#1. Mutations are...
Heritable changes in genetic information
#2. Answer true or false:
Mutations occur more frequently if there is a need for them.
False, Mutations occur spontaneously at a more or less constant frequency, regardless of selective pressure. Once they occur, however they can be selected for or against, depending on the advantage or disadvantage they confer. It is important not to confu
#3. Which of the following is not a class of mutation?
frameshift, missense, transition, transversion, or they are all classes of mutation.
They are all types of mutations.
#4. Ultraviolet usually cause mutation by a mechanism involving
A. one-strand breakage in DNA
B. light induced change of thymine to alkylated guanine.
C. induction of thymine dimers and their persistence or imperfect repair.
D. inversion of DNA segments
E
the answer is C.
#5. Not possible just study
...
#6. In the mutant strain x of E. coli, a leucine tRNA that recognizes the codon 5'- CUG-3' in normal cells has been altered so that it now recognizes the codon 5'-GUG-3'. A missense mutation that affects amino acid 10 of a particular protein is suppressed
a. What are the anitcodons of the two Leu tRNA's, and what mutational event has occurred in mutant X cells.
b. What amino acid would normally be preset at position 10 of the protein (without a missense mutations).
c. What amino acid is put in at position
Answers to Question 6:
a. if the normal codon is 5'- CUG-3', the anticodon of the normal tRNA is 5'-CAG-3'. If a mutant tRNA recognizes 5'-GUG-3', it must have an anticodon that is 5'-CAC-3'. The mutational event was a CG-to GC transversion in the gene fo
b. Since leucine- bearing (mutant) tRNA can suppress the mutation, presumably leucine is normally present at position 10?
c. The mutant tRNA recognizes the codon 5'-GUG-3', which codes for Val. In normal cells, a Val- tRNA. Valine would recognize the codo
#7. Using a eukaryotic experimental model a temperature- sensitive mutation rp11ATS, in the gene that encodes a protein subunit of RNA polymerase 2, in a missense mutation. Mutants have a recessive lethal phenotype at the higher temperature and grow norma
a. Explain how a new mutation in an interacting protein could suppress the lethality of the temperature- sensitive original mutation.
b. In addition to mutations in interacting proteins, what other type of suppressor mutations might be found?
c. Outline h
Answers to #7
a. -The sensitivity of the mutant could be due to amino acid change in the protein subunit of RNA polymerase 2 causing it be nonfunctional at the restrictive temperature.
-If it cant function is due to the structure and can't interact with o
b. a mutation in the second protein would be intergenic suppressor mutation. The original mutation could also be suppressed by reverting the missense mutation or by an interagenic supressor mutation.
-In an intragenic suppressor mutation, a particular sec
Continued from #7
d. Second- site suppressor are rare and occur at a low frequency. To suppress a defect, a very specific new mutation must occur. most mutation are induced by a mutagen will not have the compensatory ability of a suppressor mutation.
e. Since intergenic suppressor may result from mutations in interacting protein, this approach could be used to identify genes for proteins that interact during transcription.
#8. The mutant lacZ-1 was induced by treating E.coli with acridine and lacZ-2 was induced with 5BU. What kind of mutant are these likely to be. Explain. How could you confirm your predictions by studying the structure of the B-galactosidase in these cells
Acridine induces frameshift mutations, which make a completely different amino acid sequence and can become truncated at some point and can effect the structure and charge
5BU can be incorporated as T during DNA replication and read as C by DNA polymerase
#9. 5'-AUGACCCAUUGGUCUCGUUAG-3'
How many amino acids in the peptide chain?
Hydroxyamine, a mutagen is a transition AT to GC in DNA and was applied to an organism. A strain was isolated and mutation occurred at 11th position. How many amino acids long with
total of 6 amino acids
The mutation would change the bp G to A making a nonsense codon and terminating translating, therefore 3 amino acids total.
#10.
Of 94,075 births 10 were achondroplastic dwarfs( an autosomal dominant condition). 2/10 had dwarf parents, 8/10 had normal parents. What is the mutation rate at the locus?
8 new mutations in 94073 couples, 2 copies per gene so the mutation rate at the locu is 8/(96073x2) equals 4x10^-5
#11. Not possible just study.
...
#12.
a. Not possible on flash cards but know the different codes for the amino acids
b. Not possible on flash cards but know the stop codons and anticodons.
c. Will tRNA nonsense suppressors always insert the correct amino acid into the elongating polypep
No, it won't always insert the right base pair for the correct amino acid but it will insert a something better than a stop codon so the protein is mutant and just truncated.
