When designing protein libraries for screening, we must do everything possible to generate a diversity of protein variants within the physical limits of what can be screened. The goal of this chapter is to introduce the language of probability into protein engineering experiments, which in turn will answer some common questions. The source for this experiment is "A Guide to Modern Protein Engineering Experiments" [German] K.M. Arndt, K.M. Miller, eds.
Operation method
Protein library design and screening - probabilistic calculations Move This section is divided into two main subsections. In Section 3.1, we pose questions related to the design and characterization of libraries. In Section 3.2, we analyze the evaluation of the quality of the library or the significance of the library screening results. In the first two sections, we give formulas and provide numerical examples, which are illustrated by examples without detailed theoretical explanations. The full theoretical explanation can be found in the notes (see Section 8.4). For more product details, please visit Aladdin Scientific website.
3.1 Library characterization
We design a library containing n possible different theoretical variables. We randomly sample m times from this theoretical library; forming a sample set of m variables. We ask the following question.
A. How many of the selected m variables do we expect to be absent from the theoretical variables (all n)?
B. What is the probability that at least one theoretical variable (all n) of the m variables selected will not be sampled? What is the probability that at most a certain number of theoretical variables (all n) will not be sampled?
C. How many times do we expect a particular variable i to appear in the sample? More generally, what is the probability that i occurs r times?
These questions are posed as a set because the solution to the above problems is based on the same theory as described in Sections 3.1.1 and 3.1.2. We denote by a ruler the probability that a variable i occurs any number of times when we randomly sample it once from the theoretical pool.
We split into two different cases, which are treated differently in the following subsections:
( 1 ) In Section 3.1.1, the equal probability outcome, i.e., the probability that n variables are sampled is the same. Here Pi = 1/n and this example is treated as a generalization.
( 2 ) In Section 3.1.2, the results here appear to be of the non-equal probability type, i.e., some variables have a better probability of being sampled than others. By definition, it is difficult to generalize this situation, because for questions A, B, and C above, the answers depend on the probabilities themselves. As an example, we deal in detail with a non-equal probability case.
If there is no deviation in the frequency of occurrence of a given sequence or in the encoded characteristic we are interested in, it is called the equal probability case. In the library selection for this case, we consider the property we are interested in to be equiprobabilistic if it is independent of, or has a negligible effect on, the other parameters. The non-equal probability case is considered if the considered codons have preference or simplicity, or if the result affects the whole system. An example of a result affecting the whole system is as follows: when screening for high catalytic activity of enzyme X in vivo, the increased product can be toxic to the whole system, producing an adverse reaction to the detection of the enzyme's catalytic activity. Such a parameter is generally too complex in the past to be stated systematically. However we can consider the use of preference codons, where the appearance of amino acids is a non-equal probability event and other parameters, although also non-equal probability, can be quantified.
3.1.1 Equal probability scenarios
This is the simplest scenario, in which all outcomes can be considered as equal probability events: after sampling the members of the library, all n variables have the same chance of being selected, i.e., Pi = 1/n. 
Note that the expectation of the number of missing variables need not be an integer; here the expectation of the number of missing variables is 45-46 (the overall theoretical number of variables is 1 X 106 ).
If we sample m = 2 X 107 times from a pool of n = 1 X 106 theoretical variables, then λ = 0.00206. Thus, the expected value of a missing variable is either 0 or 1, and is much closer to 0. In fact, the probability of 0 or 1 missing variable is the only non-trivial probability event in this example, and 0 missing is much more probable (> 99%, see 3.1.1 Calculations in Problem B).
We can conclude that a sampling size of 20 times the theoretical pool (m = 20 n) leads to a complete sampling of the theoretical pool in principle, as shown in Figure 8.5. A 10-fold oversampling can lead to a small fraction not being sampled (which tends to 0 when the theoretical library has less than 104 variables, see Table 8.2); for many experimental purposes, knowing that a small fraction has not been sampled is sufficient. 
Note that Table 8.2 gives some counterintuitive results: if the sample size m is equal to the pool size n, we can expect that on average roughly 37% of the theoretical variables will not be sampled; even when m=2n, the average probability that a theoretical variable is not sampled is 13.5%, which is much smaller than it should be if it is sampled at all. 
For our main example (n= 1 X 106; m= 2 X 107 ), the probability that fewer than 50 out of m selections from n theoretical variables will not be sampled can be calculated by summing the probability that each 0, 1, 2, ... up to 49 will not be sampled, which is about 78%, i.e. 
Similarly, the probability that fewer than 60 are not sampled is approximately 98%. To simplify the calculation, an Excel calculation sheet can be utilized (see section 8.2 and Figure 8.1). 
i.e. less than 1% chance. Does a result that occurs 3 times for a variable suggest a bias in the system (e.g., due to selection pressure)? This issue is discussed in section 8.3.2. 
