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Characterization and immobilization of engineered sialidases from Trypanosoma rangeli for transsialylation

  • Received: 13 January 2017 Accepted: 06 April 2017 Published: 13 April 2017
  • A sialidase (EC 3.2.1.18; GH 33) from non-pathogenic Trypanosoma rangeli has been engineered with the aim of improving its transsialylation activity. Recently, two engineered variants containing 15 and 16 amino acid substitutions, respectively, were found to exhibit significantly improved transsialylation activity: both had a 14 times higher ratio between transsialylation and hydrolysis products compared to the first reported mutant TrSA5mut. In the current work, these two variants, Tr15 and Tr16, were characterized in terms of pH optimum, thermal stability, effect of acceptor-to-donor ratio, and acceptor specificity for transsialylation using casein glycomacropeptide (CGMP) as sialyl donor and lactose or other human milk oligosaccharide core structures as acceptors. Both sialidase variants exhibited pH optima around pH 4.8. Thermal stability of each enzyme was comparable to that of previously developed T. rangeli sialidase variants and higher than that of the native transsialidase from T. cruzi (TcTS). As for other engineered T. rangeli sialidase variants and TcTS, the acceptor specificity was broad: lactose, galactooligosaccharides (GOS), xylooligosaccharides (XOS), and human milk oligosaccharide structures lacto-N-tetraose (LNT), lacto-N-fucopentaose (LNFP V), and lacto-N-neofucopentaose V (LNnFP V) were all sialylated by Tr15 and Tr16. An increase in acceptor-to-donor ratio from 2 to 10 had a positive effect on transsialylation. Both enzymes showed high preference for formation α(2,3)-linkages at the non-reducing end of lactose in the transsialylation. Tr15 was the most efficient enzyme in terms of transsialylation reaction rates and yield of 3’-sialyllactose. Finally, Tr15 was immobilized covalently on glyoxyl-functionalized silica, leading to a 1.5-fold increase in biocatalytic productivity (mg 3’-sialyllactose per mg enzyme) compared to free enzyme after 6 cycles of reuse. The use of glyoxyl-functionalized silica proved to be markedly better for immobilization than silica functionalized with (3-aminopropyl)triethoxysilane (APTES) and glutaraldehyde, which resulted in a biocatalytic productivity which was less than half of that obtained with free enzyme.

    Citation: Birgitte Zeuner, Isabel González-Delgado, Jesper Holck, Gabriel Morales, María-José López-Muñoz, Yolanda Segura, Anne S. Meyer, Jørn Dalgaard Mikkelsen. Characterization and immobilization of engineered sialidases from Trypanosoma rangeli for transsialylation[J]. AIMS Molecular Science, 2017, 4(2): 140-163. doi: 10.3934/molsci.2017.2.140

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  • A sialidase (EC 3.2.1.18; GH 33) from non-pathogenic Trypanosoma rangeli has been engineered with the aim of improving its transsialylation activity. Recently, two engineered variants containing 15 and 16 amino acid substitutions, respectively, were found to exhibit significantly improved transsialylation activity: both had a 14 times higher ratio between transsialylation and hydrolysis products compared to the first reported mutant TrSA5mut. In the current work, these two variants, Tr15 and Tr16, were characterized in terms of pH optimum, thermal stability, effect of acceptor-to-donor ratio, and acceptor specificity for transsialylation using casein glycomacropeptide (CGMP) as sialyl donor and lactose or other human milk oligosaccharide core structures as acceptors. Both sialidase variants exhibited pH optima around pH 4.8. Thermal stability of each enzyme was comparable to that of previously developed T. rangeli sialidase variants and higher than that of the native transsialidase from T. cruzi (TcTS). As for other engineered T. rangeli sialidase variants and TcTS, the acceptor specificity was broad: lactose, galactooligosaccharides (GOS), xylooligosaccharides (XOS), and human milk oligosaccharide structures lacto-N-tetraose (LNT), lacto-N-fucopentaose (LNFP V), and lacto-N-neofucopentaose V (LNnFP V) were all sialylated by Tr15 and Tr16. An increase in acceptor-to-donor ratio from 2 to 10 had a positive effect on transsialylation. Both enzymes showed high preference for formation α(2,3)-linkages at the non-reducing end of lactose in the transsialylation. Tr15 was the most efficient enzyme in terms of transsialylation reaction rates and yield of 3’-sialyllactose. Finally, Tr15 was immobilized covalently on glyoxyl-functionalized silica, leading to a 1.5-fold increase in biocatalytic productivity (mg 3’-sialyllactose per mg enzyme) compared to free enzyme after 6 cycles of reuse. The use of glyoxyl-functionalized silica proved to be markedly better for immobilization than silica functionalized with (3-aminopropyl)triethoxysilane (APTES) and glutaraldehyde, which resulted in a biocatalytic productivity which was less than half of that obtained with free enzyme.


