1. Two independent E. coli mutants ( 1 and 2 ) are induced using mutagen X. Afte
ID: 3509143 • Letter: 1
Question
1. Two independent E. coli mutants (1 and 2) are induced using mutagen X. After being mutated with mutagen X, each mutant is taken and exposed to the following mutagens: Bromouracil (BU), Nitrous Acid (NA), Hydroxylamine (HA) and Proflavin. The Table below shows the reversion frequency of each mutagen. Assume all the revertants are true revertants. (4 marks)
Mutant
No mutagen
BU
NA
HA
Proflavin
1
1.5 x10-8
5 x 10-5
1.3 x 10-4
1.6 x10-8
1.6 x 10-7
2
2 x10-7
2 x 10-4
6 x 10-5
3 x10-5
1.7 x 10-7
a. Make the best prediction possible about which base changes occurred in forming the original mutations (4 marks).
Mutation Probable base Change
1
2
b. What kind of mutations are caused by mutagen X and why did you come to this conclusion? (3 marks)
2. You identify a third E. coli mutation (3) using mutagen X. You decide to map 3 and find it is close to both the GAL and BIO genes. The normal distance between the wild type GAL and BIO genes is 20 map units. You cross 3 with a gal mutant (Cross1) and with a bio mutant (Cross 2) independently and get the following results:
Cross 1
GAL 3 145
gal + 120
GAL + 4
gal 3 5
Cross 2
BIO 3 105
bio + 80
BIO + 6
bio 3 8
+ = wild-type allele; 3 = mutant allele
a. Based on the above data what is the map distance between Gene 3 and GAL gene (Show work)? (2 marks)
b. What is the map distance between Gene 3 to the BIO gene (Show work)? (2 marks)
c. What does these results suggest about the molecular basis of mutation 3 and why do you come to this conclusion? (2 marks)
d. How would you expect mutation 3 to behave in a reversion test and why do you come to this conclusion? (2 marks)
Mutant
No mutagen
BU
NA
HA
Proflavin
1
1.5 x10-8
5 x 10-5
1.3 x 10-4
1.6 x10-8
1.6 x 10-7
2
2 x10-7
2 x 10-4
6 x 10-5
3 x10-5
1.7 x 10-7
Explanation / Answer
1.a.
b. The mutations caused by mutagen X are point mutations since true revertants can revert the original base in
point mutations.
It is also a transition mutation where true revertants can revert the original base.
2.a. distance = #recombinant offspring/#total offspring=4+5(minority number)/4+5+120+145=9/274=0.03map unit
b.distance=6+8/6+8+80+105=14/199=0.07 map unit
c.The molecular basis of mutation is not recombination since the distance suggests linkage.
d.In a reversion test mutation 3 would be reversed since induced mutation can be reversed.
MUTATION PROBABLE BASE CHANGE 1. C->T 2. G->ARelated Questions
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