Finally, the probability of a variable occurring 10 times or less can be calculated by adding up the probabilities of 10 occurrences, 9 occurrences, etc., up to the probability of 0 occurrences, either by hand or by using an Excel worksheet (see 8.2 and Figure 8.1). We see that the probability of 10 or fewer occurrences in this example is about 58.3%.
3.1.2 An example of non-equivalent probabilities
We give an example of an event with non-equal probability of occurrence, where some of the n variables will occur more or less often than the average. As noted at the beginning of Section 3.1, it is difficult to derive a general rule for non-equivalent probability events, since the answer depends on what the specific probabilities are. We give some concise calculations in the examples we provide here, along with hints at the difficulties that may be encountered.
In the absence of selection pressure, non-equivalent probability events can occur in preferred DNA libraries, and such library preferences can arise by design or by accident. An example is an accidental bias that occurs during oligonucleotide primer synthesis. Such biases can be discovered after sequencing a randomly sampled DNA library. If we define theoretical libraries in terms of coding protein sequences rather than DNA sequences, then non-equivalence probabilities will still occur. For example; when " NNN" type codons are encoded, there will be a probability that each of the 6 codons encoding leucine will occur with a probability equal to the probability that the only codon encoding methionine will occur, and therefore the probability of getting a protein containing leucine at the same site will be 6 times higher than the probability of getting a protein containing methionine.
In the presence of selection pressure, it is generally not possible at the beginning of an experiment to define precisely numerically the selective advantage or disadvantage associated with certain mutations. Indeed, when there is selection pressure it is better to ask the question the other way around: rather than asking with what probability a particular variant occurs given a particular sequence preference, it is better to ask the experimental frequency of sequences that we can observe in the subset of sequences that are selected. We can determine whether such frequencies are expected as in question C) in 3.1.13, or whether there is a bias.
1 ) Characterization of our non-equivalent probability example
We consider sampling in decapeptides. If the 10 amino acids are randomly distributed (NNN) in the decapeptide, there will be 2110 = 1. 67X1013 different theoretical variants (we treat terminators the same as the other amino acids). Because of codon concatenation each amino acid is selected with non-equal probability; nevertheless, the probability of differences that can occur is the following five:
Met and Trp are encoded by one of the 64 codons; Phe, Tyr, His, Gln, Asn, Lys, Asp, Glu, and Cys are encoded by two of the 64 codons. Therefore, the amino acid species corresponding to the above five probabilities are 2, 9, 2, 5, and 3 amino acids or stop codons, respectively.
2 ) Question A
How many of the n theoretical variants, m, can we expect not to occur?
To solve this problem, the element is to consider codon parsimony. We assume that each specific decapeptide has the same probability of occurrence as any other decapeptide of the same amino acid composition (and therefore of the same simplicity), irrespective of its amino acid ordering. The answer to Problem A is then the parameter λ ( see Note 7), computed as 
The above summation numbers are summed over different orderings of the array of all 5 numbers ( n1, n2, n3, n4, n5 ) corresponding to the 5 probabilities of simplicity of a given codon. Each " 5-array" taken from the set { 0, 1, 2, ..., 10 } represents how many amino acids of each of the 5 different degrees of concurrency are present in the decapeptide. Since we only consider 10 amino acids. For example, ( n1, n2, n3, n4, n5 ) = ( 8, 0, 0, 1, 1 ) means that 8 positions in a decapeptide are encoded by non-simplex amino acids (1/64), excluding amino acids with 2/64 and 3/64 simplicity, and that 10 positions have one amino acid each with 4/64 and 6/64 simplicity.
P( n1, n2, n3, n4, n5 ) can be calculated as 
This P ( n1, n2, n3, n4, n5 ) is a new name for Pi that we briefly introduced at the beginning of Section 3.1, and implies the probability that a particular protein variant occurs any number of times when we randomly sample it once from the theoretical library.
Unfortunately, a simpler formulation for the parameter λ does not seem to exist; to compute the summation of A one must, as always, utilize computer programming.
For example, we want to randomly take m = 1014 decapeptides from the theoretical library n = 2110 (i.e., about 1.67 X 1013; see Figure 8.2).
The answer to this set of m and n is λ = 5.04 X 1012, so we expect 5.04 X 1012 random samples from a theoretical pool of 1.67 X 1013 not to occur, at an approximate rate of 30%.
3 ) Question B
What is the probability that at least one peptide variant will not be produced in the sample? What is the probability that at most 50 or 60 peptide variants are not sampled?
Such a probability can be obtained from the equation (8.2 ) in Section 3.1.1 regarding equal probability events; however, the value of the parameter λ must be calculated from equation (8.6 ). Thus, the probability that at least one peptide is not sampled can be approximated as (see Note 5).
