Based on the following data from a three point testcross: 1. Determine the gene
ID: 149689 • 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
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
From the table, phenotypes found most frequently are parental phenotypes.
Thus,
E f g - 250 and e F G - 231 are parental Phenotype.
Double cross over phenotypes arer always found in the lowest frequency.
So,
E F G - 11 and e f g - 9 are double cross over phenotypes.
Double cross over events moves the middle allele from one sister chromatid to the other. So, now we need to compare double cross over phenotype with parental phenotype in order to find the middle allele.
Parental phenotype : E f g and e F G
Double cross over : E F G and e f g
here we can find that the allele E/e is moved from one chromatid to other.
Thus allele E/e is the middle allele.
1). Hence the gene order can be written as,
g E f and G e F.
2). Now in order to calculate linkage distance, frequency of both single and double cross over should be taken. The linkage distance between the alleles can be calculated by the following formula,
=(Number of recombinant offspring/Total number of offspring) * 100.
The linkage distance between g and E :
Single cross overs - g e F(48) and G E f (53)
Double cross overs- G E F(11) and g e f (9)
= (11 + 9 + 48 + 53)/672 * 100
= 18.005 cM (cM denotes centiMorgan, unit of measurement of genetic linkage)
The linkage distance between E and f :
Single cross overs - g E F(32) and G e f (38)
Double cross overs- G E F(11) and g e f (9)
=(11 + 9 + 38 + 32)/672 * 100
=13.392 cM.
Thus the distance between g and E is 18.005 cM
and the distance between E and f is 13.392 cM.
Hence the distance between g and f is 18.005 + 13.392 = 31.397 cM
3) Linkage map with correct order of genes, allele combination on each chromatid and distance between 3 genes is given below.
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