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Hardy Weinberg Human Evolution

🎓 Class 12 Biology CBSE Theory Ch 6 – Evolution ⏱ ~14 min
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Hardy Weinberg Human Evolution

7.16 Hardy–Weinberg Principle

In any sexually reproducing population we can describe the genetic structure not by individuals but by the frequencies of alleles at each gene locus. The Hardy–Weinberg principle, formulated independently by G. H. Hardy and Wilhelm Weinberg in 1908, states that allele frequencies in a population remain constant from generation to generation in the absence of evolutionary forces. The collection of all alleles in a population is the gene pool; this remains a constant — a state called genetic equilibrium.

The Algebra

Consider a single gene locus with two alleles: A (dominant) with frequency p, and a (recessive) with frequency q. Because every individual has two alleles at this locus, the sum of all allele frequencies must equal 1:

p + q = 1

The probability that a sperm and an egg both carry A is p × p = p². Similarly, both carrying a gives , and a heterozygote (Aa) can form in two ways: A-from-mother & a-from-father (pq) or a-from-mother & A-from-father (pq) — totalling 2pq.

p² + 2pq + q² = 1

This is the Hardy–Weinberg equation — the binomial expansion of (p + q)². It gives expected genotype frequencies:

  • = frequency of homozygous dominant (AA) individuals
  • 2pq = frequency of heterozygous (Aa) individuals
  • = frequency of homozygous recessive (aa) individuals
Hardy–Weinberg Punnett Square Father's gametes → A (p) a (q) Mother's gametes A (p) a (q) AA = p² Aa = pq Aa = pq aa = q² Total: p² + 2pq + q² = 1 Genotype frequencies in equilibrium
Fig. 7.4.1: Punnett square for a gene with two alleles A (frequency p) and a (frequency q) under random mating.

7.17 Five Factors that Disturb Hardy–Weinberg Equilibrium

The Hardy–Weinberg principle is a null hypothesis: if observed genotype frequencies differ from p², 2pq, q², some evolutionary force is at work. NCERT identifies five such factors:

FactorWhat it doesExample
1. Gene Flow (Migration)Movement of individuals between populations adds new alleles to the recipient population and removes them from the source.Pollen carried by wind between two wildflower populations.
2. Genetic DriftRandom changes in allele frequencies in small populations due to chance sampling.A few seeds survive a fire; their allele mix may differ from the parent population.
3. MutationNew alleles arise spontaneously from DNA changes — the ultimate source of variation.A point mutation creates a new resistance allele in a bacterial population.
4. Genetic RecombinationCrossing over and independent assortment during meiosis shuffle existing alleles into new combinations.Each gamete carries a unique mix of parental alleles.
5. Natural SelectionNon-random survival and reproduction of heritable variants — favours certain alleles over others.Antibiotic resistance in bacteria; melanism in moths.

Genetic Drift & the Founder Effect

When the same random change in allele frequencies occurs by chance — without selection — it is called genetic drift. Drift acts most strongly in small populations. Sometimes a small subset of a population migrates to a new area; this subset carries only a sample of the original gene pool. If the change is so extreme that the new population becomes a distinct species, the original migrants are called the founders and the phenomenon is the founder effect.

Source population 10 alleles: 6 green, 2 orange, 2 blue small group migrates Founders 3 alleles: 0 green, 2 orange, 1 blue expand New population Orange-dominant gene pool
Fig. 7.4.2: Founder effect — a small migrant group establishes a new population with a very different (random) allele profile from the source.

7.18 Sample Hardy–Weinberg Calculation

Phenylketonuria (PKU) is an autosomal recessive disorder. In a population of 10,000, 16 individuals are affected (aa genotype). What are the allele frequencies and the carrier frequency?

