Five Point Circle for the Triangular Form of Double Reciprocal (Lineweaver-Burk) Plot in Enzyme Biochemistry

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Five point circle for the triangular form of double reciprocal (Lineweaver-Burk) plot in enzyme biochemistry

Vitthalrao Bhimasha Khyade1,2,3*, Seema Karna Dongare4 and Sir Andrew John Wiles5

1Sericulture Unit, Malegaon Sheti Farm, Agricultural Development Trust Baramati, Shardanagar, (Malegaon Khurd) Post Box No - 35, Baramati, Pune 413 115, Maharashtra, India. 2Head, Department of Zoology, Shardabai Pawar Mahila Mahavidyalaya, Shardanagar Tal. Baramati Dist. Pune – 413115, India. 3Dr. APIS”, Shrikrupa Residence, Teachers Society, Malegaon Colony (Baramati) Dist. Pune – 413115, India 4P.G. Student, Department of Microbiology, Fegusson college (Autonomous), Shivajinagar, Pune, Maharashtra 411004, India. 5Mathematical Institute, University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, UK.

*Corresponding author. E-mail: [email protected]

Received: April, 2020; Accepted: May, 2020

The nine circle in geometry deserve significance of correlation with circumscribed circle, inscribed circle and the Euler line for a triangle. The nine significant concyclic points on the circle recognized by the triangle include: the midpoint of each side of the triangle; the foot point of each altitude and the midpoint of the line segment from each vertex of the triangle to the orthocenter (where the three altitudes meet; these line segments lie on their respective altitudes). The present attempt is dealing with utilization of principle of “Nine point circle” in understanding the concept of relations between the concentration of substrate [S] and the velocity (v) of enzyme controlled reactions in biochemistry. The steps of the attempt include: determination of mid points of three sides of the triangular form of double reciprocal plot; foot point of each line segment forming the altitude of a triangular form of double reciprocal plot; the midpoint of the line segment from each vertex of the triangle to the point representing the orthocenter of the triangular form of double reciprocal plot and determination of center for expected circle in triangular form of double reciprocal plot. Significant feature of triangular form of double reciprocal (Lineweaver-Burk) plot in present attempt is right angled triangle. The mid points of the three sides of triangular form of double reciprocal (Lineweaver-Burk) plot are easy for location. Being a right angle triangle, the orthocenter of triangular form of double reciprocal (Lineweaver-Burk) plot in present attempt coincides with the vertex at the right angle. The right angled vertex of triangular form is serving not only as the orthocenter, but also as the foot point for the base segment and the height segment. The segment perpendicular to real form of double reciprocal (Lineweaver-Burk) plot and passing through the right angled vertex give the third altitude foot point. The mid point of base segment and height segment constitute the mid point of the segment passing from vertex to the orthocenter. The five significant points resulted on circle include: B, D, E, F and G. The x and y co-ordinates of point “B” are: 1 and (1÷Vmax) respectively. The x and y co-ordinates of point “D” are: [(1÷2)] and [(Km+2) ÷ 2Vmax)] respectively. The x and y co-ordinates of point “E” are: [(1÷1)] and [(Km+2) ÷ 2Vmax)] respectively. The x and y co-ordinates of point “F” are: (1÷2) and (1÷Vmax) respectively. The x and y co-ordinates of point “G” are: [(Vmax2) ÷ (Vmax2+Km2] and [Km(Vmax2+Km)+Vmax2]÷ [(Vmax2+Km2)Vmax] respectively. Therefore, the resulted circle is titled as, “Five Point Circle for the Triangular Form of Double Reciprocal (Lineweaver-Burk) Plot in Enzyme Biochemistry”.

Key words: Triangular form, double reciprocal plot, five point circle, substrate, enzyme.

