Respuesta :
Complete question:
You are tracking three loci in the fruit fly, loci A, B, and C, and want to know whether the loci are linked and if so what their physical distances are from one another as well as the physical order they are in. To find out this information you perform a three-point testcross with a fly that is heterozygous for all three alleles. The progeny of this cross are:
a/a B/b C/c 389
A/a b/b c/c 410
a/a b/b C/c 39
A/a B/b c/c 44
a/a b/b c/c 77
A/a B/b C/c 83
a/a B/b c/c 6
A/a b/b C/c 5
Total 1053
(a) What is the recombination frequency between loci A and B?
(b) What is the recombination frequency between loci B and C?
(c) What is the recombination frequency between loci A and C?
(d) Are these loci linked?
(e) Demonstrate the physical arrangement of the genes on the chromosome, i.e. the order of the genes on a chromosome as well as their map distances.
(f) Write out the genotypes of the parents for this cross, indicating which alleles are linked with one another for each parent?
Answer and Explanation:
The Crossing-over frequency between two genes depends on the distance between them. A short distance between genes is a very little target for crossing-over to occur, which means that only a few of them will happen, compared with the number of events between genes that are more separated between each other. Two genes that are very close will have a few recombination events and are strongly bounded. While two separated genes will have more chances of recombination and are not bound.
To know if two genes are linked, we must observe the progeny distribution. In a tri-hybrid cross, If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1:1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the heterozygote parent.
In the present example, the phenotypic ratio shows different proportions than the expected ones if they were not linked.
We can recognize the parental gametes in the descendants because their phenotypes are the most frequent,
a/a B/b C/c 389
A/a b/b c/c 410
while the double recombinants are the less frequent.
a/a B/b c/c 6
A/a b/b C/c 5
And simple recombinant gametes produced by the cross, which frequencies are intermediate.
a/a b/b C/c 39
A/a B/b c/c 44
a/a b/b c/c 77
A/a B/b C/c 83
Comparing the parental and the double recombinant we will realize that they only change in the position of the alleles C/c. This suggests that the position of the gene C is in the middle of the other two genes, A and B, because in a double recombinant only the central gene changes position in the chromatid.
A------C------B
Now we will call Region I to the area between A and C and Region II to the area between C and B.
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 A and C genes, and P2 to the recombination frequency between C and B.
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%.
So:
- P1 = (R + DR) / N = 5 + 6 + 77 + 44 + 39 / 1053 = 171/1053 = 0.162
- P2 = (R + DR)/ N = 6 + 5 + 83 / 1053 = 94/1053 = 0.089
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
0.162 + 0.089 = Pt
0.251 = 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 = 0.162 x 100 = 16.2%
- P2 = 0.089 x 100 = 8.9%
- Pt = 0.251 x 100 = 25.1 %
(a) What is the recombination frequency between loci A and B? 0.251
(b) What is the recombination frequency between loci B and C? 0.089
(c) What is the recombination frequency between loci A and C? 0.162
(d) Are these loci linked? Yes. Their recombination frequency is inferior to 50% and the phenotypic rate is different from 1:1:1:1:1:1:1:1
(e) Demonstrate the physical arrangement of the genes on the chromosome, i.e. the order of the genes on a chromosome as well as their map distances.
A-------------------C-----------------B
A--- 16.2 UM --C ---8.9 UM--B
A-------------25.1 UM--------------B
(f) Write out the genotypes of the parents for this cross, indicating which alleles are linked with one another for each parent?
a/a C/c B/b x A/a c/c b/b
