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Wize University Biology Textbook > Modern Concepts of Inheritance

Genetic Linkage

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Genetic Linkage

When genes are located on different chromosomes or far apart on the same chromosome they follow the law of independent assortment and are said to be unlinked. However, when genes are located close together on the same chromosome, they don't follow the law of independent assortment and are said to be linked.

Genes Located on Different Chromosomes: Unlinked

Genes located on different chromosomes sort independently due to the random lining up of homologous chromosomes during metaphase I.

Example: chromosome #1 may line up with the maternal chromosome on the left and the paternal chromosome on the right, while chromosome #2 may line up with the maternal one on the right and the paternal one on the left. This leads to random assortment of the homologous chromosomes during meiosis I.


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Genes Located Far Apart on the Same Chromosome: Unlinked

Genes located on the same chromosome but far apart also follow the law of independent assortment because of crossing over during prophase I.


Photo by Abbyprovenzano / CC BY

Possible Gametes:
Ab, AB, ab, aB


Genes Located Close Together on the Same Chromosome: Linked

Crossing over occurs randomly along the homologous chromosomes. When genes are located very close together, the chance of recombination occurring between them is low. Therefore, genes that are close together tend to stick together during gamete formation.


Possible Gametes: (let's say the chromosome on the left is 1 and the other is 2)
a1b1c1, a1b1c2, a2b2c1, a2b2c2


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Calculating Recombination Frequencies

Recombination frequencies can be used to determine how close together two genes are on the same chromosome.

To do this, a cross is done using a fly that is heterozygous for both genes of interest and a fly that is homozygous recessive for both genes of interest. The heterozygous individual will therefore completely determine the phenotype of the offspring.


Because the four types of offspring are not produced in equal numbers, we know these genes are linked. To calculate the recombination frequency:

Recombination Frequency (RF) = (Recombinants / Total offspring) x 100%


In the above example:

RF = (206 + 185) / (944 + 965 + 206 + 185) x 100% = 17%


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Linkage Maps

We can use the recombination frequency to determine how close together genes are located on a chromosome, allowing us to create a linkage map. Linkage maps determine the relative distance and order of genes on a chromosome.

For example, lets say we have three genes: A, C, and F and we calculate the following recombination frequencies:

RF (A - C) = 12%
RF (A - F) = 26%
RF (C - F) = 14%

Note that one % is equal to one centimorgan (cM).



Practice: Recombination Frequency Calculation

A fly which has the genotype AaBb is cross with a fly that is homozygous recessive. The following offspring are obtained:
AaBb: 953 aabb: 910 Aabb: 200 aaBb: 192 a) Are these genes linked?
b) What is the recombination frequency? (one decimal place)

Practice: Gene Order

Genes A, B, C, and D are located on the same chromosome. The distances between the genes are below:
D-A = 11
D-B = 7
D-C = 22
B-C = 15
A-B = 4

Which of the following represent the order of those genes on the chromosome?

Practice: Gene Distance

AABB and aabb individuals were crossed to each other, and the F1 generation was backcrossed to the aabb parent. The results were 775 AaBb, 715 aabb, 4 Aabb, and 6 aaBb offspring. How far apart are the A and B loci?

Practice: Gene Location

You are investigating three genes in different species of a tropical plant (genes A, B, and C). When testcrossing a certain species of the plants, you always get a 1:1 ration of parental and recombinant types for the A and B, and the A and C genes. You also get a ratio of 3:1 of parental and recombinant types for the B and C genes. What best explains this phenomena?
Extra Practice
A scientist is studying two traits in a type of seeds: texture and color. Smooth texture (S) is dominant to wrinkled texture (s), and Yellow (Y) is dominant to green (y).
A (SsYy) and a (ssyy) plants mate and produce the following offspring:
Smooth and Yellow = 345
True or False?
If an allele becomes fixed, there is no opportunity for genetic variation to exist at that locus anymore.
A fly which has the genotype AaBb is cross with a fly that is homozygous recessive. The following offspring are obtained:
AaBb: 953 aabb: 910 Aabb: 200 aaBb: 192 a) Are these genes linked?
b) What is the recombination frequency? (one decimal place)
A fly which has the genotype AaBb is cross with a fly that is homozygous recessive. The following offspring are obtained:
AaBb: 953 aabb: 910 Aabb: 200 aaBb: 192 a) Are these genes linked?
b) What is the recombination frequency? (one decimal place)
Practice: Recombination
A fly which has the genotype AaBb is cross with a fly that is homozygous recessive. The following offspring are obtained:
AaBb: 953 aabb: 910 Aabb: 200 aaBb: 192 a) Are these genes linked?
b) What is the recombination frequency? (one decimal place)
Which of the following statements is FALSE?
What phenomenon causes violates of Mendel's second law?
A large population of ducks have a locus that determines the number of feathers on their tails, lots of feathers (F) is dominant to not as many feathers (f). If you observe this population and determine that the allele frequency of F=0.8 and the allele frequency of f=0.2, but there is no discernible difference in fitness between genotypes, what will be the frequency of F and f after 5 generations?
How many potential gametes can an organism with the following genotype produce:
AaBBCcDdEEFFGgHH
Which of the following best describes how a chromosome is able to “repair itself”?
Imagine an organism with the genotype XXYyZz. If all three genes are on different
chromosomes, how many different combinations of alleles will be created by
independent assortment?
Below is a diagram showing three loci on a pair of homologous chromosomes.
Which of the following are the most likely gametes at these loci?
abC
You have identified 4 loci in the genome of a species of bird:
The B locus corresponds to wing colour where the colour black is dominant to brown.
The C locus corresponds to the number of feathers on the wings where lots of feathers is dominant to few feathers. But birds that are homozygous recessive at the B locus will always have few feathers.
Below is a diagram showing four different loci on homologous pairs.
Which of the following are the LEAST likely gametes of this individual at these loci?
ABcD
Practice: Genes
Suppose we are investigating three genes, A, B, and C. We calculated the recombination frequencies among all three possible pairs of genes (AC, AB, BC) and they were:
A-B: 14.1%
B-C: 7.5%
Practice: Genotypes
A fly which has the genotype AaBb is cross with a fly that is homozygous recessive. The following offspring are obtained:
AaBb: 953 aabb: 910 Aabb: 200 aaBb: 192 a) Are these genes linked?
b) What is the recombination frequency? (one decimal place)
A series of fruit fly matings shows that the recombination frequency between the gene for wing size and the gene for antenna length is 5% (i.e. the genetic distance between them is 5 centimorgans). List all possible recombination frequencies between the gene for color and the gene for antenna length.
A scientist is studying two traits in a type of seeds: texture and color. Smooth texture (S) is dominant to wrinkled texture (s), and Yellow (Y) is dominant to green (y).
A (SsYy) and a (ssyy) plants mate and produce the following offspring:
Smooth and Yellow = 345
Genes A, B, C, and D are located on the same chromosome. The distances between the genes are below:
D-A = 11
D-B = 7
Meiosis practice problem 1
What are the genotypes of the possible daughter cells that would result from the double crossover shown below?
A cross is made between two E. coli strains: Hfr arg+ bio+ leu+ X F- arg- bio-leu-. Interrupted-mating studies show that arg+ enters the recipient last, and so recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of bio+ and leu+. The following numbers of individuals are found for each genotype:
arg+ bio+ leu+ 320
arg+ bio+ leu- 8