The genetic code A DNA (deoxyribonucleic acid) is composed of 4 different bases; adenine (A)‚ guanine (G)‚ cytosine (C) and thymine (T). Applying these 4 bases it may contain thousands of sequences within a single strand. Each of these bases makes a specific pairing with a corresponding base whereby the double helix structure is synthesised. This interaction is called base-paring and the complementary base pairs are; T pairs only with A and C only with G. Through this simple coding language‚ the
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AP Chemistry Kinetics Lab Abstract: The purpose of this lab is to learn how to write law expressions‚ determine orders by graphs‚ and calculate rate constants. By determining the number of drops of sodium hypochlorite necessary to make a diluted food dye change to colorless in less than three minutes helps calculate the absorbance which can then be analyzed to find the pseudo rate constant and eventually leading to the rate constant and the rate law. The results show that the m and n are both 1st
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uniform disk of radius R = 0.25 m has a string wrapped around it‚ and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest. a) When the 3 kg weight is moving with a speed of 2.2 m/s‚ what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J b) If the system started from rest‚ how far has the weight fallen?
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Part B Now‚ suppose that Zak’s younger cousin‚ Greta‚ sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak’s). Instead of getting a running start‚ she asks Zak to give her a push. So‚ Zak pushes her with a force of 125 \rm N over a distance of 1.00 \rm m. If her mass is 20.0 \rm kg‚ what distance d_2 does she slide after Zak’s push ends? Remember that the frictional force acts on Greta during Zak’s push and
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signals the breakdown of fats to provide energy in a process called gluconeogenesis. This breakdown of fats produces ketones such as acetone‚ these molecules are acidic. A build-up of acetone and acetoacetate in Roberta’s blood lowers the pH of her blood. This is known as ketoacidosis. She will begin to rapidly breathe in order to compensate for the high levels of acetone and low pH of her blood. 2. She feels very thirsty despite drinking lots of water. This is a common symptom of diabetes‚ another
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Effect of Iodine on mercury concentrations in dental-unit wastewater John Michael Panganiban and Ian Jasper Ocampo Objective: This study was undertaken to determine whether iodine used to control bacteria in dental unit wastewater could increase mercury concentrations in dental wastewater. Introduction: Two of the major concerns in dentistry are biofilm in dental unit waterlines and the contamination of dental unit wastewater with mercury. Biofilms are microscopic communities that consist
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Abstract The main goal of the enzyme kinetics experiment was to see how the phosphatase-catalyzed hydrolysis of p-nitrophenyl produced p-nitrophenol in the presence of phosphate and fluoride ion inhibitors of various concentrations. The calculated Km constant was found to be 0.22 for all reactions. The Vmax values for each inhibition ion were 0.00986 for the phosphate ion and 0.00436 for the fluoride ion. The inhibitor constant‚ Ki‚ was determined to be 0.0967 for the phosphate ion. The inhibitor
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Lab : Iodine-starch Clock Reaction Pre-lab: Before the lab was conducted‚ the concentration of the Iodate ions to be in the mixture made by dissolving specific volumes of solution A with a constant concentration and water was calculated using the dissolution formula: C1V1 = C2V2 Sample Calculation 1: Concentration of the Iodate ions: For mixture 1: C2 = C1V1/ V2 = (0.020mol/L) x (0.003L)/(0.01L) = 0.006mol/L The same calculations were used in the calculating of the
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Lab #3: Method of Initial Rates: Iodine Clock Introduction The detailed system of steps in a reaction is called the reaction mechanism‚ and it is one of the principal aims of chemical kinetics to obtain information to aid in the elucidation of these mechanisms in order to better understand chemical processes. Reactions usually occur in a stepwise manner with each step proceeding at a different speed. If the rate of reaction is slow enough to measure‚ this is indicative of a step much slower
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puck of mass m initially at speed u collides head-on (without rotation) with a stationary puck of mass M. Find the velocities of both puck after the collision if: i) the collision is fully elastic ii) the collision if fully inelastic. i) momentum: kinetic energy: mu = mv+MV (+ve in direction of initial u) 1 /2 m u2 = 1/2 m v2 + 1/2 M V2 2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ
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