Exam 2006

Exam 2006 - LS4 Final Exam-Fall 2006 Name Section...

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LS4 Final Exam—Fall 2006 Name: Section: Instructions: There are 10 problems on this exam. Each is worth 15 points. 1. Be clear and concise. Circle your answers. 2. Use the back of each sheet to work out the problems. If you need more blank paper, raise your hand and we’ll bring it to you. Try to write only your final answer on the exam question itself. 3. Do not write in the blank boxes. They are for grading purposes. Score: Problem 1 Problem 6 Problem 2 Problem 7 Probl em 3 Problem 8 Problem 4 Problem 9 Problem 5 Problem 10 Total
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Problem 1. Two independent genes are located near each other on the worm genome. Tailin is required for tail-formation and Fertilin is required for fertility. Non-coding region Non-coding region Non-coding region A B C D E F A) (5 points) If you insert an extra nucleotide at position A, what will be the phenotype of an animal homozygous for this mutation? (Please do not invoke any obscure phenomena or concepts we have not talked about in class—these questions are just as straight-forward as they appear!) If you insert a nucleotide at position C, what will be the homozygous phenotype? If you insert a nucleotide at position D, what will be the homozygous phenotype? If you insert a nucleotide at position C, and take one away from position D, what will be the homozygous phenotype? If you insert a nucleotide at positions A and C, what will be the homozygous phenotype? B) (5 points) If you insert a nucleotide at position A, how might you be able to suppress the mutant phenotype at position B? If you add two nucleotides at position B, name two distinct alterations you could make at position A that would suppress the phenotype. 1. 2. C) (5 points) You find that you can suppress a mutation at position D by removing a nucleotide at position E, but not by removing one at position F. What does this tell you about the structure of the Fertilin gene? (answer concisely, but be specific) Tailin Fertilin
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Problem #2. You have performed a mutagenesis screen in Drosophila, and have found a recessive loss-of-function Drosophila mutant with an essential role in the neural pathway that allows flies to jump. A) What phenotype do you expect from a homozygous mutant fly? (2 points) B) To generate the F2 families, you crossed an F1 animal with a wild-type fly. If 1000 F2 flies were born from this cross, how many flies do you expect to be heterozygous carriers of the mutation? In how many flies do you expect to see the mutant phenotype? (2 points) 1, Hets : 2. Mutants: C) If the F2 flies are allowed to mate randomly, and they have 1000 F3 offspring, how many F3 will display the mutant phenotype? (2 points) With continued screening, you have found 5 more mutations that all cause the same mutant phenotype. You perform a complementation test and get the following results. + = wildtype, - = mutant phenotype
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Exam 2006 - LS4 Final Exam-Fall 2006 Name Section...

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