Biology Laboratory Manual BIO F110 PDF 2024-2025

LABORATORY MANUAL FOR BIOLOGY LABORATORY BIO F110 BY BIOLOGY FACULTY EDUCATIONAL DEVELOPMENT DIVISION BITS Pilani, Dubai Campus DIAC, Dubai, UAE 2024-2025  Content Pr...

LABORATORY MANUAL FOR BIOLOGY LABORATORY BIO F110 BY BIOLOGY FACULTY EDUCATIONAL DEVELOPMENT DIVISION BITS Pilani, Dubai Campus DIAC, Dubai, UAE 2024-2025  Content Preface General Laboratory Instructions Index table Experiment – 1: Microscopy and Identification of the specimen slide 1-4 Experiment – 2: Stomata density 5-6 Experiment – 3: Mitotic Index 7-11 Experiment – 4: Micrometry 12-14 Experiment – 5: Total White Blood Cell Count 15-17 Experiment – 6: Counting of Red Blood Cell 18-20 Experiment – 7: ABO blood grouping by slide agglutination test 21-23 Experiment – 8: Estimation of chlorophyll 24-27 Experiment – 9: Quantitative Estimation of Protein by Biuret method 28-30 Experiment – 10: Quantitative Estimation of Glucose by DNSA method 31-33 Experiment – 11: Qualitative analysis of different plant pigments 34-36 Experiment – 12: One Dimensional paper Chromatography 37-39 Experiment – 13: Isolation and quantitation of eukaryotic genomic DNA 40-41 Bibliography Appendix  Preface Biology laboratory manual provides details on the laboratory activities appropriate for the course. The given exercises require basic knowledge on biology and skill. Each exercise has one or two central themes which can be greatly enhanced by verbal introductions and additional discussion time. In each experiment, the introduction emphasizes the appropriate theory. This manual includes a complete set of recipes to prepare all solutions and chemicals. We encourage the students to work independently and in small group, as per the time and facilities. This manual specifically provides: 1. An introduction of important safety information. 2. Objective behind each lab exercise. 3. List of necessary materials. 4. Methods to prepare chemicals and solutions. 5. Comments on the experimental procedures and student activities.  General Laboratory Instructions The objective of these laboratory exercises is to acquaint the techniques emphasizing the need of measurements in the field of Biology. Each session is for a period of two hours. Systematic work will ensure that you complete the experiment on time. No student can leave the laboratory without showing his/her result to the instructor concerned. Students must Come with the hard copy of the journal when you come for lab. Have a lab coat of proper fit and should enter the lab with the lab coat on. Come prepared in time with the principle, theory and procedure of the experiments to be conducted on that day. Bring stationary and calculators and complete the calculations in the laboratory itself. Handle all the apparatus with utmost care while working. Report if any breakage to the Instructor immediately. Submit the journal on time as instructed in the laboratory. Present the data, diagrams, figures and tables neatly in the given space. Regularity is expected and make- ups will be granted in case of a genuine reason, as per the general institute policy.  INDEX NAME________________________________________ ID No_________________ SECTION ______ DAY___________ PRACTICAL HOUR ________________ Marks S. No Date Experiment (15) 1 2 3 4 5 6 7 8 9 10 11 12 Experiment No: 1 Microscopy and Identification of the specimen slide Objective: To study the parts of compound microscope and to understand its uses To identify and study the characteristic features of the given specimen slide Theory: The microscope is an optical instrument that enables us to see microorganisms and their intracellular structures. It contains two separate lens systems, objective lens and ocular lens. Principle: The three most important principal factors in microscopy are Magnification, Resolving power and Contrast. The main function of microscope is Magnification. Resolution or resolving power refers to the ability of lenses to distinguish between two closely related points. The distance between two points in microscopic field that can be distinguished from one another is called minimum resolvable distance. In order to be visible through a microscope, an object must possess a certain degree of contrast with the surrounding medium as a result of the fact that less light is transmitted through the object than through the medium. Contrast can be increased by adopting staining procedures. Table 1.1 Parts of the compound microscope and their functions Parts Functions Ocular lenses Magnify the image (10 times) Revolving nose piece Revolved manually to fix a desirable objective lens Multiple Objective Scanning power: 4X; Low power: 10X; High power: 40X; Oil immersion: lenses 100X Stage Supports the microscopic slide Stage manipulator To move the slide either from side to side and/or from front to back. knobs Open and close the iris diaphragm to control the amount of light striking Iris Diaphragm lever the object. Condenser Converges the light rays towards the specimen. Coarse adjustment Moves stage up and down for approximate focusing. knob Fine adjustment knob Moves stage up and down for definite focusing. Light source Illuminates the object. Rheostat Increase or decrease the light intensity. 1  Fig. 1.1 Compound Microscope Materials required: Compound microscope, Specimen slides. Procedure: Sit on a stool and place the microscope at a suitable distance to view through the eyepiece. Switch on the illumination source. With the help of the rheostat and iris diaphragm lever, adjust the intensity and amount of light. Place the slide on the stage and focus it under the lowest magnification (Scanning power) by adjusting the course adjustment screw. If further magnification is desired, revolve the nose piece to get 10X and 40X objective lenses in position. Now, use only the fine adjustment screw because the microscope is parfocal. If further magnification is needed, then use oil immersion objective in the following manner: Rotate the nose piece to get 100X objective in position. Then, place a drop of immersion oil on the slide. This oil has the same refractive index as glass (1.5).This will eliminate two refractive surfaces so that the magnification of 1000 times can be achieved while still preserving good resolution. 2  Identification of the specimen slide Diagram Characteristic feature Slide 1. _______________________________ Slide 2. _____________________________ Slide 3. _______________________________ Slide 4. _______________________________ Slide 5. _______________________________ 3 Slide 6. _______________________________ Slide 7. _______________________________ Slide 8. _______________________________ Slide 9. _______________________________ Slide 10.______________________________ 4 Experiment No: 2 Stomata density Objective: To view and compare the stomata from the leaves of monocot and dicot plant. To study the structure of stomata To study the distribution of stomata in different plants Theory: A stoma (pl. stomata) is a microscopic pore on the surface (epidermis) of land plants. It is surrounded by a pair of specialized epidermal cells called guard cells, which act as a turgor-driven valve that open and close the pores in response to given environmental conditions. The presence of countless numbers of stomata is critical for plant function. The plant epidermis is tightly sealed. Thus stomata act as a gateway for gas (carbon dioxide and oxygen) exchange and transpiration (water). Stomata helps the plant to pull and distribute water and minerals by roots and vasculature to the different parts of the plant. When a plant encounters ionic, osmotic oxidative stress, a plant hormone called abscisic acid triggers stomata to shut tightly in order to prevent plants from dehydration and wilting. Fig.2.1 Distribution of stomata in mono-cotyledon and di-cotyledon leaf Principle: Stomata density refers to the number of stomata per square millimeter. Typical densities can vary from 100 to 1000 depending on the plant species and the environmental conditions during development. More stomata are made on plant surfaces under higher light, lower atmospheric carbon dioxide concentrations and moist environments. The size and shape of stomata also vary with different plant species and environmental conditions. For example, grasses have guard cells that resemble slender dumbbells whereas trees and shrubs have guard cells that resemble kidney beans. Materials required: Leaf, grass, scalpel, glass slide, water and microscope. 