International Journal of Recent Academic Research (ISSN: 2582-158X) Vol. 01, Issue 09, pp.532-542, December, 2019

Available online at http://www.journalijrar.com

RESEARCH ARTICLE

EULER’S LINE FOR ENZYME KINETICS

1, *Vitthalrao Bhimasha Khyade, 2Avram Hershko and 3Seema Karna Dongare

1Department of Zoology, Shardabai Pawar Mahila Mahavidyalaya, Shardanagar Tal., Baramati Dist., Pune – 413115, India 2Unit of Biochemistry, The B. Rappaport Faculty of Medicine, and the Rappaport Institute for Research in the Medical Sciences, Technion-Israel Institute of Technology, Haifa 31096, Israel 3P.G. Student, Department of Microbiology, Maharashtra Education Society's, Abasaheb Garware College, Karve Road, Pune – 411004, India

ARTICLE INFO ABSTRACT

Article History: A graph of the double-reciprocal equation is also called a Line weaver-Burk, reciprocal of velocity of enzyme

Received 10th September 2019, reaction (1÷v) against reciprocal of substrate concentration [1÷S]. Lineweaver-Burk graphs are particularly useful Received in revised form for analyzing the changes in enzyme kinetics in the presence of inhibitors, competitive, non-competitive, or a 28th October 2019, mixture of the two. One more attempt is carried out for establishment of Euler’s line through the use of Line Accepted 04th November 2019, weaver-Burk plot. Line weaver-Burk plot (double reciprocal plot) is with positive value of (Km÷Vmax) as a Published online th slope. for Enzyme Kinetics is with negative value of (Km÷Vmax) as a slope. The intercept on y-axis 30 December 2019. for Line weaver-Burk plot (double reciprocal plot) for Enzyme Kinetics correspond to: (1 ÷ Vmax). The intercept on y-axis for Euler Line for Enzyme Kinetics correspond to: [(Km +2) ÷ Vmax)]. Lineweaver-Burk plot (double reciprocal plot) and Euler Line for Enzyme Kinetics are intersecting at the point, x – co-ordinate of which correspond to: (1÷2) and y- co-ordinate of which correspond to: [(Km+2) ÷ Vmax]. The for enzyme kinetics is always located between the orthocenter and the circumcenter of enzyme kinetics. The distance from the centroid to the orthocenter is always twice the distance from the centroid to the circumcenter of enzyme kinetics. Attempt may open a new avenue for three dimensional enzyme structure of and mechanism of enzyme involved reactions. *Corresponding Author: Vitthalrao Bhimasha Khyade Key Words: Centroid, Orthocenter, Circumcenter, Euler’s line.

Copyright © 2019, Vitthalrao Bhimasha Khyade. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The rate of the forward reaction from E + S to ES may be INTRODUCTION termed k1, and the reverse reaction as k-1. Likewise, for the reaction from the ES complex to E and P, the forward reaction Enzymes are protein molecules that manipulate other rate is k2, and the reverse is k-2. Therefore, the ES complex molecules the enzymes' substrates. These target molecules may dissolve back into the enzyme and substrate, or move bind to an enzyme's active site and are transformed into forward to form product. At initial reaction time, when t ≈ 0, products through a series of steps known as the enzymatic little product formation occurs, therefore the backward reaction mechanism. Enzyme kinetics is the study of the chemical rate of k-2 may be neglected. The new reaction becomes: reactions that are catalyzed by enzymes. In enzyme kinetics, the reaction rate is measured and the effects of varying the E + S ↔ ES → E + P conditions of the reaction are investigated. Studying an enzyme's kinetics in this way can reveal the catalytic Assuming steady state, the following rate equations may be mechanism of this enzyme, its role in metabolism, how its written as: activity is controlled, and how a drug or an agonist might inhibit the enzyme (Kraut, Carroll and Herschlag (2003). The Rate of formation of ES = k1[E][S] Michaelis-Menten equation arises from the general equation Rate of breakdown of ES = (k-1 + k2) [ES] and set equal to for an enzymatic reaction: each other (Note that the brackets represent concentrations).

E + S ↔ ES ↔ E + P, Therefore:

k1[E][S] = (k-1 + k2) [ES] Where E is the enzyme, Rearranging terms, S is the substrate, [E][S]/[ES] = (k-1 + k2)/k1 ES is the enzyme-substrate complex, and P is the product. The fraction [E][S]/[ES] has been coined Km, or the Michaelis constant. According to Michaelis-Menten's kinetics equations, Thus, the enzyme combines with the substrate in order to form at low concentrations of substrate, [S], the concentration is the ES complex, which in turn converts to product while almost negligible in the denominator as KM >> [S], so the preserving the enzyme. equation is essentially 533 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp.532-542, December, 2019

V0 = Vmax [S]/KM A graph of the double-reciprocal equation is also called a Lineweaver-Burk, 1/Vo vs 1/[S]. The y-intercept is 1/Vmax; Which resembles a first order reaction. the x-intercept is -1/KM; and the slope is KM/Vmax. Lineweaver-Burk graphs are particularly useful for analyzing At High substrate concentrations, [S] >> KM, and thus the term how enzyme kinematics change in the presence of inhibitors, [S]/([S] + KM) becomes essentially one and the initial velocity competitive, non-competitive, or a mixture of the two. There approached Vmax, which resembles zero order reaction. are four reversible inhibitors: competitive, uncompetitive, non- competitive and mixed inhibitors. They can be plotted on The Michaelis-Menten equation is: double reciprocal plot. Competitive inhibitors are molecules that look like substrates and they bind to active site and slow down the reactions. Therefore, competitive inhibitors increase Km value (decrease affinity, less chance the substrates can go to active site), and Vmax stays the same. On double reciprocal

plot, competitive inhibitor shifts the x-axis (1/[s]) to the right Michaelis-Menten Equation towards zero compared to the slope with no inhibitor present.