INTRODUCTION

The credit of discovery of the circle passing through the May, 1800 – 12 March, 1834) was mathematician (geometry as significant points in, on and around the triangle goes to Karl specialization) of Germany. According to David Fraivert Wilhelm von Feuerbach. Karl Wilhelm von Feuerbach (30 (2018), in very first step, Karl Wilhelm von Feuerbach AshEse J Med. Med. Res. 032

Figure 1. The real form of double reciprocal (Lineweaver-Burk) plot.

discovered the circle with six significant points on the triangle. orthocenter (where the three altitudes meet; these line segments The significant six points recognized by the circle of Karl lie on their respective altitudes). Wilhelm von Feuerbach are classified into two major groups. The aims of the present attempt are: to minimize the errors in The first group of significant points recognized by the circle of the earlier attempts of enzyme kinetics through utilization of Karl Wilhelm von Feuerbach include: the midpoints of the principle of mathematical “Nine point circle” for new three sides of the triangle. The second group of significant mathematical aspects in the form of new constants, for points recognized by the circle of Karl Wilhelm von Feuerbach example: [1÷ (Vmax-Km)], that fortify the concepts and to find include: the foot point of each altitude of a triangle. Thus, three alternative mathematical base for enzyme classification. It may midpoints (each of three sides) and three points representing fortify the concept of relations between the concentration of the foot points (each of three altitudes) constitute the substrate [S] and the velocity (v) of enzyme’s controlled significant six points recognized by the circle of Karl Wilhelm reactions. von Feuerbach. Olry Terquem (16 June, 1782 – 6 May, 1862), a French mathematician made addition of three significant points in “significant recognized by the circle of Karl Wilhelm TRIANGULAR FORM OF DOUBLE RECIPROCAL von Feuerbach”. The third group of significant points (Olry (LINEWEAVER-BURK) PLOT Terquem’s points) recognized by the circle of Karl Wilhelm von Feuerbach include: the midpoints of the line segment The very first step in this reference is to obtain the triangular passing from each vertex to the orthocenter of the triangle form of double reciprocal plot. The regular form of (Ratcliffe, 2006). This circle in geometry is recognized by Linrweaver-Burk plot or double reciprocal plot (Figure 1) is a various names of its discoverers and instances in the discovery. graphical presentation of the numerical data on substrate The names of this circle include: Feuerbach's circle, Euler's concentration [S] and the velocity [V] of biochemical reaction circle, Terquem's circle, the six points circle, the twelve points with significant role of enzymes. In this graphical presentation, circle, the n-point circle, the medioscribed circle, the mid circle inverse (reciprocal) of substrate concentration [1÷S] is or the circum-midcircle (Casey, 1886; Kocik and Solecki, considered for x- axis and the inverse (reciprocal) of velocity of 2009). But the title “Nine-Point-Circle” deserve appreciable reaction [1÷V] is considered for y- axis. Therefore, it is also popularity. Altshiller-Court (1925) listed geometrical properties called as double reciprocal plot. Slope for regular form of for the nine point circle. Accordingly, this circle is passing Linrweaver-Burk plot is (Km÷Vmax). The y- intercept for through the “Nine significant concyclic points recognized by regular form of double reciprocal plot is (1÷Vmax). Therefore, the triangle”. The nine significant concyclic points on the circle the mathematical equation for regular form of the double recognized by the triangle include: the midpoint of each side of reciprocal plot is: Y.1= (Km÷Vmax) x + (1÷Vmax)]. the triangle; the foot point of each altitude and the midpoint of Let us consider the y – intercept of regular form of double the line segment from each vertex of the triangle to the reciprocal plot as point, “A”. The x - co-ordinate of the point 033 Vitthalrao et al

Figure 2. The triangular form of double reciprocal (Lineweaver-Burk) plot.

