Name:__________________________
SBI3UC - Autosomal Linkage
Linkage group: A pair or set of genes on the SAME chromosome which tend to be inherited together.

Mendel’s law of independent assortment does not apply to these genes because they are on the
same chromosome

“Genetic linkage analysis is a statistical method that is used to associate functionality of genes to their
location on chromosomes. “
The Main Idea/usage:

Therefore, if some disease is often passed to offspring along with specific marker-genes, then it can be
concluded that the gene(s) which are responsible for the disease are located close on the chromosome
to these markers.
* The above image shows that gene 2, 3 and 4 are inherited together
Test Crossing
Backcrossing
F2
Name:__________________________
Making the gametes
Previously we have written our genotypes as for example- TtLl. This form does not represent which
alleles are linked on the same chromosome.
When we represent gene linkage we write the alleles together that are found on the same chromosome.
FL
𝑓𝑙
Here FL are linked and fl are linked on the same
chromosome
Practice:
Write the gametes for the following examples;
a)
𝑎𝑏
𝐴𝑏
b)
𝑅𝑤
𝑟𝑊
c)
𝐺𝑌
𝐺𝑌
Example with Linkage
Ex. A flower that is homozygous for purple flowers and long flower was crossed with a pure breeding red
and round flower. These traits are found on the same chromosome.
Now let’s apply this knowledge to a practice problem;
Long ears and blonde fur in the Rabid rabbit are dominant characteristics found on chromosome 5. The
corresponding recessive characteristics are small ears and black body. A rabbit homozygous for
homozygous for long and blonde is crossed with a homozygous recessive rabbit. Deduce the genotypes
and phenotypes of the offspring of a test cross on the F1 rabbits.
Name:__________________________
1.
2.
3.
Parent phenotypes
Large red
Small black
Parent genotype
Gametes
4. Write the F1 phenotype and genotype;
𝐿𝐵
𝑙𝑏
5. Test cross
Parent phenotypes
Large red
Small black
Parent genotype
Gametes
F1 genotypes and phenotypes
gametes
lb

LB
lb
Both of these do not have recombinants as they are the same as the parental type. The fact that
we get a 1: 1 ratio indicates that there is a linkage. If they were not linked we’d get a 1:1:1:1 in
the test cross.
Autosomal Linkage with Crossing over
Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over, or
_______________________________.
Note: recombinants are any combination of alleles that are ______________________ as the parental
combinations
Name:__________________________
Recombination, which occurs in prophase I of meiosis, can split the two alleles inherited from a parent,
giving recombinant types.
_____________________________________________________________.
Crossing over is rare and therefore, recombinants tend to be fewer in number than parental.
The _________________________________________________________ can give an estimate of how
_________ the two genes are to each other on the chromosome. The closer the genes are, the fewer
recombinant types should occur.
Genotypes for linked genes can be shown as:
This is an example of a parental combination
This genotype would give the same phenotype as:
BUT this is an example of a recombinant
Name:__________________________
Practice:
Write the parental gametes and the recombinant gametes for the following examples;
𝑎𝐵
𝑅𝑤
a) 𝐴𝑏
b)𝑟𝑊
𝑔𝑌
c) 𝐺𝑦
It is vital to remember that the dominant genes don’t always necessarily have to be
linked!!!!!!!!!!!!
Example of Linked genes:
From the work of William Bateson, who studied the sweet pea. He was looking at 2 genes; flower colour
(P-purple and p-red) and the shape of the pollen grain (L-long and l- round). He crossed PPLL and ppll
and then self-crossed (back crossed) the resulting F1 generation to create PpLl in the F2 generation.
According to Mendelian genetics the expected ratios would be 9:3:3:1. Instead he found an increased
number of the parental combinations PPLL and ppll.
Phenotype and Genotype
Observed
Expected from 9:3:3:1
Name:__________________________
Recombinants are recognized by:
1.
2.
3.
Example:
In wild blueberries, the genes for sweetness and size are linked. The gene for sweetness is dominant to
sourness and small berries are dominant to large berries. A heterozygous small sweet berry plant was
test crossed and the results wereSmall, sweet
215
Large, sour
212
Small, sour
16
Large, sweet
20
Large pair of number- the parental combinations
small pair of number- the recombinants
1. Let S=small , s=large; W=sweet, w=sour
2. We are told that the first parent is heterozygous and that the genes are linked. Since we have 4
phenotypes crossing over must have occurred.
3. Determine the linkage pattern via the F1 offspring. How do we know which are linked????
𝑆𝑊
𝑆𝑤
Is it 𝑠𝑤 or 𝑠𝑊 ??
Very simply by looking at the numbers. The 215 and the 212 are the parental combinations of
alleles. This means that SW are linked on one chromosome and sw are linked on the other.
4. Prove it with a Punnett square.
Parent phenotypes
Small, sweet
Large, sour
Parent genotype
𝑆𝑊
𝑠𝑤
𝑠𝑤
𝑠𝑤
Gametes
SW, sw
sw
Name:__________________________
gametes
sw
SW
sw
Sw
sW
𝑆𝑊
𝑠𝑤
𝑠𝑤
𝑠𝑤
𝑆𝑤
𝑠𝑤
𝑠𝑊
𝑠𝑤
Small, sweet
Large, Sour
Small, sour
Large, sweet
Parental
Recombinant
1. Genes P, Q, R and S are located on the same chromosome. The cross over values between them
are:
P-Q 20%
P-R 30%
P-S 15%
Q-R 15%
Q-S 30%
R-S 40%
What is the sequence of genes in the gene map of the chromosome?
A.
S, P, Q, R
B.
R, P, Q, S
C.
P, R, S, Q
D.
Q, S, R, P
2. A cross is performed between two organisms with the genotypes AaBb and aabb. What
genotypes in the offspring are the result of recombination?
A. Aabb, AaBb
B. AaBb, aabb
C. aabb, Aabb
D. Aabb, aaBb