Anatomy and Function of a Gene: Dissection Through Mutation II

  1. One way trinucleotide repeats are thought to expand and contract is through _______ _______ during DNA replication. DNA polymerase often pauses as it replicates through repeat regions, which allows one DNA strand (either the ______ _______ strand or the _______ strand) to slip relative to the other one
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    • slipped mispairing 
    • newly synthesized strand
    • template strand
  2. Because the sequence contains repeats, the _____ strand and the _____ strand can pair out of register, forming a ______. After another round of DNA replication, this slipped mispairing can result in _______ or ________ of triplet repeat number in both DNA strands
    • slipped strand
    • other strand
    • loop
    • expansion or contraction
  3. Any physical or chemical agent that raises frequency of mutations above the spontaneous rate is called a _______.
    mutagen
  4. Researchers use many different _______ to produce mutations for study. With the Watson-Crick model of DNA structure as a guide, they can understand the action of most _______ at the molecular level
    • mutagens
    • mutagens
  5. The X-rays used by Muller to induce mutations on the X chromosme, for example, can break the ______ _______ _______ of DNA strands, sometimes at the _____ position on _____ strands of the double helix.
    • sugar phosphate backbone
    • same 
    • both
  6. Multiple double strand breaks produce _____ ________, and the improper stitching back together of the fragments can cause ______ ______ or ______ _______
    • DNA fragmentation 
    • small deletion or large deletions
  7. What do the parentheses mean?
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    • 28:35
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    29:25
  9. Another molecular mechanism of mutagenesis involves mutagens known as ______ _________, which are so similar in chemical structure to the normal nitrogenous bases that the replication machinery can incorporate them into DNA
    • base analogs
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  10. Explain why base analogs can cause base substitutions on the complementary strand synthesized in the next round of DNA replication
    Because a base analog may have tautomeric forms with pairing properties different from those of the base it replaces
  11. Other chemical mutagens generate substitutions by directly altering a base's ______ ________ and __________
    • chemical structure and properties
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  12. Again, the effects of these changes become fixed in the genome when the altered bases causes incorporation of an ______ ________ base during a subsequent round replication
    incorrect complementary
  13. Yet another class of chemical mutagens consist of compounds known as intercalators (define)
    The intercalator ________ is often used in genetic research for the reason you just detailed
    • Flat, planar molecules that can sandwich themselves between successive base pairs and disrupt the machinery for replication, generating deletions or insertions of a single base pair
    • proflavin
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  14. Cells have evolved a variety of enzymatic systems that locate and repair damaged DNA and thereby dramatically ______ the high potential for mutation
    diminish
  15. The combination of these repairs sytems must be extremely efficient (why?)
    because the rates of spontaneous mutations observed for almost all genes are very low
  16. If methyl or ethyl groups were mistakenly added to guanine _________ enzymes can remove them so as to recreate the _______ base
    • alkyltransferase enzymes
    • original
  17. Other enzymes remedy other base structure alteration. For example, the enzyme ________ recognizes the thymine-thymine dimers produced by exposure to ultraviolet light 
    How does it reverse the damage?
    • photolyase
    • by splitting the chemical linkage between the thymines
  18. Many repair systems use a general (2-step) strategy of homology dependent repair:

    This strategy makes use of one of the great advantages of the double helical structure:
    • First they remove a small region from the DNA strand that contains the altered nucleotide
    • Second they use the other strand as a template to resynthesize the region removed

