2004 Exam 1 Key (pdf)

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parents half-moon female progeny derived from the cross 5kb 2kb (d). Assuming the results shown above are representative of results obtained with the remaining halfmoon female offspring, what does this information suggest about the linkage relationship between the G and H genes? If they are linked, what map distance separates the genes? There appear to be six parental types and 4 recombinant types (circled) among the half-moon female progeny. Since the recombinant types are found less often than the parental types these results are consistent with linkage between the molecular marker and the Half-moon trait. Because the RFLP corresponds to the body color trait (G), these results indicate that G and half-moon (H) are linked. The map distance separating the G and H genes is calculated as follows: 4/10(100) = 40cM (e). Draw the genetic map that is consistent with this data and that from the first part of this problem. Be sure to show the L, G, and H genes and M marker on the map and the linkage distances separating these genes/markers. Indicate the two possible locations for the half-moon gene (H) that are consistent with this data. H 40cM G 17cM M 35cM L OR G 17cM M 35cM H L 40cM Problem 4. (15 total points) You are studying aging in fruit flies and have generated six different homozygous long-lived fly mutants (you may assume that each of these mutant strains bears a mutation affecting only ONE gene). You now wish to determine how many genes these six mutants represent and proceed to set up pairwise crosses with all of the homozygous mutants. Results of this analysis are shown in the table below (where the intersection represents the phenotype of the offspring resulting from a particular cross): Mut 1 Mut 2 Mut 3 Mut 4 Mut 5 Mut 6 WT 6
Mut 1 - + - - + + + Mut 2 - + + + - + Mut 3 - - + + + Mut 4 - + + + Mut 5 - + + Mut 6 - + + indicates all offspring have normal lifespan. - indicates all offspring are long-lived. WT = a wild type strain of flies. (a). How many complementation groups do these mutations represent? These mutations represent 3 complementation groups (b). Describe which mutations fall into each complementation group. One complementation group consists of the mutations 1, 3 and 4 Another group consists of the mutations 2 and 6 The third group consists of mutation 5 (c). In more recent experiments you isolate another long-lived fly mutant (Mut 7) and proceed to cross this mutation to your previously characterized long-lived mutants with the following results: Mut 1 Mut 2 Mut 3 Mut 4 Mut 5 Mut 6 WT Mut 7 - - - - - - - + indicates all offspring have normal lifespan. - indicates all offspring are long-lived. WT = a wild type strain of flies. What are these results telling you? When you cross Mut 7 to WT you get long-lived mutants. This tells you that Mut 7 is a dominant mutation. Therefore, you would expect to see the long-lived phenotype with whatever you cross this mutant to (i.e., it fails to complement everything it is crossed to). This is why dominant mutations cannot be used in complementation experiments. ANSWER EITHER ONE OF THE FOLLOWING TWO QUESTIONS (if you answer both, you will receive the statistical average of the two scores): Problem 5. (20 total points) The following graph illustrates the time course of DNA repair in E. coli following UV-light induced DNA damage. Each curve on the graph represents the result of a particular experiment carried out using either wild type or UV-sensitive E. coli mutants. 7
Page 1 and 2: MIDTERM EXAM 1 100 points total (6
Page 3 and 4: Problem 2 (20 points) You acquire a
Page 5: G 17cM M 35cM L 52cM OR L 18cM G 17
Page 9: Problem 6. (20 points) E. coli stra

2004 Exam 1 Key (pdf)

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