Presentation on theme: ". Computational Genetics Lecture 1 This class has been edited from several sources. Primarily from Terry Speed’s homepage at Stanford and the Technion."— Presentation transcript:
. Computational Genetics Lecture 1 This class has been edited from several sources. Primarily from Terry Speed’s homepage at Stanford and the Technion course “Introduction to Genetics”. Changes made by Dan Geiger. Background Readings: Chapter 2&3 of An introduction to Genetics, Griffiths et al. 2000, Seventh Edition (CS/Fishbach/Other libraries).
5 Course Goals u Learning about computational and mathematical methods for genetic analysis. u We will focus on Gene hunting – finding genes for simple human diseases. u Methods covered in depth: linkage analysis (using pedigree data), association analysis (using random samples). u Another goal is to learn more about Bayesian networks usage for genetic linkage analysis.
6 Human Genome Most human cells contain 46 chromosomes: u 2 sex chromosomes (X,Y): XY – in males. XX – in females. u 22 pairs of chromosomes, named autosomes.
7 Genetic Information u Gene – basic unit of genetic information. They determine the inherited characters. u Genome – the collection of genetic information. u Chromosomes – storage units of genes.
8 Sexual Reproduction zygote gametes sperm egg Meiosis
10 Central Dogma Transcription mRNA Translation Protein Gene cells express different subset of the genes In different tissues and under different conditions שעתוק תרגום
11 Chromosome Logical Structure Marker – Genes, SNP, Tandem repeats. Locus – location of markers. Allele – one variant form of a marker. Locus1 Possible Alleles: A1,A2 Locus2 Possible Alleles: B1,B2,B3
12 Alleles - the ABO locus example O is recessive to A. A is dominant over O. A and B are codominant. Multiple alleles: A,B,O. GenotypePhenotype A/A, A/OA B/B, B/OB A/BAB O/OO Trait = Character = Phenotype
13 מושגים: 1.אלל רצסיבי ודומיננטי. כאשר קיים בתא גם האלל הרצסיבי וגם הדומיננטי, הפנוטיפ שקובע האלל הדומיננטי משתלט. 2.AA ו- aa הם הומוזיגוטים (Homozygote) לאלל הדומיננטי והרצסיבי, בהתאמה. Aa הוא הטרוזיגוט (Hetrozygote). 3.אללים מרובים (A,B,O),
14 X-linked u b - dominant allele. Namely, (b,b), (b,w) is Black. u w - recessive allele. Namely, only (w,w) is White. This is an example of an X-linked trait/character. For males b alone is Black and w alone is white. There is no homolog gene on the Y chromose. genotype phenotype
15 Modern genetics began with Mendel’s experiments on garden peas (Although, the ramification of his work were not realized during his life time). He studied seven contrasting pairs of characters, including: The form of ripe seeds: round, wrinkled The color of the seed albumen: yellow, green The length of the stem: long, short Mendel’s Work Mendel Gregor. 1866. Experiments on Plant Hybridization. Transactions of the Brünn Natural History Society.
16 Mendel’s first law Characters are controlled by pairs of genes which separate during the formation of the reproductive cells (meiosis) A a A a
17 P: AA X aa F1: Aa F1 X F1 Aa X Aa test cross Aa X aa Gametes: A a A AA Aa a Aa aa F2: 1 AA : 2 Aa : 1 aa ~ A a Phenotype Gametes: A a a Aa aa 1A : 1 a Phenotype: ~
18 מושגים: 2. הכלאת מבחן: הכלאת צאצאי 1F על ההורה בעל הפנוטיפ הרצסיבי. היחס בין הצאצאים המראים הפנוטיפ הדומיננטי לאלו המראים הפנוטיפ הרצסיבי הוא – 1:1 1. הכלאה של 1F על עצמו: בדור 2F היחס בין הצאצאים המראים הפנוטיפ הדומיננטי לאלו המראים הפנוטיפ הרצסיבי הוא – 3:1.
19 F2 ratioF2F1Parental Phenotype 2.96:15474 round; 1850 wrinkledRound1. Round X wrinkled seeds 3.01:16022 yellow; 2001 greenyellow2. Yellow X green seeds 3.15:1705 purple; 224 whitepurple3. Purple X white petals 2.95:1882 inflated; 299 pinchedinflated4. Inflated X pinched pods 2.82:1428 green; 152 yellowgreen5. Green X yellow pods 3.14:1651 axial; 207 terminalaxial6. Axial X terminal flowers 2.84:1787 lon; 277 shortlong7. Long X short stems Results of crosses in which parents differed for one character Mendel's First low. Conclusion, First low: The two members of a gene pair segregate from each other into the gametes.
20 דוגמא לשושלת עם מוטציה רצסיבית (נישואין של בני דודים).
