2. Population Genetics and Allelic Frequency Derivations

Population genetics shifts the focus of biological study from individual organisms to a Mendelian population — a group of sexually reproducing, interbreeding individuals of the same species inhabiting a defined geographic area.

2.1 Mathematical Derivations of Allele Frequencies

Consider a single autosomal locus with two alleles — a dominant allele \(A\) and a recessive allele \(a\) — in a sexually reproducing diploid population. Let \(p = f(A)\) and \(q = f(a)\). Because there are only two alleles at this locus, their relative frequencies must sum to unity:

$$p + q = 1 \implies q = 1 - p \quad \text{and} \quad p = 1 - q$$

Method A — the direct count method

For a population of \(N\) diploid individuals, the total number of alleles in the pool is \(2N\). Let \(n_{AA}\), \(n_{Aa}\), and \(n_{aa}\) be the counts of each genotype, so \(N = n_{AA} + n_{Aa} + n_{aa}\). Since homozygotes carry two copies of an allele and heterozygotes carry one, the allele frequencies are:

$$p = f(A) = \frac{2n_{AA} + n_{Aa}}{2N}$$ $$q = f(a) = \frac{2n_{aa} + n_{Aa}}{2N}$$

Method B — from genotype frequencies

Letting \(f(AA) = \frac{n_{AA}}{N}\), \(f(Aa) = \frac{n_{Aa}}{N}\), and \(f(aa) = \frac{n_{aa}}{N}\), the same relationship can be rewritten as:

$$p = f(A) = f(AA) + \frac{1}{2}f(Aa)$$ $$q = f(a) = f(aa) + \frac{1}{2}f(Aa)$$

2.2 Step-by-Step Worked Example (Source Page 662)

A population has the following genotypic distribution at an autosomal locus: \(AA = 114\), \(Aa = 76\), \(aa = 10\), giving \(N = 114+76+10 = 200\) individuals and \(400\) total alleles.

1. Direct count method: total \(A\) alleles \(= 2(114)+76 = 304\); total \(a\) alleles \(= 2(10)+76 = 96\). $$p = f(A) = \frac{304}{400} = 0.76$$ $$q = f(a) = \frac{96}{400} = 0.24$$ Check: \(p+q = 0.76+0.24 = 1.00\).
2. Genotype-frequency method: \(f(AA)=\frac{114}{200}=0.57\), \(f(Aa)=\frac{76}{200}=0.38\), \(f(aa)=\frac{10}{200}=0.05\). $$p = f(AA)+\tfrac{1}{2}f(Aa) = 0.57+0.19 = 0.76$$ $$q = f(aa)+\tfrac{1}{2}f(Aa) = 0.05+0.19 = 0.24$$ Both methods agree exactly.

3. The Hardy-Weinberg Principle and Baseline Equilibrium

The Hardy-Weinberg principle serves as the fundamental null model for population genetics. It states that in a large, randomly mating diploid population free from evolutionary forces, both allele and genotype frequencies remain constant from generation to generation.

3.1 Mathematical Formulation and Derivation

If gametes merge at random, the probability of any zygote carrying a specific genetic combination is the product of the individual parental allele frequencies. Mapping the gametic pool as a Punnett square of egg and sperm frequencies:

$$p^2 + 2pq + q^2 = 1.00$$

Where \(p^2\) is the expected frequency of \(AA\), \(2pq\) is the expected frequency of \(Aa\), and \(q^2\) is the expected frequency of \(aa\).

3.2 The Hardy-Weinberg Genotype Frequency Curve

3.3 The Five Key Assumptions of Hardy-Weinberg Equilibrium

1. Random mating: every individual has an equal probability of mating with any individual of the opposite sex.
   Violations — nonrandom mating: mating based on phenotypic preference or relatedness. Positive assortative mating (like mates with like) increases homozygosity for those traits; negative assortative mating (opposites attract) increases heterozygosity; inbreeding (mating between close relatives) systematically increases homozygosity across the entire genome.
2. No natural selection: all genotypes must exhibit equal viability and fertility — no genotype can have a survival or reproductive advantage.
3. No mutation: there must be no mutational conversion of alleles, or the forward mutation rate must equal the backward rate.
4. No migration (no gene flow): there must be no introduction or loss of alleles from individuals moving into or out of the population.
5. Infinite population size (no genetic drift): the population must be large enough to prevent sampling error. In small populations, random chance alone causes generation-to-generation fluctuation in allele frequencies — genetic drift — which over time fixes one allele and permanently loses the other.

