I need help with this problem about gene mapping; here is thequestion: The follo
ID: 4873 • Letter: I
Question
I need help with this problem about gene mapping; here is thequestion:
The following data was found in a three point mappingexperiment
Phenotypes
Observed
ABC
317
aBc
203
aBC
0
Abc
2
Abc
58
ABc
72
aBc
21
AbC
10
Here are the Questions: I have figured this out up to the pointof determining the phenotype frequency used to determine the mapunits. I am also confused on how to set up drawing the mapitself.
1)The parental phenotypes are [ABC] and [abc]
2)The double recombinants are [aBC] and [Abc]
3)True or false: The A gene is in the middle of this threepoint map? (YES)
4)Show a map with all three genes and their corresponding mapunits in between each of them.
Can someone help me out with this part? I know I use thenumber of recombinants divided by the total number of progeny, butI am confused on what to do with the double and singlerecombinants. A visual would help me out tremendously, sogenetic geniuses please help!
Phenotypes
Observed
ABC
317
aBc
203
aBC
0
Abc
2
Abc
58
ABc
72
aBc
21
AbC
10
Explanation / Answer
(I'm assuming that 203 observed is for abc, 58 observed isfor AbC, and 21 observed is for abC. If not, you can substitute thenumbers in the calculations below to get the correctvalues.)Knowing that gene A is in the middle, gives us the order:
B-----?-----A-----?-----C
Now, to fill in the question marks, you need to determine thedistance from gene B to gene A and from gene A to gene C.
First, let's determine the distance between gene A and gene B.To do this, we look at the parental phenotypes for genes A and B,which would be:
[AB] and [ab]
We then figure out the possible recombinations of [AB] and[ab], which would be
[Ab] and [aB]*
Then, we look at the chart that tells us the number of progenythat have this phenotype and add them together. This means that youadd up ALL progeny that have either [Ab] or [aB] in them. So, youwould add up the number of progeny for:
[AbC]+[Abc]+[aBC]+[aBc] = 58+2+0+21 = 81
Then, you divide this number by the total number of progeny,so:
81 / (317+203+0+2+58+72+21+10) 81 / 683 = 0.119
Then, take this number and multiply it by 100 to getrecombination frequency:
0.119*100 = 11.9%
This percentage is the number of map units apart B is from A,because one percent recombination frequency = one map unit:
11.9% recombination frequency = 11.9 map units
You follow the same process for gene A and gene C:
Use parental phenotypes to determine recombinantphenotypes: [Ac] and [aC]
Find the sum of the number of ALL progeny that haveeither of the recombinant phenotypes: [ABc]+[Abc]+[aBC]+[abC] 72+2+0+21 = 95
Divide the number of progeny with those particularphenotypes by the total number of progeny: 95/683 = 0.139
Convert to percentage/map units: 0.139*100 = 13.9% = 13.9 map units
Your map would then look like: B-----11.9 m.u.-----A-----13.9 m.u.-----C
* BTW, A quick way to determine these recombinations would beby seeing if the parental phenotypes are cis (meaning either bothcapital letters or lowercase letters, or both having + or lacking+) or trans (each phenotype having one of each, meaning one capitaland one lowercase; or both having one gene with + and one genewithout +). If the phenotypes are in cis, then the recombinantswould be trans. If the phenotypes are in trans, then therecombinants would be cis. Knowing that gene A is in the middle, gives us the order:
B-----?-----A-----?-----C
Now, to fill in the question marks, you need to determine thedistance from gene B to gene A and from gene A to gene C.
First, let's determine the distance between gene A and gene B.To do this, we look at the parental phenotypes for genes A and B,which would be:
[AB] and [ab]
We then figure out the possible recombinations of [AB] and[ab], which would be
[Ab] and [aB]*
Then, we look at the chart that tells us the number of progenythat have this phenotype and add them together. This means that youadd up ALL progeny that have either [Ab] or [aB] in them. So, youwould add up the number of progeny for:
[AbC]+[Abc]+[aBC]+[aBc] = 58+2+0+21 = 81
Then, you divide this number by the total number of progeny,so:
81 / (317+203+0+2+58+72+21+10) 81 / 683 = 0.119
Then, take this number and multiply it by 100 to getrecombination frequency:
0.119*100 = 11.9%
This percentage is the number of map units apart B is from A,because one percent recombination frequency = one map unit:
11.9% recombination frequency = 11.9 map units
You follow the same process for gene A and gene C:
Use parental phenotypes to determine recombinantphenotypes: [Ac] and [aC]
Find the sum of the number of ALL progeny that haveeither of the recombinant phenotypes: [ABc]+[Abc]+[aBC]+[abC] 72+2+0+21 = 95
Divide the number of progeny with those particularphenotypes by the total number of progeny: 95/683 = 0.139
Convert to percentage/map units: 0.139*100 = 13.9% = 13.9 map units
Your map would then look like: B-----11.9 m.u.-----A-----13.9 m.u.-----C
* BTW, A quick way to determine these recombinations would beby seeing if the parental phenotypes are cis (meaning either bothcapital letters or lowercase letters, or both having + or lacking+) or trans (each phenotype having one of each, meaning one capitaland one lowercase; or both having one gene with + and one genewithout +). If the phenotypes are in cis, then the recombinantswould be trans. If the phenotypes are in trans, then therecombinants would be cis. B-----?-----A-----?-----C
Now, to fill in the question marks, you need to determine thedistance from gene B to gene A and from gene A to gene C.
First, let's determine the distance between gene A and gene B.To do this, we look at the parental phenotypes for genes A and B,which would be:
[AB] and [ab]
We then figure out the possible recombinations of [AB] and[ab], which would be
[Ab] and [aB]*
Then, we look at the chart that tells us the number of progenythat have this phenotype and add them together. This means that youadd up ALL progeny that have either [Ab] or [aB] in them. So, youwould add up the number of progeny for:
[AbC]+[Abc]+[aBC]+[aBc] = 58+2+0+21 = 81
Then, you divide this number by the total number of progeny,so:
81 / (317+203+0+2+58+72+21+10) 81 / 683 = 0.119
Then, take this number and multiply it by 100 to getrecombination frequency:
0.119*100 = 11.9%
This percentage is the number of map units apart B is from A,because one percent recombination frequency = one map unit:
11.9% recombination frequency = 11.9 map units
You follow the same process for gene A and gene C:
Use parental phenotypes to determine recombinantphenotypes: [Ac] and [aC]
Find the sum of the number of ALL progeny that haveeither of the recombinant phenotypes: [ABc]+[Abc]+[aBC]+[abC] 72+2+0+21 = 95
Divide the number of progeny with those particularphenotypes by the total number of progeny: 95/683 = 0.139
Convert to percentage/map units: 0.139*100 = 13.9% = 13.9 map units
Your map would then look like: B-----11.9 m.u.-----A-----13.9 m.u.-----C
* BTW, A quick way to determine these recombinations would beby seeing if the parental phenotypes are cis (meaning either bothcapital letters or lowercase letters, or both having + or lacking+) or trans (each phenotype having one of each, meaning one capitaland one lowercase; or both having one gene with + and one genewithout +). If the phenotypes are in cis, then the recombinantswould be trans. If the phenotypes are in trans, then therecombinants would be cis.