The Anatomy Of A Pigs Heart Physical Education Essay

The Anatomy Of A Pigs Heart Physical Education Essay The aim of this experiment was to understand the external and internal structures by dissecting a pigs heart, drawing and labelling the structures. Introduction A pigs heart is covered by the thin membrane or pericardium. Myocardium exists as the muscle located below the pericardium. Most of the myocardium in the lower two chambers of the heart called ventricles (Lane, 2010a). A pigs heart has four chambers which include left atrium, left ventricle, right atrium and right ventricle (Lane, 2010a). Atrium and ventricle are separated by valves. In the entrance of the left ventricle has bicuspid valve which controls the blood flows from the left atrium to the left ventricle during diastole (Lane, 2010a). While in the entrance of the right ventricle has tricuspid valve which allows blood flow from the right atrium to the right ventricle. The function of valves is to make the blood only moving from atrium to ventricle and avoid regurgitation. (Lane, 2010a). There are some blood vessels have a significant influence on transferring the blood (Lane, 2010a). Coronary artery which lies in the groove on the front of heart carries flesh blood to the heart muscle to receive oxygen and nutrients (Lane, 2010a). Pulmonary artery which lies out of the right ventricle supplies blood with oxygen to the lungs (Lane, 2010a). Aorta which is the major vessel located near the right atria provides blood to the upper body (Lane, 2010a). Pulmonary veins carry oxygenated blood from the lungs to the left atrium (Lane, 2010a). Inferior and Superior Vena Cava which are located on left of the heart and link to the right atrium, supplies deoxygenated blood to the right receiving chamber (Lane, 2010a). Blood flows through the pig heart are similar with a humans. Deoxygenated blood moving from the superior and inferior vena cava (see Figure 1) into the right atrium which has a very low pressure (Gisbornesc, 2010). During diastole which indicates the relaxation phase, the right ventricles pressure drops down almost zero. There is a pressure gradient between the right atrium and ventricle, leading to the blood flow into the right ventricle (Gisbornesc, 2010). When the ventricle fills, the blood goes through the tricuspid valve. During systole (the contracting phase) the ventricle begins to contract, the intraventricular pressure is increased which causes the tricuspid valve to close while the pulmonary valve open (Gisbornesc, 2010). And then blood moves out of the ventricle via the pulmonary artery to the lungs (Gisbornesc, 2010). When the ventricle relaxes, intraventricular pressure drops below the pressure in the pulmonary artery, as a result, the pulmonary valve close. In this case , blood returns to the right side of the heart and is pumped back to the lungs for gas exchange (Gisbornesc, 2010). Figure 1(Biologymad, 2010) shows the blood flow within the heart. Figure 1 (Biologymad, 2010): The Blood Flow within Heart The function of a heart is to promote blood flowing to organs and tissues, providing enough blood, oxygen and various nutrients while taking away the metabolic products such as carbon dioxide, urea and uric acid. This process maintains the normal function and metabolism of the cells within the body. Method Observations Material Fresh pig heart Scalpels Container for heart Scissors Forceps Cutting board Equipment Several blank A4 sheets Pens HB pencils Erasers Rulers Sharpeners Procedure-External Anatomy A heart was placed in a dissecting pan. The size of this heart was moderate; the colour of this heart was pink and it was a little smelly. The heart was examined and the thin membrane or pericardium that still covers the heart was located. The pericardium was removed carefully. The pericardium was too thin to remove. The tip of the heart or the apex was located. The left ventricle extended all the way to the apex. The heart was placed in the dissecting pan. The major blood vessels were on the top and the apex was down. There was a groove that extended from the right side of the board end of the heart. The heart was in the pan in the position and faced my body. The left atrium, left ventricle, right atrium and right ventricle were located from this surface. Four chambers were observed unclearly, but still could be identified. While the heart was still in this position in the dissecting pan, the coronary artery, pulmonary artery, aorta, pulmonary veins and inferior superior vena cava were located at the broad end of the heart. Aorta, pulmonary artery and pulmonary veins inferior superior vena cava were observed clearly, but coronary artery was not found. A fully labelled diagram of the front heart was drawn. Figure 2 is attached. 8. A fully labelled diagram showing the back side of heart was drew. Figure 3 is attached. Procedure- Internal Anatomy The side of the pulmonary artery was cut through and continue cutting down into the wall of the right ventricle using the scissors. Cut deep enough to go through the wall of the heart chamber. (The cutting line was above parallel to the groove of the coronary artery) The heart was pushed open at the cut to examine the internal structure with fingers. There was a dried blood inside the chambers, the heart was rinsed out. The right atrium was located. The thinner muscular wall of this receiving chamber was measured by ruler. The right atrium was about 5-10 mm and very smooth. A fully labelled diagram showing the internal of right atrium was draw. Figure 4 is attached. The inferior superior vena cava enter this chamber were founded. The lack of valves was noticed. The inferior superior vena cava was easy to found. The valve between the right atrium and right ventricle was located, which is called tricuspid valve. This valve consisted of three leaflets and