1. (10pt) You have a strain of fruit flies that are black in colour ( b ) instea
ID: 92568 • Letter: 1
Question
1. (10pt) You have a strain of fruit flies that are black in colour (b) instead of the usual tan, have shortened wings (sw) relative to wild type, and pinkish eyes (p) instead of red. You cross b. sw. p females to wild type males and all the F1 are wild type.
Then you cross the F1 females to males from the original strain – b. sw. p. You obtain the following number and phenotypes of F2:
b sw p : 422
b+ sw+ p+: 403
b sw+ p: 63
b+ sw p+: 52
b sw p+: 31
b+sw+ p: 24
b+ sw p : 3
b sw+ p+: 2
a) Determine the correct gene order and explain how you know. (1 point)
b) Calculate map distances between all three pairs of genes, if appropriate (6 points)
c) Draw a map showing the arrangement of the genes (1 point)
d) Calculate the interference if any, and explain in one or two sentences what interference is (use your own words) (2 points)
Explanation / Answer
Answer:
a). Gene order = sw+--------b+--------p+
b).
Distance between sw+ & b+ = 12 map unites
Distance between b+ & p+ = 6 map unites
Distance between sw+ & p+ = 17 map unites
c). Arrangment of the genes= sw+------12 m.u.--------b+----6 m.u.----p+
d). Interference = 0.931
Interference is nothing but the reduction of cross over frequency at one spot by the nearby previous cross over.
Explanation:
b+, sw+ & p+ alleles are wild type and are dominant over b, sw & p alleles respectively.
b/b sw/sw p/p (female) X b+ sw+ p+ / Y (males)-----------Parents
b sw p / Y (recessive male) & b+/b sw+/sw p+/p (wild type female)-----Progeny.
b+/b sw+/sw p+/p (wild type female) x b sw p / Y (recessive male) -----Test cross.
1. If single cross over (SCO) occurs between b+ & sw+ and b & sw
Normal order is b+---------sw+ & b-----sw
After cross over b+-----sw & b------sw+
b+-----sw recombinants are 52+3= 55
b------sw+ recombinants are 63+2= 65
Total recombinants = 55+65=120
RF = (120/1000)*100 =12%
2. If single cross over (SCO) occurs between sw+ & p+ and sw & p
Normal order= sw+---------p+ & sw----p
After cross over= sw+-----p & sw------p+
sw+-----p recombinants are 63+24= 87
sw------p+ recombinants are 52+31= 83
Total recombinants = 170
RF = (170/1000)*100 = 17%
3. If single cross over (SCO) occurs between b+ & p+ and b&p
Normal order= b+---------p+ & b------p
After cross over= b+-----p & b------p+
b+-----p recombinants are 24+3= 27
b------p+ recombinants are 31+2= 33
Total recombinants = 60
RF = (60/1000)*100 =6%
% RF = Map unit distance
The order of genes is -----
sw+------12 m.u.--------b+----6 m.u.----p+
RF between sw+ & b+ = 12% = 0.12
RF between b+ & p+ = 6% = 0.06
Expected double cross overs = 12 * 6 = 72
Observed double cross overs (b+ sw p & b sw+ p+) = 2 + 3 = 5
Observed double cross progeny are more than the expected double cross over progeny.
Coefficent of coincidence (CC) = Obeserved dco / Expected dco
CC = 5/72 = 0.069
Interference = 1-CC = 1-0.069 = 0.931
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.