Answer :
27 different tripeptides can be formed with three different amino acids (methionine, histidine, and arginine).
This can be simply calculated by formula [tex]x^{y\\[/tex] where "[tex]x[/tex]" represents no. of amino acids & "[tex]y[/tex]" represents the length of peptide.
In this case, [tex]x[/tex]= 3 & [tex]y[/tex]=3
Hence, Total no. of tripeptides = [tex]3^{3}[/tex] = 3×3×3 = 27
All the possible different tripeptides from methionine(M), histidine(H), and arginine(R) are as follows:
Case 1: If all the three amino acids are same,
1) MMM 2) HHH 3) RRR
Case 2 : If two amino acids are same & one amino acid is different,
1)) MMH 2) MMR 3) HHM 4) HHR 5) RRM 6) RRH 7) MHM 8) HMH 9) MRM 10) HRH 11) RHR 12) RMR 13) HMM 14) RMM 15) MHH 16) RHH 17) MRR 18) HRR
Case 3: If all the amino acids are different,
1) MHR 2) MRH 3) HMR 4) HRM 5) RMH 6) RHM
So the total possibilities are = 3+18+6 = 27
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