Wednesday, May 6, 2020
Hartwell Genetics Ch. 13 Study Guide Free Essays
string(151) " of the following does not happen when an intragenic inversion occurs\? |A\) |One part of the gene is relocated to a distant region of the chromosome\." Chapter 13â€â€Chromosomal Rearrangements and Changes in Chromosome Number Reshape Eukaryote Genomes Fill in the Blank |1. |Events that reshape genomes by reorganizing the DNA sequences within one or more chromosomes are known as ____________________. | |Ans: |rearrangements | | |Difficulty:  2 | |2. We will write a custom essay sample on Hartwell Genetics Ch. 13 Study Guide or any similar topic only for you Order Now |____________________, duplications, inversions, translocations, and movements of transposable elements are all types of | | |rearrangements of chromosomal material. |Ans: |Deletions | | |Difficulty:  2 | |3. |Very large deletions are visible at the relatively low resolution of a __________________, showing up as the loss of one or more| | |bands from a chromosome. | |Ans: |karyotype | | |Difficulty:  2 | |4. |Changes in gene ____________________, the number of times a given gene is present in the cell nucleus, can create a genetic | | |imbalance. |Ans: |dosage | | |Difficulty:  3 | |5. |An unpaired bulge of one member of a homologous pair of chromosomes during prophase of meiosis I is known as a | | |____________________. | |Ans: |deletion loop | | |Difficulty:  3 | 6. |A recessive mutation in the mouse that prevents homozygous animals from walking in a straight line is known as the | | |____________________ gene. | |Ans: |shaker-1 | | |Difficulty:  3 | |7. |When repeats of a region lie adjacent to each other they are called ____________________ duplications. |Ans: |tandem | | |Difficulty:  2 | |8. |Inversions that include the centromere are termed ____________________. | |Ans: |pericentric | | |Difficulty:  2 | 9. |Inversions that exclude the centromere are termed ____________________. | |Ans: |paracentric | | |Difficulty:  2 | |10. |A crossover product that lacks a centromere is known as a(n) ____________________fragment. | |Ans: acentric | | |Difficulty:  2 | |11. |The type of large-scale mutation in which parts of two different chromosomes trade places is a ____________________ | | |translocation. | |Ans: |reciprocal | | |Difficulty:  2 | |12. |____________________ is the enzyme that catalyzes transposition. |Ans: |Transposase | | |Difficulty:  2 | |13. |Organisms with four copies of a particular chromosome (2n+2) are ________________. | |Ans: |tetrasomic | | |Difficulty:  3 | |14. |Down syndrome is also known as ____________________. |Ans: |trisomy 21 | | |Difficulty:  2 | |15. |If a tetraploid derives all of its chromosome sets from the same species, we call this kind of polyploid a(n) | | |____________________. | |Ans: |autopolyploid | | |Difficulty:  3 | Multiple Choice |16. |Which of the following are considered chromosomal rearrangements? |A) |inversions | |B) |duplications | |C) |deletions | |D) |translocations | |E) |all of the above | | |Ans:  E | | |Difficulty:  1 | |17. |Which of the following removes material from the genome? |A) |inversions | |B) |duplications | |C) |deletions | |D) |translocations | |E) |none of the above | | |Ans:  C | | |Difficulty:  1 | |18. |Which of the following adds material to the genome? |A) |inversions | |B) |duplications | |C) |deletions | |D) |translocations | |E) |none of the above | | |Ans:  B | | |Difficulty:  1 | |19. The type of chromosomal rearrange ment which reorganizes the DNA sequence within one chromosome is known as a(n): | |A) |inversion | |B) |duplication | |C) |deletion | |D) |translocation | |E) |none of the above | | |Ans:  A | | |Difficulty:  2 | |20. |In general, which of the following usually has a greater chance of lethality than the others? |A) |inversion | |B) |duplication | |C) |deletion | |D) |translocation | |E) |all have an equal chance | | |Ans:  C | | |Difficulty:  2 | |21. |Sometimes a piece of one chromosome attaches to another chromosome. This is known as a(n): | |A) |inversion | |B) |duplication | |C) |deletion | |D) |translocation | |E) |none of