Topic 3 - Biol 213

Population

A group of individuals belonging to the same species that occupy the same geographical range and actively interbreed with one another.

difference between an Individual Cross and a Population Cross on a Punnett square

An individual cross uses single-gamete probabilities along the axes (1/2$ and 1/2) to calculate specific offspring genotype probabilities (1/4). A population cross places entire allele frequencies along the axes (0.89 and 0.11) to directly calculate overall population genotype frequencies.

assumption is required to calculate genotype frequencies using a population cross

It explicitly assumes random mating across the entire population.

Traditional Ecological Knowledge (TEK) Salmon Example

Pacific Northwest Indigenous peoples observed that spring and fall Chinook runs in the same river were completely distinct populations requiring different ecological management.

Genetic Vindication of the Yurok Tribe's TEK

spring and fall run Chinook salmon have distinct genetic differences, specifically at the GREB1L gene, leading to new ecological protections and dam removals.

Genotype Frequency

The relative abundance or proportion of a specific genotype within a population (the number of 'BB')

Genotype Frequency Formula

f(BB) = (Number of BB individuals) / (Total number of individuals in the population).

Allele Frequency

The relative abundance or proportion of a specific allele at a genetic locus within a population. proportion of alleles that are B

Allele Frequency Formula

f(B) = (Number of B alleles) / (Total number of alleles in the population).

Calculating Allele Frequencies from Population Numbers (Diploid)

f(B) = [(2 × number of BB individuals) + (number of Bb individuals)] / (2 × total number of individuals).

Calculating Allele Frequencies from Genotype Frequencies (Diploid)

f(B) = f(BB) + ½ f(Bb).

Importance of Genetic Variation

the raw material required for all evolutionary change, without it, a population cannot evolve regardless of the underlying evolutionary mechanism.

Flower Size Evolutionary Example

If bees select for larger flowers, a population with genetic variation (AA, Aa, aa) will increase in size across generations, whereas a population with zero genetic variation (all aa) will not change.

The Ultimate Source of All Genetic Variation

Mutation, which introduces novel differences in DNA sequences in offspring compared to their parents (such as SNPs, insertions, and deletions).

Single Nucleotide Polymorphisms (SNPs)

A specific type of mutation involving a single nucleotide base variation at a precise location in the DNA sequence.

Three Conditions for Evolution by Natural Selection

1. Phenotypic variation must exist in the population; 2. The variation must be heritable; 3. The variation must lead to differences in fitness (differential survival or reproductive success).

Mechanisms of Evolution

1. Selection; 2. Migration (gene flow); 3. Random changes in allele frequency (genetic drift); 4. Mutation.

The Null Hypothesis for Evolution

Hardy-Weinberg Equilibrium, which states that there is no change in allele frequencies over time in a population.

Hardy and Weinberg focus

what will happened to a single trait, as a single genetic locus that is encoded by two alleles in the Absence if evolution

Hardy-Weinberg Model Assumptions

1. No mutation 2. No gene flow (migration) 3. Infinitely large population (no genetic drift) 4. No natural selection; 5. Random mating.

No mutation

single locus with two alleles does not change state between generation

No gene flow

Alleles are neither added not removed from the population

Population is infinitely large

random events and processes thus have no effect on the genetic diversity of the population

Natural selection does not affect the gene in question

the two alleles at the locus do not differentially affect survival or reproduction

Individuals mate randomly

regardless of their genotype at the locus of interest, individuals in the same population all have the same reproductive fitness

Genetic Outcome of Hardy-Weinberg Equilibrium

Genetic variation is maintained at a constant level, and allele frequencies come into equilibrium within a single generation and will not change from generation to generation. not evolving

Hardy-Weinberg Mathematical Equations

For two alleles (A and a) where f(A) = p and f(a) = q: p + q = 1, and the genotype frequencies are p² + 2pq + q² = 1.

Hardy-Weinberg Expected Genotype Frequencies

f(AA) = p², f(Aa) = 2pq, and f(aa) = q².

Apply Hardy-Weinberg as a Null Hypothesis

1. Note observed genotype frequencies; 2. Calculate allele frequencies (p and q); 3. Calculate expected genotype frequencies using p², 2pq, and q²; 4. Compare observed vs. expected values using a statistical test or margin of error.

How to Calculate a 5% Margin of Error Range for Hardy-Weinberg Comparisons

For any expected value K, the acceptable range within the margin of error is calculated as: 0.95 × K to 1.05 × K.

Departure from Hardy-Weinberg Equilibrium Indicates

It reveals that observed values do not equal expected values, meaning one or more of the five HW assumptions are being violated and evolution is currently occurring at that locus.

