Introduction: Saponins are class of chemical compounds present in many plant species. Strcucture of saponins are derived from 30 carbon precursor oxidosqualene that are attached with glycosyl residue. Saponins are divided into steroid glycosides and triterpenoids, consist with polar and non polar groups. This property shows its amphypathic nature. They consist of nonpolar aglycones part coupled with one or more monosaccharide moieties glycan parts. Extraction techniques. Recent study has reported the extraction procedure for saponins that are 1. Green extraction technologies a. Maceration extraction b. Reflux and soxhlet extraction c. Subsequent extraction 2. Conventional technologies d. Ultrasound assisted extraction (UAE) e. Microwave …show more content…
Maceration Extraction: Maceration extraction is a solid liquid extraction . The efficacy of extraction is determined by two main factors, solubility and diffusion. The solubility is determined that polar compound soluble in polar solvents and nonpolar compounds soluble in nonpolar solvents. The rate of dissolution of a solute from the plant material in the extraction solvent is determined by the rate of mass transfer of the solute from the plant material to the the solvent. Previous studies are reported that methanol and ethanol used for extract saponins from plants. Table 1. (Cheok et al., 2013) Sl.No Maceration Extraction Plant for saponin isolation 01 Allium ampeloprasum 02 Momordica charantia 03 Polygonatum odoratum 04 Ipomoea batatas 06 Panicum virgatum L. b. Reflux and Soxhlet: Reflx and soxhlet extraction involves distillation process. The process involves heating a solution to boiling and then returning the condensed vapors in to the original flask. Disadvantage of the extraction is a time consuming, ethanol is mainly used for the saponins from the plant material. Table 1. (Cheok et al., 2013) Sl.No Refulx and Soxhlet Extraction Plants for saponin isolation 1 Medicago arabica L 2 Medicago hybrid 3 Yucca schidigera Roezl 4 Medicago Arabica 5 Antonia …show more content…
Bacopa monnieri (L.) belong to family Scrophulariaceae, is commonly known as Brahmi, Jalbhrami is an important medicinal herb, an annual plant found throughout the Indian subcontinent in wet, damp and marshy areas (Majumdar et al., 2013; and reference therein ) and widely distributed in the warmer parts of Asia, Australia and America (Rahman et al., 2002). This herb has been described in Ayurvedic texts since around 800 BC and recorded as a treatment for a range of mental disorders in the ‘Carak Samhita’ (Majumdar, et al., and reference therein). Bacopa monnieri is classified as medhya rasayan, i.e., a drug that is supposed to counteract the effect of mental stress and improve intelligence and memory function that improve mental health, intellect and memory (medhya) and promote longevity and rejuvenation (rasayana) (Singh, 2013).The alcoholic extract of Bacopa monnieri contains various active alkaloids, flavonoids, glycosides triterpenoid saponins such as bacoside A and B (Charles et al., 2011; Majumdar et al., 2013). These are the secondary metabolic products of the plant. Previous studies have shown that two triterpenoid glycoside saponins viz.bacoside A and B, are pharmacologically active compounds in the plant extract (Sivaramakrishna et al., 2005). The bacoside A is a mixture of four triglycosidic saponins,
Results The lab experiment was done in two parts, one with the NAND, NOR, XOR and Hex Inverters and the other with a 7483 full adder gate, both will verify the truth table when two input bits and a carry are added together. The circuits were built by examining the 1 bits through a K-Map to create a Boolean expression for the sum and carry. The Boolean expression for the sum was A⊕B⊕C and the carry as AB+BC_in+AC_in. From these two expressions, we notice that we must use two exclusive-ORs gates in the sum inputs for A, B, and C. For the sum, we have to use NOR and NAND (the only available gates from the lab manual).
For most sequences at position 4 and 5 we observe only the nucleotides G and T, respectively. There may be rare cases where other nucleotides may also be found. To consider such observations, we need to do a process called additive smoothing or Laplace smoothing to smooth the categorical data. [9] In this case, we add 4 sequences: AAAAAAAAA, CCCCCCCCC, GGGGGGGG, TTTTTTTTT.
I need to find the area of rectangle ABCD. I know that ABCD is a rectangle with diagonals intersecting at point E. Segment DE equals 4x-5, segment BC equals 2x+6, and segment AC equals 6x. I predict that To find the area of rectangle ABCD I need to find out the base and height of the rectangle. The first step is to find what x equals. Since I know the intersecting line segments AC and DB are congruent that means when I times the equation 4x-5 for segment DE by two it will equal the equation 6x for segment AC.
Experiment 7 In this experiment we configured several DC circuits consisting of an emf and a network of resistors. The circuits were composed of a power supply, two DMMs, a circuit board, an SPST switch, and an assortment of known resistors along with one unknown resistor. We measured the current and voltage of the entire circuit as well as the potential drops across each resistor to determine the parameters of the circuit including the resistance, voltage, and current for each component.
