Question

Three recessive genes in dogs are linked on a chromosome. A dog heterozygous for all three recessive alleles is crossed with homozygous for all the recessive alleles in a three-point testcross.

(a) Determine the order of these genes on the chromosome and

(b) calculate the map distances between the genes.

The progeny of the testcross are:

r    T    W                 324

R    t   w                 319

R   t    W                880

r   T     w               887

R    T   W                2376

r    t     w                 2365

r   t     W                47

R    T     w                41

Total      7,239

Answer 1

The genotype of higher progeny number with be the genotype of parents. So here the highest is progenies 2376 and 2365 indicated the parent genotype ie, RTW and rtw.

To know the gene order we should compare the genotype of parents with double cross overs ( DCO). We have genotype of double crossovers (least number of progeny)  which are, rtW and RTw. The rest of the progeny are single cross overs (SCO).

Finding gene order :

If we take wild parent type RTW and progeny genotype RTw and rtW

• RTW/(rtW and RTw) - the gene changed from parent type here is w/W

Here the it indicates the change by crossing over is the recessive w from the wild type . So the middle gene would be W/ w. The same can be compared with the recessive parent.

a) Thus the gene order is RWT or rwt.

b) Next to find the map distance we should group same point single cross overs.

We know that map distance = recombination frequency.

Thus map distance = (Number of recombinant progeny/total progeny) * 100

That is Map distance = [(no of SCO+ Sum of DCO)/total progeny] *100

• RWT*rwt : crossover between r and w gives progeny genotype of Rwt (319) and rWT(324)

So the map distance between r and W :[(319+324+88)/7239]*100 = 10 cM

• RWT*rwt :crossover between w and t gives progeny genotype of RWt(880) and rwT(887)

Map distance between w and t :[(880+887+88)/7239]*100 = 25 cM