10 Steps of Glycolysis

10 Steps of Glycolysis Glycolysis literally means splitting sugars and is the process of releasing energy within sugars. In glycolysis, glucose (a six-carbon sugar) is split into two molecules of the three-carbon sugar pyruvate. This multistep process yields two molecules of ATP (free energy containing molecule), two molecules of pyruvate, and two high energy electron carrying molecules of NADH. Key Takeaways: Glycolysis Glycolysis is the process of breaking down glucose into two molecules of pyruvate. It produces ATP and is the first stage of cellular respiration. Glycolysis can occur with or without oxygen. In the presence of oxygen, glycolysis is the first stage of cellular respiration. In the absence of oxygen, glycolysis allows cells to make small amounts of ATP through the process of fermentation. Glycolysis takes place in the cytosol of the cells cytoplasm. However, the next stage of cellular respiration, known as the citric acid cycle, occurs in the matrix of cell mitochondria. Below are the 10 steps of glycolysis. Step 1 The enzyme hexokinase phosphorylates - adds a phosphate group to - glucose in the cells cytoplasm. In the process, a phosphate group from ATP is transferred to glucose producing glucose 6-phosphate. The equation is: Glucose (CStep 2 The enzyme phosphoglucoisomerase converts glucose 6-phosphate into its isomer fructose 6-phosphate. Isomers have the same molecular formula, but the atoms of each molecule are arranged differently. The equation for this step is: Glucose 6-phosphate (CStep 3 The enzyme phosphofructokinase uses another ATP molecule to transfer a phosphate group to fructose 6-phosphate to form fructose 1, 6-bisphosphate. The equation is: Fructose 6-phosphate (CStep 4 The enzyme aldolase splits fructose 1, 6-bisphosphate into two sugars that are isomers of each other. These two sugars are dihydroxyacetone phosphate and glyceraldehyde phosphate. The equation is: Fructose 1, 6-bisphosphate (CStep 5 The enzyme triose phosphate isomerase rapidly inter-converts the molecules dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. Glyceraldehyde 3-phosphate is removed as soon as it is formed to be used in the next step of glycolysis. The two equations for this step are: Dihydroxyacetone phosphate (CNet result for step Nos. 4 and 5: Step 6 The enzyme triose phosphate dehydrogenase serves two functions in this step. First, the enzyme transfers a hydrogen (H-) from glyceraldehyde phosphate to the oxidizing agent nicotinamide adenine dinucleotide (NAD) to form NADH. Next, triose phosphate dehydrogenase adds a phosphate (P) from the cytosol to the oxidized glyceraldehyde phosphate to form 1, 3-bisphosphoglycerate. This occurs for both molecules of glyceraldehyde 3-phosphate produced in step 5. The two equations for this step are: A. Triose phosphate dehydrogenase + 2 HB. Triose phosphate dehydrogenase + 2 P + 2 glyceraldehyde 3-phosphate (CStep 7 The enzyme phosphoglycerokinase transfers a P from 1,3-bisphosphoglycerate to a molecule of ADP to form ATP. This happens for each molecule of 1,3-bisphosphoglycerate. The process yields two 3-phosphoglycerate molecules and two ATP molecules. The equation is: 2 molecules of 1,3-bisphoshoglycerate (CStep 8 The enzyme phosphoglyceromutase relocates the P from 3-phosphoglycerate from the third carbon to the second carbon to form 2-phosphoglycerate. The equation is: 2 molecules of 3-Phosphoglycerate (CStep 9 The enzyme enolase removes a molecule of water from 2-phosphoglycerate to form phosphoenolpyruvate (PEP). This happens for each molecule of 2-phosphoglycerate. The equation is: 2 molecules of 2-Phosphoglycerate (CStep 10 The enzyme pyruvate kinase transfers a P from PEP to ADP to form pyruvate and ATP. This happens for each molecule of phosphoenolpyruvate. This reaction yields two molecules of pyruvate and two ATP molecules. The equation is: 2 molecules of phosphoenolpyruvate (CEnd Result MediaForMedica l /UIG / Getty Images A single glucose molecule in glycolysis produces a total of two molecules of pyruvate, two molecules of ATP, two molecules of NADH, and two molecules of water. Although two ATP molecules are used in step Nos. 1 through 3, two ATP molecules are generated in step No. 7 and two more in step No. 10. This gives a total of four ATP molecules produced. If you subtract the two ATP molecules used in step Nos. 1 through 3 from the four generated at the end of step No. 10, you end up with a net total of two ATP molecules produced.

