If a student has four amino acids, Glycine, Phenylalanine, Glutamine, and Aspartamine, how many unique proteins can he form without using any amino acid more than once in any protein?
Solution: the safest way to approach this question is to write out all of the chains you can create starting with one of the four amino acids. This is the more conceptual approach. For example, starting with Gly you could form: Gly-Phe-Glu-Asp, Gly-Phe-Asp-Glu, Gly-Glu-Phe-Asp, Gly-Glu-Asp-Phe, Gly-Asp-Glu-Phe, and Gly-Asp-Phe-Glu. That is six unique chains by starting with Gly. This could be done once for each of the four amino acids, so there are 6 times 4, or 24 possible proteins. Answer C is thus correct. using n! (n factorial) will also work in this case, only because each item can only be used once. the n represents the number of items and the math becomes: 4x3x2x1=24. Be careful, however, using the factorial method. We have seen many students who rely on this method use it when each item could be used more than once--in which case it gives an incorrect answer. Had thsi question allowed you to use each item multiple times (proteins such as Gly-Gly-Gly-Gly, and Phe-Phe-Phe-Phe would have been possible), you could obtain the answer using x^y where x is the number of slots, and y is the number of values possible for each slots. This only works, however , if any slot can use any of the available values (i.e., license plates). It is also pertinent to this question to remember that proteins are not reversible. In other words, A-B-C is not the same as C-B-A because in the first case A is the N-terminus and in the second case A is the C-terminus. This would also be true of combinations of nucleotides, where one side would be 5' and the other 3'