5 Procedure: Tear the dicot leaf (Bougainvillea- paper flower) diagonally to get lower epidermis layer. Cut the lower epidermis layer. Place it on a clean glass slide, add a drop of water and cover it with a cover slip. Focus at 400times magnification. Draw the leaf surface with stomata. Count all the stomata in microscopic field. Repeat counts for at least three other distinct microscopic fields. Place the monocot (Cyperus rotundus –grass) on a clean glass slide. Using a scalpel scratch off the upper epidermis and get the lower epidermis. Add a drop of water then cover it with a cover slip. Focus at 400 times magnification. Draw the leaf surface with stomata. Count the number of stomata in one microscopic field. Record the number on your data table. Repeat counts for at least three other distinct microscopic fields. Record all the counts. Determine an average number per microscopic field. Stomata per mm2 area are calculated by multiplying the average number of stomata at 400 times magnification in the microscopic field by 8. Observation: Leaf 1 (monocot) Leaf 2 (dicot) Common Name of the plant Biological name of the plant Drawing several stomata at 400 times magnification No. of Stomata in field 1 No. of Stomata in field 2 No. of Stomata in field 3 Average number of Stomata Result: Stomata density (Stomata/ mm2 area) = Avg. number of Stomata X 8 Stomata density in monocot leaf = Stomata density in dicot leaf = 6 Experiment: 3 Calculation of Mitotic Index Objective: To prepare an onion root squash using the meristematic root tip To locate the phases of Mitosis in onion root tip permanent slide To calculate Mitotic Index To calculate Duration of mitosis Theory: Culturing Onion Roots: Onion root tips are ideal to view the different phases of mitosis. The tips are used to study mitosis because they grow fast and the cells are quite large. The chromosomes of onion are larger and very dark when stained and are easily observed through a compound light microscope. Roots consist of different regions. The root cap functions in protection. The apical meristem is the region that contains the highest percentage of cells undergoing mitosis. The region of elongation is the area in distal meristem in which growth occurs. The region of maturation is where root hairs develop and where cells differentiate to become xylem, phloem, and other tissues. Mitotic index (MI) is defined as the fraction of cells in a field that are in the process of mitosis. This is of importance in determining the duration of Mitosis in the cell cycle of growing cells. Mitotic index is a measure of the proliferation status of a cell population. Significance of mitosis The presence of mitotic cells is a good indicator of active cell division. It is important that the cells divide and replace old worn-out cells Mitosis is important for growth to take place. It is also important for genetic stability. By duplicating the exact copy of the genetic material, it ensures that genetic material is stable and can carry out its function. It is important for cell regeneration. When cells get damaged it is continuously being replaced by new cell. It is also a type of asexual reproduction in unicellular organism. 7 Principle: The Cell cycle is a tightly regulated process which is controlled by various cellular and environmental signals. The cell cycle is divided into two phases, namely Interphase and Division phase. Interphase: In any population of cells, only some are in dividing phases at a given time. The remaining cells are in interphase (Non- dividing phase). During interphase, cellular organelles double in number and protein synthesis occurs. The chromosomes are not visible and the DNA appears as chromatin fiber. The G0 phase (G zero phase) is a period in the cell cycle in which cells exist in a senescent or quiescent state. G0 phase is viewed as either an extended G1 phase, where the cell is neither dividing nor preparing to divide, or a distinct quiescent. G0 is sometimes referred to as a "post- mitotic" state, since cells in G0 are in a non-dividing phase outside of the cell cycle. Division phase The cell division takes place in two stages: 1) Karyokinesis (Division of cell nucleus/genetic materials) is sub- divided into four phases viz. Prophase, Metaphase, Anaphase and Telophase. 2) Cytokinesis (Division of cytoplasm and cell organelles). G0 Fig. 3.1: cell cycle and its stages The different stages of mitotic cell division are: Prophase: The chromosomes appear and become thick. Each chromosome is formed of two sister chromatids. In this stage, the nucleus disappears and the nucleolus disintegrates (2). Metaphase: The chromosomes arrange themselves at the equatorial plane (2). Spindle fibers appear and the chromosomes get attached to them. 8 Anaphase: The chromatid of each chromosome is separated into two identical chromosomes. The sister chromosomes move towards opposite poles (2). Telophase: The sister chromosomes are at the two poles. The nuclear membrane is formed around each nucleus. The cell plate is formed between two nuclei in the equatorial plane (2). Common Nuclear Stains: Acetocarmine - nucleus and nucleic acids; Bismarck brown - gives yellow colour to acids; Nile blue - for live cell nucleus; Dapi (4, 6 di amino 2- phenyl indol) - fluorescent nuclear stain; Safranin - gives red colour to nucleus; Acridine orange - fluorescent cationic stain used to study cell cycle Materials required: Microscope, Nuclear Stains, Root tip of onion, Cover slips, Tissue paper, Slides, Forceps, Tooth pick, 10% HCl, Nuclear stain- 1% Safranin Procedure: Cut the already grown root tip from onion bulbs, about 3-4 mm from the tip at the base of the meristem and rinse briefly in distilled water. Place the tip in 10% HCl (V/V) for 15 minutes and rinse again in distilled water Put nuclear stain and keep it for 15 minutes and rinse in distilled water. Place the stained root tip in a drop of water on a slide and cover with a cover slip avoiding air bubbles. Gently tap the root tissue with a flat ended toothpick/ pencil to produce a squash having a homogenous cell suspension. Do not tap hard as it may break the cover slip and the cells. Remove the excess stain by placing a tissue paper over the cover slip. Examine under microscope at 400 times magnification. Spot a site (the field under focus) which has more number of dividing cells. In the field at least one cell should be present in each stage Count the number of cells in prophase, metaphase, anaphase and telophase Count the total number of cells in the field Record the observation 9  Figure 4.1 Different stages of Mitosis and interphase cells (1. Prophase; 2. Interphase; 3. Metaphase; 4. Telophase and 5 Anaphase.) Observation: Phase Number of % of cells cells Prophase Metaphase Anaphase Telophase Mitotic