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NAME: __________________________________________________________ GRADE: _______________________ / 125 SECTION 1: _________ / 10 SECTION 2: _________ / 11 SECTION 3: _________ / 10 SECTION 4: _________ / 24 SECTION 5: _________ / 21 SECTION 6: _________ / 49 BIOS E-­‐145 (GENETICS OF HUMAN DISEASE) MIDTERM EXAM MARCH 25, 2013 RULES FOR THE MIDTERM EXAM: 1. SIT EVERY OTHER SEAT. 2. WRITE IN PEN ONLY. ANSWERS IN PENCIL WILL NOT BE GRADED. 3. NO TALKING ONCE EXAMS ARE HANDED OUT. 4. PUT AWAY EVERYTHING OTHER THAN YOUR CALCULATOR AND YOUR PENS. 5. YOU MAY NOT LEAVE DURING THE LAST 15 MINUTES OF THE EXAM. PLEASE STAY SEATED AFTER 9:25PM. 6. READ EACH QUESTION CAREFULLY. 7. ANSWER EVERY QUESTION. 8. USE ONLY THE SPACE PROVIDED FOR SHORT ANSWERS. 9. PLEASE WRITE LEGIBLY – IF I CAN’T READ IT, I CAN’T GRADE IT! GOOD LUCK! Page 1 of 17 NAME: __________________________________________________________ PROBABILITY AND STATISTICS FORMULAS Page 2 of 17 NAME: __________________________________________________________ SECTION 1: TRUE OR FALSE (10 POINTS) SECTION 1 GRADE: _______________ / 10 For questions 1-­‐5, answer whether the following statements are true or false. If false, explain why, in no more than ONE SENTENCE. (2 POINTS EACH) 1. If you test 1000 different SNPs and find that one of them is associated with a particular trait with a p-­‐value of 0.043, that means that SNP is definitely associated with the trait. False. If you test 1000 SNPs, you would expect to see p-­‐values that low by chance alone. 2. Mendel’s Principle of Independent Assortment (which holds that the inheritance of one trait does not affect the inheritance of another trait) means that the genes for the two traits are either on different chromosomes or far apart on the same chromosome. True. 3. When looking for modifiers of Mendelian traits, you should always be able to find genetic modifiers – there’s no need to look for evidence that genetic modifiers exist before doing further studies. False. Even if you observe phenotypic variability, you still don’t know whether that variability is due to environmental or genetic factors. 4. Creutzfeldt-­‐Jacob disease (CJD) and familial fatal insomnia (FFI) are both caused by dominant mutations in the same gene. In both cases, there is the same mutation at codon 128, but individuals with CJD have a certain polymorphism at codon 129 of the same allele and individuals with FFI have a different polymorphism at codon 129. This is an example of a trans effect altering disease phenotype. False. This is an example of a cis effect – the polymorphisms are on the same allele. 5. The variable severity of cystic fibrosis due to different mutations in the CFTR gene is an example of allelic heterogeneity. True. Page 3 of 17 NAME: __________________________________________________________ SECTION 2: FILL IN THE TABLE (11 POINTS) SECTION 2 GRADE: _______________ / 11 (0.5pts each) For each disease, fill in the table with the relevant gene and the mode of inheritance. Gene options: beta globin, BRCA1, BRCA2, CFTR, FGFR3, glucocerebrosidase, HFE, huntingtin, multiple genes, PAH, NF1 Mode of inheritance options: autosomal dominant, autosomal recessive, X-­‐linked dominant, multiple contributing genes, multiple modes of inheritance. Note: do not consider inheritance of modifier genes for this table. 6. Table of diseases Disease Relevant gene(s) Mode of inheritance Achondroplasia FGFR3 Autosomal dominant Age-­‐related macular degeneration Multiple genes (complex trait) Multiple contributing genes Cystic fibrosis CFTR Autosomal recessive Gaucher disease Glucocerebrosidase Autosomal recessive Hemochromatosis HFE Autosomal recessive Huntington’s disease Huntingtin Autosomal dominant Hypogonadotropic hypogonadism/Kallmann syndrome Mendelian breast cancer syndromes Various (KAL1, GNRH, GNRHR, FGFR1, CHD7, PROK2, PROKR2, GPR54, KISS1, etc.) BRCA1, BRCA2 Multiple modes (autosomal dominant, autosomal recessive, X-­‐linked dominant) Autosomal dominant Neurofibromatosis type I NF1 Autosomal dominant Phenylketonuria PAH Autosomal recessive Sickle cell disease Beta-­‐globin Autosomal recessive Page 4 of 17 NAME: __________________________________________________________ SECTION 3: PEDIGREE INHERITANCE PATTERNS: DETERMINE THE MODE OF INHERITANCE SECTION 3 GRADE: _______________ / 10 For each of the pedigrees in questions 7 to 11, name the MOST LIKELY mode of inheritance. For all questions, assume the trait is RARE and has COMPLETE PENETRANCE. (2 POINTS EACH) 7. Pedigree 1: ______Autosomal dominant__________________________________________ 8. Pedigree 2: ________X-­‐linked recessive_________________________________________ 9. Pedigree 3: _______Autosomal recessive_________________________________________ Page 5 of 17 NAME: __________________________________________________________ 10. Pedigree 4: _____Autosomal recessive___________________________________________ 11. Pedigree 5: _______Autosomal dominant_________________________________________ Page 6 of 17 NAME: __________________________________________________________ SECTION 4: PROBABILITY, STATISTICS, AND CALCULATIONS SECTION 4 GRADE: _______________ / 24 For questions 12 to 16, answer each question in the space provided. You must show your work for full credit. 