Biology Practical on Cells - Lab Report Example

BIOLOGY PRACTICALS PART Practical Introduction to microscopy and measurement of cellular structuresusing an eye-piece graticule The microscope is a widely used research tool in biology. There are different types of microscope which differ in their magnifying capabilities, and in the source of magnification. In this cell biology experiment, the light microscope is used. The type of microscope used in this experiment is specifically the bright field microscope. It uses light from an incandescent source. The light is aimed to the lens below the stage called the condenser; it then passes through the specimen, and the objective lens that is above it. The light then passes through to the observer through the eye-piece lens. The final magnification of the details of specimen under observation is the product of both the eye-piece, and objective lenses (Caprette, 2005). For one to be able to note the details of the specimen, staining is necessary. Different substances will absorb light differently, thus transmit light in different proportions. The difference in light transmission through the different parts due to staining makes the isolation of details possible. The extent to which a microscope differentiates fine details in the specimen is called the resolution. In the light microscope, the resolution can be obtained as below: Resolution,  In this case: λ is the wavelength of the light being used to illuminate the specimen. n = refractive index of the medium which surrounds the specimen, and is between the specimen, and objective lens α is the angle which the light makes as it is coming through the objective lens from the specimen halved, 0.61 is the constant that describes the degree to which images can overlap and still be possible for the observer to tell them apart The resolution becomes smaller as the resolution of the microscope improves. The value ‘n’ is increased by using a drop of oil. The drop is placed between the objective lens and the specimen. With this type of microscope, the resolution ranges between 0.2µm and 200nm. This is because the λ of light is 450nm (Karp, 2009, p. 718). This experiment aims to measure the size of a cell, and determine its structure, using an eyepiece graticule. 1mm = 100 eyepiece units = 100 graticules 1 eyepiece unit = 1/100mm = 0.01µm When using ×10 objective: When using ×40 objective: Question: If an object has a diameter of 17 graticule units with the ×40 objective, what is the actual diameter in microns? Question: If an objective is 28 microns in length, how many eyepiece graticule units will its length be when viewed with the ×10 objective? Magnification: ×40 Magnification: ×10 Labelling: Size of cellular structure in microns: Practical 2: Haematoxylin and Eosin staining of cheek cells The intention of this part of the practical is to obtain cells from the cheek, and observe the cell structure under a light microscope. The staining was done using eosin, and haematoxylin dyes. This procedure was done while wearing a glove since it involved coming into contact with human fluid, specifically saliva. Wearing gloves was a safety measure that is a procedural requirement when dealing with specimen from animals. Procedure: 1. A cheek scrape was collected using a wooden spatula and a smear was prepared. The specimen was allowed to air-dry, and then the smear was fixed using 95% ethanol for a minute to two minutes. 2. The specimen was transferred to 50% alcohol for a few seconds. It was later washed in distilled water. 3. The smear was then stained in Harris’s haematoxylin for a period of ten minutes. 4. Then distilled water was used to wash off the excess stain. 5. The slide was washed briefly for1-2 seconds in acid alcohol in order to destain the cytoplasm. Acid alcohol consists of a mixture of 1% HCl and 70% ethanol. 6. The slide was again washed with water. 7. The cytoplasm of the specimen was counterstained in 1% eosin for 3minutes. 8. The specimen was then mounted in DPX resin. This step was carried out in the fume chamber. Results: Observation at ×10 10×10=100µm 100µm 7×10=70µm 70µm Observation at ×40 Practical 3: Observation of chromosomes undergoing mitosis in onion root tips Procedure: 1. 2mm of root tip was cut from onion. 2. The root tips were fixed in Clarke’s fluid for 10 minutes in a Bijou tube. 3. The root tip was removed from Clarke’s fluid and placed in a Petri dish with distilled water. It was washed for two minutes. 4. The root tip was placed in 1mlof 1M HCl in an Eppendorf tube. It was then incubated for five minutes at 60 0C. 5. The contents of the tube were poured into a Petri dish. The root tips were carefully picked out using a forceps and placed in an Eppendorf tube which contained aceto-orcein. The mixture was left in the dark for 10 minutes. 