biology review
QUESTIONS
1. According to the Hardy-Weinberg theorem, p + q 1 and p2 + 2pq + q2 1. What does each of these formulas mean, and how are the formulas derived? p + q 1: If you add all the dominant alleles for a gene to all the recessive alleles for the gene, you get all of the alleles for that gene, or 100% of the alleles for the gene. (Note:
This assumes the gene has only two alleles.) p2 + 2pq + q2 1: If you combine all the individuals that are homozygous dominant for a gene with all the heterozygotes and homozygous recessive individuals for that gene, you have counted or combined all the individuals in the population that carry that gene. p^2=dominant frequency q^2=recessive frequency
2pq= heterozygous frequency.
2. Assume a population is in Hardy-Weinberg equilibrium for a given genetic autosomal trait. What proportion of individuals in the population are heterozygous for the gene if the frequency of the recessive allele is 1%?
Assume that D is the dominant allele and d is the recessive allele. Because all the alleles are either d or D, if the frequency of the d alleles is 1% or 1/100 (q), then the frequency of the D alleles must be 99% or 99/100 (p). The frequency of heterozygous individuals in the population is 2pq or 2(99/100)(1/100) 198/10,000.
3. About one child in 2,500 is born with phenylketonuria (an inability to metabolize the amino acid phenylalanine). This is known to be a recessive autosomal trait.
a. If the population is in Hardy-Weinberg equilibrium for this trait, what is the frequency of the phenylketonuria allele?
Assume P is the normal allele and p is the phenylketonuria allele. The frequency of homozygous pp individuals in the population is then equal to q2, which is 1/2,500. The frequency of the p allele is the square root of 1/2,500 1/50 or 2%.
4. In purebred Holstein cattle, about one calf in 100 is spotted red rather than black. The trait is autosomal and red is a recessive to black.
a.
1. According to the Hardy-Weinberg theorem, p + q 1 and p2 + 2pq + q2 1. What does each of these formulas mean, and how are the formulas derived? p + q 1: If you add all the dominant alleles for a gene to all the recessive alleles for the gene, you get all of the alleles for that gene, or 100% of the alleles for the gene. (Note:
This assumes the gene has only two alleles.) p2 + 2pq + q2 1: If you combine all the individuals that are homozygous dominant for a gene with all the heterozygotes and homozygous recessive individuals for that gene, you have counted or combined all the individuals in the population that carry that gene. p^2=dominant frequency q^2=recessive frequency
2pq= heterozygous frequency.
2. Assume a population is in Hardy-Weinberg equilibrium for a given genetic autosomal trait. What proportion of individuals in the population are heterozygous for the gene if the frequency of the recessive allele is 1%?
Assume that D is the dominant allele and d is the recessive allele. Because all the alleles are either d or D, if the frequency of the d alleles is 1% or 1/100 (q), then the frequency of the D alleles must be 99% or 99/100 (p). The frequency of heterozygous individuals in the population is 2pq or 2(99/100)(1/100) 198/10,000.
3. About one child in 2,500 is born with phenylketonuria (an inability to metabolize the amino acid phenylalanine). This is known to be a recessive autosomal trait.
a. If the population is in Hardy-Weinberg equilibrium for this trait, what is the frequency of the phenylketonuria allele?
Assume P is the normal allele and p is the phenylketonuria allele. The frequency of homozygous pp individuals in the population is then equal to q2, which is 1/2,500. The frequency of the p allele is the square root of 1/2,500 1/50 or 2%.
4. In purebred Holstein cattle, about one calf in 100 is spotted red rather than black. The trait is autosomal and red is a recessive to black.
a.