Step-by-step:
  1. Frequency of aa = q² = 16 / 10,000 = 0.0016
  2. q = √0.0016 = 0.04
  3. p = 1 − q = 1 − 0.04 = 0.96
  4. Carriers (Aa) = 2pq = 2 × 0.96 × 0.04 = 0.0768
  5. Carrier count = 0.0768 × 10,000 = 768 individuals are carriers.
  6. Homozygous normal (AA) = p² = 0.9216 = 9,216 individuals.
Even though only 16 people show the disease, 768 carry the recessive allele — important for genetic counselling.

7.19 Brief Account of Evolution & Origin of Man

Evolution of Vertebrates — A Geological Snapshot

Period (mya)Major Event
~2,000 (2 bya)First cellular life (prokaryotes) appears
~500Invertebrates flourish in oceans
~350Jawless fish; lobefins move onto land; first amphibians
~320Seaweeds & early plants on land
~250–65Reptiles dominate — Age of Dinosaurs
~200Ichthyosaurs (fish-like reptiles)
~65Mass extinction — dinosaurs disappear; mammals diversify
~15Dryopithecus, Ramapithecus (primates)
~3–4Australopithecines in East Africa
~2Homo habilis — first "human" (brain ~650–800 cc)
~1.5Homo erectus — brain ~900 cc; tool use, meat-eating
~0.1–0.04Neanderthal man — brain ~1,400 cc; buried dead
~0.075–0.01Modern Homo sapiens; migrated from Africa; cave art (~18,000 ya)
~0.01Agriculture; permanent settlements; rise of civilisation

Key Hominid Milestones

  • Dryopithecus — ape-like; ~15 mya; common ancestor of apes.
  • Ramapithecus — more man-like; ~15 mya.
  • Australopithecines — ~2 mya, East African grasslands; upright walking; stone tools; mostly fruit-eaters.
  • Homo habilis — first true hominid; brain 650–800 cc; probably did not eat meat.
  • Homo erectus — discovered in Java (1891); ~1.5 mya; brain ~900 cc; meat-eater.
  • Neanderthal man (Homo neanderthalensis) — Near East and Central Asia, 1,00,000–40,000 ya; brain ~1,400 cc; used animal hides; buried their dead.
  • Homo sapiens — arose in Africa, migrated across continents during the ice age (75,000–10,000 ya); developed cave art ~18,000 ya (e.g., Bhimbetka rock shelter, Madhya Pradesh).
Adult human large cranium Baby chimpanzee similar to adult human! Adult chimpanzee projecting jaw Baby chimp skull resembles adult human more than it resembles adult chimp — clue about brain-development timing
Fig. 7.4.3: Skulls of adult human, baby chimpanzee and adult chimpanzee — the baby chimp's skull is closer to the adult human shape, indicating differences in developmental timing.

Interactive: Hardy–Weinberg Calculator

Enter the frequency q (recessive allele) and see the genotype frequencies (assumes equilibrium):

p = 1 − q = 0.7

AA (p²) = 0.49 = 49.0%

Aa (2pq) = 0.42 = 42.0%

aa (q²) = 0.09 = 9.0%

Sum should always equal 1.00 (or 100%).

Activity 7.4 — Predict-Observe-Explain: Bottleneck

Setup: A population of 1,000 deer has allele frequencies p(A) = 0.6 and q(a) = 0.4 at a single locus. A flood reduces the population to just 10 random survivors.

Predict: What is likely to happen to the allele frequencies in the next generation? Will the population still satisfy Hardy–Weinberg equilibrium? Why or why not?

Observation/Prediction: The allele frequencies in the 10 survivors will almost certainly differ from the original 0.6/0.4 by chance. With only 20 allele copies total (10 individuals × 2), random sampling could easily give p = 0.5 or 0.8 instead of 0.6.

Explanation: This is genetic drift — random change in allele frequencies due to small sample size. A "bottleneck" of this kind violates one of the key Hardy–Weinberg assumptions: large population size. Other assumptions also fail under stress (random mating, no selection). The future deer population will evolve from a non-random subset of the original gene pool — possibly losing rare alleles entirely.

Real example: The cheetah population went through a severe bottleneck thousands of years ago and now has extremely low genetic diversity — a problem for conservation today.

