1 / 79

Using Molecular Markers January 2009

Using Molecular Markers January 2009. Using Molecular Markers (microsatellites, enzyme digest, RFLP, electrophoresis, Northern and Southern blotting, hybridization, ASO analysis, deletion mapping, contig assembly). Microsatellite. problem: A meiotic nondis-junction.

loki
Download Presentation

Using Molecular Markers January 2009

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Using Molecular MarkersJanuary 2009

  2. Using Molecular Markers(microsatellites, enzyme digest, RFLP, electrophoresis, Northern and Southern blotting, hybridization, ASO analysis, deletion mapping, contig assembly)

  3. Microsatellite

  4. problem: A meiotic nondis-junction Fred and Mary have a child with down syndrome. PCR was done a microsatellite region of chromosome 21. There are four alleles of the microsattelite (1, 2, 3, and 4) 1 2 3 4 What do the two bands in Fred’s lane represent? A) two different alleles of the microsatellite on the same chromosome B) two different alleles of the microsatellite on homologous chromosomes C) two different alleles of the microsatellite on non homologous chromosomes

  5. I. 1 2 II. 1 3 2 I-1 I-2 II-1 II-2 II-3 Microsatellite S21 A. B. C. • Here is a pedigree of a family with Charcot-Marie-Tooth (CMT) disease and a gel with the results of PCR amplification of the S21 microsatellite. S21 is tightly linked to the gene that causes CMT when mutated. Note; bands are of equal intensity • Considering both the pedigree and the microsatellite data, what is the mode of inheritance for CMT? • A. Autosomal dominant • B. Autosomal recessive • C. X-linked dominant • D. X-linked recessive • E. More than one of the above is possible

  6. I. 1 2 II. 4 5 2 1 3 I-1 I-2 II-1 II-2 II-3 II-4 II-5 INT3 A. B. C. D. • Here is a pedigree of a family with Cardiac Valvular Dysplasia (CVD) and a gel with the results of PCR amplification of the INT3 microsatellite. INT3 is tightly linked to the gene that causes CVD when mutated. Note; bands are of equal intensity • Considering both the pedigree and the microsatellite data, what is the mode of inheritance for CVD? • A. Autosomal dominant • B. Autosomal recessive • C. X-linked dominant • D. X-linked recessive • E. More than one of the above is possible

  7. The Family: A kindred of 17 total: 13 with CAD, 9 of living had acute MI Based on the pedigree alone what is the mode of inheritance? Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive More than one of the above are possible

  8. Haplotype found in individuals with disease, absent in others What event likely lead to this haplotype? A deletion An insertion Translocation Meiotic Recombination

  9. What did happen? Spontaneous mutation Paternity issue Meiotic recombination event Adoption

  10. You are a CSI agent at the scene of a crime. You collect DNA from a blood sample and run a micosatellite analysis using primers specific for the D15SS120 microsatellite (from the MEF2A paper). What is the maximum number of bands that will appear on your gel? 1 2 23 46 Too many to count

  11. Family #1 M= dominant mutation in Mef2A m = wild-type Mef2A Microsat: D15S120 I. I-1 I-2 II-1 II-2 mm Mm 2 1 A B II. C 2 1 D Mm mm These data are consistent with linkage between MEF2A and D15S120 and imply that: The A allele of marker D15S120 is linked to MEF2A The D allele of marker D15S120 is linked to MEF2A The A allele causes the Mef2a phenotype The D allele causes the Mef2a phenotype Both B and D are correct

  12. Family #2 M= dominant mutation in Mef2A m = wild-type Mef2A Microsat: D15S87 I. I-1 I-2 II-1 II-2 Mm Mm 2 1 A B II. C 2 1 D Mm mm In the absence of recombination, these data are consistent with: Linkage between D15S87A and MEF2A Linkage between D15S87B and MEF2A Linkage between D15S87C and MEF2A Linkage between D15S87D and MEF2A Both B and C are correct.

