Chapter 11: MENDELIAN GENETIC PROBLEMS (A)

Solutions:

1. A Test Cross is used to determine the genotype of the tall pea plant. Tall plants can have the genotype TT or Tt. Short plants have the genotype tt.

The gardener needs to cross each of the tall plants with homozygous recessive short plants.

Now the gardener can determine if the tall plants are TT or Tt, based on the presence of any short plants in the phenotype of the offspring.

2. For this one, a cross is performed between a rabbit homozygous dominant for black coat color, and another rabbit homozygous recessive for brown coat color.

Gametes: BB x bb

Genotypic ratios: Bb 100%

Phenotypic ratios: Black coat color 100%

3. Try to determine the genotype of both parents if all of the offspring are to be smooth coated guinea pigs. In order for the smooth coat to show up phenotypically in the offspring, the genotype of each offspring must be rr.

4. Codominance - A situation in which both alleles of a gene contribute to the phenotype of the organism. RR genotype will yield a red flower. R'R' will yield a white flower. RR' will yield a pink flower. Let's cross a plant with red flowers and one with pink flowers.

Gametes: RR x RR'

Genotypic ratios: RR 50%, RR' 50%

Phenotypic ratios: red flowers 50%, pink flowers 50%

5. A dihybrid cross between two heterozygous parents will yield the following phenotypic ratios:   9:16  dominant, dominant,   3:16 dominant, recessive   3:16 recessive, dominant and 1:16 recessive, recessive.

In this problem the phenotypic ratios of the actual offspring obtained are listed as 20:total dominant, dominant     7:total dominant, recessive      5:total recessive, dominant.  No results of a recessive, recessive.

Axial flowers are located near the stem of the pea plant. Terminal flowers are located near the tips of the branches or lateral stems of a pea plant. Inflated pods are plump, and constricted pods are shrivled.

Gametes: AaIi x AaIi

Each of these parents must be independently assorted. This is accomplished by performing FOIL (First, Outside, Inside, Last) on each.

AaIi    x     AaIi

Phenotypically, this gives us the following probable results:

9:16  Axial flowers, Inflated pods  

3:16 Axial flowers, constricted pods

3:16 terminal flowers, inflated pods

1:16 terminal flowers, constricted pods

6.

AAIi    x    AAII

AAII    x    AAII

AaII    x    AAII

AaIi    x    AAII

AAIi    x    AAIi

AaII    x    AAIi

AaIi    x    AAIi

AAIi    x    AaII

AiIi    x    AaIi

AaII    x    AaIi

7. The phenotypic ratio for this cross is 9 tall, round; 3 tall, wrinkled; 3 short, round; 1 short, wrinkled.

8. The phenotypic ratio is 6 red, tall; 6 pink, tall; 2 red, short; 2 pink, short.

   The genotypic ratio is 4 RRTt; 4 RR'Tt; 2 RRTT; 2 RR'TT; 2 RRtt; 2 RR'tt

9. The most probable explanation for your results is that the alleles for the red flower color and yellow flower color are codominant. You could prove this hypothesis by self-pollinating the orange plants. You should get a phenotypic ratio of 1 red; 2 orange; 1 yellow which would reflect a genotypic ratio of 1 RR; 2 RR'; 1 R'R'.

10. Yes, you could get rabbits with brown curly hair in the first generation if the male rabbit was heterozygous for both curly and brown hair. The other genotypes would not produce this. The Punnett squares of the four possible crosses are a follows:

10.a              BbSs    x    bbss

Genotypes: 4 BbSs; 4 Bbss; 4 bbSs; 4 bbss

Phenotypes: 4 black, straight; 4 black, curly; 4 brown, straight; 4 brown curly

10.b             BBSS    x    bbss

Genotypes: 100 % BbSs

Phenotypes: 100 % black, straight

10.c            BbSS    x    bbss

Genotypes: 50%  BbSs; 50%  bbSs

Phenotypes: 50% Black, straight; 50% brown, straight

10.d              BBSs    x    bbss

Genotypes: 50% BbSs; 50% Bbss

Phenotypes: 50% black, straight; 50% black, curly

11. For this genetic problem, show the gametes, each independently assorted gamete, completed Punnett square, genotypes

and their ratios, phenotypes and their ratios.  Cross a homozygous short pea plant with axial flowers with a short pea plant

homozygous for axial flowers.  Tallness and having axial flowers are dominant traits in pea plants.

       Gametes                                            Independently Assorted Gametes                  Genotypes               Phenotypes

    ttAA    x    ttAA                                                ttAA            x          ttAA                            ttAA   16:16            short, axial flowers          16:16

                                                                                                   tA                            tA                              

                                                                                                    tA                            tA                          

                                                                              tA                            tA               

                                                                              tA                            tA              

12. For this genetic problem, show the gametes, each independently assorted gamete, completed Punnett square, genotypes

and their ratios, phenotypes and their ratios.  Cross two pea plants, both heterozygous for Round seeds and axial flowers.

       Gametes                                            Independently Assorted Gametes                  Genotypes               Phenotypes

    RrAa    x    RrAa                                                RrAa            x          RrAa                    RRAA   1:16            Round seeds, Axial flowers          9:16

                                                                                                                                               RRAa   2:16            Round seeds, terminal flowers     3:16

                                                                                                    RA                            RA                    RrAA    2:16            wrinkled seeds, Axial flowers        3:16

                                                                                                    Ra                            Ra                    RrAa     4:16            wrinkled seeds, terminal flowers  1:16

                                                                              rA                              rA                     RRaa    1:16

                                                                              ra                              ra                     Rraa      2:16

                                                                                                                                         rrAA       1:16

                                                                                                                                         rrAa        2:16

                                                                                                                                         rraa        1:16