Calculating the probability that "at least" or "at most" a particular number of peptides will not be sampled can be derived from the method of 8.3.1.12 ), which is not repeated here. .
For example, we again randomly remove m = 1014 decapeptides from the theoretical library n = 2110 ≈ 1.67 X 1013; and find λ = 5.04 X 1012 in subsection 2 ) of 8.3.1.1. Clearly, for the expected missing variant of λ = 5.04X 1012, it is intuitively obvious that the probability of missing at least one should be high. Using equation (8.8 ), we can indeed derive this probability to be 100%. In other words, the probability that all protein variants can be produced is 0%.
In fact, the probability that at most 50 or 60 of all peptide variants are not produced is also 0%. For such a high value of λ (5. 04X1012 ), this result is not surprising.
4 ) Problem C
How many times can we expect a decapeptide containing 7 codons with probability 3/64 and 3 codons with probability 4/64 to occur in a sample? What is the probability that such a decapeptide will occur 10 times?3 What is the probability that such a decapeptide will occur 10 times? What is the probability that it will occur 10 times or less?
Since the result will be the same for all the peptides defined in this question, there is no need to concern ourselves with the exact sequence (e.g., the order and names of the amino acids). The answers to these questions differ only slightly from those given in 8.3.1.1 (3 ). Since the probability of a decapeptide appearing after sampling an arbitrary decapeptide is P [see 8.3.1.1, 2)], the number of times we expect a decapeptide to appear is mp, where m is the number of times it is sampled. Given that some decapeptides are sampled more than once, the probability will increase with the number of samples, and the expected probability that a decapeptide occurs a certain number of times will approximately follow a Poisson distribution with parameter λ = mp (see Notes 3 and 8).
That is, we can expect to find the decapeptide λ = mp = 1014 X ( 1.21 X 10-13 ) ≈ 12.14 times in the sample. The probability of ten occurrences is
Similarly, the probability of a particular decapeptide occurring 3 times is approximated to be 0.16%, which is much smaller than the chance of it occurring 10 times ( 10% ). Adding up all the probabilities of less than 10 occurrences, we get an approximate probability of less than or equal to 10 occurrences of the decapeptide of 33% (Figure 8.3).
3.2 Tests for experimental bias
The basic problem in this section can be formulated as the problem of experiments with two or more possible observations performed multiple times, i.e., multiple sampling. The theoretical probability (expected) of each observation is known, while the experimental probability of an observation can be obtained by observation. Is the difference between the theoretical and experimental probabilities a signal that the experiment is significantly biased?
As mentioned in Section 8.1, such an analysis can be routinely applied to DNA sequencing results to test for bias before or after selection. Applying the X2 test described below allows the experimenter to determine whether the observed bias is significant or insignificant.
The problem is now posed by the following notation: a test has a theoretical probability of getting k observations A1, A2, ..., Ak with probabilities p1, p2, pk, respectively. e.g., sampling a library of proteins yields k = 2 results: functional or non-functional. This test is done n times (i.e., n proteins are randomly sampled), and the number of times the result occurs in each observation is Y1, Y2, ..., Yk times.
X2 test
The statistic in the X2 test is 
The hypothesis we are testing, called the "null hypothesis," is the average value of the test when it is conducted in a suitable manner, that is, the probabilities of the individual observations obtained in this manner are consistent with their theoretical probabilities. A large q-value indicates a significant difference between the true experimental result and the desired result, in contradiction to the null hypothesis.
The statistic q is compared with the value Xα2 obtained for the parameter k-1 at an arbitrarily chosen significance level α. The value of the parameter α is specified by the user. The value of the parameter α is specified by the user (see Note 9 for the significance level). The test is as follows:
When q < Xα2 (k-1), the original hypothesis is accepted; there is no good reason for bias.
When q > Xα2 (k-1), the hypothesis is rejected; there are good reasons to believe that bias exists.
Critical values for the X2 distribution can be found in most statistics textbooks and in many programs, such as Excel. 
The original hypothesis (that the experiment was conducted in the correct manner without bias) is accepted (Figure 8. 4 ). Figure 8.7 depicts the number of observations that characterize whether the hypothesis should be accepted (characterized by 17 to 33 observations) or rejected (all other outcomes). 
The statistic q, and the nonconformity to the expected outcome, are not large enough to conclude that there is a flaw in the student's procedure; the original hypothesis cannot be rejected. However, if we take α = 10%, then X0.102 ( 2 ) = 4.605, and the test becomes: q = 5.6 > 4.605 = X0.102 ( 2 ) Thus, we are able to reject the original hypothesis, and conclude that the test is now the desired one.