2/29/12 positional cloning the mutation that takes away pain
TRANSCRIPT
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2/29/12 Positional Cloning The mutation that takes away pain
by Michael Hopkin [email protected]
Imagine being unable to feel any pain at all. For a tiny handful of people, that is the reality and medical researchers have now pinpointed the mutation that removes their ability to perceive painful sensations. The study began when doctors in northern Pakistan examined a remarkable group of related families in which several individuals seem entirely unaffected by pain. Their attention was first attracted by one member of the clan, a locally famous boy who performed street theatre involving walking on burning coals and stabbing his arms with knives. Although it sounds like a party trick, the condition is devastating, as sufferers don't learn to know their limits. The street-performing boy killed himself on his fourteenth birthday after jumping off a house roof. The researchers studied six of his relatives, aged between 4 and 14 years. All had suffered many cuts and bruises, and injuries to lips and tongue caused by biting themselves; several had fractured bones without noticing.
This shows the importance of pain for our health and survival, notes Geoffrey Woods of the Cambridge Institute for Medical Research, UK, who led the study. "Pain is there for a jolly good reason it stops us damaging ourselves," he says. For example, the pain from a broken arm or sprained ankle encourages us to rest that body part while it recovers. The children in the study had no such safety check, causing them to be both graceless and reckless. "One girl was continually knocked down in the playground and just didn't mind at all," Woods says. BioEd Online (http://www.BioEdOnline.org) December 13, 2006 Primary publication: NATURE| Vol 444| 14 December 2006
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What gene is involved here? What is its molecular function?
è Analysis of genes at the molecular level typically requires that the gene or at least portions of the gene be available in a purified form. Cloning (in a molecular context) means identifying and “purifying” a specific gene è HOW to find a specific gene among the 3 billion pairs & twentysomeodd thousand genes that comprise the human genome? • A number of strategies are available for cloning specific human genes • What strategy you use depends to a great extent on the type of information
you have at the outset • Do you know what protein the gene produces? • Do you have a clone of a related gene from another organism?
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How do you clone a gene of interest if you have essentially no information about the gene except the mutant phenotype?
In other words, you know what happens to the organism if a mutant allele is present but you • don’t have any idea what kind of protein the gene specifies • don’t have any information on the DNA sequence of the gene • don’t have an animal model This was the case for the geneticists who wanted to clone the genes responsible for cystic fibrosis, for Huntington disease and for many other human disease states: they had no idea what kind of protein the gene coded for
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Used a strategy called positional cloning: Positional Cloning is phenotype driven
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Positional Cloning is Phenotype Driven:
Genetic disease (or other trait) with specific phenotype ê
Find map position of the Gene: Linkage to markers on a specific chromosome
Then locate to specific site on this chromosome ê
[Isolate clones of genomic DNA that correspond to the map position of the gene and look for candidate gene –only if you’re working with an organism that doesn’t yet have a sequenced genome]
ê For sequenced genomes, consult the appropriate database to find
candidate genes in the region closely linked to the disease phenotype ê
Look for mutations in the candidate genes present in individuals with the trait/disease state
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• First find map position of the gene • Then look around for candidate genes in the area • Look for mutations in affected individuals
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LOD = log of odds = ratio of probability of linkage with stated map distance over probability of independent assortment. For example, a LOD = 3 means that linkage is a 1000 X more likely than independent assortment (OPTIONAL but recommended: see pg 158-159 9th edition (not in 10th) for a nice discussion of how LOD is calculated using the basic rules of probability and linkage)
1cM = 1 centiMorgan = 1 map unit = 1% of gametes are recombinant for the two genes/loci under consideration]
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Rather than looking at linkage or independent assortment of two organismal phenotypes, these studies look linkage between a disease or variant phenotype and molecular markers that tag specific sites on each chromosome FIRST STEP: random genome-wide screen with STR & SNP markers on all 22 autosomes to determine approximate location of gene of interest Useful molecular markers: http://www.ornl.gov/sci/techresources/Human_Genome/faq/snps.shtml
