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Pedigree Inheritance Pattern

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Four patterns of inheritance and four pedigrees are shown below. Assume that individuals marrying into the family are homozygous for the wild-type allele. Match each of the inheritance patterns with a pedigree.
(a) Autosomal recessive _________
(b) Y-linked trait _________
(c) X-linked recessive _________
(d) Autosomal dominant ______

Please see attachment to view the pedigree choices of inheritance patterns.

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https://brainmass.com/biology/genetics/pedigree-inheritance-pattern-526300

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a)
An autosomal recessive disease requires two copies of a gene to be present in order for the disease to occur. Thus, both parents would have to be at least carriers in order to develop the disease. Similarly, these disorders can often skip generations, since people with the gene may not express it, but can still pass it on to their children. Therefore, this disease is represented by pedigree II, where two carriers can produce affected children.

b)
A Y-linked trait is passed from ...

Solution Summary

The pedigree inheritance patterns are examined. The homozygous for the wild-type alleles are determined.

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Pedigree Inheritance Pattern of Huntington's Disease

Please see attachment for pedigree inheritance pattern of Huntington's disease.

Huntington's disease is a late-onset disease caused by a single, dominant mutation. The following pedigree is for a family with a history of Huntington disease. Those individuals who are already suffering from the disease are shaded black. However, some additional individuals in generations II and III also have the mutant Huntington's allele and will develop Huntington disease but have not yet shown symptoms. Assume that individuals marrying into the family have no history of Huntington disease (that is, they are homozygous recessive for the gene). Also
assume that the diseased male in generation I is heterozygous for the disease gene.

(a) If individuals IIIg and IIa had a child together, what is the probability that the child would develop Huntington disease?
(b) If you were told that individual IId also developed Huntington disease, would the probability calculated in (i) change? If so, what is the new probability?

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