A man, J.B., has a sister with autosomal recessive galactosemia (OMIM 230400), and his partner, S.B., has a brother with galactosemia. Galactosemia is a serious condition caused by an enzyme deficiency that prevents the metabolism of the sugar galactose. Neither J.B. nor S.B. has galactosemia, but they are concerned about the risk that a future child of theirs will have the condition. What is the probability their first child will have galactosemia?
Ch. 2 - Transmission Genetics

Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
Chapter 2, Problem A.6
A 40-year-old woman whose father had Huntington disease currently shows no symptoms of the disease. She is newly pregnant with her first child and seeks your best estimate of the chance her child will inherit the disease. What is your estimate and how did you arrive at it? (Hint: See Figure 4.11)
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Understand that Huntington disease is an autosomal dominant disorder, meaning only one copy of the mutated gene is needed for the disease to manifest.
Since the woman's father had Huntington disease, he must have had one mutated allele (H) and one normal allele (h), so his genotype is likely Hh.
The woman currently shows no symptoms, but because Huntington disease has late onset and incomplete penetrance, she could still be a carrier with genotype Hh or be hh (normal). The chance she inherited the mutated allele from her affected father is 50%, or 1/2.
If the woman is Hh, the chance of passing the mutated allele to her child is 50%, or 1/2. If she is hh, the chance is 0%.
To find the overall chance her child will inherit the disease, multiply the probability the woman is a carrier (1/2) by the probability she passes the mutated allele (1/2), resulting in an overall probability of .

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Autosomal Dominant Inheritance
Huntington disease is inherited in an autosomal dominant pattern, meaning only one copy of the mutated gene from an affected parent is sufficient to cause the disease. Each child of an affected individual has a 50% chance of inheriting the mutation and developing the disorder.
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Variations on Dominance
Penetrance and Age of Onset
Huntington disease shows age-dependent penetrance, often manifesting symptoms in mid-adulthood. A person may carry the mutation without symptoms early in life, so an asymptomatic 40-year-old with an affected parent could still carry the gene and develop symptoms later.
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Risk Estimation Using Family History and Genetic Testing
Estimating the child's risk involves assessing the mother's probability of carrying the mutation based on her affected father and her current symptom-free status. Genetic testing can confirm her status, but without it, the risk to the child is half the mother's risk of being a carrier.
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