3.4 Quantitative Problems and Step-by-Step Solutions

Problem 1 — allele frequencies from a homozygous genotype (Source Page 664). The frequency of \(AA\) in a randomly mating population is 0.09. Find \(p\) and \(q\).

1. $$f(AA) = p^2 = 0.09$$
2. $$p = \sqrt{0.09} = 0.30$$
3. $$q = 1-p = 1-0.30 = 0.70$$

Problem 2 — carrier frequency for a rare recessive disorder (Source Page 664). An autosomal recessive disorder occurs in 1 in 10,000 individuals. Find the carrier frequency.

1. $$f(aa) = q^2 = \frac{1}{10{,}000} = 0.0001$$
2. $$q = \sqrt{0.0001} = 0.01$$ $$p = 1-q = 0.99$$
3. $$2pq = 2 \times 0.99 \times 0.01 = 0.0198 \; (1.98\%)$$

Problem 3 — validity testing via chi-square analysis (Source Pages 665–666). In a sample of 1000 individuals: observed \(AA=800\), \(Aa=185\), \(aa=15\). Test whether the population is in Hardy-Weinberg equilibrium.

1. Observed allele frequencies: $$p = \frac{2(800)+185}{2000} = \frac{1785}{2000} = 0.8925$$ $$q = \frac{2(15)+185}{2000} = \frac{215}{2000} = 0.1075$$
2. Expected genotype counts: \(Expected(AA) = p^2 N \approx 796.5\), \(Expected(Aa) = 2pqN \approx 192.0\), \(Expected(aa) = q^2 N \approx 11.5\).
3. Chi-square goodness of fit: $$\chi^2 = \sum \frac{(Observed-Expected)^2}{Expected}$$

4. Advanced Extensions of the Hardy-Weinberg Principle

4.1 Multiple Alleles (Trinomial Expansion Model)

When an autosomal locus carries three alleles (such as \(A_1, A_2, A_3\), or the human ABO alleles \(I^A, I^B, i\)), their frequencies \(p, q, r\) satisfy \(p+q+r=1.0\). Expanding \((p+q+r)^2=1\) gives the expected genotype frequencies:

$$p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.0$$

Homozygous genotypes: \(p^2 (A_1A_1)\), \(q^2(A_2A_2)\), \(r^2(A_3A_3)\). Heterozygous genotypes: \(2pq(A_1A_2)\), \(2pr(A_1A_3)\), \(2qr(A_2A_3)\).

Worked problem — ABO blood type (Source Page 667)

In a randomly mating human population, \(f(I^A)=0.7\), \(f(I^B)=0.2\), \(f(i)=0.1\). Let \(p=0.7\), \(q=0.2\), \(r=0.1\).

4.2 Sex-Linked (X-Linked) Genes

For genes on sex chromosomes, the heterogametic sex (human males, XY) carries only one copy of the gene (hemizygous), while the homogametic sex (human females, XX) carries two. With \(p=f(A)\) and \(q=f(a)\):

Worked problem (Source Page 667)

An X-linked recessive disease has \(q=0.02\). Assuming a 1:1 sex ratio, what is the overall disease frequency?

$$\text{Diseased males} = q = 0.02$$ $$\text{Diseased females} = q^2 = 0.0004$$

$$\text{Overall} = (0.5 \times 0.02) + (0.5 \times 0.0004) = 0.01 + 0.0002 = \mathbf{0.0102}$$

4.3 Polyploidy

The Hardy-Weinberg principle generalizes to polyploid organisms via the binomial expansion \((p+q)^n\), where \(n\) is the ploidy number. For a tetraploid organism (\(n=4\)):

$$p^4 + 4p^3q + 6p^2q^2 + 4pq^3 + q^4 = 1$$

Where \(AAAA=p^4\), \(AAAa=4p^3q\), \(AAaa=6p^2q^2\), \(Aaaa=4pq^3\), and \(aaaa=q^4\).

5. Inbreeding, Pedigree Path Analysis, and Population Substructure

Inbreeding is mating between individuals closely related by common ancestry.

