Inheritance for Huntington's disease, an autosomal dominant traitis illustrated in the page Note - The disease...

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Inheritance for Huntingtons disease, an autosomal dominant traitis illustrated in the page Note - The disease allele is not

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Answer 1

Answer #1

Part 1:

The answer provided in the question is right.Since the offspring of two affected individual shows no disease, it demonstrates autosomal recessive inheritance

Part 2:

Individual I.1= Aa

Individual I.2= aa

Individual II.2= Aa

Individual III.12= Aa

Individual IV.9= AA or Aa probabilty is 1:2

For determining probability of individual IV.9

	A	a
A	AA	Aa
a	Aa	aa

AA: 25%, Aa: 50 %. Therefore AA:Aa = 25:50= 1:2

Part 3:

Affected male has unaffected sibling. So affected male is either Aa or AA and sibling is aa. Affected males parents show disease but they an affected off spring. So both the parents are Aa (heterozygous)

Individual (female) IV.9 is either AA or Aa (previosly determined in part 2)

Affected Male is AA or Aa.

So there are four probable crosses

Female (Genotypes)	Male (Genotypes)	Offspring
AA	AA	4 -all are affected
Aa	AA	4-all are affected
AA	Aa	4-all are affected
Aa	Aa	4-3 affected and 1 not affected

The probability that they will have an unaffected child is 1/16

Part 4:

Individual III.14 (Female) = Aa ( Since one parent is affected and one is unaffected. Also they produce unaffected siblings, so the affected parent is heterozygous)

According to the question this female has children with normal male. So:

Female (Aa) x Male (aa).

Female gametes : A and a

Male gametes: a and a

	a	a
A	Aa (affected)	Aa(affected)
a	aa (not affected)	aa (not affected)

The probability that two children will show the trait is 2/4 = 1:2 (50%)

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