Downregulation of long noncoding RNA SNHG1 inhibits cell proliferation, metastasis, and invasion by suppressing the Notch‐1 signaling pathway in pancreatic cancer
Authors: Long Cui
Yangsui Liu
Yadong Dong
Xinhui Zhang
Xin Zhao
Xinchun Jiang
Chenchen Kong
Xiaochuan Wang
Journal: Journal of Cellular Biochemistry (2019)
Discipline: MOLECULAR BIOLOGY & GENETICS
Keywords: cell proliferation and metastasis; long noncoding RNA (lncRNA) SNHG1; pancreatic cancer (PC).
Affiliations: Department of Hepatopancreatobiliary Surgery, Central Hospital of Xuzhou, Affiliated to Dongan University, Jiangsu, China., Department of Hepatopancreatobiliary Surgery, Henan People's Hospital, Henan, Zhengzhou, China.

Abstract:  Background: Pancreatic cancer (PC) has become the fourth most lethal among human cancers. Long noncoding RNAs (lncRNAs) have been reported to play a role in the progression of a variety of cancers. However, the role of lncRNA SNHG1 in PC is not clear. Methods: Real-time Quantitative PCR Detection System (qPCR) was used to detect the expression of SNHG1 in PC cells. Then, the SNHG1 knockdown cell was constructed with si-SNHG1. AsPC-1 and PANC1 cells were used to analyze the ability of cell proliferation, invasion, and migration. MTT assay was used to analyze the proliferation ability. Transwell experiments and wound healing experiments were used to detect the capacity of invasion and migration. Finally, Western blot analysis was used to explore the mechanism of SNHG1 in PC. Results: SNHG1 was significantly upregulated in PC cells. Knockdown of SNHG1 could obviously suppress cell proliferation, invasion, and migration. Furthermore, SNHG1 knockdown inhibited the activation of the Notch-1 signaling pathway and inhibited the expression of N-cadherin, Hes1, Vimentin, Notch-1. The inhabitation was reversed when Notch-1 was overexpressed in si-SNHG1 cells. Conclusion: The lncRNA SNHG1 promotes cell growth and metastasis in PC through activation of the Notch-1 signaling pathway in PC. 

