Journal of Bacteriology and Infectious Diseases

Research Article - Journal of Bacteriology and Infectious Diseases (2017) Volume 1, Issue 1

Analysis of antibody by Gamma function

Bin Zhao*, Aibing Li, Lichun Liang

College of Science, Northwest A&F University, Yangling, Shaanxi, China

*Corresponding Author:
Bin Zhao
College of Science
Northwest A&F University
Yangling
Shaanxi
China
Tel: +86 13028517572
E-mail:
[email protected]

Accepted Date: October 30, 2017

Citation: Zhao B, Li A, Liang L. Analysis of antibody by Gamma function. J Bacteriol Infec Dis. 2017;1(1):1-8

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Abstract

 Along with the rapid development of genetic engineering technology and antibody engineering technology, humanized monoclonal antibody has been rapidly developed and gradually replaces the rat sourced monoclonal antibody. In this paper, we establish two new logarithmically completely monotonic functions involving the gamma function according to two preferred interaction geometries, necessary and sufficient conditions are presented for one of them to be logarithmically completely monotonic. As a consequence, a sharp inequality involving the gamma function is deduced to solve the problems of genetically engineered antibody.

Keywords

Gamma function, Genetically engineered antibody, Logarithmically completely monotonic, Inequality, Psi function

Introduction

Antibodies have been proven to be indispensable tools for biomedical applications. Different engineered antibodies have been development for various purposes according to the amino acid sequence and/or spatial structure of protein (Figure 1). At present, it is still difficult to predict the optimal structure of antibodies. Topology knowledge can be important in antibody application as well as transformation. Theoretically, we can obtain desired antibodies by using protein/gene engineering technology. For instance, we can transform the complementarity determining region (CDR) to promote the affinity of the antibody to antigen. Similarly, we could also transform any domain of antibody to make it bind with any desired target. Under this vision, topology is a powerful tool to predict the structure of protein and it will serve to antibody engineering. Our present work tries to explain, and predict, if possible, the change of structure, size and function of antibodies as well as their fragments from a topological perspective.

biomedical-imaging-engineered-antibodies

Figure 1: Different antibody formats. a: different antibody or engineered antibodies; b: different shape of antibody.

For Re (z) > 0 the classical Euler’s gamma function G and psi (digamma) function y are defined by

equation (1.1)

respectively. The derivatives ψ(n)(z) for n ∈equation are known as polygamma functions.

For ψ(z) [1], the following series representations are established:

equation (1.2)

equation (1.3)

equation (1.4)

Where, g = 0.577215664901...denotes the Euler’s constant.

We next recall that a function f is said to be completely monotonic on an interval I , if f has derivatives of all orders on I which alternate successively in sign, that is,

(−1)nf(n) (x) ≥ 0 , (1.5)

for all x ∈ Iand for all n ≥ 0 . If inequality (1.5) is strict for all x ∈ Iand all n ≥ 0 , then f is said to be strictly completely monotonic [2-5]. The classical Bernstein–Widder theorem [6] states that a function f is completely monotonic on (0, ¥) if and only if it is a Laplace transform of some nonnegative measure m , that is,

equation (1.6)

Where, m (t ) is non-decreasing and the integral converges for x > 0.

We recall also that a positive function f is said to be logarithmically completely monotonic on an interval I if f has derivatives of all orders on I and

(−1)n [ln f(x)](n) ≥ 0, (1.7)

for all x ∈ I and for all n ≥ 1. If inequality (1.7) is strict for all x ∈ I and all n ≥ 1, then f is said to be strictly logarithmically completely monotonic [7-9].

The antibody structure will be changed when it binds certain target (Figure 2a), i.e.,: antigen, receptor. How to describe the changes in the view of topology? The following cases will explain it in detail. It was proved explicitly by Berg C [8] and other articles that a logarithmically completely monotonic function must be completely monotonic.

biomedical-imaging-FcRY-monomer

Figure 2: Model of pH-dependent conformational change of FcRY and structures for the FcRY monomer and dimer. a: FcRY has an extended conformation at pH 8 (s*= 7.2 S) with no predicted interaction between the CysR-FNII domains and the CTLDs. At pH 6 the CysR-FNII region folds back and binds to the CTLDs, resulting in a more compact conformation (s*= 7.9 S) that is able to bind IgY. b: Likely orientations of FcRY and FcRY–IgY on a membrane. The two FcRY monomers on the right are shown in an orientation that would allow formation of a 2:1 FcRY–IgY complex.