Uncompetitive inhibitors can bind close to the active site but In this equation: don't occupy the active site. As a result, uncompetitive

inhibitors lower Km (increase affinity) and lower Vmax. On V is the initial velocity of the reaction. 0 double reciprocal plot, x-axis (1/[s]) is shifted to the left and up V is the maximal rate of the reaction. max on the y-axis (1/V) compared to the slope with no inhibitor. [Substrate] is the concentration of the substrate. Non-competitive inhibitors are not bind to the active site but somewhere on that enzyme which changes its activity. It has K is the Michaelis-Menten constant which shows the m the same Km but lower Vmax to those with no inhibitors. On concentration of the substrate when the reaction velocity is the double reciprocal plot, the slope goes higher on y-axis equal to one half of the maximal velocity for the reaction. It (1/V) than the one with no inhibitor. Km value is numerically can also be thought of as a measure of how well a substrate equal to the substrate concentration at which the half of the complexes with a given enzyme, otherwise known as its enzyme molecules are associated with substrate. km value is an binding affinity. An equation with a low K value indicates a m index of the affinity of enzyme for its particular substrate. large binding affinity, as the reaction will approach V more max The velocity (v) of biochemical reaction catalyzed by the rapidly. An equation with a high K indicates that the enzyme m enzyme vary according to the status of factors like: does not bind as efficiently with the substrate, and V will max concentration of the substrate [S]; hydrogen ion concentration; only be reached if the substrate concentration is high enough to temperature; concentration of the respective enzyme; activators saturate the enzyme. As the concentration of substrates and inhibitors. There is no linear response of velocity (v) of increases at constant enzyme concentration, the active sites on biocatalyzed reaction to the concentration of the substrate [s]. the protein will be occupied as the reaction is proceeding. This may be due to saturable nature of enzyme catalyzed When all the active sites have been occupied, the reaction is biochemical reactions. If the initial velocity (v) or rate of the complete, which means that the enzyme is at its maximum enzyme catalyzed biochemical reaction is expressed in terms capacity and increasing the concentration of substrate will not of substrate-concentration of [S], it appears to increase. That is increase the rate of turnover. Here is an analogy which helps to to say, initial velocity (v) of the enzyme catalyzed biochemical understand this concept easier. reaction get increase according to the increase in the

concentration of substrate [S]. This tendency of increase in Vmax is equal to the product of the catalyst rate constant (kcat) initial velocity (v) of the enzyme catalyzed biochemical and the concentration of the enzyme. The Michaelis-Menten reaction according to the increase in the concentration of equation can then be rewritten as V= Kcat [Enzyme] [S] / (Km substrate [S] is observed up to certain level of the + [S]). Kcat is equal to K2, and it measures the number of concentration of substrate [S]. At this substrate concentration substrate molecules "turned over" by enzyme per second. The [S], the enzyme exhibit saturation and exert the initial velocity unit of Kcat is in 1/sec. The reciprocal of Kcat is then the time (v) of the biocatalyzed reaction to achieve maximum velocity required by an enzyme to "turn over" a substrate molecule. The (V ). higher the Kcat is, the more substrates get turned over in one max second. MATERIALS AND METHODS Km is the concentration of substrates when the reaction reaches half of Vmax. A small Km indicates high affinity since The medians, altitudes and bisectors are the requirements for it means the reaction can reach half of Vmax in a small number establishment of Euler’s line for a . The intersection of of substrate concentration. This small Km will approach Vmax the three medians yields the point of “Centroid” for a triangle. more quickly than high Km value. The intersection of the three altitudes yields the point of “Orthocenter” for a triangle. The intersection of the three When Kcat/ Km, it gives us a measure of enzyme efficiency bissectors yields the point of “Circumcenter” for a triangle. with a unit of 1/(Molarity*second)= L/ (mol*s). The enzyme Material and methods for the present attempt is divided into efficiency can be increased as Kcat has high turnover and a the steps, which include: (A) Establishment of a Right angled small number of Km. Taking the reciprocal of both side of the Triangle Through the Linewever-Burk Plot (line Y.1); (B) Michaelis-Menten equation gives: Establishment of Geometrical Centroid of Right Angled Triangle; (C) Establishment of Geometrical Orthocenter of = + Right Angled Triangle; (D) Establishment of Geometrical 534 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp. 532-542, December, 2019

Circumcenter of Right Angled Triangle and (E) Establishment y1= = + of Euler’s line for a right angled triangle.

This Lineweaver–Burk plot deserve wide applicability. It is (A). Establishment of a Right angled Triangle Through the useful for the determination of K , the most significant factor Linewever-Burk Plot (line Y.1): m in enzyme kinetics. The intercept on y – axis of “Lineweaver–

