We are interested in detremining the genetic map of three different house finch traits: beak color (dark vs light), song length (short vs long) and patch size (reduced vs normal). We have done a large experiment involving the usual P0 crossed, with the F1 test crossed. The following are the data for the offspring from the test cross: Beak Shade Patch Size Song Length Number Observed
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
Choose the map below that best fits the data
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

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.

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)

Double recombinant)

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.

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 = (81+83+8+8)/1306

P1 = 180/1306

P1 = 0.138

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                    

i think the answer is g1

hope i

-lavira: )