Linkage Recombination and the Mapping of Genes on Chromosomes III

  1. For diploid organisms such as Drosophila, mice, peas, humans etc, we do not know which, if any, of the parents' other progeny arose from gametes created in the same _______. Because of this limitation, the analysis of random products of meiosis in diploid organisms must be based on: 
    pg 149
    • meiosis
    • Statistical samplings of large populations
  2. In contrast, various species of fungi provide a unique opportunity for genetic analysis because they house all four haploid products of each meiosis in a sac called an ______. These haploid cells, or _________ can germinate and survive as viable haploid individuals that perpetuate themselve by _______
    • ascus (asci plural)
    • ascospores (halspores)
    • mitosis
  3. The phenotype of such haploid fungi is a direct representation of their genotype, without complications of _______
  4. The normally unicellular baker's yeast (Saccharomyces cerevisiae), is sold in supermarkets and contributes to the texture, shape, and flavor of bread; it generates ______ ascopores with each meiosis. The other, Neurospora crassa, is a mold that renders the bread on which it grows inedible; it too generates ______ ascopores with each meiosis. What changes at completion of meiosis?
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    • four 
    • four
    • At completion of meiosis, Neurospora crassa immediately divides once by mitosis to yield four pairs, for a total of eight haploid cells
  5. The two cells in each pair of Neurospora ascopores have the same genotype, because they arose from ________
  6. Haploid cells of both yeast and Neurospora normally reproduce vegetatively (________) by mitosis.
  7. However, sexual reproduction is possible for Neurospora. How?
    Because the haploid cells come in two mating types, and cells of opposite mating types can fuse to form a diploid zygote
  8. In yeast, these diploid cells are stable and can reproduce through successive ______ cycles. Stress, such as that caused by a scarcity of essential nutrients, induces what? What happens to the diploid zygote of the bread mold?
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    • successive mitotic cycles
    • Induces the diploid cells of yeast to enter meiosis. .
    • In bread mold the diploid zygote instead immediately undergoes meiosis, so the diploid state is only transient
  9. Mutations in haploid yeast and mold affect many different traits, including the appearance of the cells and their ability to grow under particular conditions. (2-story-example)
    • Yeast cells with the his4 mutation are unable to grow in the absence of the amino acid histidine
    • Yeast with the trp1 mutation cannot grow without an external source of the amino acid tryptophan
  10. Geneticists devised a system of representing genes that in which they use capital letters (HIS4) to designate ______ alleles and lowercase letters (his4) to represent _______ alleles
    • dominant 
    • recessive
  11. Most of the time, wild-type alleles are ______ and may also be represented by the alternative shorthand ____, while the symbol for the recessive alleles remains the lowercase abbrv (his4).
    • dominant
    • +
    • *Remember, however, that dominance or recessiveness is relevant only for diploid yeast cells, not for haploid cells that carry only one alle
  12. The assemblage of four ascopores (pairs) in a single ascus is called a _______
    • tetrad
    • *Note that this is a second meaning for tetrad, CH4 had tetrads as the four homologous chromatids vs here where it is the four products of a single meiosis held together in a sac
  13. In yeast, each tetrad is _______ (meaning) . In Neurospora crassa, each tetrad is ________ (meaning)
    • unordered: the four meiotic products (spores) are arranged at random within the ascus
    • ordered: with the four pairs, or eight haplospores, arranged in a line
  14. How do researchers analyze both unordered and ordered tetrads? (3-story)
    • Release the spores of each ascus
    • Induce the haploid cells to germinate under appropriate conditions 
    • Analyze the genetic makeup of the resulting haploid cultures
  15. With the aid of a dissecting microscope, how do the researchers recover the ascopores? What is the point?
    In the order in which they occur within the ascus and and thereby obtain additional info that is useful for mapping
  16. Tetrads can be characterized as (3)?
    • Parental ditypes (PDs)
    • Nonparental ditypes (NPDs)
    • Tetratypes (Ts)
  17. When diploid yeast cells heterozygous for each of two genes are induced to undergo meiosis, ______ tetrad types can be produced whether the two genes are on the same chromosome or different chromosomes.