  13. Family #1 M= dominant mutation in Mef2A m = wild-type Mef2A Microsat:D15S120 I. I-1 I-2 II-1 II-2 mm Mm 2 1 A B II. C 2 1 D Mm mm In individual II-1, there is A. Linkage between the MEF2A gene and marker D15S120 B. No linkage between the MEF2A gene and marker D15S120

  14. Family #2 M= dominant mutation in Mef2A m = wild-type Mef2A Microsat:D15S87 I. I-1 I-2 II-1 II-2 Mm Mm 2 1 A B II. C 2 1 D Mm mm In individual II-1, there is A. Linkage between the MEF2A gene and marker D15S87 B. No linkage between the MEF2A gene and marker D15S87

  15. From lecture 20, identifying the heart disease gene Family #1 M= dominant mutation in Mef2A m = wild-type Mef2A Microsat:D15S120 I. I-1 I-2 II-1 II-2 mm Mm 2 1 A B II. C 2 1 D Mm mm In individual II-2, which allele of MEF2A is linked to which allele of D15S120? A. There is no linkage B. Wild type MEF2A and allele B C. Wild type MEF2A and allele D D. Wild type MEF2A and allele B on one homolog, mutant MEF2A and allele D on the other homolog E. Wild type MEF2A and allele D on one homolog, mutant MEF2A and allele B on the other homolog

  16. The DS15S20 microsatellite marker and the MEF2A gene, a gene that has • been implicated in heart disease, are linked • 1. The DS15S20 microsatellite marker causes heart disease • True • False 2. The DS15S20 microsatellite marker is a byproduct of having heart disease A. True B. False

  17. Suppose that MEF2A gene was found to be 5 cM away from marker D15S120. If you were to perform linage analysis within you own family, The MEF2A-D15S120 distance should also be 5 cM in your family B. The MEF2A-D15S120 distance should be at a distance other than 5 cM in your family

  18. I-2 I-2 II-1 II-2 II-3 III-1 III-2 1 2 I 1 A 1 2 3 B II C III D 1 2 What is the mode of inheritance? A) autosomal dominant B) autosomal recessive C) X-linked dominant D) X-linked recessive

  19. I-2 I-2 II-1 II-2 II-3 III-1 III-2 1 2 I 1 A 1 2 3 B II C III D 1 2 Which marker is linked to the disease phenotype? A) A B) B C) C D) D

  20. I-2 I-2 II-1 II-2 II-3 III-1 III-2 1 2 I 1 A 1 2 3 B II C III D 1 2 Which individual is a recombinant? A) II-2 B) II-3 C) III-1 D) III-2

  21. Is the son a “carrier”? YES NO

  22. What would be the STS markers in a child without the disease or the mutation? A/C A/D B/C B/D

  23. Two questions about the fetus. 1. Will the fetus get sickle cell anemia? A - Yes B - No 2. Is the fetus a carrier of sickle cell allele? A - Yes B - No

  24. Enzyme Digest

  25. Here is a bacterial plasmid and a piece of human genomic DNA that contains the FFI gene. You want to ligate the entire human FFI gene and the bacterial plasmid. What enzyme should you use to cut both the plasmid and the genomic DNA? Human genomic DNA Plasmid DNA BamHI PvuII EcoRII PvuII HindIII PvuII EcoRII BamHI EcoRII Lac Z gene AmpR gene FFI gene PvuII Unique restriction sites BamHI HindIII Ori PvuII BamHI HindII EcoRII BamHI or EcoRII

  26. Here is a bacterial plasmid and a piece of human genomic DNA that contains the FFI gene. You want to ligate half of human FFI gene and the bacterial plasmid. What enzyme should you use to cut both the plasmid and the genomic DNA? Human genomic DNA Plasmid DNA BamHI PvuII EcoRII PvuII HindIII PvuII EcoRII BamHI EcoRII Lac Z gene AmpR gene FFI gene PvuII Unique restriction sites BamHI HindIII Ori PvuII BamHI HindII EcoRII BamHI or EcoRII

  27. Which statement regarding restriction endonucleases is NOT correct? A) They recognize a specific base sequence in the DNA. B) They are produced by bacterial cells as a primitive immune system. C) They digest DNA by removing nucleotides from a free 3' end. D) They often generate short single stranded sequences.

  28. RFLP

  29. A genomic region shows 2 alleles of RFLP for EcoRI as shown below Which individuals are heterozyotes for RFLP alleles 1 and 2? II-1, II-1 and II-4 I-1, II-3 and I-2 Not enough information

  30. ? These individuals you just saw are members of a family in which an autosomal dominant disease allele is segregating. The disease gene is linked to the RFLP marker. While allele of RFLP did the mother give to II-1, II-2 & II-4? Allele 1 (10 kb) Allele 2 (6 & 4 kb) Both neither

  31. ? These individuals you just saw are members of a family in which an autosomal dominant disease allele is segregating. The disease gene is linked to the RFLP marker. While allele of RFLP in the mother is linked to the disease allele? Allele 1 (10 kb) Allele 2 (6 & 4 kb) Both neither

