How on earth do you draw a linkage map?
There are three linked genes associated with coat colour in rats. A male rat heterozygous for the three linked genes, genotype AaBbCc, was test crossed with a number of females that were aabbcc. The linkage groups of these mice are:
Male Females
A B c a b c
---------- ----------
a b C a b c
1000 offspring were obtained. The number that received each of the gamete types from the heterozygous male are given in the table below (the alphabetical order given is not necessarily the order of the genes on the chromosome).
ABc 305
abC 310
aBC 42
Abc 43
ABC 140
abc 145
aBc 6
AbC 9
Analyse these results to draw a linkage map of the genes, including distance in map units.
Justify your map with relevant calculations and discuss any discrepancies within the data
This is a question from a past exam paper I have been studying, for which I can find no answers and therefore cannot mark my own ones. I have concuded for this question that obviously the genes A, B and C are linked and are therefore found relatively close together on the chromosome. I have also concluded (I think) that the Genes A and B are found closer together than the space between A and C and B and C. How do I show this however on a linkage map? I have google imaged this, however I don't still know how to draw one. Help much appreciated!!!
Thanks
There are three linked genes associated with coat colour in rats. A male rat heterozygous for the three linked genes, genotype AaBbCc, was test crossed with a number of females that were aabbcc. The linkage groups of these mice are:
Male Females
A B c a b c
---------- ----------
a b C a b c
1000 offspring were obtained. The number that received each of the gamete types from the heterozygous male are given in the table below (the alphabetical order given is not necessarily the order of the genes on the chromosome).
ABc 305
abC 310
aBC 42
Abc 43
ABC 140
abc 145
aBc 6
AbC 9
Analyse these results to draw a linkage map of the genes, including distance in map units.
Justify your map with relevant calculations and discuss any discrepancies within the data
This is a question from a past exam paper I have been studying, for which I can find no answers and therefore cannot mark my own ones. I have concuded for this question that obviously the genes A, B and C are linked and are therefore found relatively close together on the chromosome. I have also concluded (I think) that the Genes A and B are found closer together than the space between A and C and B and C. How do I show this however on a linkage map? I have google imaged this, however I don't still know how to draw one. Help much appreciated!!!
Thanks
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You can find that out by looking at the ones that have made the cross-over. So, we need to focus on recombinant which are the ones that did make a cross over between two genes or three genes.
For example, you need to focus on the following:
aBc 6
AbC 9
Single Crossover between A and others (15 / 1000)
aBC 42
Abc 43
Double cross over (85 / 1000)
ABC 140
abc 145
Single Crossover between C and others (285 / 1000)
Percentage of this recombinant between A and B = (15 / 1000) + ( 85 / 1000) = 10%
Percentage of this recombinant between B and C = (285 / 1000) + ( 85 / 1000) = 37%
Percentage of this recombinant between A and C is not (285 / 1000) + (15 / 1000)
"Because the double recombinants restored the parental configuration, they were missed in the scoring. So the two rare classes of double recombinants need to be added (twice) to the data."
Percentage of this recombinant between A and C is (285 / 1000) + (15 / 1000) + ( 85 / 1000) + ( 85 / 1000) = 470 / 1000 = 47%
Linkage map
A---10cM-----B----37cM-----C
<--------------47cM-------------->
For example, you need to focus on the following:
aBc 6
AbC 9
Single Crossover between A and others (15 / 1000)
aBC 42
Abc 43
Double cross over (85 / 1000)
ABC 140
abc 145
Single Crossover between C and others (285 / 1000)
Percentage of this recombinant between A and B = (15 / 1000) + ( 85 / 1000) = 10%
Percentage of this recombinant between B and C = (285 / 1000) + ( 85 / 1000) = 37%
Percentage of this recombinant between A and C is not (285 / 1000) + (15 / 1000)
"Because the double recombinants restored the parental configuration, they were missed in the scoring. So the two rare classes of double recombinants need to be added (twice) to the data."
Percentage of this recombinant between A and C is (285 / 1000) + (15 / 1000) + ( 85 / 1000) + ( 85 / 1000) = 470 / 1000 = 47%
Linkage map
A---10cM-----B----37cM-----C
<--------------47cM-------------->