had long fibbers of connective tissue. The thickness of the right ventricle and its smooth was felt by fingers. The network of irregular muscular cords was noticed on the inner wall of this chamber. The right ventricle was about 20mm and very smooth. The septum was found on the right side of the right ventricle. The septum was thick which separated the right and left ventricles from each other. The pulmonary artery was located inside the right ventricle. The one-way valve called the pulmonary valve was found. A fully labelled diagram showing the back side of heart was drew. Figure 3 is attached. The heart was continued to cut open using scissors. A cut was started on the outside of the left atrium downward into the left ventricle cutting toward the apex to the septum at the centre groove. The heart was pushed open at this cut with fingers any dried blood was rinsed out with water. The heart was difficult to cut downward into the left ventricle because the knife was not sharp. The left atrium was examined. The openings of the pulmonary veins were found from the lungs. The pulmonary veins were hard to observe. Bicuspid valve was looked for inside left atrium. This valve consisted of two leaflets. The left ventricle was examined. The thickness of the ventricular wall was noticed. The left ventricle was found and the thickness of the ventricular was about 20mm. The left ventricle was cut across the left ventricle toward the aorta continues cutting to expose the valve using scissors. The three flaps or leaflets were counted on this valve called aortic valve. This valve had half-moon shape. Notes to all the diagrams were added relating the labelled structures to their functions. Results Figure 2 shows the front view of heart. In this figure, the aorta is clearly visible at the top, with left and right atrium on either side, while the ventricles are in the bottom. This figure also shows the aortic and pulmonary arteries as well as the pulmonary veins and superior vena cava. 2. Figure 3 shows the back side of heart. In this figure, the superior vena cava and aorta can be seen. 3. Figure 4 shows the internal view of the right side. In this figure, the tricuspid valve can be seen and the right ventricular outflow tract which includes the pulmonary artery. 4. Figure 5 shows the internal view of the left side In this figure, the bicuspid valve and aorta can be seen. Discussion Possible errors may have arisen during this experiment. Firstly, the heart was cut difficultly, because the equipment given was not sharp enough. Secondly, some blood vessels can not be found such as coronary artery, because the pigs heart was not fresh enough, may be damaged by the butcher. Thirdly, human errors may lead to the results not precise. For instance, when cut deeply to go through the wall of the heart chamber, the cutting line was not parallel to the groove of the coronary, as a result, the chamber can not observed carefully. A future experiment should be improved in following ways. Firstly, the equipments should be given sharp enough in order to cut the heart efficiently according to the instruction. Secondly, the pig heart should be provided fresh enough and completely. In this case, all the organs can be observed and described. Finally, when cutting line is required parallel, it is better to use the ruler to measure. Conclusion In conclusion, after doing the experiment, the external and internal structure was examined. A pigs heart is four chambered pump which includes left atrium, left ventricle, right atrium and right ventricle. Valves, which separate atrium and ventricle, prevent blood from flowing backwards. A pig has double system which can make blood circulate the whole body via the vessels.

Glory vs Shawshank Redemption :: essays papers

Glory vs Shawshank Redemption Two of the well known movies that were the abolishinst movement in Amerrica were Glory and The Shawshank Redemption.On the surface, the movies the shawshank redemption and glory seem to be completely different.But, as the movies upfold, it is evident that both the prisoners of Shawshank and the soldiers of the 54th Massachusetts are in search of the same thing. For the prisoners of Shashank, normalcy is freedom. For the soldiers of the 54th, normalcy is equality. In the Shawshank Andy uses his influences to give prisoners of shawshank tastes of freedom. The solodiers of the 54th strives to be normal, but before they can be normal, they must be equal. They wanted freedom too. One point of intrest that shows Andy's spirit is indeed unbreakable was when he offered Captain Hadley some financial advice on his recent inheritance. He tells captain Hadley that if he trusts his wife he can tell the I.R.S that he is giving the inhertance as a gift from Andy to his wife. By doing this the money can not touched by the I.R.S, so Captain Hadley gains all the money without any tax taken out.In return andy Dufresne( tim robbins) asked for beer for his " Co-workers.His friend ., Red simply states," we sat and drank with the sun on our shoulders and felt like free men.Hell, we could have been tarring the roof of one of our own houses. We were the lords of all creation. As for andy he spent that break hunkered in the shade, a strange little smile on his face, watching us drink his beer." This quote shows how little things teh prisoners get can make them happy.Another example would be when he used the hammer to escape from prison. It started when he asked Red to get him a rok hammer, which he said he would use to shape rocks. He calms Red's conscious as he tells it would take him a thousand years to break out of prison with a rosk hammer.When he received the rosk hammer he started to shape rocks as soon as he could and hidden that hole with a poster.After he had the hole big enough to crawl throught, he asked Heywood for a six-foot piece of rope.

