review problems 1905 william bateson and r.c. punnett red petals, round pollen (rr,ss) purple...
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Review Problems
1905 William Bateson and R.C. Punnett
Red petals, round pollen(rr,ss)
Purple petals, long pollen(RR,SS)
X
F1 Purple petals, long pollen (Rr,Ss)
Question
• If two genes are tightly linked, such that no crossing over occurs between them:
a. All progeny will be parentals.
b. All progeny will be nonparentals.
c. All progeny will be recombinants.
d. Progeny will be 50% parental, 50% nonparental.
e. Progeny will be 25% nonrecombinant, 75% recombinant.
a. All progeny will be parentals.
Xyellow (Gg), round (Ww) yellow (Gg), round (Ww)
+oo
GWGGWW
GgWw
GgWwGgWw
GgWw
GgWw
GgWwGGWW
GGWW
ggww
ggwwggww
ggwwGgWw
GGWW GgWw
Generation
F2
GW GW gw gw
gw
gw
GW
F1 selfed (Rr,Ss) X (Rr,Ss)
215Purple, long
24red, round
71red, long
71Purple, round
Expected F2
If they assort independently (they are not linked)
9 3 3 1
Xyellow (Gg), round (Ww) yellow (Gg), round (Ww)
GW Gw gW gw+o
o
GW
Gw
gW
gw
GGWW
GgWw
GgWwGgWW
GgWw
GgWW
GgWwGGWw
GGWw
ggWw
ggWwggWW
ggwwGgww
GGww Ggww
Generation
F2
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. Are homologsb. Are genetically linked and would cosegregate during
meiosisc. Are separated by no more than 1 cMd. Are located on the same chromosomee. Are located on separate chromosomes
d. Are located on the same chromosome
284Purple, long
55red, round
21red, long
21Purple, round
Results
F1 selfed (Rr,Ss) X (Rr,Ss)
215Purple, long
24red, round
71red, long
71Purple, round
Expected F2
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
1905 William Bateson and R.C. Punnett
Red petals, round pollen(rr,ss)
Purple petals, long pollen(RR,SS)
X
F1 Purple petals, long pollen (Rr,Ss)
The observation
284Purple, long
55red, round
21red, long
21Purple, round
Results
F1 selfed (Rr,Ss) X (Rr,Ss)
216Purple, long
24red, round
72red, long
72Purple, round
Expected F2
Symbols and terminology
Cis AB/ab or ++/abTrans Ab/aB or +b/a+
AB alleles on the same homolog, no punctuationA/a alleles on different homologs, slashA/a; B/b genes known to be on different
chromosomes, semicolonA/a . B/b genes of unknown linkage, use a period
Thomas Hunt Morgan & Drosophilia
Red eyes, normal(pr+/pr+ . vg+/vg+)
Purple eyes, vestigal(pr/pr . vg/vg)
X
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)
Purple eyes, vestigal(pr/pr . vg/vg)
X
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+)
305/2839 = 10.7 percentpr+
vg
vg+
pr
cis or trans ?
Red eyes, vestigal(pr+/pr+ . vg/vg)
Purple eyes, normal(pr/pr . vg+/vg+)
X
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+)
Initial cross
304/2335 = 12.9 percentpr+
vg+
vg
pr
Morgan proposes Linkage and Crossing Over
Fig. 4-3
Crossing-over of the chromosomes.
A chiasma is formed.
Genetic recombination.
Occurs at Prophase I (tetrad stage)
Microscopic evidence for chromosome breakage and gene recombination
Harriet Creighton and Barbara McClintock, 1931
Wx
wx c
C Wx
wx
c
C
Fig. 4-8
For linked genes, recombinant frequencies are less than 50% in a testcross.
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
QuestionYou 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 questionb. Recombination frequenciesc. Measuring the distance in a scanning EM
micrographd. Independent assortment
b. Recombination frequencies
Question
• Referring to the cM numbers in the last question, what is the relative gene order of these three genes?
a. Z-X-Yb. Y-X-Zc. X-Y-Zd. a and b
a. Z-X-Yb. Y-X-Z
Summary
• Gene linkage
• Crossing over
• Recombinant mapping
Morgan proposes Linkage and Crossing Over
Fig. 4-3
Fig. 4-8
For linked genes, recombinant frequencies are less than 50% in a testcross.
Map distances are additive.
Fig. 4-9
Review Problems
A B
a b
1. A plant of genotype
is test crossed.
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 B14% for B and C24% for A and C 4% for B and D10% 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 B14% for B and C24% for A and C 4% for B and D10% 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 MarkersChapter 4, continued.
What is a molecular marker
SNP = single nucleotide polymorphisms
• Silent SNPs• SNP that cause phenotype• SNP in polygenes• SNP in intergenic regions• RFLPs (restriction fragment length polymorphisms)
AAGGCTCATTTCCGAGTA
AAGACTCATTTCTGAGTA
RFLPs
SNPs that introduce a restriction enzyme site.
GGATTCCCTAAG
GAATTCCTTAAG
EcoR1 site
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 nucleotidesACACACACACACAC
Minisatellites
Fig. 4-18
Microsatellites
Amplified by polymerase chain reaction.
CACACACACAGTGTGTGTGT
primer 1
primer 2St.
CACACACACACACAGTGTGTGTGTGTGT
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
142 A.B parental133 a.b parental113 A.b recombinant112 a.B recombinant
Total 500
Observed Expected
Using recombinant maps with physical maps
Summary
• Mapping using molecular markers– SNPs, RFLP mapping, haplotypes– SSLP
• Minisatellites
• Microsatellites