9. Consider the following pedigree below for a rare autosomal trait. Be sure to

Question

9. Consider the following pedigree below for a rare autosomal trait. Be sure to take into account the inbreeding convention, when needed. (2 pts)What is the probability that individual B is heterozygous? a. b. (2 pt) What is the inbreeding coefficient of individual E? (3 pts) After divorce, individual "A"decides to marry at randomn outside the pedigree. If the population allele frequency of the rare autosomal trait is 0.01 what are the chances of having an affected child for his first born of his second marriage? c. E"

Explanation / Answer

Hi,
this is a rare autosomal disease. This means only few members in a family may have it. It only occurs when two copies of recessive alleles are present together in the individual. Let us call the allele for disease as 'm'.
The normal healthy allele be M. So healthy person can be either MM or Mm. The diseased person is mm.
The parents and grand parents of individual B are normal. but b has a diseased relative in family born to his grandmother. That means grandmother was a heterozygote.
Now, His grandfather is assumed to be homozygous normal individual. So his genotype is MM,
when MM and Mm mate, we get MM , Mm combinations. Now only 1/2 i heterozygous
Probability that his mother is heterozygous is = 1/2.
Again B father is assumed to be healthy and homozygous with MM. When B is born again there is chance of being 1/2 heterozygote. So total probability that B is heterozygote = 1/2 x 1/2 =1/4

b. inbreeding coefficient is calculated by counting number of individuals who have contributed genes to person E. The E is born from two different families who share a common ancestor. The number of people directly relating E to Common ancestor are = go on calculating from E upwards on right side, track parents and their parent ( count only one parent, ther other is not the common gene between to families) till you reach common grandmother, now track down till E from left side. Do not count E. we get 8 people or 8nodes .
Inbreeding coefficient = (1/2)^n ; n = no. of nodes
= (1/2)^8 = 0.0039

c. The frequency of diseased allele in family is 0.012. so q = 0.012 ; p = 1-0.012 = 0.988.
The person A is marrying outside the family, this means we can say that even in that family the allele frequency = 0.012.
So the probability of having diseased child is 0.012 x 0.012 x 0.25 (because probability to get diseaed child when two heterozygotes meet is 0.25)
= 0.000036.

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