la. What type of inheritance does the above early onset Alzheimer\'s pedigree di
ID: 195831 • Letter: L
Question
la. What type of inheritance does the above early onset Alzheimer's pedigree display in generations I, II and III ? b. The inherited Alzheimer's trait does not appear until middle age and the individuals in generation IV are not old enough to know if they have the trait (thus the ?). i) What is the probability that a mating between IV-1 and IV-2 would produce a child who would later develop the trait? ii) What is the probability that a mating between IV-1 and IV-3 would produce a child who would later develop the trait? 2 10 7 8 9 10 2a. The above pedigree is an autosomal recessive trait. What is the probability that individual II1-4 is a carrier? b. If IV-2 and IV-4 were to mate, what is the probability that their first child will be afflicted with the recessive trait?Explanation / Answer
Solution 1a) We have a pedigree showing early onset Alzheimer's inheritance. we know that it follows an autosomal inheritance. in this case, it does not skip a generation which indicates that it must be a dominant inheritance. also, it has early-onset which means that affected individuals carry the genes and unaffected individuals do not carry a gene because they were enough old at the time of DNA profiling. therefore, the given pedigree is showing Autosomal dominant inheritance pattern for early-onset Alzheimer's disease.
b i)) Let's say A is the disease allele for Alzheimer and a is the normal allele.
we can see that IV-1 is a son from III-2(aa) and III-3(Aa).
the probability of carrying an affected allele to IV-1 would be; 1/2 Aa (from the parental cross of aa x Aa)
IV-2 is the daughter of III-5(aa) and III-6(aa). so IV-2 is a normal daughter (aa) from the aa x aa cross.
now, the cross of IV-1 with the IV-2 would be Aa x aa and would show 1/2 probability of getting Alzheimer carrying offspring (Aa)
so, the probability of offspring of IV-1 and IV-2 showing Alzheimer = probability that IV-1 has Alzheimer x probability of a child getting Alzheimer
= 1/2 x 1/2 = 1/4 = 0.25
ii) Similarly, probability of IV-1 carrying Alzheimer's allele = 1/2
Probability of IV-3 carrying Alzheimer's allele = 1/2
so, the probability of a child of IV-1 and IV-3 carrying Alzheimer = probability of IV-1 carrying Alzheimer's allele x probability of IV-3 carrying Alzheimer's allele x probability that child inherits the Alzheimer's allele
= 1/2 x 1/2 x 3/4 = 3/16 = 0.18
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.