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

This study guide covers IB Diploma Programme Biology, sections D3.2.16 to D3.2.21 of the second part of D3.2 Inheritance, as set out in the IB Biology guide, first assessment 2025. All six statements are in the additional higher level (AHL) part of the guide, so everything here is HL only. The single-gene material in D3.2.1 to D3.2.15 is not repeated.

Use it with the revision notes and the practice questions. Course hub: IB DP Biology. Checklist: printable checklist. To find weak spots first, try the free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do	Tier
D3.2.16	Link chromosome movements in meiosis to the outcome of dihybrid crosses of unlinked genes	HL only
D3.2.17	Use Punnett grids; explain how 9:3:3:1 and 1:1:1:1 arise; Mendel’s second law as a “law” with exceptions	HL only
D3.2.18	Search databases for where human genes sit and which polypeptides they code for; pick gene pairs on different chromosomes and on the same one	HL only
D3.2.19	Draw linked crosses with alleles beside vertical lines for homologues; explain why linked alleles can fail to assort independently	HL only
D3.2.20	Work out outcomes of a double heterozygote crossed with a double homozygous recessive; identify recombinants in gametes, genotypes and phenotypes	HL only
D3.2.21	Use a chi-squared test on dihybrid data: significance, p = 0.05, null and alternative hypotheses, observed versus expected	HL only

Segregation and independent assortment (D3.2.16)

Alleles of one gene are on homologous chromosomes. In anaphase I of meiosis the homologues of each pair separate, so each gamete receives one allele of each gene. This is segregation.

Each homologous pair lines up at the equator of the cell in metaphase I. Which way round each pair faces the poles is random and does not affect the orientation of any other pair. Genes on different chromosomes therefore assort independently: the allele of gene 1 a gamete gets tells you nothing about its allele of gene 2.

A cell heterozygous for two unlinked genes, AaBb, can orient its two bivalents in two equally likely ways:

• A-bearing and B-bearing chromosomes face the same pole: gametes AB and ab.
• A-bearing and b-bearing chromosomes face the same pole: gametes Ab and aB.

Across many meioses all four gamete types appear in equal numbers, each ¼. With n pairs of homologous chromosomes there are 2ⁿ possible orientations. For n = 3 that is 8; for the 23 pairs of a human cell it is 2²³ = 8 388 608, before crossing over adds more variety.

Dihybrid crosses and Punnett grids (D3.2.17)

A dihybrid cross follows two genes at once. Take the fruit fly Drosophila melanogaster. Ebony body (e) is recessive to grey (E), and vestigial wings (n) are recessive to normal wings (N). The two genes are on different chromosomes. (The letters are simplified for this guide.)

Worked example 1: heterozygote × heterozygote. Cross EeNn × EeNn.

Step 1: gametes from each parent are EN, En, eN, en, each ¼.

Step 2: write one parent’s gametes across the top and the other’s down the side of a 4 × 4 grid. Fill each of the 16 cells with the combined genotype.

Step 3: read off the phenotypes.

Phenotype	Genotypes	Cells
grey body, normal wings	E_ N_	9
grey body, vestigial wings	E_ nn	3
ebony body, normal wings	ee N_	3
ebony body, vestigial wings	ee nn	1

The phenotypic ratio is 9:3:3:1. There are nine different genotypes in the grid, in the ratio 1:2:1:2:4:2:1:2:1, with EeNn the commonest at 4 of 16.

Why 9:3:3:1 arises. Each gene on its own gives a 3:1 ratio from Ee × Ee. Because the genes assort independently, you multiply: the probability of grey and normal is ¾ × ¾ = 9/16. Grey and vestigial is ¾ × ¼ = 3/16. Ebony and normal is 3/16. Ebony and vestigial is ¼ × ¼ = 1/16.

Worked example 2: test cross. Cross EeNn × eenn. The tester makes only en gametes, so each offspring’s phenotype shows the gamete from the heterozygote. The four gamete types are equally likely, so the offspring ratio is 1:1:1:1: grey normal, grey vestigial, ebony normal, ebony vestigial.

Nature of science. Both the 9:3:3:1 ratio and the 1:1:1:1 ratio follow from what is called Mendel’s second law, the law of independent assortment. It holds only when genes are on different chromosomes, or far enough apart on one chromosome that recombination reaches 50%. Expect exceptions to any biological “law” under certain conditions. Linkage is the exception in this unit.

Finding gene loci in databases (D3.2.18)

A locus is the position of a gene on a chromosome. The guide asks you to look up genes, and the polypeptides they code for, in online databases. Open-access examples include NCBI Gene, OMIM and UniProt. For a gene symbol, a database record normally gives the chromosome and band (for example 7q31.2: chromosome 7, long arm), plus the name and function of the product.

Facts to practise with, checked against public gene databases:

Gene	Polypeptide product	Locus
HBA1	alpha-globin (haemoglobin subunit alpha 1)	16p13.3
HBB	beta-globin	11p15.4
HBD	delta-globin	11p15.4
CFTR	a membrane channel for chloride and bicarbonate ions	7q31.2

You are asked to find pairs of genes:

• Different chromosomes: HBA1 (chromosome 16) and HBB (chromosome 11) both contribute to adult haemoglobin, yet they are unlinked and should assort independently.
• Close together on one chromosome: HBB and HBD are neighbours in a cluster of beta-like globin genes on chromosome 11. Genes this close are expected to be inherited together almost always.

Being on the same chromosome does not by itself prove tight linkage. What matters is the distance between the loci.