  18. As we can see, the cross in which haploid cells of opposite mating types and with alternate alleles of two genes mate to form an Aa Bb diploid. State the three possibilities
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    • One possibility is that all four spores in the resulting tetrad is called a parental ditype (PD)
    • The second kind of tetrad, called a nonparental ditype (NPD), contains contains four recombinant spores, two of each type
    • The final possibility is a tetratype (T) tetrad, which contains four different kinds of spores (two recombinants [one of each type] and two parentals [one of each type])
    • *Note that the spores in each yeast ascus are not arranged in any particular order. The classification of a tetrad as PD, NPD, or T is based on the number of parental and recomb. spores in the ascus
  19. In order to determine RF between the two genes in the pic, you could simply (3):
    What does the RF equal?
    Image Upload 4
    • Break open all the spore cases
    • Pool the spores
    • Analyze them to determine which ones are parentals and which ones are recomb.

    The RF equals the number of recomb. spores divided by the total number of spores (parental plus recombs.) counted
  20. Alternatively, as you determine the genotype of each spore, you could: (2)
    • Keep track of which spores came from the same ascus 
    • Count instead the number of each type of tetrad (PD, NPD, or T)
  21. When the latter method is used, what is the formula for RF?
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  22. Analyzing the products of a fungal cross as tetrads has several advantages. Name two
    • In some fungi tetrad analysis enables you to determine the distance between genes and centromeres
    • *Analysis of tetrads enables you to develop a deeper appreciation for the events of meiosis
  23. What is the best way to understand tetrad analysis?
    Examine how the different tetrad types are generated when the two genes in a dihybrid are on different chromosomes and when they are on he same chromosome
  24. Consider a mating between haploid strain of yeast of mating type a, carrying the his4 mutation and the wild-type allele of the TRP1 gene, and a strain of the opposite mating type α that hass the genotype HIS4 trip1. The resulting a/α diploid cells are his4/HIS4 trp1/TRP1. Three different types of meiosis can take place, each of which produce a different tetrad type, name them:
    • A PD tetrad will result from one of the two random alignments of homologous chromosomes during meiosis I 
    • The equally likely alternative chromosome alignment produces an NPD tetrad 
    • T-Type tetrads are produced from a crossover between only one of the genes and the centromere of the chromosome on which it is found 
    • *See figure 5.21 on pg 152
  25. The results from the meioses reveal two important facts about tetrads:
    • First, because meiosis events (b) and (c) are equally likely, the number of PDs will equal the number of NPDs when the two genes are on different chromosomes
    • Second, as expected for alleles of genes on different chromosomes, RF = 50%
  26. If two genes in doubly heterozygous diploid yeast cells are on different chromosomes (unlinked), ______________________ ≃ _____________________
    • the number of PD tetrads ≃ the number of NPD tetrads
    • *not true in linked genes
  27. When yeast dihybrid for linked genes sporulate (that is undergo ________), the number of ____ produced far exceeds the number of ______
    • meiosis
    • PDs
    • NPDs
  28. Tell the story of this shot: 
    When genes are linked, _____ exceed ______
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    • PDs exceed NPDs
    • A haploid yeast strain containing the arg3 and ura2 mutations was mated to a wild-type ARG3 URA2 haploid strain
    • When the resultant diploid was induced to sporulate, the 200 tetrads produced 127 PD tetrads which far outnumbered the 3 NPD tetrads 
    • The distribution suggested that the two genes are linked
  29. If no crossing over occurs between the two genes, the resulting tetrad will be _____.  A single crossover between ARG3 and URA2 will generate a ________
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    • PD 
    • T tetrad
  30. In the case of double crossovers, there are actually four different possiblities, depending on which chromatids participate, and each of the four should occur with ______ frequency
  31. Tell the story of this shot (2)
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    • A double crossover involving only two chromatids generates a PD tetrad in (c)
    • Three-strand double crossovers results in a T tetrad
  32. Tell the story of this shot: (2)
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    • Three-strand double crossover (different type) resulting in a T tetrad (e)
    • (f) All four chromatids take part in the two crossovers (one crossover involves two strands and the other crossover, the other two strands), the resulting tetrad is NPD
  33. If two genes are linked, the only way to generate an NPD tetrad is trough a:
    four-strand double exchange
  34. Explain why if two genes are linked, the number of PDs must greatly exceed the number of NPDs
    When two genes are close together on a chromosome, meioses with one of the four kinds of double crossovers are much rarer than those with no crossing over or single crossovers, which produce PD and T tetrads, respectively
  35. Calculate the RF from the data in the figure, using the RF equation
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    • Image Upload 11
    • RF = [3 + (1/2)70]/200 * 100 = 19 m.u.
