) because in that case the number of unique combinations would not suffice to code for 20 different amino acids (32 = 9).
Case 3) Five different type of nucleotides:
In this case, the minimum codon size to cover 20 amino acids is 2. Number of possible combinations with a codon size of C = 2 and number of nucleotides N = 5 is NC = 52 = 25. It can't be less than 2 (for example 1) because in that case the number of unique combinations would not suffice to code for 20 different amino acids (51 = 5).
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