23 Mendel’s second law When two or more pairs of genes segregate simultaneously, they do so independently. A a; B b A BA bA ba Ba ba b P AB = P A P B P Ab =P A P b P aB =P a P B P ab =P a P b
25 Mendel's second low. A dihybrid cross for color and shape of pea seeds P wrinkled and yellow X round and green rrYYRRyy F1 round yellow Rr Yy X Rr Yy F2 round yellow 315 round green 108 wrinkled yellow 101 wrinkled green 32 a. Check segregation pattern for each allele in F2: 416 yellow : 140 green (2.97:1) 423 round : 133 wrinkled (3.18:1) Conclusion: both traits behave as single genes, each carrying two different alleles. 556
26 Question: Is there independent assortment of alleles of the different genes? Probability to get yellow is 3/4; probability to get round is 3/4; probability to get yellow round is 3/4 X 3/4, namely 9/16 Probability to get yellow is 3/4; probability to get wrinkled is 1/4; probability to get yellowwrinkled is 3/4 X 1/4, namely 3/16 Probability to get green is 3/4; probability to get round is 3/4; probability to get green round isX 3/4, namely 3/16 Probability to get green is 1/4; probability to get wrinkled is 1/4; probability to getgreen wrinkled is1/4 X 1/4, namely1/16. 1/4
27 A standard presentation in terms of counts expected expected observed yellow round 9 312.75 315 yellow wrinkled 3 104.25 101 green round 3 104.25 108 green wrinkled 1 34.75 32 Total 16 556 556 Conclusion, second law: Different gene pairs assort independently in gamete formation
28 “Exceptions” to Mendel’s Second Law Morgan’s fruit fly data (1909): 2,839 flies Eye colorA: reda: purple Wing lengthB: normalb: vestigial AABB x aabb AaBb x aabb AaBb Aabb aaBb aabb Expected 710 710 710 710 Observed 1,339 151 154 1,195 The pair AB stick together more than expected from Mendel’s law.
29 Morgan’s explanation A A BB a a b b F1: A a Bb a a b b F2: A a Bb a a b b A a bb a a B b Crossover has taken place
30 Parental types:AaBb, aabb Recombinants: Aabb, aaBb The proportion of recombinants between the two genes (or characters) is called the recombination fraction between these two genes. It is usually denoted by r or . For Morgan’s traits: r = (151 + 154)/2839 = 0.107 If r < 1/2: two genes are said to be linked. If r = 1/2: independent segregation (Mendel’s second law).
31 Recombination Phenomenon (Happens during Meiosis) Recombination Haplotype Male or female תאי מין: ביצית, או זרע The recombination fraction Between two loci on the same chromosome Is the probability that they end up in regions Of different colors
32 כרומוזומים מזווגים המראים כיאסמתה הכיאסמתה היא הביטוי הציטולוגי לשחלוף.
33 Example: ABO, AK1 on Chromosome 9 Hardy-Weinberg law of population genetics permits calculation of genotype frequencies from allele frequencies P(a)= frequency of “a” in the population P(ab) =2P(a)P(b) Hardy-Weinberg equilibrium corresponds to a random union of two gamets, called zygote. 2 4 5 1 3 A A 1 /A 1 O A 2 /A 2 A A 1 /A 2 O A 1 /A 2 A A 2 /A 2 O A 1 A 2 A O A 1 A 2 A | O A 2 | A 2 O A 2 Recombinant Phase inferred
34 Example: ABO, AK1 on Chromosome 9 Recombination fraction is 12/100 in males and 20/100 in females. One centi-morgan means one recombination every 100 meiosis. One centi-morgan corresponds to approx 1M nucleotides (with large variance) depending on location and sex. 2 4 5 1 3 A A 1 /A 1 O A 2 /A 2 A A 1 /A 2 O A 1 /A 2 A A 2 /A 2 O A 1 A 2 A O A 1 A 2 A | O A 2 | A 2 O A 2 Recombinant Phase inferred
36 Maximum Likelihood Principle What is the probability of data for this pedigree, assuming a recessive mutation ? What is the probability of data for this pedigree, assuming a dominant mutation ? Maximum likelihood principle: Choose the model that maximizes the probability of the data.
37 Linkage Equilibrium u Linkage Equilibrium =haplotype frequency is the product of the underlying allele’s frequencies: independence. u Exceptions occur for tightly linked loci.