12. An individual with genotype Aa BB Cc Dd EE is crossed with an individual with genotype aa bb cc dd ee. Assuming all loci undergo independent assortment, what fraction of the progeny are expected to have genotype Aa Bb cc dd Ee? (2 POINTS) (1/2)(1)(1/2)(1/2)(1) = 1/8 – independent assortment at each locus 13. In Drosophila, the dominant allele Cy results in curly wings (wildtype wings are straight), but homozygous Cy/Cy is lethal (the embryos do not survive). On another unlinked gene, B, a dominant mutation B results in brown eyes rather than the wildtype red eyes. In a cross of Cy/+ B/+ × Cy/+ B/+, where the + sign denotes the wildtype alleles of both genes, what is the expected ratio of curly-­‐wing, red-­‐eyed: straight-­‐wing, brown-­‐eyed flies? (4 POINTS) Wing shape: 2/3 curly-­‐wing (Cy/+), 1/3 straight-­‐wing (+/+) – Cy/Cy die Eye color: ¾ brown-­‐eyed (B/B and B/+), ¼ red-­‐eyed (+/+) Curly-­‐wing and red-­‐eyed: 2/3 x ¼ = 2/12 = 1/6 Straight-­‐wing and brown-­‐eyed: 1/3 x ¾ = 3/12 = ¼ Remember that the two genes show independent assortment. (Also acceptable to answer ratio of 2 curly/red to 3 straight/brown.) 14. This pedigree is for an autosomal recessive trait with a penetrance of 75%. The bands in the gel are restriction fragments used to identify the wild-­‐
type allele A (longer band) and the mutant allele a (shorter band). Using all of the information in the pedigree, what is the probability that the fetus III-­‐1 will be unaffected? Show all of your work. (5 POINTS) Probability that III-­‐1 is Aa and unaffected: (1/2)(1) = ½ Probability that III-­‐1 is aa and unaffected: (1/2)(1-­‐
3/4) = 1/8 Total probability = ½ + 1/8 = 5/8 Page 7 of 17 NAME: __________________________________________________________ 15. This pedigree is for a rare autosomal recessive trait with complete penetrance. What is the probability that at least one of IV-­‐1, IV-­‐2, and IV-­‐4 is a carrier? (5 POINTS) Because IV-­‐3 is affected, we know that III-­‐2 and III-­‐3 must both be carriers (Aa). Because IV-­‐1, -­‐2, and -­‐4 are all unaffected, we know they must be either Aa or AA – they’ll be 2/3 Aa and 1/3 AA. The probability that at least one is a carrier is 1 minus the probability that they’re all non-­‐carriers – it’s 1 – (1/3)3. 16. This pedigree is for an extremely rare autosomal recessive trait with complete penetrance. (8 POINTS TOTAL) a. Using only the information in generations I and II, what is the probability that individual II-­‐2 is a heterozygous carrier? (2 POINTS) 2/3 (we know that II-­‐2 is unaffected, so can’t be aa, and we know that the gen I parents are both heterozygous carriers because II-­‐5 is affected) b. Ignoring the sibship in generation IV, what is the probability that individual III-­‐1 is a heterozygous carrier? (2 POINTS) P(III-­‐1 carrier) = P(II-­‐2 is carrier)*P(II-­‐2 transmits disease allele given carrier) = 2/3 x ½ = 1/3 c. Ignoring the sibship in generation IV, what is the probability that individuals III-­‐1 and III-­‐2 are both heterozygous carriers? (2 POINTS) P(III-­‐1 and III-­‐2 both carriers) = P(III-­‐1 is carrier)*P*(III-­‐2 is carrier) = 1/3 x 1/3 = 1/9 Carrier probabilities are the same for both III-­‐1 and III-­‐2 – it’s the same inheritance pattern d. Does the sipship in generation IV increase, decrease, or not affect the probability that individuals III-­‐1 and III-­‐2 are both heterozygous carriers? (2 POINTS) Decrease – having 3 unaffected children makes it less statistically likely that the parents in gen III are both carriers Page 8 of 17 NAME: __________________________________________________________ SECTION 5: SHORT ANSWER SECTION 5 GRADE: ________________ / 21 For questions 17 to 21, briefly answer each question in the space provided. 