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DSE Biology | Genetics Teaching Guide (for tutors) | DNA Base Pairing, Meiosis and the 9:3:3:1 Ratio
Students can recite that A pairs with T but freeze when asked how long 3.2 billion base pairs are. This lesson uses three numbers: base ratios, two to the power 23, and a sixteen-square grid.
Objectives — Students can: (1) use complementary base pairing to derive all four base percentages and compute the physical length of DNA; (2) explain how two meiotic divisions halve 46 chromosomes and compute the 2^23 combinations; (3) use Punnett squares for monohybrid and dihybrid crosses and explain where 9:3:3:1 and the 1/16 come from.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Cell division | Main difference between mitosis and meiosis | Chromosome number halves |
| Probability | How to combine two independent events | Multiply |
| Percentages | One quarter as a percentage | Twenty-five |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | DNA and ratios | A equals T, G equals C, total 100% | G is 20% |
| 10-24 min | Length and packing | Base pairs times 0.34 nm, then metres | 1.09 m |
| 24-36 min | Meiosis | Two divisions, four haploid cells | 2 to the power 23 |
| 36-46 min | Monohybrid cross | Tt by Tt, four squares | 3:1 and 75% |
| 46-56 min | Dihybrid cross | Sixteen squares, 320 offspring | 20 double recessive |
| 56-60 min | Exit check | Three items marked live | All three right |
03 Teaching script (read it out)
Scene 1: DNA is a chain with ratios, not a string of letters
The human genome holds about 3.2 billion base pairs; A pairs with T and G with C, so the A share always equals the T share.
If A is 30%, T is also 30%, leaving 40% shared equally by G and C at 20% each.
Length: each base pair spans 0.34 nm, so 3.2 billion times 0.34 nm is 1.09 metres, which must fit a nucleus about 6 micrometres across, hence the wrapping around proteins.
At ten base pairs per turn the chain makes about 3.2 hundred million turns.
If A is 30%, T is also 30%, leaving 40% shared equally by G and C at 20% each.
Length: each base pair spans 0.34 nm, so 3.2 billion times 0.34 nm is 1.09 metres, which must fit a nucleus about 6 micrometres across, hence the wrapping around proteins.
At ten base pairs per turn the chain makes about 3.2 hundred million turns.
Ask students: if A is 30%, what is G (20%).
Scene 2: Meiosis is two divisions and one halving
The first division separates homologous chromosomes, taking the count from 46 to 23; the second separates sister chromatids and keeps 23, giving four haploid cells.
Each gamete carries one of 2 to the power 23 = 8,388,608 combinations, and crossing over adds more, so siblings are never identical.
Mitosis makes two cells with 46 chromosomes unchanged, and papers love asking which division halves the number.
Crossing over happens in prophase one, so it reshuffles genes on the same chromosome pair.
Each gamete carries one of 2 to the power 23 = 8,388,608 combinations, and crossing over adds more, so siblings are never identical.
Mitosis makes two cells with 46 chromosomes unchanged, and papers love asking which division halves the number.
Crossing over happens in prophase one, so it reshuffles genes on the same chromosome pair.
Ask students: how does the chromosome number change in meiosis (46 to 23).
Scene 3: Ratios are counted in the grid, not memorised
Tt by Tt gives TT, Tt, Tt and tt, so phenotypes are 3:1, that is 75% tall; among the tall plants only TT breeds true, which is 1/4 of all offspring and 1/3 of the tall ones.
A dihybrid RrYy by RrYy fills sixteen squares giving 9:3:3:1; both recessive traits appear in one square only, 1/16 = 6.25%, so 320 offspring give about 20.
For a sex-linked trait, a carrier mother XCX with a normal father XY has half her sons colour-blind (50%) and no affected daughters, though 50% are carriers.
Always state the denominator: 1/16 without one is an incomplete answer.
A dihybrid RrYy by RrYy fills sixteen squares giving 9:3:3:1; both recessive traits appear in one square only, 1/16 = 6.25%, so 320 offspring give about 20.
For a sex-linked trait, a carrier mother XCX with a normal father XY has half her sons colour-blind (50%) and no affected daughters, though 50% are carriers.
Always state the denominator: 1/16 without one is an incomplete answer.
Ask students: how many of 320 offspring are double recessive (20).
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Base percentages | A equals T, G equals C, four sum to 100 | G = 20% |
| Calculation | DNA length | Base pairs times 0.34 nm, then convert | 1.09 m |
| Explanation | Meiosis and variation | Two divisions, 2^23, crossing over | One sentence each |
| Genetic diagram | Dihybrid ratio | Sixteen-square grid with phenotypes | 9:3:3:1, double recessive 1/16 |
Sanity check (teach this): A equals T and G equals C, and the four percentages must total 100%; a total of 90% means one base was missed.
Convert nanometres to metres by dividing by a billion, so 1.09 metres is not 1.09 millimetres.
Phenotype and genotype ratios differ: 3:1 is phenotypic, 1:2:1 is genotypic.
9:3:3:1 holds only when the two genes are on different chromosomes with complete dominance; linkage shifts it.
Convert nanometres to metres by dividing by a billion, so 1.09 metres is not 1.09 millimetres.
Phenotype and genotype ratios differ: 3:1 is phenotypic, 1:2:1 is genotypic.
9:3:3:1 holds only when the two genes are on different chromosomes with complete dominance; linkage shifts it.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| G share when A is 30% | 20% | (100 minus 2 times 30) over 2 |
| Double recessive among 320 offspring | 20 individuals | 320 divided by 16 |
| Colour-blind sons from a carrier mother | 50% | Sons take X only from the mother |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Pairing A with G | A pairs with G, so they are equal | A pairs with T and G with C; a wrong pairing ruins everything after it. |
| Skipping the unit change | DNA length given as 1.09 millimetres | Nanometres divide by a billion, so the answer is 1.09 metres. |
| Mixing the two ratios | Tt by Tt gives 1:2:1 phenotypes | 1:2:1 is genotypic; phenotypes merge into 3:1. |
| Leaving out the denominator | The double recessive is 1 | It is 1 in 16, so 20 out of 320. |
07 Homework (paper, section and question numbers)
- Practice: genetic diagrams and ratio calculation — Draw the grid before writing ratios
- Practice: DNA structure and base pairing — Four percentages must total 100
- All lesson-prep guides — Compare the ten-section structure
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| G share when A is 30% | 20% | All three right: move to mutation |
| Double recessive in 320 offspring | 20 individuals | One wrong: redraw the grid |
| Colour-blind sons of a carrier mother | 50% | Two wrong: back to act three |
09 Tutor FAQ
Must students memorise every genetic disease?
No. Papers give the pedigree; the skill is inferring genotypes then computing probability.
Why does 9:3:3:1 sometimes fail?
Linkage, lethal genotypes or small samples all skew it, and answers should name the cause.
What is the point of the DNA length?
It gives scale: 1.09 metres folded into a micrometre nucleus explains packing and replication.
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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.
