Hardy-Weinberg Equilibrium Tournament
Meiosis packages the alleles; Hardy-Weinberg squares them. Ten H2 rounds: 2pq, the 1-in-10 000 carrier calculation, counting p from genotypes, the five assumptions, 42% heterozygotes, and why inbreeding does not change allele frequency.
Where the Meiosis fits in Cell Cycle
Meiosis is where the alleles in p and q are packaged into gametes. Hardy-Weinberg is what happens when those gametes pair at random. Click the highlighted meiosis panel to enter the tournament.
Click the highlighted Meiosis box to continue.
What this tournament tests
Each task maps to a distinct GCE A-Level cognitive demand. The first two orient you in the broader topology; the next 8 test the high-yield mechanism, regulation, sequence and quantitative reasoning that consistently appear on test day.
The Bigger Picture
See meiosis as the source of the gametes HWE squares.
Whole-Pathway Overview
Pan the meiosis / variation figure.
Fill in the Blank
The heterozygote term is 2pq.
Disruptor
Incidence 1/10 000 → carrier ≈ 1/50.
Sequence Ordering
Counts → p → expected genotypes.
Match the Pairs
Five assumptions and their violations.
Numeric Input
2pq as a percent when p=0.7.
Select All That Apply
Null model, mating vs p, chi-squared.
Odd One Out
Inbreeding reshuffles, it does not change p.
Variation Disruptor
Polymorphism is not a HWE violation.
Public leaderboard
Your score posts to a global, persistent leaderboard scored by points first, time as tiebreaker.
Hardy-Weinberg in 60 seconds
(p + q)² = p² + 2pq + q². p is the allele, not the dominant phenotype.
Rare recessive: q = √incidence, carrier frequency 2pq ≈ 2q.
Five assumptions. Break selection, mutation, migration, or drift and p moves. Break random mating and only genotype frequencies move.
The classic 9744 trap: a dominant phenotype frequency is p² + 2pq, not p. Count alleles from genotypes (or take √q² for a rare recessive) before you write 2pq.
FAQ
How do I get carrier frequency from a rare recessive disease?
If incidence is q², then q = √incidence and carriers are 2pq ≈ 2q when p ≈ 1. For 1 in 10 000, q = 0.01 and 2pq ≈ 0.02 — about 1 in 50. Do not treat the dominant phenotype as p.
Does inbreeding change allele frequency p?
No. Inbreeding is non-random mating. It raises homozygosity and lowers 2pq, but the allele count is unchanged. Selection, mutation, migration, and drift are the forces that move p.
What are the five Hardy-Weinberg assumptions?
No selection, no mutation, no migration, no drift (infinite population), and random mating. Break any of the first four and p changes. Break only random mating and genotype frequencies change while p stays put.
Is this a calculator or a tournament?
A timed tournament. You still do the 2pq arithmetic yourself — the point is retrieval under a clock, not a black-box solver.
Keep going
Where HWE thinking meets a real evolving pathogen.
Different paper, same exam — the ecology cluster.
Overview diagram: Reactome Pathway R-HSA-1640170, licensed CC BY 4.0. Reference figure: WikiPathways WP4655.