IB DP Biology: Dihybrid crosses, gene linkage and the chi-squared test (HL) (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology sections D3.2.16 to D3.2.21 (dihybrid crosses, linkage, recombinants and the chi-squared test) from the IB Biology guide, first assessment 2025. Every statement here is additional higher level, so all of it is HL only. For full explanations and worked examples, read the study guide first.

Links: IB DP Biology hub, printable checklist, practice questions and free 10-minute diagnostics.

Key terms

Term	Meaning
Segregation	Homologous chromosomes separate in anaphase I, so each gamete gets one allele of each gene
Independent assortment	The orientation of one bivalent in metaphase I does not affect any other, so alleles of unlinked genes combine at random
Dihybrid cross	A cross that follows two genes
Locus	The position of a gene on a chromosome
Linked genes	Genes with loci on the same chromosome
Autosome	A chromosome that is not a sex chromosome
Parental type	A gamete or offspring carrying the allele combinations already on the parent’s chromosomes
Recombinant	A gamete or offspring carrying a new combination, made by crossing over
Recombinant frequency	recombinants ÷ total offspring × 100

D3.2.16: meiosis explains the ratios

• Segregation: homologues separate in anaphase I.
• Independent assortment: random orientation of bivalents at the equator in metaphase I.
• AaBb on different chromosomes gives AB, Ab, aB, ab, each ¼.
• Combinations of chromosomes from orientation alone = 2ⁿ for n pairs.

D3.2.17: Punnett grids and ratios

Cross	Offspring phenotype ratio
AaBb × AaBb (both heterozygous, unlinked)	9:3:3:1
AaBb × aabb (test cross, unlinked)	1:1:1:1

Method in steps

1. Write the genotypes and the gametes each parent makes (four from AaBb).
2. Draw a 4 × 4 grid for two heterozygotes (a test cross has a 4 × 1 grid).
3. Fill in the cells and count by phenotype.
4. Check: the numbers add to 16 (or 4).

Why 9:3:3:1: (3:1) × (3:1) = 9:3:3:1. Multiply independent probabilities: ¾ × ¾ = 9/16, ¾ × ¼ = 3/16, ¼ × ¼ = 1/16.

Shortcut grid for a test cross (alleles A/a and B/b, unlinked):

Gamete from AaBb	Tester gamete ab gives	Phenotype
AB	AaBb	both dominant
Ab	Aabb	A only
aB	aaBb	B only
ab	aabb	both recessive

Result: 1:1:1:1.

For linked genes the same four offspring classes appear, but in unequal numbers.

Genotype counts in the 16 cells: nine genotypes, 1:2:1:2:4:2:1:2:1. Four of 16 are heterozygous for both genes. Four of 16 are homozygous for both.

Nature of science: the ratios rest on Mendel’s second law. It fails when genes are close together on one chromosome. All biological “laws” have exceptions under some conditions.

D3.2.18: gene loci in databases

Gene	Product	Locus
HBA1	alpha-globin	16p13.3
HBB	beta-globin	11p15.4
HBD	delta-globin	11p15.4
CFTR	membrane channel for chloride and bicarbonate ions	7q31.2

• Different chromosomes (HBA1 and HBB): expect independent assortment.
• Same chromosome, close together (HBB and HBD): expect linkage.
• Method: search the gene symbol, read the chromosome and band, read the product, then compare two genes.

D3.2.19: linkage

• Write linked alleles beside two vertical lines for the two homologues, for example SH | sh.
• Linked alleles travel together because whole chromosomes are moved to poles.
• They separate only by crossing over in prophase I, between non-sister chromatids at a chiasma.
• Closer loci: fewer crossovers, fewer recombinants.
• Maximum recombinant frequency is 50%, which is the value for unlinked genes.

D3.2.20: recombinants in a test cross

Method in steps

1. Write the heterozygote with alleles beside the vertical lines, for example RT | rt or Rt | rT.
2. List its gametes: two parental, two recombinant.
3. The tester gives only recessive alleles, so offspring phenotypes mirror the heterozygote’s gametes.
4. Label the two largest classes parental and the two smallest recombinant (check against step 1).
5. Recombinant frequency = (sum of recombinant offspring) ÷ total × 100.

Worked reminder. An invented fish with RT | rt is crossed with rt | rt. Of 300 offspring: RT 104, rt 96, Rt 52, rT 48. Recombinants = 52 + 48 = 100. Frequency = 100/300 × 100 = 33.3%. This is below 50%, so linkage is indicated, though loosely.