“A” is zero. The y - co-ordinate of the point “A” is (1÷Vmax). triangular form of double reciprocal plot; foot point of each line Likewise, the x - co-ordinate of point “B” is one and the y - co- segment forming the altitude of a triangular form of double ordinate of point “B” is (1÷Vmax). According to the regular reciprocal plot; the midpoint of the line segment from each form of double reciprocal (Linrweaver-Burk) plot, Y.1= vertex of the triangle to the point representing the orthocenter (Km÷Vmax) x + (1÷Vmax)], when the reciprocal of substrate of the triangular form of double reciprocal plot and concentration (1÷S) is one, the y- value tends to: [(Km +1) ÷ determination of center for the circle in expection of the Vmax]. The point with the x - co-ordinate of one and y – co- principle of “Nine point circle” in triangular form of double ordinate of [(Km +1) ÷ Vmax] is designated as the point “C”. reciprocal plot. Therefore, the co-ordinates of the point “A”, “B” and “C” can be written as: [A (0, 1÷Vmax)]; [B (1, 1÷Vmax) and [C (1, (Km+1) ÷Vmax] respectively. The line segment, “AB” is going Determination of mid points of three sides of the triangular to serve as the base for the triangular form of double reciprocal form of double reciprocal plot plot. The line segment, “BC” is going to serve as the height for the triangular form of double reciprocal plot. The line segment, The attempt towards the determination of the mid points of the “AC” is going to serve as the hypotenuse for the triangular three sides of geometric triangle by Johnson (1929) and Wells form of double reciprocal plot. That is to say, joining the point (1991) is similar to the determination of the geometric centroid “A” to the point “B”; the point “B” to the point “C” and the (center of mass) of a triangle. The intersection of three medians point “C” to “A” yields right angled triangle (Figure 2). A yields the centroid. That is to say, the centroid is the point right-angled triangle is a triangle in which measure of one formed through the intersections of the three medians of the angle is ninety degree (or a right angle). The figure thus triangle. This may be the reason for recognizing the centroid as obtained is titled as, “Triangular form of double reciprocal plot the median point of a triangle. The centroid of a triangle is with in enzyme biochemistry”. equivalent center function. By geometrical definition, median The triangular form of double reciprocal plot (Figure 2) is of a triangle is line segment established through joining the now eligible for application of “Nine point circle”. According vertex point of a triangle to the midpoint of the opposite side. to Altshiller-Court (1925), requirements for the nine point The line segment representing the median has to bisect the side circle in a triangle include: the midpoint of each line segment of triangle into two equal subsegments. forming the side of a triangle; the foot point of each line Figure 3 deals with three median lines (Y.2; Y.3 and Y.4) for segment forming the altitude of a triangle and the midpoint of the determination of mid points of the triangular form of double the line segment from each vertex of the triangle to the point reciprocal plot (∆ ABC). The slope for the line Y.2; Y3 and representing the orthocenter. This clearly indicate that, the Y.4 respectively correspond to: [-(Km÷Vmax)]; attempt towards the establishment of nine point circle for the [(Km÷2Vmax)] and [(2Km÷Vmax)]. The y - intercept for the triangular form of double reciprocal plot should include the line Y.2; Y3 and Y.4 respectively correspond to: [(Km+1) ÷ steps: determination of mid points of three sides of the (Vmax)]; [(1÷Vmax)] and [-(Km-1) ÷ (Vmax)]. Therefore, the AshEse J Med. Med. Res. 034

Figure 3. The triangular form of double reciprocal (Lineweaver-Burk) plot (with the three medians).