the above | | |Ans:  D | | |Difficulty:  1 | |22. |Sometimes a part of the genome moves from chromosome to chromosome. This is known generally as a(n): | |A) |inversion | |B) |duplication | |C) |deletion | |D) |translocation | |E) |transposable element | | |Ans:  E | | |Difficulty:  2 | |23. |Rearrangements and changes in chromosome number may affect gene activity or gene transmission by altering the | | |________________________ of certain genes in a cell. |A) |position | |B) |order | |C) |number | |D) |all of the above | | |Ans:  D | | |Difficulty:  2 | |24. |Karyotypes generally remain constant within a species because: | |A) |rearrangements occur frequently. | |B) |changes in chromosome number occur infrequently. | |C) |genetic instabilities produced by genomic changes usually are at a selective disadvantage. | |D) |genetic imbalances are often at a selective advantage. | |Ans:  C | | |Difficulty:  2 | |25. |Despite selection against chromosomal variations: | |A) |related species almost always have the same karyotype. | |B) |related species almost alway s have a different karyotype. | |C) |closely related species diverge by many chromosomal rearrangements. | |D) |distantly related species diverge by only a few chromosomal rearrangements. | | |Ans:  B | | |Difficulty:  2 | |26. In higher organisms, using genetic analysis is usually difficult to distinguish small deletions in one gene from: | |A) |heterozygotes. | |B) |small duplications. | |C) |monosomies. | |D) |point mutations. | | |Ans:  D | | |Difficulty:  1 | |27. |For an organism to survive a deletion of more than a few genes, it must carry a nondeleted homolog of the deleted chromosome. | | |This is known as: | |A) |a deletion heterozygote. | |B) |a deletion homozygote. | |C) |dosage compensation. | |D) |a triplolethal chromosome. | |Ans:  A | | |Difficulty:  2 | |28. |Individuals born heterozygotes for certain deletions have a greatly increased risk of losing both copies of certain genes and | | |developing cancer. One such disease is: | |A) |triplolethal. | |B) |scarlet eyes. | |C) |retinoblastoma. | |D) |cataracts. | |Ans:  C | | |Difficulty:  1 | |29. |During the pairing of homologs in prophase of meiosis I, the region of a normal, nondeleted chromosome that has nothing with | | |which to recombine forms a so-called: | |A) |inversion loop. | |B) |deletion heterozygote. | |C) |crossover suppressor. | |D) |deletion loop. | | |Ans:  D | | |Difficulty:  2 | |30. Using Drosophila polytene chromosomes and small deletions, geneticists have been able to: | |A) |map the shaker-1 gene in Drosophila. | |B) |assign genes to regions of one or two polytene chromosome bands. | |C) |assign genes to regions of 100kb or less of DNA. | |D) |all of the above | | |Ans:  D | | |Difficulty:  2 | |31. |Which of the following molecular techniques could a scientist use to help locate genes on cloned fragments of DNA with deletion | | |mutants? |A) | In situ hybridization | |B) |Crossover analysis | |C) |Southern blot analysis | |D) |all of the above | |E) |both a and c | | |Ans:  E | | |Difficulty:  3 | |32. |Duplications arise by: | |A) |chromosomal breakage and faulty repair. | |B) |unequal crossing over. | |C) |errors in replication. |D) |all of the above | | |Ans:  D | | |Difficulty:  1 | |33. |During the pairing of homologs in prophase of meiosis I, the region of a chromosome bearing extra copies of a particular | | |chromosomal region that has nothing with which to recombine forms a so-called: | |A) |inversion loop. | |B) |deletion heterozygote. | |C) |duplication loop. | |D) |deletion loop. | | |Ans:  C | | |Difficulty:  2 | |34. An inversion may result from: | |A) |a half-circle rotation of a chromosomal region following two double-strand breaks in a chromosome’s DNA. | |B) |the action of a transposable element. | |C) |a crossover between DNA sequences present in two positions on the same chromosome in inverted orientat ion. | |D) |all of the above | |E) |none of the above | | |Ans:  D | | |Difficulty:  2 | |35. Inversions may be hard to detect because they: | |A) |never visibly change chromosome banding patterns. | |B) |increase