  32. d d D d ? D=dominant disease allele, d = recessive normal Will the fetus get the disease? Yes No

  33. Restriction Fragment Length Polymorphism 1 2 3 Southern Blot RFLP analysis was done on two parents and their child. Which lane represents the child’s DNA? A) 1 B) 2 C) 3

  34. Restriction Fragment Length Polymorphism Mother Child Male 1 Male 2 Could male 1 be the child’s father?A) yes B) no

  35. Restriction Fragment Length Polymorphism Mother Child Male 1 Male 2 Could male 2 be the child’s father?A) yes B) no

  36. Restriction Fragment Length Polymorphism Mom Child Male 1 Male 2 Which male could be the father of the child? A) Male 1 B) Male 2 C) Both males could be the father D) Neither male could be the father

  37. Gel Electrophoresis

  38. 250 150 E B 600 EcoRI 850 600 400 150 BamHI If you cut a given piece of DNA with either EcoRI OR BamHI you get: 400 bp 600 bp EcoRI fragments: BamHI fragments: What fragments would you get if you cut the same DNA with both EcoRI AND BamHI? 600, 400, 850 and 150 600, 250 and 150 600 and 150 850 and 400 850 bp 150 bp

  39. Northern and Southern Blotting

  40. Genomic DNA 1000 bp You run a Southern blot using human genomic DNA and a probe against the human FFI gene. You get this result: 500 bp Muscle Brain Next, you run a Northern blot using RNA from human brain and muscle tissue, and a probe against the human FFI gene. You get this result: 1000 bp 500 bp Which of the following is a reasonable conclusion from your results? The FFI gene is expressed in both the brain and muscle The FFI gene is not expressed in brain and muscle The FFI gene is expressed in the brain but not in the muscle The FFI gene is expressed in the muscle but not in the brain

  41. Hybridization

  42. Which probe would hybridize to the target sequence 5’…ATTCGACATT…3’ 5’…ATTCGACATT…3’ 5’…TTACAGCTTA…3’ 5’…AATGTCGAAT…3’ 5’…TAAGCTGTAA…3’

  43. Which probe would hybridize to the target sequence 5’…AGTCGACAGC…3’ 5’…GCTGTCGACT…3’ 5’…TCAGCTGTCG…3’ 5’…AGTCGACAGC…3’ 5’…CGACAGCTGT…3’

  44. ASO Analysis

  45. Mom Dad Jared Jane Sara Row 1 Row 2 Jared is homozygous for the most common missense mutation that causes sickle cell anemia. Genomic DNA from Jared, his parents who do not have sickle cell anemia, and his new twin sisters (Jane and Sara) was analyzed by Allele Specific Oligonucleotide (ASO) hybridization. Which row was hybridized with an ASO for the wild type allele of the hemoglobin gene? Row 1 Row 2 Both rows Neither row

  46. Detecting the CF deletion allele with ASO(it’s autosomal recessive) 1 2 3 4 5 6 Using only the ASO data: A) Individuals 1, 2, 3, 4, and 6 definitely do not have cystic fibrosis B) Individual 5 is the only individual that can have cystic fibrosis C) Individual 2 definitely has two copies of the wild type CFTR gene D) all of the above E) none of the above

  47. A dominant missense mutation in the blue gene causes smurfs to be red. Specific oligonucleotides were designed that hybridize specifically either to the wild type or mutant blue gene in the region of the missense mutation. Below is an ASO assay of a family of smurfs with a history of red skin. I-1 I-2 II-1 II-2 II-3 II-4 wild type mutant • Which individuals definitely have red skin? • II-4 only • I-2 and 1-4 • 1-1, II-1, II-3, and II-4 • all individuals

  48. A dominant autosomal missense mutation in the blue gene causes smurfs to be red. Specific oligonucleotides were designed that hybridize specifically either to the wild type or mutant blue gene in the region of the missense mutation. Below is an ASO assay of a family of smurfs with a history of red skin. I-1 I-2 II-1 II-2 II-3 II-4 wild type mutant • I-1 and I-2 are Smurfette and her husband and generation II are their • children. Since smurfette is the only adult female smurf, she is courted by a lot of • smurfs. Using only the ASO data, has Smurfette been faithful to her Smurf? • yes • no • can’t be determined

  49. Detecting the CF deletion allele with ASO(it’s autosomal recessive) Will this work for other CF alleles? A. Yes B. No

  50. Microarray analysis - Other individuals Individual 1 Individual 2 Individual 3 Who is a heterozygote carrier (+/-)? Individual 1, 2 or 3?

More Related