    If one strand sustains damage, cells can use complementary base pairing with the undamaged strand to re-create the original sequence
  19. Base excision repair (5-story)
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    • Homology-dependent mechanism in which specific enzymes (glycosylase) cleave an altered base from the sugar of its nucleotide to create an apurinic or apyrimidinic (AP) site in the DNA chain
    • AP endonuclease creates a nick in the DNA backbone at the AP site.
    • Surrounding nucleotides are removed from the vicinity of the nick by DNA exonucleases
    • DNA polymerase fills in this gap by copying the undamaged strand, restoring the original DNA sequence
    • DNA ligase seals the nick
  20. In base excision repair, different ________ enzymes cleave specific damaged bases. Base excision repair is particularly important in the removal of _______ from DNA (which results from the _______ of cytosine)
    • glycosylase
    • uracil
    • deamination
  21. In this repair process, after the enzyme uracil-DNA glycosylase has removed ______ from its sugar, leaving an AP site, the enzyme _____ __________ makes a nick in the DNA backbone at the ____ _____
    • uracil 
    • AP endonuclease
    • AP site
  22. Other enzymes (known as DNA _________) attack the nick created by AP endonuclease and remove __________ from their vicinity to create a gap in the previously damaged strand. _____ ______ fills in the gap by copying the ________ strand, restoring the original nucleotide in the process. Finally, _____ _______ seals up the backbone of the newly repaired DNA strand
    • exonuceleases 
    • nucleotides 
    • DNA polymerase
    • undamaged 
    • DNA ligase
  23. Nucleotide excision repair (5-story)
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    • Homology-dependent mechanism that removes DNA alterations/errors, such as thymine-thymine dimers that base excision repair cannot take care of.
    • Depends on process-specific enzyme complexes (UvrA + UvrB) that patrol the DNA for irregularities
    • (UvrB + UvrC) nicks the damaged strand in two places that flank the error, releasing a short single-stranded region containing the alteration.
    • DNA polymerase fills the gap in
    • DNA ligase seals he nick
  24. How does Nucleotide excision repair remove alterations that base excision cannot repair?
    because the cell is not dependent a DNA glycosylase to recognizes or excise the problem
  25. Nucleotide excision repair depends on enzyme complexes containing more than one ______ molecule. In E. coli, these complexes are made of two out of three possible ______:
    • protein
    • proteins
    • UvrA, UvrB and UvrC
  26. One of the comlexes (UvrA + UvrB) patrols the DNA for irregularities detecting lesions that disrupt ______ ________and thus ______ the double helix (like thymine-thymine dimers not yet corrected via photorepair)
    • base pairing
    • distort
  27. A second complex (UvrB + UvrC) cuts the damaged strand in ____ places that _____ the damage. This _______ ________ excises a short region of the damaged strand and leaves a gap that will be filled in by _____ ________ and sealed with ______ ______
    • two places 
    • flank
    • double-cutting
    • DNA polymerase
    • DNA ligase
  28. X-rays can cause ______ ______ _______, in which ______ strands of the double helix are broken at nearby sites
    • double strand breaks
    • both
  29. Double strand breaks represent a particularly dangerous kind of DNA lesion (explain)
    because if not repaired properly, such chromosomal breakages can lead not only to point mutation, but also to large deletions and other kinds of chromosomal rearrangements
  30. Fortunately, cells can use the process of _______ _______ to repair most double-strand breaks accurately through complementary base pairing
    homologous recombination
  31. The first step of meiotic recombination is the formation of a ______-strand break, and that through strand invasion, cells undergoing _________ eventually repair this ______-strand break using the ________ chromosome as a template
    • double
    • recombination 
    • double
    • homologous
  32. ______ cells can employ much of the same enzymatic machinery for homologous recombination to repair _______ _______ breaks caused by X-ray exposure, using either a _______ chromosome, or more often a ______ _______,  as the template for repair
    • Mitotic
    • double strand 
    • homologous 
    • sister chromatid
  33. If the homologous chromosome serves as the template (during mitosis), repair of the break results in ______ ________. However, finding a homolog is _______, so repair through _________ usually occurs instead between _______ ________ during the G2 phase of the cell cycle (post replication). In this case, repair of the break does not produce ________ ________
    • mitotic recombination
    • inefficient
    • recombination 
    • sister chromatids
    • mitotic recombination
  34. Nonhomologous end-joining (NHEJ)
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    Mechanism for stitching back together ends formed by double-strand breaks. NHEJ relies on proteins that bind to the ends of the broken DNA strands and bring them close together
  35. NHEJ is especially important for the repair of double strand breaks formed during the _____ phase of the cell cycle
    G1 phase (that is, before a sister chromatid is available to serve as a template for homologous recombination)
  36. The proteins participating in NHEJ bind to DNA ends at the site of the ______ and protect the ends from _______. The NHEJ protein also _______ the two ends, allowing them to be _______ together by DNA _______ enzyme
    • breakage
    • nucleases
    • bridge
    • stitched
    • ligase
  37. Methyl-directed mismatch repair
    DNA repair mechanism that corrects mistakes in replication, discriminating between newly synthesized and parental DNA by the methyl groups on the parental strand
  38. How does bacteria solve this problem: (8-story)

