Transcript Bio 2970 Lab 5: Linkage Mapping

Bio 2970 Lab 5:
Linkage Mapping
Sarah VanVickle-Chavez
Gender
The case of the white-eyed mutant
Character
Eye colour
Traits
Red eye (wild type)
White eye (mutant)
P Phenotypes
Wild type (red-eyed) female x White-eyed male
F1 Phenotypes
All red-eyed
Red eye is dominant to white eye
Hypothesis
A cross between the F1 flies should give us: 3 red eye :
1 white eye
F2
Phenotypes
Numbers
So far so good
Red eye
White eye
3470
82%
782
18%
An interesting observation
F2
Phenotypes
Redeyed
males
Redeyed
females
Whiteeyed
males
White-eyed
females
Numbers
1011
2459
782
0
24%
58%
18%
0%
© 2007 Paul Billiet ODWS
A reciprocal cross
Morgan tried the cross the other way round
white-eyed female x red-eyed male
Result
All red-eyed females and all white-eyed males
This confirmed what Morgan suspected
The gene for eye colour is linked to the X chromosome
A test cross
Phenotypes
F1 Red-eyed female x White-eyed male
Expected result
50% red-eyed offspring: 50% white-eyed offspring
Regardless of the sex
Observed Results
Red-eyed
Males
Red-eyed
Females
White-eyed
Males
White-eyed
Females
132
129
86
88
Determining if a mutant is dominant or
recessive, and if it is X-linked or autosomal
• To determine if a mutant is dominant or recessive, and if it is X-linked or
autosomal, you perform a pair of reciprocal crosses (where the gender of
the parents is reversed).
• If the gene is autosomal  identical results in both crosses.
• If the gene is X-linked results of the two crosses are different.
• The expected results for reciprocal crosses is summarized in the table
below.
Mutants – Linkage Mapping
Phenotype
b pr rw
+
+
+
+ pr rw
b
+
+
b
+ rw
+ pr
+
b
pr
+
+ + rw
b pr rw
+++
Mutants
1. Record offspring with each of 8
phenotypes:
• E.g., cross using three genes A, B, and C
Phenotype
+
A
B
C
A;B
A;C
B;C
A;B;C
TOTAL:
Number
3223
632
1051
226
229
1019
617
3091
10088
2. Rearrange into pairs, identifying
double recombinants, single
recombinant, and nonrecombinant
(parental)
Phenotype
+
A
B
C
A;B
A;C
B;C
A;B;C
TOTAL:
Number
3223 (NR)
632 (SR)
1051 (SR)
226 (DR)
229 (DR)
1019 (SR)
617 (SR)
3091 (NR)
10088
3. Compare the NR class with the DR
class to determine the one gene that is
switched in the double crossover class.
This is the gene that is in the middle.
Phenotype
+
A
B
C
A;B
A;C
B;C
A;B;C
TOTAL:
Number
3223 (NR)
632 (SR)
1051 (SR)
226 (DR)
229 (DR)
1019 (SR)
617 (SR)
3091 (NR)
10088
In this example, it is (C), so
the order is (A-C-B).
You could also state the gene
order as the reverse (B-C-A).
4. Each of the remaining classes is a
single crossover class between a gene
on the end and the gene in the middle:
either (B-C singles) or (C-A singles).
Phenotype
+
A
B
C
A;B
A;C
B;C
A;B;C
TOTAL:
Number
3223 (NR)
632 (SR) = C-A
1051 (SR) = B-C
226 (DR)
229 (DR)
1019 (SR) = B-C
617 (SR) =C-A
3091 (NR)
10088
Compare each singles class
with the nonrecombinant
class to see which gene is
switched to determine which
one is the B-C singles class,
and which is the C-A
singles class.
5. Calculate the distance between each
pair of genes.
• B-C =
• C-A =
6. Calculate the expected double
crossover frequency, the obtained
double crossover frequency, the
coefficient of coincidence, and the
interference.