Respuesta :
To know the distances between linked genes, first we need to know their order in the chromsome and then the distance between each other. In this example, a) 12 MU, b) 5MP, c) 17 MP, d) D, e) P.
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In this problem, we need to find the distances between genes and their order in the chromosome.
To do it, we have the following data about the progeny of the cross between tomato plants.
Note: I suggest you check if the provided data here is the same as yours. If not, use your data following the same procedure as here.
Three diallelic genes involved: M, D, and P
Phenotype Genotype Number
Solid, Normal, Smooth MmDdPp 420
Solid, Normal, Peach MmDdpp 21
Solid, Dwarf, Smooth MmddPp 2
Solid, Dwarf, Peach Mmddpp 52
Mottled, Normal, Smooth mmDdPp 62
Mottled, Normal, Peach mmDdpp 4
Mottled, Dwarf, Smooth mmddPp 23
Mottled, Dwarf, Peach mmddpp 416
N = 1000 plants
Assuming that the genes are linked, we can calculate the order of the genes and the genetic distances between them.
GENES ORDER
To know their order in the chromosome, we need to figure out which gene changes positions when comparing the genotypes of the parental gametes with the ones of the double recombinants.
Among the descendants, the parental gametes are the ones with the highest phenotypic frequency, while the double recombinants are the less frequent. So:
Parental)
- Solid, Normal, Smooth -----MmDdPp ------ 420
- Mottled, Dwarf, Peach ---- mmddpp ------ 416
Double recombinant)
- Solid, Dwarf, Smooth ----- MmddPp ------ 2
- Mottled, Normal, Peach ------ mmDdpp ------4
Comparing them we will realize that between
- Mm Dd Pp (parental)
- Mm dd Pp (double recombinant)
and
- mm dd pp (Parental)
- mm Dd pp (double recombinant)
They only change in the position of the alleles Dd/dd.
This suggests that the position of the gene D is in the middle of the other two genes, M and P, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
---- M ---- D ----- P ----
GENE DISTANCES
Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side.
We will call Region I to the area between M and D and Region II to the area between D and P.
First we will calculate the recombination frequencies, and we will do it by region.
We will call P1 to the recombination frequency between M and D genes, and P2 to the recombination frequency between D and P.
- P1 = (R + DR) / N
- P2 = (R + DR)/ N
Where:
- R is the number of recombinants in each region,
- DR is the number of double recombinants in each region, and
- N is the total number of individuals.
So:
P1 = (R + DR) / N
P1 = (52+62+4+2)/1000
P1 = 120/1000
P1 = 0.12
P2 = (R + DR) / N
P2 = (21+23+4+2)/1000
P1 = 50/1000
P1 = 0.05
Now, to calculate the recombination frequency between the two extreme genes, M and P, we can just perform addition or a sum:
P1 + P2= Pt
0.12 + 0.05 = Pt
0.17=Pt
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.
Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:
- GD1= P1 x 100 = 0.12 x 100 = 12 MU
- GD2= P2 x 100 = 0.05 x 100 = 5 MU
- GD3=Pt x 100 = 0.17 x 100 = 17 MU
So, summing up
- The correct order of the genes is M, D, P
- The distance between M and D is 12 MU
- The distance between D and P is 5MU
- The distance between M and P is 17MU
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