Transcript Chapter 4 - Mapping eukaryotic chromosomes by recombination

Review Problems
1905 William Bateson and R.C. Punnett
Red petals, round pollen X
(rr,ss)
Purple petals, long pollen
(RR,SS)
F1 Purple petals, long pollen (Rr,Ss)
Question
•
If two genes are tightly linked, such that no
crossing over occurs between them:
a.
b.
c.
d.
e.
will
bebe
parentals.
a.All
Allprogeny
progeny
will
parentals.
All progeny will be nonparentals.
All progeny will be recombinants.
Progeny will be 50% parental, 50% nonparental.
Progeny will be 25% nonrecombinant, 75%
recombinant.
X
yellow (Gg), round (Ww) yellow (Gg), round (Ww)
o
+
GW
o
GW
GW
gw
gw
F2
GGWW
GGWW
GgWw
GgWw
GGWW
GGWW
GgWw
GgWw
GgWw
ggww
ggww
GgWw
ggww
ggww
GW
gw
GgWw
gw
GgWw
G
e
n
e
r
a
t
i
o
n
If they assort independently (they are not linked)
F1 selfed
(Rr,Ss) X (Rr,Ss)
Expected F2
215
Purple, long
9
71
Purple, round
3
71
red, long
3
24
red, round
1
X
yellow (Gg), round (Ww) yellow (Gg), round (Ww)
o
+
GW
o
GW
Gw
gW
gw
F2
GGWW
GGWw
GgWW
GgWw
GGWw
GGww
GgWw
Ggww
GgWW
GgWw
ggWW
ggWw
GgWw
Ggww
ggWw
ggww
Gw
gW
gw
G
e
n
e
r
a
t
i
o
n
Question
•
If two nuclear genes in a diploid eukaryote are
physically linked by DNA sequence data, but we
have no additional data other than this, we can
say with confidence that they:
a.
b.
Are homologs
Are genetically linked and would cosegregate during
d. meiosis
Are located on the same chromosome
c. Are separated by no more than 1 cM
d. Are located on the same chromosome
e. Are located on separate chromosomes
F1 selfed
(Rr,Ss) X (Rr,Ss)
Expected F2
215
Purple, long
71
Purple, round
71
red, long
24
red, round
Results
284
Purple, long
21
Purple, round
21
red, long
55
red, round
Gene linkage, Recombination and Mapping
Chapter 4
Why map the genome ?