Autosomal gene linkage (D3.2.19)

Linked genes have loci on the same chromosome. An autosome is any chromosome other than a sex chromosome, so autosomal linkage means both loci are on one of chromosomes 1 to 22 in humans.

Alleles of linked genes are written beside vertical lines that stand for the two homologues. This tells you which alleles are on the same chromosome:

	Alleles on the same chromosome	Alleles on opposite chromosomes
Homologue 1	S and H	S and h
Homologue 2	s and h	s and H
Written as	SH	sh

Why linked alleles can fail to assort independently. Whole chromosomes, not single genes, are moved to the poles in anaphase I. Two alleles on the same chromosome travel together to the same gamete unless the chromosome is broken and rejoined. That happens through crossing over in prophase I, when non-sister chromatids of a bivalent exchange segments at chiasmata. The closer two loci are, the smaller the chance that a chiasma forms between them, so the more strongly the alleles stay together.

Why recombinants cannot exceed 50%. Suppose the loci are so far apart that a chiasma forms between them in every meiosis. Of the four chromatids, two are recombinant and two are parental. So 50% of gametes are recombinant at most, which matches independent assortment.

Recombinants (D3.2.20)

A recombinant carries a combination of alleles not present on either chromosome of the parent that made it. The guide asks about one cross type: a double heterozygote × a double homozygous recessive. You identify recombinants at three levels: gametes from the heterozygote, genotypes of offspring, and phenotypes of offspring.

Worked example 3: linked genes, alleles on the same chromosome. In an invented plant, S (serrated leaf margin) is dominant to s (smooth), and H (hairy stem) is dominant to h (smooth stem). A plant SH/sh is crossed with sh/sh. Of 500 offspring:

Offspring phenotype	Genotype	Number	Class
serrated, hairy	SH/sh	214	parental
smooth, smooth	sh/sh	205	parental
serrated, smooth	Sh/sh	43	recombinant
smooth, hairy	sH/sh	38	recombinant

• Gametes from the heterozygote: parental SH and sh; recombinant Sh and sH.
• Recombinants: 43 + 38 = 81 of 500.
• Recombinant frequency = 81/500 × 100 = 16.2%.

If the genes were unlinked, you would expect 125 in each class, so recombinants would be 50%.

Worked example 4: alleles on opposite chromosomes. Now the heterozygote is Sh/sH, crossed with sh/sh. Of 500 offspring: serrated smooth 228, smooth hairy 221, serrated hairy 27, smooth smooth 24. The parental gametes are now Sh and sH, so serrated hairy and smooth smooth are the recombinants: 27 + 24 = 51, giving 51/500 × 100 = 10.2%. Always work out the parental arrangement before deciding which class is recombinant. The commonest phenotypes are not always the same two.

Chi-squared test (D3.2.21)

Do the observed counts differ from the expected ratio by more than chance? A chi-squared test gives an objective answer.

• Null hypothesis: there is no significant difference between observed and expected results; any difference is due to chance.
• Alternative hypothesis: there is a significant difference, so something other than chance is involved.

Formula: χ² = Σ (O − E)² / E, where O is observed and E is expected. Use counts, not percentages. Degrees of freedom = number of classes − 1, so 3 for four phenotype classes.

Statistical significance at the p = 0.05 level means: if the null hypothesis were true, a difference this large or larger would arise by chance less than 5% of the time. Compare χ² with the critical value for p = 0.05:

Degrees of freedom	1	2	3	4
Critical value	3.84	5.99	7.81	9.49

If χ² is less than the critical value, you do not reject the null hypothesis. If it is equal to or greater, you reject it.

Worked example 5: F2 of an unlinked dihybrid cross. Test whether 400 F2 flies fit 9:3:3:1. Observed: 218, 83, 71, 28.

Expected = 400 × 9/16, 3/16, 3/16, 1/16 = 225, 75, 75, 25.

χ² = 49/225 + 64/75 + 16/75 + 9/25 = 0.218 + 0.853 + 0.213 + 0.360 = 1.64

Degrees of freedom = 3, critical value 7.81. Since 1.64 < 7.81, do not reject the null hypothesis. The results are consistent with 9:3:3:1 and the genes are consistent with independent assortment.

Worked example 6: detecting linkage. Use worked example 3. Expected for 1:1:1:1 from 500 is 125 in each class.

χ² = (214 − 125)²/125 + (205 − 125)²/125 + (43 − 125)²/125 + (38 − 125)²/125 = 63.37 + 51.20 + 53.79 + 60.55 = 228.9

This far exceeds 7.81, so reject the null hypothesis. The offspring do not fit independent assortment, which supports linkage.

Nature of science. Statistical testing often uses a sample to stand for a population. Here the sample is the F2 generation, or the offspring of the test cross. A larger sample is a better representative. In other experiments the sample is a set of repeated measurements.

A non-significant result does not prove the hypothesis; it shows the data do not contradict it. Expected values should not be very small (many courses use 5 as a rough minimum).

Common errors

• Writing four gamete types for a pair of linked genes as if they were equally likely.
• Calling the two commonest offspring classes recombinants without checking the arrangement in the parent.
• Using percentages instead of counts in χ².
• Using degrees of freedom of 4 for four classes.
• Saying a significant result “proves” linkage. It rejects chance as the sole explanation.
• Saying “the null hypothesis is accepted” instead of “not rejected”.

Next steps

Condense with the revision notes, then test yourself with the practice questions.

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).