38 One locus: founder probabilities Founders are individuals whose parents are not in the pedigree. They may of may not be typed (namely, their genotype measured). Either way, we need to assign probabilities to their actual or possible genotypes. This is usually done by assuming Hardy-Weinberg equilibrium (H-W). If the frequency of D is.01, then H-W says: pr(Dd ) = 2x.01x.99 Genotypes of founder couples are (usually) treated as independent. pr(pop Dd, mom dd ) = (2x.01x.99)x(.99) 2 D d dd 1 21
39 One locus: transmission probabilities Children get their genes from their parents’ genes, independently, according to Mendel’s laws; also independently for different children. D d d 3 21 pr(kid 3 dd | pop 1 Dd & mom 2 Dd ) = 1/2 x 1/2
40 One locus: transmission probabilities - II D d pr(3 dd & 4 Dd & 5 DD | 1 Dd & 2 Dd ) = (1/2 x 1/2)x(2 x 1/2 x 1/2) x (1/2 x 1/2). The factor 2 comes from summing over the two mutually exclusive and equiprobable ways 4 can get a D and a d. d D 1 4 5 3 2
41 One locus: penetrance probabilities Pedigree analyses usually suppose that, given the genotype at all loci, and in some cases age and sex, the chance of having a particular phenotype depends only on genotype at one locus, and is independent of all other factors: genotypes at other loci, environment, genotypes and phenotypes of relatives, etc. Complete penetrance: pr(affected | DD ) = 1 Incomplete penetrance) pr(affected | DD ) =.8 DD
42 One locus: penetrance - II Age and sex-dependent penetrance (liability classes) pr( affected | DD, male, 45 y.o. ) =.6 D D (45)
43 דוגמא למוטציה דומיננטית בה הפנוטיפ המוטנטי לא תמיד מתבטא A healthy daughter transmits The mutation to her daughter.. חדירות חלקית:
44 One locus: putting it all together Assume penetrances pr(affected | dd ) =.1, pr(affected | Dd ) =.3 pr(affected | DD ) =.8, and that allele D has frequency.01. The probability of data for this pedigree assuming penetrances of 1 =0.1 and 2 =0.3 is the product: D d d D 1 4 53 2 (2 x.01 x.99 x.7) x (2 x.01 x.99 x.3) x (1/2 x 1/2 x.9) x (2 x 1/2 x 1/2 x.7) x (1/2 x 1/2 x.8) This is a function of the penetrances. By the maximum likelihood principle, the values for 1 and 1 that maximize this probability are the ML estimates.
45 Fully penetrant Recessive Disease Let q be the probability of the disease allele. The probability of data for this pedigree assuming full penetrance is the product: 1 4 53 2 L = (1-q) x q x (1-q) x q (3/4)(3/4)(1/4) Exercise: write the likelihood for a fully penetrant dominant disease.
47 Crossing Over u Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over. u New combinations are obtained, called the crossover products.
48 Recombination During Meiosis Recombinant gametes
49 Linkage u 2 genes on separate chromosomes assort independently at meiosis. u 2 genes far apart on the same chromosome can also assort independently at meiosis. u 2 genes close together on the same chromosome pair do not assort independently at meiosis. u A recombination frequency << 50% between 2 genes shows that they are linked.
50 Two Loci Inheritance Recombinant 2 1 A B a b A a B b 34 a b A a b 56 A a B b
51 Linkage Maps u Let U and V be 2 genes on the same chromosome. u In every meiosis, chromatids cross over at random along the chromosome. u If the chromatids cross over between U & V, then a recombinant is produced. The farther apart U & V are the greater the chance that a crossing over would occur between them the greater the chance of recombination between them.
52 Recombination Fraction The recombination fraction between two loci is the percentage of times a recombination occurs between the two loci. is a monotone, nonlinear function of the physical distance separating between the loci on the chromosome.
53 Centimorgan (cM) u 1 cM (or 1 genetic map unit, m.u.) is the distance between genes for which the recombination frequency is 1%.
54 Interference Crossovers in adjacent chromosome regions are usually not independent. This interaction is called interference. u A crossover in one region usually decreases the probability of a crossover in an adjacent region.
55 Building Genetic Maps u At first: only genes with variant alleles producing detectably different phenotypes were used as markers for mapping. u Problem: the chromosomal intervals between the genes were too large the resolution of the maps wasn’t high enough. u Solution: use of molecular markers (a site of heterozygosity for some type of silent DNA variation not associated with any measurable phenotypic DNA variation).
56 Linkage Mapping by Recombination in Humans. u Problems: l It’s impossible to make controlled crosses in humans. l Human progenies are rather small. l The human genome is immense. The distances between genes are large on average.
57 Lod Score for Linkage Testing by Pedigrees The results of many identical matings are combined to get a more reliable estimate of the recombination fraction. 1. Calculate the probabilities of obtaining a set of results in a family on the basis of (a) independent assortment and (b) a specific degree of linkage. 2. Calculate the Lod score = log(b/a). A Lod score of 3 is considered convincing support for a specific recombination fraction. A Lod score of 3 is considered convincing support for a specific recombination fraction.