cells (cells in Prophase + Metaphase + Anaphase + Telophase ) Interphase cells Total number of cells (Mitotic cells + Interphase cells) 10 Formula: 1. Mitotic Index = Total Number of mitotic cells in the field/ Total number of cells in the field 2. Duration of mitosis= Mitotic Index x Duration of cell cycle (Cell cycle duration is 19 hours) 3. Index of cells in a particular phase = Number of cells in that phase/ Number of mitotic cells 4. Duration of cells in a particular phase= Duration of mitosis X Index of cells in that phase Calculation: Mitotic Index = Duration of Mitosis = Prophase Index = Duration of Prophase = Metaphase Index = Duration of Metaphase = Anaphase Index = Duration of Anaphase = Telophase Index = Duration of Telophase = Result: 11 Experiment No: 4 Micrometry Objective: To calculate the calibration constant of ocular micrometer scale To calibrate the microbial object’s measurements using micrometry. Theory: Micrometry can be defined as measurement of minute objects using a micrometer. The science of Micrometry involves taking measurements through a microscope, of objects too small to be seen with the naked eye. It involves the use of both eyepiece with ocular micrometer and stage micrometers. Ocular micrometers require calibration for accurate measurements. The reason is that there are inevitably some systematic errors in any measurements due to fluctuations in the environment when the measurements are made, the electronic and/or mechanical components in the instrument. Some instruments require a single point calibration, others require two point calibrations. Calibration is just a way of improving accuracy. This tends to cancel systematic errors. Calibration of micrometer: Micrometer slide consists of two scales, which are: Stage Micrometer scale Eyepiece or Ocular or retical micrometer scale. In the stage micrometer 1 mm divided into 100 parts. The minimum distance that can be measured using a scale is known as the least count. So in the stage micrometer the least count is 0.01 mm (1/100). The eyepiece scale also has graduations. In that 10 mm divided into 100 parts. The Calibration Constant is calculated by the formula: No. of divisions in stage micrometer between two adjacent coinciding points X Least count Calibration Contstant = Corresponding no. of divisions on ocular micrometer Actual measurement = Calibration Constant X No. of divisions the microbial object occupies on eyepiece scale Materials required: Microscope, Ocular and stage micrometers, specimen slides. 12 Procedure: Place the Calibration Slide on the stage of the microscope. View the slide under 4X and then at 10X magnifications. Then, place the ocular scale on one of the grooves of the binocular tube and observe the calibration slide. As you bring the calibration slide into focus, you will be able to see both the stage micrometer scale and the ocular scale simultaneously. Observe which divisions on the eyepiece scale exactly coincides with the corresponding division on the stage micrometer. Take 2 points of coincidence and count the number of division between the two points both in stage and ocular micrometer. Calculate the calibration constant. Multiply the calibration constant value with the parameter measured which will give the reading in micrometers (µm). Ocular micrometer Stage micrometer a and b are the two adjacent Fig 4.1 Stage & Ocular Micrometers Fig 4.2 Coinciding points Coincidence points Draw the points of coincidence Identification and measurement of Microorganisms with a Micrometer: A micrometer can be used to measure the length, Diameter and thickness of a microscopic object. 13 Calculation: No. of divisions between the points of coincidence on stage micrometer = Corresponding No. of divisions on eyepiece scale = Least count of stage micrometer = 0.01mm = 10µm Therefore, The calibration constant = (Calculated under _____________ times total magnification) Observations: Name of the Actual Shape of the Parameters S.No specimen and measurement Drawing specimen Measured magnification in µm Result: 14 Experiment No: 5 Total White Blood Cell Count Objective: To study the structure of counting chamber in hemocytometer To understand the WBC counting chamber Perform a total WBC count Hemocytometer - Counting Chamber: Hemocytometer is a device used for counting the number of cells in suspension. It is a thick glass slide with two ruled surfaces separated from each other by a transverse grove. The hemocytometer consists of two raised platforms called elevated bars. The elevated bars are provided in the slide so as to support the cover slip. Cover slips for counting chambers are thicker than those for conventional microscopy, since they must be heavy enough to overcome the surface tension of a drop of liquid. The raised edges of the hemocytometer hold the cover slip 0.1 mm off the marked grid. This gives each square a defined volume. The cover slip is placed over the counting surface prior to putting on the cell suspension. Figure 5.1 Components of a counter chamber The ruled area of the hemocytometer consists of 3types of 1mm x 1 B mm (1 mm2) squares. The area of the small squares are 0.25 x 0.25 mm (0.0625 mm2) – 16*4 W B B 0.25 x 0.20 mm (0.05 mm2) – 20*4. it is further subdivided B into 0.25 x 0.05 mm (0.0125 mm2) squares- 4*5*20- B 0.20 x 0.20 mm (0.04 mm2) - 5*5. it is further subdivided Figure 5.2 Blood Cell Counting grids (R – RBC;into W-0.05 WBC,x 0.05 mm (0.0025 mm2) squares- 5*5*25- R B-Bacteria) 15 Theory: White blood cells, or leukocytes, are cells of the immune system defending the body against both infectious disease and foreign materials. Measures of the total white blood cell count and the relative proportion of each type of white blood cell change in a characteristic way in different disease state. The white blood cells in our body are classified into two types: Granulocytes (lobed nuclei and granular cytoplasm) are divided into Neutrophils, Basophils and Eosinophil. Agranulocytes (non- lobed nuclei and non-granular cytoplasm) are divided into T lymphocytes and B lymphocytes. Deficiencies of WBCs are referred to as Leucopenia, while elevated levels of particular type of WBC in general (during a disease state) is described as Leukocytosis. Principle: The main objective of diluting a sample of blood with a suitable fluid (0.1% Methylene Blue in 2% acetic acid or 1% hydrochloric acid) is to count the cells in a small portion of this under the microscope. Methylene blue is a basic dye which stains the nuclei of WBCs and the acid lyses all the blood cells. Materials / equipment: Blood sample, Thoma diluting pipettes with white bead, Neubauer haemocytometer, 70% Alcohol, Sterile lancets, Microscope, and Cotton swabs. WBC Dilution fluid - Dissolve 100mg Methylene Blue in 100ml of 1% hydrochloric acid Procedure: Clean the ring finger with a cotton swab soaked in 70% spirit and let it air dry. Prick the finger with a sterile lancet. Draw blood till 0.5 mark in thoma dilution pipette (With the white bead). Then draw the WBC dilution fluid up to 11 mark. Homogenize the sample for 2-3 minutes. Discard the first few drops of sample present at tip as it contains only the dilution fluid. Charge the hemocytometer with the sample by gently touching the tip of the pipette on the surface of the counting platform. In contact with the edge of the cover slip small amount of fluid will flow automatically by surface tension and capillary action. Wait for the fluid and cells to settle ( wait for at least 5 min) The WBCs are counted in the four corners each having an area of 1 sq. mm. Each of these squares is further divided into 16 (4X4) small squares. The cells are counted under the low power (10X) objective. To avoid counting the cells which lie on the demarcating lines twice, consider only those cells which lie on the lower and the left borders of each square, while ignoring the cells on the right and the top borders of the squares. 