recombination in heterozygotes. | |C) |do not usually cause an abnormal phenotype. | |D) |normally are removed immediately in natural populations. | | |Ans:  C | | |Difficulty:  3 | |36. |Which of the following does not happen when an intragenic inversion occurs? |A) |One part of the gene is relocated to a distant region of the chromosome. You read "Hartwell Genetics Ch. 13 Study Guide" in category "Papers" | |B) |One part of the gene stays at its original site. | |C) |Homozygotes for the inversion do not survive. | |D) |The gene’s function is not disrupted. | | |Ans:  D | | |Difficulty:  3 | |37. |When a crossover occurs within the inversion loop of a pericentric inversion each recombinant chromatid will have: | |A) |a single centromere. | |B) |a duplication of one region. | |C) |a deletion different from the one of duplication. |D) |all of the above | | |Ans:  D | | |Difficulty:  2 | |38. |Robertsonian translocations result from which of the following? | |A) |Breaks at or near the centromeres of two acrocentric chromosomes followed by the reciprocal exchange of broken parts. | |B) |A part of one chromosome becomes attached to a non-homologous chromosome. | |C) |Unequal crossing over during meiosis. | |D) |The fusion of two small chromosomes end-to-end such that a double centromere occurs. | | |Ans:  A | | |Difficulty:  2 | |39. |Which of the following does not usually show a problem during meiosis? |A) |translocation heterozygotes | |B) |translocation homozygotes | |C) |paracentric inversion | |D) |pericentric inversion | | |Ans:  B | | |Difficulty:  1 | |40. |Of the following segregation patterns, which one is most likely to result in a normal zygote? |A) |alterna te | |B) |adjacent-1 | |C) |adjacent-2 | |D) |nondisjunction | | |Ans:  A | | |Difficulty:  2 | |41. |The condition of semisterility is most closely associated with: | |A) |chromosomal duplications. | |B) |pericentric inversions. | |C) |translocation heterozygotes. | |D) |translocation homozygotes. | |Ans:  C | | |Difficulty:  2 | |42. |Translocations can help: | |A) |determine linkage groups. | |B) |aid in the diagnosis and treatment of certain cancers. | |C) |map important genes. | |D) |all of the above | | |Ans:  D | | |Difficulty:  1 | |43. Down Syndrome can result from: | |A) |three copies of chromosome 21. | |B) |a translocation of a part of chromosome 21. | |C) |a reciprocal translocation between any two autosomes. | |D) |a and b | |E) |a, b, and c | | |Ans:  D | | |Difficulty:  2 | |44. |Which of the following do translocations and inversions not have in common? |A) |don’t alter the amount of DNA in the genome | |B) |ability to alter gene function | |C) |use of inversion loops during crossing over | |D) |catalysts of speciation | | |Ans:  C | | |Difficulty:  2 | |45. |A transposition is considered a cytologically invisible sequence rearrangement. With which of the following does it share this | | |property? |A) |small deletion | |B) |large duplication | |C) |inversion | |D) |translocation | | |Ans:  A | | |Difficulty:  2 | |46. |Barbara McClintock is most closely associated with which of the following? | |A) |The initial discovery of genetic transposition. | |B) |The discovery of transposable elements in corn. | |C) |The mutation rate in translocation heterozygotes. | |D) |The demonstration of the presence of transposable elements in polytene chromosomes. | | |Ans:  B | | |Difficulty:  1 | |47. |Transposable elements have many things in common. Which of the following is not a usual characteristic of them? | |A) |Typically smaller than 50 bp. | |B) |May be present in a genome from one to thousands of times. | |C) |Are found only in a select group of organisms. | |D) |Need not be sequences that do something for the organism. | | |Ans:  A | | |Difficulty:  3 | |48. |Retroposons and retro-viruses have structural parallels. Which of the following also shares structural parallels with them? |A) |tRNA | |B) |DS-DNA | |C) |rRNA | |D) |mRNA | | |Ans:  D | | |Difficulty:  2 | |49. |Which of the following is a possible effect that a transposable element may have on a gene? | |A) |Shift the reading