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Cytogenetics Practice Problems Help (page 2)

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By — McGraw-Hill Professional
Updated on Aug 23, 2011

Practice 3

Assume that an autotetraploid of genotype AAaa has the A locus 50 ormore map units from the centromere, so that the equivalent of a single crossover always occurs between the centromere and the A locus. In this case, the chromatids will assort independently. Further assume that random assortment of chromatids to the gametes occurs by twos. Determine (a) the expected genotypic ratio of the progeny that results from selfing this autotetraploid, and (b) the expected increase in the incidence of heterozygous genotypes compared with selfed diploids of genotype Aa.

Solution 3

  1. Let each of the genes of the duplex tetraploid be labeled as shown in the illustration below. All capital letters represent identical dominant genes; all lowercase letters represent identical recessive alleles.
  2. Cytogenetics Solved Problems

    As in Practice 2, let us first use a Punnett square to determine the kinds and frequencies of different combinations of alleles (in pairs) expected in the gametes of the autotetraploid. Note in the illustration that (for example) alleles A1 and A2, originally on sister chromatids, can enter the same gamete if crossing over occurs between the centromere and the A locus. Likewise, any other pair of alleles could enter a gamete by the same mechanism (chromatid assortment). Random assortment (by twos) of eight chromatids of the autotetraploid during meiosis is shown in the following table. Note that the diagonal of the table represents the nonsense union of any given allele with itself (e.g., A1 with A1). The table is symmetrical on either side of the diagonal.

    Cytogenetics Solved Problems

    Ignoring the superscript identification of alleles, the expected ratio of possible diploid gametes is 6 AA: 16 Aa: 6 aa or 3 : 8: 3. Using these gametic expectations, we can now construct a zygotic checkerboard to generate the expected progeny. Summary of progeny genotypes: 9 AAAA (quadruplex) : 48 AAAa (triplex) : 82 AAaa (duplex) : 48 Aaaa (simplex) : 9 aaaa (nulliplex).

    Cytogenetics Solved Problems

  3. The total of the numbers (3 + 8 + 3) along the top or side of the preceding table is 14. Thus, the total of all crossproducts in the table is 14 × 14 = 196. Nine genotypes are homozygous AAA, and another nine are homozygous aaaa.Thus, all other genotypes (196 – 18 = 178) are heterozygotes. Selfing the autotetraploid produces 178/196 = 91% heterozygous progeny. Selfing a diploid of genotype Aa produces 50% heterozygous progeny. The increase from 50 to 91% is 41/50 = 82%

Practice 4

Pericarp is the outermost layer of the corn kernel and is maternal in origin. A dominant gene B produces brown pericarp, and its recessive allele b produces colorless pericarp. Tissue adjacent to the pericarp is aleurone (triploid). Purple pigment is deposited in the aleurone when the dominant gene C is present; its recessive allele c results in colorless aleurone. Aleurone is actually a single specialized layer of cells of the endosperm. The color of endosperm itself is modified by a pair of alleles. Yellow is governed by the dominant allele Y and white by the recessive allele y. Both C and Y show xenia to their respective alleles. A plant that is bbCcYy is pollinated by a plant of genotype BbCcYy. (a) What phenotypic ratio is expected among the progeny kernels? (b) If the F1 is pollinated by plants of genotype bbccyy, in what color ratio will the resulting F2 kernels be expected to appear?

Solution 4

  1. If pericarp is colorless, then the color of the aleurone shows through. If aleurone is also colorless, then the color of the endosperm becomes visible. Since the maternal parent is bb, the pericarp on all F1 seeds will be colorless. Any seeds with C will have purple aleurone. Only if the aleurone is colorless (ccc) can the color of the endosperm be seen.
  2. Cytogenetics Solved Problems

  3. Half of the F1 embryos is expected to be Bb and will thus lay down a brown pericarp around their seeds in the F2; the other half is expected to be bb and will envelop its seeds with a colorless pericarp. Thus, half of the seeds of the F1 plants will be brown. Of the remaining half that has colorless pericarp, we need show only as much of the genotype as is necessary to establish the phenotype.
  4. Cytogenetics Solved Problems

      Summary of F2 seed colors : 1/2 brown : 1/4 purple: 1/8 yellow: 1/8 white.

Practice 5

In 1931, Curt Stern found two different translocations in Drosophila from which he developed females possessing heteromorphic X chromosomes. One X chromosome had a piece of the Y chromosome attached to it; the other X was shorter and had a piece of chromosome IV attached to it. Two sex-linked genes were used as markers for detecting crossovers; the recessive trait carnation eye color (car) and the dominant trait bar eye (B). Dihybrid bar females with heteromorphic chromosomes (both mutant alleles on the X portion of the X-IV chromosome) were crossed to hemizygous carnation males with normal chromosomes. The results of this experiment provided cytological proof that genetic crossing over involves an actual physical exchange between homologous chromosome segments. Diagram the expected cytogenetic results of this cross showing all genotypes and phenotypes.

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