1. Introduction to Genetic Analysis
Griffiths • Wessler • Carroll • Doebley
Introduction to
Genetic Analysis
ELEVENTH EDITION
CHAPTER 4
Mapping Eukaryote Chromosomes by Recombination
© 2015 W. H. Freeman and Company

2. CHAPTER OUTLINE 4.1 Diagnostics of linkage
4.2 Mapping by recombinant frequency
4.3 Mapping with molecular markers
4.4 Centromere mapping with linear tetrads
4.5 Using the chi-square test to infer linkage
4.6 Accounting for unseen multiple crossovers
4.7 Using recombination-based maps in conjunction with physical maps
4.8 The molecular mechanism of crossing over

3. A recombination-based map of one of the chromosomes of Drosophila

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.

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

6. 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)

7. 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

8. 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 dot or period
Cis AB/ab or ++/ab
Trans Ab/aB or +b/a+

9. 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.

10. 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+)

11. 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 ?

12. 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+)
303/2335 = 13.0 percent
pr+
vg+ vg pr

13. Linked alleles tend to be inherited together

14. Crossing over produces new allelic combinations

15. Chiasmata are the sites of crossing over
Chiasmata are the sites of crossing over. Occurs in Prophase I of meiosis.

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

17. Microscopic evidence for chromosome breakage and gene recombination
Harriet Creighton and Barbara McClintock, 1931
-studied two genes in corn (on chromosome 9) for seed color (C and c) and endosperm composition (Wx and wx) Wx C Wx c wx c wx C

18. Page 133 Note the knob on the C end and the longer piece of chromosome on the Wx end.

19. Recombinants are produced by crossovers

20. For linked genes, recombinant frequencies are less than 50 percent in a testcross

21. 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

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

23. A map of the 12 tomato chromosomes

24. Map distances are additive.

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. Longer regions have more crossovers and thus higher recombinant frequencies

28. Double recombinants arising from two crossovers

29. Different gene orders give different double recombinants

30. Phenotypic ratios in progeny reveal the type of cross

31. Summary Gene linkage Crossing over Recombinant mapping
Be sure to read p and p in order to do Chi-Square (c2) tests for some homework problems

32. Chi-Square (c2) Test
We must apply the χ2 analysis to test a hypothesis of no linkage. If that hypothesis is rejected, we can infer linkage. (We cannot test a hypothesis of linkage directly because we have no way of predicting what recombinant frequency to test.) The calculation for testing lack of linkage is as follows:
Observed (O)
Expected (E)
O–E
(O – E)2
(O – E)2/E 60 50 10
100
2.00 37
–13
169
3.38 41 –9 81
1.62 62 12
144
2.88
χ2 = Σ (O – E)2/E for all classes = 9.88
Since there are four genotypic classes, we must use 4 − 1 = 3 degrees of freedom. Consulting the chi-square table in Chapter 3, we see our values of 9.88 and 3 df give a p value of ~0.025, or 2.5 percent. This is less than the standard cut-off value of 5 percent, so we can reject the hypothesis of no linkage.

33. 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

34. 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.

35. 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.

36. 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

37. Mapping with Molecular Markers
Chapter 4, continued.
SNPs-single nucleotide polymorphisms
SSLPs-simple sequence length polymorphisms
SSLPs also called VNTRs-variable number tandem repeats
Molecular markers can be used instead of phenotype to map genes

38. 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)

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

40. Simple sequence length polymorphisms
(SSLPs)
Also known as VNTRs (variable number tandem repeats)
Repeats of DNA sequence, with different numbers of repeats occurring in different individuals; used for DNA fingerprinting.
Minisatellites (DNA fingerprints)
– Repeating units of nucleotides
Microsatellites (DNA fingerprints)
– Repeating units of 2-15 nucleotides
[e.g., ACACACACACACAC]

41. A microsatellite locus can show linkage to a disease gene

42. Alignment of physical and recombination maps

43. Summary Mapping using molecular markers SNPs, RFLP mapping SSLPs/VNTRs
Minisatellites
Microsatellites