#13 & #14. Not possible on flash cards, knowing the amino acids and their codon sequences are essential
...
#15. Not possible on flash cards, same as before
...
#16. Ames Test can effectively evaluate whether compounds or their metabolites are mutagenic.
a. What type of genetic selection is used by the Ames test and why does this type of selection a highly sensitive test.
b. How to use an Ames test on to assess w
a. Ames test test measures the rates of reversion of his auxotrophs to wild type. it selects his+ revertants by spreading his cells on a medium without histidine or rodent enzymes in the presence of a mutagen.
-Since the spontaneous mutation rate is v. lo
c. Herbicides decays to compounds that are not identical to its animal metabolites. How does this effect Ames test results. How could addition concerns be addressed?
Ames test could provide safe test results whenever it hazardous. This can be partly addressed by performing aditio Ames tests on extracts of plant and soil material treated with the herbicide. It is also possible that the hervicide is mutagenic in the Ame
#17.
a. How does DNA polymerase attempt to correct missmatches during DNA replication?
b. What mechanism is used to repair such mismatches if they escape detection by DNA polyermase?
c. How does a mismatched base in the newly synthesized strand distinguis
a. they have proofreading capability that allows them to stall at a mismatch. The will switch 3 to 5' and excise mismatched bases to be replaced.
b. MutS, MutL, and MutH proteins brings a unmethylated DNA strand to the 5'-GATC-3' closest to the mistake an
#18. Compare thymine dimers repair systems photoreactivation and excision (dark) repair.
Photoreactivation is executed by photlyase and activated by a photon with the wavelength 320-370nm.
Dark repair does not need light but mutiple enzymes.
-uvrABC- makes double stranded cuts
-Exonuclease excises 12bps segment btn nicks from uvrABC
-DNA poly
#19.
a. What response has E. coli developed to large amounts of DNA damage by mutagens? How is it controlled?
b. Why is the response itself a mutagenic system?
c. What effects does loss of function mutagens have on the genes recA and lexA have on the resp
a. Lg amount of damage triggers the SOS response. RecA activates and stimulates LexA protein functions. LexA stimulates the repair mechanisms used in SOS.
b. Bc the DNA polymersase for translesion DNA synthesis inputs one or more nucleotides across the le
#20.
a. In terms of Watson and Crick model, diagram a series of steps by which 5BU may be produced the mutants.
b. Assuming that the revertants were not caused by a suppressor mutation, indicate the steps by which nitrous acid may have produced the back m
a. in the normal state bromouracil performs an alternate with T, pairing with A
but in its rare state it resembles C and can pair with G. Draw your own diagram
b. Nitrous acid can deaminate C to U, resulting in a CG to TA transition.
#21. Holy cow no way. Check it out
...
#22.
a. What is the complementary strand of the following piece the hypothetical strand.
b. Indicate which way it replicates.
5'- T-HX-U-A-G-BU-enol-2AP- C-BU-X-2AP-C-BU-X-2AP-imino-3'
c. concepts on mosaicism
a & b:
3'-A-C-T-C---G----T--G-A-C-----C-5'
<-------replication
c.Products of DNA replication at the first mitotic division consist of one normal an done mutant helix, this produces mosaic individual.
#23. What happens to the template with DNA polymerase, dATP, dGTP, dCTP, dTTP, and Mg2+ come together.
Template:
5'-ATACGT-3' with backbone perfect
This will produce the expected complete complimentary strand of DNA.
#24. What happens with the template, DNA polymerase, dATP, dGMP, dCTP, dTTP, and Mg2+ come together.
Template:
5'-ATACGT-3' with backbone perfect
The absence of dGTP blocks polymerization after the first two bases.
#25. What happens with the DNA Template, DNA polymerase, dATP, dGMP, dCTP, dTTP, and Mg2+ come together.
Template:
5'-ATACGT-3' with backbone perfect
dHTP can be substituted for dGTP, but no dCPT is there to proceed after the first base.
#26. What happens when the DNA template is pretreated with HNO2, DNA Polymerase, dATP, dGTP, dCTP, dTTP, and Mg2+.
