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3. 3 A gene is composed of strings of bases (A,G, C, T) held together by a sugar phosphate backbone. Reminder - nucleotides are the building blocks of genes

4. 4 From www.kadets. D20.co.edu/~lundberg/dna.html Mutations can occur along the sequence so that two individuals Have different bases at sequence positions. Usually only one strand of dna is read (codes for a protein)

5. 5 Human Genetic Studies: 1. Our Objective: To map the position of disease susceptibility or trait genes relative to genes of known location (markers). 2. A single gene can vary in its composition (due to mutations, insertions or deletions), so that two randomly selected people may have biochemically distinct forms of the gene. These distinct forms are called alleles. We are also interested in finding the alleles which lead to increase risk of disease. 3. Example of traits affected by genes with multiple alleles. (a) Some alleles in genes may increase the risk of disease (qualitative trait) like diabetes (b) Allelic differences may result in different enzyme levels or activities (quantitative traits) like insulin levels.

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7. 7 Single Base Mutations in the genome 1.4 million locations single base DNA differences = single nucleotide polymorphisms = SNPs Example: AAGGCTAA to ATGGCTAA once every 100 to 300 bases predispose to disease influence drug response Wellcome News, Issue 20 Q3 1999

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9. 9 How to Gene Map? Use pedigrees. Pedigrees (families) provide a powerful source of information in genetics 6 1 2 3 4 7 3-5 4-4 4-5 4-7 5 Phenotype = an individuals observable (measurable) values. Examples: blue/brown eyes, diabetes (yes, no), cholesterol concentration Genotype = the often unobservable state of an individual’s genes at a particular locus that relates to the phenotype of interest. If alleles are the same then the genotype is homozygous, otherwise heterozygous. By Mendel’s law 1 (independent assortment of alleles at a locus). Person 1 is equally likely to transmit a 3 or a 5 to his child at the marker locus. Person 2 must transmit a 4. Suppose the marker has alleles, 2, 3, …, 7. What are the possible marker genotypes for person 5? How about person 6?

10. 10 standard12 Increasing Repeat number 6 5 4 3 7 2 Marker genotype: 3 4 7 4-5 4-7 3-54-4

11. 11 The pedigree can be expressed in a file, if the appropriate fields are recorded 6 1 2 3 4 7 3-5 4-4 4-5 4-7 5 Fam. Subj mom dad sex Status trait geno- type 110 0 M alive brown 3-5 12 0 0 F alive blue 4-4 132 1 F alive blue 4-5 142 1 F alive brown 4-5 1 52 1 F dead blue 0-0 160 0 M dead unknown 0-0 173 6 F alive blue 4-7

12. 12 Phenotype, Genotype and Penetrance: Example ABO blood group ABO blood group: A and B are antigens (proteins that cause an immune response) expressed on the surface of blood cells. Phenotypes detected by antibody reaction. O is the absence of an antigen. A phenotype reacts with the A antibody but not B antibody B phenotype reacts with the B antibody but not A antibody AB phenotype reacts with both A and B antibodies O phenotype does not react with either antibody Phenotype Genotype AA/A, A/O, or O/A ABA/B or B/A BB/B, B/O or O/B OO/O

13. 13 A masks the presence of O so we don’t know if person 1 is A-A or A-O. A is dominant to O O is recessive to A B is dominant to O A and B are both detectable - codominant Penetrance = probability of observing a phenotype given a genotype. If given the genotype we know the phenotype with 100% certainty then the trait is fully penetrant. Example ABO: P(A|A-O)=1 P(B|A-O)=P(O|A-O)=P(AB|A-O)=0. Most trait or disease genes are not so penetrant. Having the apoE4-apoE4 genotype elevates lifetime risk of Alzheimer’s disease but P(lifetime risk of Alzheimer’s disease | apoE4/apoE4) < 0.50 likewise BRCA1 mutated allele and breast cancer P(lifetime risk of Breast Cancer | BRCA1/brca1) =.30-.80.

14. 14 Genotypes: Example ABO Since there are three alleles, there are 3 2 = 9 ordered genotypes. By ordered genotype we mean that we know the parental origins of the alleles (the phase). AO 1 2 5 4 3 A AB B We know the phase for person’s 3 and 5 genotypes. Person 4 is heterozygous but phase is unknown. Also note that we can infer person 2’s genotype with certainty but not person 1’s.

15. 15 B-b B-B B-b b-b

16. 16 B-bb-b B-b b-b

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19. 19 Keep this formula in mind when we look the equations for segregation analysis and genetic model based linkage.

20. 20 Final Probability Rule to remember: Independence: A and B are independent events if If Pr(B) >0, then the independence of A and B is equivalent to Pr(A|B)=Pr(A). That is, knowing that B occurred does not help us predict whether A has occurred.

21. 21 Application: Using probability rules and Hardy Weinberg equilibrium to calculate genotype frequencies from allele frequencies ABO blood group: Suppose we know that the allele frequencies are p A = 0.21, p B = 0.05, and p O =.74 What is the probability that a randomly selected individual has the unordered genotype AO? Hardy Weinberg Equilibrium (HWE): Suppose mating is random with respect to the gene in question, there is no selection, migration or mutation, then the allele frequencies will remain constant from generation to generation and we can calculate the genotype frequencies from the allele frequencies (and vice versa). P(AO unordered genotype) = P(A/O or O/A ordered genotype) = P(A/O)+P(O/A) = 2P(A/O). P(A/O) = P( A allele and O allele) =p A p O (HWE is a form of independence) P(AO unordered genotype) =2p A p O = 0.3108 Important: HWE explains why multiple alleles persist in the population

22. 22 35 Bb bb 44 34 bb 54 Bb 34 Bb 54 bb P = 1/4 Mendel’s law #2 - Independent assortment of loci

23. 23 What if the two loci are on the same chromosome? One might expect that the two loci are always transmitted together. Bb 35 b b 44 B b 43 b b 45 B b 43 b b 45 P = 1/2 The set of alleles arranged by parental origin are called haplotypes. Here we observe three haplotypes B3 and b5 and b4.

24. 24 One often observes more than the parental haplotypes in the children: eg seeB5, B3, b5, and b3 Bb 35 b b 4 4 B b 43 b 4 b 4 b b 45 b 3 B 5 P = (1-  )/2 P =  /2 1. The two paternal copies of the chromosome can crossover and recombine during the formation of the gametes (sperm). Recombination can also occur maternal chromosomes but it is not observable in this case.  is the recombination fraction and equals the probability of recombination between the two loci.  is a function of the distance between the two loci.  =1/2 when loci are far apart or on different chromosomes.

25. 25 The distortion in the expected segregation pattern of 1/4:1/4:1/4:1/4 depends on the frequency of crossovers between the two loci. The frequency is dependent on the distance between the two loci. A2 B1 A2 B2 A1 B1 B2 45% 5% Recombination Fraction  = 10% in gametes The two members of the same autosomal pair, duplicate and pair up B1 A1 B1 A2 B2 Let the crossover frequency be 10% A1 B1 A2 B2

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28. 28 But it’s not as easy as it seems.. The degree that genes play a role in trait susceptibility varies from trait to trait –Some traits are primarily determined by genes –Some traits are primarily determined by the environment –Some traits are a combination of genes and environment Single Gene Genes and EnvironmentEnvironment Cystic Fibrosis diabetes cancer infections accidents Huntington’s heart disease We need accurate information about family history, life styles, environment, genetic data, to determine first whether it is worth attempting gene mapping and then to map them.

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