State the Hardy-Weinberg principle and assumptions

WHAT is it?

WHY do we care? Because it is a null hypothesis. If real genotype numbers match Hardy-Weinberg predictions → the gene is probably not evolving. If they don't match → some force (selection, drift, migration, etc.) is at work, and that deviation is what tells us evolution is happening.

HOW do we derive it from first principles?

Don't memorise $p^2 + 2pq + q^2$ . Build it.

Step 1 — Define the gene pool. Imagine all the gametes (eggs + sperm) of the whole population dumped into one bucket. A fraction $p$ carry allele $A$ ; a fraction $q$ carry allele $a$ . Why this step? Random mating means offspring are formed by drawing two gametes at random from this bucket — so genotype frequencies become a probability problem.

Step 2 — Draw two gametes independently. Probability the offspring is:

$AA$ = P(first is $A$ ) × P(second is $A$ ) = $p \times p = p^2$
$aa$ = $q \times q = q^2$
$Aa$ = $A$ then $a$ , or $a$ then $A$ = $pq + qp = 2pq$

Why the factor of 2? Because heterozygotes can form two ways — you can get $A$ from mum & $a$ from dad, or the reverse. Miss this and your frequencies won't add to 1.

Step 3 — Check they sum to 1. $p^2 + 2pq + q^2 = (p+q)^2 = (1)^2 = 1 \checkmark$ Why this step? It's the algebraic guarantee we've covered every possible offspring — nothing is missing.

Step 4 — Show it's stable (the actual "equilibrium"). Compute the allele frequency of $A$ in the offspring generation. Every $AA$ contributes 2 copies of $A$ ; every $Aa$ contributes 1: $p' = \frac{2p^2 + 2pq}{2} = p^2 + pq = p(p+q) = p(1) = p$ So $p' = p$ . The allele frequency did not change. That constancy — generation after generation — is Hardy-Weinberg equilibrium.

Figure — State the Hardy-Weinberg principle and assumptions

The 5 Assumptions (the "no-force" conditions)

Equilibrium only holds if all five are true. Each failed assumption is the definition of an evolutionary force.

Assumption	If violated →
No mutation ( $A \leftrightarrow a$ doesn't happen)	new alleles / changed $p,q$
No gene flow (migration) — no alleles in/out	frequencies shift toward migrants
Very large population (ideally infinite)	random drift changes $p,q$ by chance
Random mating (no mate preference)	genotype ratios distort (e.g. inbreeding ↑ homozygotes)
No natural selection — all genotypes survive & reproduce equally	favoured alleles rise

Worked Examples

Worked example Example 1 — From recessive phenotype to allele frequency

In a population, 9% of people show a recessive disorder ( $aa$ ). Find $p$ , $q$ , and the % of carriers.

Step 1: $q^2 = 0.09$ . Why? Only $aa$ shows the recessive phenotype, and $aa$ has frequency $q^2$ . Step 2: $q = \sqrt{0.09} = 0.3$ . Why? $q$ is the allele frequency; take the square root of the genotype frequency. Step 3: $p = 1 - q = 0.7$ . Why? Because $p+q=1$ (only two alleles). Step 4: carriers $= 2pq = 2(0.7)(0.3) = 0.42 = \mathbf{42\%}$ . Why? Carriers are heterozygous $Aa$ , frequency $2pq$ . Sanity check: $p^2 + 2pq + q^2 = 0.49 + 0.42 + 0.09 = 1.00 \checkmark$

Common Mistakes

Recall Feynman: explain it to a 12-year-old

Imagine a giant jar of red and blue marbles (red = allele $A$ , blue = allele $a$ ). To make a baby, you close your eyes and grab two marbles at random. If nobody adds, removes, repaints, or cheats by picking favourites, then the fraction of red vs blue marbles in the jar stays the same forever. And you can predict exactly how many babies get two reds, two blues, or one of each — just by multiplying the chances. That "nothing changes" jar is Hardy-Weinberg. When the real jar does change, you know someone's been messing with the marbles — that "messing" is evolution.

Active Recall Flashcards

State the Hardy-Weinberg equation for genotype frequencies.

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

, with

p+q=1

What does

2pq

represent and why the factor of 2?

Frequency of heterozygotes (

Aa

); the 2 is because

Aa

can form two ways (

A

a

a

A

List the 5 assumptions of Hardy-Weinberg equilibrium.

No mutation, no gene flow/migration, very large (infinite) population, random mating, no natural selection.

If 16% of a population is

aa

, what is

q

q=\sqrt{0.16}=0.4

Under HW, what happens to allele frequencies across generations?

They stay constant (

p'=p

q'=q

Why is Hardy-Weinberg called a "null hypothesis"?

It predicts frequencies with NO evolution; deviations from it reveal that an evolutionary force is acting.

Does dominance make an allele more frequent under HW?

No — dominance affects phenotype masking, not allele frequency.

How do you calculate allele frequency

p

from genotype counts?

p=\frac{2(AA)+(Aa)}{2\times \text{total individuals}}

Which violated assumption corresponds to genetic drift?

Small population size (finite population).

Show that

p^2+2pq+q^2

sums to 1.

It equals

(p+q)^2=1^2=1

Connections

Allele and Genotype Frequencies
Natural Selection — violates the "no selection" assumption
Genetic Drift — violates "large population"
Gene Flow and Migration — violates "no gene flow"
Mutation as Source of Variation — violates "no mutation"
Non-random Mating and Inbreeding — distorts genotype ratios
Speciation — happens when populations diverge away from shared HW equilibrium
Punnett Square — the single-cross version of the same gamete-drawing logic

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Hardy-Weinberg principle basically ek "kuch nahi badal raha" wali situation hai. Socho ek badi population hai jahan mating random hai, koi mutation nahi, koi bahar se aa-ja nahi raha (migration), aur koi allele dusre pe selection ka pressure nahi daal raha. Aisi ideal condition mein allele frequencies ( $p$ aur $q$ ) generation se generation same rehti hain — bilkul constant. Isko yaad rakho jaise physics ka "no force, no acceleration" rule.

Formula simple hai: $p+q=1$ aur $p^2 + 2pq + q^2 = 1$ . Yahan $p^2$ matlab $AA$ (homozygous dominant), $q^2$ matlab $aa$ (homozygous recessive), aur $2pq$ matlab $Aa$ (heterozygous). Yeh formula hum derive karte hain gene pool se — do gametes randomly uthao, probability multiply karo, bas ho gaya. $2pq$ mein jo "2" hai woh isliye hai kyunki $Aa$ do tareeke se ban sakta hai (mummy se $A$ + papa se $a$ , ya ulta).

Ab iska asli fayda kya hai? Yeh ek null hypothesis hai. Agar real population ke genotype numbers is formula se match karte hain, matlab woh gene evolve nahi ho raha. Agar match nahi karte, matlab koi na koi force (selection, drift, migration, mutation, ya non-random mating) kaam kar raha hai — aur wahi deviation humein batata hai ki evolution ho raha hai. Exam tip: agar recessive phenotype ka percentage diya ho, woh $q^2$ hota hai, uska square root lo to $q$ milega. Aur factor of 2 kabhi mat bhoolna warna sab galat!

Test yourself — Population Genetics & Speciation

Connections