Linkage vs independent assortment (basic recombination)
When two genes are on the same chromosome, they may be linked, meaning they tend to be inherited together.
If two genes are on different chromosomes (or far apart on the same chromosome), they behave as if they assort independently.
This calculator compares these two ideas using a standard testcross and a tunable recombination frequency.
Core setup used in this calculator (testcross)
The most calculator-friendly experimental design for linkage is the classic testcross:
a dihybrid heterozygote crossed with a double-recessive tester.
The tester is always ab/ab, so it contributes only ab gametes.
That means the offspring phenotype/genotype classes directly reflect the gametes produced by the heterozygote.
\[
\text{Heterozygote (phase chosen)} \times \frac{ab}{ab}
\]
The heterozygote can be in one of two chromosome arrangements (phases):
coupling (cis) or repulsion (trans).
\[
\text{Coupling (cis): } \frac{AB}{ab}
\qquad\qquad
\text{Repulsion (trans): } \frac{Ab}{aB}
\]
What “recombination frequency” means
Recombination frequency, written r, is the fraction of gametes that are recombinant due to crossing over between the two loci.
In basic genetics, r ranges from 0 to 0.5 (0% to 50%).
Values above 50% are not possible in this two-locus model because at most half of gametes can be recombinant.
\[
0 \le r \le 0.50
\]
Interpretation:
if r \approx 0, the genes are very tightly linked (almost all gametes are parental).
If r \approx 0.50, the genes behave like they assort independently (parental and recombinant classes become equally common).
Independent assortment as a special case
In a testcross, independent assortment produces four gamete types with equal probability.
This is equivalent to setting r = 0.50, because recombinant gametes make up half of the gametes, and each of the two recombinant classes gets half of that.
\[
r = 0.50 \;\;\Rightarrow\;\; P(AB)=P(ab)=P(Ab)=P(aB)=0.25
\]
Expected offspring classes in a testcross
In a testcross, the offspring classes are:
AB, ab (the two parental classes) and Ab, aB (the two recombinant classes),
but which two are “parental” depends on whether the heterozygote is in coupling or repulsion phase.
Step-by-step: from r to offspring proportions
The calculator uses a simple and standard rule:
the total parental fraction is 1 − r, and the total recombinant fraction is r.
Since there are two parental classes and two recombinant classes, each class gets half of its category total.
\[
P(\text{each parental class}) = \frac{1-r}{2}
\qquad\qquad
P(\text{each recombinant class}) = \frac{r}{2}
\]
Coupling phase: AB/ab
If the heterozygote is AB/ab, the parental gametes (and thus the high-frequency offspring classes) are AB and ab.
The recombinant classes are Ab and aB.
\[
\text{Coupling: } \frac{AB}{ab}
\]
\[
P(AB)=\frac{1-r}{2},\quad P(ab)=\frac{1-r}{2},\quad
P(Ab)=\frac{r}{2},\quad P(aB)=\frac{r}{2}
\]
Repulsion phase: Ab/aB
If the heterozygote is Ab/aB, the parental classes are Ab and aB.
The recombinant classes become AB and ab.
\[
\text{Repulsion: } \frac{Ab}{aB}
\]
\[
P(Ab)=\frac{1-r}{2},\quad P(aB)=\frac{1-r}{2},\quad
P(AB)=\frac{r}{2},\quad P(ab)=\frac{r}{2}
\]
Converting proportions to expected counts
If the total number of offspring is N, the expected count in each class is the probability times N.
These are expected values (averages) and do not need to be integers.
\[
E(\text{count of class }i) = N \cdot P(i)
\]
Reading the live bar chart
The four-bar chart displays the predicted distribution across the four offspring classes.
Parental classes are highlighted separately from recombinant classes.
As you move the recombination slider:
the parental bars shrink as r increases, and the recombinant bars grow.
At r = 50% all four bars become equal (25% each), matching independent assortment.
Common interpretation checkpoints
If the observed offspring distribution has two high-frequency classes and two low-frequency classes, that pattern suggests linkage.
A quick estimate of recombination frequency from observed counts (for a testcross) is:
add up the two recombinant classes and divide by the total offspring count.
\[
\hat{r} = \frac{\text{recombinant count total}}{N}
\]
If the estimated value is near 0.50, the genes behave like they assort independently.
If it is near 0, the genes are tightly linked.
