Problem Set 3 KEY

Problem Set 3 KEY - Biology 202 Problem Set 3 2/9/07 This...

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Biology 202 Problem Set 3 2/9/07 This problem set covers material from January 29 th through February 9 th From Griffiths et al.: Chapter 2, problems: 34, 35, 36 (and the previously assigned complementation problems) Chapter 5, problems: 1, 2, 4, 6 Chapter 6, problems 6, 18, 26, 30, 35, 39 Hand in answers to the following questions, due on February 18 th 1. You are working on a small lobster-like creature found off the coast of North Carolina which is a striking light blue color. Typically the creatures have two enlarged limbs with prominent pincers. While collecting individuals in the wild you notice a male lacking the characteristic large arms and pincers. You collect him, take him home to the lab and cross him to a female with large arms and pincers in an attempt to understand the genetic basis of this trait. All of the F1 progeny lacked the large arms and pincers. You mate two F1 progeny and see 766 that lack the large arms and pincers, 196 that have large arms and pincers and 62 that have large arms but lack pincers. a.) assign genotypes to the P, F1 and F2 generations parent with arms: a/a;B/B parent without arms: A/A;b/b F1: A/a;B/b F2: 9 A_;B_ no arms or pincers 3 A_;bb no arms or pincers 3 a/a;B_ arms and pincers 1 a/a;b/b only arms b.) propose a model to explain these data The F2 data indicate that the trait is segregating like a modified dihybrid. The F1 data indicate that lack of arms and pincers is dominant to having arms and pincers. The F2 phenotypic classes are very close to a 12:3:1 ratio which suggests a case of dominant epistasis in which individuals that are homozygous recessive for the first gene (aa) can develop arms. Gene B enables the arms to develop pincers (B_ ). The dominant A allele prevents arm development. Without arms no pincers can develop so A is epistatic to B c.) use a statistical test to confirm your hypothesis Observed Expected X2 12 766 768 0.005208 3 196 192 0.083333 1 62 64 0.0625 SUM 1024 1024 0.151042
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There are three phenotypic classed therefore there are 2 degrees of freedom (df). This Chi Square value with 2 df corresponds to a P value greater than 0.05 therefore we can conclude that the slight difference between the observed data and the expected phenotypic ratio typical of dominant epistasis can be explained by chance. Therefore we conclude that our data are reflective of a pair of genes one of which is dominantly epistatic to the other. 2. You are working in a lab that studies ladybugs. You have isolated 10 different ladybug mutants with big antenna (antenna size is important for ladybug self-esteem). It turns out that a buddy of yours working at Duke has also isolated 10 ladybug mutants with big antenna and you decide to pool your resources and try to publish a paper together. You cross each of your mutants to each of your buddies and include wild-type controls to generate the following data table (+ indicates you observe the wt phenotype, - indicates you observe the mutant phenotype) your ladybug mutants
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This note was uploaded on 04/02/2008 for the course BIOL 202 taught by Professor Kieber-hogan during the Spring '08 term at UNC.

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Problem Set 3 KEY - Biology 202 Problem Set 3 2/9/07 This...

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