Worked Examples

Worked Example 1: In a population of 400, 64 individuals show the recessive trait (aa). Find p, q, and number of heterozygotes.

Step 1: q² = 64/400 = 0.16
Step 2: q = √0.16 = 0.4
Step 3: p = 1 − 0.4 = 0.6
Step 4: Heterozygote frequency = 2pq = 2 × 0.6 × 0.4 = 0.48
Step 5: Number of heterozygotes = 0.48 × 400 = 192 individuals

Verification: AA = 0.36 × 400 = 144; Aa = 192; aa = 64. Total = 400 ✓

Worked Example 2: If 9% of a population has a recessive disorder, what is the carrier frequency?

q² = 0.09, so q = 0.3 and p = 0.7.
Carrier frequency = 2pq = 2 × 0.7 × 0.3 = 0.42 = 42%.

A striking finding — nearly half the population are carriers even though only 9% show the disorder. This is typical and underlies the importance of recessive-disease genetic counselling.

Worked Example 3: Order the following hominids from earliest to most recent: Homo erectus, Australopithecus, Homo sapiens, Homo habilis, Neanderthal.

Order (earliest → most recent):
  1. Australopithecus — ~2 mya
  2. Homo habilis — ~2 mya (overlap with Australopithecus); brain 650–800 cc
  3. Homo erectus — ~1.5 mya; brain ~900 cc; first to use fire and migrate from Africa
  4. Neanderthal (Homo neanderthalensis) — 1,00,000–40,000 ya; brain ~1,400 cc
  5. Homo sapiens — appeared in Africa ~200,000 ya; modern form ~75,000–10,000 ya during ice age
Note: brain size increased roughly with time but is not the only marker of human evolution — tool sophistication, language and social complexity also developed.

Competency-Based Questions

Q1. The Hardy–Weinberg equation is: L1 Remember

  • (a) p² + q² = 1
  • (b) p² + 2pq + q² = 1
  • (c) p + q = 0
  • (d) p² − 2pq + q² = 1
Answer: (b) p² + 2pq + q² = 1. This is the binomial expansion of (p + q)². The three terms give the expected frequencies of AA, Aa and aa genotypes in a population at equilibrium.

Q2. Which is NOT a factor that disturbs Hardy–Weinberg equilibrium? L2 Understand

  • (a) Gene flow
  • (b) Genetic drift
  • (c) Natural selection
  • (d) Random mating
Answer: (d) Random mating. Random mating is one of the assumptions of Hardy–Weinberg equilibrium. The five disturbing factors are: (1) gene flow, (2) genetic drift, (3) mutation, (4) genetic recombination and (5) natural selection.

Q3. Calculation: In a population, 4% individuals are homozygous recessive (aa). What is q? L3 Apply

q² = 0.04, therefore q = √0.04 = 0.2.
p = 1 − 0.2 = 0.8.
Heterozygote frequency = 2pq = 2(0.8)(0.2) = 0.32 = 32%.
AA frequency = p² = 0.64 = 64%.

Q4. Analyse: Why is the founder effect a special case of genetic drift? L4 Analyse

Answer: The founder effect occurs when a small group of individuals migrates from a large source population to establish a new colony. The founding group, being small, is a non-random sample of the source gene pool — by pure chance some alleles will be over-represented and others under-represented or missing.

This is genetic drift because the change in allele frequencies is driven by random sampling, not by selection. It is a "special case" because the sampling event is dramatic (sudden migration of a few individuals) rather than spread across many generations. Classic examples: the Amish in Pennsylvania, Easter Island populations.

