O Question 5: Phenylketonuria (PKU) is a rare, mendelian, recessive genetic disorder where individuals lack the enzyme phenylalanine hydroxylase, which is necessary for conversion of the amino acid phenylalanine to tyrosine. In individuals with PKU, phenylalanine builds up in the circulatory system to levels that are toxic. Assuming that the population is in equilibrium, if 1 in 10,000 individuals have PKU, what proportion of the population is a carrier (i.e., is heterozygous) for the PKU allele?

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Question 5: Phenylketonuria (PKU) is a rare, mendelian, recessive genetic
disorder where individuals lack the enzyme phenylalanine hydroxylase, which is
necessary for conversion of the amino acid phenylalanine to tyrosine. In
individuals with PKU, phenylalanine builds up in the circulatory system to levels
that are toxic. Assuming that the population is in equilibrium, if 1 in 10,000
individuals have PKU, what proportion of the population is a carrier (i.e., is
heterozygous) for the PKU allele?
Transcribed Image Text:Question 5: Phenylketonuria (PKU) is a rare, mendelian, recessive genetic disorder where individuals lack the enzyme phenylalanine hydroxylase, which is necessary for conversion of the amino acid phenylalanine to tyrosine. In individuals with PKU, phenylalanine builds up in the circulatory system to levels that are toxic. Assuming that the population is in equilibrium, if 1 in 10,000 individuals have PKU, what proportion of the population is a carrier (i.e., is heterozygous) for the PKU allele?
Question 1. Is the population in H-W equilibrium?
Observed genotype frequencies
●
■
●
f[A₁A₁] =0.16
f[A₁A₂] =0.48
f[A₂A₂] =0.36
O Question 2. Is the population in H-W equilibrium?
Observed genotype frequencies
f[A₁A₁] =0.04
f[A₁A₂] =0.12
● f[A₂A₂] =0.84
●
Question 3. Is the population in H-W equilibrium?
Observed genotype numbers
A₁A₁ =40
A1A2 =20
● A₂A2 =40
O Question 4: Is the population in H-W equilibrium?
Observed number of each genotype
A1A1=36
●
● A1A2=28
A2A2=36
Note that raw numbers are given. The question is whether observed
frequencies = expected frequencies. You must first compute the
genotype frequencies.
Transcribed Image Text:Question 1. Is the population in H-W equilibrium? Observed genotype frequencies ● ■ ● f[A₁A₁] =0.16 f[A₁A₂] =0.48 f[A₂A₂] =0.36 O Question 2. Is the population in H-W equilibrium? Observed genotype frequencies f[A₁A₁] =0.04 f[A₁A₂] =0.12 ● f[A₂A₂] =0.84 ● Question 3. Is the population in H-W equilibrium? Observed genotype numbers A₁A₁ =40 A1A2 =20 ● A₂A2 =40 O Question 4: Is the population in H-W equilibrium? Observed number of each genotype A1A1=36 ● ● A1A2=28 A2A2=36 Note that raw numbers are given. The question is whether observed frequencies = expected frequencies. You must first compute the genotype frequencies.
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