Tay-Sachs disease is a rare autosomal recessive disease in humans. Homozy- gous recessive individuals (aa) lack an enzyme called hexosaminidase A and, therefore, accumulate gangliosides in their nervous system leading to paralysis, epilepsy, blind- ness, and eventual death. Heterozygotes (Aa) are phenotypically normal. A second gene independently assorting from the A locus exhibits autosomal dominant expression in the form of dramatically shortened fingers, a condition known as brachydactyly. B individuals are brachydactylous and bb are normal. In a marriage between the follow- ing two individuals AaBb x Aabb    find the probability that the first child has Tay-Sachs disease   find the probability that the first child has Tay-Sachs and brachydactyly  find the probability that the first child is a carrier (Aa) for Tay-Sachs given they appear normal (A )

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Tay-Sachs disease is a rare autosomal recessive disease in humans. Homozy- gous recessive individuals (aa) lack an enzyme called hexosaminidase A and, therefore, accumulate gangliosides in their nervous system leading to paralysis, epilepsy, blind- ness, and eventual death. Heterozygotes (Aa) are phenotypically normal. A second gene independently assorting from the A locus exhibits autosomal dominant expression in the form of dramatically shortened fingers, a condition known as brachydactyly. B individuals are brachydactylous and bb are normal. In a marriage between the follow- ing two individuals AaBb x Aabb

 

  1.  find the probability that the first child has Tay-Sachs disease

  2.   find the probability that the first child has Tay-Sachs and brachydactyly

  3.  find the probability that the first child is a carrier (Aa) for Tay-Sachs given they appear normal (A )

  4.  find the probability that the first two children are normal (A bb). Assume inde- pendence here the two children are not twins

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thank you for the question, According to out honor code can answer only three sub parts at a time, as you did not mention which parts are you looking for solved first three parts.  Kindly find the solution to first three parts. If you have any doubt in remaining part kindly make a new request with the part which you need help on.

Given that
Tay-Sachs disease is with individuals with (aa)
The individuals with (Aa) are normal
The individuals with (AA) and (aA) are also normal.
The individuals with (Bb) have Brachydactyly 
The individuals with (bb) are normal.

The parents are AaBb x Aabb

1)
The probability that first child has Tay-Sachs disease is when the child is with (aa)

The total possible pairs =4 aa,aA,AA,Aa
The number of pairs with Tay-Sachs =1aa
The probability that child is with Tay-Sachs =14

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