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Review Problems

Purple petals, long pollen (Rr,Ss). F 1. Review Problems. 1905 William Bateson and R.C. Punnett. X. Red petals, round pollen (rr,ss). Purple petals, long pollen (RR,SS). Question. If two genes are tightly linked, such that no crossing over occurs between them:

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Review Problems

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  1. Purple petals, long pollen (Rr,Ss) F1 Review Problems 1905 William Bateson and R.C. Punnett X Red petals, round pollen (rr,ss) 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. o o + X yellow (Gg), round (Ww) yellow (Gg), round (Ww) GW GW gw gw F2 GW GgWw GGWW GGWW GgWw Generation GW GGWW GgWw GGWW GgWw gw GgWw GgWw ggww ggww gw GgWw ggww GgWw ggww

  4. 24 red, round 71 red, long 215 Purple, long 71 Purple, round If they assort independently (they are not linked) F1 selfed (Rr,Ss) X (Rr,Ss) Expected F2 9 3 3 1

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

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

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

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

  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. Red eyes, normal wings (pr+/pr . vg+/vg) F1 Thomas Hunt Morgan & Drosophilia X Red eyes, normal (pr+/pr+. vg+/vg+) Purple eyes, vestigal (pr/pr . vg/vg) Instead of selfing the population, he did a test cross.

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

  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 vg pr cis or trans ?

  17. Test cross with pr/pr . vg/vg Red eyes, normal wings (pr+/pr . vg+/vg) F1 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+ vg 304/2335 = 12.9 percent vg+ pr Initial cross X Red eyes, vestigal (pr+/pr+. vg/vg) Purple eyes, normal (pr/pr . vg+/vg+)

  18. 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 c wx 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. 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 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

  32. 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.

  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 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

  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 GAATTC CTTAAG GGATTC CCTAAG 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 15-100 nucleotides Microsatellites – repeat of 2-3 nucleotides ACACACACACACAC

  44. Minisatellites Fig. 4-18

  45. CACACACACACACA GTGTGTGTGTGTGT Microsatellites Amplified by polymerase chain reaction. primer 1 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

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