Question

CC DD and cc dd individuals were crossed to each other, and the F1 generation was backcrossed to the cc dd parent. 903 Cc Dd, 897 cc dd, 98 Cc dd, and 102 cc Dd offspring resulted.

1. How far apart are the c and d loci?

2. What progeny and in what proportions would you expect to result from testcrossing the F1 generation from a CC dd x cc DD cross to cc dd?

Answer 1

A. Parents genotype given CCDD*ccdd

Here the offspring would inherit 1 dominant and 1 recessive allele for the 2 traits.The resultant genotype will be CcDd

Now, backcross is done between CcDd * ccdd

The distance between the c and d loci will be

RF (recombination frequency) =(Total no. of recombinant/Total no. of progeny) 100

(98+102/903+897+98+102)100= 10% or 10 m.u/ map units between c and d.

Map units are defined as the tendency of the 2 loci to undergo crossing over. The genes located far have a high probability of crossing over and vice versa.

[Given that-903 (parent) Cc Dd

897 (parent) cc dd,

98 Cc dd-recombinant

and 102 cc Dd – recombinant]

B. The test cross of F1 generation will give the following progenies

CCDD*ccDD (Parents)

Cd     cD (gametes)

     CcDd (F1 generation)

F1 generation(CcDd) * ccdd (test cross)

         CD Cd cD cd           cD (gametes)

The F2 generation will be

Gametes	CD	Cd	cD	cd
cd	CcDd	Ccdd	ccDd	ccdd
cd	CcDd	Ccdd	ccDd	ccdd

Frequency of CcDd=No. of progenies with CcDd genotype/Total progenies= 2/8=1/4

     Similarly the other genotypes frequency will be

Frequency of Ccdd=2/8=1/4 and frequency of ccDd and ccdd will also be ¼.

RF=(98+102/1800)*100=10%m.u, Thus the probability of an offspring with one of the parental genotypes is 90% [Cross between CcDd * ccdd parents and offsprings –Ccdd and ccDd)