LabReport - The Effect of Different Concentrations of...

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The Effect of Different Concentrations of Sugars on the Hormone Response of a Seed Embryo Aubrey McMillan Introduction This experiment was performed to answer the question: How does the concentration and type of sugar available to the seed embryo alter the hormone response? A barley seed consists of three primary structures - the starch endosperm, the aleurone, and the embryo. The starchy endosperm is a food reserve for the embryo, which is alive in early seed development and dead by the time the grain separates from its mother plant. The aleurone is living tissue that surrounds the starchy endosperm, which secretes enzymes once germination has begun. Finally, the embryo is the progeny that develops after germination (Kaska and Bush, 2009). Once the embryo is in contact with water, Gibberellic Acid is produced, which causes the aleurone to produce amylase. Amylase is a hormone that breaks down starch into the monosaccharide glucose once it is secreted into the endosperm, which may then nourish the embryo (Abdul-Baki 1969). Thus, the production of amylase depends heavily on what is available to the embryo. Considering the nature of germination, we decided to investigate the effect of sugar (glucose and sucrose) concentration on the amylase response. Methods The initial volumes of sucrose and glucose were calculated (as well as the volume of water needed), given that an initial 1 M stock supply of each sugar was available. It was decided that four different concentrations of glucose would be used (1 M, 0.5 M, 0.15 M, and 0.01M), as well as the same four concentrations of sucrose. A control, with 3 mL of water instead of sugar,
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would also be accounted for. The calculations are provided below, using the formula C 1 V 1 =C 2 V 2 , also taking into consideration that a final volume ( V 2 ) of 3 mL was needed for incubation: For 1 M of glucose/sucrose ( C 2 = 1M) (1M)( V 1 ) = (1M)(3 mL) V 1 = 3 mL of glucose/sucrose (3 mL of medium) - (3 mL of glucose/sucrose) = 0 mL of H 2 O For 0.5 M of glucose/sucrose ( C 2 = 0.5 M) (1M)( V 1 ) = (0.5M)(3 mL) V 1 = 1.5 mL of glucose/sucrose (3 mL of medium) - (1.5 mL of glucose/sucrose) = 1.5 mL of H 2 O For 0.15 M of glucose/sucrose ( C 2 = 0.15M) (1M)( V 1 ) = (0.15M)(3 mL) V 1 = 0.45 mL of glucose/sucrose (3 mL of medium) - (0.45 mL of glucose/sucrose) = 2.55 mL of H 2 O For 0.01 M of glucose/sucrose ( C 2 = 0.01M) (1M)( V 1 ) = (0.01M)(3 mL) V 1 = 0.03 mL of glucose/sucrose (3 mL of medium) - (0.03 mL of glucose/sucrose) = 2.97 mL of H 2 O For 0 M of glucose/sucrose (control - C 2 = 0M) (1M)( V 1 ) = (0M)(3 mL) V 1 = 0 mL of glucose/sucrose (3 mL of medium) - (0 mL of glucose/sucrose) = 3 mL of H 2 O Nine test tubes were obtained - four for the various glucose solutions, four for the various sucrose solutions, and one for the control. The test tubes were labeled according to which sugar it would soon contain, as well as its concentration. Using a P20, a P200, and a P1000 micropipette (depending on the amount of liquid needed), the respective amounts of glucose
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LabReport - The Effect of Different Concentrations of...

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