bis101samplefinal - 1. The Ames Test is a rapid and...

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1. The Ames Test is a rapid and convenient way to assess the mutagenicity of common substances in our environment. The table below shows the results of a typical test of three compounds; X, Y and Z. These compounds were tested on three his - bacterial strains that were generated with mutagens known to cause: AT to GC transitions (Strain A); CG to AT transversions; (Strain B) and a single base pair insertion (Strain C). About 10 9 his - bacterial cells were plated on each minimal agar plate lacking histidine. The sterile water control indicates the rate of spontaneous reversion. Number of his + Revertants Strain A Strain B Strain C Control (water)10 14 1 Compound X 11 13 1 Compound Y 12 1350 0 Compound Z 9 12 97 What do the results of these tests tell you about the mutagenicity of compounds X, Y and Z. (A) Which of these compounds are mutagenic? — yes or no (9 pts.) Compound X __ No _____ Compound Y __ Yes ___ Compound Z __ Yes ____ (B) How mutagenic are these compounds in terms of approximated fold- increase over the spontaneous reversion rate? (9 pts.) Compound X __ O ___ Compound Y __ 100 ____ Compound Z __ 100 ____ (C) Describe the kinds of mutations produced by these compounds (e.g., GC to TA transversion). (9 pts.) Compound X __ None _____________________________________ Compound Y __ A/T to C/G transitions _________________________ Compound Z ___ single base pair deletion ______________________
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Name____________________________ 2 (D) Two compounds, A and B, were found to be mutagenic. To determine relative risk of these compounds to human health, the HERP% (Human Exposure/Rodent Potency x 100) was calculated from the following values. Human Exposure/Yr Rodent Potency in mg/kg Compound A 150mg 1.5mg Compound B 0.0005mg 0.005mg Which of these compounds poses the greatest threat to human health and therefore should be regulated in the environment? Explain your answer. (10 pts.) Compound A because it has a HERP% of 10,000 2. The following X-linked recessive traits are found in fruit flies: vermillion eyes ( v ) are recessive to normal (red eyes), miniature wings ( m ) are recessive to long wings and sable body ( s ) is recessive to gray body. A cross was made between true breeding wild type males and females homozygous recessive for all three alleles. The progeny of this cross, the F1, was crossed with a homozygous recessive tester and the following progeny were obtained: 1320 vermillion eyes/miniature wings/sable body 1346 red eyes/long wings/gray body 101 vermillion eyes/miniature wings/gray body 90 red eyes/long wings/sable body 43
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This note was uploaded on 12/18/2009 for the course BIS 52191 taught by Professor Marksanders during the Spring '09 term at UC Davis.

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bis101samplefinal - 1. The Ames Test is a rapid and...

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