Syllabus sections » |
3.2 Chromosomes
Description
Nature of science: Developments in research follow improvements in techniques—autoradiography was used to establish the length of DNA molecules in chromosomes. (1.8) |
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Understandings:
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International-mindedness:
Syllabus and cross-curricular links: Biology Topic 1.6 Cell division Aims:
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Directly related questions
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17N.1.HL.TZ0.10:
The image shows a karyogram.
[Source: https://commons.wikimedia.org/wiki/File:Karyotype_of_sheep_(Ovis_aries).png, by M. Singh, X. Ma, E. Amoah and G. Kannan]
What information can be determined from this karyogram?
A. The sex is female.
B. The haploid number is 54.
C. Disjunction occurred during meiosis.
D. The species is not human.
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17N.1.HL.TZ0.10:
The image shows a karyogram.
[Source: https://commons.wikimedia.org/wiki/File:Karyotype_of_sheep_(Ovis_aries).png, by M. Singh, X. Ma, E. Amoah and G. Kannan]
What information can be determined from this karyogram?
A. The sex is female.
B. The haploid number is 54.
C. Disjunction occurred during meiosis.
D. The species is not human.
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21M.1.SL.TZ1.17:
A two-cell sea urchin (Echinoidea) embryo was physically separated by scientists into two cells. Each cell, through further embryonic development, became an adult sea urchin.
[Source: Clker-Free-Vector-Images/Pixabay.]
What is the relationship between the two adult sea urchins?
A. They are equivalent to non-identical twins.
B. Half of the genes would be the same.
C. Both adults would have haploid cells.
D. They are clones.
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21M.1.SL.TZ1.17:
A two-cell sea urchin (Echinoidea) embryo was physically separated by scientists into two cells. Each cell, through further embryonic development, became an adult sea urchin.
[Source: Clker-Free-Vector-Images/Pixabay.]
What is the relationship between the two adult sea urchins?
A. They are equivalent to non-identical twins.
B. Half of the genes would be the same.
C. Both adults would have haploid cells.
D. They are clones.
- 21M.1.SL.TZ2.14: What feature of eukaryotic chromosomes distinguishes them from the chromosomes of...
- 21M.1.SL.TZ2.14: What feature of eukaryotic chromosomes distinguishes them from the chromosomes of...
- 21M.1.HL.TZ2.11: What feature of eukaryotic chromosomes distinguishes them from the chromosomes of...
- 21M.1.HL.TZ2.11: What feature of eukaryotic chromosomes distinguishes them from the chromosomes of...
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22M.2.HL.TZ2.2b.ii:
Explain Cairns’s technique to measure the length of the DNA molecule.
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22M.2.HL.TZ2.2b.ii:
Explain Cairns’s technique to measure the length of the DNA molecule.
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22M.2.HL.TZ2.b.ii:
Explain Cairns’s technique to measure the length of the DNA molecule.
- 22M.1.HL.TZ2.10: Where can the entire genome of an organism be found? A. In the DNA present in plasmids of a...
- 22M.1.HL.TZ2.10: Where can the entire genome of an organism be found? A. In the DNA present in plasmids of a...
- 22M.1.SL.TZ1.14: Chromosome numbers vary between species. Which statement refers to humans? A. An egg cell has 22...
- 22M.1.SL.TZ1.14: Chromosome numbers vary between species. Which statement refers to humans? A. An egg cell has 22...
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22M.1.SL.TZ2.14:
The karyogram shown belongs to a human being.
[Source: Reproduced from Tennakoon J, Kandasamy Y, Alcock G, Koh TH. Edwards syndrome with double trisomy. Singapore Med J. 2008 Jul;49(7):e190-1. PMID: 18695855.]
What can be deduced from this karyogram?
A. The person is a male with Down syndrome.
B. The person is a female with Down syndrome.
C. The person is a male with a genetic disorder.
D. The person is a female with a missing chromosome.
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22M.1.SL.TZ2.14:
The karyogram shown belongs to a human being.
[Source: Reproduced from Tennakoon J, Kandasamy Y, Alcock G, Koh TH. Edwards syndrome with double trisomy. Singapore Med J. 2008 Jul;49(7):e190-1. PMID: 18695855.]
What can be deduced from this karyogram?
A. The person is a male with Down syndrome.
B. The person is a female with Down syndrome.
C. The person is a male with a genetic disorder.
D. The person is a female with a missing chromosome.
- 18M.1.SL.TZ1.14: What is the same in all parts of homologous chromosomes? A. Base pair sequence B. Alleles C....
- 18M.1.SL.TZ1.14: What is the same in all parts of homologous chromosomes? A. Base pair sequence B. Alleles C....