Anderson GD et al. [10] proved that the function

g( x) = x(lnx − ψ( x)) (1.8)

is strictly decreasing and strictly convex on (0, ¥), with two limits

equation (1.9)

From (1.9) and the monotonicity of g ( x), then the double inequalities

equation (1.10)

holds for all x > 0 .

By using the well-known Binet’s formula, Alzer H [11] generalized the monotonicity and convexity of g (x), that is, the function

ga(x) = xa(lnx − ψ(x)) (1.11)

is strictly completely monotonic on (0, ∞) if and only if α ≤ 1

Kershaw D and Laforgia A [12], proved that the function [Γ (1 + 1 / x)]x is decreasing on (0, ∞) and x[Γ (1 + 1 / x)]x is increasing on (0, ∞). These are equivalent to the function [Γ (1 + 1 / x)]1/x being increasing and [Γ (1 + 1 / x)]1/x / x being decreasing on (0, ∞) , respectively.

F Qi and Chp Chen [13] generalized these functions. They obtained the fact that for all x > 0 the function xr [Γ (1 + x)]1/x is strictly increasing for r ≥ 0 and strictly decreasing for r ≤ −1, respectively.

After the papain digestion, the remained antibody functional part (usually the Fab domain), will be smaller and the structure is also changed (Figure 1b). These changes can be revealed vividly using topology. Recently Qi F et al. [14] established another excellent result, which states that for given y ∈(−1, ∞) and α ∈(−∞, ∞) , let

equation (1.12)

The function (1.12) is logarithmically completely monotonic with respect to x ∈(−y − 1, ∞) if and only if α ≥ max{1,1 / ( y + 1)} and if α ≤ min{1,12 / ( y + 1)}, the reciprocal of the function (1.12) is logarithmically completely monotonic with respect to x ∈(−y − 1, ∞) .

Antibodies occur spontaneously gathering and forming dimer, polymer, which will influence their functions (Figure 2b). In antibody engineering practice, it urgently needs some measures to overcome this difficulty. From topology perspective, we could understand this issue as follow.

Stimulated by the above results, we put forward the function as follows: for given y ∈(0, ∞) and real number a , let the function fα,y (x) be defined by:

equation (1.13)

Our first result is contained in the following theorem.

Theorem 1. For the function (1.13), then the following statements are true:

(1) For any given y≥1, the function (1.13) is strictly logarithmically completely monotonic with respect to x Î(- y, ∞) \ {0}if and only if α ≥1 ;

(2) For any given 0 < y < 1, if α ≥ e−(1− y) / y , then the function (1.13) is strictly logarithmically completely monotonic with respect to x ∈(− y,∞) \ {0};

(3) For any given y > 0, the reciprocal of the function (1.13) is strictly logarithmically completely monotonic with respect to x ∈(− y,∞) \ {0} if and only if α ≤ 0.

Our second result is presented in the following theorem.

Theorem 2. For any given y ∈[1,∞] , let the function hy ( x) be defined on (0, ∞) by

equation (1.14)

Where γ denotes the Euler’s constant, then the function (1.14) is strictly logarithmically completely monotonic with respect to x on (0, ∞).

The following corollary can be derived from Theorems 2 immediately.

Corollary 1. For any given y ≥ 1, the inequality

equation (1.15)

holds for all x > 0.

Lemma

In order to prove our main results, we need the following lemmas.

It is well known that Bernoulli polynomials Bk (x)and Euler polynomials Ek (x) are defined by

equation

equation (2.2)

respectively. The Bernoulli numbers Bn are denoted by Bn = Bn (0) , while the Euler numbers En are defined by En = 2 En (1/2).

The following summation formula is given:

equation (2.3)

for any nonnegative integer k , which implies

equation (2.4)

In particular, it is known that for all n Î equation we have

equation (2.5)

equation (2.6)

And the first few nonzero values are

equation

E0 = 1, E2 = -1, E4 = 5,

The Bernoulli and Euler numbers and polynomials are generalized [15-21].

Lemma 1. For real number x > 0 and natural number m [22,23], then

equationequation equation (2.7)

equation, equation (2.8)

equation, equation (2.9)

equationequation (2.10)

Remark 1. θ1 , θ2 , θ3 , θ4 only depend on natural number m.

