Exam Review Notes

Water Potential

• Net flow direction is determined by water potential.
• Dimensional analysis from chemistry is applicable: same units on top and bottom cancel out.
• If units work out, the setup is likely correct.
• Example:
  • Solute potential of sugar water: -0.15 \times 10 = -1.5 bars
  • Water potential equals solute potential (no pressure potential).
  • If cell and beaker have the same water potential, there is zero net flow.
  • Zero net flow does not mean no movement, but equal movement in both directions.

Hardy-Weinberg Equilibrium

• Two major equations:
  • p^2 + 2pq + q^2 = 1
  • p + q = 1
• p and q represent alleles; p^2, 2pq, and q^2 represent genotypes.
• p = frequency of the dominant allele.
• q = frequency of the recessive allele.
• p^2 = percentage of homozygous dominant genotype (e.g., big A big A).
• q^2 = percentage of homozygous recessive genotype (e.g., little a little a).
• 2pq = percentage of heterozygous genotype (e.g., big A little A).
• Example Problem:
  • Population of biology instructors: 396 red-sided, 557 tan-sided.
  • Red is totally recessive.
  • Calculate allele frequencies:
    • q^2 = percentage of recessive individuals = 396 / (396 + 557) = 396 / 953 = 0.42
    • q = \sqrt{0.42} = 0.64
    • p = 1 - q = 1 - 0.64 = 0.36
  • Expected genotype frequencies:
    • Big A Big A (p^2) = (0.36)^2 = 0.1296 \approx 0.13
    • Big A Little a (2pq) = 2 \times 0.36 \times 0.64 = 0.4608 \approx 0.45
    • Little a Little a (q^2) = 0.42
  • Number of heterozygous individuals:
    • Frequency of heterozygotes (0.45) multiplied by the total population (953) = 0.45 \times 953
  • Expected phenotype frequencies:
    • Tan: heterozygous and homozygous dominant.
    • Red: homozygous recessive.
• Hardy-Weinberg Conditions:
  • Population not evolving.
  • No natural selection.
  • Random mating.
  • No gene flow.
  • No genetic drift.
  • Allele frequencies remain constant.
• If Hardy-Weinberg conditions are met, allele frequencies remain the same.
  • New population size: 1,245
  • Number of red-sided individuals: 0.42 \times 1245

Simpson's Diversity Index

• Measure of community diversity.
• Community: group of different species living in the same area.
• Equation: D = 1 - \sum (n/N)^2
  • D = Simpson's diversity index.
  • n = number of organisms of a particular species.
  • N = total number of organisms of all species in the community.
• Example:
  • Calculate Community 1:
    • Species A: 5 individuals
    • Species B: 5 individuals
    • Species C: 5 individuals
    • Total individuals: 20
    • \sum (n/N)^2 = (5/20)^2 + (5/20)^2 + (5/20)^2 = 0.25
    • D = 1 - 0.25 = 0.75
• Key Points:
  • Calculate Community 2 result: 0.35
  • Higher diversity index = healthier ecosystem = more resilient community.
  • Major difference: even distribution of species in healthier communities.

Surface Area to Volume Ratio

• Larger surface area to volume ratio = healthier cell.
• Larger ratios found in smaller cells.
• Cube A: 12:1 ratio (healthiest).
• Cube B: 6:1 ratio.
• Cube C: 3:1 ratio.
• Cube D: 1.5:1 ratio.
• Larger surface area allows efficient export of waste and import of nutrients, oxygen, etc.

Cellular Respiration

• Main purpose: create ATP (mitochondria = powerhouse of the cell).
• ATP produced at each stage.
• 80-90% of ATP made during oxidative phosphorylation via the electron transport chain (ETC).
• Glycolysis and Krebs cycle produce ATP through substrate-level phosphorylation.
• Glycolysis: breakdown of glucose into pyruvate.
• Pyruvate enters mitochondria and becomes acetyl CoA.
• Each glucose molecule yields two acetyl CoA molecules; Krebs cycle runs twice per glucose.
• High-energy electrons carried by NADH and FADH2 to the ETC to power ATP production.

Heredity & Pedigrees

• Refer to Unit 3 Day 3 for pedigree analysis.
• Videos available explaining different types of pedigrees (autosomal vs. sex-linked, dominant vs. recessive).