Diffusion and Osmosis of Solutes and Water Across a Membrane

Diffusion and Osmosis of Solutes and Water Across a Membrane Brittany Bacallao Nova Southeastern University Abstract: This experiment gave a visual understanding of osmosis and diffusion. The first experiment proved that solutes would move down a concentration gradient if permeable to the selective membrane. The second experiment proved different solute concentrations affect the movement of water, depending on the solute concentration inside the cell. The purpose of this lab was to look for different solutes that can cross an artificial membrane and to observe the effect of different concentrations of sucrose on the mass of a potato cell.Results for Part One suggested that the molecular weight of albumin and starch was too large to pass through the dialysis tube, but glucose and sodium sulfate molecules were small enough to pass through the dialysis tube. Also, a decrease in water weight occurred due the dialysis tube being placed in a hypertonic solution. Results for Par Two showed the potato cell having a molar concentration of 0. 2734, which caused sucrose concentrations above 0. 2 M to have a decrease in mass. Inversely, sucrose concentrations below 0. 2 M caused an increase in mass.Diffusion is the random movement of molecules spreading evenly into available space (Cain, Jackson, Minorsky, Reece, & Urry, 2011). Movement of water also follows a similar concept, however, water can act as a shield for solutes and become unavailable to diffuse while in other cases water is free and will move to an area of low solute concentration to an area of high solute concentration: this processes is better known as osmosis (Keith, Messing, Schmitt, & Feingold, 2010). Osmosis and diffusion can occur along a permeable membrane or selective membrane.A cell with a selective membrane allows small molecules and ions to pass through but excludes others; also, substances that are able to pass through the membrane do so at different rates. On the other hand, permeable membranes al low nonpolar molecules, such as hydrophobic molecules (water fearing), to dissolve in the lipid bilayer, which allows the molecule to easily cross the membrane. However, molecules such as glucose can pass through the lipid bilayer, but not as rapidly as nonpolar molecule (Cain et al. , 2011).Understanding the concept of osmosis helps explain why lakes cannot have an increase in salinity. If saltiness of a lake increases, species living in the lake could die. This occurs when the lake water becomes hypertonic solution, which causes the animal cells to lose an excessive amount of water forcing the cell to shrivel up and die (Cain et al. , 2011). On the contrary, understanding the concept of diffusion can help explain why after spraying perfume in one area of the room, then after several minutes, the perfume is smelled throughout the room.This is because particles of the perfume move randomly and eventually spread out evenly throughout the room. Moreover, in the experiment performed, d iffusion and osmosis was observed using artificial systems (plastic membranes) and potato cells. The null hypothesis for Part One of the experiment is that the concentration gradient has no effect on the weight of the dialysis tube. The alternate hypothesis is that the weight of the dialysis tube will be affected by the concentration gradient.The null hypothesis for Part Two of the experiment is that the increase of sucrose concentration has no effect on the mass of the potato cell. The alternate hypothesis is that the difference in sucrose concentration will affect the mass of the potato cell. This experiment tests all hypotheses and helps to explain the concepts of diffusion and osmosis. Materials and Methods: Part One: Gloves were used to obtain a 20 cm section of dialysis tube that had soaked in a beaker of distilled water prior to the experiment. The dialysis tube was cleaned with distilled water and then tied off to form a pouch.Once the pouch was formed, 3 mL of starch and so dium sulfate solution was placed inside the tube, and then tied off and weighed. The weight obtained was recorded as initial weight. While weighing the dialysis tube with the solution of starch and sodium sulfate, eight test tubes were obtained and solution of starch/sodium sulfate was added to two test tubes labeled bag start (Keith et al. , 2010). After weighing dialysis tubing of starch/sodium sulfate and adding the solution to two test tubes, the tubing was placed in a beaker containing a solution of albumin and glucose.Next, 1. 0 mL of albumin and glucose were then placed in two test tubes labeled solution start. The tubing in the albumin/glucose solution was kept inside the solution for 75 minutes. Every 15 minutes the solution and tube was mixed (Keith et al. , 2010). At the end of the 75 minutes, two 1. 