Burk-Plot” is the reciprocal of V or (1/ V ). And intercept Hans Lineweaver and Dean Burk (1934) suggested the double max max on X – axis of “Lineweaver–Burk-Plot” is the reciprocal of - reciprocal plot for presenting the information in the form of K or (−1/K ). Reciprocals of both, [S] and (v) are utilized in readings or the data on the concentration of substrate [S] and m m the Lineweaver-Burk plot. That is to say, this plot is pertaining rate or velocity (v) of the biocatalyzed reaction. In enzyme and . Therefore, “Lineweaver–Burk-Plot” is also termed as kinetics, double reciprocal plot suggested by Hans Lineweaver and Dean Burk is well esteemed graphical presentation of the a double reciprocal graph. This attempt through the data on concentration of the substrate [S] and velocity (v) of “Lineweaver-Burk-Plot”, is giving quick, concept or idea of the biocatalyzed reaction recognized as, the “Lineweaver–Burk the biochemical reaction. It also allow to understand the plot”. This plot deserve wide applicability. The most mechanism of activation of enzyme and inhibition of enzyme. significant application of Lineweaver–Burk plot lies in the Researchers including authors of present attempt designating determination of concentration of substrate [S] which is the double reciprocal plot as a Nobel Plot. Most of researchers responsible for achievement of the half the maximum rate or entertaining the enzyme kinetics through this double reciprocal the velocity (Vmax ÷ 2) of the biochemical reaction catalyzed plot are non-mathematical academicians. Present attempt is by the enzymes. The “Km” or Michaelis constant is the trying it’s best to minimize the errors in understanding the concentration of substrate [S] responsible for yield of the concepts in enzyme kinetics through modification in the reaction rate, which is corresponding to exactly half the rate or “Lineweaver-Burk-Plot”. Each and every method is with velocity of maximal (Vmax ÷ 2) for enzyme involved positive and negative points of advantages. According to biochemical reaction. For practical purposes, this “Km” or Hayakawa, et al (2006), there is distortion of error structure Constant of Michaelis is the reading pertaining [S] that allows through this double reciprocal plot of “Lineweaver-Burk-Plot”. velocity to achieve half with reference to maximum rate or It is therefore, method of graphical presentation of velocity (Vmax). The affinity of enzymes for their substrate “Lineweaver–Burk-Plot” (double-reciprocal-plot) appears to vary. Generally, the enzyme with a higher Km value has little attempt to minimize errors. This may yield easier method of bit lower affinity for its substrate. According to Keith J. calculation of constants or parameters of enzyme kinetics. On Laidler (1997), enzymes with lower affinity for their substrate, this line of improvement of method of calculation of constants requires a greater volume of substrate or substrate or parameters of enzyme kinetics, much more work is already concentration for the purpose to achieve maximum rate or exist. According to Hayakawa, et al (2006), methods of velocity of enzyme involved biochemical reactions. The wide improvement in the calculation of constants or parameters of range of applicability is the distinguishing feature of enzyme kinetics are under the title, “non-linear regression or Lineweaver–Burk plot. In the past, there was no computer alternative linear forms of equations”. And they include: the facilities as today. In such a critical situation, the parameters of plot of “Hans-Woolf”; the plot of “Eadie-Hofste”; such and the enzyme kinetics, Km and Vmax served a lot through this others (Greco and Hakala, 1979). Lineweaver-Burk plot for fortified concept of enzyme kinetics. In this Lineweaver-Burk plot, reading the inverse of maximum Dick (2011) explained type of inhibition of enzyme activity or velocity of biocatalyzed reaction (1÷ Vmax) take the position stoping the working of enzymes. Of course, this discussion is of y-intercept (Fig.1). The negative value of inverse of Km based exclusively on “Lineweaver–Burk-Plot” (reciprocals ob (1÷Km) take the position of x-intercept. The quick visual both the axes) is able to group or classify the inhibitors of impression of the inverse form of substrate concentration and actions of enzymes. Accordingly, the inhibitors of enzyme can rate or velocity of reaction is one more advantage of basically be grouped into the types like: The “Competitive Lineweaver-Burk plot. And this feature help for understanding Inhibitors”; “Non-competitive inhibitors” and “uncompetitive the concept of enzyme inhibition. Accordingly, mathematical inhibitors”. The inhibitors of enzyme of “Competitive” class equation suggested by Lineweaver and Dean Burk (1934) can deserve one and the same point of intersection on the Y-axis. It be written as: = + clearly means, inhibitors of enzyme of “Competitive” class are not affecting on maximal rate or velocity of reaction

(competitive inhibitors provide protection the maximum velocity Vmax . They keep this maximum velocity Vmax non- affected). But, slopes of equations are not same. Slopes are different slopes. The inhibitors of enzyme of “Non- competitive” class deserve one and the same point of intersection on the X-axis. It clearly means, inhibitors of enzyme of “Non-competitive” class are not affecting on the Km, the [S] for half the maximal rate or velocity of reaction (Km is remains unaffected by non-competitive inhibitors. The inverse of Km doesn't change). The non-competitive inhibition produces plots with the same x-intercept (−1/Km) as uninhibited enzyme (Km is unaffected) but different slopes and y-intercepts. Uncompetitive inhibition causes different intercepts on both the y- and x-axes (Berg, et al, 2002). John E.

