Alleles of the IGF-1 gene in dogs, encoding insulin-like growth factor, largely determine whether a domestic dog will be large or small. Dogs with an ancestral dominant allele are large, whereas dogs homozygous for the mutant recessive allele are small. Chondrodysplasia, a short-legged phenotype (as in dachshunds and basset hounds), is caused by a dominant gain-of-function allele of the FGF4 gene. The MSTN gene encodes myostatin, a regulator of muscle development. Dogs with a dominant ancestral allele of the MTSN gene have normal muscle development, while dogs homozygous for recessive mutants in the MTSN gene are 'double muscled' and have trouble running quickly. However, dogs heterozygous for the mutant allele run faster than either of the homozygotes.
You breed a pure-breeding small basset hound of normal musculature with a pure-breeding 'bully' whippet, a double-muscled large dog with normal legs.
If the F₁ of this cross is interbred, what proportion of the F₂ are expected to be fast runners and what proportion normal-speed runners?
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
2. Mendel's Laws of Inheritance
Dihybrid Cross
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Two organisms with the genotypes Aa bb Cc Dd Ee and Aa Bb Cc dd Ee were crossed. Use the branch method to determine the proportion of the following genotypes in the offspring. I. aa bb cc dd ee
A
1/256
B
1/64
C
1/16
D
1/4

1
Identify the genotypes of each parent for each gene locus. Parent 1 has Aa bb Cc Dd Ee and Parent 2 has Aa Bb Cc dd Ee.
Determine the possible gametes for each gene locus separately using the branch method. For example, for the 'A' locus, both parents can produce gametes with 'A' or 'a'.
Calculate the probability of each offspring genotype by multiplying the probabilities of inheriting each allele from the parents. For example, the probability of 'aa' is the product of the probabilities of inheriting 'a' from both parents.
Repeat the process for each gene locus ('b', 'c', 'd', 'e') to find the probability of the offspring having the genotype 'aa bb cc dd ee'.
Multiply the probabilities obtained for each locus to find the overall probability of the offspring having the genotype 'aa bb cc dd ee'.
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