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The “Dirty Dozen” Objectives and Results

Laboratory 1: Diffusion and Osmosis

Overview

In this lab you will investigate the movement of water across semi-permeable membranes. You will also examine the effect of solute
concentrations on water potential as it relates to living plant tissues.

Objectives

• Describe the mechanisms of diffusion and osmosis
• Describe how solute size and molar concentration affect the process of diffusion through a selectively permeable membrane
• Design an experiment to demonstrate and measure water potential
• Relate osmotic potential to solute concentration and water potential (Mr. Knight’s Notes on Water Potential)
• Describe how pressure affects the water potential of a solution
• Describe the effects of water gain or loss in animal or plant cells
• Calculate the water potential of living plant cells from experimental data

Results

When a solution such as that inside a potato cell is separated from pure water by a selectively permeable membrane water will move
by osmosis from the surrounding area where the water potential is higher into the cell where water potential is lower due to the
presence of solute. The movement of water into the cell, causes the cell to swell and the cell membrane pushes against the cell wall
to produce an increase in pressure (turgor). This process will continue until the water potential of the cell equals the water
potential of the pure water outside the cell. At this point, a dynamic equilibrium is reached and net water movement will cease.

Sample Multiple – Choice Questions

1. A dialysis bag is filled with a 3% starch solution. The bag is immersed in a beaker of water containing a 1% IKI solution.
All of the following observations are correct EXCEPT:

A. When the bag is first placed in the beaker, the water potential inside the bag is negative.

B. When the bag is first placed in the beaker, the solution in the beaker is yellow brown.

C. The starch solution inside the bag is hypertonic relative to the solution in the beaker.

D. After fifteen minutes, the solution in the bag turns blue.

E. After fifteen minutes, the mass of the dialysis bag has decreased.

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Laboratory 2: Enzyme Catalysis

Overview

In this lab you will measure the rate of a reaction in the presence and absence of a catalyst. The catalyst, (catalase), is an enzyme in cells that
catalyzes the breakdown of toxic H202.

Objectives

• Graph data from an enzyme experiment
• Determine the rates of enzymatically catalyzed reactions
• Discuss the method for determining enzyme activity
• Discuss the relationship between dependent and independent variables
• Discuss the effect of initial reaction rates produced by changes in temperature, pH, enzyme concentrations, and substrate concentrations
• Design an experiment to measure the effect of enzyme activity produced by changes in temperature, pH, enzyme concentrations,
  and substrate concentrations

Results

In the first few minutes of an enzymatic reaction, the number of substrate molecules is usually so large compared to the number of
enzyme molecules that changing the substrate concentration does not (for a short period at least) affect the number of successful collisions
between substrate and enzyme. During this early period, the enzyme is acting on substrate molecules at a constant rate. The slope of the
graph line during this early period is called the initial velocity of the reaction. The initial velocity, or rate, of any enzyme catalyzed reaction
is determined by the characteristics of the enzyme molecule. It is always the same for an enzyme and its substrate as long as temperature
and pH are constant and substrate is present in excess. Also, in this experiment the disappearance of the substrate, H202 is essential in this
reaction. Once all the H202 has reacted, any more KMnO4 added will be in excess an will not be decomposed.

Sample Multiple – Choice Questions

1. Which of the following is LEAST likely to increase the forward rate of an enzyme-mediated reaction?

1. 1. An increase in the substrate concentration
   2. An increase in the enzyme concentration
   3. An increase in the product concentration
   4. An increase in pH
   5. An increase in the temperature

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Laboratory 3: Mitosis and Meiosis

Overview

In this lab you will study plant mitosis using prepared slides of onion root tips and will calculate the relative period of the phases of mitosis
in the meristem of root tissue. You will also study the crossing over and recombination that occurs during meiosis.

Objectives

• Compare the events of mitosis in plant cells with those of animal cells
• Demonstrate the procedure to stain tissue for the identification of cells in the various stages of mitosis
• Calculate the relative duration of mitosis

Results

The relative length of mitotic stages are, 53.4% of prophase, 17.4% of metaphase, 16.8% of anaphase and 12.4% of telophase. Meiosis
is important for sexual reproduction because it reduces the chromosome number by half and it also results in new combinations of genes
through independent assortment and crossing over, followed by the random fertilization of eggs by sperm.

