BiologyChapter 14: Evolution
Hardy-Weinberg Named
A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put.
Learning objectives
- Explain hardy-weinberg named in clear biological language.
- Distinguish 1. assumptions from 2. little sums.
- Use the key terms to answer a short Federal Board exam question.
Definition
A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put. p + q = 1. p2 + 2pq + q2 = 1 for two alleles. p^2 = homozygous dominant frequency, 2pq hetero, q^2 recessive phenotype if complete dominance. Real populations break the assumptions. Use it to estimate carrier 2pq from q^2 if the question allows.
Mathematical form
q = 0.1
p = 0.9
q = 1. p2 + 2pq + q2
q = 0.1 p
Important terms
- HARDY-WEINBERG NAMED
- A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put.
- 1. ASSUMPTIONS
- Large, random mates, no pick, no new alleles, no one arrives.
- 2. LITTLE SUMS
- If q^2 = 0.01, q = 0.1, p = 0.9, 2pq = 0.18 carriers.
Relationship
- p + q = 1
- p2 + 2pq + q2 = 1
Overview
Equilibrium as a foil:
Assumptions
Large, random mates, no pick, no new alleles, no one arrives. Any real force (selection) moves the numbers. That is the point: measure change against this baseline.
Examples p = freq of A · q = freq of a · Not a claim that life is still
Little Sums
If q^2 = 0.01, q = 0.1, p = 0.9, 2pq = 0.18 carriers. Show the square root step. Do not invent data. q² = frequency of aa (recessive homozygotes). q = √q². p = 1 − q. p² = AA frequency. 2pq = Aa carrier frequency. Show all working.
Examples q = sqrt(q^2) · p = 1 - q · 2pq carriers
WHAT — HARDY-WEINBERG NAMED
A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put. p + q = 1. p2 + 2pq + q2 = 1 for two alleles. p^2 = homozygous dominant frequency, 2pq hetero, q^2 recessive phenotype if complete dominance. Real populations break the assumptions. Use it to estimate carrier 2pq from q^2 if the question allows.
WHY — Why this topic matters
HARDY-WEINBERG NAMED is an independent FBISE learning unit because students must connect its structures and processes with their biological functions and examples.
HOW — How it works
ASSUMPTIONS: Large, random mates, no pick, no new alleles, no one arrives. Any real force (selection) moves the numbers. That is the point: measure change against this baseline. Examples: p = freq of A; q = freq of a; Not a claim that life is still. LITTLE SUMS: If q^2 = 0.01, q = 0.1, p = 0.9, 2pq = 0.18 carriers. Show the square root step. Do not invent data. q² = frequency of aa (recessive homozygotes). q = √q². p = 1 − q. p² = AA frequency. 2pq = Aa carrier frequency. Show all working. Examples: q = sqrt(q^2); p = 1 - q; 2pq carriers. Worked: recessive phenotype 1% (q^2 = 0.01). q = 0.1 p = 0.9 2pq = 2 x 0.9 x 0.1 = 0.18 Carriers about 18%.
Worked example
- recessive phenotype 1% (q2 = 0.01).
- q = 0.1
- p = 0.9
- 2pq = 2 x 0.9 x 0.1 = 0.18
- Carriers about 18%.
In simple words
Algebra of frequencies. Not a eugenics tool.
Key points
- p + q = 1
- p2 + 2pq + q2 = 1
- five school assumptions
- q from the recessive class
Exam preparation
Exam Focus
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.Hardy-Weinberg Named
Sticky Notes · Quick Revision
Core Idea
- A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put. p + q = 1. p2 + 2pq + q2 = 1 for two alleles. p^2 = homozygous dominant frequency, 2pq hetero, q^2 recessive phenotype if complete dominance. Real populations break the assumptions. Use it to estimate carrier 2pq from q^2 if the question allows.
1. ASSUMPTIONS
- Large, random mates, no pick, no new alleles, no one arrives. Any real force (selection) moves the numbers. That is the point: measure change against this baseline.
2. LITTLE SUMS
- If q^2 = 0.01, q = 0.1, p = 0.9, 2pq = 0.18 carriers. Show the square root step. Do not invent data. q² = frequency of aa (recessive homozygotes). q = √q². p = 1 − q. p² = AA frequency. 2pq = Aa carrier frequency. Show all working.
Exam Example
- Worked: recessive phenotype 1% (q^2 = 0.01).
- q = 0.1
- p = 0.9
- 2pq = 2 x 0.9 x 0.1 = 0.18
- Carriers about 18%.
Common Misconception
- Do not treat 1. ASSUMPTIONS and 2. LITTLE SUMS as the same idea; each explains a different part of the topic.
FBISE Exam Check
- Define hardy-weinberg named, explain the two key parts, and use the correct biological terms.
- Exam reminder: Worked: recessive phenotype 1% (q^2 = 0.01).
60-second revision
- A null model: if no selection, no mutation, no migration, no chance wobble, and random mating, allele frequencies stay put.
- 1. ASSUMPTIONS: Large, random mates, no pick, no new alleles, no one arrives.
- 2. LITTLE SUMS: If q^2 = 0.01, q = 0.1, p = 0.9, 2pq = 0.18 carriers.
- Exam reminder: Worked: recessive phenotype 1% (q^2 = 0.01).
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