Math Review 1: Literal equations

What is the frequency of white flowers (homozygotes for CWCW) in a population if frequency of the allele CW is 0.32 and the population is in H-W equilibrium?

0.10 Explanation: The frequency of homozygotes is simply the square of the allele frequency for that allele. Since frequency of CW is 0.32, frequency of CWCW is 0.32^2 or 0.10.

Eight percent of a population shows a recessive trait. What would be the frequency of the heterozygotes for that trait if the population is in Hardy=Weinberg equilibrium?

0.40 Explanation: If 8% of individuals have the recessive trait, then q^2=0.08 and q=squareroot(0.08)=0.28. Therefore p=0.72 (since p+q=1). The frequency of heterozygotes is 2pq=2*0.72*0.28=0.40.

What is the frequency of plants with red flowers (homozygotes for CRCR) in a population if the frequency of the CW allele is 0.32 and the population is in H-W equilibrium? Assume that CW and CR are the only 2 alleles for flower color.

0.46 Explanation: If the CW allele has a frequency of 0.32, then the CR allele has a frequency of 0.68 (because p+q=1). Therefore, the CRCR genotype has a frequency of 0.46 (0.68^2).

In a population with two alleles, B and b, the allele frequency of B=0.59. What would be the frequency of the heterozygotes if the population is in Hardy-Weinberg equilibrium?

0.48 Explanation: Since allele frequency of B is 0.59, then the allele frequency of b is 0.41 (since p+q=1). The genotype frequency of heterozygotes is 2pq=2*0.59*0.41=0.48.

In a population that is in Hardy-Weinberg equilibrium for two alleles, D and d, 72% of the population shows the dominant trait. What is the frequency of the recessive allele, d?

0.53 Explanation: Since frequency of DD and Dd combined is 72%, then frequency of dd has to be 28%. (p^2+2pq+q^2=1; 0.72+q^2=1; q^2=0.28). Therefore q (frequency of d allele) is the square root of 0.28, which is 0.53.

A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA genotype?

0.64 Explanation: If allele a has a frequency of 0.2, then allele A has a frequency of 0.8 (since p+q=1). Therefore, the frequency of the AA genotype is q2 or (0.8)2, which is 0.64.

In a population that is in Hardy-Weinberg equilibrium for two alleles, D and d. D shows complete dominance over d (i.e. heterozygotes Dd have the same phenotype as homozygotes DD). 72% of the population shows the dominant trait. What proportion shows the recessive trait?

28% Explanation:Since frequency of DD and Dd combined is 72%, then frequency of dd has to be 28%. (p^2+2pq+q^2=1; 0.72+q^2=1; q^2=0.28).

A population has the following genotype frequencies: AA= 0.36, Aa = 0.48, and aa= 0.16. What are the allele frequencies?

A=0.6; a=0.4 Explanation: To obtain allele frequency, we can simply take the square root of the genotype frequency for homozygotes for that trait. Since AA=0.36, A^2=0.36 therefore A= squareroot (0.36) = 0.6. Since aa=0.16, a^2=0.16 therefore a = squareroot (0.16)=0.4.