equation of the line Y.2; Y3 and Y.4 respectively correspond form of double reciprocal plot. The point “F” is thus the mid to: Y.2= [-(Km÷Vmax)] X + [(Km+1) ÷ (Vmax)]; Y.3 = point of segment “AB”. The co-ordinates of the point “A”, [(Km÷2Vmax)] X + [(1 ÷ Vmax)] and Y.4= [(2Km÷Vmax)] X point “B” and point “F” are known. Therefore, the co-ordinates + [-(Km-1) ÷ (Vmax)]. of the point “A”, “B” and “F” can be written as: [A (0, In the triangular form of double reciprocal plot (fig.3), the 1÷Vmax)]; [B (1, 1÷Vmax)] and [F (0.5, (1 ÷Vmax)] line segment “BD” (belong to line Y.2); the segment “AE” respectively. The point “E” constitutes the mid point of the (belong to line Y.3) and the segment “CF” (belong to line Y.4) segment “BC”. Therefore, the co-ordinates of the point “E” can are the three medians of triangular form of double reciprocal be written as: E [(1, (Km+2) ÷ 2Vmax]. The point “D” plot. The point, “O” is the point of intersection of three constitutes the mid point of the segment “AC”, the hypotenuse medians (segment “BD”, segment “AE” and segment “CF”). of the triangular form of double reciprocal plot. Therefore, the The point, “O” is centroid of triangular form of double co-ordinates of the point “D” can be written as: D [0.5, (Km+2) reciprocal plot (∆ ABC) (Fig.3). ÷ 2Vmax]. The centroid is a point of concurrency of triangular form of Lineweaver-Burk plot (double reciprocal plot) for enzyme kinetics (Vitthalrao et al., 2019). It is the point of saturation of Determination of foot point of each line segment forming the enzyme substrate reaction. The distinguishing features the altitude of a triangular form of double reciprocal plot (geometrical properties) of centroid of a triangle include: In a triangle, intersection of the base (or extended base) and the - Intersection of the three medians exert to form the altitude is called the foot of the altitude. Altitude is centroid; perpendicular to the line containing the base. The base line - The centroid is one of the points of concurrency of a segment of triangle is the side opposite the vertex. The line triangle; containing the opposite side is called the extended base of the - The centroid is always located inside the triangle; altitude. Foot of altitude is the intersection of the extended base - The centroid divides each median in a ratio of 2:1. and the altitude. Many times the word “altitude” is used to - The centroid will always be 2/3 of the way along any denote the length of the altitude. Altitude is the distance given median. between the extended base and the vertex. Dropping the altitude is nothing but the “process of establishment of The point “O” in Figure 3 is representing centroid of the perpendicular line segment passing from the vertex of a triangular form of double reciprocal plot (∆ ABC). Attempt triangle to the foot”. Altitude segment is perpendicular to the towards the establishment of the three medians is helping not opposite side of each vertex of a triangle. Altitudes of triangle only for knowing the point of centroid but also for the can be utilized for the computation of the area of a triangle. determination of mid points of three sides of the triangular The area of the triangle is one half of the product of length of 035 Vitthalrao et al

Figure 4. The triangular form of double reciprocal (Lineweaver-Burk) plot (with the medians, altitudes, centroid and orthocenter).

