Skip to main content
Biology LibreTexts

1.9: Practice - Introduction to heredity

  1. Last updated
  2. Save as PDF
  • Page ID
    73826

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Query \(\PageIndex{1}\)

    Step-by-step solution
    1. Individuals with two of the same allele are considered homozygous, while individuals with two different alleles are heterozygous.

    2. If the dominant allele is present, then the dominant trait will be shown:

      WW - homozygous dominant (wide wings)
      Ww - heterozygous (wide wings)
      ww - homozygous recessive (thin wings)

    3. The correct answer is

      Ww - homozygous dominant

    Query \(\PageIndex{2}\)

    Step-by-step solution

    1. The genotype of an organism is represented by two letters (ex: LL, Ll, or ll).

      An organism's phenotype is its physical characteristic (ex: lactase persistence or lactose intolerance).

    2. An individual with a homozygous dominant genotype (LL) would show the dominant phenotype (lactase persistence).

      An individual with a heterozygous dominant genotype (Ll) would show the dominant phenotype (lactase persistence).

      An individual with a homozygous recessive genotype (ll) would show the recessive phenotype (lactose intolerance).

    3. The correct answer is

      ll - homozygous recessive - lactose intolerant

    Query \(\PageIndex{3}\)

    Step-by-step solution

    1. The only possible genotype for wacky wings is ww (homozygous recessive).

      The possible genotypes for normal wings are WW (homozygous dominant) and Ww (heterozygous).

    2. The only possible genotype for flashy feathers is ff (homozygous recessive).

      The possible genotypes for normal feathers are FF (homozygous dominant) and Ff (heterozygous).

    3. The correct answer is

      wwff; WWFF

    Query \(\PageIndex{4}\)

    Step-by-step solution
    1. Individuals that are heterozygous for a trait are called carriers. They have both the dominant and recessive allele in their genotype.
    2. The presence of at least one dominant allele will result in the expression of the dominant phenotype in the individual.

    3. The correct answer is

      Dominant and recessive alleles can be found in the same individual

      Dominant alleles mask the expression of recessive alleles

    Query \(\PageIndex{5}\)

    Step-by-step solution

    1. A dominant allele is denoted by an uppercase letter such as H.

      A recessive allele is denoted by a lowercase letter such as h.

    2. A homozygous genotype contains two identical alleles. An individual can either be homozygous dominant (ex: HH) or homozygous recessive (ex: hh).

      A heterozygous genotype contains two different alleles (ex: Hh).

    3. The correct answer is

      Heterozygous; one dominant allele and one recessive allele

    Query \(\PageIndex{6}\)

    Step-by-step solution

    1. The genotype of an organism is represented by two letters (ex: AA, Aa, or aa).

      An organism's phenotype is its physical characteristic (ex: black feathers or red feathers).

    2. A chicken with a homozygous dominant genotype (AA) would show the dominant phenotype (black feathers).

      A chicken with a heterozygous dominant genotype (Aa) would show the dominant phenotype (black feathers).

      A chicken with a homozygous recessive genotype (aa) would show the recessive phenotype (red feathers).

    3. The correct answer is

      aa - genotype; red feathers - phenotype

    Query \(\PageIndex{7}\)

    Step-by-step solution
    1. In order to have fancy fins, the possible genotypes are FF (homozygous) and Ff (heterozygous).
    2. In order to have shiny scales, the possible genotypes are SS (homozygous) and Ss (heterozygous).

    3. The correct answer is

      FFSS; FfSs

    Query \(\PageIndex{8}\)

    Step-by-step solution
    1. In a heterozygous organism, one of the alleles may completely determine the phenotype of the organism, masking the presence of the other allele.
    2. The allele that masks the presence of the other allele is the dominant allele. The masked allele is the recessive allele.

    3. In this example, both parents are heterozygous for brown coats (Bb x Bb).

      When these two dogs are crossed, some of the puppies have black coats because they inherited two recessive (b) alleles. Others have brown coats because they have at least one dominant (B) allele.

    4. The correct answer is

      The allele for brown coat is dominant to the allele for black coat.

    Query \(\PageIndex{9}\)

    Step-by-step solution
    1. Individuals with two of the same allele are considered homozygous, while individuals with two different alleles are heterozygous.

    2. If the dominant allele is present, then the dominant trait will be shown:

      TT - homozygous dominant (taster) Tt - heterozygous (taster)
      tt - homozygous recessive (non-taster)

    3. The correct answer is

      TT - homozygous dominant

    Query \(\PageIndex{10}\)

    Step-by-step solution

    1. The genotype of an organism is represented by two letters (ex: SS, Ss, or ss).

      An organism's phenotype is its physical characteristic (ex: smooth pods or constricted pods).

    2. A pea plant with a homozygous dominant genotype (SS) would show the dominant phenotype (smooth pods).

      A pea plant with a heterozygous dominant genotype (Ss) would show the dominant phenotype (smooth pods).

      A pea plant with a homozygous recessive genotype (ss) would show the recessive phenotype (constricted pods).

    3. The correct answer is

      ss - genotype; constricted pod - phenotype

    Query \(\PageIndex{11}\)

    Step-by-step solution
    1. In a heterozygous organism, one of the alleles may completely determine the phenotype of the organism, masking the presence of the other allele.
    2. The allele that masks the presence of the other allele is the dominant allele. The masked allele is the recessive allele.

    3. In this example, one plant is homozygous for the red allele (RR), while the other is homozygous for the white allele (rr).

      When these two plants are crossed, the offspring will all be heterozygous (Rr), with one of each allele. Because of the presence of the dominant R allele, the flowers will all be red.

    4. The correct answer is

      The allele for red flowers is dominant to the allele for white flowers.

    Query \(\PageIndex{12}\)

    Step-by-step solution
    1. Individuals that are heterozygous for a trait are called carriers. They have both the dominant and recessive allele in their genotype.
    2. The presence of at least one dominant allele will result in the expression of the dominant phenotype in the individual.

    3. The correct answer is

      An individual can have both dominant and recessive alleles

      Recessive traits are expressed only if the dominant allele is absent

    Contributors and Attributions

    • Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)

    1.9: Practice - Introduction to heredity is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts.