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Linkage and Chromosome Mapping Practice Test (page 4)

By — McGraw-Hill Professional
Updated on Apr 25, 2014

Recombination Mapping With Tetrads Questions

Construct a map that includes these two genes.

  1. Given the adjoining meiotic metaphase orientation in Neurospora, determine the simplest explanation to account for the following spore patterns.
  2. A certain strain of Neurospora cannot grow unless adenine is in the culture medium. Adenineless is a recessive mutation (ad). Another strain produces yellow conidia (ylo). Below are shown the results from crossing these two strains. Calculate the map distance between these two genes.
  3. A riboflavineless strain (r) of Neurospora is crossed with a tryptophaneless strain (t) to give
  4. Two of the genes s, t, and u are linked; the third assorts independently and is very tightly linked to its centromere. Analyze the unordered tetrads produced by the cross (stu) × ( + + + ). (Hint: See Problem 6.10.)

Answers

Vocabulary

  1. tetrad analysis
  2. chiasma, or X (chi)
  3. crossing over
  4. coupling phase
  5. repulsion phase
  6. centimorgan
  7. coincidence, coincidence coefficient, or coefficient of coincidence
  8. interference
  9. balanced lethals
  10. tetrad analysis

Multiple-Choice

  1. d
  2. b
  3. d
  4. a
  5. b
  6. c
  7. d
  8. b
  9. c
  10. c

Recombination Among Linked Genes

  1. 63
  2. 30 units

Genetic Mapping

  1. net   (4)   ho   (7) ed(1)   ft(1)   dp (3.5)   cl (14.5)   d   (10)   J
  2. (a) sh c/sh c(shrunken; colorless) × sh+ c+/sh+ c+ (plump; colored)   (b) sh+ c+/sh c(coupling phase)   (c) 3.6 map units
  3. (a) 2/5   (b) 1/5
  4. (a) All open symbols = ee, all solid symbols = Ee   (b) re (II1, II2, III6), RE (II5, II9, III2-5, (7)   (c) All 10 cases   (d) 5 each   (e) 1/1024   (f) Yes   (g) 1/512   (h) RE/re (coupling phase)
  5. (a) 17 map units   (b) b-cn = 8.9 map units; cn-vg = 9.5 map units   (c) No. In part (a), two-point testcross cannot detect double crossovers that then appear as parental types, thus underestimating the true map distance.   (d) 0.89
  6. (a) c   (3.5)   sh     (18.4)     wx   (b) 86.1%
  7. (a) f   (b)Repulsion phase   (c)Coupling phase   (d)X chromosome (sex-linked)   (e) g–f = 12.0, fod = 2.5   (f) None

Linkage Estimates From F2 Data

  1. (a) Sl S/sl s (slow, silver males) × sl s/W (rapid, gold females)   (b) Coupling phase   (c) 14.2%
  2. 50%
  3. 20%
  4. (a) 52.5 rex : 17.5 normal   (b) Yes   (c) R1r2/r1R2, normal   (d) r1r2/r1r2, rex   (e) 17.14 map units

Use of Genetic Maps

  1. (a) 1408 yellow, mottled : 1408 rusty : 16 yellow, rusty : 16 mottled : 76 yellow, mottled, rusty : 76 wild type. Note: Rounding errors may allow one whole individual difference in each of these phenotypic classes.   (b) 32,651
  2. (a) 19,964 rough-eyed : 36 wild type   (b) Approximately 1 : 555   (c) Approximately 1 : 1.289
  3. (a)   (b)  (c)

Crossover Suppression

  1. (d) All three crosses = 25% mortality
  2. (a) 55% wild type : 20% black : 20% vestigial : 5% black, vestigial   (b),   (c) 50% wild type : 25% black : 25% vestigial
  3. 250 glueless, golden : 125 glueless : 125 golden
  4. 13 map units

Recombination Mapping With Tetrads

  1. A dominant trait may appear suddenly in a population by mutation of a recessive wild-type gene to a dominant allelic form. Such a mutant individual would be heterozygous.   (a) No   (b) No   (c) Yes   (d) No   (e) Chromosome 3   (f) Chromosome 4   (g) X chromosome (sex-linked)   (h) Chromosome 2
  2. (a) (1, 3)   (b) (2, 4)   (c) (1, 3)(1, 3) or (1, 3)(2, 4)   (d) (1, 3)(2, 3) or (1, 3)(1, 4)   (e) (2, 3)(1, 3) or (1, 4)(1, 3)   (f) (2, 3)(2, 3) or (1, 4)(2, 3)   (g) (2, 4)(1, 3) or (2, 4)(2, 4)   (h) (2, 4)(2, 3) or (2, 4)(1, 4)   (i) (1, 3)   (j) (2, 3)   (k) (2, 4)   (l) (1, 4)
  3. 5.83 map units
  4. (1.8)   r   (16.4)   t
  5. Genes u and s are linked on the same side of the centromere with u closest to the centromere. Centromere to u = 16 units, u to s - 14 units. Gene t is on another chromosome and very closely linked to its centromere.
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