Question

Three loci are found to be located (linked) on Chromosome 2 in fruit flies. At locus 1, wild is dominant over tall (a). For locus 2, wild is dominant over red (b) and for locus 3, wild is dominant over bent (c). Assume that a standard trihybrid backcross is made and the following progeny are produced: tall red bent 322 wild wild wild 306 wild red bent 6 tall wild wild 10 tall wild bent 32 tall red wild 153 wild red wild 41 wild wild bent 130 The distance between the locus for red (r) and the locus for bent (c) is ________ map units

Answer 1

Answer:

38.8 MU  

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

• At locus 1, wild (a+) is dominant over tall (a).  

• For locus 2, wild (b+) is dominant over red (b)  

• For locus 3, wild (c+) is dominant over bent (c).  

A standard trihybrid backcross is made and the following progeny are produced:  

• tall red bent, a b c   322

• wild wild wild, a+ b+ c+  306  

• wild red bent, a+ b c   6  

• tall wild wild, a b+ c+   10  

• tall wild bent, a b+ c    32  

• tall red wild, a b c+  153  

• wild red wild, a+ b c+   41  

• wild wild bent, a+ b+ c  130

The total number of individuals, N, is 1000.

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 genotypes of the parental gametes 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)

• tall red bent, a b c   322

• wild wild wild, a+ b+ c+  306  

Double recombinant)

• wild red bent, a+ b c   6  

• tall wild wild, a b+ c+   10  

Simple recombinant)

• tall wild bent, a b+ c    32  

• tall red wild, a b c+  153  

• wild red wild, a+ b c+   41  

• wild wild bent, a+ b+ c  130

By comparing them we will realize that between  

a b c   (parental)

a+ b c   (double recombinant)

and  between

a+ b+ c+  (Parental)

a b+ c+   (double recombinant)

They only change in the position of the alleles a/a+. This suggests that the position of the gene “a” is in the middle of the other two genes, “b” and “c”, because in a double recombinant only the central gene changes position in the chromatid.  

So, the order of the genes is:

---- b ---- a -----c ----

Now we will call Region I to the area between b and a, and Region II to the area between a and c.

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 b and a genes, and P2 to the recombination frequency between a and c.

• P1 = (R + DR) / N

• P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  So:  

Parental)

•  b a c  

•  b+ a+ c+    

Double recombinant)

• b a+ c    

• b+ a c+    

Simple recombinant)

•  b+ a c      

•  b a c+  

• b a+ c+  

•  b+ a+ c

P1 = (R + DR) / N

P1 = (32+41+6+10)/1000    

P1 = 89/1000

P1 = 0.089

P2= = (R + DR) / N

P2 = (153+130+6+10)/1000

P2 = 299/1000

P2 = 0.299

Now, to calculate the recombination frequency between the two extreme genes, b and c, we can just perform addition or a sum:

P1 + P2= Pt

0.089 + 0.299 = Pt

0.388=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 every 100 meiotic products, only 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 = 0.089 x 100 = 8.9 MU

GD2= P2 x 100 = 0.299 x 100 = 29.9 MU

GD3=Pt x 100 = 0.388 x 100 = 38.8 MU