Gene position important to build complex
genomes

To determine the structure and function
of a gene

To determine the evolutionary
relationships and potential mechanism.
Two types of maps ?
 Recombination-based maps*
 Physical maps
The observation
1905 William Bateson and R.C. Punnett
Red petals, round pollen X
(rr,ss)
Purple petals, long pollen
(RR,SS)
F1 Purple petals, long pollen (Rr,Ss)
F1 selfed
(Rr,Ss) X (Rr,Ss)
Results
284
Purple, long
21
Purple, round
21
red, long
55
red, round
72
Purple, round
72
red, long
24
red, round
Expected F2
216
Purple, long
Symbols and terminology
AB
A/a
A/a; B/b
A/a . B/b
alleles on the same homolog, no punctuation
alleles on different homologs, slash
genes known to be on different
chromosomes, semicolon
genes of unknown linkage, use a period
Cis
Trans
AB/ab or ++/ab
Ab/aB or +b/a+
Thomas Hunt Morgan & Drosophilia
Red eyes, normal
(pr+/pr+ . vg+/vg+)
X
Purple eyes, vestigal
(pr/pr . vg/vg)
F1 Red eyes, normal wings (pr+/pr . vg+/vg)
Instead of selfing the population, he did a test cross.
Test cross
Red eyes, normal
(pr+/pr . vg+/vg)
X
Purple eyes, vestigal
(pr/pr . vg/vg)
1339 Red eyes, normal wings (pr+ . vg+)
1195 Purple eyes, vestigal (pr . vg)
151 Red eyes, vestigal (pr+. vg)
154 Purple eyes, normal wings (pr . vg+)
Test cross
1339 Red eyes, normal wings (pr+ . vg+)
1195 Purple eyes, vestigal (pr . vg)
151 Red eyes, vestigal (pr+. vg)
154 Purple eyes, normal wings (pr . vg+)
pr+
305/2839 = 10.7 percent
pr
vg+
vg
cis or trans ?
Initial cross
Red eyes, vestigal
(pr+/pr+ . vg/vg)
X
Purple eyes, normal
(pr/pr . vg+/vg+)
F1 Red eyes, normal wings (pr+/pr . vg+/vg)
Test cross with pr/pr . vg/vg
157 Red eyes, normal wings (pr+ . vg+)
146 Purple eyes, vestigal (pr . vg)
965 Red eyes, vestigal (pr+. vg)
1067 Purple eyes, normal wings (pr . vg+)
pr+
304/2335 = 12.9 percent
pr
vg
vg+
Morgan proposes
Linkage and Crossing Over
Fig. 4-3
Occurs at Prophase I (tetrad stage)
Crossing-over of the
chromosomes.
A chiasma is formed.
Genetic recombination.
Microscopic evidence for chromosome
breakage and gene recombination
Harriet Creighton and Barbara McClintock, 1931
Wx
C
wx
c
Wx
wx
c
C
For linked
genes,
recombinant
frequencies
are less than
50% in a
testcross.
Fig. 4-8
Mapping by Recombinant Frequency
Morgan set his student Alfred Sturtevant to the project.
“In the latter part of 1911, in conversation with Morgan,
I suddenly realized that the variations in strength of
linkage, already attributed by Morgan to differences in
the spatial separation of genes, offered the possibility of
determining sequence in the linear dimension of a
chromosome. I went home and spent most of the night
(to neglect of my undergraduate homework) in
producing the first chromosome map.” Sturtevant
Frequency of
crossing over,
indicates the
distance between
two genes on the
chromosome.
Map distances are additive.
Fig. 4-9
Question
You construct a genetic linkage map by following
allele combinations of three genes, X, Y, and Z.
You determine that X and Y are 3 cM apart, and X
and Z are 3 cM apart, and that Y and Z are 6 cM
apart. These cM numbers are most likely based
on:
a. DNA sequencing of the region in question
b. Recombination
b.
Recombination frequencies
frequencies
c. Measuring the distance in a scanning EM
micrograph
d. Independent assortment
Question
•
Referring to the cM numbers in the last
question, what is the relative gene order of
these three genes?
a.
b.
c.
d.
Z-X-Y
a.Y-X-Z
Z-X-Y
b. Y-X-Z
X-Y-Z
a and b
Summary
• Gene linkage
• Crossing over
• Recombinant mapping
Morgan proposes
Linkage and Crossing Over
Fig. 4-3
For linked
genes,
recombinant
frequencies
are less than
50% in a
testcross.
Fig. 4-8
Map distances are additive.
Fig. 4-9
Review Problems
1. A plant of genotype
is test crossed.
A
B
a
b
If the two loci are 14 m.u. apart, what proportion
of progeny will be AB/ab ?
43% AB, 43% ab, 7% Ab, 7% aB
Review Problems
2. A plant of genotype A/a . B/b is test crossed.
The progeny are
74 A/a . B/b
76 a/a . b/b
678 A/a . b/b
672 a/a . B/b
Explain.
A and B are linked in trans and are 10 m.u. apart.
Review Problems
3. You have analyzed the progeny of a test cross to a tetrahybrid.
The results indicate that
10% of the progeny are recombinant for A and B
14% for B and C
24% for A and C
4% for B and D
10% for C and D
14% for A and D
Provide a linear map for the chromosome.
Review Problems
3. You have analyzed the progeny of a test cross to a tetrahybrid.
The results indicate that
10% of the progeny are recombinant for A and B
14% for B and C
24% for A and C
4% for B and D
10% for C and D
14% for A and D
Provide a linear map for the chromosome.
|----------|----|----------|
A 10 B 4 D 10 C
Mapping with Molecular Markers
Chapter 4, continued.
What is a molecular marker
 SNP = single nucleotide polymorphisms
AAGGCTCAT
TTCCGAGTA
AAGACTCAT
TTCTGAGTA
• Silent SNPs
• SNP that cause phenotype
• SNP in polygenes
• SNP in intergenic regions
• RFLPs (restriction fragment length
polymorphisms)
RFLPs

SNPs that introduce a restriction enzyme site.
GGATTC
CCTAAG
EcoR1 site
GAATTC
CTTAAG
digest with EcoR1
RFLP analysis
Fig 4-15a
RFLP analysis
Fig 4-15b
RFLP analysis
Fig 4-15c
Using combinations of SNPs
A haplotype is a chromosomal segment defined by a
specific array of SNP alleles.
Using
haplotypes
to deduce
gene
position
Fig. 4-16
Simple sequence length polymorphisms
(SSLPs)
VNTRs (variable number tandem repeats)
Repeats of DNA sequence, with different numbers of
repeats occurring in different individuals.
Minisatellites (DNA fingerprints)
– Repeating units of 15-100 nucleotides
Microsatellites – repeat of 2-3 nucleotides
ACACACACACACAC
Minisatellites
Fig. 4-18
Microsatellites
Amplified by polymerase chain reaction.
primer 1
CACACACACACACA
CACACACACA
GTGTGTGTGTGTGT
GTGTGTGTGT
primer 2
St. M
M’
Fig. 4-19
Molecular markers can be used
instead of phenotype to map genes.
Chi-square
A/A . B/B X a/a . b/b
A/a . B/b
Test cross to a/a . b/b
Observed
142 A.B parental
133 a.b parental
113 A.b recombinant
112 a.B recombinant
Total 500
Expected
Using recombinant maps with physical maps
Summary
• Mapping using molecular markers
– SNPs, RFLP mapping, haplotypes
– SSLP
• Minisatellites
• Microsatellites