  36. The RF value is not an accurate reflection of the actual number of crossover events when two genes are far enough apart that DCOs occur (NPDs appear). State an example
    For example, the equation does not count PDs, and some of them are generated from DCO meioses
  37. In tetrad analysis, just as in en masse linkage analysis, two genes may be so far apart on the same chromosome that they will be ____________ from two genes on different chromosomes. What is equivalent in both cases
    • indistinguishable 
    • PD = NPD
  38. If two genes are sufficiently far apart on the chromosome, at least one ________ occurs between them during every meiosis. Under such circumstances, no meioses are NCOs, and thereore all PD tetrads as well as all NPD tetrads come from: (?)
    • crossover
    • Equally frequent kinds of DCOs
  39. Whether two genes are assorting independtly because they are on different chromosomes or because they are far apart on the same chromosome, the end result is the same:
    PD = NPD and RF = 50%
  40. Analyses of ordered tetrads, such as those produces by the bread mold Neurospora crassa, allow you to map the centromere of a chromosome relative to other genetic markers, infomration that you cannot normally obtain from _________ ______ __________
    unordered yeast tetrads
  41. Immediately after specialized haploid Neurospora cells of different mating types fus at fertilization, the _______ ________ undergoes meiosis within the confines of a narrow ascus (review 5.19 on pg 150)
    diploid zygote
  42. At the completion of meiosis, each of the _____ _______ meiotic products divides once by mitosis , yielding an ______ of _____ haploid ascopores. Dissection of the the ascus at this point allows one to determine the ________ of each of the eight haploid cells
    • four haploid
    • octad 
    • eight
    • phenotype
  43. The cross-sectional diameter of the ascus is so small that cells cannot:
    slip past each other
  44. During each division after feritlization, the microtubule fibers of the spindle extend outward from the __________ parallel to the long axis of the ascus
    These facts have two important repercussions: 
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    • centrosomes
    • First, when each of the four products of meiosis divides once by mitosis, the two genetically identical cells that result lie adjacent to each other 
    • Second, from the precise positioning of the four ascopore pairs within the ascus, you can infer the arrangement of the four chromatids of each homologous chromosome pair during the two meiotic divisions
  45. What allows you to count the octad of ascospores as four cell pairs of and analyze it as a tetrad?
    When each of the 4 products of meiosis divides once by mitosis, the resulting cells (two genetically identical) lie next to each other
  46. To understand the genetic consequences of the geometry of the ascospores, it is helpful to consider:
    what kinds of tetrads you would expect from the segregation of two alleles of a single gene
  47. The mutant ______ _____ allele (ws) alters ascospore color from wild-type black to white. In the absence of _________, the two alleles (___ and ____) separate from each other at the first meiotic division because the _________ to which they are attached separate at that stage
    • white-spore alleles
    • recombination
    • ws+ and ws
    • centromeres
  48. The second meiotic division and subsequent mitosis create ______ in which the top _____ ascospores are of one ________ (for instance ws+) and the bottom four of the others (ws)
    • asci 
    • four
    • genotype
  49. Whether the top four are ws+ and the bottom four ws, or vice versa, depends on what exactly?
    The random metaphase I orientation of the homologs that carry the gene relative to the long axis of the developing ascus
  50. The segregation of two alleles of a single gene at the first meiotic division is thus indicated by an ______ in which an imaginary line drawn between the ______ and the ______ ascospores of the octad cleanly separates ______ products bearing the two alleles. The figure display which segregation pattern?
