Download Gene mapping: linkage

Course Overview
Gene mapping
February 6
The dihybrid situation
• In dogs dark coat colour (C) is dominant to
albino and short hair (S) is dominant over long
hair. C/C;S/S X c/c;s/s gave C/c;S/s F1
• A backcross to the homozygous recessive is
C/c;S/s X c/c;s/s
• The ratio of progeny from a backcross to the
homozygous recessive is
–
–
–
–
138 CS
135 cs
112 Cs
115 cS
• Parental vs recombinant
Outline
Week
1
2
3
4
5
6
7
8
9
10
11
12
Topic
Course objectives and Introduction to genetics
Human Pedigrees
Patterns of Inheritance: sex-linkage
Chromosomal basis of inheritance
Changes in chromosome number
Gene Mapping
Gene to Phenotype
Modified Mendelian ratios
Model organisms and mutants
Genetics of Plant Development (Arabidopsis)
Genetics of Animal Development (Drosophila)
Behaviour Genetics/Quantitative genetics
Chapter
Ch. 1 & Ch. 2
Ch. 2
Ch. 2
Ch. 3
Ch. 15
Ch. 4 (Ch. 16)
Ch. 6
Ch. 6
Ch. 6 (Ch. 16)
Ch. 18
Ch. 18
Ch. 16 + papers
Chi-Square (Χ2 ) Test eg.
1. H0: based on Mendel’s first law of equal segregation the
observed progeny fits a 1:1:1:1 ratio of phenotypes
CS:cs:Cs:cS (coat colour C= dark and hair length S= short).
2. Calculate chi-square
Class
CS
cs
CS
cS
df= 4-1
O
138
135
112
115
500
E
125
125
125
125
(O-E)2
(O-E)2/E
225
1.35
100
0.8
225
1.35
100
0.8
Chi-square = 4.3
1
Chi-Square (Χ2 ) Test eg. 2
Recombinant frequency
• The ratio of progeny from a testcross
C/c;S/s X c/c;s/s is
• Χ2 =4.3 , df = 3, p>10%, rejection level =5%
• Conclusion: given the chi-square =5.2 and df=3 a
deviation from the expected 1:1:1:1 ratio at least this
large would occur by chance alone 10% of the time so
we fail to reject the null hypothesis that the observed
ratio of CS:cs:Cs:cS is 1:1:1:1.
Drosophila genetics
• Vestigial (vg)
– 138 CS
– 135 cs
– 112 Cs
– 115 cS
• Parental vs recombinant
• 227/500 = 45.4% recombinant frequency
Linkage?
• T. H. Morgan crossed Drosophila
melanogaster wildtype flies with pure
breeding purple-eyed, vestigial winged
flies
• He then crossed F1 females heterozygous
for purple-eyes (pr) and vestigial wings
(vg) with tester males homozygous
recessive for pr and vg
• i.e. pr+/pr · vg+/vg X pr/pr · vg/vg
2
F1 of testcross
Class Phenotype
pr+ vg+ Wildtype eyes
and wings
pr vg Purple eyes,
vestigial wings
pr+ vg Wildtype eyes,
vestigial wings
pr vg+ Purple eyes,
wildtype wings
Total
O
1339
Type
Parental
1195
Parental
151
Recombinant
154
Recombinant
2839
Two autosomal genes
• H0: based on Mendel’s first law of equal
segregation the observed progeny fits a 1:1:1:1
ratio of phenotypes
pr+· vg+: pr ·vg+ : pr+ · vg : pr · vg
Class
O
E
(O-E)2
+
+
pr vg
1339 709.75 (629.25)2
1195 709.75 (485.25)2
pr vg+
151
709.75 (-558.75)2
pr+ vg
pr vg+
154
709.75 (-555.75)2
Total
2839
Chisquare=
For 4 phenotypic classes df= 4-1= 3
Linked?
• So df =3, chi-square =1764.82
• Rejection level p=0.05 the Χ2 =7.815. In fact on
the chi-square table for p=0.005 the Χ2 = 12.838
• Conclusion: given the chi-square =1764.82 and
df=3 a deviation from the expected 1:1:1:1 ratio
at least this large would occur by chance alone
much less than 5% of the time so we reject the
null hypothesis that the observed ratio of pr+·
vg+: pr ·vg+ : pr+ · vg : pr · vg is 1:1:1:1.
