Chapter 15. Population Genetics and Genetic Diversity
Assume that a population is in Hardy-Weinberg equilibrium and the frequency of a rare autosomal recessive disease allele is q, then the frequency of disease carriers can be estimated as:
According to the Hardy Weinberg distribution:
(p2 + 2pq + q2 = 1)
Thus the genotype frequency of a carrier (heterozygous) of the recessive allele q is 2q(1-q) which is approximately 2q when q is rare as noted in the question.
Mr and Mrs Smith have come to your clinic. Mrs Smith is 9 weeks pregnant and has a sister with cystic fibrosis (CF), an autosomal recessive disease. They are concerned about whether their baby will also be born with the same disease. Mr and Mrs Smith are Caucasian (incidence for CF is 1 in 2500). What is the probability that Mrs Smith is a carrier of CF?
Mrs Smith is a healthy sibling of an affected person. Half of the siblings will be healthy and carriers. One fourth (25%) will be healthy and not carriers and the remaining one fourth will have disease. Therefore, of the healthy siblings the chance of being a carrier is 2/3. This type of question is commonly asked. Be sure you completely understand this. Drawing out a Punnett square may help you visualize the proportions; the answer refers only to the phenotypically normal individuals in the Punnett square.
It is recognized that parents and children share ½ (half) of their genes and that siblings share ½ (half) of their genes. Which one of the following is the correct number of shared genes for first cousins?
First cousins will have one of their parents who are siblings. Siblings share ½ ...