- 18M.1.HL.TZ2.16: For what purpose is a karyogram used? A. To identify gene loci in a species B. To identify...
- 18M.1.HL.TZ2.16: For what purpose is a karyogram used? A. To identify gene loci in a species B. To identify...
- 18M.2.SL.TZ2.2b.ii: Distinguish between the chromosomes of eukaryotic cells and prokaryotic cells.
- 18M.2.SL.TZ2.2b.ii: Distinguish between the chromosomes of eukaryotic cells and prokaryotic cells.
- 18M.2.SL.TZ2.b.ii: Distinguish between the chromosomes of eukaryotic cells and prokaryotic cells.
- 19M.2.SL.TZ1.6a: Identify, with a reason, the sex of this individual.
- 19M.2.SL.TZ1.6a: Identify, with a reason, the sex of this individual.
- 19M.2.SL.TZ1.a: Identify, with a reason, the sex of this individual.
- 19M.2.SL.TZ1.6b: Identify the chromosome that is affected by a trisomy in this individual, naming the condition...
- 19M.2.SL.TZ1.6b: Identify the chromosome that is affected by a trisomy in this individual, naming the condition...
- 19M.2.SL.TZ1.b: Identify the chromosome that is affected by a trisomy in this individual, naming the condition...
- 19M.2.SL.TZ2.2b: Outline the use of a karyogram during pregnancy.
- 19M.2.SL.TZ2.2b: Outline the use of a karyogram during pregnancy.
- 19M.2.SL.TZ2.b: Outline the use of a karyogram during pregnancy.
- 19M.3.SL.TZ2.2b: Determine, with a reason, the nucleotide base that was marked with...
- 19M.3.SL.TZ2.2b: Determine, with a reason, the nucleotide base that was marked with...
- 19M.3.SL.TZ2.b: Determine, with a reason, the nucleotide base that was marked with...
- 19M.3.SL.TZ2.2a: Estimate the length of the molecule of DNA shown in the autoradiogram between the two...
- 19M.3.SL.TZ2.2a: Estimate the length of the molecule of DNA shown in the autoradiogram between the two...
- 19M.3.SL.TZ2.a: Estimate the length of the molecule of DNA shown in the autoradiogram between the two...
- 19N.2.SL.TZ0.3c: List three characteristics of eukaryotic homologous chromosomes.
- 19N.2.SL.TZ0.3c: List three characteristics of eukaryotic homologous chromosomes.
- 19N.2.SL.TZ0.c: List three characteristics of eukaryotic homologous chromosomes.
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19N.1.SL.TZ0.14:
A pregnant woman had fetal cells removed by chorionic villus sampling and tested. The following karyogram was produced.
[Source: Mediscan / Alamy Stock Photo]
What does this show?
A. The child is female with Down syndrome.
B. The child is female without Down syndrome.
C. The child is male with Down syndrome.
D. The child is male without Down syndrome.
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19N.1.SL.TZ0.14:
A pregnant woman had fetal cells removed by chorionic villus sampling and tested. The following karyogram was produced.
[Source: Mediscan / Alamy Stock Photo]
What does this show?
A. The child is female with Down syndrome.
B. The child is female without Down syndrome.
C. The child is male with Down syndrome.
D. The child is male without Down syndrome.
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18N.1.HL.TZ0.11:
The image shows chromosomes from an insect (2 n = 8).
Which pair of chromosomes are the sex chromosomes of this insect species?
A. c and e
B. a and h
C. f and h
D. a and e
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18N.1.HL.TZ0.11:
The image shows chromosomes from an insect (2 n = 8).
Which pair of chromosomes are the sex chromosomes of this insect species?
A. c and e
B. a and h
C. f and h
D. a and e
- 19M.1.HL.TZ1.11: What technique was used by John Cairns to measure the length of the DNA molecule in Escherichia...
- 19M.1.HL.TZ1.11: What technique was used by John Cairns to measure the length of the DNA molecule in Escherichia...
- 19M.2.HL.TZ1.2a: Distinguish between the structure of the chromosomes of prokaryotes and eukaryotes.
- 19M.2.HL.TZ1.2a: Distinguish between the structure of the chromosomes of prokaryotes and eukaryotes.
- 19M.2.HL.TZ1.a: Distinguish between the structure of the chromosomes of prokaryotes and eukaryotes.
- 19M.2.HL.TZ1.2c.i: Identify, with a reason, the sex of this individual.
- 19M.2.HL.TZ1.2c.i: Identify, with a reason, the sex of this individual.
- 19M.2.HL.TZ1.c.i: Identify, with a reason, the sex of this individual.
- 19N.2.HL.TZ0.6b: Outline the structural and genetic characteristics of eukaryotic chromosomes.
