A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be - PowerPoint PPT Presentation

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A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be

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A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be a. if the tall plant was TT? – PowerPoint PPT presentation

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Title: A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be


1
  • A tall pea plant is crossed with a recessive
    dwarf pea plant. What will the phenotypic and
    genotypic ratio of offspring be
  • a. if the tall plant was TT?
  • b. if the tall plant was Tt?
  • 2. A true-breeding tall, purple-flowered pea
    plant (TTPP) is crossed with a true-breeding
    dwarf, white flowered plant (ttpp).
  • a. What is the phenotype of the F1 generation?
  • b. What is the genotype of the F1 generation?
  • c. What four types of gametes are formed by the
    F1 plants?
  • d. Create and solve a punnett square to show the
    offspring of the F2 generation.
  • d. List the phenotypes and ratios found in the
    F2 generation.

2
  • 3. In the following cross, what is the
    probability of obtaining offspring that show all
    three dominant traits, A_B_C_ ( _ indicates that
    the second allele can be either dominant or
    recessive without affecting the phenotype
    determined by the first dominant allele)?
  • AaBbcc x AabbCC
  • a. probability of offspring that are A_B_C_ ?
  • b. What is the probability that the offspring
    of this AaBbcc x AabbCC cross will show at least
    two dominant traits?

3
  • 5. A dominant allele M is necessary for the
    production of the black pigment melanin, mm
    individuals are white. A dominant allele B
    results in the deposition of the a lot of pigment
    in an animals hair, producing a black color. The
    genotype bb produces brown hair. Two black
    animals heterozygous for both genes are bred.
    Using MmBb x MmBb
  • a. List the three phenotypes and possible
    genotypes
  • b. List the ratios for each of the three
    phenotypes.
  • Phenotype Genotype Ratio

4
Linked Genes
  • Given b gray vg normal wings
  • b black vg vestigial
    wings
  • Cross (Male) b b vg vg X (Female) b b vg
    vg
  • Mendels law of independent assortment would
    produce which phenotypes?
  • The actual results pg.280

5
  • 6. Determine the sequence of genes along a
    chromosome based on the following recombination
    frequencies
  • A-B, 8 map units
  • A-C, 28 map units
  • A-D, 25 map units
  • B-C, 20 map units
  • B-D, 33 map units

6
  • 7. A man with hemophilia (a recessive, sex linked
    condition) has a daughter of normal phenotype.
    She marries a man who is normal for the trait.
  • What is the probability that a daughter of this
    mating will be a hemophiliac?
  • That a son will be hemophiliac?
  • If the couple has four sons, what is the
    probability that all four will be born with
    hemophilia?

7
  • 8. Red-green colorblindness is caused by a SL
    recessive allele. A CB man marries a woman with
    normal vision whose father was CB. What is the
    probability of having a daughter who is also CB?
    What is the probability that their first son will
    be CB?
  • 9. A wild-type fruit fly (heterozygous for gray
    body color and normal wings) is mated with a
    black fly with vestigial wings. The offspring
    have the following phenotypic distribution wild
    type 778, black vestigial 785, black normal 158,
    gray vestigial 162. What is the recombination
    frequency between these genes for body color and
    wing size?
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