Fig. 1. Regular form of Linrweaver-Burk Plot (Double Reciprocal Plot) Dowd and Douglas Briggs (1965) reviewed the literature on “Estimates of Michaelis – Menten kinetic constants through 535 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp.532-542, December, 2019 the use of different linear transformation” and listed some any triangle that is not equilateral. It is a central line of the problems with Lineweaver–Burk plot (double reciprocal plot). triangle, and it passes through several important points Accordingly, Lineweaver–Burk plot (double reciprocal graph) determined from the triangle, including the orthocenter, the is appearing in most of the new as well older books of circumcenter, the centroid, the Exeter point and the center of biochemistry. It seems in having prone to error. There may be the nine-point circle of the triangle. The centroid is also called mistake in understanding the expected for researchers. The as geometric center of a plane figure. It is the arithmetic mean readings of inverse of “v” are on Y – axis. The readings of position of all the points in the figure. Altshiller-Court, Nathan inverse of “[S]” are on X – axis. The lower values of both the (1925) informally defined the centroid as the point at which a readings (inverse of “v” and inverse of “[S]”) are occupying cutout of the shape could be perfectly balanced on the tip of a higher (signifiacant) position in graph. And… and… higher pin. The barycenter is a synonym used for centroid. The values of both the readings (inverse of “v” and inverse of centroid of a triangle is the point of intersection of it’s three “[S]”) are occupying lower (non-significant) position in graph. medians. It is a point of concurrency of the triangle. It is the Both the conditions may be interpreted wrongly. point, where all the three medians intersect. It is often described as the center of gravity of triangle. The geometrical properties of centroid of a triangle include: Intersection of the three medians exert to form the centroid; The centroid is one of the points of concurrency of a triangle; The centroid is always located inside the triangle; The centroid divides each median in a ratio of 2:1. The centroid will always be 2/3 of the way along any given median. According to Johnson (1929) and Wells (1991), the geometric centroid (center of mass) of the of a triangle is the point formed through the intersections of the three medians of the triangle. The point of centroid, is therefore also called as the median point. It has equivalent functions. Orthocenter of triangle is the point of intersection of the altitudes. Each leg in a right triangle forms . In a right-angled triangle, the orthocenter lies at the vertex containing the right angle. Altitude of a triangle is a line Fig. 2. Right angled triangle through the use of “Regular form of segment through a vertex and perpendicular to a line Linrweaver-Burk Plot (Double Reciprocal Plot) (Y.1) containing the base (the side opposite the vertex). This line

In a Regular form of Linrweaver-Burk Plot (Double Reciprocal containing the opposite side is called as the extended base of altitude. The circumcenter of a triangle is the point of Plot) [ y1= (Km÷Vmax) x + (1÷Vmax)]; when the value of x is one, the y value correspond to [(Km +1)÷Vmax]. The y – intersections of the three perpendicular bisectors of the sides of intercept of the regular form of Linrweaver-Burk Plot (Double that particular triangle intersects. In other words, the point of Reciprocal Plot) correspond to (1÷Vmax)]. In figure 2; the y – concurrency of the bisector of the sides of a triangle is called intercept of the regular form of Linrweaver-Burk Plot (Double the circumcenter. Every triangle has exactly three medians, one Reciprocal Plot) is designated as point: “A”. The co-ordinates from each vertex, and they all intersect each other at the of the point “A” are zero and reciprocal of maximum velocity triangle's centroid. For getting the point of centroid in right of biochemical reaction involving the interplay of the enzymes. angled triangle (∆ ABC) (figure – 2) in present attempt, it is It can be written as: [A (0, 1÷Vmax)]. The line segment necessary to establish the three medians, one from each vertex. perpendicular to y – axis and passing through as point: “A” up to the point: “B” is considered as one of the side of triangle. At (B.1). Establishment of the line Y.2 (For one of the median the point “B”, x equals to one and y equals to the reciprocal of of (∆ ABC) (Fig.3): maximum velocity of biochemical reaction involving the Let us first consider the line y2. The slope and intercept on y- interplay of the enzymes (figure 2). Therefore the co-ordinates axis of this y line are considered as: and of the point “B” can be written as: [A (1, 1÷Vmax). As stated 2 earlier, when the value of x is one, the y value of Regular form respectively. Therefore, the mathematical equation in “Slope- of Linrweaver-Burk Plot (Double Reciprocal Plot) (y1) intercept” form (in the form of typical y= m x +c) of this line correspond to: [(Km +1)÷Vmax]. The point (with x – value y3 can be written as: equals to one) in regular form of Linrweaver-Burk Plot (Double Reciprocal Plot) (y1) claiming the y value is Y = x + 2 designated as “C”. The co-ordinates of the point “C” can be written as: C [ 1, (Km +1) ÷Vmax]. Joining the point “A” to This equation may also be writte as: the point “B”; the point “B” to the point “C” and the point “C” to “A” yields right angled triangle (figure – 2). A right-angled Y = x + + 2 triangle is a triangle in which measure of one angle is ninety degree (or a right angle). The above equation contain the term: (Km÷Vmax) for two times. (B). Geometrical Centroid of Right Angled Triangle: Let us take out one of the term: (Km÷Vmax) as a common The relation between the sides and angles of a right angled factor and let us simplyfy the above equation. triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the Y = [ + ] + 2 right angle are called legs. In geometry, the Euler line, named Y = [ ] + after Leonhard Euler. The Euler line is a line determined from 2 536 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp. 532-542, December, 2019

If we replace the “x” by (1÷ S) If we replace the “x” by (1÷ S) and substitute the Km = [S (Vmax –v) ÷ v]; the mathematical equation for the line y2 is going to transform into: Y2 = [ ] + () = + Let us now substitute the Km = [S (Vmax –v) ÷ v]; the mathematical equation for the line y2 is going to transform into: Simplification of this equation is going to yields into

() () = + Y2 = [ ] + . ()() = = + = Simplification will yields into

()() It definitely means for plotting the y3; it is necessary to = + consider X = (1÷ S) and

()() == Y =

It definitely means for plotting the y2; it is necessary to Through replacing the values of Vmax; V and S, it is possible consider X = (1÷ S) and to calculate the respective values of Y. This is going to serve the purpose of plotting this new line (y3) along with y1 and y2 ()() Y = This new line (y3) is passing from the point “A” (one of the vertex of ∆ ABC) and attain half the measurement of the Through replacing the values of Vmax; V and S, it is possible segment “BC” at point “F”. The line (y3) is dividing the to calculate the respective values of Y. This is going to serve segment BC into two equal parts. Therefore, the segment “AF” the purpose of plotting this new line (y2) along with y1. This is designated as median of ∆ ABC drawn from the vertex “A” new line (y2) is passing from the point “B” (one of the vertex on the side segment “BC”. of ∆ ABC) and intersecting the line y1 at the point “D” and dividing the segment AC into two equal parts. Therefore, the segment BD is designated as median of ∆ ABC drawn from the vertex “B” on the side segment “AC”.

