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
Problem 32
Textbook Question
Two parents who are each known to be carriers of an autosomal recessive allele have four children. None of the children has the recessive condition. What is the probability that one or more of the children is a carrier of the recessive allele?
![](/channels/images/assetPage/verifiedSolution.png)
1
<span>Step 1: Understand the genetic scenario. Each parent is a carrier of an autosomal recessive allele, meaning their genotype is heterozygous (Aa), where 'A' is the dominant allele and 'a' is the recessive allele.</span>
<span>Step 2: Determine the possible genotypes of the children. When two heterozygous parents (Aa) have children, the possible genotypes for each child are: AA (homozygous dominant), Aa (heterozygous carrier), and aa (homozygous recessive).</span>
<span>Step 3: Calculate the probability of each genotype for a single child using a Punnett square. The probabilities are: 1/4 for AA, 1/2 for Aa, and 1/4 for aa.</span>
<span>Step 4: Since none of the children have the recessive condition, we know none of them are aa. We need to find the probability that at least one child is a carrier (Aa). First, calculate the probability that a child is not a carrier (AA), which is 1/4.</span>
<span>Step 5: Use the complement rule to find the probability that at least one child is a carrier. Calculate the probability that none of the children are carriers (all are AA) and subtract from 1. The probability that all four children are AA is (1/4)^4. Subtract this from 1 to find the probability that at least one child is a carrier.</span>
Recommended similar problem, with video answer:
![](/channels/images/assetPage/verifiedSolution.png)
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.
Autosomal Recessive Inheritance
Autosomal recessive inheritance occurs when a trait or condition is expressed only when an individual has two copies of a recessive allele. Carriers, who possess one dominant and one recessive allele, do not exhibit the condition but can pass the recessive allele to their offspring. In this scenario, both parents are carriers, meaning each child has a chance of inheriting the recessive allele.
Recommended video:
Guided course
Autosomal Pedigrees
Punnett Square
A Punnett square is a diagram used to predict the genetic makeup of offspring from two parents. By organizing the possible alleles from each parent, it allows for the calculation of probabilities for various genotypes and phenotypes in the offspring. In this case, it can help determine the likelihood of the children being carriers of the recessive allele.
Recommended video:
Guided course
Chi Square Analysis
Probability of Inheritance
The probability of inheritance refers to the likelihood that a specific genetic trait will be passed from parents to offspring. In the context of this question, it involves calculating the chances that at least one of the four children inherits the recessive allele from the carrier parents. This can be determined using the probabilities derived from the Punnett square and the principles of independent assortment.
Recommended video:
Guided course
Probability
Related Videos
Related Practice