|Molisch reagent | |agar-agar‚ gum arabic‚glycogen‚ cotton‚ |I2 in KI solution (Lugol’s iodine reagent) | |starch |Benedict’s reagent | |10% HOAc solution‚ |Barfoed’s reagent
Premium Glucose Iodine Carbohydrate
Antony R 9:00Am-11Am Unknown Number 94 Unknown Results Streptococcus salivarius The first step in my unknown identification was to carry out a gram stain to observe the cell shape and arrangement. My unknown turned out to be gram positive bacteria cocci shaped with long chain cell arrangement. The unknown was gram positive because the cells stained blue-violet ‚this indicate that my unknown has ability to retain crystal violet-iodine after alcohol application on the cells which is a gram
Premium Bacteria Streptococcus Agar plate
5) Provide the structure of the major organic product in the reaction below. (CH3)2CHCH2CH2CHO → 6) Provide the structure of the major organic product in the reaction below. 7) Provide the reagents necessary to carry out the transformation shown below. 8) Provide the reagents necessary to carry out the transformation shown below. 9) 2-Methylpentan-3-ol is classified as __________. A) a primary alcohol B) a secondary alcohol C) a tertiary alcohol D) a phenol
Premium Alcohol Oxygen
Materials • Samples of different types of milk • Benedict’s Reagent • Biuret Reagent • Sudan III • Water bath • Pipettes/syringes • Test tubes • Microscopic slides and cover slips • Microscope Method (testing for reducing sugars) 1. Add 3cm³ of whole milk‚ by using a pipette or syringe to the test tube. 2. Add 5cm³ of Benedict’s reagent and place it in the boiling water bath for 8 minutes. Do the same for semi-skimmed milk and skimmed milk. 3. Once all 3 of the test tubes are
Premium Glucose Disaccharide Carbohydrate
Biuret reagent was made by the mixing of two chemicals. 10. The reaction with Biuret reagent was observed and recorded. 11. The solutions were disposed as stated in the 8th procedure; the lab equipment and chemicals were returned to the teacher. CONCLUSIONSThrough this lab‚ we learned how to determine protein‚ lipids‚ glucose and starch with their indicators from unknown samples. During the lab‚ I have noticed that the protein solution reacted most dramatically with the Biuret reagent. When adding
Free Glucose Carbohydrate Starch
FOUNDATION OF MEDICAL SCIENCE BIOLOGY-FGS0044 LAB REPORT 3.1 : CARBOHYDRATE DETECTION TEST LAB REPORT 3.2 : PROTEIN DETECTION TEST LAB REPORT 3.3 : LIPID DETECTION TEST GROUP MEMBERS : CONTENTS NO. | TITLE | PAGE | 1 | INTRODUCTION | 3 | 2 | LAB 3.1 : CARBOHYDRATES DETECTION TEST | 5 | 3 | LAB 3.2 : PROTEINS DETECTION TEST | 8 | 4 | LAB 3.3 : LIPIDS DETECTION TEST | 12 | 5 | REFERENCES | 14 | INTRODUCTION Carbohydrates (saccharides) are molecules consist of carbon
Premium Amino acid
the reagent used to prepare the standard. Reagents used as standards are divided into primary reagent and secondary reagent. A primary reagent can be used to prepare a standard containing an accurately known amount of analyte. A primary reagent must have a known stoichiometry‚ a known purity (or assay) and be stable during a long term storage both in solid and solution form. The purity of a secondary reagent in a solid form or the concentration of a standard prepared from a secondary reagent must
Premium Titration Hydrochloric acid Sodium carbonate
negative result because biuret reagent is blue to begin with. If a solution is pinkish purple‚ or purple‚ then the test for protein is positive. The test solution for starches is yellowish brown. If any substance that is yellowish brown when mixed with iodine‚ then the test for starches is negative. If the solution turns purple‚ or a dark purplish black‚ then the test for starches is positive. If DI water‚ and sucrose are tested for proteins using biuret reagent‚ then they will show a negative
Premium Milk Glucose Carbohydrate
minutes. Cool and filter the mixture. Add enough fresh 2M HCl to wash the residue and then bring the volume of the filtrate to 5ml. Place 1ml each of the filtrate to 2 test tubes as follows: 1. To 1ml of the filtrate‚ add a few drops of Mayer’s Reagent. Result: Light brown solution with
Premium Acetic acid Water Brown
two new σ bonds are formed. • Alkenes are electron rich‚ with the electron density of the π bond concentrated above and below the plane of the molecule. • Because alkenes are electron rich‚ simple alkenes do not react with nucleophiles or bases‚ reagents that are themselves electron rich. Alkenes react with electrophiles. Electrophilic Addition Energy Diagram: • The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions. In step 1‚ the alkene is the Lewis
Premium Alkene Hydrogen Electron