frame. | |B) |Diminish the efficiency of splicing. | |C) |Provide a transcription stop signal. | |D) |all of the above | | |Ans:  D | | |Difficulty:  2 | |50. Which of the following is not an aneuploidy? | |A) |monosomy | |B) |tetraploid | |C) |trisomy | |D) |tetrasomy | | |Ans:  B | | |Difficul ty:  1 | |51. |The most common human aneuploidy is trisomy 21, Down syndrome. All of the effects listed below may be seen in this syndrome | | |except: | |A) |death always by age 25. | |B) |mental retardation. | |C) |skeletal abnormalities. | |D) |heart defects. | |E) |increased susceptibility to infection. | | |Ans:  A | | |Difficulty:  2 | |52. |Which of the following sex chromosome aneuploidies is not usually seen in live births? | |A) |XO |B) |XXY | |C) |YO | |D) |XXX | |E) |None of the above | | |Ans:  C | | |Difficulty:  2 | |53. |Turner syndrome, XO, is a sex chromosome aneuploidy. Of the effects listed below, which one is not usually seen in this | | |syndrome? |A) |unusually short stature | |B) |infertility | |C) |skeletal abnormalities | |D) |unusually long limbs | | |Ans:  D | | |Difficulty:  2 | |54. |In Drosophila, a gynandromorph, which is composed of equal parts male and female tissue, results from: | |A) |an XX female losing one X chromosome during the first mitotic division after fertilization. | |B) |an egg carrying an X chromosome fertilized by a Y-carrying sperm. | |C) |a normal egg fertilized by both an X-carrying sperm and a Y-carrying sperm. | |D) |the fusion of a female embryo with a male embryo. | |Ans:  A | | |Difficulty:  3 | |55. |Which of the following is not an example of a euploid condition? | |A) |triploidy | |B) |diploidy | |C) |Down syndrome | |D) |tetraploidy | | |Ans:  C | | |Difficulty:  1 | |56. Triploid organisms usually result from: | |A) |the union of haploid and diploid gametes. | |B) |unequal disjunction during embryogenesis. | |C) |propagation of fused cell lines. | |D) |fusion of three gametes simultaneously. | | |Ans:  A | | |Difficulty:  2 | |57. |During mitosis, if the chromosomes in a diploid tissue fail to separate after replication, the resulting daughter cells will be:| |A) |monoploid. | |B) |tetrasomic. |C) |triploid. | |D) |tetraploid. | | |Ans:  D | | |Difficulty:  2 | |58. |Hybrids in which the chromosome sets come from two distinct, though related, species are known as: | |A) |autopolyploids. | |B) |allopolyploids. | |C) |amphiploids. | |D) |bivalents. | | |Ans:  B | | |Difficulty:  2 | |59. |The genus Triticale is a new genus of the various allopolyploid hybrids between wheat and rye. Some of the members of this genus| | |show agricultural promise because: | |A) |wheat has a high yield. | |B) |rye adapts well to unfavorable environments. | |C) |wheat has a high level of protein. | |D) |rye has a high level of lysine. | |E) |all of the above | | |Ans:  E | | |Difficulty:  2 | |60. |Which of the following rarely, if ever, results in a positive force for evolution? |A) |polyploidy | |B) |allopolyploidy | |C) |trisomy | |D) |amphidiploidy | | |Ans:  C | | |Difficulty:  2 | Matching | | Match the following descriptions with the terms below a. inversion b. duplication c. deletion d. translocation e. transposable element |61. |A piece of genetic material that moves from place to place in the genome. | |Ans: |e | | |Difficulty:  2 | |62. |A change in the genome whereby new material is added to the genome. | |Ans: |b | | |Difficulty:  1 | |63. |A change in the genetic material where a DNA sequence changes direction. |Ans: |a | | |Difficulty:  1 | |64. |A decrease of genetic material in the genome. | |Ans: |c | | |Difficulty:  1 | |65. |A piece of chromosome attaches to another chromosome. | |Ans: |d | | |Difficulty:  2 | | | Match the following descriptions with the terms below a. retroposon b. transposon c. transposable element d. transposase |66. |Any DNA segment that moves about in the genome. |Ans: |c | | |Difficulty:  3 | |67. |Moves in the genome with the aid of an RNA intermediate. | |Ans: |a | | |Difficulty:  2 | |68. |Moves DNA directly. | |Ans: |b | | |Difficulty:  3 | |69. |An enzyme that catalyzes a transposition event. | |Ans: |d | | |Difficulty:  1 | True or False |70. |When comparing mouse and human Giemsa-stained karyotypes, we see no conservation of banding patterns. | |Ans:  True | | |Difficulty:  2 | |71. |Karyotypes generally remain constant within a species because rearrangements and changes in chromosome number occur | | |infrequently. | | |Ans:  False | | |Difficulty:  3 | |72. |Changes in chromosome number include aneuploidy, monoploidy, polyploidy, and duplications. | |Ans:  False | | |Difficulty:  2 | |73. |Deletion may arise from errors in replication, from faulty meiotic or mitotic recombination, and from exposure to X-rays. | | |Ans:  True | | |Difficulty:  2 | |74. |Homozygosity for a deletion is often, but not always, lethal. | |Ans:  True | | |Difficulty:  2 | |75. |Recessive mutations can often be covered by deletions in heterozygotes. | | |Ans:  True | | |Difficulty:  2 | |76. |Most duplications have no obvious phenotypic consequences and can be detected only by cytological or molecular means. | |Ans:  False | | |Difficulty:  2 | |77. |Duplication of chromosomal segments rarely has an effect on the evolution of genomes. | | |Ans:  True | | |Difficulty:  2 | |78. |Crossing-over within an inversion loop produces aberrant recombinant chromatids. | |Ans:  False | | |Difficulty:  3 | |79. |Reciprocal translocations are usually phenotypically abnormal because they have neither lost nor gained genetic material. | | |Ans:  True | | |Difficulty:  1 | |80. |A hallmark of transposons is that their ends are inverted repeats of each other. | |Ans:  False | | |Difficulty:  3 | |81. |The mouse genome has high synteny with the human genome since about 170 DNA blocks are simply rearranged betwee n the two | | |genomes. | | |Ans:  True | | |Difficulty:  2 | |82. |Euploid cells contain only incomplete sets of chromosomes. | |Ans:  False | | |Difficulty:  2 | |83. |Down syndrome is an example of triploidy. | | |Ans:  False | | |Difficulty:  2 | |84. |Genetic imbalance results from polyploidy. | |Ans:  False | | |Difficulty:  1 | |85. |An acentric fragment is an inversion cross-over product lacking a centromere. | | |Ans:  True | | |Difficulty:  1 | Short Answer |86. |Explain how data from the linkage groups of the mouse can be used as a resource for assessing human linkage groups. |Ans: |Because virtually all genes cloned from the mouse genome are conserved in the human genome and vice versa, it is | | |possible to construct linkage maps for the two genomes from the same set of markers. Comparisons of the mouse and human| | |linkage groups allow one to see a picture somewhere between complete correspondence and unrelatedness. Genes closely | | |linked in the mouse tend to be closely linked in humans, but genes that are less tightly linked in one species tend not| | |to be linked at all in the other. This shows that even though mice and humans diverged about 65 million years ago, the | | |DNA sequences in many regions are very similar. | |Difficulty:  4 | |87. |Explain the differences between chromosomal rearrangements and changes in chromosome number. Cite at least one example of each. | |Ans: |Chromosomal rearrangements reorganize the DNA sequences within one or more chromosomes. Changes in chromosome number | | |involve losses or gains of entire chromosomes or sets of chromosomes. (Student may cite as an example of | | |rearrangements: deletion, duplication, inversion, translocation, and transposable elements. For changes in chromosome | | |number student may cite an aneuploidy such as a monosomy or trisomy, monoploidy, or polyploidy. | | |Difficulty:  4 | |88. |Describe how an inversion heterozygote can reduce the number of recombinant progeny. | |Ans: |When inversion heterozygotes have chromosomes pair up during meiosis, an inversion loop is formed to allow the tightest| | |possible alignment of homologous regions. This always produces aberrant recombinant chromatids. Two inversion cases are| | |possible – pericentric and paracentric. In a pericentric crossover within the inversion loop each recombinant will | | |carry a duplication of one region and a deletion of another. This abnormal