Template:
5'-ATACGT-3' with backbone perfect
The HNO2 deaminates or changes G to X, C to U, and A to H. However, X will pair with C, U will pair with A, and H pairs with C. This changes causes mutations that appear as:
3'-CACACA-5'
5'-HTHUXT-3'
#27. What happens when the DNA template, DNA polymerase, dATP, dGMP, dHTP,dCTP, dTTP, and Mg2+ come together.
Template:
5'-ATACGT-3' with backbone perfect
The dHTP will substitute for the dGTP's and pair with C.
3'-TATHCA-5'
5"-ATACGT-3'
#28. Table with BU AP, HNO2, and HA mutagens and how they change base pairs.
Not a good flash card.
#29.
a. Explain why after a wild type bacteria is treated with HNO2 a amino acid for methionine is substituted with valine.
b. Could mutants in part (a) be back mutated by using hydroxylamine. Hydroxylamine adds an OH group to cytosine causing it to pair
a. Met= AUG the anitcodon is TAC. In this case A was deaminated to form hypoxanthine and paired with C in the template strand and paired with C in the template strand. the new template was CAC and the new codon was GUG, Valine.
b. The CAC template strand
#30. Not a good flash card question, check it out though.
...
#31.
...
#32. There are evolutionary similarities between the bacteria and humans. What are the features that are shared between bacteria and human DNA repair systems.
Human gene hMSH2 is homologous to mutS and hMLH1, hPMS1, and hPMS2 are homologous to mutL
Xeroderma pigmentosum and Fanconi anemia both effect the DNA repair of UV damage just as it does in E. coli.
#33. E. Coli was plated on Mac Conkey- lactose medium produce red colonies, cells that can't metabolize lactose produce white colonies, but sometimes they produce a white colony having a red sector, that vary in size.
a. Why is there pink sectors in the w
a. The original cells was lac but as replication took place a mutation took place that created a lac+. The reproduction of this cell will create the pink sector.
B. Mutator mutant increase mutation rates. The mutants would revert and these mutants will re
#34. What are the difference between prokaryotic insertion elements and transposons.
*holy shit terribly long
IS elements are simpler in structure and have a transposase gene flanked by perfect or nearly inverted repeat sequences. Both can transpose into site of DNA that are not homologous.
-Composite transposons- have a central gene region flanked by IS elements
#35. What properties do bacterial and eukaryotic transposable elements have in common?
A difference
The structure, function of transposable elements, and integration events involve non-homologous recombination
-some eurkaryotic transposons mve through intermediate RNA unlike prokaryotes.
#36. An IS element inserted into lacZ gene, 40 bp deletion occurred next to the left border of the IS element. It removed 10 lacZ bp including the left target site and 30 left most bp's of the IS element. What are the consequences?
The left inverted repeat of the IS element has been remove, so that the two ends of this IS element are no longer homologous. Th element will not be able to move out of this location and insert into another site.
#37.a
Use an example to illustrate different transposition mechanisms that require
i.) DNA replication of the element
ii.) no DNA replication of the element
iii.) an RNA intermediate
i.)Tn3 requires DNA replication. The transposable elements gets replicated with the DNA and the donor and recipient both get a copy of the transposan. This is an example of a cointegrate.
ii). Tn10 and corn's Ac elements transpose without DNA replication.
#37b. What evidence is there that the inverted or direct terminal repeat sequences found in trasspoable elements are essential for transposition.
Their transposition comes from teh observation that mutation altering these sequences eliminate the ability an element to transpose.These sequences are recognized by transposease during a transposition event.
#37c. Do all transposable elements generate a target-site duplication after insertion.
yes
#38. In addition to single gene mutations, the frequency of chromosomal abbreviations such as deletion or inversions can be increased when transposable elements are present. How?
Deletions, inversions, and translocations can occur when homogolous recombination occurs between two identical transposons inserted in different location in a genome.
#39. Its incredibly long but maybe worth checking out.
...
#40. Two yeast transposable elements, A & B. Each contain an intron that transposes to a new location. In the new locations A has no intron and B does. What can you conclude about the two elements from the movement.
Since introns are spliced out at the RNA level, it had to occur during an RNA intermediate. Thus A moved though RNA intermediate. The lack removal during B suggests it used a DNA to DNA transposition mechanism.
#41.
a. What is the value of developing a transformation vector of an insect pest?
b. What is the basic information about the Minos element would need to be used for gathered before it could be used for germ-line transformation?
a. It could develop a better strategy for pest control. It could also help reduce the amount of parasites that are introduced to human or other animals from insects.
b. To understand the mode of transport of the element would be needed and the essential f