Based on the following data from a three point testcross: 1. Determine the gene
ID: 149328 • Letter: B
Question
Based on the following data from a three point testcross:
1. Determine the gene order and phase for the genes E, F and G.
2. Determine the distance between genes E, F and G.
3. Draw a linkage map with the correct gene order, allele combinations on each chromatid and distance between the three genes.
For this exercise, phenotypes with capital letters are dominant and phenotypes in lower case letters are recessive. For example, “phenotype A” means the dominant phenotype for the gene and “phenotype a” means the recessive phenotype for that gene.
Show all your calculations. Remember that the three genes can be in any order. Remember that the dominant alleles do not necessarily need to all be on the same chromosome in the triple heterozygote parent.
Phenotypes
N
Efg
250
eFG
231
EFg
32
efG
38
EFG
11
efg
9
EfG
53
eFg
48
Total
672
Phenotypes
N
Efg
250
eFG
231
EFg
32
efG
38
EFG
11
efg
9
EfG
53
eFg
48
Total
672
Explanation / Answer
We can find out the gene order by observing the results only. We do following process.
The parental genotype is Efg and eFG. Here only e is odd one.
The least number of progenies are EFG and efg, which are homozygous. Therefore, the ‘e’ gene is in between F and G because EFG and efg are double crossover products and that can only possible when a centered gene alters.
Now we will calculate gene distance in between e and F.
Gene distance = [(53+48) + (11+9)]/ 672
Gene distance = 0.18
Now we will calculate gene distance in between G and e.
Gene distance = [(11+38) + (11+9)]/ 672
Gene distance = 0.10
Now the gene map is
F---------0.18---------e-------0.1---------G
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