Biology Lab Notes

LABORATORY ONE: Measures and Statistics

  • Introduction
    • Biological phenomena exhibit variation, posing challenges in scientific analysis due to their qualitative nature.
    • Statistics is essential for assessing biological variation and describing relationships among biological parameters.
    • Statistics is a mathematical branch that models data to decipher patterns and processes.
  • Assumptions of Statistics:
    • Objects measured are randomly selected and independent.
    • Variation among objects fits a normal curve.
    • Numbers used in statistics are gained by measuring.
  • Statistical techniques derive numerical values compared to standardized distributions.
    • This comparison helps determine if observed variation is due to a cause rather than randomness.
  • Lab Procedure
    • Divide into lab groups.
    • Collect two leaves from the same shrub (one from the top surface, one from deep inside).
    • Remove leaves with stems attached, then disconnect the leaf from the stem.
    • Weigh each leaf to the nearest one-thousandth of a gram and record the weight.
    • Approximate leaf surface area using the formula: (major\ axis/2)(minor\ axis/2)(\pi) = area

Chi Square

  • Chi-square (x^2) is used to measure the association between variables and perform probability-of-occurrence analysis.
  • Procedure
    • Use class data to fill in 2 x 2 tables.
    • Record number of leaves of above-average weight versus smaller than average for both surface and interior leaves.
    • Calculate totals for each row and column.
    • Calculate the Chi-Square value using the formula: x^2 = \sum \frac{(O-E)^2}{E}, where E = \frac{(row\ total \times column\ total)}{overall\ total}
    • Degrees of freedom for a 2 x 2 table = 1.
    • Significance value is determined using a probability < 0.05, which corresponds to a value of 3.84.
  • Example Data and Calculations
    • Mass average: 0.30
    • Area average
    • Calculated x^2 value for both mass and area: 1.06
  • Hypothesis
    • Leaves from the exterior would be bigger due to sunlight exposure.
  • Results
    • For both mass and area, the x^2 value (1.06) is less than 3.84, indicating no significant difference.
    • The hypothesis was not supported.
  • Possible Reasons for Hypothesis Not Being Supported
    • Different areas of the shrub might have similar sunlight exposure.
    • The area is better for determining the difference in leaf size because the more surface area the more sunlight exposure it has.

Probability and Correlation

  • Probabilities are represented by values from 0-1 (0 = never happens, 1 = must happen).
  • A probability of 0.5 does not guarantee an event will happen if repeated twice due to the margin of error.
  • Probability of Event A: P = \frac{number\ of\ successes}{number\ of\ trials}
  • Example: Whitetip Reef Shark Sightings
    • Data table showing sightings in reef vs. seagrass bed across different months.
  • Calculate the correlation (using Excel) between the number of sharks observed and the two sites.
    • Reef area: 0.945 (very strong positive relationship
    • Seagrass bed: -0.008 (negligible relationship).
  • Determine which categories these values fall into using the correlation table.
  • Check if probabilities and correlations support each other.
  • Determine which month is best for seeing whitetip reef sharks based on the highest probability.
  • Correlation Table Interpretation
    • +0.70 or higher: Very strong positive relationship
    • +0.40 to +0.69: Strong positive relationship
    • +0.30 to +0.39: Moderate positive relationship
    • +0.20 to +0.29: Weak positive relationship
    • +0.01 to +0.19: No or negligible relationship
    • 0: No relationship [zero correlation]
    • -0.01 to -0.19: No or negligible relationship
    • -0.20 to -0.29: Weak negative relationship
    • -0.30 to -0.39: Moderate negative relationship
    • -0.40 to -0.69: Strong negative relationship
    • -0.70 or higher: Very strong negative relationship

Laboratory Two: Organelles of Eukaryotic Cells

  • All eukaryotic cells contain organelles, small structures with specific functions.
  • Organelles contain proteins and enzymes essential for cell function.
  • Some organelles (nuclei, chloroplasts, mitochondria) are visible via light microscopy; others require staining or electron microscopy.
  • Membranes
    • Each organelle is enveloped by a membrane that controls molecule transport.
    • Semi-permeability gives membranes selective control, crucial for organelle function.
  • Part B: Functional Model of an Organelle
    1. Soak cellophane tubing in water.
    2. Smash and grind potato to create potato extract.
    3. Seal one end of the tubing with a twist-tie or clamp.
    4. Pour potato extract into the tube (() 2/3 full).
    5. Remove air from the tube and seal the top end.
    6. Describe the tube membrane.
    7. Prepare a beaker with 15 ml of H2O2 and 200 ml of water.
    8. Place the tubing into the beaker for 10 minutes.
    9. Record observations.
    10. Note that H2O2 breaks down into water and oxygen slowly under normal conditions, but catalase in potato cells speeds up the breakdown.
    11. Explain how catalase comes in contact with H2O2 through osmosis and the semi-permeable membrane.
    12. Potato cells contain catacholase that forms melanin in the presence of oxygen, causing color change when cut.
    13. Identify what the various components of the experiment represent (beaker as cell, water in beaker as cytoplasm, etc.).
  • Part C: Organelle Identification
    • Examine models of animal and plant cells.
    • Identify labeled parts of the animal cell.
    • Identify labeled parts of the plant cell.
    • Name a key organelle that plants have but animal cells lack (cell wall).
    • State the purpose of the cell wall (structure and shape).
    • Name organelles common between plants and animals (mitochondria, nucleus, lysosomes).

Laboratory Four: Photosynthesis

  • Photosynthesis is the method by which photoautotrophs produce energy in chloroplasts.
  • It's a primary production converting inorganic carbon (CO_2) into organic carbon, with oxygen as a waste product.
  • Aquatic plants absorb dissolved CO_2 from water, which acts as a carbon sink.
  • Increased dissolved CO2 increases water acidity, forming Carbonic acid: CO2 + H2O \rightleftharpoons H2CO_3
  • Indicators like bromothymol blue (blue at neutral pH, yellow in acid) show the presence/absence of a solution.
  • Spectrophotometers measure light absorption/transmission to quantify substances; higher CO_2 concentration results in lower light absorption (~625 nm).
  • Experiment Procedure
    1. Hypothesis
    2. Add 25 drops of bromothymol blue to 15 mL of water in a large test tube (Plant Tube 1 or P1). Bubble CO_2 until color change.
    3. Put a piece of aquatic plant into the water bubbled, and seal it completely with parafilm. Place the tube in the sunlight for 25 minutes.
    4. In another tube (No Plant Tube 1 or NP1), add 25 drops of bromothymol blue to 15 mL of water, and bubble the water until it changes color. Seal the top of the tube and place it in the sun.
    5. Prepare another tube (No Plant Tube 2 or NP2) in the same way as NP1, DO NOT bubble.
    6. Prepare one more tube (Plant Tube 2 or P2) the same as P1, but place this tube (with the plant in it) in the dark room.
    7. Place samples from your test tubes into the spectrophotometer as instructed. Record the absorption of each sample.
  • Results
    • P1: 0.116
    • P2: 0.134
    • NP1: 0.08
    • NP2: 0.767
  • Which plant tube used up the most CO_2? P2
  • Was the hypothesis supported by the results of the indicator test? no
  • Was it supported by the spectrophotometer test? no
  • What are the controls of this experiment? NP1 NP2