Answer the quesion. please ignore all the pencil marks. Treat it as a new questi

Question

Answer the quesion. please ignore all the pencil marks. Treat it as a new question

2. Xerodermia pigmentosa is a genetic diso repair, There are multiple types of hcreditary X P aused by mutations on differen Pedigrees from two families with XP are diag Norder associated with mutations in nucleotide excision below The following pedigree is from a family with Type 1 XP. Aa an How is type I XP inherited? Be specific. (2 point) What is the genotype of the III-4 individual? (2 points) What is the probability that 1III-4 and III-5 have a child with XP? (3 pts.) Aa The following pedigree is from a family with Type IIXP. r Ari How is type II XP inherited? Be specific. (2 point) What is the genotype of the III-4 individual? (2 points) What would be the probabilty that IL-4 and II15 have a child with XP? (8 points) 2.

Explanation / Answer

2) I and II ) Type I XP is an example of Autosomal recessive inheritance and individual III4 can either be normal (homozygous dominant) or carrier(heterozygous)

2) III) for instance we are considering that A is the dominant and a is the recessive allele.

As Individual III5's father has type I XP therefore, III5 must be a carrier. Individual III4 could be a carrier as one of her brothers has type 1 XP which signify that their parent must be a carrier of this inheritance

therefore, the probability of III4 to be carrier = 2/3

hence the probability of a child of III4 and III5 will have XP when the probability of III4 for being a carrier is 2/3 will be

= (1/4)x(2/3)= 1/6

2)I and II ) Type II XP is an example of Autosomal dominant inheritance and individual III4 is homozygous recessive for typeII XP

III) In order to have a child with XP either or both of III4 and III5 should have at least one dominant allele for XP but as this is the case of Autosomal dominant if either had dominant allele they should have inherited with type II XP. but as this is not the case, therefore the probability of either of III4 and III5 have dominant allele= 0

therefore the probability of having a child with XP when the probability of either of III4 and III5 have dominant allele is 0

=0

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