Genetics Question?
How do i use results from a F2 generation to calculate the map units of distance between two gene gene locis?
Answers:
It is quite straightforward for just 2 loci and the calculation is at the end. Hopefully, if you read the explanation it will make it clearer how the calculation works.
If you started with a parental generation with two genotypes:
AABB or aabb then your F1 generation will all have the genotype AaBb.
The F1 will have one chromosome of AB and another of ab.
There are two possible non-recombinant gametes AB and ab, which are called parental types. Alternatively, if crossing over (recombination) occurs you could get chromsomes with Ab and aB.
In your test cross you will now cross your F1s with a recessive homozygote aabb with all gametes ab.
In the F2 the following combinations are possible:
AB x ab = AaBb = parental genotype (F1).
ab x ab = aabb = parental genotype (testcross)
Ab x ab = Aabb = recombinant genotype
aB x ab = aaBb = recombinant genotype
By using a recessive homozygote as the test cross then each genotype will have a different phenotype. Two phenotypes will be the same has one of the parents and the other two will be new combinations. Parental phenotypes will be the majority unless loci are very far apart.
What you need to do is count up the F2 progeny find the number of each type and the total number of progeny. Then the percentage recombination is simply:
No recombinant progeny/ Total no. progeny x 100. This then gives you the map distance in centimorgans (1% recombination = 1 cM.
So from above if nos parental phenotypes were 52 and 36 and recombinant phenotypes were 2 and 10 then the map distance between A and B is:
2 + 10/2+10+52+36
=12/100 * 100% = 12cM = map distance
if i had the slightest idea of what you were talking about perhaps i could help..sorry
split the genes
Hmmmm. I have no clue- but you sure sound smart just asking the question.
Wow. How old school.
In real life, you would go to ensembl, and just look it up.
But check out the middle of this page, when it says "Solving linkage problems"
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Answers:
It is quite straightforward for just 2 loci and the calculation is at the end. Hopefully, if you read the explanation it will make it clearer how the calculation works.
If you started with a parental generation with two genotypes:
AABB or aabb then your F1 generation will all have the genotype AaBb.
The F1 will have one chromosome of AB and another of ab.
There are two possible non-recombinant gametes AB and ab, which are called parental types. Alternatively, if crossing over (recombination) occurs you could get chromsomes with Ab and aB.
In your test cross you will now cross your F1s with a recessive homozygote aabb with all gametes ab.
In the F2 the following combinations are possible:
AB x ab = AaBb = parental genotype (F1).
ab x ab = aabb = parental genotype (testcross)
Ab x ab = Aabb = recombinant genotype
aB x ab = aaBb = recombinant genotype
By using a recessive homozygote as the test cross then each genotype will have a different phenotype. Two phenotypes will be the same has one of the parents and the other two will be new combinations. Parental phenotypes will be the majority unless loci are very far apart.
What you need to do is count up the F2 progeny find the number of each type and the total number of progeny. Then the percentage recombination is simply:
No recombinant progeny/ Total no. progeny x 100. This then gives you the map distance in centimorgans (1% recombination = 1 cM.
So from above if nos parental phenotypes were 52 and 36 and recombinant phenotypes were 2 and 10 then the map distance between A and B is:
2 + 10/2+10+52+36
=12/100 * 100% = 12cM = map distance
if i had the slightest idea of what you were talking about perhaps i could help..sorry
split the genes
Hmmmm. I have no clue- but you sure sound smart just asking the question.
Wow. How old school.
In real life, you would go to ensembl, and just look it up.
But check out the middle of this page, when it says "Solving linkage problems"
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