Osmosis and Diffusion Lab Report Introduction Essays

Osmosis and Diffusion Lab Report Introduction Essays Osmosis and Diffusion Lab Report Introduction Paper Osmosis and Diffusion Lab Report Introduction Paper While molecules in diffusion move down a concentration gradient, molecules during osmosis both move down a concentration gradient as well as across it. Both diffusion, and osmosis are types of passive transport, which do not require help. When the concentration of the environment outside of the cell is lower than the inside of the cell, this is called a hypotonic solution. In hypotonic solutions, when water moves into the cell they burst, which is known as lists. A hypersonic solution is when the concentration of outside the cell is higher than the inside of the cell. In hypersonic solutions, water moves out of the cell, making the cell lose water and shrink; this is called creation. An isotonic solution is when the concentration of dissolved molecules is the same both inside and outside of the cell. Thus there is no net movement. The purpose of this experiment is to investigate whether osmosis occurs across the membrane of potato cells. Hypothesis The potato that will be more hypersonic to the sucrose solution will gain weight, whereas the potato when hypotonic to the solution will lose weight. When the potato is most isotonic the similarity of the potato will be 0. . Materials 1 potato 7 250-ml Beakers Sucrose Solutions: 0. 1 (M) Razor Blade Denizen Water (0 molar) Balance that weighs to the nearest 0. 01 Cork Borer Paper Towels Calculator Metric Ruler Methods To perform this experiment seven 250-ml beakers were obtained, rinsed thoroughly and dried with a paper towel. Next a small amount of tape was applied on the beakers and labeled accordingl y to the sucrose solution molarities. Each beaker received 100-ml of the various solutions. However one beaker received 100 ml of Denizen water with a molarities of 0. . Afterwards a cork borer was pushed through the potato and was twisted back and forth. Once the borer was filled it was removed from the potato. Pushing the potato cylinder out of the borer, this this step was repeated six more times in order to get seven undamaged potato cylinders. Using a sharp razor blade, the potato cylinders were both cut to a uniform length of about CM, and were removed of their potato skins. The potato pieces were also cut in half to give the cells a greater reface area in which it was easier to absorb the solution. After the cylinders were weighed on a balance and the data was recorded in Table 4 Using the razor blade each potato was cut lengthwise into two long halves. Then the potato pieces were transferred to the water beaker and the time they were submerged was recorded. This step was repeated for all potato cylinders in which the pieces were placed in solutions O. 1 to 0. 6 M. The potatoes were incubated for ninety minutes. At the end of the incubation period the time was recorded. Then the potato piece was removed form the first sample. Next potato pieces were weighed the and the final weight was recorded in Table 4. This procedure was repeated until all samples had been weighed and recorded in the chronological order they were initially placed in the test solution. Afterwards the table was completed by recording the weight change and percentage change. In order to calculate the final percentage change of weight the formula that would be needed is (weight change)/(initial change) X 100. The data received at the end was not normal; therefore each potato cylinder was weighed again, and recorded on two different scales.

Georgia Cancer Foundation Research Paper Example | Topics and Well Written Essays - 750 words

Georgia Cancer Foundation - Research Paper Example In midst of such cancer crisis, Georgia Cancer Foundation (2010) is one of the non-profit organizations that are playing a significant and momentous role in preventing and reducing the number of cancer patients on this planet, specifically in the United States. In specific, 1975 witnessed establishment of this non-profit organization with the name of Southeastern Cancer Foundation (GCF, 2010), which later changed into its present name, Georgia Cancer Foundation. Since then, it has been momentous in putting efforts to plan and arrange provision of education and awareness facilities to residents of Georgia in the United States. Besides planning and arranging educational programs, the foundation is responsible for providing facilities of early detection of cancer agents in the human body that has been very significant in reducing the number of cancer patients that reach the level of untreatable cancer. One of the noteworthy attributes of Georgia Cancer Foundation (GCF, 2010) is its huge scope that includes all types of cancer, and thus, the foundation caters individuals coming from different and diverse backgrounds while fulfilling their requirements related with the disease of cancer. In order to cater all the residents of Georgia, the foundation runs a support group network that consists of different cancer services, especially of women that have been very beneficial for the organization to endeavor for the cause of cancer. With such networking of cancer services, the Georgia Cancer Foundation (GCF, 2010) has been successful in proving itself as the leader in the field of cancer care in the US state of Georgia. Statistics (GCF, 2010) have indicated that only in Georgia, approximately six thousand women get their breast cancer diagnosed in the state that points out the urgent need for an organization to focus on the women at risk of getting breast cancer. For this