6. The root tip was placed on a slide in a drop of 45% acetic acid, and covered with a cover slip. 7. The softened root tip, which by now had softened, was squashed. It was then stained by lightly tapping lightly on the cover slip using a pencil. On doing this the root tip spread out as a pink mass. 8. The specimen was viewed using ×400 light microscopy Results: 1/25 = 0.038 mm At ×10 × 1000 = 38µm 1 Epm = 38 µm Observation at ×40 Nucleus 1- .4×40 =1.6 Nucleus 2- .4×40=1.6 Nucleus 3- .3×40=1.2 Cell 1- .26×40 = Cell 2 - .19×40 = Cell 3- .28×40 = PART 2: WRITE-UP Practical 4: Measurement of enzyme activity in different sub cellular fraction in liver (Over 2 weeks) Summary: Enzymes are proteins that are distributed in all cellular compartments. In definition, enzymes are proteins which catalyze biological processes. This means that all enzymes are protein in nature. Enzymes are found in different parts of the cell. Those that are found as part of cell membranes exist in the inner side of the membrane. The only exception to this is some digestive enzymes which include disaccharidases, isomaltase, and specific peptidases. Enzymes that membrane-associated are found in all cell organelles. They are, however, in highest concentration in the mitochondria (Gropper, Smith, & Groff, 2008, p. 15). Different enzymes in the cell interact in such a way that the product of one enzymatic reaction forms the substrate that is required by another enzyme in the cell. This being the case, enzymatic reactions tends to follow a sequential order. Reactions that follow a certain order that is metabolically beneficial are known as a multienzyme system. In the cell, multienzyeme reactions occur in a specific manner. That is, there is one enzymatic reaction that will take place in the cytoplasm, another in the mitochondria, yet another in the nucleolus. This experiment aims to fractionate liver cells, and determine the cellular location of some important metabolic enzymes. Introduction The liver is a very important organ found in vertebrates. It is also found in some other animals. This organ serves a variety of functions including detoxification, metabolism, and protein synthesis. The major role of this organ is metabolism. The metabolic functions of the liver aid in control protein, lipid and carbohydrate are synthesis and utilization (Bowen, 2001). In humans, the liver is found in the abdominal-pelvic region, below the diaphragm. It has tissues that are highly specialized to play the different roles it serves. These tissues control the different biochemical reactions which are needed for normal vital processes. The liver has an impressive number of enzymes. By saying that metabolism is the main role of the liver, it means that most of the processes that occur therein involve energy. Cells of the liver are known as hepatocytes. The cells pf the liver, hepatocytes, play a critical role in the synthesis of molecules which find use elsewhere in the body. These molecules that have been synthesized are used up in support of homeostasis, regulating energy balances, and conversion of molecules from one type to another. The major metabolic functions of the liver can be divided into three main categories; carbohydrate, fat, and protein metabolism. Metabolic pathways are by definition sequential, enzyme-mediated reactions that are driven by energy (Starr, Evers, & Starr, 2010, p. 76). Carbohydrate metabolism is critical for all animals. It serves to maintain the concentrations of glucose in blood within a normal, narrow range. This regulation is done over both long and short term basis. There are two main processes that help in serving this function. The first is glucogenesis which is the synthesis of glucose from amino acids and other carbohydrates that are non-hexose. The other process can occur in two ways depending on glucose concentrations in the blood. When there is excess glucose in the blood, such as after meals, glycogenesis occurs. This enables the glucose to be quickly taken up by the liver. When the glucose concentration in the blood is low, such as during starvation, glycogenolysis takes place. In this process, pathways are activated in the body that facilitate glycogen to be converted to glucose. These pathways also enable the transportation of this glucose to the blood, for transportation to all the other tissues (Lanham-New, Macdonald, & Roche, 2011, p. 448). The role of the liver in lipid metabolism is carried out in several ways. The first is oxidation of triglycerides in order to generate energy. This is also the place where lipoproteins are synthesized. Conversion of excess proteins and carbohydrates into triglycerides and fatty acids also takes place in the liver. These, among many other processes that enable fat metabolism, include specific enzymes. This translates to many different enymes for the different pathways. Different parts of the liver will, therefore, have differing concentrations of varying enzymes. Protein metabolism takes place in the liver in many ways. One crucial way in which this occurs is the synthesis of urea so as to remove ammonia from the body. Non-essential amino acids are also synthesized in the liver, in addition to conversion of amino acids to glucose or lipids. Most of the plasma proteins are also synthesized by hepatocytes (Bowen, 2001). Cell fractionation is a method in cell biology that provides insight regarding the enzymes that are present in a cell. For mammalian liver cells, fractionation can be done via differential centrifugation (Arias & et.al, 2009, p. 141). The type of enzyme can then be analyzed following specified tests. Methods Dispersion of liver in isotonic media by mild procedures leaves the mitochondria and cell nuclei intact. By fractionation, sub cellular structures can be centrifuged out of the homogenate, leaving all soluble components of the liver in the supernatant. Cytoplasmic enzymes are included in the supernatant since they are soluble. The supernatant can be analyzed to ascertain the presence of different enzymes (Mullins, 2005, p. 3). Biuret reaction is used in the analysis of the resultant mixture. In this reaction, the polypeptide concentration is estimated from the colour of the chelate that is formed at room temperature, between the nitrogen atoms of peptide bonds, and copper in alkaline solution. Since each protein has a unique amino acid concentration, and composition every unit mass of polypeptide gives a slightly varying colour. When the concentration of the protein is unknown, the assay gives results which are interpreted in terms of equivalent concentration. 