Two fruit bats are crossed in a captive breeding program. It is a testcross invo
ID: 163432 • Letter: T
Question
Two fruit bats are crossed in a captive breeding program. It is a testcross involving three loci that control for different traits: Fur color (Ff = purple, ff = blue), Ear size (Ll = large, ll = small), and Fruit Preference (Aa = avocados, aa = mangos). The phenotypes and numbers of offspring in each category are shown below.
Phenotype
Genotype
# of Offspring
Class
Purple, large, avocado-loving
Ff, Ll, Aa
1
Double crossover
Purple, large, mango-loving
Ff, Ll, aa
558
Parental
Purple, small, avocado-loving
Ff, ll, Aa
56
Single crossover
Purple, small, mango-loving
Ff, ll, aa
70
Single crossover
Blue, large, avocado-loving
ff, Ll, Aa
62
Single crossover
Blue, large, mango-loving
ff, Ll, aa
53
Single crossover
Blue, small, avocado-loving
ff, ll, Aa
497
parental
Blue, small, mango-loving
ff, ll, aa
3
Double crossover
a.Calculate the distance from the middle gene to each flanking gene.
b.How many double crossover events do you expect among these bat offspring?
Phenotype
Genotype
# of Offspring
Class
Purple, large, avocado-loving
Ff, Ll, Aa
1
Double crossover
Purple, large, mango-loving
Ff, Ll, aa
558
Parental
Purple, small, avocado-loving
Ff, ll, Aa
56
Single crossover
Purple, small, mango-loving
Ff, ll, aa
70
Single crossover
Blue, large, avocado-loving
ff, Ll, Aa
62
Single crossover
Blue, large, mango-loving
ff, Ll, aa
53
Single crossover
Blue, small, avocado-loving
ff, ll, Aa
497
parental
Blue, small, mango-loving
ff, ll, aa
3
Double crossover
Explanation / Answer
Answer:
a). Middle gene is A
b). Expected double crossover events= 12
Explanation
Genotypes (short form)
Number of progeny
FLA
1
FLa
558
FlA
56
Fla
70
fLA
62
fLa
53
flA
497
fla
3
Total
1587
Parental combinations are more so, the parental combinations FLa/flA
1. If single cross over (SCO) occurs between F & L
Normal order= F --------- L & f ----- l
After cross over= F----- l & f------ L
F ----- l recombinants are 70+56= 126
f ------ L recombinants are 62+53=115
Total recombinants = 241
RF = (241/1300)*100 =18.54%
2. If single cross over (SCO) occurs between L&a
Normal order= L --------- a & l --------- A
After cross over= L ---------A & l ------ a
L --------- A recombinants are 1+62= 63
l ------ a recombinants are 3+70=73
Total recombinants = 136
RF = (136/1300)*100 = 10.46%
3. If single cross over (SCO) occurs between F & a
Normal order= F --------- a & f ----- A
After cross over= F ----- A & f ------ a
F ----- A recombinants are 1+56=57
f ------ a recombinants are 3+53=56
Total recombinants = 113
RF = (113/1300)*100 = 8.69%
% RF = Map unit distance
The order of gene is -----
F ------8.69m.u.------- A -----------10.46 m.u.----------L
Expected double cross over frequency = (RF of F&A) RF of A & L)
= 0.0869 * 0.1046= 0.0091
Expected double cross overs= Expected double cross over frequency * total progeny
= 0.0091*1300
= 12
Genotypes (short form)
Number of progeny
FLA
1
FLa
558
FlA
56
Fla
70
fLA
62
fLa
53
flA
497
fla
3
Total
1587
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