Why Hardy-Weinberg Equilibrium is a Powerful Test for Evolution

It provides a non-evolving 'baseline' to compare against, is primarily used in experiments to test for natural selection, and helps estimate genotype frequencies when population data is limited.

Why is Hardy-Weinberg equilibrium useful?

It provides a null hypothesis for comparison.

The locus A has two alleles: A and a. The frequency of A is 0.6. What is the frequency of the Aa phenotype?

0.48. If f(A) = p = 0.6, then f(a) = q = 1 - 0.6 = 0.4. The frequency of the heterozygous Aa genotype/phenotype under HWE is 2pq = 2(0.6)(0.4) = 0.48

What best describes Hardy-Weinberg equilibrium?

Allele frequencies remain constant when specific assumptions are met.

Which of the following is not an assumption of a population at Hardy-Weinberg equilibrium?

Equal frequencies of the two alleles.

A population of 447 individuals is comprised of 208 individuals homozygous for one allele, 91 heterozygous individuals, and 148 individuals homozygous for the other allele. There are only two alleles at the locus in question. Is this population in Hardy-Weinberg equilibrium? [Answer only 'yes' or 'no']

no. (Observed frequencies: f(AA) = 0.465, f(Aa) = 0.204, f(aa) = 0.331. Allele frequencies: p = 0.465 + 0.102 = 0.567; q = 0.433. Expected frequencies under HWE: p^2 \approx 0.321, 2pq =0.491, q^2 = 0.187. The observed heterozygotes (0.204) drastically deviate from the expected value of 0.491, exceeding the typical 5% margin of error).

A population of 312 flowering plants has the following genotype counts: AA - 113, Aa - 150, aa - 49. (a) Calculate the observed frequency for the dominant allele (p).

0.603 (or 0.60). Calculated as: [2(113) + 150] / [2(312)] = 376 / 624 = 0.60256...

A population of 312 flowering plants has the following genotype counts: AA - 113, Aa - 150, aa - 49. (b) What is the expected frequency for the aa genotype?

yes. (The observed genotype frequencies are f(AA) = 0.362, f(Aa) = 0.481, and f(aa) = 0.157[cite: 846]. The expected genotype frequencies calculated from your allele frequencies are p² = 0.364, 2pq = 0.479, and q² = 0.158[cite: 805, 834, 835, 836]. Because the observed values match the expected baseline almost perfectly, they fall well within the 5% margin of error

Why is Hardy-Weinberg equilibrium useful?

It provides a null hypothesis for comparison.

The locus A has two alleles: A and a. The frequency of A is 0.6. What is the frequency of the Aa phenotype?

0.48. If f(A) = p = 0.6 , then f(a) = q = 1 - 0.6 = 0.4. The frequency of the heterozygous Aa genotype/phenotype under HWE is 2pq = 2(0.6)(0.4) = 0.48

What best describes Hardy-Weinberg equilibrium?

Allele frequencies remain constant when specific assumptions are met.

A population of 447 individuals is comprised of 208 individuals homozygous for one allele, 91 heterozygous individuals, and 148 individuals homozygous for the other allele. There are only two alleles at the locus in question. Is this population in Hardy-Weinberg equilibrium?

no. (Observed frequencies: f(AA) = 0.465, f(Aa) = 0.204, f(aa) = 0.331. Allele frequencies: p = 0.567; q = 0.433. Expected frequencies under HWE: p^2 = 0.321, 2pq =0.491, q^2 = 0.187. The observed heterozygotes drastically deviate from the expected value).

A population of 312 flowering plants has the following genotype counts: AA - 113, Aa - 150, aa - 49. (a) Calculate the observed frequency for the dominant allele (p). 0.603 (or 0.60).

[2(113) + 150] / [2(312)] = 376 / 624 = 0.60256...

A population of 312 flowering plants has the following genotype counts: AA - 113, Aa - 150, aa - 49. (b) What is the expected frequency for the aa genotype?

0.158 (or 0.16). If p = 0.603, then q = 1 - 0.603 = 0.397. The expected frequency of aa is q^2 = (0.397)^2 = 0.1576

A population of 312 flowering plants has the following genotype counts: AA - 113, Aa - 150, aa - 49. (c) Is this population at HWE? [Answer 'yes' or 'no']

yes. (The observed genotype frequencies are f(AA) = 0.362, f(Aa) = 0.481, and f(aa) = 0.157. The expected genotype frequencies calculated from your allele frequencies are p² = 0.364, 2pq = 0.479, and q² = 0.158. Because the observed values match the expected baseline almost perfectly, they fall well within the 5% margin of error).