%% Init % clear all; close all; Fs = 4e3; Time = 40; NumSamp = Time * Fs; load Hd; x1 = 3.5*ecg(2700). ' ; % gen synth ECG signal y1 = sgolayfilt(kron(ones(1,ceil(NumSamp/2700)+1),x1),0,21); % repeat for NumSamp length and smooth n = 1:Time*Fs '; del = round(2700*rand(1)); % pick a random offset mhb = y1(n + del) '; %construct the ecg signal from some offset t = 1/
Suppose we have a single-hop RCS where there is one AF relay that amplifies the signal received from a transmitter and forwards it to a receiver. Assume that the transmitter sends over the transmitter-to-relay channel a data symbol ${s_k}$, from a set of finite modulation alphabet, $S={S_1, S_2,ldots,S_{cal A}}$, where ${cal A}$ denotes the size of the modulation alphabet. The discrete-time baseband equivalent signal received by the relay, $z_k$, at time $k$ is given by egin{equation} z_k = h_{1,k}s_k + n_{1,k},~~~~for~~k=1,2,ldots,M label{relaySignal} end{equation} where $n_{1,k}sim {cal N}_c(0,sigma_{n1}^2)$ is a circularly-symmetric complex Gaussian noise added by the transmitter-to-relay channel, $h_{1,k}$ denotes the transmitter-to-relay channel, and
(a) 3Mbps / 150Kbpa =3 X 1024 / 150 = 3072 / 150 =20.48 20 Users can be supported 150Kbps dedicated. (b)
Figure shows the intersection of line joining the camera center and image points ${\bf x}$ and ${\bf x'}$ which will be the 3D point ${\bf X}$.\\ \end{figure} The ‘gold standard’ reconstruction algorithm minimizes the sum of squared errors between the measured and predicted image positions of the 3D point in all views in which it is visible, i.e.\\ \begin{equation} {\bf X=\textrm{arg min} \sum_{i} ||x_i-\hat{x_i}(P_i,X)||^2} \end{equation} Where ${\bf x_i}$ and ${\bf \hat{x_i}(P_i,X)}$ are the measured and predicted image positions in view $i$ under the assumption that image coordinate measurement noise is Gaussian-distributed, this approach gives the maximum likelihood solution for ${\bf X}$. Hartley and Sturm [3] describe a non-iterative
Using the data provided in each one of these tests it can be assumed that one has done the steps to be able to determine the magnitude and orientation of the charges of the tape in each test, thus, allowing them to apply the same principle to any object they so desired. Their results would line up with the following; that if the two pieces of tape are torn from the same 40 centimeter strip then the tops of both pieces of tape would be positive and the bottoms of both pieces of tape would be negative and that if they would double the tape the attraction or repulsion in general would lower due to the increased density. Their data would also show that two pieces of tape ripped from each other would result in one piece being entirely positive and the other being entirely negative, they would also be able to state that the orientation of how the tape is paired up doesn’t matter.
1. The test subjects will prepare for sleep by acquiring everything needed for the subjects’ sleep preferences. 2. The test subjects will all set alarms on their smartphones for approximately 6, 8, and 10 hours after the subjects’ enter the resting period (Subjects may wake during the resting period for the bathroom, but they must not stay awake for more than ten minutes at a time to prevent as much deviation as possible.). 3.
1. What area/aspect of this setting is the most challenging? 2. In the setting, you work in, is there a certain population of patients you see more? How does this affect you?
1. Identify the range of senses involved in communication • Sight (visual communication), Touch (tactile communication), Taste, Hearing (auditory communication), Smell (olfactory communication) 2. Identify the limited range of wavelengths and named parts of the electromagnetic spectrum detected by humans and compare this range with those of THREE other named vertebrates and TWO named invertebrates. Figure 1: the electromagnetic spectrum source: www.ces.fau.edu Vertebrates Human Japanese Dace Fish Rattlesnake Zebra Finch Part of electromagnetic spectrum detected ROYGBV (visible light) detected by light sensitive cells in the eye called rods and cones.
Unknown Lab Report Unknown # 25 By: Jenna Riordan March 19, 2018 Bio 2843 1. Introduction Microbiology is the study of microorganisms found in all different environments throughout Earth, from the hot thermal vents at the bottom of the ocean to the ice at the top of a mountain.
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
Lab 1 helps create a better understand of the changes in crystal structures when the annealing and quenching process is applied to 1020 and 1080 steel. The numbered steel refers to the ASTM grain-size number. Formula 1 is used to solve for the grain size. n=2^(G-1) Equation (1) at 100x magnification Crystal structures change shapes which changes the strength of the material and its properties. The metal might become soft, brittle, hard, or ductile.