dosage of some genes will result in abnormal| | |gametes and if they fertilize normal gametes, zygotes may die because of genetic imbalance. In a paracentric crossover | | |within the inversion loop the recombinant chromatids will be unbalanced in both gene dosage and centromere number. | | |(Student may then explain how centromere number can result in genetically unbalanced gametes such as what acentric and | | |dicentric chromatids would produce. ) | | |Difficulty:  4 | |89. Discuss the several effects that translocations and inversions have in common. | |Ans: |Both translocations and inversions change genomic position without affecting the total amount of DNA. If a breakpoint | | |of either one is within a gene, the gene function may be altered or lost. Both types may produce genetically imbalanced| | |gametes that may negatively affect a zygote or developing embryo. (Student may explain at this point the differences | | |between how the imbalanced gametes are produced. ) Because both reduce viable progeny and heterozygotes, they may play a| | |role in speciation and evolution. | | |Difficulty:  4 | |90. Explain the possible effects that a transposable element may have on a gene. | |Ans: |Insertion of a transposable element near or within a gene can affect gene expression and alter phenotype. For example, | | |a B type hemophilia occurs after insertion of Alu into the gene encoding clotting factor IX. Secondly, the effect of | | |insertion depends on what the element is and where the insertion point is. If insertion is into a protein-coding exon, | | |the reading frame may shift or a stop codon may be introduced. Insertion into an intron may lower the efficiency of | | |splicing, which may result in removal from the transcript that could lower production of a normal polypeptide. A stop | | |signal could also affect genes downstream. Upstream insertion into a regulatory gene could affect gene function in | | |various ways also. | | |Difficulty:  4 | |91. |Explain the mechanism by which aneuploidy occurs. | |Ans: |Aneuploidy occurs because of meiotic nondisjunction either in meiosis I or meiosis II. In meiosis I if homologs do not | | |separate all gametes produced will contain an error. Two of the gametes will contain both homologs and two will contain| | |neither. When fertilization of a normal gamete occurs by either of these abnormal gametes, aneuploidy results. Half of | | |the zygotes will be trisomic and half will be monosomic. Meiotic nondisjunction during meiosis II will produce two | | |normal and two abnormal gametes. If fertilization occurs with either of the abnormal gametes, aneuploid zygotes are | | |produced. | | |Difficulty:  4 | |92. |Discuss why triploid organisms are almost always sterile. | |Ans: |(Student may explain how triploids occur. ) Triploids are almost always sterile because meiosis produces mostly | | |unbalanced gametes. During the first meiotic division in a triploid germ cell, three sets of chromosomes must segregate| | |into two daughter cells. Most likely one daughter will end up with two chromosomes and the other will have only one of | | |any one set of homologs. Some cells will have two of some chromosomes and the normal one of others. Many combinations | | |of incorrect number of chromosomes will occur with very little chance of the normal amount. Most gametes will be | | |aberrant and will have a reduced chance of producing viable offspring. | | |Difficulty:  4 | |93. |Discuss how deletions and duplications may contribute to evolution. |Ans: |General examples of how chromosomal rearrangements might contribute to evolution: | | |Deletions – a small deletion that moves a coding sequence of one gene next to a promoter or other regulatory element of| | |an adjacent gene may, rarely, allow expression of a protein at a novel time in development or in a novel tissue. If the| | |new tim e or place of expression is advantageous to the organism, it might become established in the genome. | | |Duplications – a duplication will provide at least two copies of a gene. If one copy maintains the original function, | | |the other could conceivably