16  False high results: False low results: 1. Not wiping away the blood at the tip of the pipette 1. Excessive dilution 2. Blood drawn above the mark. 2. Blood drawn below mark 3. Diluting fluid taken below the mark. 3. Insufficient loading 4. Improper mixing 4. Clumping of cells or clotting 5. Not allowing the cells to settle after application of blood W1 W2 W3 W4 Total no. Of WBC in W1 = W2= W3= W4= Calculation: Total no. of WBC’s/ / µl = DXN/V Unit of blood = 0.5 Unit of dilution fluid = 10 D: dilution factor = 10 /0.5 = 20 times V: volume of Neubauer chamber = (l X b X h) X4=0.4ul (l = 1 mm; b= 1 mm; h= 1/10mm) N: number of cells (WBC) counted = + + + = Total no. of WBC’s = 20 x N / 0.4 = N x 50 Cells / µl = X 50 = Cells / µl of blood Normal range of white blood cells is 5000 to 10,000 cells/ mm3 of human blood. Result: 17 Experiment No: 6 Counting of Red Blood Cell Objective: To understand the RBC counting chamber Perform a total RBC count Theory: Counting of RBC in a blood sample Red blood corpuscles are enucleated biconcave cells that contain hemoglobin which is essential for the carriage of oxygen and carbon dioxide. The normal RBC count is 3.6 million to 5.0 million cells per cubic mm of blood for females and 4.0 to 5.4 million cubic mm of blood for males. A measure of total number of RBCs in a cubic millimeter of blood i.e. RBC count is important in the diagnosis of diseases of red blood cells, especially anemia and polycythemia. For erythrocytes, the central most square with a volume of 0.1cu.mm is used which in turn is divided into 400 equal small squares of area 0.0025 sq. mm. After placing the cover slip, the depth of the counting chamber will be 0.1mm and the volume of each small square will be 0.00025 mm3. Figure 6.1. RBC counting area Principle: The main objective of diluting a sample of blood with a suitable isotonic fluid (Hayem's solution) is to count the cells in a small portion of this under the microscope. The most important items in the apparatus are the counting chamber and the diluting system. Materials/ Equipment required: Blood sample, Hayem’s solution, Thoma diluting pipette with red bead, Neubauer haemocytometer, 70% alcohol, sterile lancets and lancet jet, and Microscope. Hayem's solution: This is an isotonic fluid which consists of the following: 0.5gm Sodium chloride (maintains tonicity of the fluid), 0.5gm Sodium sulphate (discourages 18 clumping of the erythrocytes) and 2.5 gm Mercuric chloride (acts as a preservative) in 100 ml sterile distilled water. Procedure: Clean the ring finger with a cotton swab soaked in 70% spirit and let it air dry. Prick the finger with the help of a sterilized lancet needle. Hold the red blood cell diluting thoma pipette (With the red bead) horizontally with its tip just dipped in the drop of blood oozing out of the site and draw the blood gently up to the 0.5 mark. Remove the pipette from the site and wipe off the extra blood adhering to the tip. Draw and fill in Hayem’s fluid up to the 101 mark and homogenize the blood with the fluid thoroughly by rotating the pipette in the horizontal position. The red bead aids in mixing. Discard the first few drops of the sample as this contains only the dilution fluid without blood. Prepare the counting chamber and the cover slip by cleaning with soft tissue paper. Place the cover slip on the counting chamber Charge the chamber with the sample by gently touching the tip of the pipette on the surface of the counting platform in contact with the edge of the cover slip. A small amount of fluid will be drawn in automatically by surface tension and capillary action. There should not be any air bubbles or overflow into the central depression. Once the chamber is correctly charged, place the hemocytometer on the stage of the microscope. Focus under 4X then under 10X to check if the distribution of cells is uniform. The center square is used for RBC counting. The central square is divided into 25 squares in 5 rows of 5 squares each. Out of the 5 squares, the 4 corner squares and 1 central square are used for RBC counting. Each of these squares is further sub divided into 16 equal smaller squares. Counting of RBCs is done under the high power (40X). To avoid counting the cells which lie on the demarcating lines twice, consider only those cells which lie on the lower and the left borders of each square, while ignoring the cells on the right and the top borders of the squares. False high results: False low results: Not wiping away the blood at the tip of the pipette. Excessive dilution. Blood drawn above the mark. Blood drawn below mark. Diluting fluid taken below the mark. Insufficient loading Improper mixing. Hemolysis Not allowing the cells to settle after application Clumping of cells or clotting of Overloading. blood. 19  Observations: R5 R1 R2 R3 R4 Calculation: Total no. of RBC’s = D* N/ V D: dilution factor Unit of blood = 0.5 Total unit of volume = 100 Dilution factor = 100 / 0.5 = 200 N: number of cells (RBC) counted = V: volume of Neubauer chamber = l*b*h (l = 1/20 mm; b= 1/20 mm; h= 1/10mm) = 1/20 mm*1/20mm* 1/10mm*16*5 = 0.02mm3 Total no. of RBC’s = N*200/ 0.02 = N *10,000 Cells / µl = = Result 20 Experiment: 7 Determination of blood group by slide agglutination test Objectives: To determine A, B, O & AB Blood types To learn about the importance of Blood Groups Theory: The differences in human blood are due to the presence or absence of antigens on the Red blood cells’ surface and antibodies in the blood plasma. Individuals have different types and combinations of these molecules. Non-compatible blood mixing can lead to heme agglutination (clumping of cells due to reaction between antigens and antibodies). The clumped red cells can cause toxic reactions and have fatal consequences. Karl Landsteiner discovered that blood clumping was an immunological reaction which occurs when the receiver of a blood transfusion has antibodies against the donor blood cells. Karl Landsteiner's work made it possible to determine blood types and thus paved the way for blood transfusions to be carried out safely. For this discovery he was awarded the Nobel Prize in Physiology in 1930. There are more than 20 genetically determined blood group systems known today, but the ABO and Rh systems are the most important ones used for blood transfusions. Rh factor blood grouping system: Rh factor or D-Antigen present on the red blood cell surface indicates Rh+. Those who do not have the antigen are called Rh-. A person with Rh- blood does not have Rh antibodies naturally in the blood plasma. But can develop Rh antibodies in the blood plasma if he/she receives blood from an Rh+ person. Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh- blood without any problems. Rh incompatibility arises when a Rh negative person receives blood from an Rh positive person. Group Antigen Antibody Can donate blood to Can receive blood from AB A&B - AB AB, A, B, O A A B A & AB A&O B B A B & AB B&O O - A&B AB, A, B & O O Table 7.1 Antigen, antibody, donor and recipient of ABO blood grouping Principle: A blood type is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells (RBCs). Several of these red blood cell surface antigens that stem from one allele (or very closely linked genes), collectively form a blood group system. 