Dr. Peter Gerhardt: Transition to Adulthood for Young Adults with ASD
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Dr. Peter Gerhardt: Transition to Adulthood for Young Adults with ASD
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Question 1 (Applying): Tay-Sachs disease is an autosomal recessive disorder. A couple, Jabari (he/him) and Olivia (she/her) have a son, Kai (they/them) who is diagnosed with Tay-Sachs. Jabari has Tay-Sachs but Olivia does not. Who among Olivia, Kai and Jabari is homozygous or heterozygous for the disease allele? Explain your answer.
Applying Question 2 & 3 (Applying): You are studying the genetics of a new (hypothetical) cat breed. Assume whisker and tail length in these cats are each controlled by a single gene. Long whiskers (W) and tails (T) are dominant phenotypes and short whiskers (w) and tails (t) are recessive phenotypes. Question 2: What are the gametes of a cat with a long tail and whiskers? Assume the cat is heterozygous for each trait and the traits are linked. Question 3: Compare and contrast the cat's gametes for whisker and tail length if the traits are linked or unlinked. Are the gametes the same? Are the gametes different? Why? Feel free to attach a drawing of homologous chromosomes to help explain your answer.
Question 4a-c (Applying): X-linked hypophosphatemia is an X-linked dominant disorder that is associated with rickets-like leg deformities. A mutation in gene H is associated with the dominant form of the disease. Given this information, please answer the following questions:
A. If a mother is heterozygous for X-linked hypophosphatemia (Hh) and the father does not have X-linked hypophosphatemia what are the chances their male child will have X-linked hypophosphatemia?
B. If a mother is homozygous for X-linked hypophosphatemia (HH) and the father does not have X-linked hypophosphatemia what are the chances their male child will have X-linked hypophosphatemia?
C. If a father has X-linked hypophosphatemia (H) and the mother does not have X-linked hypophosphatemia, what are the chances their male children will have X-linked hypophosphatemia?
Tay-Sachs disease is an autosomal recessive disorder, which means that both copies of the gene must be mutated in order for the disease to manifest. Since Jabari has Tay-Sachs, he must be homozygous for the recessive allele. Olivia does not have Tay-Sachs, so she must be heterozygous for the recessive allele. Kai is the offspring of two heterozygous parents, so he has a 25% chance of being homozygous for the recessive allele, a 50% chance of being heterozygous, and a 25% chance of being homozygous for the dominant allele.
Answer
- Jabari: Homozygous recessive (tt)
- Olivia: Heterozygous (Tt)
- Kai: 25% homozygous recessive (tt), 50% heterozygous (Tt), 25% homozygous dominant (TT)