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.

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

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.

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.

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.

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