Human Genetics 31

Overview of Population Genetics and Frequency Calculations

• The primary goal of this unit is to predict allele and genotype frequencies in populations and understand how they change over time.

• There are three main types of calculations used to understand population genetics:     - Allele Frequencies: The frequency of a specific allele, such as the probability of selecting the $A$ allele versus the $a$ allele within a population.     - Genotype Frequencies: The frequency of a specific pair of alleles, such as the frequency of the homozygous dominant ($AA$), heterozygous ($Aa$), or homozygous recessive ($aa$) combinations.     - Phenotype Frequencies: The frequency of observable traits, such as the incidence of the PKU (Phenylketonuria) illness in a population.

• Data Sources for Analysis:     - Genetic analysis can look at Single Nucleotide Differences (SNPs) for alleles.     - Analysis can also involve non-gene regions, such as repeated sequences (used in DNA fingerprinting).     - Gene vs. Non-gene regions: Researchers may choose non-gene regions because genes are subject to evolutionary selective pressures (such as antibody selection or immune regions) which can interfere with frequency data. Non-gene regions are often more neutral.

• The complexity of these calculations increases significantly when looking at multiple genes, multiple alleles at a single locus, or multiple phenotypes at once. For this introductory course, calculations are limited to two-allele, three-genotype systems.

Population Variation and Ancestry

• Genetic frequencies vary significantly between different populations based on geography and historical interbreeding.

• Tay-Sachs Disease: This condition has a much higher allele frequency in the Ashkenazi Jewish population compared to African Americans.

• Sickle Cell Allele: Higher in African Americans and African populations because of the heterozygote advantage; individuals who are heterozygous ($Aa$) are protected from malaria and are thus selected for in regions where malaria is or was prevalent.

• Ancestry Data and Biases:     - Accurate ancestry data relies on reference data. Currently, most reference data is derived from Europe and North America.     - Current data is not representative of the Earth’s full population, though it is improving as more people (such as through 23andMe) are sequenced.

• Phenotype Frequency Example (PKU):     - PKU is a metabolic condition where an individual cannot break down phenylalanine.     - Turkish populations: Highest incidence of PKU in the world.     - Japanese individuals: Lowest incidence of PKU in the world.

Microevolution and Macroevolution

• Microevolution: Involves small, incremental steps of genetic change over time. An example is the slow growth of a phenotypic feature through small stages.

• Macroevolution: Refers to large-scale evolutionary changes that occur more rapidly or represent the accumulation of sufficient microevolutionary changes to cause significant phenotype differences.

• Speciation: Occurs when genetic changes are great enough that two individuals can no longer interbreed to produce fertile offspring. If a mom and dad produce offspring that are not fertile, they are considered different species.

• Human vs. Ape Evolution Case Study:     - Significant differences exist in the $FOXP2$ gene, which relates to communication and speech.     - Jaw Muscle Attachment: In apes, the jaw muscle attachment to the skull is very large and strong, requiring a thick, strong skull bone. Humans have a mutation that caused a smaller jaw muscle attachment. While this made chewing less efficient, it allowed the human skull to be thinner, providing space for brain expansion.

Calculating Allele Frequencies ($p$ and $q$)

• Allele frequencies are denoted by variables:     - $p$ = Frequency of the dominant allele (e.g., $A$).     - $q$ = Frequency of the recessive allele (e.g., $a$).

• The sum of all allele frequencies in a population must equal 1:     - $p + q = 1.0$.

• Calculation Formula:     - $p = \frac{2 \times (\text{Number of homozygous dominant individuals}) + 1 \times (\text{Number of heterozygotes})}{\text{Total number of alleles in the population}}$.     - Note: The total number of alleles is twice the number of individuals in a diploid population.

• Example calculation:     - Population: 20 $AA$, 30 $Aa$, 50 $aa$ (Total = 100 people, 200 alleles).     - Number of $A$ alleles in homozygous dominant: $20 \times 2 = 40$.     - Number of $A$ alleles in heterozygotes: $30 \times 1 = 30$.     - Total $A$ alleles: $40 + 30 = 70$.     - Allele frequency $p = \frac{70}{200} = 0.35$.     - Allele frequency $q = \frac{130}{200} = 0.65$.     - Verification: $0.35 + 0.65 = 1.0$.

