Ch. 24

Hardy-Weinberg Principle: A model to study genetic variation in a population under ideal conditions where allele and genotype frequencies remain constant from generation to generation in a large, randomly mating population. This principle provides a foundational framework for understanding the genetic structure of populations and serves as a baseline for detecting evolutionary changes.
Formula: $p^2 + 2pq + q^2 = 1$ where:
- $p$ = frequency of the dominant allele (B) in the population.
- $q$ = frequency of the recessive allele (b) in the population.
- $p^2$ = expected frequency of the homozygous dominant genotype (BB).
- $2pq$ = expected frequency of the heterozygous genotype (Bb).
- $q^2$ = expected frequency of the homozygous recessive genotype (bb).
- The values of $p$ and $q$ relate as $p + q = 1$ , indicating that the sum of the frequencies of all alleles for a given gene in a population must equal 100%.

Example Problems

Recessive Phenotype Calculation
- Given: 98 out of 200 individuals display a recessive phenotype (bb).
- Calculate the frequency of the recessive phenotype:
  - $q^2 = 98/200 = 0.49$
  - Therefore, $q = ext{sqrt}(0.49) = 0.7$ which represents the frequency of the recessive allele.
- Calculate the frequency of the dominant allele:
  - $p = 1 - q = 1 - 0.7 = 0.3$
- Calculate the frequency of heterozygotes in the population:
  - $2pq = 2(0.3)(0.7) = 0.42$
Birds with Tail Feathers
- Given: 16 red-tailed birds (recessive phenotype, bb), 34 blue-tailed birds (dominant phenotype, BB or Bb).
- Total: 50 birds in this population.
- Calculate the frequency of the red tail allele ( $q$ ):
  - $q^2 = \frac{16}{50} = 0.32$
  - Therefore, $q = ext{sqrt}(0.32) = 0.57$
- Calculate the frequency of the blue allele ( $p$ ):
  - $p = 1 - 0.57 = 0.43$
- Frequency of heterozygotes ( $2pq$ ):
  - $2(0.43)(0.57) = 0.49$
- Frequency of homozygous blue (BB):
  - $p^2 = (0.43)^2 = 0.18$
Blood Type Frequency in Population
- Given: 6840 AA, 2860 AB, 300 BB individuals out of 10,000 individuals total in a population.
- Calculate the frequency of each genotype:
  - $f(AA) = 6840/10000 = 0.684$
  - $f(AB) = 2860/10000 = 0.286$
  - $f(BB) = 300/10000 = 0.03$
- Frequency of A allele:
  - $f(A) = rac{2<em>6840 + 2860}{2</em>10000} = 0.785$
- Frequency of B allele:
  - $f(B) = rac{2<em>300 + 2860}{2</em>10000} = 0.215$
- Project BB in the next generation (with an estimated population of 25,000 individuals):
  - Expected $BB = 0.03 \times 25000 = 750$
ABO Blood Group Alleles
- Two subtypes for blood groups: IA1 and IA2.
- Expected genotype frequency among IA individuals can be classified as follows:
  - For $IA1IA1$ : 75% expected frequency among IA individuals.
  - For $IA1IA2$ : 25% expected frequency among IA individuals.
  - For $IA2IA2$ : 0% as it was not mentioned in the genotypic distribution.
Male Pattern Baldness
- Given a sample of 1000 men (360 bald, 480 heterozygous, 160 homozygous for the baldness gene).
- Phenotype ratios calculated show a ratio of balding to non-balding individuals:
  - Total balding (360) and non-balding (640). Heterozygous (Bb): 480, homozygous (BB): 160
- Allele frequencies:
  - Total $B$ alleles = 480 (from heterozygotes) + 160 \times 2 (from homozygotes) = 800.
  - Total $b$ alleles = 360 (bald) + 480 (heterozygous) = 840.
  - Frequency of allele $B = \frac{800}{(800 + 840)} = 0.4$ , frequency of allele $b = \frac{840}{(800 + 840)} = 0.6$ .
- Expected genotype frequencies based on calculated allele frequencies:
  - $B/B = p^2 = (0.4)^2 = 0.16$ ,
  - $B/b = 2pq = 2(0.4)(0.6) = 0.48$ ,
  - $b/b = q^2 = (0.6)^2 = 0.36$

Speciation and Species Definition

Biological Species Concept: Defines a species as a group of organisms capable of interbreeding in nature and producing fertile offspring. This concept emphasizes the role of reproductive isolation as a key factor in speciation.
Reproductive Isolation: Factors that prevent different species from interbreeding and producing viable offspring can be classified into two main categories:
- Prezygotic Barriers: These barriers prevent mating or fertilization between species and include:
  - Temporal Isolation: Species breed at different times (seasonal or daily).
  - Habitat Isolation: Species occupy different habitats and rarely encounter each other.
  - Behavioral Isolation: Differences in mating behaviors prevent interbreeding.
- Postzygotic Barriers: These barriers affect the viability or reproductive capacity of hybrid offspring and include:
  - Reduced Hybrid Viability: Hybrids fail to develop properly or are frail.
  - Reduced Hybrid Fertility: Hybrids may be sterile (e.g., mules).

Types of Speciation

Allopatric Speciation: Occurs when a population is geographically separated by physical barriers such as mountains or rivers, leading to divergent evolution of isolated populations.
Sympatric Speciation: Occurs in overlapping populations and can arise due to mechanisms like polyploidy, sexual selection, or habitat differentiation.
- Example: The apple maggot fly diverges based on available food sources, leading to reproductive isolation despite living in the same geographic area.

Hybrid Zones

Hybrid zones are regions where distinct species meet and mate, producing hybrid offspring. The outcomes of hybrid zones can include:
- Reinforcement: Strengthening of reproductive barriers, limiting hybrid formation over time.
- Fusion: Weakening of barriers, which leads to the merging of species into a single lineage.
- Stability: Maintenance of persistent hybrid populations, balancing the number of hybrids and parent species.

Evolutionary Mechanisms

Punctuated Equilibrium: A model of evolution that suggests species remain relatively stable for long periods, with rapid changes occurring in relatively short bursts driven by environmental events or changes in the genetic structure of the population.
Gradualism: The theory that evolution occurs slowly and gradually through the accumulation of small genetic changes over extended periods.