Table of contents
- 1. Introduction to Genetics51m
- 2. Mendel's Laws of Inheritance3h 37m
- 3. Extensions to Mendelian Inheritance2h 41m
- 4. Genetic Mapping and Linkage2h 28m
- 5. Genetics of Bacteria and Viruses1h 21m
- 6. Chromosomal Variation1h 48m
- 7. DNA and Chromosome Structure56m
- 8. DNA Replication1h 10m
- 9. Mitosis and Meiosis1h 34m
- 10. Transcription1h 0m
- 11. Translation58m
- 12. Gene Regulation in Prokaryotes1h 19m
- 13. Gene Regulation in Eukaryotes44m
- 14. Genetic Control of Development44m
- 15. Genomes and Genomics1h 50m
- 16. Transposable Elements47m
- 17. Mutation, Repair, and Recombination1h 6m
- 18. Molecular Genetic Tools19m
- 19. Cancer Genetics29m
- 20. Quantitative Genetics1h 26m
- 21. Population Genetics50m
- 22. Evolutionary Genetics29m
7. DNA and Chromosome Structure
Bacterial and Viral Chromosome Structure
2:54 minutes
Problem 13
Textbook Question
Textbook QuestionAssume that a viral DNA molecule is a 50-µm-long circular strand with a uniform 20-Å diameter. If this molecule is contained in a viral head that is a 0.08-µm-diameter sphere, will the DNA molecule fit into the viral head, assuming complete flexibility of the molecule? Justify your answer mathematically.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Dimensions
Understanding the dimensions of both the viral DNA and the viral head is crucial. The DNA is a circular strand measuring 50 µm in length and 20 Å in diameter, while the viral head is a sphere with a diameter of 0.08 µm. Converting these measurements into consistent units is essential for accurate comparison.
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Geometric Fitting
Geometric fitting involves determining whether one shape can fit within another based on their dimensions. In this case, we need to assess if the circular DNA strand can be accommodated within the spherical viral head, considering the DNA's flexibility and the head's volume.
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Mathematical Justification
Mathematical justification requires applying formulas to calculate the volume of the viral head and the effective space occupied by the DNA. By comparing these values, we can determine if the DNA can fit inside the viral head, thus providing a quantitative basis for the conclusion.
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