base and length of altitude. Significant feature of triangle lies in the altitude segments is helping not only for knowing the point occurrence of the shortest side perpendicular to the longest of orthocenter but also for the determination of foot points of altitude. Mitchell (2005) established the relation between the three altitudes of the triangular form of double reciprocal plot. altitudes and the sides of the triangle through trigonometric The point “B” in fig-4 is the foot point for the altitude segment functions. There is only one point in a triangle for intersection “AB” as well as the altitude segment “CB”. When the triangle of three altitudes. The point of intersection of three altitudes is right angle, the feet of two of the altitudes coincide with the forms “Orthocenter” of a triangle. The orthocenter lies inside vertex of the right angle of the triangle. In addition, the feet of the triangle if and only if the triangle is acute (i.e. does not two of the midpoints of segments drawn from the vertices to have an angle greater than or equal to a right angle). If one the orthocenter coincide with the midpoints of the legs of the angle is a right angle, the orthocenter coincides with the vertex right triangle (Amy Bell, 2006). The point “G” in Figure 4 is at the right angle (Dorin and Dan S ̧ tefan, 2017). the foot point for the altitude segment “GB” Figure 4 deals with three median lines (Y.2; Y.3 and Y.4); Altitude is perpendicular to the line containing the base. The altitudes (segment AB, Segment CB and segment BG); base line segment of triangle is the side opposite the vertex. centroid (point “O”) and orthocenter (point “B”). In The line containing the opposite side is called the extended comparison with fig. 3, there is additional line (Y.5) in Figure base of the altitude. Foot of altitude is the intersection of the 4. The slope for the line Y.5 corresponds to: [-(Vmax÷Km)]. extended base and the altitude. Many times the word “altitude” The y - intercept for the line Y.5 correspond to: [(Vmax2 + is used to denote the length of the altitude. Altitude is the Km) ÷ (Vmax.Km)]. Therefore, the equation of the line Y.5 distance between the extended base and the vertex. Dropping corresponds to: Y.5= [-(Vmax÷Km)] X + [(Vmax2 + Km) ÷ the altitude is nothing but the process of drawing the altitude (Vmax.Km)]. In Figure 4, the segment “AB” constitutes from the vertex point to the foot point. Altitudes can be utilized altitude dropped from the vertex point “A” on the base segment for the computation of the area of a triangle. The area of the “BC”. In fig–4, the segment “CB” constitutes altitude dropped triangle is one half of the product of length of base and length from the vertex point “C” on the base segment “AB”. In Figure of altitude. Significant feature of triangle lies in occurrence of 4, the segment “BG” constitutes altitude dropped from the the shortest side perpendicular to the longest altitude. Mitchell vertex point “B” on the base segment “AC”. Therefore, the (2005) established the relation between the altitudes and the point “B” in the right angled triangle (∆ ABC) (Figure 4) is the sides of the triangle through trigonometric functions. There is intersection of the three altitudes (line segment AB, line only one point in a triangle for intersection of three altitudes. segment CB and the line segment BG). The point “B” in the The point of intersection of three altitudes forms “Orthocenter” triangular form of double reciprocal plot (right angled triangle of a triangle. The orthocenter lies inside the triangle if and only ∆ ABC) (fig. 4) is eligible for designation as the orthocenter if the triangle is acute (i.e. does not have an angle greater than (Vitthalrao et al., 2019). Attempt towards the establishment of or equal to a right angle). If one angle is a right angle, the AshEse J Med. Med. Res. 036

Figure 5. The triangular form of double reciprocal (Lineweaver-Burk) plot (with the center for nine point circle).

orthocenter coincides with the vertex at the right angle (Dorin orthocenter of triangular form of double reciprocal plot (which and Dan S ̧ tefan, 2017). are the fundamental requirements of geometrical nine point circle) (Figures 4 and 5). The mid points of the three sides of triangular form of double reciprocal plot include: D, E and F. Determination of midpoint of the line segment from each The feet points of altitudes in of triangular form double vertex of the triangle to the point representing the reciprocal plot include: B and G. The mid point of segment orthocenter of the triangular form of double reciprocal plot passing from vertex to orthocenter of triangular form of double reciprocal plot include: E and F. Thus, the five significant The Figure 4 is going to serve the purpose of determination of points of triangular form of double reciprocal plot include: B, midpoint of the line segment from each vertex to the point D, E, F and G (Figures 4 and 5). Let us proceed for the attempt representing the orthocenter of the triangular form of double on determination of center for “Nine point circle” in triangular reciprocal plot. The segment “AB” in Figure 4 is passing from form of double reciprocal plot. the vertex point “A” to the point “B” representing the Figure 5 deals with the determination of point of center for orthocenter of the triangular form of double reciprocal plot (∆ the nine point circle for triangular form of double reciprocal ABC). The point “F” is the mid point of the segment “AB” in plot of enzyme kinetics. The method used for the determination Figure 4. Likewise, segment “CB” in fig. 4 is passing from the of point of center for the nine pont circle in present attempt vertex point “C” to the point “B” representing the orthocenter belongs to A. S. Ozdemir (2012). In Figure5, let us consider the of the triangular form of double reciprocal plot (∆ ABC). The segment DE and segment EF. Now imagine their midpoints and point “E” is the mid point of the segment “CB” in Figure 4. their perpendicular bisectors (a line perpendicular to each Thus the point “E” and the point “F” are the midpoints of the respective segment through its midpoint). The intersection of line segment from each vertex of the triangle to the point the perpendicular bisectors will give us the center of our representing the orthocenter of the triangular form of double expected circle. Let's label this center O9. The lengths of reciprocal plot. segment O9F; segment O9B; segment O9E; segment O9D and segment O9D in fig. 5 are equal and representing the radius of expected circle. Determination of center for the circle in expection of the The center for expected circle can also be determined principles of “Nine point circle” in triangular form of through the line equations. In Figures 5 and 6; the segment FB, double reciprocal plot the segment BE, the segment DE and the segment DF are forming the rectangle. The segment BD and segment EF are The results of the earlier attempts (I, II and III) include the forming the diagonals, intersecting at the center of rectangle (as yield of triangular form; mid points of the three sides; feet of well as the expected circle). The segment BD and segment EF altitudes and the mid point of segment passing from vertex to belong to the line Y.2 and Y.6 respectively. 037 Vitthalrao et al

