Chapter 31.
Answer B, D, and E are true statements, whereas A and C are false.
Since each person has 2 chromosomes, manifest disease necessitates the presence of 2 genes, that is q2 = 10-4 , or q = 10-2 . Accordingly, there is one gene per 100 chromosomes, and the gene is found in 1 out of 50 persons.
1. Pancreatic cystic fibrosis is a recessive genetic disease caused by dysfunction of exocrine glands. The defect is in a transmembrane regulator protein called the cystic fibrosis transmembrane conductance regulator (CFTR). The CFTR represents a b-adrenergic gated chloride channel, which is normally opened by elevated intracellular cAMP. The patients have a minimal chloride excretion and thus as minimal excretion of salt and water into the duct systems. This is what makes all exocrine secretions viscid, the duct systems are occluded and dilated, and finally the ducts are destroyed (eg. chronic respiratory disease and pancreatic insufficiency).
2. The frequency of the abnormal homozygote is q2 and of the abnormal cystic fibrosis gene, q. q2 = 1/1600 , so q = 1/40 or 0.025.
3. p = (1 - q). Accordingly, p is (1 - 0.025) or 0.975. The frequency of the normal homozygote is p2 or 0.9752 = 0.95.
4. 2pq = (2 * 0.025 * 0.975) = 0.049. Approximately 5 out of 100 live births are heterozygous carriers.
1. Albinism or amelanosis is inherited as an autosomal recessive disorder of melanin synthesis. The biosynthesis of the enzyme tyrosinase is defective, which results in lack of melanin. Amelanosis is manifest by white hair, pink-white skin, blue eyes and photophobia.
2. The genotype of the female is a0 and of the male aa. Thus the 4 gene combinations are: aa, a0, a0, and aa. Accordingly, the probability of having an albino child is 50%.
3. As seen above this probability is also 50%.
1. Phenylketonuria (PKU) is also an autosomal recessive disorder. There is a defect conversion of phenylalanine to tyrosine and thus hypopigmentation. The genetic defect results in lack of the enzyme phenylalanine 4-hydroxylase. PKU must be diagnosed and treated soon after birth in order to avoid severe mental retardation. PKU patients almost never reproduce. PKU occurs once in 25000 live births in the population.
2. There are 4 possibilities: AA, Aa, aA, and aa. The normal phenotype of the brother eliminates the possibility of being homozygous recessive for PKU (aa). Thus the probability is 2/3 – of the remaining 3 possibilities.
3.It appears likely that the female is normal (eg, homozygous dominant). Accordingly, the probability of having PKU children is close to zero (1/300).
4. The PKU gene is maintained in and transmitted by heterozygous, who occasionally get a PKU child. The loss of recessive genes in PKU patients is obviously balanced by mutations as long as the incidence is constant in the population.
1. The diagnosis is Duchenne Muscular Dystrophy (DMD).
2. Creatine phosphokinase is a soluble protein in the sarcolemma. Muscular cell death with destruction of the sarcolemma results in liberation of the enzyme to the blood.
3. DMD is confined to boys, because females have two X chromosomes. The X chromosome in the girl originating from her father must contain a normal dystrophin gene, since chromosomes with the mutation cannot survive in males by adolescence.
4.The proximal limb muscles are weak, so any attempt of locomotion has to activate synergistic muscles elsewhere (including the calves). This is why there is a relative hypertrophy or pseudohypertrophy.
5.The sister with a defect dystrophin gene on one of her two chromosomes has a 50% risk of transferring this gene to a boy, and approximately a 50% likeliness of having boys as opposed to girls. The combined risk of having a child with manifest DMD is 25%.