Fig. 4. Line Y.2 and Y.3 as the two medians in Right angled triangle (∆ ABC)

(B.3). Establishment of the line Y.4 (For one of the median of (∆ ABC) (Fig.5):

Let us now consider the line y4 (Fig.5). The slope and intercept ()

on y- axis of this y4 line are considered as: and Fig. 3. Line Y.2 as one of the median in Right angled triangle (∆ ABC) respectively. Therefore, the mathematical equation in “Slope- (B.2). Establishment of the line Y.3 (For one of the median intercept” form (in the form of typical y= m x +c) of this line of (∆ ABC) (Fig.4): y4 can be written as:

Let us consider the line y3 (Fig.4). The slope and intercept on Y = x - 4 y- axis of this y line are considered as: and 3 respectively. Therefore, the mathematical equation in “Slope- If we replace the “x” by (1÷ S) and substitute the Km = [S intercept” form (in the form of typical y= m x +c) of this line (Vmax –v) ÷ v]; the mathematical equation for the line y4 is y3 can be written as: going to transform into:

() [()] Y2 = x + = - 537 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp.532-542, December, 2019

Simplification of this equation is going to yields into The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. () [()] = - Every triangle has exactly three medians, one from each vertex, and they all intersect each other at the triangle's

()[()] centroid. In a right angled triangle in figure 5, the point “O” is = the median, the x and y– coordinates of which correspond to: 2÷3 and [(Km +1) ÷ 3 Vmax] respectively. [()][()] = (C). Establishment of Geometrical Orthocenter of Right Angled Triangle: It definitely means for plotting the y4; it is necessary to consider X = (1÷ S) and Altitude of a triangle is a line segment through a vertex and perpendicular to the line containing the base. The base segment [()][()] Y = of triangle is the side opposite the vertex. The line containing the opposite side is called the extended base of the altitude. Through replacing the values of Vmax; V and S, it is possible The intersection of the extended base and the altitude is called to calculate the respective values of Y. This is going to serve the foot of the altitude. The length of the altitude, often simply the purpose of plotting this new line (y4) along with y1 ; y2 and called "the altitude". It is the distance between the extended y3. This new line (y4) is passing from the point “C” (one of the base and the vertex. The process of drawing the altitude from vertex of ∆ ABC) and attain half the measurement of the the vertex to the foot is known as dropping the altitude at that segment “AB” at point “E”. The line (y4) is dividing the vertex. Altitudes a triangle can be used in the computation of segment AB into two equal parts. Therefore, the segment “CE” the area of a triangle. One half of the product of length of base is designated as median of ∆ ABC drawn from the vertex “B” and length of altitude equals the area of the triangle. The on the side segment “AB”. The three medians of right angled longest altitude is perpendicular to the shortest side of the triangle in Figure – 5 (and Figure – 6 too) are intersecting at a triangle. The altitudes are also related to the sides of the common point “O”. triangle through the trigonometric functions (Mitchell, Douglas W. 2005). The three (possibly extended) altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle if and only if the triangle is acute (i.e. does not have an angle greater than or equal to a right angle). If one angle is a right angle, the orthocenter coincides with the vertex at the right angle (Dorin Andrica and Dan S ̧ tefan Marinescu, 2017).

Let us now shift to the determination of orthocenter of Right angled triangle (∆ ABC) with (Real Form of Lineweaver-Burk Plot); line Y.2 (First Median); Y.3 (Second Median) and Y.4 (Third Median). An altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side opposite the vertex). This line containing the opposite side is called the extended base of the altitude. The intersection of the extended base and the altitude is called the foot of the altitude. The length of the altitude, often simply called "the altitude", is the distance between the extended base and the vertex. The Fig. 5. Centroid (The point “O”) of Right angled triangle (∆ ABC) process of drawing the altitude from the vertex to the foot is with (Real Form of Lineweaver-Burk Plot); line Y.2 (First known as dropping the altitude at that vertex. It is a special Median); Y.3 (Second Median) and Y.4 (Third Median) case of orthogonal projection. In Fig. 6; the line Y.6 is passing from the vertex point “B” and the point “G” on hypotenuse of (B.4). Establishment of Centroid (The point “O”) of Right right angled triangle (∆ ABC). The mathematical equation for angled triangle (∆ ABC) obtained through the Linewever- the line Y.6 correspond to: Y.6 = - (Vmax ÷ Km) x + Burk Plot (line Y.1): 2 [(Km+Vmax )÷(Vmax.Km)]. The line segment BG is

