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To form a tetrapeptide sequence with two phenylalanines (F) and two glycines (G), we will need to determine the total number of unique combinations for these amino acids. This can be achieved using the formula for permutations without repetition: \[ \frac{n!}{n_1!n_2!} \] where n is the total number of amino acids, and \(n_1\) and \(n_2\) are the number of occurrences for each amino acid. In our case, n = 4 (since there are a total of 4 amino acids in the tetrapeptide), \(n_1 = 2\) (two phenylalanines), and \(n_2 = 2\) (two glycines). Plugging these values into the formula, we get: \[ \frac{4!}{2!2!} = \frac{24}{(2)(2)}=6 \] So there are 6 unique tetrapeptide sequences that can be formed with two phenylalanines and two glycines. Now, let's list all the possible combinations: 1. FFGG 2. FGFG 3. FGGF 4. GFFG 5. GFGF 6. GGFF Therefore, there are 6 unique tetrapeptide sequences that contain two phenylalanines and two glycines.

For part b, we need to form a tetrapeptide sequence with two phenylalanines (F), one glycine (G), and one alanine (A). Following the same logic as in part A, we will use the formula for permutations without repetition: \[ \frac{n!}{n_1!n_2!n_3!} \] Here, n = 4 (since there are a total of 4 amino acids in the tetrapeptide), \(n_1 = 2\) (two phenylalanines), \(n_2 = 1\) (one glycine), and \(n_3 = 1\) (one alanine). Plugging these values into the formula, we get: \[ \frac{4!}{2!1!1!} = \frac{24}{(2)(1)(1)}=12 \] So there are 12 unique tetrapeptide sequences that can be formed with two phenylalanines, one glycine, and one alanine. Now, let's list all of the possible combinations: 1. FFGA 2. FFGG 3. FGFG 4. FGFA 5. FGAG 6. FGAF 7. GFFA 8. GFFG 9. GFAG 10. GFAF 11. AGFF 12. AGGF Therefore, there are 12 unique tetrapeptide sequences that contain two phenylalanines, one glycine, and one alanine.