Genetic Map + distances between genes
---TSN---
║---6.4%---║16.3%---║
║22.7%║
Explanation:
Available data:
three diallelic genes are linked on the same chromosome: T gene codifies for height, S gene codifies for skin texture, and N gene codifies for fruit shape.Tall (T-) is dominant to dwarf (tt)Skin that is smooth (S-) is dominant to skin that is peachy (ss)Fruit with a normal tomato shape (N-) is dominant to oblate shape (nn)A plant that is true-breeding for the dominant traits was crossed to a dwarf plant with peachy skin and oblate fruitF1 plants were then testcrossed F2: 151 tall, smooth, normal T-S-N-33 tall, smooth, oblate T-S-nn11 tall, peach, oblate T-ssN-2 tall, peach, normal T-ssnn155 dwarf, peach, oblate ttssnn29 dwarf, peach, normal ttssN-12 dwarf, smooth, normal ttS-N-0 dwarf, smooth, oblate ttS-nn
We can recognize the parental gametes in the descendants because their phenotypes are the most frequent,
151 tall, smooth, normal TSN155 dwarf, peach, oblate tsn
while the double recombinants are the less frequent.
2 tall, peach, normal TsN0 dwarf, smooth, oblate tSn
And simple recombinant gametes produced by the cross, which frequencies are intermediate.
33 tall, smooth, oblate TSn11 tall, peach, oblate Tsn29 dwarf, peach, normal tsN12 dwarf, smooth, normal tSN
Comparing the parental and the double recombinant we will realize that they only change in the position of the alleles S. This suggests that the position of the gene skin texture is in the middle of the other two genes, because in a double recombinant only the central gene changes position in the chromatid.
---TSN---
Now we will call Region I to the area between T and S and Region II to the area between S and N.
---TSN---
RI RII
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. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between T and S genes, and P2 to the recombination frequency between S and N.
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. To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
N = 151 + 33 + 11 + 2 + 155 + 29 + 12 + 0 = 393
So:
P1 = (R + DR) / N = 2 + 0 + 11 + 12 / 393 = 0.064
P2 = (R + DR)/ N = 2 + 0 + 33 + 29 / 393 = 0.163
Now, to calculate the recombination frequency between the two extreme genes, A and B, we can just perform addition or a sum:
P1 + P2= Pt
0.064 + 0.163 = Pt
0.227 = 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.
P1 = 0.064 x 100 = 6.4%
P2 = 0.163 x 100 = 16.3%
Pt = 0.227 x 100 = 22.7%
Genetic Map
---TSN---
║---6.4%---║16.3%---║
║22.7%║