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    • ascus
    • fourth and fifth
    • haploid
    • First division (MI) segregation pattern
  51. What if during meisois I, a crossover occurs in a heterozygote between the white-spore gene and the centromere of the chromosome on which it travels. The fig shows that this can lead to _____ equally possible ascospores arrangements, each one depending on a particular ________ of the four chromatids during the two meiotic divisions
    Image Upload 14
    • four
    • orientation
  52. In all four cases, both ws+ and ws spores are found on both sides of the imaginary line drawn between ascospores 4 and 5. (Why?)
    Octads carrying configurations of spores display a _____ ______ ____ segregation pattern
    • Because cells with only one kind of allele do not arise until the end of the second meiotic division
    • second division (MII) segregation pattern
  53. Why is it that the relative number of asci with this pattern can be used to determine the gene ↔ centromere distance?
    Because MII patterns result from meioses in which there has been a crossover between a gene and its centromere
  54. In an ascus showing MII segregation, one half of the ascospres are derived from where?
    The remaining half arise from where?
    • Chromatids that have exchanged parts
    • Chromatids that have not participated in crossovers leading to recombination 
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  55. Calculate the distance between a gene and its centromere (MII octads)
    gene ↔ centromere distance = (1/2) MII/ totals tetrads * 100
  56. In one experiment, a thr+ arg+ wild-tpe strain of Neurospra was crossed with a thr arg double mutant. What does each need to grow?
    • The thr mutants cannot grow in the absence of the amino acid threonine
    • The arg mutants cannot grow without a source of agrinine
    • Cells carrying the wild-type alleles of both genes can grow in a medium that contains neither amino acid
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  57. From this cross, ______ octads, considered here as tetrads, were obtained. These tetrads, were obtained. These tetrads were classified in seven different groups A, B, C, D, E, F, and G. What can we now find with these two genes? and How?
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    For each of the two genes, we can now find the distance between the gene and the centromere of the chromosome on which it is located

    • We count the number of tetrads with an MII pattern for that gene 
    • Drawing an imaginary line through the middle of the tetrads, we see that those in groups B, D, E, and G are the result of MII segregations for thr, while the remainder show MI pattenrs
  58. What is the centromere ↔ thr distance?
    Half the percentage of MII patterns = [(1/2) (16 + 2 + 2 + 1)/105] * 100 = 10 m.u.
  59. Similarly, the MII tetrads forate arg gene are in groups C, D, E, and G, so the distance between arg and its centromere is
    [(1/2) (11 + 2 + 2 + 1)/105] * 100 = 7.6 m.u.
  60. Is PD > NPD? Is it linked or unlinked?
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    • We find that groups A and G are PD, because all the ascospores show parental combinations, while groups E and F, with four recombinant spores, are NPD. PD is thus 72 +1 = 73, while NPD is 1+2 = 3 
    • The two genes are linked
  61. What is the map distance between thr and arg?
    • For this calculation we need to find the numbers of T and NPD tetrads. Tetratypes are found in groups B,C, and D, and wea lready know that groups E and F carry NPDs. Using the same formula for map distances as  the one previously used for yeast,
    • RF = [(3) + (1/2) (16 + 11 + 2 )/105] * 100 = 16.7 m.u.
  62. Why must the centromere lie between thr and arg, yielding the map
    arg ↔ thr (16.7 m.u.)
    arg ↔ centromere (7.6 m.u.)
    centromere ↔ thr (10 m.u.)
    Because the distance between thr and arg is larger than that separating either gene from the centromere
  63. The distance between the two genes calculated by the formula above (16.7 m.u.) is smaller than the sum of the two gene ↔ centromere distances (10.0 + 7.6 = 17.6 m.u.) (Why?)
