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Announcements 1. Answers to Ch. 3 problems 6, 7, 8, 12, 17, 22, 32, 35 posted - 230A. 2. Problem set 1 answers due in lab this week at the beginning of lab. Bring calculators to lab this week. 3. Getting to know Flylab and testcross (lab 2) - printout of assignments from “notebook”, due this week at the beginning of lab. 4. Confusion with X-linked crosses: 1 cross or 2? 5. Seminar this Thursday - faculty research interests Review of last lecture 1. Genetic ratios are expressed as probabilities. Thus, deriving outcomes of genetic crosses relies on an understanding of laws of probability, in particular: the sum law, product law, conditional probability (likelihood that one outcome will occur, given a particular condition), and the binomial theorum (used to determine particular combinations). Expand the binomial OR use factorial general formula to solve. 2. Statistical analyses (Chi square) - used to test the validity of experimental outcomes. In genetics, some variation is expected, due to chance deviation. Outline of Lecture 6 I. Chi-square revisited II. Pedigree analysis- recessive vs. dominant traits - solving pedigree problems Chi-square formula 2 (o e) 2 X e where o = observed value for a given category, e = expected value for a given category, and sigma is the sum of the calculated values for each category of the ratio • Once X2 is determined, it is converted to a probability value (p) using the degrees of freedom (df) = n- 1 where n = the number of different categories for the outcome. Chi-square - Example 1 Phenotype Expected Observed A 750 760 a 250 240 1000 1000 Null Hypothesis: Data fit a 3:1 ratio. 2 2 2 o e 760 750 240 250 2 750 250 e 2 0.53 degrees of freedom = (number of categories - 1) = 2 - 1 = 1 Use Fig. 3.12 to determine p - on next slide X2 Table and Graph Unlikely: Reject hypothesis likely unlikely Likely: Do not reject Hypothesis 0.50 > p > 0.20 Figure 3.12 Interpretation of p • 0.05 is a commonly-accepted cut-off point. • p > 0.05 means that the probability is greater than 5% that the observed deviation is due to chance alone; therefore the null hypothesis is not rejected. • p < 0.05 means that the probability is less than 5% that observed deviation is due to chance alone; therefore null hypothesis is rejected. Reassess assumptions, propose a new hypothesis. Conclusions: • X2 less than 3.84 means that we accept the Null Hypothesis (3:1 ratio). • In our example, p = 0.48 (p > 0.05) means that we accept the Null Hypothesis (3:1 ratio). • This means we expect the data to vary from expectations this much or more 48% of the time. Conversely, 52% of the repeats would show less deviation as a result of chance than initially observed. X2 Example 2: Coin Toss I say that I have a non-trick coin (with both heads and tails). Do you believe me? 1 tail out of 1 toss 10 tails out of 10 tosses 100 tails out of 100 tosses Tossing Coin - Which of these outcomes seem likely to you? Compare Chi-square with 3.84 (since there is 1 degree of freedom). a) Tails 1 of 1 b) Tails 10 of 10 c) Tails 100 of 100 1 1 1 1 1 1 0 2 2 2 2 2 2 1 1 1 a) 2 2 Chi-square b) 2 2 c) 2 2 10 52 0 52 5 50 Reject 10 100 502 0 502 Don’t reject 100 Reject X2 - Example 3 F2 data: 792 long-winged (wildtype) flies, 208 dumpywinged flies. Hypothesis: dumpy wing is inherited as a Mendelian recessive trait. Expected Ratio? X2 analysis? What do the data suggest about the dumpy mutation? II. Pedigree analysis The complex study of human genetics - we don’t control human matings! Instead, we study family trees (pedigrees) to identify how traits are inherited. Importance of Pedigrees • Genetic counselors use them to identify risk of inherited illness. • Genetic researchers use them to identify genes responsible for genetic disorders. • If you or a relative keep good family records, they may some day be useful in tracking down a genetic illness in your family. Recessive vs. Dominant Traits Autosomal Recessive Traits • Example: The albino (aa) mutation inactivates the gene for tyrosinase enzyme, which normally converts tyrosine to melanin in the skin, hair and eyes. • Non-albino is AA or Aa • Autosomal recessive traits can skip generations (appear in progeny of unaffected persons) and affect both males and females equally. Albinism Trait Autosomal Dominant Traits • Example: Hypotrichosis, hair loss begins in childhood for both males and females. • Autosomal dominant traits do not skip generations and affect both males and females. – Some but not all children will be affected in every generation. • Affected individuals are usually heterozygous since mutant allele is rare. Hypotrichosis Trait Achondroplasia Trait (DD or Dd) D is a dominant allele interfering with bone growth. Most people are dd. Human Recessive and Dominant Traits OMIM: Online Mendelian Inheritance in Man (Humans) • Compiled by team at Johns Hopkins University. • Available at: www.ncbi.nlm.nih.gov/omim/ • Includes description of trait, mode of inheritance, molecular information. • GREAT resource to find good reference for your presentation topic; database is easy to search and full of interesting information • Another resource to find a topic from NPR programming: Symbols used in human pedigree analysis Male Female Mating Consanguineous marriage Parents and 1 boy, 1 girl (in order of birth) Sex unspecified Affected individuals Heterozygotes for autosomal recessive Death 3 2 Number of children of sex indicated Propositus Pedigree Problem 1 - albinism How is this trait inherited? dominant or recessive? What are most probable genotypes of each individual? Solving Pedigree Problems • Inspect the pedigree: – If trait is dominant, it will not skip generations nor be passed on to offspring unless parents have it. – If trait is recessive, it will skip generations and will exist in carriers. • Form a hypothesis, e.g. autosomal recessive. • Deduce the genotypes. • Check that genotypes are consistent with phenotypes. • Revise hypothesis if necessary, e.g. autosomal dominant. What is the Mode of Inheritance? Unaffected 2nd generation = rare, recessive trait Deduce genotypes Pedigree Example 2: p. 71, #26 Pedigree Example 3: Huntington’s Disease unaffected affected Folksinger Woody Guthrie died of Huntington’s Disease. Problem 3 Solution HD = Huntington’s allele, hd = wildtype allele; data suggests Huntington’s Disease is autosomal dominant. Inconclusive Pedigree • Try deducing the genotypes using either autosomal recessive or autosomal dominant hypothesis; either one should work.