Figure 6. The triangular form of double reciprocal (Lineweaver-Burk) plot (with the five point circle).

Y = -[(Km÷Vmax)] X + [(Km+1) ÷ (Vmax)] is the [(3÷4)] = X mathematical equation for the line Y.2. Let us proceed for the mathematical equation for the line This clearly indicates that, the x – co-ordinate of the point of Y.6. In Figures 5 and 6, the line Y.6 is parallel to the line intersection of the line Y.2 and the line Y.6 is [(3÷4)]. The y- representing the real form of double reciprocal (Lineweaver- co-ordinate for the point of intersection of the line Y.2 and the Burk) plot (Y.1 line). It means, the line Y.6 is with slope equal line Y.6 is going to yield through replacing the value of “X” as to the line Y.1, which corresponds to: [(Km÷Vmax)]. [(3÷4)] in either equation of line Y.2 or the equation of line The point “F” is on the line Y.6. The x- and y- co-ordinates Y.6. The line equation for Y.6 is: y = [(Km÷Vmax)] X - [(Km- of the point “F” are known. (1÷2) and (1÷Vmax) are 2) ÷ (2Vmax)]. Let us replace the value of “X” as [(3÷4)] in respectively the x- and y- co-ordinates of the point “F”. The x- this equation of the line Y.6. and y- co-ordinates of the point “F” are going to serve for determination of intercept of the line Y.6 on the y- axis. Let us y = [(Km÷Vmax)] X - [(Km-2) ÷ (2Vmax)] ……(The line Y.6) put the “X” value and “Y” value of the point “F” in the y = [(Km÷Vmax)] [(3÷4)] - [(Km-2) ÷ (2Vmax)] …… equation for Y.6. [y = (Km÷Vmax)] X+ C]. (Replacing “X” by 3÷4) y = [(3Km÷4Vmax)] - [(Km-2) ÷ (2Vmax)] …………. (Km÷Vmax)] X + C = y …Expected equation for the line Y.6 (Simplification) [(Km÷Vmax)] [(1÷2)] + C = (1÷Vmax) ……. Replacing “X” y = [(3Km-2Km+4) ÷ (4Vmax)]……………. (Simplification) by (1÷2) and “y” by (1÷Vmax) y = [(Km+4) ÷ (4Vmax)]……………………. (Simplification) C = (1÷Vmax) - [(Km÷Vmax)] [(1÷2)] ………(Simplification) C = - [(Km-2) ÷ (2Vmax)] …………………….(Simplification) The y- co-ordinate for the point of intersection of the line Y.2 and the line Y.6 is [(Km+4) ÷ (4Vmax)]. The segment BD and The line equation for Y.6 is: y = [(Km÷Vmax)] X - [(Km-2) ÷ segment EF (in fig. 5 and 6) are forming the diagonals, (2Vmax)]. intersecting at the center of rectangle (as well as the expected The intersection of the line Y.2 and Y.6 is going to yield the circle). Thus, the x-coordinate for the center of expected circle center of rectangle BEDF (as well as the expected circle). in the attempt corresponds to: [(3÷4)]. And, the y-coordinate Y.2 = Y.6 for the center of expected circle in the attempt corresponds to: - [(Km÷Vmax)] X + [(Km+1) ÷ (Vmax)] = [(Km÷Vmax)] X - [(Km+4) ÷ (4Vmax)]. [(Km-2) ÷ (2Vmax)]

