Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a
wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w),
and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs).
You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a.
What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many
progeny of each phenotype would be expected if the genes were assorting independently? c. Can
you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d.
Assuming these genes are all linked on the same chromosome, draw a genetic map showing the
order and map distances between the three genes.
Solution
Answer:
a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl
b). If the genes were assorting independently, 8 types of phenotypes are expected in equl
proportions
1:1:1:1:1:1:1:1
Green, short, hair: 1
Red, long, smooth: 1
Green, long, smooth: 1
Green, short, smooth: 1
Red, short, hairy: 1
Red, long, hairy: 1
Green, long, hairy: 1
Red, short, hairy: 1
c).
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, short, smooth
5
25
-20
400.00
16.00
Red, short, hairy
20
25
-5
25.00
1.00
Red, long, hairy
7
25
-18
324.00
12.96
Green, long, hairy
30
25
5
25.00
1.00
Red, short, hairy
32
25
7
49.00
1.96
200
200
59.92
Chi-square value = 59.92
The independent assortment at P=0.05 is rejected.
d).
Explanation
Shorthand
Number of progeny
g sw hl
44
G Sw Hl
42
g Sw Hl
20
g sw Hl
5
G sw hl
20
G Sw hl
7
g Sw hl
30
G sw Hl
32
Imagine the order is
G SW Hl / g sw hl
1. If single cross over (SCO) occurs between G & Sw
Normal order= G---------Sw & g-----sw
After cross over= G-----sw & g------Sw
G-----sw recombinants are 20+ 32 = 52
g----- Sw recombinants are 20+ 30 = 50
Total recombinants = 102
RF = (102/200.)*100 =51%
2. If single cross over (SCO) occurs between Sw & Hl
Normal order= Sw---------Hl & sw----hl
After cross over= Sw-----hl & sw------Hl
Sw-----hl recombinants are 30 + 7 = 37
sw------Hl recombinants are 32+5 = 37
Total recombinants = 74
RF = (74/200)*100 = 37%
3. If single cross over (SCO) occurs between G & Hl
Normal order= G---------Hl & g------hl
After cross over= G-----hl & g------Hl
G-----hl recombinants are 20 + 7=27
g------Hl recombinants are 20+5=25
Total recombinants = 52
RF = (52/200)*100 = 26%
% RF = Map unit distance
The order of genes is -----
G------26 m.u.--------Hl------37 m.u.---------Sw
Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene
for wings & gene for eye.
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, sho.
Genetics Homework Problem 1 You are mapping genes in a species of f.pdf
1. Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a
wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w),
and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs).
You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a.
What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many
progeny of each phenotype would be expected if the genes were assorting independently? c. Can
you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d.
Assuming these genes are all linked on the same chromosome, draw a genetic map showing the
order and map distances between the three genes.
Solution
Answer:
a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl
b). If the genes were assorting independently, 8 types of phenotypes are expected in equl
proportions
1:1:1:1:1:1:1:1
Green, short, hair: 1
Red, long, smooth: 1
Green, long, smooth: 1
Green, short, smooth: 1
Red, short, hairy: 1
Red, long, hairy: 1
Green, long, hairy: 1
Red, short, hairy: 1
c).
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
3. 5
25.00
1.00
Red, short, hairy
32
25
7
49.00
1.96
200
200
59.92
Chi-square value = 59.92
The independent assortment at P=0.05 is rejected.
d).
Explanation
Shorthand
Number of progeny
g sw hl
44
G Sw Hl
42
g Sw Hl
20
g sw Hl
5
G sw hl
20
G Sw hl
7
g Sw hl
30
G sw Hl
32
Imagine the order is
G SW Hl / g sw hl
4. 1. If single cross over (SCO) occurs between G & Sw
Normal order= G---------Sw & g-----sw
After cross over= G-----sw & g------Sw
G-----sw recombinants are 20+ 32 = 52
g----- Sw recombinants are 20+ 30 = 50
Total recombinants = 102
RF = (102/200.)*100 =51%
2. If single cross over (SCO) occurs between Sw & Hl
Normal order= Sw---------Hl & sw----hl
After cross over= Sw-----hl & sw------Hl
Sw-----hl recombinants are 30 + 7 = 37
sw------Hl recombinants are 32+5 = 37
Total recombinants = 74
RF = (74/200)*100 = 37%
3. If single cross over (SCO) occurs between G & Hl
Normal order= G---------Hl & g------hl
After cross over= G-----hl & g------Hl
G-----hl recombinants are 20 + 7=27
g------Hl recombinants are 20+5=25
Total recombinants = 52
RF = (52/200)*100 = 26%
% RF = Map unit distance
The order of genes is -----
G------26 m.u.--------Hl------37 m.u.---------Sw
Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene
for wings & gene for eye.
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
