HKDSE Biology | Evolution and Natural Selection: tutor prep guide

HKDSE Biology | Evolution and Natural Selection: tutor prep guide | allele frequencies, 2pq and speciation

2026-11-01

Tutor lesson-prep guides · Secondary Biology

Students see 2pq and multiply p by q once, and they assume a falling frequency must reach zero. This lesson uses one population of numbers to join the formula, its five conditions and selection pressure into a single chain.

Objectives — Students can: (1) state the three requirements for natural selection and give examples; (2) use p + q = 1 and 2pq to compute genotype frequencies; (3) name the five Hardy-Weinberg conditions and identify which one a question has broken.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Multiplying probabilities	Joint probability of independent events	multiply
Alleles	Different versions at the same locus	one from each parent
Dominance	Which phenotype Aa shows	the dominant one

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Three requirements	Variation, inheritance, differential survival	Give a local example
10-24 min	Allele frequency	p + q = 1	Find q when p = 0.8
24-40 min	Genotype frequency	p squared + 2pq + q squared = 1	Obtain 320 carriers
40-50 min	Breaking equilibrium	Selection shifts q	q falls from 0.20 to 0.1667
50-56 min	Speciation	Geographic then reproductive isolation	Decide if one species
56-60 min	Exit check	Three questions marked at once	Independent work

03 Teaching script (read it out)

Scene 1: Natural selection: all three or nothing

There must be (1) variation within the population, (2) heritability of that variation, and (3) a link between the variation and survival or reproduction. Only with all three does a trait frequency shift across generations. “The environment changed” is not enough: students must name the trait and the advantage it gave.

Task: use antibiotic resistance in bacteria and point to each of the three requirements.

Scene 2: Hardy-Weinberg: one equation, two unknowns

p + q = 1 and p squared + 2pq + q squared = 1. With p = 0.8 and q = 0.2 in 1000 individuals: AA = 0.64 x 1000 = 640, Aa = 2 x 0.8 x 0.2 = 0.32 giving 320, aa = 0.04 giving 40. The five conditions are no mutation, no migration, random mating, a large population and no selection. When a question asks why reality differs, one of these five has been broken.

Task: how many of the 1000 carry the recessive allele at all (360, counting heterozygotes)?

Scene 3: Selection pressure: falling is not vanishing

If every aa dies before reproducing, the next generation has q prime = q/(1 + q) = 0.2/1.2 = 0.1667, a drop of only 0.0333 in one generation. The recessive allele hides in heterozygotes, which selection cannot see, so it is never cleared completely. That is why disease alleles persist in populations. Speciation is a separate layer: geographic separation, then accumulated mutations, then reproductive isolation (no mating, or sterile offspring) before we call it a new species.

Note: “how many generations until it disappears?” is answered “never; it approaches zero”.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Genotype frequency	p squared, 2pq, q squared times N	320 Aa
Calculation	Frequency after selection	q prime = q/(1+q) for a lethal recessive	0.1667
Explanation	Conditions	Check the five against the scenario	Name the broken one
Judgement	Same species?	Look for reproductive isolation	Fertile offspring needed

Sanity check (teach this): p + q must equal 1 and the three genotype frequencies must sum to 1; the counts must add to 1000. Frequencies stay between 0 and 1, and no single generation of selection can remove a recessive allele completely.

05 Independent practice (3 questions)

Question	Answer	What to watch
Heterozygotes when p = 0.8 and N = 1000	320	2pq times N
q after one generation of lethal aa	0.1667	q/(1+q)
Two populations mate but offspring are sterile: one species?	No	Reproductive isolation already exists

06 Common mistakes and what to say

Mistake	Student writes	Say this
Writing 2pq as 2(p+q)	Multiplying 2 by 1	“p and q are frequencies; 2pq is at most 0.5.”
Forgetting to multiply by N	Answering 0.32 as a count	“0.32 is a proportion; the question asked how many.”
Expecting the allele to vanish	Saying it disappears in a few generations	“It hides in heterozygotes, so it never reaches zero.”
Calling separation speciation	Two places means two species	“You need reproductive isolation first.”

07 Homework (paper, section and question numbers)

Secondary Biology chapter index — All chapters
HKDSE Biology practice paper — Genetics and evolution

08 Exit check (last 5 minutes)

Question	Answer	Decision
q when p = 0.8	0.2	All three right: move to ecology
Heterozygotes in 1000 individuals	320	One wrong: redo 2pq next lesson
The five Hardy-Weinberg conditions	no mutation, no migration, random mating, large population, no selection	Two wrong: back to act two

09 Tutor FAQ

Why assume random mating?

Non-random mating changes genotype proportions; even with unchanged allele frequencies, 2pq no longer holds.

Must selection involve death?

No. Lower fertility or lower mating success is selection pressure too.

Is equilibrium always good?

No. It only means no evolution is happening; a changed environment breaks it.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.