0 mL samples of the albumin/glucose solution from the beaker were added to two test tubes labeled solution end. Then, the dialysis tube was removed from the beaker and rinsed off with distilled water. Once the tubing was rinsed and blotted dry the final water weight was recorded.After measuring the final water weight, the contents in the tubing was dumped into a beaker and 1. 0 mL of starch/sodium sulfate solution was added to two test tubes labeled bag end (Keith et al. , 2010). In order to test for glucose, a glucose dip and read strip was placed in the first set of test tubes that were labeled bag start, solution start, bag end, and solution end. Then, a protein dip-and-read strip was placed in the same set of test tubes and the results were recorded from both glucose and protein strips.After testing for protein, solution and bag samples were tested for sodium sulfate. To test for sodium sulfate, three drops of 2% barium chloride were added to the second set of test tubes labeled bag start, solution start, bag end, and solution end. The results were observed and then recorded (Keith et al. , 2010). To see if starch was present in the solutions, iodin e solution was added to the first set of test tubes that were used for the glucose and protein strips. Results were then observed and recorded (Keith et al. , 2010). Part two: Seven beakers were obtained. 0 mL of solution labeled 0. 2 M sucrose, 0. 4 M sucrose, 0. 6 M sucrose, 0. 8 M sucrose, 1 M sucrose, distilled water, and unknown were each placed in different beakers. Then, a potato was sliced into 28 cylinders using a cork borer. The cylinders were separated into seven groups of four and then placed under a paper towel until the group was ready to weigh the potato cylinders. Once prepared to weigh the cylinders, the weight of each group of cylinders was recorded. Four cylinders were placed into each beaker and sat for an hour (Keith et al. , 2010).After the beakers sat for an hour at room temperature, the potato cylinders were then removed and blotted dry. Final weight was recorded for each group of potato cylinders. To calculate the percentage change, the following equation wa s used (Keith et al. , 2010): Percent change =Ending mass-Starting massStarting mass? 100% Results: Part One Results for the first experiment revealed certain molecular weights were unable to pass through the selectively permeable membrane. Table 1 shows that albumin (protein) and starch were unable to pass through the selective membrane.Albumin’s molecular weight was approximately 64,000 Daltons and starch’s molecular weight was greater than 100,000 Daltons; these results were already known. Initially, glucose was present outside of the dialysis tube but in final results glucose was found in the final bag and final solution. On the contrary, sodium sulfate was initially present inside the dialysis tube but in the final results only, sulfate ion was found in the final solution. The last result was the change in water weight. Water had decreased from the initial weight. The following results are shown in Table 1.Table 1: Diffusion of solutes through an artificial membra ne after 75 minutes. | Glucose | Sulfate ion| Protein| Starch| Water weight (g)| Inside bag| Â  | Â  | Â  | Â  | Â  | Initial| -| +| -| +| 16. 59 g| Final| +| -| -| +| 16. 05 g| Outside bag| Â  | Â  | Â  | Â  | Â  | Initial| +| +| +++| -| Â  | Final| +| +| +++| -| Â  | Part Two In the second experiment results showed different concentrations of sucrose affected the potatoes’ mass. In the beakers containing 0. 0 M (distilled water) and 0. 2 M concentration of sucrose resulted in water entering the potato cell, which caused the cell to increase in mass.The beakers containing 0. 4 M, 0. 6 M, 0. 8 M, and 1 M of sucrose concentration had the opposite effect on the potatoes mass. Therefore, the higher the concentration was the greater amount of water left, causing the cell to decrease in mass. Also, the unknown concentration was found to be 0. 5 M of sucrose, which caused the mass in the potato cells to decrease as a result of water leaving the cell. The following infor mation is displayed in Table 2. Table 2: Percent change in mass of potato cells after being placed in different sucrose concentrations, also the differences in initial and final mass.Contents in beaker| Initial mass| Final mass| Mass difference| % Change in mass| a. Distilled Water| 0. 82| 0. 92| 0. 1| 12. 20%| b. 0. 2 M Sucrose| 0. 65| 0. 69| 0. 04| 6. 20%| c. 0. 4 M Sucrose| 0. 62| 0. 56| -0. 06| -9. 70%| d. 0. 6 M Sucrose| 0. 69| 0. 58| -0. 1| -15. 90%| e. 0. 8 M Sucrose| 0. 61| 0. 48| -0. 13| -21. 30%| f. 1 M Sucrose| 0. 74| 0. 57| -0. 17| -23%| g. Unknown| 0. 77| 0. 7| -0. 07| -9. 10%| The molar concentration of the potato cell was found to be 0. 