Three recessive genes on chromosome 3 of Drosophila melanoga

Three recessive genes on chromosome 3 of Drosophila melanogaster are: javelin, scarlet, and roughed. The numbers of phenotypes from a trihybrid test cross are shown below: Wild type 842 javelin. scarlet, roughed 828 roughed 2186 javelin .scarlet 2222 scarlet 192 javelin .roughed 221 javelin 1036 scarlet .roughed 1008 What is the gene order and map distance between the three genes and what is the percent interference? Please type the answer. Including full explanation of question and the solution with full details.

Solution

Javelin = J; Scarlet = S; Roughoid = R sey----------

Parental characters: R = 2186 (j-R-s); J & S = 2222 (J-r-S)

Double Cross Overs: S = 192 (j-r-S); J & R = 221 (J-R-s)

Single Cross Overs: 1). Wild type: 842; J-S-R = 828

2). J = 1036 (J-s-r); S & R = 1008 (j-S-R)

bY COMPARING pARENTAL GENOTYPES WITH DOUBLE CROSS OVERS, WE CAN FIND THE J is the middle gene. So the gene order is:

either R----J----S or S----J----R

% OF CROSSING OVER = DISTANCE BETWEEN GENES = tOTAL SINGLE CROSS OVERS + DOUBLE CROSS OVERS / TOTAL PROGENY

dISTANCE BETWEEN J & S = (1036+1008) + (192+221) / 8535 = 28% = 28 mu

dISTANCE BETWEEN J & R = (842+828) + (192+221) / 8535 = 24% = 24 mu

gENE ORDER: R-------------------------24mu----------------J--------------------------------------------------28mu---------------------------S

cOEFFICIENT OF INCIDENCE = OBSERVED DOUBLE CROSSOVERS / EXPECTED DOUBLE CROSS OVERS

EXPECTED DOUBLE CROSS OVERS = SINGLE CROSS FREQUENCIES X TOTAL PROGENY

= 0.28 X 0.24 X 8535

= 573.5

tHEN cOEFFICIENT OF INCIDENCE = 192+221 / 573.5

= 413 / 573

= 0.72

iNTERFERENCE = 1 - cOEFFICIENT OF INCIDENCE

= 1- 0.72 = 0.279