Name:__________________________
SBI3UC - Autosomal Linkage
Linkage group: A pair or set of genes on the SAME chromosome which tend to be inherited together.


These genes tend to show up together in the same combinations in the offspring.
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:
Neighboring genes on the chromosome have a tendency to stick together when passed on to offspring.
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 is a cross of an unknown dominant parent with a homozygous recessive individual to help
determine the genotype of the unknown parent depending on the phenotype of the offspring.
Backcrossing is a crossing of a hybrid with one of its parents or an individual genetically similar to its
parent, in order to achieve offspring with a genetic identity which is closer to that of the parent.
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
𝑅𝑤
𝑟𝑊
c)
Rw, rW
GY
b)
𝐺𝑌
𝐺𝑌
Stop
ab, Ab
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.
Name:__________________________
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 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.
1. The fact that the traits are both on chromosome 5, tell us these traits are linked
2. Designate variables;
a. Let L=long, l=small; B=blonde, b=black
3. Write the parent phenotypes, genotypes and gametes;
Parent phenotypes
Long blonde
Small black
Parent genotype
𝐿𝐵
𝐿𝐵
𝑙𝑏
𝑙𝑏
Gametes
LB
lb
4. Write the F1 phenotype and genotype;
𝐿𝐵
𝑙𝑏
5. Test cross
Parent phenotypes
Long blonde
Small black
Parent genotype
𝐿𝐵
𝑙𝑏
𝑙𝑏
𝑙𝑏
Gametes
LB, lb
lb
F1 genotypes and phenotypes
gametes
lb

LB
lb
𝐿𝐵
𝑙𝑏
𝑙𝑏
𝑙𝑏
Long, blonde
Small, black
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.
Name:__________________________
Autosomal Linkage with Crossing over
Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over, or
recombination.
Note: recombinants are any combination of alleles that are not the same as the parental combinations
Recombination, which occurs in prophase I of meiosis, can split the two alleles inherited from a parent,
giving recombinant types.
These tend to be in smaller proportions than the parental, linked types.
Crossing over is rare and therefore, recombinants tend to be fewer in number than parental.
The percentage of offspring showing the recombinant types can give an estimate of how close 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:
𝐴
𝐵
𝑎
𝑏
This is an example of a parental combination
This genotype would give the same phenotype as:
BUT this is an example of a recombinant
𝐴
𝑏
𝑎
𝐵
Name:__________________________
Practice:
Write the parental gametes and the recombinant gametes for the following examples;
𝑎𝐵
𝑅𝑤
a) 𝐴𝑏
𝑔𝑌
b)𝑟𝑊
Parental
a) aB, Ab
b) Rw, rW
c) gY, Gy
c) 𝐺𝑦
Recombinant
ab, AB
RW, rw
gy, 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.
Name:__________________________
Phenotype and Genotype
Purple, long (PpLl)
Purple, round (Ppll)
Red, Long (ppLl)
Red, round (ppll)
Observed
284
21
21
55
Expected from 9:3:3:1
216
72
72
24
Name:__________________________
Recombinants are recognized by:
1. Unpredicted combinations of characterisitics. (Red Long; Purple short)
2. Low frequency of new combinations of phenotype.
3. Statistical difference from ratios expected from either dihybrid and unlinked (9:3:3:1) or
monohybrid with no linkage (3:1)
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
gametes
sw
Parent genotype
𝑆𝑊
𝑠𝑤
𝑠𝑤
𝑠𝑤
Gametes
SW, sw
sw
SW
sw
Sw
sW
𝑆𝑊
𝑠𝑤
𝑠𝑤
𝑠𝑤
𝑆𝑤
𝑠𝑤
𝑠𝑊
𝑠𝑤
Small, sweet
Large, Sour
Small, sour
Large, sweet
Parental
Recombinant
Name:__________________________
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
Stop

Answer is A
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
Stop
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A normal linkage cross would result in AB//ab and ab//ab (AaBb and aabb)
Recombinants are then Ab//ab and aB//ab (Aabb and aaBb)
Therefore, the answer is D