목차
- 1. Introduction to Genetics42m
- 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
2. Mendel's Laws of Inheritance
Monohybrid Cross
Multiple Choice
Human albinism is a simple recessive trait. Determine the genotypes of the parents for each offspring combination
i. A wild-type male and albino female have 6 wild-type children
A
AA x aa
B
Aa x Aa
C
aa x aa
D
AA X AA
E
AA x Aa
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검증된 단계별 안내1
Understand that albinism is a recessive trait, meaning that the phenotype is expressed only when the individual has two recessive alleles (aa).
Recognize that a wild-type phenotype indicates the presence of at least one dominant allele (A). Therefore, a wild-type individual could be either homozygous dominant (AA) or heterozygous (Aa).
Consider the offspring: 6 wild-type children. This suggests that all children have at least one dominant allele (A), meaning they cannot be homozygous recessive (aa).
Analyze the possible genotypes of the parents: The albino female must be homozygous recessive (aa) because albinism is expressed only in the homozygous recessive state.
Determine the genotype of the wild-type male: Since all offspring are wild-type, the male must be homozygous dominant (AA) to ensure that all children receive at least one dominant allele (A) from him.
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객관식
In the context of a monohybrid cross, what is a Punnett square used for?
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