Combinatorics of aliphatic amino acids

Abstract

This study combines biology and mathematics, showing that a relatively simple question from molecular biology can lead to complicated mathematics. The question is how to calculate the number of theoretically possible aliphatic amino acids as a function of the number of carbon atoms in the side chain. The presented calculation is based on earlier results from theoretical chemistry concerning alkyl compounds. Mathematical properties of this number series are highlighted. We discuss which of the theoretically possible structures really occur in living organisms, such as leucine and isoleucine with a chain length of four. This is done both for a strict definition of aliphatic amino acids only involving carbon and hydrogen atoms in their side chain and for a less strict definition allowing sulphur, nitrogen and oxygen atoms. While the main focus is on proteinogenic amino acids, we also give several examples of non-proteinogenic aliphatic amino acids, playing a role, for instance, in signalling. The results are in agreement with a general phenomenon found in biology: Usually, only a small number of molecules are chosen as building blocks to assemble an inconceivable number of different macromolecules as proteins. Thus, natural biological complexity arises from the multifarious combination of building blocks.

Appendix

There are (at least) two ways to prove the recurrence formula (4). The “elegant” way uses Pólya’s Enumeration Theorem (Pólya 1937) (also known as Redfield-Pólya’s Theorem) to do so: Here, we first notice that instead of counting rooted trees with n nodes and at most three children at every node, we can also count rooted trees with exactly three children at every node, and n internal (non-leaf) vertices. To see this, we simply complete a rooted tree by attaching up to three leaves to every node of the original tree, so that all vertices except the new leaves have degree three (Fig. 6). Such trees are called rooted ternary trees. Then, we note that the symmetry between such trees is due to arbitrarily sorting the children of any node. The mathematical representation of this fact is the symmetric group S₃. One proceeds by computing the cycle index of S₃, which is

$$ Z({{\hbox{S}}_3}) = \frac{1}{6}\left( {{a_1}^3 + 3{a_1}{a_2} + 2{a_3}} \right). $$

Fig. 6

Illustration of the replacement of rooted trees with n nodes and at most three children at every node (a) by rooted trees with exactly three children at every node, and n internal (non-leaf) vertices (b). The nodes of the original tree (circles) become internal nodes of the new tree by adding new leaves (squares)

It is helpful to consider the generating function $T(z)$ for the number of rooted ternary trees, which is defined as $T(z) = {x_0}{z^0} + {x_1}{z^1} + {x_2}{z^2} + \ldots$, where x _n are exactly the numbers introduced above. Pólya’s Enumeration Theorem (Pólya 1937) then tells us that this function fulfils the functional equation

$$ T(z) = 1 + \frac{1}{6}z\,\left[ {T{{(z)}^3} + 3T(z)T\left( {{z^2}} \right) + 2T\left( {{z^3}} \right)} \right] $$

(8)

With the functional equation, we can now calculate the coefficient of any power ${z^n}$ in the generating function: for example, regarding $T\left( {{z^3}} \right)$, the three coefficients must add up to n − 1. Doing so, we directly reach Eq. 4. We omit all further detail and refer the reader to any textbook about generating functions (e.g. Wilf 1994).

Now, we show a direct way to prove Eq. 4. Before doing so, we note that a slightly more complicated way of computing ${x_n}$, which ultimately also results in Eq. 4, was described by Henze and Blair in 1931. We simplify their presentation to calculate x _n as follows: To any node, we may attach three trees such that subtrees have pairwise different numbers of nodes. We explicitly allow that a tree has zero nodes, in which case we attach nothing—recall that we have defined x ₀ = 1 above. For this case, we do not have to take into account symmetry considerations, since the subtrees must be pairwise different. So, we have

$$ \sum\limits_{{\begin{array}{*{20}{c}} {1 + j + k = n - 1} \\{i < j < k} \\\end{array} }} {{x_i}{x_j}{x_k} = \frac{1}{6}} \sum\limits_{{\begin{array}{*{20}{c}} {i + j + k = n - 1} \\{i \ne j,j \ne k,k \ne i} \\\end{array} }} {{x_i}{x_j}{x_k}} $$

(9)

possibilities of doing so. Next, assume that we attach two trees of the same size j, and a third tree of size i. There exist $\left( {\begin{array}{*{20}{c}} {{x_j} + 1} \\2 \\\end{array} } \right) = \frac{1}{2}\left( {{x_j} + 1} \right){x_j}$ ways to choose two trees of size j, since this is a combination with repetition. We calculate the number of tree as

$$ \sum\limits_{{\begin{array}{*{20}{c}} {i \ne j} \\{i + 2j = n - 1} \\\end{array} }} {{x_i}\left( {\begin{array}{*{20}{c}} {{x_{{j + 1}}}} \\2 \\\end{array} } \right) = \frac{1}{2}\left( {\sum\limits_{{\begin{array}{*{20}{c}} {i \ne j = k} \\{i + j + k = n - 1} \\\end{array} }} {{x_i}{x_j}{x_k} + \sum\limits_{{\begin{array}{*{20}{c}} {i \ne j} \\{i + 2j = n - 1} \\\end{array} }} {{x_i}{x_j}} } } \right)} $$

(10)

Finally, all trees may have size ${x_i}$. In this case, there exist

$$ \left( {\begin{array}{*{20}{c}} {{x_i} + 2} \\3 \\\end{array} } \right) = \frac{1}{6}\left( {{x_i} + 2} \right)\left( {{x_i} + 1} \right){x_i} = \frac{1}{6}\left( {{x_i}^3 + 3{x_i}^2 + 2{x_i}} \right) $$

(11)

ways to choose three trees of size i. We calculate

$$ \sum\limits_{{3i = n - 1}} {\left( {\begin{array}{*{20}{c}} {{x_i}} \\3 \\\end{array} } \right) = \frac{1}{6}\left( {\sum\limits_{{i = j = k,i + j + k = n - 1}} {{x_i}{x_j}{x_k} + 3\sum\limits_{{i = j,i + 2j = n - 1}} {{x_i}{x_j} + 2\sum\limits_{{3i = n - 1}} {{x_i}} } } } \right)} $$

(12)

Note that if n − 1 is not divisible by three, then all of these sums are empty and, by definition, equal zero. Now, we add these three values, and sort them: First, we put all sums of products ${x_i}{x_j}{x_k}$. One can easily see that this equals $\frac{1}{6}\sum\limits_{{i + j + k = n - 1}} {{x_i}{x_j}{x_k}}$ as desired: For the first sum on the right-hand side of Eq. 10, there are three possibilities of choosing two indices from i, j, k to be equal, as we can permute the indices. Similarly, we can reproduce the second summand of Eq. 4. Finally, the third summand directly comes from Eq. 12, which completes the proof.