Warm Up 12/2

1. Compare growth due to apical and lateral meristems in dicots.
Apical meristems allow roots and stems to elongate.  Stem apical meristems allow the production of new leaves and flowers.  Lateral meristems allow the stems and roots to grow thicker.

2. Explain the role of auxin in phototropism.
Plants use hormones to control their growth and development.  Auxin is a growth promoter, which causes the secretion of hydrogen ions into the cell walls that loosen connections between cellulose fibers and causes cell expansion.  Phototropism is the directional growth of the plant in response to the source of light.  Shoots can typically detect light.  Auxin is produced in the shoot tips and is concentrated in the shaded side of the plant.  Growth on the shaded side causes the plants to bend towards the sun.

3. Explain how water is carried by transpiration stream, including the structure of xylem, vessels transpirational pull, cohesion, adhesion and evaporation.
Transpiration causes a flow of water from the roots through the stems to the leaves of plants.  This flow is called the transpirational stream.  The process starts with evaporation of water from the cell walls of spongy mesophyll cells in the leaf.  The water is pulled out of xylem vessels and through pores in spongy mesophyll cell walls by capillary action.  Cohesion is when water molecules are weakly attracted to each other via hydrogen bonds.  Adhesion is water molecules forming hydrogen bonds with the xylem cell wall.  Low pressure or suction is created inside xylem vessels when water is pulled out.  This is called the transpirational pull.  The xylem has a specialized structure to facilitate transpiration.  The inner lining is composed of dead cells that have fused to create a continuous tube.  These cells lack a cell membrane, allowing water to enter the xylem freely.  The outer layer is perforated, allowing water to move out of the xylem and into the leaves.  The outer cell wall contains annular lignin rings which strengthens the xylem against the tension created by the transpiration stream.

Warm Up 11/13 - Plants

1. Outline the differences (at least 5) between monocots and dicots.


2. Identify and describe modification of roots, stems and leaves of different plants.  Give one example of each.
A bulb is a stem modification where leaf bases grow to form an underground food storage organ.  An example of this is onions.  Tubers are root modifications where underground stems are food storage organs.  An example of a tuber is a potato.  Storage roots are modified tap roots where swollen roots are used as food storage organs.  Carrots have storage roots.  Tendrils are leaf modifications where narrow outgrowths from leaves rotate through the air until they touch a solid support, to which they attach.  Tendrils allows plants to climb.  Sweat peas and mandevilla have tendrils.

Warm Up 11/1

1. Draw and label a diagram showing the structure of a chloroplast as seen in electron micrographs.

2. Explain photophosphorylation in terms of chemiosmosis.

Photophosphorylation is the production of ATP using the energy of sunlight and is possible because of chemiosmosis.  Chemiosmosis is the movement of ions across a selectively permeable membrane, down their concentration gradient.  During photosynthesis, light is absorbed by chlorophyll molecules.  Electrons within these molecules are then raised to a higher energy state.  These electrons then travel through Photosystem II, a chain of electron carriers and Photosystem I.  As the electrons travel through the chain of electron carriers, they release energy.  This energy is used to pump hydrogen ions across the thylakoid membrane and into the space within the thylakoid.  A concentration gradient of hydrogen ions forms within this space.  These then move back across the thylakoid membrane, down their concentration gradient through ATP synthase.  ATP synthase uses the energy released from the movement of hydrogen ions down their concentration gradient to synthesize ATP from ADP and phosphate.

3. State the final products of the two photosystems involved in the light-dependent reactions of photosynthesis.

ATP and NADPH

4. How are the products of the light-dependent reaction important to the light-independent reaction?

ATP and NADPH are oxidized to provide the energy to reduce glycerate-3-phosphate into triose phosphate.  Without ATP and NADPH, the light-independent reaction will not occur.

Warm Up 10/28

1. Why do plants need both mitochondria and chloroplasts?
Plants need both chloroplasts and mitochondria because plants produce convert solar energy into chemical energy but they also need energy for growth and metabolic processes.  The conversion of solar energy into chemical energy in the form of glucose occurs during photosynthesis, which occurs in the chloroplasts.  Plants, just like all other living organisms, need energy.  This energy is produced during cellular respiration, which takes place in the mitochondria.

2. You have a leaf from each of two very different plants.  One leaf has more pigments than the other.  Which leaf would have the greater photosynthetic rate, assuming all affecting factors are equal.
The leaf with more pigments will have a greater photosynthetic rate, because more pigments allow more photosynthesis to occur, increasing the rate of photosynthesis.

3. Outline the effect of temperature, light intensity, and carbon dioxide concentration on the rate of photosynthesis.
As temperature increases, the rate of photosynthesis increases more and more steeply until it reaches an optimum temperature.  Above the optimum temperature, the rate of photosynthesis falls steeply.  At low to medium light intensities, the rate of photosynthesis is dirtily proportional to light intensity.  At high light intensities, the rate of photosynthesis plateaus.  At very low carbon dioxide concentrations, no photosynthesis occurs.  At low to fairly high carbon dioxide concentrations, the rate of photosynthesis is positively correlated with CO2 concentration.  At very high CO2 concentrations, the rate of photosynthesis plateaus.