5.1 Inbreeding Dynamics and Heterozygosity Loss

The inbreeding coefficient \(F\) is the probability that two alleles at a locus in an individual are identical by descent. It also measures the relative reduction in heterozygosity:

$$F = \frac{H_e - H_o}{H_e} \implies H_o = H_e(1-F)$$

where \(H_e = 2pq\) is expected heterozygosity under random mating and \(H_o\) is observed heterozygosity.

Genotype frequencies with inbreeding coefficient F (Table 6.2)

$$f(AA) = p^2(1-F)+pF$$ $$f(Aa) = 2pq(1-F)$$ $$f(aa) = q^2(1-F)+qF$$

At complete inbreeding (\(F=1.0\)), the heterozygous class vanishes (\(Aa=0\)) and the homozygote frequencies default to \(AA=p\) and \(aa=q\).

Genetic implications of inbreeding

1. Inbreeding depression: the reduction in fitness, survival, and reproductive success in inbred populations, driven by increased exposure of rare, deleterious recessive alleles in homozygous form.
2. Genetic load: the accumulation of deleterious recessive genes within a population's gene pool. Inbreeding exposes this hidden genetic load.
3. Heterosis (hybrid vigor): a marked increase in fitness, growth, or size in crossbred offspring over both homozygous inbred parents, typically driven by heterozygote advantage — such as the sickle-cell allele \(HbS\) conferring malaria resistance in heterozygotes.

5.2 Quantitative Pedigree and Path Analysis

To calculate an individual \(I\)'s inbreeding coefficient (\(F_I\)) from pedigree data, the pedigree is converted into a simplified path diagram of direct gametic transfer, keeping only the common ancestors and the lines that connect them to \(I\).

Path analysis rules

1. Identify the common ancestors of \(I\)'s parents.
2. Convert the pedigree into a path diagram, removing every individual that does not contribute to \(I\)'s inbreeding.
3. Trace every path of gametic transmission from one parent, up to a common ancestor, and back down to the other parent — each path may contain only one common ancestor, and no individual may be counted twice in a single path.
4. Calculate each path's contribution: $$\text{Path contribution} = \left(\frac{1}{2}\right)^n (1+F_A)$$ where \(n\) is the number of individuals in the closed path (excluding \(I\)) and \(F_A\) is the common ancestor's own inbreeding coefficient (0 if not inbred).
5. \(F_I\) is the sum of all path contributions.

5.3 The Wahlund Effect

Mechanism: the effect is driven by population substructure, which restricts gene flow between local subpopulations. When subpopulations merge and interbreed randomly, heterozygote frequency rises to meet the overall average allele frequency.

6. Effective Population Size and Ecological Demography

In natural ecological settings, the census population size (\(N\)) — the physical number of counted individuals — is rarely the number that actually contributes gametes to the next generation.

1. Equal numbers of males and females, all capable of reproduction.
2. All individuals have an equal probability of producing offspring.
3. Mating is completely random.
4. The number of breeding individuals stays constant across generations.

In almost all natural populations, \(N_e\) is significantly smaller than \(N\). Three demographic factors skew and reduce it.

6.1 Unequal Sex Ratio

If the breeding sex ratio deviates from 1:1, the probability of genetic drift increases:

$$N_e = \frac{4 N_m N_f}{N_m + N_f}$$

Worked problem: an island bird population has 100 breeding adults — but only 10 males and 90 females due to selective predation.

$$N_e = \frac{4(10)(90)}{10+90} = \frac{3600}{100} = \mathbf{36}$$

6.2 Variation in Reproductive Success

Even with an ideal 1:1 sex ratio, not all parents produce equal numbers of viable offspring. High variation in family size decreases \(N_e\) relative to \(N\), because a few parents disproportionately pass their alleles to the next generation.

6.3 Fluctuating Population Size (The Bottleneck Constraint)

If a population fluctuates dramatically across generations, \(N_e\) over the whole timeframe is disproportionately constrained by the generations with the smallest sizes — mathematically, the harmonic mean of the temporal population sizes:

$$\frac{1}{N_e} = \frac{1}{t}\left[\frac{1}{N_1}+\frac{1}{N_2}+\dots+\frac{1}{N_t}\right]$$

$$N_e = \frac{t}{\sum_{i=1}^{t} \frac{1}{N_i}}$$

Worked problem: a butterfly population's size over four consecutive generations: \(N_1=1000\), \(N_2=10\), \(N_3=500\), \(N_4=2000\).