perpendicular to the side opposite to vertex point “B”. In a right angled triangle (∆ ABC) in figure 5, measure of Likewise, the line segment AB and CB constitute remaining angle “B” is ninety. The segment AB and segment BC are the two altitudes of right angled triangle (∆ ABC). According to two perpendicular sides. The segment AC is hypotenuse of a Smith, Geoff and Leversha, Gerry (2007), in a right angled right angled triangle (∆ ABC) in figure 5. The line segment BD triangle, the altitude from each acute angle coincides with a leg is a median of a right angled triangle. It is joining a vertex “B” and intersects the opposite side at (has its foot at) the right- to the midpoint of the opposite side (line segment AC), thus angled vertex, which is the orthocenter. The triangle (∆ ABC) bisecting that side. The line segment AF is a median of a right resulted through the present attempt (Fig. 5 and Fig. 6) is liable angled triangle in figure 5. It is joining a vertex “A” to the to avail advantage of view on altitude of right angled triangle midpoint of the opposite side (line segment BC), thus bisecting by Smith, Geoff and Leversha, Gerry (2007). And accordingly, that side. The line segment CE is a median of a right angled the line segment AB and the line segment AC are triangle in Figure 5. It is joining a vertex “C” to the midpoint perpendicular with each other and constitute the two altitudes of the opposite side (line segment AB), thus bisecting that side. of the right angled triangle (∆ ABC) (Fig. 5 and Fig. 6). The 538 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp. 532-542, December, 2019 point “B” in the right angled triangle (∆ ABC) (Fig. 5 and Fig. analysis, such as the notion of a mathematical function. He is 6) is claiming the intersection of the line segment AB and the also known for his work in mechanics, fluid dynamics, optics, line segment AC. Therefore, the point “B” in the right angled astronomy and music theory (Thiele, Rüdiger, 2005). Euler triangle (∆ ABC) (Fig. 5 and Fig. 6) is designated as the was one of the most eminent mathematicians of the eighteenth orthocenter. century and is held to be one of the greatest in history. He is also widely considered to be the most prolific mathematician (D). Establishment of Geometrical Circumcenter of Right of all time. His collected works fill 92 volumes, more than Angled Triangle: anyone else in the field. He spent most of his adult life in Saint Petersburg, Russia, and in Berlin, then the capital of Prussia According to Smith, Geoff and Leversha, Gerry (2007) the (Sandifer, C. Edward, 2007). The Euler line is a central line of perpendicular bisectors of the sides of a triangle are concurrent the triangle, and it passes through several important points (they intersect in one common point). The point of determined from the triangle, including the orthocenter, the concurrency of the perpendicular bisectors of the sides is circumcenter, the centroid, the exeter point and the center of called the circumcenter of the triangle. The point of the nine-point circle of the triangle. The concept of a triangle's concurrency is not necessarily inside the triangle. It may Euler line extends to the Euler line of other shapes, such as the actually be in the triangle, on the triangle, or outside of the quadrilateral and the tetrahedron (Kimberling Clark, 1998). triangle. The perpendicular bisectors of the sides of the Eulers line is quite popular when one is dealing with geometry. do not necessarily pass through the vertices of the It is a special line in the plane of the triangle that passes triangles. A is a circle around the outside through many important well known points. It is proving all of a figure passing through all of the vertices of the figure, in those points to be collinear. The collinearity of points is very the present attempt, passing through the three vertices of the useful in solving certain complex pure geometry problems. Let right angled triangle (∆ ABC) (Fig. 5). Since the radii of the us shift our discussion to the present attempt on Establishment circle are congruent, a circumcenter is equidistant from of Euler’s Line for Enzyme Kinetics. For this purpose, the Fig. vertices of the triangle. In a right angled triangle, the 5 and Fig. 6 in the present attempt are going to serve a lot. In perpendicular bisectors intersect on the hypotenuse. Since the the Fig. 5 and Fig. 6, the line Y.1 is real form of Lineweaver- center of the circumscribed circle lies on the hypotenuse, the Burk plot with mathematical equation: y = (Km ÷ Vmax) x + hypotenuse becomes the diameter of the circle Richinick, (1÷Vmax). The line segment AB; the line segment BC and the Jennifer (2008). line segment CA are considered for right angled triangle (∆ ABC) resulted through consideration of real form of Lineweaver-Burk plot. According to Kimberling Clark (1998), the Euler line is a central line of the triangle, and it passes through important points determined from the triangle, including the centroid, the orthocenter and the circumcenter.

RESULTS AND DISCUSSION

According to Leonhard Euler (1767), the orthocenter, circumcenter and centroid of any triangle are collinear. This unique property of triangle is also true for total nine points of center of triangle. These other six points of center of triangle had not been defined in Euler's time. Other notable points concerned with a triangle and lie on the Euler line include: de Longchamps point, Schiffler point, Exeter point, and Gossard perspector. The of a reference triangle is tangent to the latter's circumcircle at the reference triangle's Fig. 6. Circumcenter (The point “D”) of Right angled triangle (∆ ABC) vertices. The circumcenter of the tangential triangle lies on the with (Real Form of Lineweaver-Burk Plot: Y.1); line Y.2 (First Median); Euler line of the reference triangle. The center of similitude of Y.3 (Second Median); Y.4 (Third Median); Y.5 (Perpendicular Bisectors) and Y.6 (Altitude) the orthic and tangential triangles is also on the Euler line. The results of the present are explained away through the points (E). Establishment of Geometrical Euler line for the Right like: Centroid for Enzyme Kinetics; Orthocenter for Enzyme Angled Triangle Resulted through the use of Lineweaver- Kinetics; Circumcenter for Enzyme Kinetics and Euler Line Burk plot: for Enzyme Kinetics.