    In 1997, Van Hamme [19,(H.2)] proved the following supercongruence: for any prime p3(mod4),

    (p1)/2k=0(12)3kk!30(modp2), (1.1)

    where (a)n=a(a+1)(a+n1) is the rising factorial. It is easy to see that (1.1) is also true when the sum is over k from 0 to p1, since (1/2)k/k!0(modp) for (p1)/2<kp1. Nowadays various generalizations of (1.1) can be found in [8,10,11,12,13,14,16,17]. For example, Liu [12] proved that, for any prime p3(mod4) and positive integer m,

    mp1k=0(12)3kk!30(modp2). (1.2)

    The first purpose of this paper is to prove the following q-analogue of (1.2), which was originally conjectured by the author and Zudilin [10,Conjecture 2].

    Theorem 1.1. Let m and n be positive integers with n3(mod4). Then

    mn1k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk0(modΦn(q)2), (1.3)
    [5pt]mn+(n1)/2k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk0(modΦn(q)2). (1.4)

    Here and in what follows, the q-shifted factorial is defined by (a;q)0=1 and (a;q)n=(1a)(1aq)(1aqn1) for n1, and the n-th cyclotomic polynomial Φn(q) is defined as

    Φn(q)=1kngcd

    where \zeta is an n -th primitive root of unity. Moreover, the q -integer is given by [n] = [n]_q = 1+q+\cdots+q^{n-1} .

    The m = 1 case of (1.3) was first conjectured by the author and Zudilin [9,Conjecture 4.13] and has already been proved by themselves in a recent paper [11]. For some other recent progress on q -congruences, the reader may consult [2,3,4,5,6,7,8,10,15].

    In 2016, Swisher [18,(H.3) with r = 2 ] conjectured that, for primes p\equiv 3\pmod{4} and p>3 ,

    \begin{equation} \sum\limits_{k = 0}^{(p^2-1)/2} \frac{(\frac{1}{2})_k^3}{k!^3} \equiv p^2\pmod{p^5}, \end{equation} (1.5)

    The second purpose of this paper is to prove the following q -congruences related to (1.5) modulo p^4 .

    Theorem 1.2. Let n\equiv 3\pmod{4} be a positive integer. Modulo \Phi_n(q)^2\Phi_{n^2}(q)^2 , we have

    \begin{align} \sum\limits_{k = 0}^{(n^2-1)/2}\frac{(1+q^{4k+1})(q^2;q^4)_k^3}{(1+q)(q^4;q^4)_k^3}q^k \equiv \dfrac{[n^2]_{q^2}(q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}} q^{(1-n^2)/2}, \end{align} (1.6)
    \begin{align} [5pt] \sum\limits_{k = 0}^{n^2-1}\frac{(1+q^{4k+1})(q^2;q^4)_k^3}{(1+q)(q^4;q^4)_k^3}q^k \equiv \dfrac{[n^2]_{q^2}(q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}} q^{(1-n^2)/2}. \end{align} (1.7)

    Let n = p\equiv 3\pmod{4} be a prime and take q\to1 in Theorem 1.2. Then \Phi_p(1) = \Phi_{p^2}(1) = p , and

    \begin{align*} \lim\limits_{q\to 1}\dfrac{(q^{3};q^4)_{(p^2-1)/2}} {(q^5;q^4)_{(p^2-1)/2}} = \prod\limits_{k = 1}^{(p^2-1)/2}\frac{4k-1}{4k+1} = \frac{(\frac{3}{4})_{(p^2-1)/2}}{(\frac{5}{4})_{(p^2-1)/2}}. \end{align*}

    Therefore, we obtain the following conclusion.