21 Blood group determination is compulsory during Blood transfusion. For a successful blood transfusion ABO and Rh blood groups must be compatible between the donor and the recipient. Materials required: Blood sample, Glass Slides, Antiserum A, B and Anti-D, Cotton swabs, 70% alcohol, Sterile lancets and Toothpicks Procedure: Prick the finger with a sterile lancet and place a drop each in each cavity of the slide. To the first drop of blood, add anti serum A. To the second drop of blood, add anti serum B. To the third drop of blood, add anti serum D. Observe the agglutination reaction to conform the blood group and Rh factor. Observations: Draw the agglutination reaction of your blood with respective antiserum Anti – A Anti – B Anti – D Result: The given blood sample is of ________________________ blood group. Distribution of blood type in the class Blood type No. of students Percentage A +ve A -ve B +ve B -ve AB +ve AB -ve O +ve O -ve Total No. of students ------- 22  Draw the agglutination Donate Receive blood reaction Group Antigen Antibody blood to from Anti A Anti B Anti D AB- A, B - AB-, AB+ AB-, A-, B-, O- A-,A+, AB-, A- A B A-, O- AB+ B-, B+, AB-, B- B A B-, O- AB+ AB-, A-, B-, O- - A, B O-, AB+, A+, O- B+, O+ AB-, A-, B-, O-, AB+ A, B, D - AB+ AB+, A+, B+, O+ A+ A, D B A+, AB+ A-, O-, A+, O+ B+ B, D A B+, AB+ B-, O-, B+, O+ AB+, A+, O+ D A, B O+, O- B+, O+ No agglutination Agglutination 23 Experiment: 8 Quantitative Estimation of chlorophyll Objective: To understand the working principle of a UV- Visible Spectrophotometer To extract the chlorophyll from the leaf To find out the presence of different types of chlorophyll in the filtrate Theory: Spectroscopy is the measurement and analysis of electromagnetic radiation absorbed, scattered or emitted by atoms, molecules, or other chemical species. The absorption and emission of electromagnetic radiation is associated with changes in energy states of the species. The UV and Visible spectrum are from 200 nm to 400 nm and 400 nm to 800 nm respectively. A Spectrophotometer uses ultra-violet and visible electromagnetic radiation to energetically promote valence electrons in a molecule to an excited energy state. The spectrophotometer is an instrument that makes possible a quantitative measurement of light passing through a clear solution. Instrumentation: The instruments are arranged so that liquid in a quartz cuvette can be placed between the spectrometer beam and the photometer. The amount of light passing through the tube is measured by the photometer. The photometer delivers a voltage signal to a display device, normally a galvanometer. The signal changes as the amount of light absorbed by the liquid changes. Fig. 8.1. The essential parts of spectrophotometer 24 The essential parts of a spectrophotometer include a light source, a wavelength selection device such as a monochromator or filter, a sample chamber, a light detector, and a readout device. A cuvette is a kind of laboratory glassware, usually a small tube of square cross section, sealed at one end, made of optical grade quartz and designed to hold samples for spectroscopic experiments. The best cuvettes are as clear as possible, without impurities that might affect a spectroscopic reading. A cuvette may be open to the atmosphere on top or have a cap to seal it shut. Cuvettes will be clear only on opposite sides, so that they pass a single beam of light through that pair of sides; often the unclear sides have ridges or are rough to allow easy handling. The capacity of the cuvette used in our laboratory is 4ml. The cuvette should be cleaned with distilled water and soft tissue paper. Principle: A spectrophotometer is an important analytical instrument used for measuring the intensity of light of a definite wavelength transmitted by a substance or a solution, thus providing a measure of the amount of material in the solution absorbing the light. Beer- Lambert’s law states that ‘for a parallel beam of monochromatic radiation passing through homogeneous solutions of equal path length the absorbance is proportional to the concentration’. The reason why it prefers to express the law with this equation is because absorbance is directly proportional to the parameters, as long as the law is obeyed. A=εlc Where A is absorbance (no units, since A = log10 P0 / P ) ε is the molar absorptivity with units of L mol-1 cm-1 l is the path length of the sample - that is, the path length of the cuvette in which the sample is contained. We will express this measurement in centimeters. c is the concentration of the compound in solution, expressed in mol L-1 Theory: Chlorophyll absorbs sunlight and uses its energy to synthesis carbohydrates from CO2 and water. Chlorophyll traps power hence called a photoreceptor. It is found in the chloroplasts. The basic structure of a chlorophyll molecule is a porphyrin ring, co-ordinated to a central magnesium atom. There are 2 main types of chlorophyll, named a and b. They differ in the composition of a side chain (R) (in a it is -CH3, in b it is CHO). Both types of chlorophylls are very effective photoreceptors because they contain a network of alternating single and double bonds, and the orbital can delocalize stabilizing the structure. Such delocalized polyenes have very strong absorption bands in the visible regions of the spectrum, allowing the plant to absorb the energy from sunlight. 25 Principle: The different side groups in the 2 chlorophylls tune the absorption spectrum to slightly different wavelengths. Thus, these two kinds of chlorophyll complement each other in absorbing sunlight. Chlorophyll is registered as a food additive (colorant), and its E number is E140. Chlorophyll supplies the much-needed micronutrient magnesium. It also enhances the immune function. It is a powerful antioxidant, anti-carcinogenic and helps to minimize the side effects of chemotherapy. It supports the body’s detoxification processes, may reduce risk of kidney stones by preventing calcium oxalate build-up and also helps to eliminate body odor. Fig. 8.2 Structure of chlorophyll Materials required: Mortar and pestle, fresh green leaves of different plants, cuvette, centrifuge, centrifuge tube, tissue paper, spectrophotometer Procedure: Select tender green leaves. Wash them with water and dry them on a piece of blotting paper and weigh 100 mg of leaf. Chop and then grind it nicely using mortar and pestle. Add 5 ml of 80 % acetone. Mix the paste well. Centrifuge the mixture at 5000 rpm for 5 min. Turn on UV-Vis Spectrophotometer at least 20 min. before the experiment for stabilization. Use 80% acetone as blank. Measure the absorption at 663 nm and 646 nm. The formulae of Porra (2002) are based on the absorbance maxima of each pigment and are dependent on the solvent (80% acetone) used. Chlorophyll a (µg/ml) = 12.25 (A663) – 2.55 (A646) Chlorophyll b (µg/ml) = 20.31 (A646) – 4.91 (A663) Total chlorophyll (µg/ml) = 17.76 (A646) + 7.34 (A663) 26 Calculation: Chlorophyll a (µg/ml) Chlorophyll b (µg/ml) Total chlorophyll (µg/ml) Result: Common name Amount Of Total S.No. Name of the plant A663 A646 chlorophyll (µg/ml) chlorophyll a b (µg/ml) 1 Azadirachta indica Neem 2 Ocimum tenuiflorum Tulsi 3 Jasminum sambac Jasmine 4 Catharanthus rosea Periwinkle 5 Ocimum basilicum Basil 6 Phyllanthus niruri Stonebreaker 27 Experiment No: 9 Quantitative Estimation of Proteins by Biuret Method (Spectrophotometry) Objective: To measure the unknown concentration of protein. Estimation of Protein by Biuret Method: The underlying principle of protein estimation by Biuret method is that, under alkaline conditions substances containing two or more peptide bonds form a purple-colored complex with copper salts. The copper (II) ion when