Module 6 Lesson 8 - 3 (Lab)
Lesson 8 — Crossing Over Frequencies and Gene Mapping
Lab: Mapping Chromosomes
These lessons have covered the process of determining the number of recombinant types in crosses and then calculating recombination frequencies. This can be taken further, and a map or diagram of a chromosome can be drawn to show the location of the genes on it. Remember that, if you know the number of recombinants, you also know the map unit distances for those genes as well as the recombinant frequency!
Steps for Mapping Chromosomes
After you have determined the map units or have been supplied the data in a question, proceed with these steps.
Example
Genes |
Distance (mu)
|
H and G
|
4 |
G and K
|
5 |
G and J
|
15 |
K and J
|
10 |
H and K
|
9 |
-
Start with the largest map distance. Remember that does not necessarily mean they will become the two furthest points.
- For this example, start with G and J, which are 15 mu apart.
-
Choose another map distance with one of the two original letters. It works best if there are numerous stats for the letter you choose: the more, the better. Place the second choice under the initial line with the common letters lined up vertically.
-
We can choose either G or J for this example. (We will continue with G.)
- Place the second map distance of G below the first G.
- G and K are 5 mu apart, but nothing indicates whether K is on the left or right side of G.
- To determine this, we need another reference point from K.
-
K and J are 10 mu apart. This shows that K is to the right side of G.
-
We can choose either G or J for this example. (We will continue with G.)
-
Select as many map units with the original letter as possible and continue to place them vertically.
- For this example, choose all the Gs.
-
G and H are 4 mu apart, but no indication is given whether H is on the left or right side of G.
-
To determine this, we need another reference point from H.
-
H and K are 9 mu apart. This shows that H is on the left side of G.
-
Determine the final order.
- The answer is HGKJ or JKGH. Because the map unit distance provides only relative distances, we cannot determine which gene is at the beginning. Therefore, both answers are acceptable. When drawing the chromosomal map, be sure the length of the lines between genes represent the distances accurately.
Practice Questions
Gene Symbols for Drosophila
The normal allele is called wild type allele and a plus sign (+) is used to indicate the wild type. Mutations are designated by a letter or letters related to the phenotype of the mutation. Recessive mutations are written in lower case letters. For example, Drosophila can have normal (wild type, +) wings or vestigial (atrophied, vg) wings. The vestigial mutation is an autosomal recessive mutation and is represented as vg.
To describe linked genes, scientists have adopted a new method to represent the genes. To show linkage between two or more genes, alleles on the same chromosome are written together and separated from alleles on the other chromosome by a slash symbol (/). For example, vg d / vg + indicates that vg (vestigial) and d (leg length) genes are on the same chromosome (One of the chromosomes contains the alleles for vestigial wings and short legs and the other chromosome contains the alleles for vestigial wings and normal legs.).
Construct a chromosome map of three linked genes using the following information.
In Drosophila melanogaster, the mutant gene d causes short legs and the mutant gene pr causes purple eyes. The mutant gene vg causes vestigial wings.
The following crosses were performed and resulted in the F1 generations below. The recombination frequency for eye colour gene and wing type gene is 8.7%. Construct a chromosome map using this information.
|
Eye Colour and Leg Length |
Wing Type and Leg Length |
||||||||||||||||||||
Cross | pr d / + + x pr d / pr d | vg
d /
+
+ x
vg
d /
vg
d
|
||||||||||||||||||||
Result |
|
|
-
In the first cross, pr d / + + x pr d / pr d, the parental phenotypes are normal / normal (391) and purple / short (389).
The recombinants were purple / normal (115) and normal / short (105) : 115 + 105 = 220.
The percentage of recombinant F1 phenotypes = recombination frequency between two genes in question.
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Therefore, the distance between the eye colour gene and leg length gene is 22 mu.
- In the second cross, vg d / + + x vg d / vg d, the parental phenotypes are normal / normal (350) and vestigial / short (343).
The recombinants were vestigial / normal (154) and normal short (153) : 154 + 153 = 307.
The percentage of recombinant F1 phenotypes = recombination frequency between two genes in question.
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The distance between the wing type gene and leg length gene is 30.7 mu.
- The distance between eye colour gene and wing type gene is 8.7 mu.
Therefore, the chromosome map for the three genes is the following:
Wing Type -- 8.7 mu -- Eye Colour ------------------22 mu ------------------- Leg Length