Chapter 11 Answers

  1. Let tt be the genotype of a short tassels, and rr is the genotype of pathogen resistant plants. We need to start with homozygous lines with contrasting combinations of alleles, for example:
    P: RRtt (pathogen sensitive, short tassels) × rrTT (pathogen resistant, long tassels)F1: RrTt (sensitive, long) × rrtt (resistant, short)F2:  parental  Rrtt (sensitive, short) , rrTt (resistant, long)recombinant rrtt (resistant, short) , RrTt (sensitive, long)
  2. Let mm be the genotype of a mutants that fail to learn, and ee is the genotype of orange eyes. We need to start with homozygous lines with contrasting combinations of alleles, for example (wt means wild-type):P: MMEE (wt eyes, wt learning) × mmee (orange eyes, failure to learn)F1: MmEe (wt eyes, wt learning) × mmee (orange eyes, failure to learn)F2: parental  MmEe (wt eyes, wt learning) , mmee (orange eyes, failure to learn)recombinant Mmee (wt eyes, failure to learn), mmEe (orange eyes, wt learning)
  3. Given a triple mutant aabbcc , cross this to a homozygote with contrasting genotypes, i.e. AABBCC, then test cross the trihybrid progeny, i.e.:P: AABBCC × aabbccF1:              AaBbCc × aabbccThen, in the F2 progeny, find the two rarest phenotypic classes; these should have reciprocal genotypes, e.g., aaBbCc and AAbbcc. Find out which of the three possible orders of loci (i.e. A-B-C, B-A-C, or B-C-A) would, following a double crossover that flanked the middle marker, produce gametes that correspond to the two rarest phenotypic classes. For example, if the rarest phenotypic classes were produced by genotypes aaBbCc and AAbbcc, then the dihybrid’s contribution to these genotypes was aBC and Abc. Since the parental gametes were ABC and abc, the only gene order that is consistent with aBC and Abc being produced by a double crossover flanking a middle marker is B-A-C (which is equivalent to C-A-B).
  4. Based on the information given, the recombinant genotypes with respect to these loci will be Aabb and aaBb. The frequency of recombination between A-B is 1cM=1%, based on the information given in the question, so each of the two recombinant genotypes should be present at a frequency of about 0.5%. Thus, the answer is 0.5%.
    1. 4cM>
    2. Random sampling effects; the same reason that many human families do not have an equal number of boys and girls.
  5. There would be approximately 2% of each of the recombinants: (yellow, straight) and (black, curved), and approximately 48% of each of the parentals: (yellow, curved) and (black, straight).
  6. A is fur colour locus             B is tail length locus                C is behaviour locus
    fur (A) tail (B) behaviour (C) Frequency   AB AC BC
    white short normal 16 aBC R R P
    brown short agitated 0 ABc P R R
    brown short normal 955 ABC P P P
    white short agitated 36 aBc R P R
    white long normal 0 abC P R R
    brown long agitated 14 Abc R R P
    brown long normal 46 AbC R P R
    white long agitated 933 abc P P P

    B                   C           A

    |————–|———|

    4.1cM       1.5cM

    Pairwise recombination frequencies are as follows (calculations are shown below):

    A – B           5.6%                A – C     1.5%                B – C     4.1%

    AB AC BC
    16 16 0
    0 0 0
    0 0 0
    36 0 36
    0 0 0
    14 14 0
    46 0 46
    0 0 0
    112 30 82
    5.6% 1.5% 4.1%

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Open Genetics by Natasha Ramroop Singh, Thompson Rivers University is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book