Lemma 2. For real number x > 0 and natural number n, we have [24]:

equation (2.11)

Lemma 3 For real number x > 0 and natural number n, we have [1,17]:

equation (2.12)

equation (2.13)

equation (2.14)

Lemma 4. Let the sequence of functions un (x) for n ∈equation be defined on (0, ∞) by

equation (2.15)

the series equation is differentiable on (0, ∞), that is,

equation(2.16)

Proof. It is obvious that equation, therefore equation converges at x = 0. In order to prove (2.16), we need only to show that the inner closed uniform convergence of the series equation on (0, ∞). From (2.15), we have

equation (2.17)

For any interval [a,b] ⊂ (0,∞) , we have

equation (2.18)

x ∈[a,b] It is easy to check that the series equation converges, which and Weierstrass M-test implies that the seriesequation is inner closed uniformly convergent on (0, ∞). Hence the series equation is differentiable on (0, ∞) and the identity (2.17) holds for x ≥ 0 .

The lemma is proved.

Lemma 5. For 0 < α ≤1 and real number b, let the function Qa,b ( x) be defined by

equation (2.19)

If b ≥ a, then the function (2.19) satisfies

equation(2.20)

for all equation and n = 2,3,….

Proof. Taking the logarithm of Qa,b ( x) yields

lnQ (x) = xln(ax + b) −l nΓ(ax + b), (2.21)

and differentiating lnQa,b (x) , then

equation (2.22)

For given integer n ≥ 2, we get

equationequation (2.23)

and, by the identities (2.13) and (2.14), (2.23) can be written as

equation (2.24)

equation

Let

ρ(t) = (1 + bt)(1 − e−t ) −at and q(t) = (1 + t)(1 − e−t ) −t .

It is easy to check that

q' (t ) = te−t > 0, t ∈(0, ∞) (2.25)

therefore q (t ) is strictly increasing on (0, ∞), and then q (t ) > q (0) = 0.

The following two cases will complete the proof of Lemma 5.

Case 1. If 0 < a ≤ 1 ≤ b, then since q (t ) > 0 for t > 0, we have at a t < (1+ t )(1- e-t ) £ (1+ bt )(1- e-t ), (2.26)

which implies equation, and then p (t ) > 0 for all t > 0.

From (2.24), we know that the inequality (2.20) holds for x ∈(−b a,∞) and integer n ≥ 2.

Case 2. If 0 < a ≤ b ≤ 1, then we get

p' (t ) = b − a + e−t (bt + 1 − b) ≥ bte−t > 0, t ∈(0, ∞) (2.27)

Therefore p (t ) is strictly increasing on (0, ∞), and then p (t ) > p (0) = 0 .

From (2.24), we know that the inequality (2.20) holds for x ∈(-b a , ∞) and integer n ≥ 2.

The lemma is proved.

Proof of Theorems

Proof of Theorem 1. For x ≠ 0and natural number n, taking the logarithmically differential into consideration yields

equationequation (3.1)

where ψ(−1) (x + y) and ψ(0) (x + y) stand for lnΓ(x + y) and ψ(x + y) respectively.

Furthermore, differentiating equation directly gives

equationequation (3.2)

Making use of (2.11) and (2.13) shows that for all n ∈ equation and any fixed y > 0 ,the double inequality

equationequationequation (3.3)

holds for all x ∈(− y,∞) \ {0} and α∈ (−∞,∞) .

For any fixed y ∈(0,∞), let u (t ) and v (t ) be defined on (−∞,∞) by

equationrespectively.

Differentiating u (t ) and v (t ) directly, we obtain

equation(3.4)

v'(t) = e−yt (1 − y − yt) (3.5)

Therefore, for given y ∈(0,∞)we have

equation (3.6)

and

equation (3.7)

From (3.6) and (3.7), we conclude that for all t > 0 we obtain u(t) > 0, (3.8)

and

equation (3.9)

From (3.3) and (3.8)-(3.9), it is easy to see that

equation (3.10)

For all n ∈ and all x∈equation (−y,∞) \ {0}.

On the one hand, if x ∈(0, ∞), then the inequalities (3.10) can be equivalently changed into

equation (3.11)

and

equation (3.12)

for k ∈equation .

From (3.1), then simple computation shows that equation (3.13)

for all n ∈equation and any given y ∈(0,∞). As a result,

equation (3.14)

and

equation (3.15)

for all k ∈equation and all x > 0.

Therefore, (3.14) and (3.15) imply

equation (3.16)

for all n ∈equation and all x > 0.

Hence, if either α ≥ e−(1− y) / y for given 0 < y < 1 or α ≥ 1 for given y ≥ 1, the function (1.13) is strictly logarithmically completely monotonic with respect to x on (0, ∞), and if α ≤ 0 for given y > 0, so is the reciprocal of the function (1.13).