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Laboratory 4: Plant Pigments and Photosynthesis

Overview

In this lab you will separate plant pigments using paper chromatography. You will also measure the rate of photosynthesis in isolated
chloroplasts using a measurement technique that involves the reduction of the dye, DPIP. The transfer of electrons during the
light-dependent reactions of photosynthesis reduces DPIP and changes its color from blue to colorless.

Objectives

• Understand the principles of chromatography
• Calculate Rf values
• Design an experiment in which chromatography is used as a separation technique
• Describe a technique for determining photosynthetic rate
• Understand the relationship between dependent and independent variables
• Describe how light intensity, light wavelength, and temperature can affect photosynthesis
• Design an experiment to measure how light intensity, light wavelength, and temperature can affect photosynthetic rates

Results

The solvent moves up the paper by capillary action, which occurs as a result of the attraction of solvent molecules to the paper and the
attraction of solvent molecules to one another. As the solvent moves up the paper, it carries along any substances dissolved in it, in
this case pigments. The pigments are carried along at different rates because they are not equally soluble in the solvent and because
they are attracted, to different degrees, to the cellulose in the paper through the formation of hydrogen bonds. Also, as the DPIP is
reduced and becomes colorless, the resultant increase in light transmittance is measured over a time course using a spectrophotometer.

Sample Multiple – Choice Questions

1. In a paper chromatography procedure, molecules with which of the following characteristics migrate the fastest up the
chromatography paper?

A. High solubility in solvent and weak hydrogen bonding to cellulose.

B. High solubility in solvent and strong hydrogen bonding to cellulose.

C. Low solubility in solvent and strong hydrogen bonding to cellulose.

D. Low solubility in solvent and weak hydrogen bonding to cellulose.

E. Insoluble in solvent.

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Laboratory 5: Cell Respiration

O2 Graph and CO2 Graph

Overview

In this lab you will measure oxygen consumption during respiration as a change in gas volume in germinating and nongerminating peas
at two different temperatures.

Objectives

• Discuss the gas laws as they apply to the function of a respirometer
• Interpret data related to the effects of temperature on cell respiration
• Interpret data related to the effects of germination or nongermination on cell respiration
• Explain or determine the significance of a control
• Explain the relationship between dependent and independent variables
• Calculate a rate of cell respiration by utilizing graphed data
• Design an experiment to use a respirometer to measure cellular respiration

Results

Germinating peas respire and need to consume oxygen in order to continue the growing process. Pea seeds are nongerminating and do not
respire actively. These seeds are no longer the site of growth and thus do not need oxygen for growth. In consideration to temperature,
at higher temperatures more oxygen is consumed which means more respiration is occurring. 686 kilocalories are released during respiration.
When temperature decreases molecular motion slows down and respiration decreases because less energy is made available.

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Laboratory 6: Molecular Biology

Overview

In this lab you will investigate some basic principles of genetic engineering. Plasmids containing specific fragments of foreign DNA will be
used to transform E. coli cells, conferring antibiotic (ampicillin) resistance and the Lac + phenotype (ability to metabolize lactose).
Restriction enzyme digests of phage lambda DNA will also be used to demonstrate techniques for separating and identifying DNA fragments
using gel electrophoresis.

Objectives

• Discuss the principles of bacterial transformation
• Describe how to prepare competent E. coli cells
• Discuss the mechanisms of gene transfer using plasmid vectors
• Discuss the transfer of antibiotic resistance genes and tell how to select positively for transformed cells that are antibiotic resistant
• Discuss the mechanisms of action for restriction endonucleases
• Discuss how a plasmid can be engineered to include a piece of foreign DNA that alters the phenotype of the transformed cells
• Understand and be able to explain the principles of electrophoresis as they pertain to separating and identifying DNA fragments

Results

Bacterial Transformation-Ampicillin resistance: In this exercise, we will introduce competent E. Coli cells to take up the plasmid pAMP
which contains a gene for ampicllin resistance. Normally, E. Coli cells are destroyed by the antibiotic ampicillin, but E. Coli cells that
have been transformed will be able to grow on agar plates containing ampicillin. Thus, we can select for transformants; those cells
that are not transformed will be killed by ampicillin; those that have been transformed will survive.

