A pea plant that has the genotype RrGgwwdd is crossed to a plant that has the rrGgWwDd genotype. The R gene controls round versus wrinkled seed, the G gene controls yellow versus green seed, the W gene controls purple versus white flower, and the D gene controls tall versus short plants. Determine the following;
What are the phenotypes of each plant?
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
Probability and Genetics
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
In a family of six children, where both parents are heterozygous for albinism, what is the probability that four are normal and two are albinos?
A
0.50, 50%
B
0.25, 25%
C
0.30, 30%
D
0.10, 10%

1
Identify the genetic makeup of the parents. Since both parents are heterozygous for albinism, their genotype is Aa, where 'A' is the normal allele and 'a' is the allele for albinism.
Determine the possible genotypes of the children using a Punnett square. The possible genotypes are AA, Aa, Aa, and aa. The probabilities are: 1/4 for AA, 1/2 for Aa, and 1/4 for aa.
Calculate the probability of a child being normal (AA or Aa). Since both AA and Aa result in a normal phenotype, the probability of a child being normal is 3/4.
Calculate the probability of a child being albino (aa). The probability of a child being albino is 1/4.
Use the binomial probability formula to find the probability of having exactly four normal children and two albino children in a family of six. The formula is: P(X=k) = (n choose k) * (p^k) * ((1-p)^(n-k)), where n is the total number of children, k is the number of normal children, and p is the probability of a child being normal.
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Probability and Genetics practice set