- 19N.2.HL.TZ0.6b: Outline the structural and genetic characteristics of eukaryotic chromosomes.
- 19N.2.HL.TZ0.b: Outline the structural and genetic characteristics of eukaryotic chromosomes.
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20N.1.HL.TZ0.13:
Testing the chromosomes of a girl with Turner syndrome produced the following karyogram.
[Source: Turner’s syndrome karyotype 45,XO. This female lacks the second X chromosome present in the
normal karyotype. Symptoms include short stature, neck webbing, elbow deformity, widely spaced nipples with shield chest, primary amenorrhea, sexual infantilism and sterility. The ovaries are reduced to fibrous streaks. Also known as XO syndrome or ovarian short-stature syndrome. Credit: Wessex Reg. Genetics Centre. Attribution 4.0 International (CC BY 4.0).]The condition can result from non-disjunction occurring in anaphase I of meiosis in an egg cell. Two cells result from the first division, one of which would lead to Turner syndrome. Which chromosomes will be in the other cell (polar body) at the end of meiosis I?
A. 44 autosomes and X
B. 44 autosomes and XX
C. 22 autosomes and X
D. 22 autosomes and XX
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20N.1.HL.TZ0.13:
Testing the chromosomes of a girl with Turner syndrome produced the following karyogram.
[Source: Turner’s syndrome karyotype 45,XO. This female lacks the second X chromosome present in the
normal karyotype. Symptoms include short stature, neck webbing, elbow deformity, widely spaced nipples with shield chest, primary amenorrhea, sexual infantilism and sterility. The ovaries are reduced to fibrous streaks. Also known as XO syndrome or ovarian short-stature syndrome. Credit: Wessex Reg. Genetics Centre. Attribution 4.0 International (CC BY 4.0).]The condition can result from non-disjunction occurring in anaphase I of meiosis in an egg cell. Two cells result from the first division, one of which would lead to Turner syndrome. Which chromosomes will be in the other cell (polar body) at the end of meiosis I?
A. 44 autosomes and X
B. 44 autosomes and XX
C. 22 autosomes and X
D. 22 autosomes and XX
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21M.2.HL.TZ2.7a:
Describe the structure of the DNA molecule.
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21M.2.HL.TZ2.7a:
Describe the structure of the DNA molecule.
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21M.2.HL.TZ2.a:
Describe the structure of the DNA molecule.
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21M.1.HL.TZ2.12:
The image shows tetrads in the anther of a lily. A tetrad is a group of four cells, produced when one mother cell divides by meiosis. The tetrad indicated by the arrow contains a total of 48 chromosomes.
[Source: [Tetrad], 2012. [image online] Available at: https://www.iasprr.org/old/iasprr-pix/lily/tetrad.jpg [accessed: 4 April 2019]. Photo courtesy of Professor Scott D. Russell.]
What is the diploid number of the plant?
A. 12
B. 24
C. 48
D. 96
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21M.1.HL.TZ2.12:
The image shows tetrads in the anther of a lily. A tetrad is a group of four cells, produced when one mother cell divides by meiosis. The tetrad indicated by the arrow contains a total of 48 chromosomes.
[Source: [Tetrad], 2012. [image online] Available at: https://www.iasprr.org/old/iasprr-pix/lily/tetrad.jpg [accessed: 4 April 2019]. Photo courtesy of Professor Scott D. Russell.]
What is the diploid number of the plant?
A. 12
B. 24
C. 48
D. 96
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21N.1.SL.TZ0.13:
John Cairns used the technique of autoradiography to produce photographs of DNA from the bacterium E. coli.
[Source: © Cold Spring Harbor Laboratory Press. Autoradiography of bacterium E. coli DNA - micrograph, The Chromosome of
Escherichia coli Cairns, J.P., 1963. Cold Spring Harbor Symposia, Quantitative Biology, 28(44).]Which conclusion was drawn from his experiments?
A. The DNA in all organisms is circular.
B. DNA in E. coli naturally contains thymidine.
C. DNA replication is conservative.
D. The DNA in E. coli is 900 μm in length.
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21N.1.SL.TZ0.13:
John Cairns used the technique of autoradiography to produce photographs of DNA from the bacterium E. coli.
[Source: © Cold Spring Harbor Laboratory Press. Autoradiography of bacterium E. coli DNA - micrograph, The Chromosome of
Escherichia coli Cairns, J.P., 1963. Cold Spring Harbor Symposia, Quantitative Biology, 28(44).]Which conclusion was drawn from his experiments?
A. The DNA in all organisms is circular.
B. DNA in E. coli naturally contains thymidine.
C. DNA replication is conservative.
D. The DNA in E. coli is 900 μm in length.