DSE Biology — Genetics and DNA Technology: Lesson Prep Guide (Tutor Edition)

2026-11-03

Tutor lesson-prep guides · Secondary Biology

Students can recite the 3:1 ratio but freeze at a test cross or a PCR cycle count. This lesson puts inheritance ratios and DNA technology on one causal chain: genotype decides phenotype, and the technology only cuts, copies and reads.

Objectives — Students can: (1) turn 200 offspring and a 3:1 ratio into 150 dominant and 50 recessive, and explain why a test cross gives 1:1; (2) build 64 codons from 4 bases in triplets, note that 61 code for 20 amino acids and 3 are stop signals, and explain why synonymous mutations are tolerated; (3) treat a 6-base recognition site as one cut every 4096 bases on average, scale 20 PCR cycles to about 1,048,576 copies, and multiply locus frequencies 0.12, 0.05 and 0.02 to a match chance near 1 in 8,333.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Meiosis	How many alleles does a gamete carry per locus?	One, because the two parental alleles separate
Protein synthesis	How many bases code for one amino acid?	Three, called a codon
Probability	How do you combine two independent events?	Multiply, never add
Enzymes	Where does a restriction enzyme cut?	At a specific base sequence, usually leaving sticky ends

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Ratio warm-up	Give 200 offspring and ask for a 3:1 split	150 and 50
10-22 min	Test cross	Work back from a 1:1 result to the parent genotype	The tester must be homozygous recessive
22-32 min	Codons	Build 64 triplets from 4 bases	61 sense and 3 stop
32-44 min	Sites and PCR	A 6-base site and 20 cycles	One per 4096 bases, about 1,048,576 copies
44-52 min	Question drill	One ratio, one mutation, one site, one profile question	Eight minutes unaided
52-60 min	Exit check	Three written questions marked on the spot	Done independently

03 Teaching script (read it out)

Scene 1: 3:1 is a probability, not a law

In an Aa self-cross each offspring has four equally likely combinations, giving three dominant to one recessive. With 200 offspring the expectation is 150 to 50, but four offspring will not land exactly on 3:1, and the smaller the sample the wider the spread. A test cross mates the unknown with a homozygous recessive: if the parent is Aa, half the offspring are aa, so 1:1; if all show the dominant trait, the parent is AA. That is exactly why examiners like the test cross: it turns an invisible genotype into a visible ratio.

Ask the student to restate why the tester must be homozygous recessive.

Scene 2: Sixty-four codons for twenty amino acids

4 to the power 3 is 64; 3 are stop signals and 61 code for 20 amino acids, so several codons share one amino acid. That redundancy means a single base substitution often changes nothing, while an insertion or deletion shifts the reading frame and changes everything downstream. Students must separate the two: substitution depends on position, insertion and deletion depend on frame.

Ask the student to restate why one inserted base hurts more than one substituted base.

Scene 3: Technology only cuts, copies and reads

A 6-base site appears about once every 4096 bases in random DNA, so choosing the right enzyme gives predictable fragments; gel electrophoresis then sorts by length and short fragments travel furthest. PCR doubles each cycle, so 20 cycles give 2 to the power 20, about 1,048,576 copies, enough to read a few dozen cells. DNA profiling does not read the whole genome: it reads several repeat loci, and 0.12 x 0.05 x 0.02 gives roughly 1 in 8,333, which is why more loci make a match harder to fake.

Ask the student to restate why a DNA profile needs several loci instead of one.

04 Question types and worked steps

Type	Example	Steps	Answer
Ratio calculation	200 offspring with a 3:1 ratio	1) Divide by four 2) Multiply by three and one 3) State it as an expectation	150 to 50
Test cross	Offspring show 1:1, so what is the parent	1) The tester always gives a 2) Half the offspring are recessive 3) The unknown is Aa	Aa x aa
Codon and mutation	What happens when one base is inserted	1) Reading frame shifts 2) Later amino acids change 3) A stop codon usually appears early	Frameshift, far broader than a substitution
DNA technology	20 PCR cycles and three locus frequencies	1) Raise two to the cycle count 2) Multiply the frequencies 3) Invert for the odds	1,048,576 copies, about 1 in 8,333

Sanity check (teach this): Sanity checks: never mix genotype ratio with phenotype ratio; offspring counts are whole numbers and 150 plus 50 must equal 200; sense codons (61) must be fewer than 64; PCR doubles rather than adds, so 20 cycles is not 20 times; short fragments travel furthest in a gel; multiplying probabilities always gives a smaller number than any single frequency.

05 Independent practice (3 questions)

Question	Answer	What to watch
200 offspring with a 3:1 ratio: how many dominant?	150	Check the student divides by four before multiplying by three
How many triplets can 4 bases form?	64	Check the student cubes four rather than multiplying by three
How many copies after 20 PCR cycles?	About 1,048,576	Check the student uses a power of two

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating 3:1 as guaranteed	Four offspring must be three tall and one short	3:1 is an expectation; only around 200 offspring approach 150 to 50.
Mixing up the two ratios	An Aa self-cross gives a 3:1 genotype ratio	Genotypes are 1 AA to 2 Aa to 1 aa; 3:1 is the phenotype ratio.
Assuming 64 codons mean 64 amino acids	Sixty-four codons means sixty-four amino acids	Only 20 amino acids exist; 3 codons are stops and the rest repeat.
Treating PCR as linear	Twenty cycles means twenty times as much DNA	Each cycle doubles, so 20 cycles give 2 to the 20, about 1,048,576 copies.

07 Homework (paper, section and question numbers)

Secondary Biology chapter index — Copy three keywords from the genetics chapter listed there and define each in one sentence
Cell cycle and mitosis (S4 chapter) — Redo practice question 2 and state how meiosis differs from mitosis in chromosome number
Enzymes and their factors (S4 chapter) — Use the temperature curve there to explain why restriction enzymes need a set temperature

08 Exit check (last 5 minutes)

Question	Answer	Decision
200 offspring with a 3:1 ratio: how many recessive?	50	Wrong answer 66 means re-teach the four quarters
How many sense codons are there?	61	Wrong answer 64 means re-teach the stop signals
Which fragment travels furthest in a gel?	The shortest one	Wrong answer “the longest” means re-teach charge and mesh size

09 Tutor FAQ

Which year is this for?

S6 genetics; S5 classes can start with ratios and codons and leave the technology for later.

Is 60 minutes enough?

Yes; if probability is weak, spend ten minutes on quarters and products and set the profiling question as homework.

Is there practice attached?

The homework links to in-site chapters whose practice questions carry answers.

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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.