coordinating with peptide bonds, changes the colour of the solution from blue to purple. This colour change depends on the number of peptide bonds present in the sample solution. So, the intensity of the colour increases with concentration of protein in the sample. The unknown protein concentration will be estimated based on the absorbance value of the known protein concentration. This method is the most preferred way of estimating proteins in a given solution because of the following reasons: The method involves only a single incubation of 20 min so less time consuming The end product is a coloured liquid substance. There are very few interfering agents e.g. ammonium salts. More reliable method than titration and gravimetric method. More accurate. Up to 10-7 molar concentration. Fig. 9.1 Structure of the Purple colored coordination complex Materials required: Protein sample- Bovine Serum Albumin, Biuret reagent, UV-Visible Spectrophotometer, 7 test tubes, test tube stand, 1ml pipette, 10ml pipette, Beakers, Cuvette, Soft Tissue paper and Distilled water. BSA Stock solution: 1g of Bovine Serum Albumin (BSA) is dissolved in 100ml of distilled water. Biuret reagent: 3 g of Copper Sulphate and 9 g of Sodium-Potassium Tartrate in 500 ml of 0.2 N Sodium hydroxide (4 g of sodium hydroxide dissolved in 500 ml distilled water). 28 Procedure: Stabilize the instrument by switching it ON before 20 min. Set the absorbance at 540nm. Label the test tubes 1, 2,3,4,5 and B for Blank. Add about 0.1, 0.2, 0.3, 0.4 and 0.5 ml of BSA in the labelled test tubes. Add distilled water to each of the 5 test tubes to make the volume to 1ml. Add 2ml of biuret regent to all the tubes. Prepare a blank with 1 ml of distilled water and 2 ml of biuret reagent. No BSA to be added. In the 7th test tube, you will be provided with 1ml protein sample of unknown concentration. To this test tube, add 2ml of biuret reagent only. Auto-zero the spectrophotometer using blank solution Calibrate the absorbance value with increasing concentrations of the protein standards. Read the absorbance for the unknown protein sample. Record your result. Draw a standard graph for protein estimation by Biuret method. Plot Concentration of BSA (mg/ml) on the X- axis vs. absorbance A540 on the Y- axis. Find out the concentration of protein present in the unknown using the standard graph by extrapolating the absorbance value to the concentration. Observations: BSA D/W Conc. Of BSA Biuret reagent Test tube Absorbance (ml) (ml) (mg/ml) (ml) 1 0.1 0.9 1 2.0 2 0.2 0.8 2 2.0 3 0.3 0.7 3 2.0 4 0.4 0.6 4 2.0 5 0.5 0.5 5 2.0 B (Blank) 0.0 1.0 - 2.0 0.000 U1 (Unknown) ? ? ? 2.0 U2 (Unknown) ? ? ? 2.0 U3 (Unknown) ? ? ? 2.0 U4 (Unknown) ? ? ? 2.0 29 U5 (Unknown) ? ? ? 2.0 Result: The concentration of protein present in the unknown sample was____________________ 30 Experiment: 10 Quantitative Estimation of Glucose by Dinitrosalicylic Acid Method (Spectrophotometry) Objective: To quantitatively estimate the concentration of glucose present in the given sample by Dinitrosalicylic acid method Theory: Glucose, a monosaccharide (or simple sugar), is an important carbohydrate in biology. The living cell uses it as a source of energy and metabolic intermediate. Glucose is one of the main products of photosynthesis and starts cellular respiration in both prokaryotes and eukaryotes. Glucose (C6H12O6) contains six carbon atoms. This method tests for the presence of free carbonyl group (C=O), the so- called reducing sugars. This involves the oxidation of the aldehyde functional group present in glucose and the ketone functional group in fructose. Simultaneously 3, 5-dinitrosalicylic acid (DNS) is reduced to 3-amino, 5-nitrosalicylic acid under alkaline conditions. Principle: The exact reactions of glucose are as follows: Oxidation Aldehyde group (glucose) Carboxyl group (gluconic acid) Reduction 3, 5-Dinitrosalicylic acid 3-Amino 5-nitrosalicylic acid The dissolved oxygen can interfere with glucose oxidation. Sodium sulfite, which itself is not necessary for the color reaction, is added in the reagent to absorb the dissolved oxygen. The above reaction scheme shows that one mole of sugar will react with one mole of 3, 5- dinitrosalicylic acid. Chemicals required: Dinitrosalicylic Acid Reagent Solution 1% (prepare freshly)- Dissolve 10 g of Dinitrosalicylic acid, 0.5 g of Sodium sulfite and 10 g of Sodium hydroxide in 1 liter of distilled water. Potassium sodium tartrate solution 40% (Rochelle solution) - 40g of Potassium sodium tartrate dissolved in 100ml of distilled water. Glucose standard 1% - 1g glucose dissolved in 100ml of distilled water. Materials required: Test tubes, Spectrophotometer, Cuvette, Tissue paper, Wash bottles, Pipettes-1ml and 10ml 31 Procedure: Stabilize the spectrophotometer at 575 nm, by switching it ON at least 20 min prior. Take 7 test tubes and mark them as B, S1, S2, S3, S4, S5 and T (B-blank, S-standard and T- test solution). Fill the test tubes with reagents as follows: Potassium Conc. of Test Glucose D/W DNSA sodium S.No glucose tube (ml) (ml) (ml) tartrate Absorbance (mg/ml) (ml) Heat the mixture at 80º C for 5-15 minutes to develop the reddish brown color 1 0.1 0.9 1 0.5 S1 1 Cool the solutions to room temperature in a cold water 0.2 0.8 1 0.5 2 S2 2 0.3 0.7 1 0.5 3 S3 3 Read the absorbance at 575 nm. 0.4 0.6 1 0.5 4 S4 4 0.5 0.5 1 0.5 5 S5 5 Blank 1ml 1 0.5 0.000 6 B 0 Not Test 7 T1 - 1 0.5 known solution Not Test 8 T2 - 1 0.5 known solution Not Test 9 T3 - 1 0.5 known solution Not Test 10 T4 - 1 0.5 known solution Not Test 11 T5 - 1 0.5 known solution In S1, S2, S3 S4 and S5 test tubes add 0.1ml, 0.2ml, 0.3ml, 0.4ml and 0.5ml of standard glucose solution. Make up the volume in each tube to 1ml by adding distilled water. 32  In test tube B, add 1ml of distilled water. In test tube T, add 1ml of test solution which is already given. Add 1ml of DNSA reagent in each test tube. Heat the mixture at 80º C for 5-15 minutes to develop the reddish-brown color. Cool the solutions to room temperature in cold water. Add 0.5 ml of 40% Potassium sodium tartrate (Rochelle salt) solution to stabilize the color. Read the absorbance at 575 nm wavelength light. Use blank solution to adjust zero setting on the instrument. Prepare standard graph. Take concentration of glucose (mg/ml) in X axis and Absorbance in Y axis. This graph should pass through origin. It is a linear graph. Match the Absorbance of the test sample with the corresponding concentration value in x axis. That is the concentration of glucose (mg/ml) present in the given sample. Result: 33 Experiment: 11 Qualitative analysis of different plant pigments Objective: To prepare a leaf extract and perform the qualitative analysis of different plant pigments by paper chromatograpgy. Principle - Plant pigments are macromolecules produced by the plant, and these pigments absorb specified wavelengths of visible light to provide the energy required for photosynthesis. Chlorophyll is necessary for photosynthesis, but accessory pigments collect and transfer energy to chlorophyll. Although pigments absorb light, the wavelengths of light that are not absorbed by the plant pigments are reflected back to the eye. The reflected wavelengths are the colors we see in observing the plant. (Example: green pigments reflect green light) Plants contain different pigments, and some of the pigments observed include: chlorophylls (greens) carotenoids (yellow, orange red) anthocyanins (red to blue, depending on pH) xanthophylls (red or yellow) Paper chromatography involves the use of mobile phase and stationery phase. The process of chromatography separates molecules on