STR fingerprint loci: highly polymorphic and easy to assay SNP = single nucleotide polymorphism (biallelic if using restriction enzymes to assay)
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SNP = single nucleotide polymorphism (pronounced snip) http://www.ornl.gov/sci/techresources/Human_Genome/faq/snps.shtml
About one in every 900-1000 nucleotides in the human genome is the site of a single-nucleotide polymorphism
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Detecting SNPs with restriction enzymes RFLP= restriction fragment length polymorphism
D allele GAATTC
B1
CTTAAG
d allele GACTTC
B2
CTGAAG
SNP detected by PCR with primers 1 and 2followed by digestion with the restrictionenzyme EcoRI which only cuts atGAATTC
NOTE: the SNP sequence variation is linked tobut otherwise unrelated to the disease gene
P1
P2
P1
P2
D allele GAATTC
B1
CTTAAG
d allele GACTTC
B2
CTGAAG
SNP detected by PCR with primers 1 and 2followed by digestion with the restrictionenzyme EcoRI which only cuts atGAATTC
NOTE: the SNP sequence variation is linked tobut otherwise unrelated to the disease gene
P1
P2
P1
P2
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Figure from text: Chromosome 1: 356cM & 247 Mbps 1 cM = 1 map unit
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Does the dominant disease phenotype appear linked to SNP A (located on chromosome 15) or SNP B (located on chromosome 2) or neither? SNP A is a site on chromosome 2 that represents a single nucleotide polymorphism For example: Allele 1 (A1) = GC base pair Allele 2 (A2) = AT base pair at this position Keep in mind that for the study to be useful, the polymorphic locus must be heterozygous in the individuals carrying the mutant allele
A1A2 A1A1
A1A2
A1A2
A1A2 A1A2A2A2 A1A2A2A2 A2A2 A2A2
A2A2
B1B2 B2B2
B1B2
B1B1
B2B1 B1B1B1B1 B2B1B2B1 B2B1 B2B1
B1B1
A1 & A2 = SNP
B1 & B2 = SNP
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Using LOD (logarithm of odds) scores to assess linkage in human pedigrees
(see end of this lecture) LOD score for these data = ~1 so need to collect more data….. LOD = 1 means that probability of this pedigree is 10Xmore likely if the disease gene and SNP B are linked rather than assorting independently
A1A2 A1A1
A1A2
A1A2
A1A2 A1A2A2A2 A1A2A2A2 A2A2 A2A2
A2A2
B1B2 B2B2
B1B2
B1B1
B2B1 B1B1B1B1 B2B1B2B1 B2B1 B2B1
B1B1
A1 & A2 = SNP
B1 & B2 = SNP
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Linkage to chromosome 2 STR loci was established using PCR and STR primers
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Figure 4.16 Using haplotypes to deduce gene position The mutant allele remains in linkage disequilibrium only with SNPs 4 and 5. Hence, the gene must be in the vicinity of 4 and 5.
Haplotype: • a chromosomal segment defined by
the specific alleles that it carries • a group of closely linked alleles
that tend to be inherited together • The alleles may be SNPs (detected
as RLFPs), STRs (fingerprint loci) or other polymorphic loci
As illustrated in the next few figures, haplotype analysis can be used to map human disease genes very accurately by finding polymorphisms that flank the gene of interest
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Each of eight different chromosome 2 STR loci were genotyped for each person • How do these data confirm linkage to autosome 2? • How do these data define the interval on chromosome 2 that must contain
the disease gene?
The data are displayed in a standard haplotype format
• For example, individual V-1 in the third generation is heterozygous for alleles 6 and 2 for STR locus 1353, het for alleles 4 and 3 for STR locus 2370
• The loci are listed top to bottom in the order that they are found on chromosome 2
• All alleles on the left are on one homolog of 2 and all alleles on the right are on the homolog
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• Remember that the STR or SNP markers used for these haplotype studies are linked to but otherwise unrelated to the disease mutation and can be uncoupled from the disease phenotype by crossing over
D allele GAATTC
B1
CTTAAG
d allele GACTTC
B2
CTGAAG
SNP detected by PCR with primers 1 and 2followed by digestion with the restrictionenzyme EcoRI which only cuts atGAATTC
NOTE: the SNP sequence variation is linked tobut otherwise unrelated to the disease gene
P1
P2
P1
P2
D= dominant disease allele d = recessive wild-type alelle
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EXAMINE DATA ON NEXT PAGE
Taking the segregation data from all of the families together, in what portion of the haplotype is the gene mutated in erythermalgia located? Examine the haplotypes of heterozygotes in different families: why is the STR haplotype of the chromosome carrying the dominant mutant allele different in different families?
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Why is the disease allele associated with different STR haplotypes in different families?