Retracted: No

PubPeer comments:
 In Chapter 1 of this book, the authors claimed on Page 6 that $$S_{oplus}(alpha) = {alpha oplus x | xin triangle } $$ and called it the addition region of $alpha$. Then the authors claimed that **(a) Any $alpha oplus x (xin triangle)$ must fall into the area $S_{oplus}(alpha)$;** **(b) For any $beta in S_{oplus}(alpha)$, $beta ominus alpha = left(frac{mu_{beta} - mu_{alpha}}{ 1-mu_{alpha} }, frac{v_{beta}}{v_{alpha}} right)$ is still an IFN.** ![file] **Here is a counterexample.** Let $$alpha =(.5, 0)$$ and $$beta =(.8, 0)$$ Then, let $$x = (.6, 0)$$ According to definition 1.3, $$alpha oplus x = (.5, 0) oplus (.6, 0) = (.5+.6-(.5)(.6), 0) = (.8, 0)$$ ![file] This implies that $$beta in S_{oplus}(alpha)$$ According to Definition 1.5, $$beta ominus alpha =(.8, 0) ominus (.5, 0) = (0, 1)$$ However, $$left( frac{mu_{beta} -mu_{alpha}}{ 1-mu_{alpha}}, quad frac{ u_{beta}}{ u_{alpha}} right) = left(frac{.3}{.5}, frac{0}{0}right)$$ ![file] It is clear that $$(0, 1)  eq left(frac{.3}{.5}, frac{0}{0} right)$$ The results presented in the rest of this book are based on some notions of this type. , 
 #1 **Dear** **Rhipidura Albiventris**, **Thank you very much** **for your interests in our work!** **Please see** **the following** **responses:** **C.** Consider a counterexample $alpha=(0.5,0)$, $beta=(0.8,0)$ and $x=(0.6,0)$. **A.** Thank you for your comments! Firstly, it should be declared that **(1)** both of the two statements, i.e. (a) and (b), are correct; **(2)** the given example is not a counter-example for the two statements at all, much less for our theoretical framework in this book; **(3)** our work in the subsequent chapters is also well-constructed because they focus on mathematically analytical situations and deal seriously with nonanalytic cases. Next, they are explained as follows: $quad$**(1)** Firstly, both of the two statements are correct. As for the first statement (a), it tells us that any IFN with the form of $alphaoplus X$ will fall into the gray triangle of $alpha$. The example given by you mainly aim at the second one. Let us consider $betain S_oplus(alpha)$, when the expression $left(frac{mu_beta-mu_alpha}{1-mu_alpha},frac{ u_beta}{ u_alpha}right)$ with $mu_alpha eq 1$ and $ u_alpha eq 0$ is meaningful, it is easy to check $left(frac{mu_beta-mu_alpha}{1-mu_alpha},frac{ u_beta}{ u_alpha}right)$ is still an IFN because $betain S_oplus(alpha)$. As for the extreme cases with $mu_alpha= 1$ or $ u_alpha= 0$ or $beta otin S_oplus(alpha)$, the definition of  $ominus$ assigns arbitrarily an IFN  $O$ to these extreme cases. In either case, $betaominusalpha$ is an IFN. However, it needs to be underlined that these extreme situations causes directly two serious consequences, one of which is that the IFN-type value assigned arbitrarily in these extreme situations can not inherit the information of minuends and subtrahends ($alpha$ and $beta$). Another of them forces us to encounter the nonanalytic  $frac{0}{0}$ . Both of the two serious consequences are beyond all analytical framework. Remarkably, we write designedly down $betaominusalpha=left(frac{mu_beta-mu_alpha}{1-mu_alpha},frac{ u_beta}{ u_alpha}right)$ rather than $betaominusalpha$, which has actually indicated that $mu_alpha eq 1$, $ u_alpha eq 0$ and $betain S_oplus(alpha)$. Otherwise, the expression $left(frac{mu_beta-mu_alpha}{0},frac{ u_beta}{0}right)$ is not meaningful at all; $quad$**(2)** Secondly, the given example is not a counter-example at all. It is quit abnormal to write down $left(frac{mu_beta-mu_alpha}{1-mu_alpha},frac{ u_beta}{ u_alpha}right)=left(frac{0.3}{0.5},frac{0}{0}right)$ and $(0,1) eqleft(frac{0.3}{0.5},frac{0}{0}right)$ in your comments. For the equality, in the case of singular situations, like $frac{0}{0}$, it obviously belongs to the  otherwise case in the definition of  $ominus$ . The result of $betaominusalpha$ should be the arbitrarily assigned  $O$ rather than $left(frac{0.3}{0.5},frac{0}{0}right)$ with the meaningless notation  $frac{0}{0}$ acquired by misusing the formula $left(frac{mu_beta-mu_alpha}{1-mu_alpha},frac{ u_beta}{ u_alpha}right)$. Furthermore, the second inequality is also meaningless. A comparison by using  $ eq$ requires a prerequisite that both two sides of inequalities are well-defined. The concept of $(0,1)$ is clear in the left side while the  $frac{0}{0}$ of the right side makes no sense; $quad$**(3)** Our work in the subsequent chapters is also well-constructed. **(i)** Actually, just like all analytical framework, our work of calculus under the intuitionistic fuzzy environment mainly focuses on analytic situations. Under some specific problems, it is possible that non-analytic cases are treated by some mathematical technique, such as  analytic continuation , in order to remain proper analytic properties (please also see the #15 in [](http: ) https: pubpeer.com publications 763257B958790969FC1AF53012DD5B#15). **(ii)** Moreover, for some extreme (or singular) situations with  $frac{constant}{0}$ ,  $frac{constant}{1-1}$ (Please see the #14 in [](http: ) https: pubpeer.com publications 763257B958790969FC1AF53012DD5B#14),  $0^0$ (Please see the #1 in [](http: ) https: pubpeer.com publications 7803CF906784F287F8261176A52D9E#1) and two  otherwise cases in the definitions of  $ominus$ and  $oslash$ , they are similar because the arbitrarily assigned values by the definitions of  $ominus$ and  $oslash$ can not inherit the information of these minuend, subtrahend, dividend and divisor. The lose of information makes us get rid of them from our work. Similarly, the operation of $alphaoplus E$ would also lose the information of $alpha$, so it is also avoided in our study (Please see the comment #17 in https: pubpeer.com publications 763257B958790969FC1AF53012DD5B#17). In fact, it is worth noting that the first three cases  $frac{constant}{0}$ ,  $frac{constant}{1-1}$ and  $0^0$ are not allowed even in the classical calculus, while the later situations the two  otherwise cases in the definitions of  $ominus$ and  $oslash$ and $alphaoplus E$ are particular singular cases in the intuitionistic fuzzy environment. **(iii)** Finally, there is an important fact that the infinitesimal and infinity are not treated as two fixed points (or numbers) in the real number axis after the Cauchy and Weierstrass era. Instead of studying the meaningless  $frac{0}{0}$ , the mathematical analysis focuses on asymptotic properties of  $frac{infinitesimal}{infinitesimal}$ . Similarly,  $infty$ is also studied in a dynamic view rather than in a static one (Please also see #17 in https: pubpeer.com publications 763257B958790969FC1AF53012DD5B#17). In conclusion, just like other analytical framework, our work of calculus under the intuitionistic fuzzy environment focuses on analytic situations, and sometimes extend continuously it to non-analytic cases. The so-called  counter-example  $frac{0}{0}$ given is actually not allowed in almost all analytical framework (including the classical calculus). Our work has weeded out these singular situations since the beginning. In two sentences, these examples with  $frac{constant}{0}$ ,  $frac{constant}{1-1}$ or  $0^0$ are not counter-examples just as they do not violate the classical calculus. Moreover, it is also right by default that $a eq 0$, $b eq 1$ and $c eq 0$ in the case of $frac{constant}{a}$, $frac{constant}{1-b}$ and $c^0$. **Thanks again** **for your** **interests!** **Best Regards,** **Qian Lei** , 
 #2 Dear Qian Lei, Thank you very much for your very detailed comments. I am not good at classical calculus and do learn a lot. However, to eliminate possible misunderstanding and have a common stand for potential discussion, I would like to present an example of multivariate calculus in the conventional sense because you repeatedly compare your theory to the classical calculus. Consider $$f(x, y) = frac{xy}{x^2 + y^2} quad text{ if } (x, y)  eq (0, 0)$$ and $$f(x, y) = 0 quad text{ if } (x, y) = (0, 0) quad i.e., text{otherwise}$$ The partial derivative exists for both $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ at the point $x=(0, 0)$. Does this imply that $f(x,y)$ is differentiable at the point $x=(0, 0)$? Or we simply do not consider the point $x=(0, 0)$ at all as a case of u201cotherwiseu201d? ,