    Corollary 1. Let p\equiv 3\pmod{4} be a prime. Then

    \begin{align} \sum\limits_{k = 0}^{(p^2-1)/2} \frac{(\frac{1}{2})_k^3}{k!^3} &\equiv p^2\frac{(\frac{3}{4})_{(p^2-1)/2}}{(\frac{5}{4})_{(p^2-1)/2}} \pmod{p^4}, \end{align} (1.8)
    \begin{align} [5pt] \sum\limits_{k = 0}^{p^2-1} \frac{(\frac{1}{2})_k^3}{k!^3} &\equiv p^2\frac{(\frac{3}{4})_{(p^2-1)/2}}{(\frac{5}{4})_{(p^2-1)/2}} \pmod{p^4}. \end{align} (1.9)

    Comparing (1.5) and (1.8), we would like to propose the following conjecture, which was recently confirmed by Wang and Pan [20].

    Conjecture 1. Let p\equiv 3\pmod{4} be a prime and r a positive integer. Then

    \begin{align} \prod\limits_{k = 1}^{(p^{2r}-1)/2}\frac{4k-1}{4k+1}\equiv 1\pmod{p^2}. \end{align} (1.10)

    Note that the r = 1 case is equivalent to say that (1.5) is true modulo p^4 .

    We need to use Watson's terminating _8\phi_7 transformation formula (see [1,Appendix (Ⅲ.18)] and [1,Section 2]):

    \begin{align} & _{8}\phi_{7}\!\left[\begin{array}{cccccccc} a, & qa^{\frac{1}{2}}, & -qa^{\frac{1}{2}}, & b, & c, & d, & e, & q^{-n} \\ & a^{\frac{1}{2}}, & -a^{\frac{1}{2}}, & aq/b, & aq/c, & aq/d, & aq/e, & aq^{n+1} \end{array};q, \, \frac{a^2q^{n+2}}{bcde} \right] \\[5pt] &\quad = \frac{(aq;q)_n(aq/de;q)_n} {(aq/d;q)_n(aq/e;q)_n} \, {}_{4}\phi_{3}\!\left[\begin{array}{c} aq/bc, \ d, \ e, \ q^{-n} \\ aq/b, \, aq/c, \, deq^{-n}/a \end{array};q, \, q \right], \end{align} (2.1)

    where the basic hypergeometric {}_{r+1}\phi_r series with r+1 upper parameters a_1, \dots, a_{r+1} , r lower parameters b_1, \dots, b_r , base q and argument z is defined as

    \begin{equation*} {}_{r+1}\phi_{r}\!\left[\begin{matrix} a_1, a_2, \dots, a_{r+1}\\b_1, \dots, b_r \end{matrix};q, z\right]: = \sum\limits_{k = 0}^\infty \frac{(a_1;q)_k(a_2;q)_k \dots(a_{r+1};q)_k}{(q;q)_k(b_1;q)_k\cdots(b_r;q)_k} z^k. \end{equation*}

    The left-hand side of (1.4) with m\geq 0 can be written as the following terminating _8\phi_7 series:

    \begin{equation} _{8}\phi_{7}\!\left[\begin{array}{cccccccc} q^2, & q^5, & -q^5, & q^2, & q, & q^2, & q^{4+(4m+2)n}, & q^{2-(4m+2)n} \\ & q, & -q, & q^4, & q^5, & q^4, & q^{2-(4m+2)n}, & q^{4+(4m+2)n} \end{array};q^4, \, q \right]. \end{equation} (2.2)

    By Watson's transformation formula (2.1) with q\mapsto q^4 , a = b = d = q^2 , c = q , e = q^{4+(4m+2)n} , and n\mapsto mn+(n-1)/2 , we see that (2.2) is equal to

    \begin{align} &\frac{(q^6;q^4)_{mn+(n-1)/2}(q^{-(4m+2)n};q^4)_{mn+(n-1)/2}} {(q^4;q^4)_{mn+(n-1)/2}(q^{2-(4m+2)n};q^4)_{mn+(n-1)/2}} \\[5pt] &\quad\times {}_{4}\phi_{3}\!\left[\begin{array}{c} q^3, \ q^2, \, q^{4+(4m+2)n}, \ q^{2-(4m+2)n} \\ q^4, \quad q^5, \quad q^6 \end{array};q^4, \, q^4 \right]. \end{align} (2.3)

    It is not difficult to see that there are exactly m+1 factors of the form 1-q^{an} ( a is an integer) among the mn+(n-1)/2 factors of (q^6;q^4)_{mn+(n-1)/2} . So are (q^{-(4m+2)n};q^4)_{mn+(n-1)/2} . But there are only m factors of the form 1-q^{an} ( a is an integer) in each of (q^4;q^4)_{mn+(n-1)/2} and (q^{2-(4m+2)n};q^4)_{mn+(n-1)/2} . Since \Phi_n(q) is a factor of 1-q^N if and only if n divides N , we conclude that the fraction before the _4\phi_3 series is congruent to 0 modulo \Phi_n(q)^2 . Moreover, for any integer x , let f_n(x) be the least non-negative integer k such that (q^x;q^4)_k\equiv 0 modulo \Phi_n(q) . Since n\equiv 3\pmod{4} , we have f_n(2) = (n+1)/2 , f_n(3) = (n+1)/4 , f_n(4) = n , f_n(5) = (3n-1)/4 , and f_n(6) = (n-1)/2 . It follows that the denominator of the reduced form of the k -th summand

    \frac{(q^3;q^4)_k (q^2;q^4)_k(q^{4+(4m+2)n};q^4)_k(q^{2-(4m+2)n};q^4)_k} {(q^4;q^4)_k^2(q^5;q^4)_k(q^6;q^4)_k}q^{4k}

    in the _4\phi_3 summation is always relatively prime to \Phi_n(q) for any non-negative integer k . This proves that (2.3) (i.e. (2.2)) is congruent to 0 modulo \Phi_n(q)^2 , thus establishing (1.4) for m\geq 0 .

    It is easy to see that (q^2;q^4)_k^3/(q^4;q^4)_k^3 is congruent to 0 modulo \Phi_n(q)^3 for mn+(n-1)/2<k\leq(m+1)n-1 . Therefore, the q -congruence (1.3) with m\mapsto m+1 follows from (1.4).

    The author and Zudilin [11,Theorem 1.1] proved that, for any positive odd integer n ,

    \begin{equation} \sum\limits_{k = 0}^{(n-1)/2}\frac{(1+q^{4k+1})\, (q^2;q^4)_k^3}{(1+q)\, (q^4;q^4)_k^3}\, q^{k} \equiv\dfrac{[n]_{q^2}(q^3;q^4)_{(n-1)/2}}{(q^5;q^4)_{(n-1)/2}}\, q^{(1-n)/2} \pmod{\Phi_n(q)^2}, \end{equation} (3.1)

    which is also true when the sum on the left-hand side of (3.1) is over k from 0 to n-1 . Replacing n by n^2 in (3.1) and its equivalent form, we see that the q -congruences (1.6) and (1.7) hold modulo \Phi_{n^2}(q)^2 .

    It is easy to see that, for n\equiv 3\pmod{4} ,

    \dfrac{[n^2]_{q^2}(q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}} q^{(1-n^2)/2}\equiv 0\pmod{\Phi_n(q)^2}

    because [n^2]_{q^2} = (1-q^{n^2})/(1-q^2) is divisible by \Phi_n(q) , and (q^3;q^4)_{(n^2-1)/2} contains (n+1)/2 factors of the form 1-q^{an} ( a is an integer), while (q^5;q^4)_{(n^2-1)/2} only has (n-1)/2 such factors. Meanwhile, by Theorem 1.1, the left-hand sides of (1.6) and (1.7) are both congruent to 0 modulo \Phi_n(q)^2 since (n^2-1)/2 = (n-1)n/2+(n-1)/2 . This proves that the q -congruences (1.6) and (1.7) also hold modulo \Phi_n(q)^2 . Since the polynomials \Phi_n(q) and \Phi_{n^2}(q) are relatively prime, we finish the proof of the theorem.

    Swisher's (H.3) conjecture also indicates that, for positive integer r and primes p\equiv 3\pmod{4} with p>3 , we have

    \begin{equation} \sum\limits_{k = 0}^{(p^{2r}-1)/2} \frac{(\frac{1}{2})_k^3}{k!^3} \equiv p^{2r}\pmod{p^{2r+3}}. \end{equation} (4.1)

    Motivated by (4.1), we shall give the following generalization of Theorem 1.2.

