9.E: The Cellular Basis of Inheritance (Exercises)
- Page ID
- 153945
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)9.1: Sexual Reproduction
Multiple Choice
What is a likely evolutionary advantage of sexual reproduction over asexual reproduction?
A. sexual reproduction involves fewer steps
B. less chance of using up the resources in a given environment
C. sexual reproduction results in greater variation in the offspring
D. sexual reproduction is more cost-effective
- Answer
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C
Which type of life cycle has both a haploid and diploid multicellular stage?
A. an asexual life cycle
B. diploid-dominant
C. haploid-dominant
D. alternation of generations
- Answer
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D
Which event leads to a diploid cell in a life cycle?
A. meiosis
B. fertilization
C. alternation of generations
D. mutation
- Answer
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B
Free Response
Explain the advantage that populations of sexually reproducing organisms have over asexually reproducing organisms?
- Answer
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The offspring of sexually reproducing organisms are all genetically unique. Because of this, sexually reproducing organisms may have more successful survival of offspring in environments that change than asexually reproducing organisms, whose offspring are all genetically identical. In addition, the rate of adaptation of sexually reproducing organisms is higher, because of their increased variation. This may allow sexually reproducing organisms to adapt more quickly to competitors and parasites, who are evolving new ways to exploit or outcompete them.
Describe the two events that are common to all sexually reproducing organisms and how they fit into the different life cycles of those organisms.
- Answer
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The two events common to all sexually reproducing organisms are meiosis and fertilization. Meiosis reduces a diploid cell to a haploid state. The haploid cell may divide mitotically to produce an organism, some of whose cells will combine during fertilization, or the haploid cells produced by meiosis may immediately combine in fertilization to produce a diploid cell that divides to produce an organism.
9.2: Meiosis
Multiple Choice
Meiosis produces ________ daughter cells.
A. two haploid
B. two diploid
C. four haploid
D. four diploid
- Answer
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C
At which stage of meiosis are sister chromatids separated from each other?
A. prophase I
B. prophase II
C. anaphase I
D. anaphase II
- Answer
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D
The part of meiosis that is similar to mitosis is ________.
A. meiosis I
B. anaphase I
C. meiosis II
D. interkinesis
- Answer
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C
If a muscle cell of a typical organism has 32 chromosomes, how many chromosomes will be in a gamete of that same organism?
A. 8
B. 16
C. 32
D. 64
- Answer
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B
Free Response
Explain how the random alignment of homologous chromosomes during metaphase I contributes to variation in gametes produced by meiosis.
- Answer
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Random alignment leads to new combinations of traits. The chromosomes that were originally inherited by the gamete-producing individual came equally from the egg and the sperm. In metaphase I, the duplicated copies of these maternal and paternal homologous chromosomes line up across the center of the cell to form a tetrad. The orientation of each tetrad is random. There is an equal chance that the maternally derived chromosomes will be facing either pole. The same is true of the paternally derived chromosomes. The alignment should occur differently in almost every meiosis. As the homologous chromosomes are pulled apart in anaphase I, any combination of maternal and paternal chromosomes will move toward each pole. The gametes formed from these two groups of chromosomes will have a mixture of traits from the individual’s parents. Each gamete is unique.
In what ways is meiosis II similar to and different from mitosis of a diploid cell?
- Answer
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The two divisions are similar in that the chromosomes line up along the metaphase plate individually, meaning unpaired with other chromosomes (as in meiosis I). In addition, each chromosome consists of two sister chromatids that will be pulled apart. The two divisions are different because in meiosis II there are half the number of chromosomes that are present in a diploid cell of the same species undergoing mitosis. This is because meiosis I reduced the number of chromosomes to a haploid state.
9.3: Variations in Meiosis
The number, size, shape, and banding pattern of chromosomes make them easily identifiable in a karyogram and allow for the assessment of many chromosomal abnormalities. Disorders in chromosome number, or aneuploidies, are typically lethal to the embryo, although a few trisomic genotypes are viable. Because of X inactivation, aberrations in sex chromosomes typically have milder effects on an individual. Aneuploidies also include instances in which segments of a chromosome are duplicated or deleted. Chromosome structures also may be rearranged, for example by inversion or translocation. Both of these aberrations can result in negative effects on development, or death. Because they force chromosomes to assume contorted pairings during meiosis I, inversions and translocations are often associated with reduced fertility because of the likelihood of nondisjunction.
Multiple Choice
The genotype XXY corresponds to:
A. Klinefelter syndrome
B. Turner syndrome
C. Triplo-X
D. Jacob syndrome
- Answer
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A
Abnormalities in the number of X chromosomes tend to be milder than the same abnormalities in autosomes because of ________.
A. deletions
B. nonhomologous recombination
C. synapsis
D. X inactivation
- Answer
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D
Aneuploidies are deleterious for the individual because of what phenomenon?
A. nondisjunction
B. gene dosage
C. meiotic errors
D. X inactivation
- Answer
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B
Free Response
Individuals with trisomy 21 are more likely to survive to adulthood than individuals with trisomy 18. Based on what you know about aneuploidies from this module, what can you hypothesize about chromosomes 21 and 18?
- Answer
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The problems caused by trisomies arise because the genes on the chromosome that is present in three copies produce more product than genes on chromosomes with only two copies. The cell does not have a way to adjust the amount of product, and the lack of balance causes problems in development and the maintenance of the individual. Each chromosome is different, and the differences in survivability could have to do with the numbers of genes on the two chromosomes. Chromosome 21 may be a smaller chromosome, so there are fewer unbalanced gene products. It is also possible that chromosome 21 carries genes whose products are less sensitive to differences in dosage than chromosome 18. The genes may be less involved in critical pathways, or the differences in dosage may make less of a difference to those pathways.
Contributors and Attributions
Remixed and/or curated from the following works:
Fowler, S., Roush, R., & Wise, J. (2013). 7.E The Cellular Basis of Inheritance ( Exercises). In Concepts of Biology. OpenStax (CC BY 4.0; Access for free at https://openstax.org/books/concepts-biology/pages/1-introduction(opens in new window)).