In geometry, the Euler line, named after Leonhard Euler is a Centroid for Enzyme Kinetics line determined from any triangle that is not equilateral. According to Fraser, Craig G. (2005), Leonhard Euler (15 The median of a triangle is a line segment joining a vertex to April, 1707 – 18 September, 1783) was a Swiss the midpoint of the opposite side, thus bisecting that side. mathematician, physicist, astronomer, geographer, logician and Every triangle has exactly three medians, one from each engineer. He made important and influential discoveries in vertex, and they all intersect each other at the triangle's many branches of mathematics, such as infinitesimal calculus centroid. In the Fig. 5 and Fig. 6, the line Y.2; the line Y.3 and and graph theory, while also making pioneering contributions line Y.4 are representing the medians of a right angled triangle to several branches such as topology and analytic number (∆ ABC). The line Y.2 is with negative slope correspond to: - theory. He also introduced much of the modern mathematical (Km ÷Vmax) and y – intercept correspond to: terminology and notation, particularly for mathematical [(Km+1)÷Vmax]. The mathematical equation for line Y.2 539 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp.532-542, December, 2019 correspond to: y = - (Km ÷ Vmax )x + [(Km+1)÷Vmax]. The Vmax)]. This point “B” with co-ordinates: (1) and [(1 ÷ line Y.2 is passing through the point vertex “B” and the Vmax)], herewith labeled as “Orthocenter for Enzyme midpoint of hypotenuse (midpoint: “D” on line segment: AC). Kinetics”. The line Y.3 is with positive slope correspond to: (Km ÷ 2 Vmax) and y – intercept correspond to: [1÷Vmax]. The Circumcenter for Enzyme Kinetics mathematical equation for line Y.3 correspond to: y = (Km ÷ 2Vmax) x + [1÷Vmax]. The line Y.3 is passing through the The perpendicular bisectors of the sides of a triangle are point vertex “A” and the midpoint of the vertical side of right concurrent (they intersect in one common point). The point of angled triangle (∆ ABC) (midpoint: “F” on line segment: BC). concurrency of the perpendicular bisectors of the sides is The line Y.4 is with positive slope correspond to: (2Km ÷ called the circumcenter of the triangle (Smith, Geoff and Vmax) and y – intercept correspond to: [(Km – 1) ÷ Vmax ]. Leversha, Gerry, 2007). The point of concurrency is not The mathematical equation for line Y.4 correspond to: y = necessarily inside the triangle. It may actually be in the (2Km ÷ Vmax) x + [(Km-1)÷Vmax]. The line Y.4 is passing triangle, on the triangle, or outside of the triangle. The through the point vertex “C” and the midpoint of the horizontal perpendicular bisectors of the sides of the triangles do not side of right angled triangle (∆ ABC) (midpoint: “E” on line necessarily pass through the vertices of the triangles. A segment: AB). All the three line segments representing circumscribed circle is a circle around the outside of a figure medians of right angled triangle (∆ ABC) (line segment: BD; passing through all of the vertices of the figure, in the present line segment: AF and line segment: CE) are intersecting the attempt, passing through the three vertices of the right angled common point: “O”. Therefore, the point: “O” inside the right triangle. Since the radii of the circle are congruent, a angled triangle (∆ ABC) is considered as the centroid. The x- circumcenter is equidistant from vertices of the triangle. In a co-ordinate of the point “O” correspond to: (2÷3). The y- co- right angled triangle, the perpendicular bisectors intersect on ordinate of the point “O” correspond to: [(Km+3) ÷ (3 Vmax)]. the hypotenuse. Since the center of the circumscribed circle This point “O” with co-ordinates: (2÷3) and [(Km+3) ÷ (3 lies on the hypotenuse, the hypotenuse becomes the diameter Vmax)], herewith labeled as “Centroid for Enzyme Kinetics”. of the circle Richinick, Jennifer (2008). The line segment (BD) representing median drawn to the hypotenuse has the measure half the hypotenuse (line segment In the Fig. 6, the line Y.5 is representing the bisector of AC). hypotenuse (line segment: AC). The slope of line Y.5 is minus (vmax ÷Km). The intercept of line Y.5 on y – axis correspond Orthocenter for Enzyme Kinetics to: [(km2+2Km+Vmax2) ÷ 2KmVmax].

According to Douglas W. (2005), the three (possibly extended) .. Y5 = x + altitudes intersect in a single point, called the orthocenter of . the triangle. The orthocenter lies inside the triangle if and only In the Fig. 5 and Fig. 6, the line segment DF is the bisector of if the triangle is acute (i.e. does not have an angle greater than the line segment BC of the right angled triangle (∆ ABC). In or equal to a right angle). If one angle is a right angle, the the Fig. 5 and Fig. 6, the line segment DE is the bisector of the orthocenter coincides with the vertex at the right angle (Dorin line segment AB of the right angled triangle (∆ ABC). The Andrica and Dan S ̧ tefan Marinescu, 2017). An altitude of a three bisectors (bisector line segment DE; bisector line triangle is a line segment through a vertex and perpendicular to segment DF and the line Y.5) are intersecting at a common a line containing the base (the side opposite the vertex). This point: “D”. Therefore, the point: “D” of the right angled line containing the opposite side is called the extended base of triangle (∆ ABC) is considered as the circumcenter. The the altitude. The length of the altitude, often simply called "the lengths of line segment BD; line segment AD and line segment altitude", is the distance between the extended base and the CD are equal. According to Euler, Leonhard (1767), the vertex. The process of drawing the altitude from the vertex to circumcenter is equidistant from the three vertices of the the foot is known as dropping the altitude at that vertex. It is a triangle and lies on the perpendicular bisector. The x- co- special case of orthogonal projection (Bryant and Bradley, ordinate of the point “D” correspond to: (1÷2). The y- co- 1998). In the Fig. 5 and Fig. 6, the line segment: CB is the ordinate of the point “B” correspond to: [(Km+2) ÷ 2Vmax]. altitude through a vertex (the point: “C”) and perpendicular to This point “D” with co-ordinates: (1÷2) and [(Km+2) ÷ the line segment: AB. In the Fig. 5 and Fig. 6, the line 2Vmax], herewith labeled as “Circumcenter for Enzyme segment: AB is the altitude through a vertex (the point: “A”) Kinetics”. and perpendicular to the line segment: BC. In the Fig. 6, the