Q5. HOT (Apply): A geneticist studying a Pacific island population finds q² = 0.04 for a recessive disorder. The mainland population has q² = 0.0001 for the same disorder. Suggest reasons. L6 Create

Hypotheses:
  1. Founder effect: The island was colonised by a small group that happened to include one or more carriers of the recessive allele. q in the founders was much higher than mainland q. This is the most common explanation.
  2. Genetic drift in small population: Over generations, drift increased q on the island; the small population size prevents averaging.
  3. Inbreeding: Limited mate choice on a small island leads to more matings between relatives, increasing homozygosity (more q² individuals from the same q).
  4. Differential selection: Less likely — but possible — that the allele confers some advantage on the island (e.g., disease resistance) that does not apply on the mainland.
  5. Reduced gene flow: Isolation prevents the mainland gene pool from "diluting" the island's elevated q.
Real example: Several rare disorders are unusually common in Finland — a population that was small and isolated for centuries.

Assertion–Reason Questions

Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.

A: If a population is at Hardy–Weinberg equilibrium, no evolution is occurring at that locus.

R: Equilibrium requires that mutation, gene flow, drift, selection and non-random mating all be absent.

Answer: (A). Both true; R explains A. Hardy–Weinberg equilibrium is defined as the state in which all five evolutionary forces are absent. If allele frequencies are not changing, by definition evolution is not happening at that locus.

A: Genetic drift is more important in small populations than large ones.

R: The effects of random sampling are larger when sample sizes are small.

Answer: (A). Both true; R explains A. Just as flipping a coin 10 times can easily give 7 heads (chance deviation), but 10,000 flips will give very close to 5,000 heads, allele sampling deviates more from expectations in small populations.

A: Homo erectus had a smaller brain than Neanderthal.

R: Homo erectus appeared earlier (~1.5 mya) than Neanderthal (~1,00,000 ya), and brain capacity broadly increased over time.

Answer: (A). Both true; R explains A. Homo erectus brain ~900 cc; Neanderthal ~1,400 cc. The trend of increasing brain size with time is a general pattern in human evolution (though not perfectly linear).

Frequently Asked Questions - Hardy Weinberg Human Evolution

What is the main concept covered in Hardy Weinberg Human Evolution?
In NCERT Class 12 Biology Chapter on Evolution, "Hardy Weinberg Human Evolution" covers the core biological structures, processes, and pathways students need for board exam success. The MyAiSchool lesson explains the topic with definitions, labelled diagrams, comparison tables, and interactive simulations. Scientific terminology and physiological/genetic significance are highlighted throughout to build conceptual depth aligned with CBSE 2025-26 syllabus.
How is Hardy Weinberg Human Evolution useful in real-life or applied biology?
Real-life applications of "Hardy Weinberg Human Evolution" from NCERT Class 12 Biology Evolution include medical diagnostics, agriculture, biotechnology, public health, evolutionary insights, and ecological monitoring. The MyAiSchool lesson links every biological concept to a tangible application so students see biology as a problem-solving framework for living systems and real-world challenges.
What are the key terms students should memorize for Hardy Weinberg Human Evolution?
Key terms in "Hardy Weinberg Human Evolution" (NCERT Class 12 Biology Evolution) are tabulated in the MyAiSchool key-terms grid. Students should memorize each term with its precise definition, function, and example. Terminology is high-yield in CBSE board exams — 1-mark MCQs and 2-mark short answers test definitions directly. The Summary section provides a printable quick-reference card.
How does this part connect to other parts of the chapter?
NCERT Class 12 Biology Evolution is structured so each part builds biological understanding sequentially. "Hardy Weinberg Human Evolution" connects to neighbouring parts via shared mechanisms, structural hierarchies, and physiological processes. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected biological story rather than disconnected fragments.
What types of CBSE board questions come from Hardy Weinberg Human Evolution?
CBSE board questions from "Hardy Weinberg Human Evolution" typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer differences/comparisons, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Hardy Weinberg Human Evolution" lesson allows students to explore biological processes, classifications, or pathways using selectors and sliders, with live visual feedback. To use it effectively: (1) explore each option/state, (2) compare with textbook diagrams, (3) note the function/outcome changes, (4) try the integrated practice quiz. The simulation reinforces visual-spatial understanding that pure text-based study cannot.
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