If the heterozygote is Rt | rT, the parental gametes are Rt and rT, so the recombinants are RT and rt.

D3.2.21: chi-squared test

Step	What to write
1	Null hypothesis: no significant difference between observed and expected; chance only
2	Alternative hypothesis: there is a significant difference
3	Expected counts from the ratio × total
4	χ² = Σ (O − E)² / E
5	Degrees of freedom = classes − 1
6	Compare with the critical value at p = 0.05
7	Conclude in words about the biology

Degrees of freedom	1	2	3	4
Critical value, p = 0.05	3.84	5.99	7.81	9.49

• χ² below the critical value: do not reject the null hypothesis; differences are consistent with chance.
• χ² at or above it: reject the null hypothesis; the difference is significant.
• p = 0.05 means a 5% chance of results this extreme if the null hypothesis were true.
• The sample is the F2 generation (or the test-cross offspring), standing for the population.

Worked reminder. 300 F2 offspring against 9:3:3:1. Observed 170, 52, 54, 24. Expected 168.75, 56.25, 56.25, 18.75. χ² = 0.01 + 0.32 + 0.09 + 1.47 = 1.89. With 3 degrees of freedom this is below 7.81, so do not reject the null hypothesis.

Worked reminder: linkage test. Take the fish cross above: 104, 96, 52, 48 against 1:1:1:1. Expected 75 each. χ² = 841/75 + 441/75 + 529/75 + 729/75 = 11.21 + 5.88 + 7.05 + 9.72 = 33.9. This is above 7.81, so reject the null hypothesis. The genes are not assorting independently, which fits the 33.3% recombinant frequency.

Model wording for a conclusion. “Calculated χ² is 33.9 and the critical value for 3 degrees of freedom at p = 0.05 is 7.81. Because 33.9 is greater, the difference between observed and expected is significant, so the null hypothesis is rejected. The two genes are probably linked.”

Must-know distinctions

• Segregation vs independent assortment: one gene separating vs two genes combining at random.
• Parental vs recombinant: present on a parental chromosome vs a new combination.
• 9:3:3:1 vs 1:1:1:1: two heterozygotes crossed vs a heterozygote crossed with a double recessive.
• Reject vs accept: you reject, or fail to reject, the null hypothesis.
• Observed vs expected: counts from the experiment vs counts predicted by the ratio.

Quick self-test

1. How many orientation combinations are possible in metaphase I of a cell with six pairs of homologous chromosomes?
2. What fraction of gametes from an AaBb individual with unlinked genes is aB?
3. Give the phenotypic ratio from AaBb × aabb for unlinked genes.
4. In a cross AaBb × AaBb with unlinked genes, what fraction of offspring is aabb?
5. Explain why HBA1 and HBB should assort independently.
6. Define a recombinant.
7. State the maximum recombinant frequency and give the reason.
8. A test cross gives 380 offspring, of which 57 are recombinant. Calculate the recombinant frequency.
9. State the degrees of freedom and the critical value (p = 0.05) for a test with four phenotype classes.
10. A test cross gives 58, 52, 47, 43 in four classes, expected 1:1:1:1. Calculate χ² and give the conclusion.
11. Why is “accept the null hypothesis” poor wording?

Answers

1. 2⁶ = 64.
2. ¼.
3. 1:1:1:1.
4. ¼ × ¼ = 1/16.
5. They are on different chromosomes (16 and 11), which move independently at metaphase I.
6. An individual or gamete with a combination of alleles not on either chromosome of the parent, produced by crossing over.
7. 50%; even with a chiasma between the loci in every meiosis, only two of four chromatids are recombinant.
8. 57/380 × 100 = 15.0%.
9. 3 degrees of freedom; critical value 7.81.
10. Expected 50 each. χ² = (64 + 4 + 9 + 49)/50 = 2.52, below 7.81, so do not reject the null hypothesis: no significant departure from 1:1:1:1.
11. A non-significant result shows the data do not contradict the hypothesis; it does not prove it.

Where marks are usually lost

• Writing gametes for AaBb as two types rather than four.
• Putting alleles of linked genes in the wrong arrangement beside the vertical lines.
• Labelling the biggest classes as recombinants.
• Using percentages instead of counts in χ².
• Forgetting to state the null hypothesis, or stating it as “the genes are linked”.
• Miscounting degrees of freedom.
• Concluding “significant” without comparing with the critical value.
• Stopping at “reject the null hypothesis” without stating what that means for the genes.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Sections D3.2.16 to D3.2.21 (additional higher level).