On the other hand, if x ∈(− y, 0) for any given y > 0, then (3.10) implies

equation (3.17)

for n ∈equation.

In view of (3.13), we can conclude that

equation(3.18)

for n ∈equation . It is obvious that (3.18) is equivalent to that (3.14) and (3.15) hold for any given y > 0 and x ∈(- y,0).Therefore, it is easy to prove similarly that (3.16) is also valid on x ∈(- y,0) for any given y > 0 and all n ∈equation .

The amino acid of antibody/protein possesses different preferences. Thus we can conduct site-directed mutation to promote the affinity and/or hydrophilic with the prediction of topology. For example, bovine antibodies have an unusual structure comprising a β-strand ‘stalk’ domain and a disulphidebonded ‘knob’ domain in CDR3 (Figure 3). Attempts have been made to utilize such amino acid preference for antibody drug development.

biomedical-imaging-IgG-antibodies

Figure 3: Unique structural domain in bovine IgG antibodies and application

Consequently, the function (1.13) is the same logarithmically completely monotonicity on (- y,0) as on (0, ∞), that is, if either α ≥ e–(1−y) / y for given 0 < y < 1 or α ≥ 1for given y ≥ 1 the function (1.13) is strictly logarithmically completely monotonic with respect to x on (- y, 0) , and if α ≤ 0 for given y > 0, so is the reciprocal of the function (1.13).

Conversely, we assume that the reciprocal of the function (1.13) is strictly logarithmically completely monotonic on (− y,∞) \ {0} for any given y > 0. Then we have for any given y > 0 and all x > 0

equation (3.19)

which implies

equation (3.20)

By L’Hˆospital’s rule, we have equation (3.21)

for any given y > 0. By virtue of (3.20) and (3.21), we conclude that the necessary condition for the reciprocal of the function (1.13) to be strictly logarithmically completely monotonic is α ≤ 0.

If the function (1.13) is logarithmically completely monotonic on (− y,∞) \ {0}

for any given y > 0, then the inequality (3.19) and (3.20) are reversed for any given y > 0 and all x > 0.

By utilizing (2.7) and (2.8), it is easy to see that

equation (3.22)

for any given y > 0. In fact, it is not difficult to show that the necessary condition for the function (1.13) to be strictly logarithmically completely monotonic is α ≥ 1.

The proof of Theorem 1 is completed.

Proof of Theorem 2. Taking the logarithm of hy ( x) gives

equationequation (3.23)

Let

equation (3.24)

equationequation (3.25)

then

lnhy (x) = μ(x) + ω(x) . (3.26)

In view of Lemma 4, straightforward calculation gives

equationequationequation (3.27)

By virtue of (1.2), the identity (3.27) is equivalent to

equationequationequation (3.28)

By Lemma 5, we know that μ'(x) is strictly increasing on (0, ¥), which and (1.10) imply the limit of μ'(x) equals 1 as x → ∞ therefore

μ'(x) < 1 (3.29)

holds for all x > 0.

We know that g ( x) is strictly completely monotonic on (0, ¥), where g ( x) defined by (1.8), hence for given integer n ≥ 0 , the inequality.

(−1)n+1 (ω' ( x))(n) > 0 (3.30)

holds for all x > 0.

And then by using inequality (1.9) and (1.10), we get

−2 < ω' ( x) < −1 (3.31)

for all x > 0 .

From (3.29) and (3.31), we conclude that

(lnhy( x)) = μ( x) +ω( x) < 0 (3.32)

for all that x > 0. Utilizing Lemma 5 and (3.30), for given integer n ≥ 2, it is easy to see that equation for all x > 0.

Theorem 2 follows from (3.32) and (3.33).

Thus the proof of Theorem 2 is completed [25-31].

Conclusion

In conclusion we establish two new logarithmically completely monotonic functions involving the gamma function according to two preferred interaction geometries, and a sharp inequality involving the gamma function is deduced to solve the problems of genetically engineering antibody. It is necessary to address, many other aspects (such as thermal condition, alkalinity or acidity, adhesion of antibody) are also playing key roles in antibody functioning, which could be also understood from bio-mathematical perspective, and such knowledge will be in return useful for biomedical application of antibodies as well as proteins [25-31].

Acknowledgements

This work was supported by the Ph.D. programs foundation of ministry of education (20130204110023); the national nature science foundation (31572556); the international scientific and technological cooperation and exchange program in Shaanxi province (S2015YFKW0002), and the Key Construction Program of International Cooperation Base in S&T, Shaanxi Province (2015SD0018), China.

References