Restriction Enzyme Cleavage of DNA: Restriction endonuclease recognizes specific DNA sequences in double-stranded DNA and digests
the DNA at these sites. The result is the production of fragments of DNA of various lengths corresponding to the distance between
identical DNA sequences within the chromosome. By taking DNA fragments and systematically reinserting the fragments into an
organism with minimal genetic material, it is possible to determine the function of particular gene sequences

Electrophoresis: fragments of DNA can be separated by gel electrophoresis when any molecule enters the electrical field, the
mobility or speed at which it will move is influenced by the charge (negative charges travel to positive/top pole of gel), the density
of the molecule, (the smaller the molecule, the faster it travels), the strength of the electrical field, and the density of the medium (gel)
which it is migrating.

Sample Multiple – Choice Questions

1. After growth on ampicillin to select bacteria transformed with a mixture of recombinant DNA containing plasmids, you must identify
   a clone containing a specific gene sequence. You would:

1. Blot transfer clones to membranes, and screen using a radioactive probe complementary to the gene.
2. Re-grow bacteria in ampicillin where only transformants containing the gene of interest can grow.
3. Culture bacteria in both ampicillin and tetracycline to select for bacteria containing the gene of interest.
4. Digest DNA from the plasmid to isolate the gene fragment.
5. Do a restriction map of plasmid DNA to identify the correct clone

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Laboratory 7: Genetics Of Drosophila

Overview

In this lab you will use the fruit fly Drosophila melanogaster to do genetic crosses. You will learn how to collect and manipulate fruit flies,
collect data from F1 and F2 generations, and analyze the results from a monohybrid, dihybrid or sex link cross.

Objectives

• Conduct a genetics experiment for a number of generations
• Compare predicted results with actual results
• Explain the importance of Chi-Square analysis
• Design genetic crosses in an experiment to illustrate independent assortment and sex linkage
• Discuss the life cycle of the fruit fly, recognize the sex of fruit flies, and recognize several types of classic mutations

Results

From this lab, you will be able to find genotypes and phenotypic expression within a fruit fly. Also, recessive genes and mutations will be
revealed as the student crosses a variety of Drosophila alleles. For example, if a female carrier for an x-linked, recessive trait, was crossed
with a male without the recessive trait the results would be:

½ males with x-linked trait ½ males without

½ females carriers ½ females without

0 females express sex linked traits

Sample Multiple – Choice Questions

7. A population consists of 20 individuals of which 64% are homozygous dominant for a particular trait and the remaining individuals
   are all heterozygous. All of the following can explain the situation except

1. Genetic drift is occurring
2. The recessive allele is deleterious
3. All homozygous recessive individuals emigrate
4. The population is very small
5. Only heterozygous individuals mate

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Laboratory 8: Population Genetics and Evolution

Overview

In this lab you will learn about Hardy-Weinberg law of genetic equilibrium and study the relationship between evolution and changes
in allele frequency by using your class as a sample population.

Objectives

• Calculate allele and genotype frequencies using the Hardy-Weinberg theorem
• Discuss the effect of natural selection on allelic frequencies
• Explain and predict the effect of allelic frequencies of selection against the homozygous recessive
• Discuss the relationship between evolution and changes in allele frequencies, as measured by deviations from the Hardy-Weinberg
  law of genetic equilibrium

Results

Assuming that Hardy-Weinberg equilibrium is maintained allele and genotype frequencies should remain constant from generation to
generation. For this to happen the five following situations must all occur:

1. Population is very large. The effects of chance on changes in allele frequencies is thereby greatly reduced.
2. Individuals show no mating preference, i.e. random mating.
3. There is no mutation of alleles.
4. No differential migration occurs, (no immigration or emigration).
5. All genotypes have an equal chance of surviving and reproducing, i.e. there is no natural selection.

