Table of contents
- 1. Introduction to Biology2h 42m
- 2. Chemistry3h 40m
- 3. Water1h 26m
- 4. Biomolecules2h 23m
- 5. Cell Components2h 26m
- 6. The Membrane2h 31m
- 7. Energy and Metabolism2h 0m
- 8. Respiration2h 40m
- 9. Photosynthesis2h 49m
- 10. Cell Signaling59m
- 11. Cell Division2h 47m
- 12. Meiosis2h 0m
- 13. Mendelian Genetics4h 44m
- Introduction to Mendel's Experiments7m
- Genotype vs. Phenotype17m
- Punnett Squares13m
- Mendel's Experiments26m
- Mendel's Laws18m
- Monohybrid Crosses19m
- Test Crosses14m
- Dihybrid Crosses20m
- Punnett Square Probability26m
- Incomplete Dominance vs. Codominance20m
- Epistasis7m
- Non-Mendelian Genetics12m
- Pedigrees6m
- Autosomal Inheritance21m
- Sex-Linked Inheritance43m
- X-Inactivation9m
- 14. DNA Synthesis2h 27m
- 15. Gene Expression3h 20m
- 16. Regulation of Expression3h 31m
- Introduction to Regulation of Gene Expression13m
- Prokaryotic Gene Regulation via Operons27m
- The Lac Operon21m
- Glucose's Impact on Lac Operon25m
- The Trp Operon20m
- Review of the Lac Operon & Trp Operon11m
- Introduction to Eukaryotic Gene Regulation9m
- Eukaryotic Chromatin Modifications16m
- Eukaryotic Transcriptional Control22m
- Eukaryotic Post-Transcriptional Regulation28m
- Eukaryotic Post-Translational Regulation13m
- 17. Viruses37m
- 18. Biotechnology2h 58m
- 19. Genomics17m
- 20. Development1h 5m
- 21. Evolution3h 1m
- 22. Evolution of Populations3h 52m
- 23. Speciation1h 37m
- 24. History of Life on Earth2h 6m
- 25. Phylogeny2h 31m
- 26. Prokaryotes4h 59m
- 27. Protists1h 12m
- 28. Plants1h 22m
- 29. Fungi36m
- 30. Overview of Animals34m
- 31. Invertebrates1h 2m
- 32. Vertebrates50m
- 33. Plant Anatomy1h 3m
- 34. Vascular Plant Transport1h 2m
- 35. Soil37m
- 36. Plant Reproduction47m
- 37. Plant Sensation and Response1h 9m
- 38. Animal Form and Function1h 19m
- 39. Digestive System1h 10m
- 40. Circulatory System1h 57m
- 41. Immune System1h 12m
- 42. Osmoregulation and Excretion50m
- 43. Endocrine System1h 4m
- 44. Animal Reproduction1h 2m
- 45. Nervous System1h 55m
- 46. Sensory Systems46m
- 47. Muscle Systems23m
- 48. Ecology3h 11m
- Introduction to Ecology20m
- Biogeography14m
- Earth's Climate Patterns50m
- Introduction to Terrestrial Biomes10m
- Terrestrial Biomes: Near Equator13m
- Terrestrial Biomes: Temperate Regions10m
- Terrestrial Biomes: Northern Regions15m
- Introduction to Aquatic Biomes27m
- Freshwater Aquatic Biomes14m
- Marine Aquatic Biomes13m
- 49. Animal Behavior28m
- 50. Population Ecology3h 41m
- Introduction to Population Ecology28m
- Population Sampling Methods23m
- Life History12m
- Population Demography17m
- Factors Limiting Population Growth14m
- Introduction to Population Growth Models22m
- Linear Population Growth6m
- Exponential Population Growth29m
- Logistic Population Growth32m
- r/K Selection10m
- The Human Population22m
- 51. Community Ecology2h 46m
- Introduction to Community Ecology2m
- Introduction to Community Interactions9m
- Community Interactions: Competition (-/-)38m
- Community Interactions: Exploitation (+/-)23m
- Community Interactions: Mutualism (+/+) & Commensalism (+/0)9m
- Community Structure35m
- Community Dynamics26m
- Geographic Impact on Communities21m
- 52. Ecosystems2h 36m
- 53. Conservation Biology24m
12. Meiosis
Genes & Alleles
Problem 6`
Textbook Question
Plant species A has a diploid chromosome number of 12. Plant species B has a diploid number of 16. A new species, C, arises as an allopolyploid from A and B. The diploid number for species C would probably be
a. 14
b. 16
c. 28
d. 56

1
Understand the concept of allopolyploidy: Allopolyploidy occurs when two different species hybridize and the resulting offspring have a combination of the chromosome sets from both parent species.
Identify the diploid chromosome numbers of the parent species: Species A has a diploid number of 12, and species B has a diploid number of 16.
Calculate the haploid chromosome numbers: For species A, the haploid number is half of the diploid number, which is 6. For species B, the haploid number is 8.
Combine the haploid numbers to determine the chromosome number of the allopolyploid species C: Add the haploid numbers from species A and B (6 + 8) to get the total chromosome number for species C.
Double the combined haploid number to find the diploid number for species C: Since species C is an allopolyploid, it will have a diploid number that is twice the sum of the haploid numbers from species A and B.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
1mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Diploid Chromosome Number
The diploid chromosome number refers to the total number of chromosomes present in a cell, which includes two sets, one from each parent. In the context of the question, species A has 12 chromosomes and species B has 16 chromosomes, indicating their respective diploid numbers.
Recommended video:
Guided course
Homologous Chromosomes
Allopolyploidy
Allopolyploidy is a form of polyploidy where a new species is formed from the combination of chromosomes from two different species. This process involves hybridization followed by chromosome doubling, resulting in a new species with a chromosome number that is the sum of the parent species' diploid numbers.
Recommended video:
Polyploidy
Chromosome Doubling
Chromosome doubling is a key mechanism in polyploidy, where the chromosome number of a hybrid organism is doubled, stabilizing the genome and allowing for successful reproduction. In the case of species C, the chromosome number would be the sum of species A and B's diploid numbers, doubled, leading to a diploid number of 28.
Recommended video:
Guided course
Homologous Chromosomes
Related Videos
Related Practice