17. Mendel’s Principle of Segregation requires a number of assumptions. Name three of them. (3 POINTS) Single gene 2 alleles Homozygous parents Sex of parent doesn’t matter Complete dominance Mendelian segregation Equal survival of gametes Complete penetrance Correct classification Sufficiently large numbers 18. Name and briefly describe two approaches (one sentence or less per approach) that can be used to identify genes that cause Mendelian diseases. (3 POINTS) Linkage approaches: scan markers across the genome in family-­‐based studies – look for areas that are shared between affected individuals Fluorescence in-­‐situ hybridization (FISH) for translocations: this method allows visualization of the specific location of translocation breakpoints Mouse models: look for clues from mice with similar phenotypes Exome sequencing: sequencing the coding regions of the genes to look for mutations (0.5 point for the name of the method, 1 point per description) 19. Explain in three sentences or less (bullet points are fine!) how you identify linkage between two genes or between a gene and a molecular marker. (4 POINTS) Observe that parental types are more common than non-­‐parental types -­‐ with independent assortment, expect 1:1 ratio Do a chi-­‐sq test for linkage—is the ratio actually 1:1? If chi-­‐sq test significant, suggests linkage Page 9 of 17 NAME: __________________________________________________________ 20. Name and briefly define/explain three different sources of phenotypic heterogeneity in the expression of a disease. NO MORE THAN ONE SENTENCE PER SOURCE, BULLET POINTS OK. (6 POINTS) Genetic heterogeneity of the primary factor (either locus or allelic) Locus heterogeneity: different genes involved in different subdiseases Example: hypertrophic cardiomyopathy – Myh7 mutations are more severe than MyBPC3 mutations Allelic heterogeneity: different mutations/alleles have different phenotype CF – variable severity of mutations in CFTR Another variant in the gene, in either cis or trans Trans effect: expression depends on normal allele This is what they were looking for in that NF1 study – variation in the normal allele that affected disease phenotype Cis effect: clinical phenotype varies depending on the exact haplotype that carries the disease allele Creutzfeldt-­‐Jakob and famililal fata insomnia: get one disease or the other depending on which polymorphism they have at codon 129 (have same mutation at codon 128) Modifier genes of disease expression, either multifactorial or “monogenic-­‐like” Monogenic-­‐like: single modifier gene shows almost fully penetrant mutations that explain much of the variability in disease expression (digenic/oligogenic inheritance) Multifactorial: multiple modifier genes each with small effect (what we’ve seen so far for most of these diseases) Environmental factors Other plausible explanations were also accepted for full credit. 2 points per explanation. 21. Discuss in no more than three sentences some of the considerations necessary when designing a study to search for modifiers of Mendelian diseases. Bullet points ok. (5 POINTS) Pick your study population Look at extreme phenotypes? Limit to a single mutant allele? Define your phenotype Which disease feature are you going to study? What’s the variability like like in general? What covariates should be used to adjust? (ie, age, sex, etc.) Find some evidence for genetic modifiers! Page 10 of 17 NAME: __________________________________________________________ SECTION 6: DATA ANALYSIS SECTION 6 GRADE: ________________ / 49 For questions 22 to 28, briefly answer each question in the space provided. 22. This figure is from Linus Pauling’s 1949 paper identifying the molecular basis of sickle cell disease. The experiment performed to generate the data here looked at the mobility of different types of hemoglobin molecules from normal individuals, individuals with sickle cell anemia, and individuals with sickle cell trait (heterozygotes). The arrow in each part of the figure gives a basis for comparison (you can think of it like an axis on a graph). PLEASE ANSWER EACH PART OF THE QUESTION IN ONE SENTENCE. (6 POINTS TOTAL) a. From this figure, what do you conclude about the mobility of normal vs. sickle cell hemoglobin? (2 POINTS) Normal hemoglobin moves more quickly than sickle cell hemoglobin. b. From this figure, what do you observe regarding the hemoglobin from individuals with sickle cell trait? (2 POINTS) Individuals with sickle cell trait appear to have a mixture of normal and sickle cell hemoglobin. c. Compare the mobility profile of hemoglobin from individuals with sickle cell trait and the 50-­‐50 mixture of normal and sickle cell hemoglobin. (2 POINTS) The distribution is different – individuals with sickle cell trait have more normal hemoglobin than sickle cell hemoglobin. Page 11 of 17 NAME: __________________________________________________________ 