In humans, several genetic diseases have been well characterized. Some of these diseases are controlled by a single allele where the
homozygous recessive genotype has a high probability of not reaching reproductive maturity. If this were to occur both the homozygous
dominant and heterozygous individuals will survive while the homozygous recessive will become extinct.

Sample Multiple – Choice Questions

1. Which of the following generates the formation of adaptations?

1. Genetic drift
2. Mutations
3. Gene flow
4. Sexual reproduction
5. Natural selection

2. All of the following are examples of evolution, except:

1. Mutations in an individual
2. Changes in an allele frequency in a population
3. Changes in an allele frequency in a species
4. Divergence of a species into two species
5. Adaptive Radiation

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Laboratory 9: Transpiration

Overview

In this lab you will apply what you learned about water potential from Lab 1 (Diffusion and Osmosis) to the movement of water within
the plant. You will study the organization of the plant stem as it relates to these processes by observing sections of fresh tissue.

Objectives

• Describe how differences in water potential affect the transport of water from roots to stems to leaves
• Relate transpiration to the overall process of water transport in plants
• Discuss the importance of properties of water – including hydrogen bonding, adhesion, and cohesion – to the transport of water in plants
• Quantitatively demonstrate the effects of different environmental conditions on the rate of transpiration in plants
• Identify the vascular tissues of the plant stem and describe their functions

Results

Conditions that cause a decreased rate of water loss from leaves result in a decreased water potential gradient from stem to leaf and
therefore in a decreased rate of water movement up the stem to the leaves. Conditions that cause an increased rate of water loss from
leaves result in an increase in the water potential gradient from stem to leaf and therefore in an increase in the rate of water movement
up the stem to the leaves.

1. 1. Room Conditions

1. • • 1. When you expose a plant to room conditions nothing is supposed to happen. The reasoning for this is room conditions
          don’t cause drastic changes in the plants environment for major transpiration or even water gain to occur. The plant
          under room conditions is considered to be your control

1. 1. Floodlight

1. • • 1. When light is absorbed by the leaf, some of the light energy is converted to heat and remember that transpiration
          rate increases with temperature. We learned in Unit One of the Campbell’s edition that when the temperature of
          liquid water rises, kinetic energy of the water molecules increases. As a result, the rate at which liquid water is
          converted to water vapor increases. When the water is turned into water vapor, it easily passes out through stomata
          out into the outer atmosphere.
       2. The floodlight is an example of a plant near the sun (which is why the plant is one meter away from the light) Due
          to the aforementioned properties of plants you should see a loss of water

1. 1. Fan

1. • • 1. An increase in wind speed results in an increase in the rate of leaf water loss because increased wind decreases
          the boundary layer of still air at the leaf surface. This boundary layer acts to slow leaf water loss. Increased
          wind also causes the rapid removal of evaporating water molecules from the leaf surface. This results in
          a low water potential in the air immediately and the water level should drop.

1. 1. Mist

1. • • 1. Increased humidity in the air surrounding the leaf decreases the water potential gradient between the saturated
          air in the leaf air spaces and the air surrounding the leaf, resulting in a decreased rate of leaf water loss.
          However, when the humidity of the air surrounding the leaf if very low, the water potential of the air is low.
          Therefore, the water potential gradient between the air spaces of the leaf and the surrounding air is high, and
          the rate of leaf water loss increases.

Or maybe this explanation is better

1. • • 1. When there is a great amount of humidity, transpiration decreases because of water potential. When the humidity is
          at a low or normal, the mesophyll cells in the plant are much higher in water potential than the relatively drier
          surrounding air. Due to the properties of water potential, which states that water tends to evaporate from the leaf
          surface moving from an area of higher water potential to an area of lower water potential, transpiration occurs.
          But, because of the high humidity, the surrounding air has a higher water potential than the mesophyll cells and water
          loss is at a minimum.

