Question

8. [12 pts total] Huntington disease is a rare degenerative autosomal disorder that determined by a dominant allele. The disorder is typically manifested after the age of 45. A young man has learned that his father has developed the disease; his mother does not carry the dominant allele for the condition [4 pts] A. What is the probability that the young man will later develop the disease? (4 pts] B. If the young man has a child with a woman who does not carry the dominant disease allele, what is the probability that their first child will be affected by Huntington disease as an adult? [4 pts] C. Ifthe young man has a child with a woman who is heterozygous for the gene responsible for Huntington's disease, what is the probability that their first child will be affected as an adult? EXTRA-CREDIT (BONUS) PROBLEM A. [3 pts] In Drosophila, an X-linked recessive mutation, scalloped (sd), causes irregular wing margins. If a scalloped female is crossed with a wildtype male, what would be the genotypes of the male and female progeny (please use Xd and X' to represent the scalloped and wildtype alleles respectively)? Fi females: Fi males: B. [3 pts] If the above Fi individuals were crossed, what would be the four expected genotypes for the F generation? F2 females: Fz males C. 3 pts] If a scalloped male is crossed with a homozygous wildtype female, whiat would be the genotypes of the male and female progeny? -F, females:-- F, males: B. [3 pts] If the Fi individuals from part C were crossed, what would be the four expected genotypes for the F: generation? F females F males:

Answer 1

8. Let's say the genotype of father of the young man is AA^h

where A is the normal.allele and A^h  is the allele for Huntington disease

The genotype for the mother will then be AA

AA^h x AA

F1: A A

A AA AA

A^h   AA^h AA^h

thus there are 50% chances that the man will have the disease

B. The chances for the man's child to carry the disease will be also 50% since here too like in the previous cross, the man Carries the dominant disease allele and the woman doesn't carry any allele. The inheritance will follow the same course

C. If the man marries a woman who is heterozygous for the allele then the make would be AA^h and since the disease is autosomal dominant, the woman would carry the disease.

CROSS

A A^h

A AA AA^h

A^h  AA^h   A^h A^h

Here the chances are 75%

or 3/4

Credits

A. Female genotype:since its X linked recessive two alleles must be present for the trait to be visible- X^sd X^sd

Male is wild type, Male genotype will be X⁺ 0

The cross.  

X⁺   0

X^sd X^SdX⁺ X^sd 0

X^sd X^SdX⁺   X^Sd0

All females are X^​​​sdX⁺ i.e. the females will all show wild phenotype

All males are X^Sd0 or will show scalloped mutated phenotype

B. F2 generation

X^sd X⁺

X^Sd X^Sd​​​​​​X^sd X^sdX⁺

0 X^sd0 X⁺0

f2 females-  X^Sd X^Sd  and X^sdX⁺

^{F2 males-} X^Sd0 and X⁺0

C. Wild type.homozygoyse female X⁺X⁺

Scalloped Male X^sd0

X⁺ X⁺

X^sd X^Sd​​​​​X⁺ X^Sd​​​​​​X⁺

⁰ X⁺0 X⁺0

F1 females, 50% X^sdX^ad and other 50% are X^{​​​​sd}X⁺

^{F1 makes, 100%} X⁺0

C.

X^Sd   X⁺

X⁺   X^SdX⁺   X⁺​​​​​​X⁺

⁰ X^Sd0 X⁺​​​​​​0