23. This figure is from the 1996 paper by Feder et al mapping the HFE gene to hereditary hemochromatosis. The authors genotyped 45 markers in 101 patients with hereditary hemochromatosis (HH) and 64 controls. They calculated Pexcess, which is a measure of linkage disequilibrium, at each marker. This figure (part a) shows the map of Pexcess against the physical location for each marker. In part b of this figure, the deviation from Hardy-­‐Weinberg equilibrium is mapped. (9 POINTS TOTAL) a. In one sentence, what was the goal of this figure? (2 POINTS) The goal was to refine the location of the hereditary hemochromatosis gene. b. In part a, the authors calculated Pexcess, a measure of linkage disequilibrium. In part b, the authors calculated F, the proportionate excess of homozygotes, which is greatest at the disease locus. In one sentence, how do the regions identified by the two approaches compare? (2 POINTS) The two regions appear to be similar. c. What do these two approaches tell us about the location of the HH gene? Do they define precise boundaries? How might you narrow down the region further? NO MORE THAN 4 SENTENCES – BULLET POINTS OK. (5 POINTS) 1 point – they localize the gene to the same ~600kb region 1 point – no, they don’t define precise boundaries 3 points – discussion of an approach for narrowing down the region One possible approach would be to look at haplotypes in individuals with and without the disease – if there is a shared haplotype among the disease chromosomes, that suggests the location of the gene. Page 12 of 17 NAME: __________________________________________________________ 24. This table is from the 1989 paper by Kerem et al identifying the gene for cystic fibrosis. It looks at the distribution of CF and non-­‐
CF chromosomes with and without the 3-­‐bp ΔF508 deletion. They also compared CF with pancreatic insufficiency (CF-­‐PI) and CF with pancreatic sufficiency (CF-­‐PS). (8 POINTS TOTAL) a. In one sentence, what was the goal of this figure? (2 POINTS) To look at the frequency of the ΔF508 mutation in CF and non-­‐CF chromosomes. (They were also interested in looking at individuals with and without good pancreatic function as well.) b. In no more than two sentences, describe the main findings of this figure. How often is the ΔF508 mutation seen in CF chromosomes vs. non-­‐CF chromosomes? (2 POINTS) The mutation is not seen in non-­‐CF chromosomes. It is seen in 68% of CF chromosomes. c. In no more than two sentences, what does this figure suggest about the relationship between the ΔF508 mutation and cystic fibrosis? What conclusion can you draw and why? (4 POINTS) This figure suggests that the mutation is specific to CF (because it isn’t seen in non-­‐CF chromosomes) and that this mutation is the major mutation that causes CF (because it is seen in 68% of CF mutations). Page 13 of 17 NAME: __________________________________________________________ 25. This table is from a 1995 paper by Bellus et al looking at mutations in a gene implicated in achondroplasia. They presented both new mutation analysis and data from a previous study looking at specific mutations in the transmembrane domain of the gene. (5 POINTS TOTAL) a. In one sentence, what can you conclude from the data presented in this figure? (2 POINTS) Specific mutations in FGFR3 are seen in patients with achondroplasia – mutations in the TM domain of FGFR3 lead to achondroplasia. b. What data are not presented in this figure that you might otherwise expect to see? Why are such data a critical part of an experiment? PLEASE ANSWER IN NO MORE THAN 2 SENTENCES. (Note: these data are presented elsewhere in the paper.) (3 POINTS) No control data are presented. You need to compare mutation frequency in both cases and controls in order to see if there’s a difference and draw conclusions about causality. Page 14 of 17 NAME: __________________________________________________________ 26. This table is from the 1993 identification of the gene for Huntington’s disease (HD) by the Huntington’s Disease Collaborative Research Group. The gene implicated in HD has a polymorphic trinucleotide repeat. The authors compared the number of trinucleotide repeats seen in normal chromosomes to the number of trinucleotide repeats seen in HD chromosomes. (6 POINTS TOTAL) a. In one sentence, what was the goal of this figure? (2 POINTS) To see if the lengths of the trinucleotide repeats differed between normal and HD chromosomes. b. In one sentence, what are the results of this analysis? Please be specific (give numbers). (2 POINTS) Normal chromosomes have fewer trinucleotide repeats than HD chromosomes – normal chromosomes