“line segment: BG” on line Y.6 is representing the third Euler Line for Enzyme Kinetics altitude. The slope of line Y.6 is minus (Vmax ÷Km). The intercept of line Y.6 on y – axis correspond to: [(Km+Vmax2) Euler line, named after Leonhard Euler. The Euler line is a line ÷ (KmVmax)]. The mathematical equation for line Y.6 can be determined from any triangle that is not equilateral. It is a written as: central line of the triangle, and it passes through several

. important points determined from the triangle, including the Y = x + 6 . orthocenter, the circumcenter, the centroid, the Exeter point and the center of the nine-point circle of the triangle. The All the three altitudes of right angled triangle (∆ ABC) (line orthocenter, the circumcenter and the centroid are the three segment: AB; line segment: CB and line segment: BG) (Fig. 6) “centers” of the triangle lie on one straight line, called the are intersecting the common point: “B”. Therefore, the point: Euler line. The point with co-ordinates: (2÷3) and [(Km+3) ÷ “B” of the right angled triangle (∆ ABC) is considered as the (3 Vmax)] in Lineweaver-Burk plot is “Centroid for Enzyme orthocenter. The x- co-ordinate of the point “B” correspond to: Kinetics” (The point: “O” in Fig. 5 and 6). 1. The y- co-ordinate of the point “B” correspond to: [(1 ÷ 540 International Journal of Recent Academic Research, Vol. 01, Issue 09, pp.532-542, December, 2019

The point with co-ordinates: (1) and [(1 ÷ Vmax)] in Burk plot is “Centroid for Enzyme Kinetics”. The point with Lineweaver-Burk plot is “Orthocenter” for Enzyme Kinetics” co-ordinates: (1) and [(1 ÷ Vmax)] in Lineweaver-Burk plot is (The point: “B” in Fig. 5 and 6). The point with co-ordinates: “Orthocenter” for Enzyme Kinetics”. The point with co- (1÷2) and [(Km+2) ÷ Vmax] in Lineweaver-Burk plot is ordinates: (1÷2) and [(Km+2) ÷ Vmax] in Lineweaver-Burk “Circumcenter” for Enzyme Kinetics” (The point: “D” in Fig. plot is “Circumcenter” for Enzyme Kinetics”. The line passing 5 and 6). The line passing through the point of centroid (The through the point of centroid; Orthocenter and Circumcenter of point: “O” in Fig. 5 and 6); Orthocenter (The point: “B” in Fig. right angled triangle resulted through Lineweaver-Burk plot is 5 and 6) and Circumcenter (The point: “D” in Fig. 5 and 6) of named herewith as: “Euler Line for Enzyme Kinetics”. right angled triangle resulted through Lineweaver-Burk plot is Lineweaver-Burk plot (double reciprocal plot) and Euler Line named herewith as: “Euler Line for Enzyme Kinetics”. for Enzyme Kinetics are with equal and opposite slope. Lineweaver-Burk plot (double reciprocal plot) is with positive Properties of Euler Line for Enzyme Kinetics value of (Km÷Vmax) as a slope. Euler Line for Enzyme Kinetics is with negative value of (Km÷Vmax) as a slope. The 1. Lineweaver-Burk plot (double reciprocal plot) and intercept on y-axis for Lineweaver-Burk plot (double Euler Line for Enzyme Kinetics are with equal and reciprocal plot) for Enzyme Kinetics correspond to: (1 ÷ opposite slope. Lineweaver-Burk plot (double Vmax). The intercept on y-axis for Euler Line for Enzyme reciprocal plot) is with positive value of (Km÷Vmax) as Kinetics correspond to: [(Km +2) ÷ Vmax)]. Lineweaver-Burk a slope. Euler Line for Enzyme Kinetics is with plot (double reciprocal plot) and Euler Line for Enzyme negative value of (Km÷Vmax) as a slope. Kinetics are intersecting at the point, x – co-ordinate of which 2. The intercept on y-axis for Lineweaver-Burk plot correspond to: (1÷2) and y- co-ordinate of which correspond (double reciprocal plot) for Enzyme Kinetics to: [(Km+2) ÷ Vmax]. The centroid for enzyme kinetics is correspond to: (1 ÷ Vmax). The intercept on y-axis for always located between the orthocenter and the circumcenter Euler Line for Enzyme Kinetics correspond to: [(Km of enzyme kinetics. The distance from the centroid to the +2) ÷ Vmax)]. orthocenter is always twice the distance from the centroid to 3. Lineweaver-Burk plot (double reciprocal plot) and the circumcenter. Euler Line for Enzyme Kinetics are intersecting at the point, x – co-ordinate of which correspond to: (1÷2) and Acknowledgements y- co-ordinate of which correspond to: [(Km+2) ÷ Vmax]. 31 December is the birthday of Hon. Dr. Avram Hershko 4. The centroid for enzyme kinetics is always located (Hungarian-born Israeli biochemist and Nobel laureate). His between the orthocenter and the circumcenter of kind self is well known for ubiquitin mediated degradation of enzyme kinetics. the proteins. Through the best compliments from India, the 5. The distance from the centroid (for enzyme kinetics) to present attempt is wishing Hon. Dr. Avram Hershko Happy the orthocenter (for enzyme kinetics) is always twice Birthday. Academic support and inspiration received from the the distance from the centroid (for enzyme kinetics) to Agricultural Development Trust, Baramati for the present the circumcenter (for enzyme kinetics). studies exert a grand salutary influence. 6. The line segment (BD) (of line: Y.2) representing median drawn to the hypotenuse has the measure half REFERENCES the hypotenuse (line segment AC) (of line: Y.1).

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