Genotype Frequencies and the Hardy-Weinberg Equation

• Genotype frequencies are expressed as decimals.

• Using the population from the previous example (100 people total):     - Frequency of $AA = \frac{20}{100} = 0.2$.     - Frequency of $Aa = \frac{30}{100} = 0.3$.     - Frequency of $aa = \frac{50}{100} = 0.5$.

• The Hardy-Weinberg Equation:     - Derived from the normal distribution: $(p + q)^2 = p^2 + 2pq + q^2 = 1.0$.     - $p^2$: Frequency of homozygous dominant ($AA$).     - $2pq$: Frequency of heterozygotes ($Aa$).     - $q^2$: Frequency of homozygous recessive ($aa$).

Hardy-Weinberg Equilibrium (HWE)

• A population in HWE maintains the same allele frequencies generation after generation.

• Five Assumptions for HWE:     1. Random mating.     2. No migration (gene flow).     3. No natural selection.     4. No genetic drift (requires a large population).     5. No mutation.

• Generally, these assumptions are never fully met in nature (especially selection and mutation). Therefore, allele frequencies usually change over time.

• Non-gene regions: Parts of the genome that do not affect phenotype (like repeated DNA segments) often follow Hardy-Weinberg equilibrium because they are not subject to natural selection.

Practical Applications and Problem Solving

Generation-to-Generation Example (Fingers)

• Trait: $D$ (Normal fingers), $d$ (Short middle finger).

• Initial Allele Frequencies: $p = 0.7$, $q = 0.3$.

• Calculation of Genotype Frequencies:     - $DD (p^2) = (0.7)^2 = 0.49$.     - $Dd (2pq) = 2 \times 0.7 \times 0.3 = 0.42$.     - $dd (q^2) = (0.3)^2 = 0.09$.     - Verification: $0.49 + 0.42 + 0.09 = 1.0$.

• Gamete Frequency Calculation:     - To find the next generation's alleles ($p$ and $q$):     - Big $D$ contribution from $DD$: $0.49$.     - Big $D$ contribution from $Dd$: $\frac{1}{2} \times 0.42 = 0.21$.     - Total next-gen $p = 0.49 + 0.21 = 0.70$.     - Recessive $d$ contribution from $Dd$: $\frac{1}{2} \times 0.42 = 0.21$.     - Recessive $d$ contribution from $dd$: $0.09$.     - Total next-gen $q = 0.21 + 0.09 = 0.30$.

Cystic Fibrosis (Carrier Frequency)

• Cystic Fibrosis is autosomal recessive. Phenotype incidence is easier to track than genotype.

• Example: Incidence of phenotype ($q^2$) is $0.0005$.

• Step 1: Find allele frequency $q = \sqrt{0.0005} \approx 0.022$.

• Step 2: Find $p = 1 - 0.022 = 0.978$.

• Step 3: Find carrier frequency ($2pq$). $2 \times 0.978 \times 0.022 \approx 0.043$, which is approximately $1$ in $23$.

• Health Care Implications: Insurance companies and healthcare systems use these statistics to determine if genetic testing or screening (like IVF with embryo selection) is cost-effective. For example, colonoscopies were lowered to age 45 not out of kindness, but because it was financially beneficial for insurance companies to screen earlier.

X-Linked Traits (Hemophilia)

• In males, genotype equals phenotype because they only have one X chromosome.

• If $1$ in $10,000$ males has hemophilia, then the allele frequency $q = 0.0001$.

• This data allows the calculation of the carrier frequency in females using $2pq$.

• This also allows for the prediction of affected females ($q^2$); for hemophilia, this is roughly $1$ in $100,000,000$.

Rare Gene Shortcut (Tay-Sachs)

• For rare autosomal recessive diseases, a shortcut for carrier frequency can be used: once you find $q$ (the square root of the incidence), the carrier frequency ($2pq$) is approximately $2q$ because $p$ is so close to $1.0$.

• Example: Tay-Sachs incidence is $1$ in $3,600$. $q = \sqrt{\frac{1}{3600}} = 0.017$. Carrier frequency is roughly $2 \times 0.017$.