acquire a new function that would probably be related to the original function. Many | | |examples can be seen in higher plants and animals. (Students may also write about the evolutionary contributions of the| | |other chromosomal rearrangements and might even mention the role of changes in chromosome number. ) | | |Difficulty:  4 | |94. |Why do inversions act as cross-over suppressors? | |Ans: |Inversions act as cross-over suppressors because only progeny that do not recombine within an inversion loop will | | |survive. | | |Difficulty:  4 | |95. |What is a balancer chromosome? |Ans: |A balancer chromosome is a special chromosome often created by the use of X-rays for the purpose of genetic | | |manipulation; these chromosomes often carry multiple, overlapping inversions that enable researchers to follow them | | |through crosses, and a recessive lethal mutation that prevents the survival of homozygotes. | | |Difficulty:  4 | |96. |What is the difference between alternate and an adjacent-1 segregation or an adjacent-2 segregation pattern? | |Ans: |An alternate segregation pattern results in balanced chromosomes while adjacent 1or 2 patterns yield chromosomes that | | |are unbalanced. | |Difficulty:  4 | Experimental Design and Interpretation of Data |97. |We now know that several organisms have a high degree of synteny at the genomic level. You wish to test the hypothesis that the| | |laboratory mouse and human share genomic similarities. What tests would you complete and given that we now know that the mouse | | |and human genomes are highly syntenic, what results would you expect? | |Ans: |Karyotype analysis can be used to test the hypothesis of genomic similarities however, only animals that have high | | |homology will show similar banding patterns. Therefore, FISH (fluorescence in situ hybridization) would be a more | | |useful technique to determine synteny. The mouse and human genomes are similar in that approximately 170 similar | | |fragments an average length of about 18 Mb are simply rearranged (this is not visible in a karyotype). | | |Difficulty:  4 | |98. |You are mapping traits in your favorite organism but unbeknownst to you, your laboratory model organism contains a rare | | |deletion. How will your mapping results be affected? |Ans: |The mapping distance will appear smaller than the actual physical distance in the wild-type organism. | | |Difficulty:  4 | |99. |You have discovered an altered phenotype and cloned the gene responsible. However, the gene you cloned appears to have an | | |unusual sequence in it. In order to determine the chromosomal location of your new gene, you perform FISH, using only the | | |unusual sequence, on several animals. To your surprise, the FISH results suggest that each an imal contains the gene on a | | |different chromosome. How would you interpret your results. |Ans: |The unusual sequence is a transposon and your â€Å"new†phenotype arose via the disruption of its gene by the transposon. | | |Difficulty:  4 | |100. |You are a master gardener and your favorite tomato plant is very sensitive to a pesticide called DEADBUG. You wish to make your| | |special tomato plants resistant to the pesticide which you spray on other bushes in your garden. Using microbial techniques | | |give sufficient and complete details of how you would do this (include ploidy status). | |Ans: |Haploid pollen grains are cold treated and plated on agar plates. The resulting embryoids are treated with hormone in | | |liquid culture and eventually grown as a monoploid plant. The plant is treated with a mutagen to induce mutations that | | |can result in insensitivity to the pesticide. Somatic cells are removed from the treated plant and plated on agar | | |containing DEADBUG. Only cells resistant to DEADBUG will grow. Again the embryoid is hormone treated and grown into a| | |resistant monoploid plant. Treatment with colchicine will allow duplication of chromosomes without separation | | |resulting in a normal diploid plant. | | |Difficulty:  4 | How to cite Hartwell Genetics Ch. 13 Study Guide, Papers
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