    Suppose that a G-C base pair, has been copied to produce two daughter molecules, one of which has the correct G-C pair and the other an incorrect G-T. The mismatch repair system can easily recognize the incorrectly matched G-T base pair because the improper base pairing distorts the double helix, resulting in abnormal bulges and hollows. But how does the system know whether to correct the pair to a G-C or to an A-T?
    • pg 222-
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  39. Eukaryotic cells also have a mismatch correction system, but we do not yet know how this system distinguishes ______ from newly replicated strands
    templates
  40. Cells sometimes become exposed to levels or types of mutagens that they cannot handle with these high-fidelity repair systems. Strong does of UV light, for example, might make more ______ ______ dimers than the cell can fix.
    thymine-thymine dimers
  41. Any unrepaired damage has _______ consequence for cell divisions; the DNA polymerases normally used in replication will ______ at such lesions, so cells cannot _________
    • severe
    • stall
    • proliferate
  42. These cells can initiate ________ responses that may allow them to overcome these problems and thus survive and divide, but their ability to proceed in such circumstances comes at the expense of introducing _____ _____ to the genome
    • emergency 
    • new mutations
  43. One type of emergency repair in bacteria, called the ____ ________ (hint Morse Code), relies on _______ ______ (or _______) DNA polymerases
    • SOS system 
    • error-prone (or sloppy) DNA polymerases
  44. These sloppy DNA polymerases are not available for normal DNA replication; they are produced only in the presence of _____ ______
    DNA damage
  45. The damage induced, error-prone DNA polymerases are attracted to replication forks that have become _______ at sites of unrepaired, damage nucleotides. There, the enzymes add _______ nucleotides to the strand being synthesized _________ the damaged bases.
    • stalled 
    • random 
    • opposite
  46. The SOS polymerase enzyme thus allows the cell with damaged DNA to divide into _____ _______ cells, but because at each position the sloppy polymerases restore the proper nucleotide only ____-______ of the time, the genomes of these _______ cells carry _____ mutations
    • two daughter 
    • one-quater
    • daughter 
    • new
  47. In bacteria, the mutagenic effect of many mutagens either depends on, or is enhanced by the ______ ______
    SOS system
  48. Microhomology-mediated end-joining (MMEJ)
    DNA double-strand break repair process similar to NHEJ except that DNA ends are resected, which results in small deletions in rejoined DNA
  49. Another kind of ________ repair system microhomology-mediated end-joining (MMEJ), deals with dangerous ______ ______ breaks  that have not been corrected by ________ _________ or _______.
    • emergency 
    • double strand
    • homologous recombination 
    • NHEJ
  50. However in MMEJ, the broken DNA ends are cut back on _______ side of the break (resected) by ________. The resection exposes ______ _____ ______ regions of complementary DNA sequence ("_________") on ________ side of the break that help in bringing the ends together
    • either
    • enzymes
    • small single stranded regions
    • microhomology
    • either
  51. Why does MMEJ typically result in small deletions in the rejoined DNA?
    because nucleotides are removed at the sites of the double-stranded breaks during resection
  52. Although differences of detail exist between the DNA repair systems of various organisms, DNA repair mechanisms appear in some form in virtually _____ species
    all
  53. The many known human hereditary diseases associated with the defective repair of DNA damage reveal how crucial these mechanisms are for survival. In one example, the cells of patients with xeroderma pigmentosum lack the ability to conduct _______ ______ ______
    nucleotide excision repair
  54. The people who suffer from xeroderma pigmentosum are ________ for mutations in any one of ________ genes encoding enzymes that normally function in this repair system. As a result, the ______ ______ ______ caused by UV light cannot be removed efficiently
    • homozygous 
    • seven 
    • thymine thymine dimers
  55. Unless these people avoid all exposure to sunlight, their skin cells begin to accumulate _______ that eventually lead to _____ _____
    • mutations
    • skin cancer
  56. In yet another example, the breast cancer genes ______ and ________ encode ________ that function in _______ _______ break repair via _________ ________
    • BRCA1 and BRCA2
    • proteins 
    • double strand break repair 
    • homologous recombination
  57. The wide-ranging variation in the genetic makeup of the human population and other populations is the result of which three factors?
    • The continuous introduction of new mutations
    • The loss of deleterious mutations because of the selective disadvantage they impose on the individuals that carry them
    • The increase in frequency of rare mutations that either provide a selective advantage to the individuals carrying them or that spread through a population by other means
  58. 2 vs 4 compare somatic mutations to germline mutations
    Somatic mutations:

    • Somatic mutations affect only the individual and will not be passed on to offspring
    • Can lead to disease alleles or cancer depending on the mutation

    Germ-line mutations:

    • Germline mutations will be passed on to offspring
    • Affected by natural selection
    • Many mutations silent or slightly deleterious
    • Few mutations beneficial
  59. In sexually reproducing multicellular organisms, only _______ mutations that can be passed on to the next generation play a role in evolution. Nevertheless, mutations in ______ cells can still have an impact on the well-being and survival of individuals
    • germ-line
    • somatic
  60. Somatic mutations in genes that help regulate the cell cycle may, for example, lead to _______. The U.S. Food and Drug Administration tries to identify potential cancer-causing agents (aka _______) by using the ______ test to screen for chemicals that cause _______ in bacterial cells
    • cancer
    • carcinogens
    • Ames test
    • mutations
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  61. This test asks whether a particular chemical can induce _______ ________ of a special _______ mutant strain of the bacterium Salmonella typhimurium
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    • histidine+ (his+) revertants
    • histidine- (his-) mutants
  62. This his+ revertants can synthesize all the _____ they need from simple compounds in their environment, whereas the original his- mutants cannot make ________, so they can survive only if _______ is supplied
    • hisitidine 
    • histidine 
    • histidine
  63. To increase the sensitivity of mutation detection, the his- strain, used in the Ames test system, contains a second mutation that inactivates the ________ _____ ______ system and thereby prevents the ready repair of mutations caused by the potential mutagen.
    nucleotide excision repair
  64. The bacteria also carry a third mutation causing defects in the cell wall that allows tested chemicals _______ ________ to the cell interior
    easier access
  65. Researchers commonly define a gene as a _______ ______ that directs the appearance of a ______ _______ that, in turn, contributes to a particular _______. They can use this definition to determine whether two mutations are in the ______ gene or in _______ genes
    • functional unit 
    • molecular product 
    • phenotype 
    • same gene or in different genes
  66. If two homologous chromosomes in an individual each carry a mutation recessive to wild type, a normal phenotype will result if the mutations are in _______ genes. Why does the normal phenotype occur?
    • different
    • because almost all recessive mutations disrupt a gene's function.
  67. The dominant wild-type alleles on each of the two homologs can make up for, or ________, the defect in the other chromosome by generating enough of both gene products to yield a normal phenotype
    • complement
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  68. In contrast, if the recessive mutations on the two homologous chromosomes are in the same gene, no _____ _______ allele of that gene exists in the individual and neither mutated copy of the gene will be able to perform the _____ function. As a result, no __________ will occur and no ______ gene product will be made, so a _______ phenotype appear
    • wild type
    • normal 
    • complementation
    • normal 
    • mutant
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  69. Ironically, a collection of mutations that do not complement each other is known as a __________ ______.
    complementation group
  70. Geneticists often use "complementation group" as a synonym for "gene" (why?)
    because the mutations in a completation group all affect the same unit of function, and thus, the same gene
  71. Complementation testing (2-story)
    Hint: A simple test based on the idea of a gene as a unit of function can determine whether or not two recessive mutations are alleles of the same gene:
    page 226 bottom left
  72. Complementation testing has, in fact, shown that garnet, ruby vermilion, and carnation pigmentation are caused by mutations in ______ genes. But chromosomes carrying mutations yielding white, cherry coral, apricot, and buff phenotypes fail to _______ each other. These mutations therefore constitute _______ alleles of a single gene
    • separate
    • complement 
    • different
  73. visit the box 226 upper right
  74. The table illustrates how researchers collate data from many complementation tests in a _________ table. Such a table helps visualize the relationships among a large group of ______
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    • complementation table 
    • mutants
  75. In the late 1950s, Seymore Benzer used recombination analysis to show that two different mutations that did not complement each other and were therefore known to be in the _____ gene can in fact change _______ parts of that gene
    • same 
    • different
  76. He reasoned that if a gene is composed of ______ mutable subunits, then it should be possible for recombination to occur within a gene that happens to be _______ these subunits. Therefore, crossovers between homologous chromosomes carrying different _______ known to be in the same gene could in theory generate a ______ _____ allele
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    • separately
    • between
    • mutations
    • wild-type
  77. Why is it necessary to examine a very large number of progeny (of mutant parents that resulted in wild-type progeny) to see if even one crossover event occurred between them?
    because mutations affecting a single gene are likely to lie very close together
  78. Working with bacteriophage T (4-story)
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    pg 229
  79. Phenotypic properties of rII- mutants of bacteriophage T4 (2-story)
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    pg 229
  80. A customized complementation test between rII- mutants of bacteriophage T4 (2-story)
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    pg 230
  81. Detecting recombination between two mutations in the same gene (2-story)
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    pg 230
Author
chikeokjr
ID
335489
Card Set
Anatomy and Function of a Gene: Dissection Through Mutation II
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7.2-
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