A dihybrid cross follows two genes at once. The maths looks heavier — sixteen boxes instead of four — but the biology is the same, plus one new idea: chromosomes line up randomly in meiosis, so the two genes are shuffled independently.
📚 What you need to know
Unlinked genes sit on different chromosomes (not on homologous copies of the same one).
In metaphase I each bivalent lines up independently of the others. Which way round it faces is random.
That random alignment decides which combination of alleles ends up in each gamete — independent assortment.
Mendel’s law of independent assortment: inheriting a particular allele of one gene does not affect which allele of another gene you inherit.
Write genotypes with both alleles of one gene first, then both of the other: BbEe, never BEbe.
A double heterozygous cross (BbEe × BbEe) gives 9 : 3 : 3 : 1. A double heterozygote crossed with a double homozygous recessive gives 1 : 1 : 1 : 1.
Unlinked genes and independent assortment
Two genes are unlinked when they are on different chromosomes. In humans, the gene for trypsin is on chromosome 7 and the gene for growth hormone is on chromosome 17 — completely separate, so they travel separately.
Here is the mechanism. During metaphase I, homologous pairs line up along the middle of the cell. Each pair lines up on its own, with no reference to the others: the chromosome from your mother could face either pole, and so could every other one.
In anaphase I the pairs are pulled apart. Because the alignment was random, the combinations that end up together in a gamete are random too. With two genes there are two possible arrangements, and both are equally likely, so a BbEe individual makes four kinds of gamete in equal numbers.
The two genes here are on different chromosome pairs. Which way one pair faces has no effect on the other — that independence is the whole law.
Exam answers often say “the alleles separate randomly”. Push it one step further: it is the bivalents that align randomly in metaphase I, and that is what decides which alleles end up together. Naming the stage is usually worth a mark.
Working out the gametes
A double heterozygote BbEe makes four kinds of gamete. Take one allele from each gene, in every possible combination:
🧩 Getting all four, every time
Split the genotype into its two genes: Bb and Ee.
Pair the first allele of gene 1 with each allele of gene 2 → BE, Be.
Pair the second allele of gene 1 with each allele of gene 2 → bE, be.
Check the count. Heterozygous at both genes = 4 gametes. Heterozygous at one = 2. Homozygous at both = 1.
Keep the gene order the same in every gamete, so you can compare them at a glance.
Quick check before you draw the grid. Count the gamete types first. If a parent is BbEe you need four columns; if a parent is bbEe you need only two. Getting this wrong means redrawing the whole grid, so spend the ten seconds.
The 9 : 3 : 3 : 1 cross
Take horses with two unlinked genes: coat colour (B black, dominant to b chestnut) and eye colour (E brown, dominant to e blue). Cross two double heterozygotes, BbEe × BbEe.
Each parent makes four gametes, so the grid is 4 × 4 = 16 boxes.
Gametes
BE
Be
bE
be
BE
BBEE black, brown
BBEe black, brown
BbEE black, brown
BbEe black, brown
Be
BBEe black, brown
BBee black, blue
BbEe black, brown
Bbee black, blue
bE
BbEE black, brown
BbEe black, brown
bbEE chestnut, brown
bbEe chestnut, brown
be
BbEe black, brown
Bbee black, blue
bbEe chestnut, brown
bbee chestnut, blue
Count the phenotypes, not the genotypes:
9 black coat, brown eyes — at least one B and at least one E
3 black coat, blue eyes — at least one B, but ee
3 chestnut coat, brown eyes — bb, but at least one E
1 chestnut coat, blue eyes — bbee, the only double recessive
The second cross is a test cross. Because bbee makes only one kind of gamete (be), every offspring shows the alleles it received from the BbEe parent directly. Four gametes, four phenotypes, equal numbers.
Worked examples
WORKED EXAMPLE 1
State all the gamete types produced by an organism with genotype BbEe, and by one with genotype bbEe.
Step 1: BbEe — split it
Gene 1 gives B or b. Gene 2 gives E or e.
Step 2: combine every optionBE, Be, bE, be — four typesStep 3: bbEe — split it
Gene 1 can only give b. Gene 2 gives E or e.
bE, be — two typesBbEe → 4 gamete types; bbEe → 2 gamete typesEach heterozygous gene doubles the number of gamete types. Two heterozygous genes give 2 × 2 = 4.
WORKED EXAMPLE 2
Two double heterozygous horses (BbEe) are crossed and produce 240 foals. Calculate the expected number in each phenotype class.
Step 1: state the expected ratio9 : 3 : 3 : 1, which is 16 parts in total
Step 2: find one part240 ÷ 16 = 15Step 3: multiply out9 × 15 = 135 3 × 15 = 45 3 × 15 = 45 1 × 15 = 15135 black/brown : 45 black/blue : 45 chestnut/brown : 15 chestnut/blueCheck your answer adds back to 240. If it does not, you have divided by the wrong total.
When the law breaks down
Mendel worked all this out before anyone knew DNA existed, and he happened to choose genes that were on different chromosomes. His law only holds when the genes are unlinked.
Genes on the same chromosome tend to travel together into the same gamete, so the four gamete types are not equal and the offspring ratios are not 9 : 3 : 3 : 1. That is linkage, and it gets its own page.
A useful habit of mind. Biological “laws” nearly always come with conditions attached. Knowing when a law fails is a stronger sign of understanding than knowing the law itself.
💡 Exam tip
Write gametes in circles and keep the same gene order throughout.
Draw the grid big enough. Sixteen cramped boxes are where the counting errors happen.
Fill in the phenotype under each genotype as you go — counting at the end is much faster.
Learn 9 : 3 : 3 : 1 and 1 : 1 : 1 : 1 and what causes each. You are expected to recall them.
You do not need to memorise the genotype ratio — read it off the grid if a question asks.
Name the stage (metaphase I) when explaining independent assortment.
⚠ Common mix-up
Writing a gamete as BbE. A gamete carries exactly one allele of each gene, so two letters here.
Mixing the alleles of two genes in a genotype, like BEbe. Keep them as pairs: BbEe.
Giving a double heterozygote two gamete types. It makes four.
Confusing 9 : 3 : 3 : 1 with 1 : 1 : 1 : 1. The first is heterozygote × heterozygote; the second is a test cross.
Counting genotypes when the question asks for phenotypes. The nine “black, brown” foals include four different genotypes.
Applying independent assortment to linked genes. Check first whether the genes are on different chromosomes.
Up next: Statistics of Dihybrid Crosses — using a chi-squared test to decide whether your results really do differ from the expected ratio.
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