    Because the formula does not account for all of the double crossovers. As always calculating map positions by adding shorter intervals produces the most accurate genetic maps. There gene ↔ centromere distances are shorter and are therefore more accurate than the thr/arg distance calculation in this example
  64. In 1936, the Drosophila geneticist Curt Stern inferred the existence of mitotic recombination from observations of "twin spots" in fruit flies. (Define twin spots)
    Twin spots: are adjacent islands of tissue that differ both from each other and from the tissue surrounding them
  65. The distinctive patches of twin spots arise from homozygous cells with a ________ phenotype growing amid a generally heterozygous cell population displaying the _______ phenotype
    • recessive
    • dominant
  66. In Drosophila, the yellow (y) mutation changes body color from normal brown to yellow, while the singed bristles (sn) mutation causes body bristles to be ______ and ______ rather than long and straight. Both of these genes are ________
    • short and curled 
    • X-linked
  67. In his experiments, Stern examined Drosophila females of genotype y sn+/y+ sn. These double heterozygous were generally wild type in appearance, but Stern noticed that some flies carried patches of _______ body color, others had small areas of ______ bristles and still others displayed twin spots: *context*(________________________________)
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    • yellow 
    • singed bristles
    • adjacent patches of yellow cells and cells with singed bristles
  68. What was Stern's assumption about the mistakes in mitotic division
    That they accompanied fly development and could have led to these mosaic animals containing tissues of different genotypes
  69. Individual yellow or singed patches could arise from _______ _____ or by ________ ________. These errors in mitosis would yield XO cells containing only y (but not y+) or sn (but not sn+) alleles; such cells would show one of the ________ phenotypes
    • chromosome loss or mitotic nondisjunction
    • recessive
  70. The twin spots must have a different origin. Stern reasoned that they represented the reciprocal products of mitotic crossing-over between the sn gene and the centromere. 
    Image Upload 20
    Image Upload 21
    • During mitosis in a diploid cell, after chromosomes occasionally (rarely) pair up with each other
    • While the chromosomes are paired, nonsister chromatids can exchange parts by crossing-over
    • The pairing is transient, and the homologous chromosomes soon resume their independent positions on the mitotic metaphase palte
    • There, the two chromosomes can line up relative to each other in either of two ways 
    • One of these orientations would yield two daughter cells that remain heterozygous for both genes and are thus indistinguishable from the surrounding wild-type cells
    • The other orientation, however, will generate two homozygous daughter cells: one y sn+/ y sn+, the other y+ sn / y+ sn
    • Because the two daughter cells would lie next to each other, subsequent mitotic divisions would produce adjacent patches of y and sn tissue (that is, twin spots)
    • *Note that if crossing over occurs between sn and y single spots of yellow tissue can form but a reciprocal singed spot cannot be generated in this fashion
  71. Diploid yeast cells that are heterozygous for one or more genes exhibit mitotic recombination in the form of _______ (define)
    sectors: portions of a growing colony that have a different genotype than the remainder of the colony
  72. If a diploid yeast cell of genotype ADE2/ade2 is placed on a petri plate, its _______ descendants wil grow into a colony. Usually, such colonies will appear white (Why?)
    • mitotic
    • The dominant wild type ADE2 allele specifies that color
  73. However, many colonies will contain red sectors of diploid ade2/ade2 cells. These cells are red because?
    A block in the adenine biosynthesis pathway causes them to accumulate red pigment. The red sectors arose as a result of mitotic recombination events between the ADE2 gene and it s centromere
  74. Homozygous ADE2/ADE2 cellls will also be produced by the same event, but they cannot be distinguished from heterozygotes. Why is that?
    Both types of cells are white
  75. The size of the red sectors indicates when ________ __________ took place. If they are large, it happened ______ in the growth of the colony. How much time does this give the resulting daughter cells to proliferate? What if they are small?
    • mitotic recombination 
    • early 
    • A long time
    • The recombination happened later
Card Set
Linkage Recombination and the Mapping of Genes on Chromosomes III