2734 M. The molecular weight was found by looking for the x-intercept on the graph below (Figure 2. . Figure 2: Percent change in mass of potato cells put in different concentrations of sucrose. Discussion: Part One of the experiment indicated that the dialysis tube was selectively permeable and only molecular weight fewer than 64,000 Daltons were able to pass through the membrane. This explains why albumin and starch were unable to pass through the membrane because their molecules were too large. Conversely, glucose was able to pass through the selectively permeable membrane due to its relatively small molecular weight.However, because glucose was present in both the final bag and final solution this meant that glucose had evenly distributed its molecules by complying with the concept of diffusion. Sulfate ions present outside the dialysis tube in the final results show that sulfate ions were also able to diffuse through the selective membrane into the final solution. A decrease in water weight from initial weight shows that the dialysis tube was placed in a hypertonic solution causing more of the inside solution to diffuse to the outside leading to a decrease in the final weight of the bag.The null hypothesis is rejected in Part One of the experiment because the concentration gradient did affect the weight of t he dialysis tube. This is due to the fact that sodium sulfate completely left the bag, thus causing the bag to decrease in weight. The reason why sodium sulfate left is because there was no sodium sulfate in the solution; therefore, molecules went to an area of lower concentration. The alternate hypothesis is not rejected because the concentration gradient did affect the weight of the dialysis tube.This is proven by a decrease in initial weight due to sodium sulfate leaving the tube. Part Two of the experiment showed that the potato cells had some kind of change in their mass after being placed in different sucrose concentrations. The change in mass occurred because water either left the cell or entered the cell depending on the sucrose concentration. This explains why distilled water had the greatest increase in mass, because water wanted to go to an area (potato cell) of higher concentration from an area of low concentration.On the other hand, 1 M of sucrose concentration had the greatest decrease in mass because water wanted to leave the cell to move to an area of higher concentration. Therefore, if the concentration was greater than the molar concentration of the potato cell than water left the cell at a faster rate. The null hypothesis for Part Two of the experiment proved to be wrong because an increase in sucrose concentration did have an affect on the change in mass of the potato cell.Increased sucrose concentration changed the mass of the cell because the concentration was higher than the molar concentration of the potato cell. Thus, the alternate hypothesis is proven correct. The difference in sucrose concentration will affect the mass of the potato cell. References: Cain, M. L. , Jackson, R. B. , Minorsky, P. V. , Reece, J. B. , & Urry, L. A. (2011). Biology (9th Edition ed. ). San Francisco: Pearson Education, Inc. Keith, E. , Messing, C,. Schmitt, E. , Feingold, J. (2010). Laboratory Exercises in Biology (3rd ed. ). Dubuque, IA: Kendall Hunt Publi shing Company.

The Death Of Patients With Pneumococcal Pneumonia

The leading cause of death in patients diagnosed with pneumococcal pneumonia is respiratory failure. Other causes include respiratory and systemic organ failure (3). The Streptococcus pneumoniae bacterium has a lethal role in causing respiratory failure in patients with pneumonia. This disease is infectious in humans of all ages, but children and the elderly are at higher risk (5). Pneumonia can be classified based on how you can contract it. The basic types of pneumonia include; community-acquired pneumonia, hospital-acquired pneumonia, aspiration pneumonia, and opportunistic pneumonia. Out of the four different types, community-acquired pneumonia is the most common(2). Streptoccocus pneumoniae is not a stranger to the human body.†¦show more content†¦Specific ATP-binding cassette transporters (ABC transporters) are one group of transporters that provide the mechanics for substrate transport in pneumococci. They can import carbon amino acid substrates, carbohydrates, a nd other nutrients needed by the cell (1, 4). They can also export material such as the outer surface adhesins, degradation enzymes and synthetic capsular components. Additionally, these transporters are vital to the cell’s ability to take up DNA and they function as efflux pumps to counter antibiotics. When describing the mechanism, it is important to note that the ABC transporter has two ATPases attached to it to help generate energy, two permeases that intersect the membrane and a substrate binding protein that acts as a receptor. In the presence of ATP, the permease causes a conformation change in the transporter that allows the substrate binding protein to release the substrate (1). Hydrolysis of ATP causes the substrate to enter the cell. Once ADP and Pi is released, the transporter resets to the resting state(1). Substrate transport as described above is important to the existence of pneumococci in hosts. 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