HKDSE Biology | Genetics and Mendel | Punnett Squares, Test Crosses and Sex Linkage

2026-11-01

Tutor lesson-prep guides · Secondary Biology

Students know the big T is dominant, then guess their way through a pedigree. This lesson drills the same three steps, parent genotype to gametes to Punnett square, until it runs like a times table.

Objectives — Students can: 1) work out possible genotypes from a phenotype; 2) use a Punnett square to get offspring ratios; 3) explain what a test cross proves and calculate the 1 to 1 result; 4) decide from a pedigree whether a trait is autosomal or X-linked.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Cell division	Meiosis and gametes	Chromosome number halves
DNA and genes	A gene codes for a trait	Alleles come in pairs
Probability	Independent events multiply	Half times half is a quarter

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Vocabulary	Genotype, phenotype, dominant, recessive	Name the phenotype of Tt
10-24 min	Monohybrid cross	Punnett square for Tt by Tt	3 to 1, that is 300 to 100
24-34 min	Test cross	Tt by tt gives 1 to 1	100 to 100
34-46 min	Two traits	Independent assortment, 9:3:3:1	[180, 60, 60, 20]
46-56 min	Sex linkage	X-linked, males have one X	50 per cent of sons affected
56-60 min	Exit check	Three questions collected and marked	Done independently

03 Teaching script (read it out)

Scene 1: Parent to offspring in three steps

Every genetics question runs the same three steps: write the parent genotypes, write the gametes, then pair them in a Punnett square. Tt by Tt gives gametes T and t in equal numbers, so the square gives TT to Tt to tT to tt as 1 to 2 to 1, and the phenotypes as 3 to 1. Out of 400 offspring that is about 300 tall and 100 dwarf.
The key sentence: a ratio is a probability, not a promise. Four hundred offspring will not land exactly on 300, only close.

Ask the student: TT by tt gives all Tt, so why is the ratio not 1 to 1?

Scene 2: Test cross: forcing the genotype into the open

An unknown tall plant may be TT or Tt. Crossing it with tt settles it: TT by tt gives all Tt, all tall, while Tt by tt gives 1 Tt to 1 tt, one tall to one dwarf. With 200 offspring split 100 and 100, the parent must be Tt.
For two traits, independent assortment gives 9:3:3:1 from TtRr by TtRr. Out of 320 offspring that is 180 to 60 to 60 to 20.

Ask the student to compute: the double recessive is one sixteenth, so 320 divided by 16 is 20.

Scene 3: Sex linkage: count sons and daughters separately

With an X-linked recessive trait such as colour blindness the key is that a male has only one X. A carrier mother X-B X-b and a normal father X-B Y give sons the mother X and the father Y, so 50 per cent of sons are affected. Daughters always receive the father X-B, so 0 per cent are affected and 50 per cent are carriers.
Three pedigree clues: affected males far outnumber affected females, an affected father has no affected daughters, and every affected male has a carrier or affected mother. Two of the three means X-linked recessive.

Reminder: do not write that the sons have a 50 per cent chance; write that each son has a 50 per cent chance.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculate	Phenotypes among 400 offspring of Tt by Tt	3 to 1	300 tall, 100 dwarf
Deduce	Tall by dwarf gives 1 to 1	The test cross result	Parents are Tt by tt
Calculate	320 offspring of TtRr by TtRr	9:3:3:1	180:60:60:20
Explain	Affected sons from a carrier mother	X-linked, sons get the mother X	50 per cent

Sanity check (teach this): A monohybrid phenotype ratio is always 3 to 1 or 1 to 1, and a ratio near 2 to 1 means TT and Tt were counted separately. The four parts of 9:3:3:1 must add to the total (320). With a normal father no daughter can be affected, so more affected daughters than sons means the trait is autosomal, not X-linked.

05 Independent practice (3 questions)

Question	Answer	What to watch
Genotype ratio from Tt by Tt	1 TT to 2 Tt to 1 tt	Fill the four squares
Phenotype ratio from Tt by tt	1 to 1	Test cross
Affected sons from a carrier mother and normal father	50 per cent	Sons take the mother X

06 Common mistakes and what to say

Mistake	Student writes	Say this
Mixing genotype and phenotype	Calls Tt the tall gene	Tt is the genotype, tall is the phenotype.
Forgetting independent assortment	Treats TtRr as one gene	Sort the two pairs first, then multiply.
Calling a carrier affected	Says the mother is ill so the son is ill	A carrier is healthy but can pass the allele on.
Running a sex-linked cross as autosomal	Uses one probability for sons and daughters	Males have one X, so split the sexes.

07 Homework (paper, section and question numbers)

Secondary Biology chapter index — All chapters
Prep guide master index — Other chapters in this subject

08 Exit check (last 5 minutes)

Question	Answer	Decision
400 offspring of Tt by Tt	300 tall, 100 dwarf	All three right: move to molecular genetics
What a test cross is for	Separating TT from Tt	One wrong: reteach gametes
Chance an affected son is born	50 per cent	Two wrong: restart scene three

09 Tutor FAQ

Why is the ratio 3 to 1 and not 2 to 1?

Tt arises two ways, from father T with mother t and the reverse, so genotypes are 1 to 2 to 1, and TT plus Tt all look tall.

Must the test cross use a recessive individual?

Yes. A recessive partner supplies only t gametes, so any dwarf offspring proves the unknown parent carries t.

Why are X-linked conditions commoner in males?

One recessive allele on their single X is enough, while a female needs two.

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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.

Practice papers: Paper 1 (alleles and the 3:1 ratio) · Paper 2 (probability and dihybrid crosses) · Paper 3 (pedigrees and inherited disorders) · Paper 4 (probability calculations)