    Theorem 4.1. Let n and r be positive integers with n\equiv 3\pmod{4} . Then, modulo \Phi_{n^{2r}}(q)^2\prod_{j = 1}^{r}\Phi_{n^{2j-1}}(q)^2 , we have

    \begin{align} \sum\limits_{k = 0}^{(n^{2r}-1)/2}\frac{(1+q^{4k+1})(q^2;q^4)_k^3}{(1+q)(q^4;q^4)_k^3}q^k \equiv \dfrac{[n^{2r}]_{q^2}(q^3;q^4)_{(n^{2r}-1)/2}} {(q^5;q^4)_{(n^{2r}-1)/2}} q^{(1-n^{2r})/2}, \end{align} (4.2)
    \begin{align} [5pt] \sum\limits_{k = 0}^{n^{2r}-1}\frac{(1+q^{4k+1})(q^2;q^4)_k^3}{(1+q)(q^4;q^4)_k^3}q^k \equiv \dfrac{[n^{2r}]_{q^2}(q^3;q^4)_{(n^{2r}-1)/2}} {(q^5;q^4)_{(n^{2r}-1)/2}} q^{(1-n^{2r})/2}. \end{align} (4.3)

    Proof. Replacing n by n^{2r} in (3.1) and its equivalent form, we see that (4.2) and (4.3) are true modulo \Phi_{n^{2r}}(q)^2 . Similarly as before, we can show that

    \begin{align*} \dfrac{[n^{2r}]_{q^2}(q^3;q^4)_{(n^{2r}-1)/2}} {(q^5;q^4)_{(n^{2r}-1)/2}} q^{(1-n^{2r})/2} \equiv 0\pmod{\prod\limits_{j = 1}^{r}\Phi_{n^{2j-1}}(q)^2}. \end{align*}

    Further, by Theorem 1.1, we can easily deduce that the left-hand sides of (4.2) and (4.3) are also congruent to 0 modulo \prod_{j = 1}^{r}\Phi_{n^{2j-1}}(q)^2 .

    Letting n = p\equiv 3\pmod{4} be a prime and taking q\to1 in Theorem 4.1, we are led to the following result.

    Corollary 2. Let p\equiv 3\pmod{4} be a prime and let r\geq 1 . Then

    \begin{align} \sum\limits_{k = 0}^{(p^{2r}-1)/2} \frac{(\frac{1}{2})_k^3}{k!^3} &\equiv p^{2r}\frac{(\frac{3}{4})_{(p^{2r}-1)/2}}{(\frac{5}{4})_{(p^{2r}-1)/2}} \pmod{p^{2r+2}}, \end{align} (4.4)
    \begin{align} [5pt] \sum\limits_{k = 0}^{p^{2r}-1} \frac{(\frac{1}{2})_k^3}{k!^3} &\equiv p^{2r}\frac{(\frac{3}{4})_{(p^{2r}-1)/2}}{(\frac{5}{4})_{(p^{2r}-1)/2}} \pmod{p^{2r+2}}. \end{align} (4.5)

    In light of (1.10), the supercongruence (4.4) implies that (4.1) holds modulo p^{2r+2} for any odd prime p .

    It is known that q -analogues of supercongruences are usually not unique. See, for example, [2]. The author and Zudilin [10,Conjecture 1] also gave another q -analogue of (1.2), which still remains open.

    Conjecture 2 (Guo and Zudilin). Let m and n be positive integers with n\equiv 3\pmod{4} . Then

    \begin{align} \sum\limits_{k = 0}^{mn-1}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}\, q^{2k} &\equiv 0 \pmod{\Phi_n(q)^2}, \\ \sum\limits_{k = 0}^{mn+(n-1)/2}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}\, q^{2k} &\equiv 0 \pmod{\Phi_n(q)^2}. \end{align} (4.6)

    The author and Zudilin [10,Theorem 2] themselves have proved (4.6) for the m = 1 case. Motivated by Conjecture 2, we would like to give the following new conjectural q -analogues of (1.8) and (1.9).

    Conjecture 3. Let n\equiv 3\pmod{4} be a positive integer. Modulo \Phi_n(q)^2\Phi_{n^2}(q)^2 , we have

    \begin{align*} \sum\limits_{k = 0}^{(n^2-1)/2}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k} &\equiv \dfrac{[n^2](q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}}, \\[5pt] \sum\limits_{k = 0}^{n^2-1}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k} &\equiv \dfrac{[n^2](q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}}. \end{align*}

    There are similar such new q -analogues of (4.4) and (4.5). We omit them here and leave space for the reader's imagination.

    The author is grateful to the two anonymous referees for their careful readings of this paper.

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