[(Km+1) ÷ (Vmax)] + [(Km-2) ÷ (2Vmax)] = [(Km÷Vmax) + Properties of five point circle for the triangular form of (Km÷Vmax)] X double reciprocal (Lineweaver-Burk) plot in enzyme [(2Km+2+Km-2)] = [(2Km÷Vmax)] X biochemistry [(3Km) ÷ (2Vmax)] = [(2Km÷Vmax)] X [(3Km) ÷ (2Vmax)] x [(Vmax÷2Km)] = X In biochemistry, the five point circle can be constructed for [(3Km) ÷ (2Vmax)] x [(Vmax÷2Km)] = X triangular form of double reciprocal (Lineweaver-Burk) plot. AshEse J Med. Med. Res. 038

This circle is passing through five significant concyclic points formed is appearing to analogue higher. defined from the triangular form of double reciprocal 3. The next points: G, O and D in the attempt of the “five (Lineweaver-Burk) plot. The five significant concyclic points points circle of enzyme kinetics” are explaining the extent to of this circle include: the midpoint of each side (Base: segment speed up the rate of reaction as the concentration of substrate “AB”; Height: segment “BC” and Hypotenuse: segment “AC”) increases. These are the points are nothing but the events in of the triangular form of double reciprocal (Lineweaver-Burk) process of saturation of the active sites of the enzyme with plot; the point of right angle vertex and the point of intersection substrate. of the line perpendicular to real form of double reciprocal 4. The Point “O” is position of center for the “five points circle (Lineweaver-Burk) plot and passing through the right angle of enzyme kinetics”. This point “O” is representing the part and vertex. partial of the enzyme structure concerned with controlling the The mid-point of base segment and height segment constitute rate of reaction. It may lead to postulate the controlling role of the mid-point of the segment passing from vertex to the the point “G”, point “O” and point “D” on “five points circle of orthocenter. enzyme kinetics” in the attempt. These three points in “five The five significant points resulted on circle include: B, D, E, F points circle of enzyme kinetics” may be regarded as enzymatic and G. The x and y co-ordinates of point “B” are: 1 and GOD. (1÷Vmax) respectively. 5. The Lineweaver–Burk plot is classically used in older texts, but is prone to error, as the y-axis takes the reciprocal of the The x and y co-ordinates of point “D” are: [(1÷2)] and rate of reaction – in turn increasing any small errors in [(Km+2) ÷ 2Vmax)], respectively. measurement. Also, most points on the plot are found far to the The x and y co-ordinates of point “E” are: [(1÷1)] and [(Km+2) right of the y-axis. Large values of [S] (and hence small values ÷ 2Vmax)], respectively. for 1/[S] on the plot) are often not possible due to limited The x and y co-ordinates of point “F” are: (1÷2) and solubility, calling for a large extrapolation back to obtain x- and (1÷Vmax), respectively. y-intercepts. The attempt on “five points circle of enzyme The x and y co-ordinates of point “G” are: [(Vmax2) ÷ kinetics” is trying it’s best to minimize the errors. (Vmax2+Km2] and [Km(Vmax2+Km)+Vmax2]÷ [(Vmax2+Km2)Vmax], respectively. The x and y co-ordinates of point of center (O9) are: [(3÷4)] Limitations and [(Km+4) ÷ (4Vmax)], respectively. The attempt on “five points circle of enzyme kinetics” is unable Therefore, the resulted circle is titled as, “Five Point Circle for to give the value for the substrate concentration for velocity of the Triangular Form of Double Reciprocal (Lineweaver-Burk) reaction to attain maximum velocity. Plot in Enzyme Biochemistry”.

