krit.club logo

Genetics and Evolution (AHL) - Gene pools and speciation

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A gene pool consists of all the genes and their different alleles, present in an interbreeding population. Evolution involves a cumulative change in the allele frequency in a population's gene pool over successive generations.

•

Allele frequency is the proportion of a specific allele (e.g., AA or aa) relative to all alleles of that gene in a population, represented as a value from 00 to 11.

•

Directional selection occurs when phenotypes at one extreme are favored (e.g., antibiotic resistance in bacteria), while stabilizing selection favors intermediate phenotypes (e.g., human birth weight).

•

Disruptive selection favors both extremes of a phenotypic range and acts against intermediate variants, which can lead to the split of a single gene pool into two.

•

Reproductive isolation is the primary mechanism for speciation and can be: Geographic (physical barriers), Behavioral (differences in courtship rituals), or Temporal (reproducing at different times of the day or year).

•

Speciation can be explained by two theories: Gradualism suggests species diverge slowly through small, continuous changes, while Punctuated Equilibrium suggests long periods of stability (stasis) are interrupted by rapid bursts of evolutionary change.

•

Polyploidy is the condition where an organism has more than two complete sets of chromosomes (e.g., 3n3n, 4n4n, or 8n8n). This can lead to instantaneous speciation because polyploids often cannot produce fertile offspring with the original diploid population.

•

The genus AlliumAllium provides a clear example of polyploidy, where chromosome numbers vary significantly between species (e.g., 2n=162n = 16 or 3n=243n = 24), leading to reproductive isolation and diverse species of onions and leeks.

📐Formulae

p+q=1p + q = 1

p2+2pq+q2=1p^2 + 2pq + q^2 = 1

💡Examples

Problem 1:

In a population of 10,00010,000 individuals, the frequency of a recessive genetic condition (aaaa) is 0.040.04. Calculate the expected number of heterozygous carriers (AaAa) assuming the population is in Hardy-Weinberg equilibrium.

Solution:

Given q2=0.04q^2 = 0.04. Therefore, q=0.04=0.2q = \sqrt{0.04} = 0.2. Since p+q=1p + q = 1, then p=1−0.2=0.8p = 1 - 0.2 = 0.8. The frequency of carriers is 2pq=2(0.8)(0.2)=0.322pq = 2(0.8)(0.2) = 0.32. The number of carriers is 0.32×10,000=3,2000.32 \times 10,000 = 3,200.

Explanation:

We use the frequency of the homozygous recessive genotype (q2q^2) to find the recessive allele frequency (qq). We then find the dominant allele frequency (pp) and calculate the frequency of the heterozygotes (2pq2pq).

Problem 2:

Identify the type of selection occurring if a population of birds has a range of beak sizes, but environmental changes mean only those with very small beaks (to eat small seeds) or very large beaks (to crack large nuts) survive.

Solution:

Disruptive selection.

Explanation:

Disruptive selection favors the two phenotypic extremes over the intermediate phenotypes, which often leads to a bimodal distribution and can eventually result in speciation.

Problem 3:

A plant species with a diploid number of 2n=142n = 14 undergoes a non-disjunction event during meiosis, leading to a tetraploid offspring. What is the chromosome number of this polyploid individual?

Solution:

4n=284n = 28.

Explanation:

A tetraploid individual has four sets of chromosomes. If the original diploid set (2n2n) is 1414, then n=7n = 7. Thus, 4n=4×7=284n = 4 \times 7 = 28.

Gene pools and speciation - Revision Notes & Key Diagrams | IB Grade 12 Biology