Adaptations to reduce leaf water loss include a reduced number of stomates, an increase in the thickness of the leaf cuticle, a decrease
in leaf surface area, and adaptations that decrease air movements around stomates, such as dense hairs and sunken stomates. Because leaves
are all different in size, reporting the water loss without considering a unit area would provide non-comparable data.

Sample Multiple – Choice Questions

9. Which of the following series of terms correctly indicates the gradient of water potential from lowest water potential to highest
   water potential?

1. 1. Air, leaf, stem, root, soil
   2. Soil, root, stem, leaf, air
   3. Root, leaf, stem, air, soil
   4. Air, soil, root, leaf, stem
   5. Stem, leaf, root, soil, air

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Laboratory 10: Physiology of the Circulatory System

Overview

You will learn how to measure blood pressure and measure pulse rate under different physiological conditions: standing, reclining,
after the baroreceptor reflex, and during and immediately after exercise. The blood pressure and pulse rate will be analyzed and
related to a relative fitness index. You will also measure the effect of temperature on the heart rate of the Daphnia magna, and
calculate a Q10 for the relationship between temperature and heart rate.

Objectives

• Measure pulse rate
• Measure blood pressure
• Describe the relationship between the changes in heart rate and blood pressure relative to changes in body position
• Describe the relationship between changes in heart rate and exercise
• Determine the “fitness index” for an adult human
• Perform statistical analysis on class data
• Define Q10
• Determine the Q10 of heart rate in a living organism such as Daphnia

Results

The sphygmomanometer measures the blood pressure. The blood pressure cuff is inflated so that blood flow stops to through the brachial
artery in the upper arm. A stethoscope is used to listen to blood flow entering the brachial artery. When blood first enters the artery,
snapping sounds called the sounds of Korotkoff are generated.

1. 1. Blood pressure and heart rate increase when you move from a reclining to a standing position counteracting gravitational pull
      on the blood
   2. Elevated arterial blood pressure indicates increased arterial resistance to blood flow
   3. Fit individuals can pump a larger volume of blood with each contraction and deliver more oxygen to muscle tissue than the
      hearts of unfit individuals. As a result, blood pressure and heart rate increases are smaller for fit individuals, and the time
      required to return to normal conditions is shorter for fit individuals than unfit individuals.
   4. For the Daphnia, remember that ectothermic animals use behavior to regulate their body temperatures and that Q10
      cannot be determined for endothermic animals because body temperatures remain constant regardless of environmental temperatures.

Sample Multiple – Choice Questions

9. A Q10 value of 3 in an ectothermic animal means that the metabolic rate

1. 1. Triples when body temperature triples
   2. Triples when body temperature increases by 10° C
   3. Doubles when the body temperature increases by 3° C
   4. Doubles when the body temperature increases by 10° C
   5. Triples when the body temperature decreases by 10° C

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Laboratory 11: Behavior: Habitat Selection

Overview

In this lab, you will examine the habitat preferences of the brine shrimp, Artemia. You will use controlled experimentation to determine
the thermal, pH, and light environments selected by Artemia. Based on your experience with this lab, you will design an experiment that
could be used to survey other variables and other organisms.

Objectives

• Describe the relationship between dependent and independent variables
• Discuss the value of comparing experimental results with control results
• Graph an interpret histogram data
• Measure the volumes, distances, and temperature using metric scales
• Design and conduct an experiment to measure the effect of environmental variables on habitat selection

Results

When conducting this experiment, a couple of things should be understood.

1. 1. Artemia are able to survive in a wide range of salty environments by they tend to live in really salty environments.
   2. Three variables are tested: light, temperature and pH. The control is exposed to room light, room temperature, and neutral pH is
      also prepared. For each of the variables, a gradient is established providing a continuous variation from weak to strong intensities. Each habitat variable is applied to a 100 cm clear plastic tube filled with water and brine shrimp. The ends of the plastic tube
      possess extremes of a continuous variation in the habitat property. After the brine shrimp have been exposed to the habitat
      condition, they are divide into four groups by tightening three clamps around the tube at equal intervals. Each group is individually removed from the tube at equal intervals. Each group is individually removed from the tube and the number of live shrimp is counted.
   3. The histograms are prepared showing the number of brine shrimp in each of the four intensities. A histogram is prepared for each
      of the three variables and the control. From the data in the histograms, conclusions cab be made describing the habitat preferences
      of the brine shrimp