Conclusion Applications of the five point circle in enzyme biochemistry The triangular form of double reciprocal (Lineweaver-Burk) 1. The five points on the circle are representing the significant plot is right angled triangle. The mid points of the three sides of events in the biochemical reaction catalyzed by the enzymes in triangular form of double reciprocal (Lineweaver-Burk) plot biochemistry. The point “B” and point “E” on the five points are easy for location. Being a right angle triangle, the on the circle are representing initial stage of enzyme involved orthocenter of triangular form of double reciprocal biochemical reaction. At low concentration of substrate, there (Lineweaver-Burk) plot in present attempt coincides with the is a steep increase in the rate of reaction. The five point circle is vertex at the right angle. The right angled vertex of triangular giving the extent of steepness of initial stage of enzyme form is serving not only as the orthocenter, but also as the foot involved biochemical reaction. point for the base segment and the height segment. The 2. For the regular Lineweaver-Burk plot, when the (1÷S) segment perpendicular to real form of double reciprocal correspond to one unit of substrate, the (1÷v) attains the value (Lineweaver-Burk) plot and passing through the right angled of: [(Km+1) ÷ Vmax]. For the same value of (1÷S), that is for vertex give the third altitude foot point. The mid-point of base one unit of substrate, the five points on the circle is giving the segment and height segment constitute the mid-point of the (1÷v) that equals to: [(Km+2) ÷ 2Vmax]. The attempt on five segment passing from vertex to the orthocenter. The five point circle is sufficient to explain the fact about mechanism significant points resulted on circle include: B, D, E, F and G. for the initial stage of the enzyme involved biochemical The x and y co-ordinates of point “B” are: 1 and (1÷Vmax) reaction, at low concentration of substrate; there is a steep respectively. The x and y co-ordinates of point “D” are: [(1÷2)] increase in the rate of reaction. The five point circle is giving and [(Km+2) ÷ 2Vmax)] respectively. The x and y co-ordinates the extent of steepness of initial stage of enzyme involved of point “E” are: [(1÷1)] and [(Km+2) ÷ 2Vmax)] respectively. biochemical reaction. At initial stage of reaction, the catalytic The x and y co-ordinates of point “F” are: (1÷2) and (1÷Vmax) site of the enzyme is empty. It is waiting for substrate to bind, respectively. The x and y co-ordinates of point “G” are: for much of the time, and the rate at which product can be [(Vmax2) ÷ (Vmax2+Km2] and [Km(Vmax2+Km)+Vmax2]÷ 039 Vitthalrao et al

[(Vmax2+Km2)Vmax], respectively. Therefore, the resulted University of Oxford, specializing in number theory. He is best circle is titled as, “Five Point Circle for the Triangular Form of known for proving Fermat's Last Theorem, for which he was Double Reciprocal (Lineweaver-Burk) Plot in Enzyme awarded the 2016 Abel Prize and the 2017 Copley Medal by Biochemistry”. The concept on five point circle for the the Royal Society. With the best compliments from India, the triangular Form of Double Reciprocal (Lineweaver-Burk) plot attempt on “Five Point Circle for the Triangular Form of may open a new avenue in enzyme biochemistry. Double Reciprocal (Lineweaver-Burk) Plot in Enzyme Biochemistry” is wishing Sir Andrew John Wiles “Happy Birthday”. Academic support and inspiration received from the Acknowledgements Agricultural Development Trust, Baramati exert a grand salutary influence. 11 April is the birthday of Sir Andrew John Wiles (born 11 18 April is the birthday of Joseph Leonard Goldstein (born April 1953). Sir Andrew John Wiles is an English 18 April 1940). Joseph Leonard Goldstein is an American mathematician and a Royal Society Research Professor at the biochemist. He received the Nobel Prize in Physiology or AshEse J Med. Med. Res. 040

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