the stationery phase because of the different solubilities of the molecules in the mobile phase. In paper chromatography, paper marked with an unknown, such as plant extract, is placed in a developing chamber with the mobile phase. The mobile phase carries the dissolved pigments as it moves up the paper. The pigments are carried at different rates because they are not equally soluble. A pigment that is the most soluble will travel the greatest distance and a pigment that is less soluble will move a shorter distance. The distance the pigment travels is unique for that pigment in set conditions and is used to identify the pigment. Distance traveled by the substance Rf value = Distance traveled by the solvent Materials and Chemicals required: Chromatography chamber, chromatography paper, scale, pencil, scissors, capillary tube, mortar pestle, leaves, acetone Mobile phase: Petroleum ether : acetone (9:1) 34 Procedure Cut out small pieces of leaves of using scissors. Add them to the mortar. Accurately measure 5ml acetone using a measuring cylinder and add it into the mortar. With the help of mortar and pestle, grind the leaves into a smooth paste. Horizontally trace a line with a scale and a pencil that is 2 to 3 cm apart from the bottom. Using a capillary tube, add 1 drop of the leaf extract in the midsection of the line. Let the drop dry. Repeat the same process of adding a drop and allowing it to dry for 3-4 times. Suspend the strip in the chamber. The loading spot remains about 1 cm above the level of the solvent. Let the chamber remain uninterrupted for a while. We can notice that the solvent passes along the paper scattering various pigments of the blend to different distances. Once the solvent reaches 3/4th of the strip, carefully take the strip off. Allow the strip to dry. Calculate the Rf values of bands obtained for different pigments. Standard Rf values range – Chlolrophyll a (yellow colour) – Chlorophyll b (green) – Anthocyanins (red to blue) – Xanthophylls (red to yellow)- Calculation- Distance travelled by mobile phase = Distance travelled by pigment 1 ( ) = Distance travelled by pigment 2 ( ) = Distance travelled by pigment 3 ( ) = Distance travelled by pigment 4 ( ) = Rf values of Chlolrophyll a = Rf values of Chlorophyll b = Rf values of Anthocyanins = Rf values of xanthophylls = 35 Result and discussion Calculation – Experiment: 12 One Dimensional Chromatography – Separation of amino acid Objective: To study the principle behind chromatography To perform radial / one dimensional ascending paper chromatography To find out the different amino acids present in the sample To find out the Rf values of the aminoacids Principle: Paper chromatography is a partition chromatography, where mobile phase is a liquid and the stationary phase is coated on a solid support. Paper chromatography is a technique for separating and identifying complex mixtures of similar compounds. The sample is absorbed onto the paper. The flow of mobile phase is by capillary action as a result of the differential adsorption of the solvent molecules to the paper. As the solvent rises through the paper it meets and dissolves the sample mixture, which will then travel up the paper with the solvent. Different compounds in the sample mixture travel at different rates due to differences in solubility, and differences in their attraction to the paper. The more soluble the component the further it goes. Rƒ value is defined as the ratio of the distance traveled by the substance to the distance traveled by the solvent. If Rƒ value of a solution is zero, the solute remains in the origin. If Rƒ value = 1 then the solute has no affinity for the stationary phase and travels with the solvent front. Fig. 12.1. Ascending paper chromatography Fig. 12.2. Chromatogram 36 The amino group of α - amino acid undergoes oxidation with one molecule of ninhydrin to form hydrintantin, carbon-di-oxide, and aldehyde. A second equivalent of ninhydrin reacts with the reduced ninhydrin (hydrintantin) and ammonia to produce purple color di keto hydrinta (Ruheman’s blue) Reaction between amino acid and ninhydrin Materials and Chemicals required: Chromatography chamber, chromatography paper, scale, pencil, scissors, capillary tube, sprayer, amino acid solution, butanol, acetic acid, distilled water, ninhydrin, acetone and acetate buffer 1 % Amino acid solution: 10 mg of amino acid dissolved in 1ml of 0.1 N Hydrochloric acid Mobile phase: N - Butanol, acetic acid and water (4:1:1). 8 ml of N - butanol, 2 ml acetic acid and 2 ml distilled water mixed together. 1% Ninhydrin (2, 2-Dihydroxyindane-1, 3-Dione): 2 grams of ninhydrin dissolved in 25 ml of acetone. Then 25 ml of 0.2N acetate buffer (pH 5.5) added. Store it in an amber color bottle. Procedure: Take 6 ml of the mobile phase in the chromatography chamber. Close the lid and wait 10 min for saturation of the chamber with vapors. Take whatmann paper of 6*4 cm2 size. On that, draw a line with a pencil, 1cm from the bottom for marking origin. Mark three spots with proper placing on the line, on which amino acid samples and the unknown mixture are to be spotted. Label them A1, A2 and U. Using a capillary tube, spot the amino acids and the given unknown mixture on the spots. Air dry the spot and repeat the procedure 3 more times. Keep the paper inside the chromatography chamber in such a way that origin line should be above the solvent level. The solvent will slowly rise through the paper. When solvent reaches 4/5 of the paper remove it from the chamber. Immediately mark the solvent front using pencil. Dry the paper. Spray ninhydrin on the paper. 37  Wait for the color development. (Or keep it in the oven at 50C to fasten drying). Mark the center of the coloured spot. Measure the distance between the origin and solvent front (distance traveled by the solvent). Measure the distance traveled by the amino acid (from the origin line to the center of the spot). Calculate the Rf values for different amino acids. Distance traveled by the substance Rf value = Distance traveled by the solvent Stick the chromatogram paper / draw the chromatogram Calculation: Distance travelled by mobile phase = Distance travelled by amino acid 1 ( ) = Distance travelled by amino acid 2 ( ) = Rf values of amino acid 1 = Rf values of amino acid 2 = Rf values of spot 1 = Rf values of spot 2 = Distance travelled by spot 1 in unknown mixture = Distance travelled by spot 2 in unknown mixture = Result: 38 Experiment No: 12 Isolation of Eukaryotic genomic DNA with high yield and purity Objectives: To become familiar with the physical properties of DNA by isolating it from living tissue To learn the purpose of each step in the isolation procedure as it relates to the physical and biochemical characteristics of the genetic material Principle: Before DNA can be released from the nuclei of the banana tissues, the cell walls, plasma membranes, and the nuclear material must first be broken down. This step can be accomplished by homogenizing the tissues. The detergent solution causes the cell membrane to break down and emulsifies the lipids and denatures the proteins of the cell by disrupting the polar interactions that hold the cell membrane together. The DNA can then be separated from the chromosomal proteins by the chemical components of the homogenizing medium which will cause the proteins to precipitate out of solution. Requirements: 60⁰ C water bath, ice bucket, ice cold 95 % ethanol, Banana, sodium lauryl sulfate, sodium chloride, sodium citrate, ethylene diamine tetra acetic acid, deionized water, cuvette, centrifuge tube, spectrophotometer, tissue paper, waste collection beaker, wash bottle Preparation of homogenization medium: Sodium lauryl sulfate (SDS or SLS) 50.000 g (emulsifies lipds), sodium chloride 8.770 g (precipitates proteins by salting out), sodium citrate 4.410 g (stabilizes DNA by attaching to phosphate backbone), ethylene diamine tetra acetic acid (EDTA) 0.292 g (chelating agent – binds to Mg2+ and block the activity of DNAse) dissolved in 1L deionized water. Protocol: A. Homogenization: 39 1. Wearing plastic gloves, cut banana into cubes of 3 mm size. Plastic gloves prevent DNAse enzymes on your hands from degrading the DNA into small fragments. 