Sample Multiple – Choice Questions

9. Artemia brine shrimp are rarely found in bodies of water with salt concentrations below 5%.

1. 1. This is probably because the brine shrimp prefer low levels of salt concentration
   2. The brine shrimp prefer high levels of salt concentration
   3. The brine shrimp cannot survive in fresh water
   4. The brine shrimp cannot survive in temperatures found in bodies of water with fresh water or water with low salinity
   5. Predators of the brine shrimp are common in fresh water and water with low salinity

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Laboratory 12: Dissolved Oxygen and Aquatic Primary Productivity

Overview

You will measure and analyze the dissolved oxygen concentration in water samples using the Winkler technique. You will also measure and
analyze the primary productivity of natural waters or lab cultures

Objectives

• Describe the physiological importance of carbon and oxygen in an ecosystem
• Understand the physical and biological factors that affect the solubility of dissolved gases in aquatic ecosystems
• Describe a technique for measuring dissolved oxygen
• Define primary productivity
• Describe the relationship between dissolved oxygen and the processes of photosynthesis and respiration as they affect primary
  productivity in an ecosystem
• Design an experiment to measure primary productivity in an aquatic ecosystem
• Understand the effect of light and nutrients on photosynthesis

Results

The amount of oxygen dissolved in natural water samples is measured and analyzed to determine the primary productivity of the sample.
The amount of dissolved oxygen is dependent upon many factors.

A. Temperature

B. Salinity

C. Photosynthesis

D. Respiration

Primary productivity is a measure of the amount of biomass produced by autotrophs through photosynthesis per unit time. It can be
examined by the following factors:

1. 1. Gross Primary Productivity
   2. Net Primary Productivity
   3. Respiratory Rate

These determine primary productivity:

1. 1. The Winkler Method—this is use to measure dissolved oxygen using a titration technique. Titration is the process of adding a substance of known concentration to a solution containing a substance of unknown concentration until a specific reaction is completed and a color change occurs
   2. The light and dark bottle method
   3. Initial bottle
   4. Light bottle
   5. Dark bottle

Sample Multiple – Choice Questions

12. The net primary productivity for a temperate forest was measured at 2000 mg C/m2/day. The respiratory rate of the community
    was determined to be 1000 mg C/m2/day. The gross primary productivity for this community is

1. 1. 1000 mg C/m2/day
   2. 2000 mg C/m2/day
   3. 3000 mg C/m2/day
   4. 4000 mg C/m2/day
   5. 5000 mg C/m2/day

Answers

2. (C) Since enzyme-mediated reactions are reversible (they convert product back to substrate), increasing the concentration of the
product will slow the forward direction of the reaction and accelerate the reverse reaction. Conversely, and increase in the
substrate concentration will increase the forward rate of the reaction. Increasing the enzyme concentration will not slow the
reaction rate but may increase it if the substrate concentration is high enough to utilize additional enzyme. An increase
in pH or temperature may change the rate of reaction, but the nature of the enzyme must be known in order to determine whether
the rate is increased or decreased

9. (A) Water potential is highest in soil, decreases from root to leaf, and is lowest in the air. Water moves from the soil into the roots
   and through the plant an transpires from the leaf because water moves from the area of greatest water potential to the area of lowest
   water potential
10. (B) The Q10 is the ratio of the metabolic rate at one temperature to the metabolic rate at a temperature 10° colder. A Q10 equal
    to 3 indicates that he metabolic rate triples when the body temperature increases by 10° C
11. (E) The predators of brine shrimp cannot survive in bodies of water with high concentrations of sale. Thus, brine shrimp survive in
    bodies of water with a high salt concentration because predators are absent. In waters with low concentrations of salt, predators
    eliminate the brine shrimp
12. (C) The gross primary productivity is the sum of the net primary productivity and the respiratory rate.