BIOL202 week6 conference 1. The figures below are plots of the parental height vs offspring height of Goldenseal (Hydrastis canadensi) grown in two different soil types (clay and loam). All individuals were pure breeding lines that were selfed, and the same lines were used in both soil types. Best fit linear regression lines have also been provided. Clay Loam 10 16 9 14 7 12 6 10 5 4 f(x) = 0.97x + 0.33 R² = 0.92 3 2 1 Offspring Height Offspring Height 8 8 6 f(x) = -0.04x + 11.71 R² = 0 4 2 0 0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 Parental Height Parental Height 1) What is the narrow heritability (h2) value of height when grown in clay? Loam? 2) How would you describe the norm of reaction for this trait? 3) If height is a highly heritable trait when grown in clay, why might parental height not predict offspring height in loam? 2. You have spent time gathering data on style length in a population of Geum triflorum and the trait appears to vary continuously. From the original population you’ve created a pure-breeding long-styled line (avg. length 5 cm) and a pure breeding short-styled line (1 cm avg. length). You conduct a cross between individuals from each line and get an F1 with intermediate style lengths (3 cm avg.). You have style length frequency distributions for each population. From those distributions you calculate variances and compile them into the following table. Generation P(short) P(long) F1 Original Variance 4 cm2 4 cm2 7 cm2 56 cm2 From these values calculate the broad heritability, H2, of style length in G. triflorum. What exactly does this value suggest about variation in style length in the original population? A colleague of yours across the country repeats the above experiment using siblings of the short and long-styled parents and gets a much lower H2, how would you interpret this result? 10 BIOL202 week6 conference 3. You are researching a remote small village within a region of Papua New Guinea because of extremely rare undocumented cases of patients showing a disease with several dysmorphic features. On top of the dysmorphic features, generalized (not one but several specific forms of cancer) seems to present itself in relatively high frequencies at a very early age of onset leading to death by the age of 20. You also find that cancer runs especially high in many of the “normal” siblings although the age of onset is generally later. You document one of the families and draw up the following pedigree. a) Would this most likely be caused by a tumour suppressor gene or an oncogene? b) What is the term used to describe a single gene that can have multiple phenotypic effects? You head back to the lab in Montreal with DNA samples from everyone and do a linkage analysis on all of the families you collected from and identify a stop codon in an as yet uncharacterized gene. However, the sequence seems to align with a known gene whose function is to identify DNA mismatches to arrest the cell cycle. c) Does this support your original hypothesis? d) Why might the siblings be more likely to develop cancer? 4. Zack, a graduate student is studying a type colon cancer only seen in zebra’s. You are an undergraduate student helping with Zack’s research. Part of your job is to determine what type of protein (oncogene or tumor-suppressor) is mutated in the colon cancer. What would you conclude given the following observations? - Loss in regulation of cell growth, specifically cells from the colon have been removed and plated on nutrient rich media, after 24 hours there is an irregular exponential increase in the number of cells. - Affected zebras give rise to affected offspring. 5. You suspect the discrete trait you are studying might be a threshold trait “T” (ex: heart attack), so you decide to measure the suspected liability trait “L” (ex: blood pressure). What would be the best mapping strategy? A) Linkage analysis of T B) QTL analysis of T, if H2(T)>H2(L) C) QTL analysis of L D) All of the above would provide useful information E) Not enough information to choose a mapping strategy BIOL202 week6 conference 16 14 12 10 Height Answers: 1. 1) H2 = 0.97 (clay); H2 = 0 (loam) 2) When grown in clay, height is nearly fully heritable, whereas, in loam all individuals are tall, regardless of genotype. Ie. The genotypes have different slopes -> *note this figure is given as an example only and could not be directly derived from the data given since the genotypes were not labeled. 3) This is because of the norm of reaction described in 2. It is also important to note that heritability is not the opposite of phenotypic plasticity, and a character may have perfect heritability and still be subject to large changes due to environmental variation. 8 6 4 2 0 Clay Loam Soil type 2. The P(short), P(long) and F1 variances must all be due to environmental effects since the plants within those populations are genetically identical. Averaging their variances gives a value of 5 cm2 which we can call se2. There are multiple genotypes within the original population and therefore the variance is due to both genotypic and environmental effects so we can call this variance sp2. Because sp2 = sg2 + se2, the genotypic variance must then be 51 cm2. Given that H2 = sg2/ sp2 , the broad heritability is thus 0.91. Following the definition of broad heritability this value suggests that 91% of the variation in style length in the original population is attributable to genetic variation. Your estimate of H2 is specific to your population in the environments you provided. Your colleague may have subjected their plants (which should be genetically identical to yours) to different environmental conditions thereby affecting that component of heritability. 3. The pedigree shows a recessive disorder, it is most likely a tumour suppressor a) Pleiotropy b) Yes, a gene that would arrest the cell cycle would be considered a tumour suppressor gene c) It is either haplo-insufficiency (one copy of the gene can’t quite meet the demand of binding all mismatches, or the result of two hits…a new mutation occurring within a somatic cell that already carries one inherited mutation. 4. The protein is a mutated oncogene, since it is a dominant mutation (affected zebras give rise to affected individuals). 5. Answer: “C” Not A: The multiple genetic variants of a threshold trait would be found in the “unaffected” population, greatly reducing the statistical power of the analysis. Not B: QTL analysis can only be performed on quantitative traits, which T is not. Not D: Because “B” is incorrect. Not E: Nothing is specifically preventing you from choosing “C”.