2. Weigh out 5 g of banana. Transfer the banana pieces to a zip lock bag and smash. Transfer all of the mashed material to a glass beaker. 3. Add 10 ml of homogenizing medium and incubate the beaker in a 60°C water bath for 15 min. This heat treatment softens the banana tissue and allows penetration of the homogenization solution. It also denatures many enzymes that could interfere with the isolation procedure. 4. Quickly cool your preparation to 15-20°C in an ice bath for 6 min. It prevents the denaturation of DNA due to overheating. 7. Filter the homogenate through cheesecloth into a 50 ml beaker, taking care to leave the foam behind. B. Precipitation of DNA: The homogenate contains only DNA and the components of the homogenizing medium. When ice- cold ethanol is added to the homogenate, all the components of the homogenizing medium stay in solution-except DNA. DNA precipitates as a thick, stringy, white mass that may be spooled out by winding it on a glass rod. If the DNA has been damaged, it will still precipitate. 8. Place your beaker with its filtered homogenate into an ice bath. Let it cool until it reaches 10-15°C (about 10-15 minutes). 9. Measure out 10 ml of ice-cold ethanol into a cold graduated cylinder. Slowly add the ethanol down the side of your beaker until the white, stringy DNA precipitate appears. 10. Spool out or wind up the stringy DNA onto a glass rod by rotating the rod in one direction. 11. Gently ease DNA off the end of the glass rod into a centrifuge tube filled with 1 ml water. C. Quantitative and qualitative estimation: Find the optical density (OD) in spectrophotometer at 260 and 280 nm. Dilute the sample if necessary. Use water as blank. For pure DNA, the ratio of the optical densities OD260/OD280 is 1.7 to 2. About 50 µg/ml of DNA has OD of 1. 12. The concentration of DNA (µg/ml) is calculated by the formula: Concentration of DNA (µg/ml) = OD260 * 50 µg/ml DNA yield (µg) = Concentration of DNA * total sample volume (ml) 40 OD260 = OD280 = Purity of DNA = Concentration of DNA (µg/ml) = DNA yield (µg) = Bibliography 1. Cambell, N. and Reece, J. 2004. Biology, 2002. Biology (6th Edition), Pearson Education, Inc. 2. Enger, E.D. and Ross, F.C 2000. Concepts in Biology (11th Edition), McGraw Hill Publishers, USA. 3. Green.N.P.O. and Stout.G.W. 1980. Biological science (3rd Edition), Cambridge University Press, Great Britain. 4. L. Stryer Biochemistry, W.H. Freeman and Co, San Francisco, 1975. 5. Miller, G.L., Use of dinitrosalicylic acid reagent for determination of reducing sugar, Anal. Chem., 31, 426, 1959. 6. Reid. M.E. and Lomas-Francis. C. 1997. The blood group antigen facts book, Academic Press, London / San Diego. 7. Streitweiser and Heathcock Introduction to Organic Chemistry, MacMillan, New York, 1981. 8. Taylor, M.R. 2000. Student study guide for biology. 9. www.answers.com/topic/micrometry 10. www.biology.about.com 11. www.biology.arizona.edu 12. www.biosci.ohiou.edu 13. www.cellsalive.com/mitosis 14. www.changbioscience.com 15. www.chm.davidson.edu 16. www.eng.umd.edu/~nsw 17. www.hyclone.com 18. www.labtestsonline.org 19. www.licor.com/6400XT 20. www.microbiologyprocedure.com 21. www.microscopestation.com 22. www.molecularstation.com/protein 23. www.rnceus.com 41  24. www.search.com 25. www.sci.sdsu.edu/multimedia/mitosis 26. www.scienceisnthard.com/Droop.jpg) 27. www.spectrophotometer.com 28. www.arlis.org/docs/vol1/52386062/large_fish_diagram.jpg&ir 30. faculty.washington.edu/ktorii/stomata.html 31. http://www.biologyjunction.com/leaf_stomata_lab.htm 32. http://www.topendsports.com/testing/bodycomp.htm Appendix Salient features of the specimens Microscopic view Schematic diagram Frog Intestine (CS): The principal organ of digestion and absorption of digested food. The small intestine contains small finger-like projections of tissue called villi which increase the surface area of the intestine. It contains specialized cells that transport substances into the bloodstream. Rhizobium: is a common soil bacterium. It is not toxic to humans, plants, or animals. It is one of the most beneficial bacteria to agriculture. They play a very important role in agriculture by inducing nitrogen-fixing nodules on the roots of legumes such as peas, beans, clover etc. This symbiosis can relieve the requirements for added nitrogenous fertilizer during the growth of leguminous crops. They cannot independently fix nitrogen. Daphnia is small aquatic crustaceans commonly called water fleas. The most prominent features are the compound eyes, the second antennae, and a pair of abdominal setae. The carapace is translucent. The life cycle based on "cyclical parthenogenesis". A popular live food in tropical and marine fish keeping. Radiolaria: are holoplanktonic protozoa and form part of the zooplankton, they are mostly non-motile they may be solitary or colonial. It is composed of a skeleton made up of hollow silica bars joined by organic material. The degree of preservation of Radiolaria depends on the robustness of the skeleton, depositional and burial conditions and diagenesis. They are wholly marine. 42 Aspergillus: Species of Aspergillus are important medically and commercially. Some species can cause infection in humans and other animals. Aspergillus is a genus consisting of a few hundred mold species found in various climates worldwide. Plant cells are eukaryotic cells. Its distinctive features include cell wall, plastids and vacuole. A large central vacuole, a water-filled volume enclosed by a membrane known as the tonoplast that maintains the cell's turgor controls movement. Cell wall composed of cellulose and hemicelluloses, pectin and lignin Paramecium: is a genus of unicellular ciliates. Paramecia are widespread in freshwater, brackish, and marine environments. The body of the cell is enclosed by a stiff but elastic membrane (pellicle), uniformly covered with simple cilia, hair like organelles which act like tiny oars to move the organism in one direction. Mammalian skin cell: is the soft outer tissue covering In mammals, the skin is an organ of the integumentary system made up of multiple layers of ectodermal tissue. All mammals have some hair on their skin It is the first line of defense from external factors. Hair follicle is a dynamic organ found in mammalian skin. The skin consists of three layers of tissue: the epidermis, dermis, and subcutis Taenia solium: is commonly called pork Tapeworm, an intestinal zoonotic endoparasite. The adult worm is found in humans and has a flat, ribbon-like body, which is white in color and measures 2 to 3 m in length. Its distinct head, the scolex, contains suckers and hooks as an organ of attachment. The main body consists of a chain of segments known as proglottids. It completes its life cycle in humans as the definitive host and pigs as intermediate host. 43 Trypanosoma is unicellular parasitic flagellate protozoa. It causes Human African trypanosomiasis, also known as sleeping sickness, is a vector-borne parasitic disease. It was pleomorphic ranging from 27 to 50 μm in length in blood smears. 44