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LOD = log of odds = ratio of probability of linkage with stated map distance over probability of independent assortment. For example, a LOD = 3 means that linkage is a 1000 X more likely than independent assortment
~1 X 106 bp (1 Mb)/ centimorgan (cM)
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Finding candidate genes in 2q31-32
Go to chromosome 2 and examine this interval (next page) http://www.ncbi.nlm.nih.gov/mapview/map_search.cgi?chr=hum_chr.inf&query
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Many genes in this interval and a gene called SCN9A nearby Genes
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RED= BRAGGING RIGHTS Gain-of-function mutation in Nav1.7 in familial erythromelalgia induces bursting of sensory neurons Brain (2005), 128, 1847–1854 Erythromelalgia is an autosomal dominant disorder characterized by burning pain in response to warm stimuli or moderate exercise. We describe a novel mutation in a family with erythromelalgia in SCN9A, the gene that encodes the Nav1.7 sodium channel. Nav1.7 produces threshold currents and is selectively expressed within sensory neurons including nociceptors. We demonstrate that this mutation, which produces a hyperpolarizing shift in activation and a depolarizing shift in steady-state inactivation, lowers thresholds for single action potentials and high frequency firing in dorsal root ganglion neurons. Erythromelalgia is the first inherited pain disorder in which it is possible to link a mutation with an abnormality in ion channel function and with altered firing of pain signalling neurons.
Mutation in exon 23 Genomic DNA from the proband and control subjects was isolated from venous blood samples and used as template to amplify all known exons of SCN9A and compare the sequence with Nav1.7 cDNA (Klugbauer et al., 1995). Proband and control templates produced similar amplicons which were purified and sequenced. Sequence analysis identified a T-to-G transversion in exon 23 (E23), corresponding to position 4393 of the reference sequence (see Supplementary material). This mutation substitutes phenylalanine (F) by valine (V) at position 1449 of the polypeptide, located at the N-terminus of loop 3 which joins domains III and IV. F1449 is invariant in all known mammalian sodium channels (Fig. 2). Restriction digestion analysis (see Supplementary material) confirmed the presence of the F1449V mutation in 17 out of 17 affected individuals, and its absence in five out of five unaffected family members, three out of three unaffected spouses and 100 ethnically matched control chromosomes. Segregation of the T4393G mutation with disease was confirmed by DNA sequencing of E23 in all family members.
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Voltage-gated sodium channels are essential for the initiation and propagation of action potentials in neurons. The sodium channel α subunits are large, transmembrane proteins with approximately 2,000 amino acid residues, composed of 4 homologous domains containing well-characterized voltage sensor and pore regions (Figure 1). The transmembrane segments are highly conserved through evolution. The 4 domains associate within the membrane to form a sodium-permeable pore, through which sodium ions flow down a concentration gradient during propagation of an action potential. The transmembrane sodium gradient is subsequently restored by the activity of the ATP-dependent sodium/potassium pump.
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Missense Mutation in Exon 23 coding for Loop 3 (see figure previous page)
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Nonsense mutations in the SCN9A gene take pain away http://fire.biol.wwu.edu/trent/trent/nopain.pdf
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OH, Really?
This article (10.1056/NEJMoa0907006) was published on May 5, 2010, at NEJM.org. l-Histidine Decarboxylase and Tourette’s Syndrome
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Using LOD (logarithm of odds) scores to assess assess linkage in human pedigrees
•Logarithm of odds score (LOD): a way of supporting a linkage hypothesis with small sample size (typical in human pedigree analysis.) •Are these loci linked? 2 of 6 offspring are recombinants – a recombination frequency of 33%. •Are these loci unlinked? Maybe small sample size has skewed our interpretation – so calculate the probability of obtaining 2 recombinant and 4 parental configurations under the condition that the RF=50% •When you calculate a LOD score, you express the odds of linkage under a certain RF and the odds of getting a particular combination of progeny under independent assortment as a ratio, and take the logarithm of that ratio. •Logarithms are exponents of base 10, so a LOD score of 3 means an RF value that is 103 as likely as no linkage. LOD = log10 of [likelihood if linked]/[likelihood if the loci are unlinked]
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Lod score calculation for pedigree data on pg 13: 7 parental offspring and one recombinant between gene D and SNP B First start with a hypothesis about linkage: – maybe they are linked and 10 cM (10 map units apart) • What is the probability of a non-recombinant gamete? 1-r = prob(no recombination) = 0.90 Remember there are two different parental genotypes. If you keep track of individual genotypes, then each has probability (1-r)/2 so, (1-0.9)/2 = 0.45 • Probability of this sibship (7 parental and one recombinant genotype in given birth order) if genes are linked (at 10 cM) is: (0.45)7 (0.05)= 1.8 x 10-4
• Probability if not linked is (0.25)8 = 1.53 X10-5 remember, r=0.5 when loci are not linked • Lod score = log10 (1.8 x 10-4/1.53 X10-5) = log10 (~10) = 1 The pattern of 7 parental and 1 recombinant (in the given birth order) is 10 times more likely if the genes are linked at 10 cM than if they were unlinked BUT LOD must be >3.0 to be “significant”