Albinism, caused by a mutational disruption in melanin (skin pigment) production, has been observed in many species, including humans. In 1991, and again recently in 2017, the only documented observations of an albino humpback whale (named 'Migaloo') were observed near New South Wales. Recently, Polanowski and coworkers (Polanowski, A., S. Robinson-Laverick, and D. Paton. (2012). Journal of Heredity 103:130–133) studied the genetics of humpback whales from the east coast of Australia, including Migaloo. Assuming that Migaloo's albinism is caused by a rare dominant gene, what would be the likelihood of the establishment of a natural robust subpopulation of albino white humpback whales in this population?
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?

1
Determine the genotype of the parents. Since both parents are carriers of an autosomal recessive allele, their genotypes are heterozygous (Aa).
Use a Punnett square to calculate the probabilities of the offspring's genotypes. Cross Aa (parent 1) with Aa (parent 2). The resulting genotypes are: 25% AA (homozygous dominant), 50% Aa (heterozygous carrier), and 25% aa (homozygous recessive).
Since none of the children has the recessive condition, the probability of each child being aa (homozygous recessive) is excluded. This means the possible genotypes for each child are either AA or Aa.
Calculate the probability of a child being a carrier (Aa). From the Punnett square, the probability of being Aa is 50%, and the probability of being AA is 50%. Since aa is excluded, the probabilities are normalized: P(Aa) = 50% / (50% + 50%) = 50%.
To find the probability that one or more of the four children is a carrier, use the complement rule. First, calculate the probability that none of the children is a carrier (all are AA). Then subtract this value from 1 to find the probability of at least one child being a carrier.

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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.
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Autosomal Pedigrees
Punnett Square
A Punnett square is a diagram used to predict the genetic makeup of offspring from two parents. By arranging the alleles of each parent, it allows for the visualization of possible combinations in their children. In this case, a Punnett square can help determine the probabilities of the children being carriers or expressing the recessive condition.
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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.
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