Warm Up 10/14

1. Describe chemiosmosis as it relates to oxidative phosphorylation.
Chemiosmosis occurs at the inner membrane.  It involves the movement of protons (hydrogen ions) across a membrane (down its concentration gradient) to provide energy so that oxidative phosphorylation (ATP synthesis) can occur.  Hydrogen ions build up in the intermembrane space, creating a high concentration and a proton force.  This allows the movement of hydrogen ions through the enzyme ATP synthase.  The enzyme uses energy from the hydrogen flow to couple phosphate with ADP to produce ATP.

2. What is the role of oxygen in the ETC?
Oxygen is the final electron acceptor in the ETC.  Oxygen also accepts hydrogen ions to form water.  If oxygen is not available, the electron flow along the ETC stops, but glycolysis can still occur.

Warm Up 10/10

1. What is denaturation?
Denaturation is changing the structure of an enzyme (or other protein) so that it can no longer carry out its function.  Enzymes are denatured at very high temperatures and extreme pHs.

2.What does a graph for increasing substrate concentration of an enzyme look like? Describe it...
At low substrate concentrations, enzyme activity increases steeply as substrate concentration increases.  Random collisions between substrate and active site happen more frequently with higher substrate concentrations.  At high substrate concentrations, most of the active sites are occupied, so raising the substrate concentration has little effect on enzyme activity and the graph begins to plateau.

3. Compare competitive/non-competitive inhibition.
Competitive inhibition is when an inhibitor similar to the substrate binds to the active site, blocking the active site and preventing substrate binding.  The presence of a competitive inhibitor reduces the rate of reaction, but increasing the concentration of substrate can reduce the effect of the inhibitor.
Non-competitive inhibition is when an inhibitor not similar to the substrate binds to a site other than the active site (allosteric site) and changes the active site, preventing the substrate from binding to the active site.

4. Discuss lock-and-key vs. induced-fit.
The lock-and-key hypothesis states that enzymes are substrate-specific and the substrate fitting in the enzyme is like a key fitting in a lock.  This does not explain the broad specificity of some enzymes.  The induced-fit model helps explain how substrates change to fit an enzyme.  The induced-fit model says the active site may undergo conformational change to better fit the substrate.  The active site may change shape or configuration.  This change is induced by the substrate.

Warm Up - 10/4

Explain the process of aerobic cellular respiration.

Aerobic cellular respiration begins with glycolysis, the conversion of a molecule of glucose into two pyruvate molecules and two ATP molecules.  The first step of glycolysis is phosphorylation, where hexokinase and then phosphofructokinase add two phosphates to the glucose molecule, creating a hexose biphosphate.  The second step of glycolysis is lysis, the splitting of hexose biphosphate into two 3-carbon sugars by the enzyme aldolase.  These 3-carbon sugars are called G3P, PGAL, or triose phosphate.  The third step of glycolysis is oxidation where each G3P molecule loses two atoms of hydrogen by converting NAD+ into NADH + H+.  The creation of NADH releases energy to add a phosphate group to both G3Ps.  The resulting 3-carbon molecules with extra phosphate are called PGA.  In the fourth step of glycolysis, ATP formation, ADP undergoes substrate-level phosphorylation, producing ATP.  The remaining carbon and phosphate molecule is turned into pyruvate.

Fatty acids can also be used to bypass glycolysis.  Since fatty acids are a long chain of carbon molecules, coenzyme CoA oxidizes this chain and breaks it down, making Acetyl CoA with two carbons.

After glycolysis, the link reaction occurs.  Pyruvate is decarboxylated to form an acetyl group (2C) and CO2 is released.  Then the pyruvate is oxidized and NAD+ accepts a hydrogen to form NADH + H+.  The acetyl group combines with a coenzyme (CoA) to form Acetyl CoA, which enters the Krebs cycle.

In the Krebs Cycle, Acetyl CoA combines the acetyl group (2C) with oxaloacetate (4C), forming citrate (6C).  Citrate undergoes oxidative decarboxylation, converting NAD+ into NADH + H+ and producing CO2.  This forms a 5 carbon molecule, which undergoes oxidative decarboxylation again, making a 4 carbon molecule.  Substrate level phosphorylation occurs, adding a phosphate to ADP to create ATP.  FAD2 is converted into FADH2 and NAD+ is converted into NADH + H+.  After these reactions, the 4 carbon molecule is converted into oxaloacetate, which is used in the next turn of the Krebs Cycle.  The products of the Krebs Cycle are used to power the Electron Transport Chain.

Warm Up 2 - 10/2/13

1(a)(i) Mitochondria
      (ii) The structure labelled I above is the cristae.  Cristae folds are shelf-like projections of the inner membrane which increase the surface area available for oxidative phosphorylation.

Warm Up 1 - Cellular Respiration - 9/30/13

1. What is oxidative decarboxylation?
Oxidative decarboxylation is the removal of a hydrogen atom (oxidation) and the removal of carbon dioxide (decarboxylation).

2. What is substrate-level phosphorylation?
Substrate-level phosphorylation is a way of producing ATP by adding a phosphate to ADP.

3. What are the products of Krebs? Glycolysis?
The Krebs cycle produces 2 ATP, 6 NADH, 2 FADH2, Oxaloacetate and 4 CO2 per molecule of glucose.  These products are the result of two turns of the Krebs cycle.  Glycolysis produces 4 ATP (for a net gain of 2 since 2 ATP are used up), 2 NADH molecules, and 2 pyruvate (3C) molecules.