D. Beak Shade - 13.78cM - Patch Size - 5.67cM - Song Length
Explanation:
We have the number of descendants of each phenotype product of the tri-hybrid cross.
Light Reduced Long 83 Light Reduced Short 8 Light Normal Short 581
Light Normal Long 28 Dark Normal Long 8 Dark Normal Short 81 Dark Reduced Short 30
Dark Reduced Long 487
The total number of individuals is N = 1306.
In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:
Parental)
Light Normal Short (581 individuals)
Dark Reduced Long (487 individuals)
Double recombinant)
Light Reduced Short (8 individuals) Dark Normal Long (8 individuals)
Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for patch size.
They only change in the position of Normal and Reduced. This suggests that the position of the gene patch size is in the middle of the other two genes, beak color and song length, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
Beak color patch size song length
Now we will call Region I to the area between Beak color and patch size, and Region II to the area between patch size and song length.
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 Beak color and patch size genes, and P2 to the recombination frequency between patch size and song length.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of simple recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals. So:
Region I
Light Normal Short (Parental) 581
Dark Normal Short (Simple Recombinant) 81
Light Reduced Short (Double Recombinant) 8
DarkReduced Long (Parental) 487
Light Reduced Long (Simple Recombinant) 83
Dark Normal Long (Double recombinant) 8
Region II
Light Normal Short (Parental) 581
Light Normal Long (Simple Recombinant) 28
Light Reduced Short (Double Recombinant) 8
DarkReduced Long (Parental) 487
Dark Reduced Short (Simple Recombinant) 30
Dark Normal Long (Double recombinant) 8
In each region, the highlighted traits are the ones that suffered recombination.
P1 = (R + DR) / N
P1 = (81+83+8+8)/1306
P1 = 180/1306
P1 = 0.138
P2= = (R + DR) / N
P2 = (28+30+8+8)/1306
P2 = 74/1306
P2 = 0.057
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 every 100 meiotic products, one 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 = 0.138 x 100 = 13.8 MU = 13.8 cM
GD2= P2 x 100 = 0.057 x 100 = 5.7 MU = 5.7 cM
Options:
A. Patch Size - 13.78cM - Beak Shade - 5.67cM - Song Length
B. Patch Size - 5.67cM - Song Length - 13.78cM - Beak Shade
C. Beak Shade - 5.67cM - Patch Size - 13.78cM - Song Length
D. Beak Shade - 13.78cM - Patch Size - 5.67cM - Song Length
E. Patch Size - 13.78cM - Song Length - 5.67cM - Beak Shade
F. Patch Size - 5.67cM - Beak Shade - 13.78cM - Song Length
