1. Review Problems 1905 William Bateson and R.C. Punnett
Red petals, round pollen
(rr,ss) X
Purple petals, long pollen
(RR,SS) F1
Purple petals, long pollen (Rr,Ss)

2. Question
If two genes are tightly linked, such that no crossing over occurs between them:
All progeny will be 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.
a. All progeny will be parentals.

3. X GW GW gw gw + F2 GW GGWW GGWW GgWw GgWw Generation GW GGWW GGWW GgWw
yellow (Gg), round (Ww)
yellow (Gg), round (Ww) GW GW gw gw o + o F2 GW
GGWW
GGWW
GgWw
GgWw
Generation GW
GGWW
GGWW
GgWw
GgWw gw
GgWw
GgWw
ggww
ggww gw
GgWw
GgWw
ggww
ggww

4. If they assort independently (they are not linked)
F1 selfed (Rr,Ss) X (Rr,Ss)
Expected F2
215
Purple, long 24
red, round 71
red, long
Purple, round

5. X GW Gw gW gw + F2 GGWW GGWw GgWW GW GgWw Generation Gw GGWw GGww GgWw
yellow (Gg), round (Ww)
yellow (Gg), round (Ww) GW Gw gW gw o + o F2
GGWW
GGWw
GgWW GW
GgWw
Generation Gw
GGWw
GGww
GgWw
Ggww
GgWW
GgWw gW
ggWW
ggWw
GgWw gw
Ggww
ggWw
ggww

6. 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:
Are homologs
Are genetically linked and would cosegregate during meiosis
Are separated by no more than 1 cM
Are located on the same chromosome
Are located on separate chromosomes
d. Are located on the same chromosome

7. F1 selfed (Rr,Ss) X (Rr,Ss)
Expected F2
215
Purple, long 24
red, round 71
red, long
Purple, round
284
Purple, long 55
red, round 21
red, long
Purple, round
Results

8. Gene linkage, Recombination and Mapping
Chapter 4

9. 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.

10. Two types of maps ?
Recombination-based maps*
Physical maps

11. The observation 1905 William Bateson and R.C. Punnett
Red petals, round pollen
(rr,ss) X
Purple petals, long pollen
(RR,SS) F1
Purple petals, long pollen (Rr,Ss)

12. F1 selfed (Rr,Ss) X (Rr,Ss)
Results
284
Purple, long 21
Purple, round 21
red, long 55
red, round
216
Purple, long 24
red, round 72
red, long
Purple, round
Expected F2

13. Symbols and terminology
AB alleles on the same homolog, no punctuation
A/a alleles on different homologs, slash
A/a; B/b genes known to be on different chromosomes, semicolon
A/a . B/b genes of unknown linkage, use a period
Cis AB/ab or ++/ab
Trans Ab/aB or +b/a+

14. 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.

15. Test cross Red eyes, normal (pr+/pr . vg+/vg) X Purple eyes, vestigal
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+)

16. 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+
vg+
305/2839 = 10.7 percent pr vg
cis or trans ?

17. 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+)
304/2335 = 12.9 percent
pr+
vg+ vg pr

18. Linkage and Crossing Over
Morgan proposes
Linkage and Crossing Over
Fig. 4-3

19. Occurs at Prophase I (tetrad stage)
Crossing-over of the
chromosomes.
A chiasma is formed.
Genetic recombination.

20. Microscopic evidence for chromosome breakage and gene recombination
Harriet Creighton and Barbara McClintock, 1931 Wx C Wx c wx c wx C

21. For linked genes, recombinant frequencies are less than 50% in a testcross.
Fig. 4-8

22. 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

23. Frequency of crossing over,
indicates the distance between two genes on the chromosome.

24. Map distances are additive.
Fig. 4-9

25. 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:
DNA sequencing of the region in question
Recombination frequencies
Measuring the distance in a scanning EM micrograph
Independent assortment
b. Recombination frequencies

26. Question
Referring to the cM numbers in the last question, what is the relative gene order of these three genes?
Z-X-Y
Y-X-Z
X-Y-Z
a and b
a. Z-X-Y
b. Y-X-Z

27. Summary
Gene linkage
Crossing over
Recombinant mapping

28. Linkage and Crossing Over
Morgan proposes
Linkage and Crossing Over
Fig. 4-3

29. For linked genes, recombinant frequencies are less than 50% in a testcross.
Fig. 4-8

30. Map distances are additive.
Fig. 4-9

31. Review Problems 1. A plant of genotype A B is test crossed. 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

32. Review Problems 2. A plant of genotype A/a . B/b is test crossed.
The progeny are 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.

33. 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.

34. Review Problems |----------|----|----------| A 10 B 4 D 10 C
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 B 4 D C

35. Mapping with Molecular Markers
Chapter 4, continued.

36. 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)

37. RFLPs SNPs that introduce a restriction enzyme site. EcoR1 site GGATTC
CCTAAG
GAATTC
CTTAAG
digest with EcoR1

38. RFLP analysis
Fig 4-15a

39. RFLP analysis
Fig 4-15b

40. RFLP analysis
Fig 4-15c

41. Using combinations of SNPs
A haplotype is a chromosomal segment defined by a specific array of SNP alleles.

42. Using haplotypes to deduce gene position
Fig. 4-16

43. 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 nucleotides
Microsatellites – repeat of 2-3 nucleotides
ACACACACACACAC

44. Minisatellites
Fig. 4-18

45. Microsatellites Amplified by polymerase chain reaction. primer 1
CACACACACACACA
GTGTGTGTGTGTGT
CACACACACA
GTGTGTGTGT
primer 2
St. M M’

46. Fig. 4-19

47. 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
Expected
A.B parental
133 a.b parental
113 A.b recombinant
112 a.B recombinant
Total 500

48. Using recombinant maps with physical maps

49. Summary Mapping using molecular markers SNPs, RFLP mapping, haplotypes
SSLP
Minisatellites
Microsatellites