mostly have fewer than 24 repeats while HD chromosomes have at least 42. c. In one sentence, what do you conclude about trinucleotide repeat length and Huntington’s disease from this figure and why do you make this conclusion? (2 POINTS) An increased number of trinucleotide repeats in this gene is correlated with HD – the increased repeat number is seen only in HD chromosomes and never in normal chromosomes. Page 15 of 17 NAME: __________________________________________________________ 27. An association between a variant in the gene LIPC and age-­‐related macular degeneration (AMD) was reported by Neale et al in 2010. The LIPC variant associated with AMD is a functional variant in the promoter that influences LIPC expression and serum HDL levels. The minor (less common) allele of the LIPC variant increases HDL (“good cholesterol”) levels and has a protective effect on the development of both wet and dry AMD. In their study, Neale et al also for associations between AMD and other previously reported HDL genetic loci. (7 POINTS TOTAL) Chr, chromosome; position, base-­‐pair position in NCBI 36.1; HDL-­‐raising allele, allele reported in Kathiresan et al. (25) as the allele responsible for raising HDL; MAF, minor-­‐allele frequency; Z score, weighted average and direction of minor allele signal; meta P, Pvalue for the association between the minor allele and AMD; OR, odds ratio for the minor allele; 95% CI, confidence interval for the OR. a. In one sentence, explain why it’s a reasonable hypothesis that previously reported HDL genetic loci might be associated with AMD given the new association between LIPC and AMD. (2 POINTS) Because LIPC is involved in HDL cholesterol function, it is an indicator of a possible new pathway that may be involved in AMD pathogenesis. b. Given the data in this figure, what do you conclude about the association between AMD and previously reported HDL genetic loci – do you see a consistent trend for association between these loci and AMD? Is there a difference in between HDL increasing or decreasing loci? What does this suggest about the mechanism of action of the LIPC variant identified by Neale et al? Explain your answer in no more than 3 sentences. (5 POINTS) There is no consistent trend for association between the HDL genetic loci and AMD (2 points). Neither the increasing nor decreasing loci show an affect on AMD (2 points). This suggests that the LIPC variant may act through some mechanism other than its effect on HDL levels (1 point). Page 16 of 17 NAME: __________________________________________________________ 28. Neurofibromatosis type 1 (NF1) shows a great deal of clinical variability. Sabbagh et al (2009) studied the genetic component of the variable expressivity of the disease. As part of their study, they looked at five quantitative traits that are clinical features of NF1 and estimated the heritability of the traits. The table below is from their paper. (8 POINTS TOTAL) a. What is heritability? What does it tell us? Please answer in no more than 2 sentences. (2 POINTS) Heritability is the proportion of variability in a population attributable to genetic factors (1 point for definition). It gives an indication of how strong the genetic influence on a particular trait is (1 point for what it tells). b. Looking at the data in this table, is there any evidence for a genetic contribution of the mutation in the NF1 gene itself? For which trait(s)? What indicates that? How strong is the evidence? Please answer in no more than 2 sentences. (2 POINTS) There is weak evidence for variation due to NF1 mutations for CAL spots (of both large and small size) – the sigma-­‐squared-­‐G values indicate that, but the error is large (and crosses 0). (Data are not particularly strong, so also acceptable to say no.) c. Which trait(s) have the strongest evidence for genetic effects? What leads you to that conclusion? Please answer in no more than 2 sentences. (2 POINTS) CAL spots of small size have the highest heritability (0.62), although the additive genetic component of variance is higher for subcutaneous and plexiform neurofibromas. (1 point for trait(s), 1 point for why) d. Based on the data from this table, do you think there are likely to be genetic modifiers of NF1? Which trait would you study first and why? Please answer in no more than 2 sentences. (2 POINTS) Yes – there’s strong heritability for these traits (1 point). I would probably start by looking for modifiers of neurofibromas (0.5 points) because they are more likely to be clinically relevant (0.5 points). (Ok to pick other traits to study first as long as an explanation is given.) Page 17 of 17