Warm Up 2/24

11. Identify all of the cellular and molecular layers that a molecule would have to pass through in order to be ultrafiltered and then reabsorbed into the bloodstream within a single nephron

·      Capillary bed called the glomerulus

·      Basement membrane

·      Podocyte

·      Filtrate

·      Outer lining of Bowman’s capsule

22. Some, but not all, substances can be tested for in the urine of a patient.  Why do some substances present in the bloodstream show up in urine samples and others do not?

·      Not all substances are filtered out and exit the body through urine.

33.    

a.     Predict the relative amount of ADH produced in a person who has been drinking lots of water and has not been exercising recently.  Justify your prediction.

·      ADH levels will be low because the intake of water and lack of exercise results in proper water levels in the body

b.     Predict the relative amount of ADH produced in a person who has been exercising vigorously and has not had chance to completely hydrate themselves.  Justify your prediction.

·      ADH levels will be high because the lack of water in the bloodstream triggers a release of ADH

44. The filtering action of the kidneys does not eliminate urea from the bloodstream.  Why is complete elimination of urea not necessary?

·      Urea is non-toxic, so not all of it needs to be eliminated.  Urea can be reabsorbed by the body, broken down, and used for other things.

Warm Up 2/12

1. Describe the steps of blood clotting
Clotting is a mechanism that prevents the loss of blood from broken vessels.  Damaged cells and platelets release helical signals called clotting factors which trigger a coagulation cascade.  Clotting factors convert the inactive zymogen prothrombin into the activated enzyme thrombin.  Thrombin catalyses the conversion of the soluble plasma protein fibrinogen into an insoluble form (fibrin).  Fibrin forms an insoluble mesh of fibers that trap blood cells at the site of damage.  Clotting factors also cause platelets to become sticky, which then adhere to the damaged region to form a solid plug called a clot.  The clot prevents further blood loss and blocks entry to foreign pathogens

2. Outline the immune response to an antigen
An antigen is a virus or bacteria.  Surface molecules called epitopes are recognized by leukocytes and trigger the immune response.  The antigens' epitope is detected by a large white blood cell called a macrophage.  The macrophage engulfs the antigen, digests it and then incorporates the epitope into its own cell membrane.  Then the macrophage migrates to the lymph nodes.  The macrophage presents the antigens' epitope and selects a B lymphocyte that has a complementary surface epitope.  As the macrophage may have a number of molecules in its membrane, it is possible for multiple B Lymphocytes to be selected (polyclonal selection).  Corresponding B or T lymphocytes are selected.  A Helpter T cell with the corresponding epitope is also selected.  The Helper T cell works with the B and T lymphocytes to stimulate the B and T lymphocyte to divide rapidly.  B lymphocytes rapidly divide to form a clone of Plasma cells and B memory cells.  The T lymphocytes also rapidly clone and produce other cells called T Killer cells and T Killer memory cells.

3. Discuss Active vs. Passive Immunity
Active immunity is immunity due to the production of antibodies by the organism after the body's defense mechanisms are stimulated by antigens.  Passive immunity is immunity due to the acquisition of antibodies from another organism in which active immunity has been stimulated.  This passive acquisition of antibodies can be achieved via the placenta, colostrum or by injection (ex: blood transfusions).

Warm Up 2/4 - Genetics

1. Homozygous - same alleles
2. Phenotype - physical characteristics
3. Carrier - has one allele but it doesn't affect the person, have ability to pass it on
4. Can a mom with A blood and a dad with B blood have children with O blood?
Yes
5. Can a mom with hemophilia and a dad that is normal have children with hemophilia?
Yes

Warm Up 1/31

List the steps for germination of a starchy seed. (outline)
1. Imbibition - absorption of water due to low water potential in the dry seed, which rehydrates living cells in the seed and allows the cells to become metabolically active; embryo increases respiration
2. Formation of gibberellin - gibberellic acid is released
3. GA stimulates amylase, a digestive enzyme that breaks down starch, to be produced
4. Digestion of starch into maltose in the food stores of the seed
5. Maltose is transported from food stores to the growth regions of the seedlings, including the embryo root and the shoot tissue.
6. Maltose is converted to glucose, which is used in aerobic cell respiration as a source of energy, or is used to synthesize cellulose or other substances needed for growth.

2. Explain how flowering is controlled in long-day and short-day plants, including the role of phytochrome.

• Flowering is controlled by phytochrome, which is affected by light (photoperiodicity)
• Phytochrome exists in two forms:
  •  A red (P_r) form absorbs red light (~660 nm) and is converted into a far red form (P_fr)
  • A far red (P_fr) form absorbs far red light (~730 nm) and is converted into a red form (P_r)
• The P_fr form is the active form of phytochrome, while the P_r form is the inactive form of phytochrome
• Sunlight contains more red light, so the P_fr form is predominant during the day, with the gradual reversion to the P_r form occurring at night
• In long day plants, the active P_fr form is a promoter of flowering and so flowering is induced when the night period is less than a critical length and P_fr levels are high
• In short day plants, the active P_fr form is an inhibitor of flowering and so flowering is induced when the night period is greater than a critical length and P_fr levels are low

Warm Up 1/27 - Plant Science

1. Of what values are tubers, type of stem modification, to the survival of plant species?
Tubers allow plants to store more carbohydrates.

2. When moving a plant from one place to another, why is it important to leave original soil around the roots?
The original soil should be left around the roots to ensure that all root hairs are transferred to the new location.  The original soil also contains the necessary nutrients for the new plant.  With some of the old soil, plants will ensure that they have enough of the necessary nutrients and can slowly adjust to the nutrients found in the new soil.  A direct change to the new soil could cause the plant go into shock and die.

3. What is the usual cause when a plant wilts?
Plants wilt when the water content within their cells becomes too low.  This often happens when the solid around a plant gets too dry or too salty, as this prevents water from moving into the plant.

4. Explain when a seed would be a sink and when it would be a source.
A seed would be a sink when it uses or stores sugar.  A seed would be a source when it produces sugar.

5. Why is it necessary that veins are relatively close together in leaves?
Phloem is located in all of the veins of the leaf, so veins help transport minerals and nutrients.  The distance between veins shows how well the veins are supplying nutrients to the leaf.  The more veins there are, the closer together the veins will be, and the more nutrients the leaf will be supplied with.

Warm Up 1/14

1. What is the importance of standard deviation with regards to the mean?
Two sets of data can have the same mean but be totally different.  The standard deviation shows the range of the data and helps more accurately compare two sets of data with the same mean.

2. What do error bars indicate?
Error bars are a graphical representation of the variability of data that show the degree of uncertainty relative to each data marker in a series.  They show the spread of measurements around a central tendency on a graph.  Error bars usually show the range of data, standard deviation or the 95% confidence intervals.

3. What percentage of values fall within 1 standard deviation of the mean?  And 2?
68% of the values fall within one standard deviation of the mean.  95% fall within two standard deviations.

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.