1. 30g of fresh liver was homogenized in isotonic KCl (9ml per gram of tissue). 2. The homogenate was strained through several layers of surgical gauze. 3. Five standards were prepared with their concentrations varying from 0 to 10 mg/ml. The initial concentration of the solution used to prepare this standard was 10mg/ml. 4. 2ml of each standard was mixed with 3ml of Biuret reagent and the mixture incubated for 30 minutes. The absorbance of the resultant mixture was read at 600 nm. 5. The concentration of the protein was calculated in each ml of each fraction. Results Pipette volume 0 2 4 6 8 10 H2O (µl) 200 160 120 80 40 0 2 1.6 1.2 0.8 0.4 0 Albumin Standard 0 0.4 0.8 1.2 1.6 2 LDH is 0.1 ml of Lactate MDH is 0.1 ml of Maltate LDH 1(A) 0.375 – 0.026 = LDH 2 (B) 0.333 – 0.026 = LDH 3(C) 0.210 – 0.026 = Control LDH = 0.026 MDH 1 (A) 0.109 – 0.019 = MDH 2 (B) 0.291 – 0.019 = MDH 3 (C) 0.179 – 0.019 = Control MDH 0.019 Control Absorbance A – Cytosol 0.673 B – Intract mitochondria 0.598 C 0.472 Discussion The main thing to note in this experiment, the numbers are obtained as a result of enzyme activity. Enzyme activity is also illustrated by the resultant colour of the product (Bolsover & et.al, 2011, p. 215). Since the enzyme activity differs, then it is expected that the activity observed for each case is different. When prepared in isotonic solution, mitochondria clump together. The sediment in this case which consists essentially of mitochondria is prepared using a centrifuge. Mitochondria that have been prepared using this method show a disparity in the cell behaviour. (Chiras, 2011, p. 109). Conclusion The liver plays three main functions; metabolism, secretion and excretion, and vascular functions. The metabolic function entails the regulation of how proteins, lipids, and carbohydrates are synthesized and utilized in the body. Enzymes are substances that are protein in nature, and are used to catalyze biological reactions. Biological reactions, without the involvement of enzymes, would take place in infinitesimally slow rates. These processes are necessary for the support of life so if they were to take place slowly, life would be made impossible. Metabolic processes are regulated by enzymes. This means that without enzymes, metabolic processes would be hampered. Metabolic processes occur within cells, and are show high levels of specificity. The site, substrate, and conditions have to be at certain levels for the enzyme to function optimally. When enzymes are catalysing processes that are interactive, the product from the action of one enzyme usually becomes the substrate of another enzyme. Fractionation separates the components of a biological specimen. Soluble cellular component form part of the supernatant. Those components of the cell that are not soluble, sediment in granular form. Fractionation is done by the use of centrifuging. Hormones are soluble, therefore, when centrifuging is done, the hormones are part of the supernatant. The type,and quantity of the hormone in a sample can then be analysed by caryingout tests on the supernatant. Use of a chelating agent is a method that reveals both the type, and concentration of the hormone in the sample. The identifying factor here is the colour of the mixture, on addition of the chelating agent. The absorbance can also be used to test for the same. Different hormones will absorb and/or transmit radiation at different regions of the spectrum. In addition, different concentrations will show varying absorbance or transmittance properties. With this type of evidence, inference, and later conclusions can be made in regards to both concentration and type of the enzyme. References Arias, I., & et.al, A. W. (2009). The Liver: Biology and Pathobiology. Hoboken: John Wiley & Sons. Bolsover, S. R., & et.al, E. A. (2011). Cell Biology: A Short Course. Hoboken: John Wiley & Sons. Bowen, R. (2001, March 10). The Liver: Introduction and Index. Retrieved January 18, 2012, from http://www.vivo.colostate.edu/hbooks/pathphys/digestion/liver/index.html Caprette, D. R. (2005, May 13). Retrieved January 18, 2012, from Experimental Biosciences: http://www.ruf.rice.edu/~bioslabs/methods/microscopy/microscopy.html Chiras, D. D. (2011). Human Biology. Sudbury: Jones & Bartlett Learning. Gropper, S. S., Smith, J. L., & Groff, J. L. (2008). Advanced nutrition and human metabolism. Stamford: Cengage Learning. Karp, G. (2009). Cell and Molecular Concepts and Experiments. Hoboken: John Wiley and Sons. Lanham-New, S. A., Macdonald, I. A., & Roche, H. M. (2011). Nutrition and Metabolism. Hoboken: John Wiley & Sons. Mullins, C. (2005). The biogenesis of cellular organelles. New York: Springer. Starr, C., Evers, C. A., & Starr, L. (2010). Biology: Concepts and Applications. Stamford: Cengage Learning. Read More