Question

2. A trihybrid individual heterozygous for three corn seed traits is testcrossed.

Parents: CCssWW (CsW/CsW)       X       ccSSww (cSw/cSw)
Trihybrid: CcSsWs (CsW/cSw)         X       ccssww (csw/csw)
Test cross offspring
Seed trait	Gamete from trihybrid	Number
Red, shrunken, normal	CsW	2777
White, plump, waxy	cSw	2708
Red, plump, waxy	CSw	116
White, shrunken, normal	csW	123
Red, shrunken, waxy	Csw	643
White, plump, normal	cSW	626
Red, plump, normal	CSW	4
White, shrunken, waxy	csw	3
Total number of progeny:		7000

a) Identify parental, SCO (single crossover) and DCO (double crossover) classes

b) Determine the gene order

c) Calculate the map distance for each gene interval

Answer 1

Analyses: Given corn has three loci, under observation, i.e. 'C' stands for seed colour (Red dominant and white recessive), 'S' stands for seed shape (Plump dominant and shrunken recessive) and 'W' stands for seed coating (Normal dominant and waxy recessive); controlling loci.

Given, trihybrid cross is: CsW/cSw X csw/csw

parental combination will be: CsW and cSw ... the respective offspring no is 2777 and 2708.

while some recombinant offspring will also be produced due to the cross over (CO) between loci.

1) CO between C and S loci : resulting phenotype of offspring will be as : CSw and csW (from single CO between C & S) and CSW and csw (from Double CO between C & S and S & W, simultaneously), according to the question the numbers are 116, 123, 4, 3 respectively, among 7000 offsprings. Thus frequency of CO between C ands S loci is = {(116+123+4+3)/7000} = (246/7000) = 0.0351. This frequency also denotes the relative distance between these two loci as centiMorgan (cM) unit, when converted to percentage. Thus the relative distnace between C and S loci is approx 3.51 cM.

2) CO between S and W loci : resulting phenotype of offsprings : Csw and cSW (from single CO between S and W) and CSW and csw (from Double CO between C & S and S & W, simultaneously), according to the question the numbers are 643, 626, 4, 3. Thus the frequency of CO = {(643+626+4+3)/7000} = (1276/7000) = 0.1823. Thus the relative distance between s and v locus is approx 18.23 cM.

3) CO between C and W loci : resulting phenotypes of offsprings : CSw, csW, Csw cSW, according to the question the respective numbers are 116, 123, 643, 626. Thus the frequency of CO = {116+123+643+626)/7000} = (1508/7000) = 0.2154. Thus the relative distance between C and W loci is approx 21.54 cM.

According to the results from 1), 2) and 3) we can say the distance between C and W is highest (21.54 cM) among all the three and locus S is lying betwen these two. Thus the speculated relative distances and arrangement of these three given loci, is going to be as shown below:

C----S------------------W

3.5cM 18.2cM

Answers: Now depending on the above analyses the answers of the given question will be :

a)

No	classes	Offspring genotype	Offspring No
1	Parental	CsW and cSw	2777 and 2708
2	SCO	CSw, csW, Csw and cSW	116, 123, 643 and 626
3	DCO	CSW and csw	4 and 3

b) The order (or, sequence) of the genes are as follows upon a chromosome : --C--S------W--

and

c) The map distances are i) between C and S = 3.51 cM, ii) between S and W = 18.23 cM and iii) between C and W = 21.54 cM.