(O-E)2/E
557.88
331.76
439.87
435.16
1764.82
Linked
•
•
•
•
•
Parental vs recombinant
1339+1195 vs 151+154
Recombinant frequency = (151+154)/2839
305/2839 = 10.7% recombinant frequency
Linked in cis conformation
– pr+ vg+ / pr vg
3
F1 of another testcross
Class Phenotype
pr+ vg+ Wildtype eyes
and wings
O
157
Type
Recombinant
pr vg
Purple eyes,
vestigial wings
146
Recombinant
pr+ vg
Wildtype eyes,
vestigial wings
Purple eyes,
wildtype wings
965
Parental
pr vg+
Total
Linked
•
•
•
•
•
Parental vs recombinant
965+1067 vs 157+146
Recombinant frequency = (157+146)/2839
303/2335 *100 = 13% recombinant frequency
Linked in trans conformation
– pr vg+ / pr+ vg
• So the parents must have been
1067
Parental
– pr /pr · vg+/ vg+ X pr+/ pr+ · vg/ vg
– And F1 is pr vg+ / pr+ vg crossed to the tester males
2335
4
Chiasmata
• Chiasmata are the visible
manifestations of
crossovers
• Where do chiasmata
occur?
• A crossover is the
breakage of two DNA
molecules at the same
position and their
rejoining in two reciprocal
nonparental combinations
Parentals vs recombinants
• The gene combinations originally contributed to
the F1 by the parental flies represent in the most
frequent class of F1 from a testcross
• The original gene arrangement is called the
parental combination
• Some newer (and less frequent) combinations
that result from crossovers are called
nonparental combinations or simply
recombinants
1:1:1:1 ratio
• A mouse cross:
• A/a;B/b x a/a;b/b gave:
–
–
–
–
25% A/a;B/b
25% A/a;b/b
25% a/a;B/b
25% a/a;b/b
• What is the probability of
getting a mouse that
resembles either parent?
5
Bateson and Punnet: sweet pea
inheritance
• Bateson and Punnet crossed sweet peas
that were purple and long with those
plants that were red and short
– P= purple, p= red
– L= long, l=round
• F1 x F1 did not give the expected 9:3:3:1
result
Class
Observed
P_L_
P_ll
pp L_
pp ll
284
21
21
55
381
Expected (from
9:3:3:1 ratio)
215
71
71
24
• Bateson and Punnet suggested a physical
coupling of P + L and p + l in the original
parental gametic types might have prevented
their independent assortment in the F1 .
6
F1
Fly genetics
If you repeated their experiments and got
a similar result what would you do to test
if these two genes are linked?
1. Cross the F1 PpLl to a tester ppll
2. Do a chi-square to test whether the
genes assort independently
3. Calculate the recombinant frequency
• Genes on the same
chromosome are
separated by commas or
alternatively by spaces.
• For example, the genes
cn and bw are on the
second chromosome and
encode eye pigments.
• The genotype of a fly
homozygous for
mutations in these two
autosomal genes and w
is written: w; cn bw ; +;+
4
3
2
1
Genes on the X chromosome
X+
Xw
Xw
Xw Xw
Xw
Y
X+
(red ♀)
Xw Y (white ♂)
Xw X+ (red ♀)
Xw Y (white ♂)
• If we want to know how far apart genes are on
the X-chromosome we do not need a test cross
because the mother’s gametic genotype will be
the son’s phenotype
7
Fungal genetics
Single gene
• Neurospora crassa is a haploid organism
• The products of the meiotic divisions reflect the
order of the genes
• In N. crassa the four products of meiosis remain
together in groups of four called tetrads
• For linked genes A and B the tetrads contain
genotypes that could only be explained if
crossovers occurred at the two-chromosome
stage
Neurospora crassa
• For linked genes A and B the tetrads contain
genotypes that could only be explained if
crossovers occurred at the four-chromatid stage
8
Two crossovers
• ABCxabc
–ABc
–AbC
–aBC
–abc
• Which chromatids participated in the
crossovers?
• Two crossovers involving three chromatids
Two crossovers
• ABCxabc
–ABc
–Abc
–aBC
–abc
• Which chromatids participated in the
crossovers?
• Two crossovers involving four chromatids
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10
Introduction to genetic analysis
Griffiths, A., Wessler, S.R., Lewontin,R.C., Gelbart, W.M.,Suzuki, D.T.
and Miller, J.H.
Eighth Edition, W.H. Freeman and Company NY
•
•
•
•
Part I Transmission genetic analysis
Chapter 1: all questions p. 24-26
Chapter 2: all the questions p. 62-72
Chapter 3: questions #1-12,18,19, 22, 25-27, 29,
30, 32, 40-42.
• Chapter 4: 1-4, 6-13, 15-22, 24-43.
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