Chapter 23 Group Exercise: The Hardy-Weinberg Principle

Considering the same population of cats as in Part A, what is the expected frequency of each genotype (TLTL, TLTS, TSTS ) based on the equation for Hardy-Weinberg equilibrium? Keep in mind that you just learned in Part A that: The allele frequency of TL is 0.4. The allele frequency of TS is 0.6. The equation for Hardy-Weinberg equilibrium states that at a locus with two alleles, as in this cat population, the three genotypes will occur in specific proportions: p2+2pq+q2=1 Enter the values for the expected frequency of each genotype: TLTL, TLTS, and TSTS. Enter your answers numerically to two decimal places, not as percentages.

0.16 0.48 0.36 *The TL allele accounts for 160 of the alleles (60 × 2 = 120 in TLTL cats, plus 40 × 1 = 40 in TLTS cats). Therefore, the TL allele makes up 40% (160/400) of the alleles in the population, so the value of p is 0.4. The allele frequencies of the population must add up to one (in other words, p+q=1); therefore, since the value of p is 0.4, the value of q is 0.6. According to the Hardy-Weinberg equation, the expected frequencies of the genotypes should add up to 1. p2+2pq+q2=1 0.42+2(0.4)(0.6)+0.62=1 0.16+0.48+0.36=1*

A hypothetical population of 300 wolves has two alleles, FB and FW, for a locus that codes for fur color. The table below describes the phenotype of a wolf with each possible genotype, as well as the number of individuals in the population with each genotype. Which statements accurately describe the population of wolves? Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFB genotype is 40. Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFB genotype is 12. Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFW genotype is 40. Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFW genotype is 96. The population may be evolving because the actual number of individuals with each genotype differs from the expected number of individuals with each genotype. The population is not evolving because it is at Hardy-Weinberg equilibrium. The population is not at Hardy-Weinberg equilibrium.

Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFB genotype is 12. Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFW genotype is 96. The population may be evolving because the actual number of individuals with each genotype differs from the expected number of individuals with each genotype. The population is not at Hardy-Weinberg equilibrium. *The FB allele accounts for 120 of the alleles (40 × 2 = 80 in FBFB wolves, plus 40 × 1 = 40 in FBFW wolves). Therefore, the FB allele makes up 20% (120/600) of the alleles in the population, so the value of p is 0.2. The allele frequencies of the population must add up to one (in other words, p+q=1); therefore, since the value of p is 0.2, the value of q is 0.8. According to the Hardy-Weinberg equation, the expected frequencies of the genotypes should add up to 1. p2+2pq+q2=1 0.22+2(0.2)(0.8)+0.82=1 0.04+0.32+0.64=1 To predict the number of individuals with each genotype, multiply the expected frequency of each genotype by the number of individuals in the population. 0.04×300=12 FBFB individuals 0.32×300=96 FBFW individuals 0.64×300=192 FWFW individuals The wolf population may be evolving because the expected number of individuals with each genotype, calculated with the Hardy-Weinberg equation, does not equal the actual number of individuals with each genotype.

A hypothetical population of 200 cats has two alleles, TL and TS, for a locus that codes for tail length. The table below describes the phenotypes of cats with each possible genotype, as well as the number of individuals in the population with each genotype. Which statements about the population are true? Heterozygotes make up 20% of the population. Homozygotes make up 80% of the population. Homozygotes make up 30% of the population. In the entire cat population, 60% of the alleles are TS. In the entire cat population, the frequency of the TS allele is 0.5. In the entire cat population, the frequency of the TL allele is 0.4. Assuming random mating, each gamete has a 50% chance of having a TL allele and a 50% chance of having a TS allele. Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.

Heterozygotes make up 20% of the population. Homozygotes make up 80% of the population. In the entire cat population, 60% of the alleles are TS. In the entire cat population, the frequency of the TL allele is 0.4. Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele. * The gene pool for tail length in the population is composed of alleles TL and TS . The allele frequency of TL is 0.4 (160/400). Another way to express this is that 40% of the alleles are TL . The allele frequency of Ts is 0.6 (240/400). Another way to express this is that 60% of the alleles are Ts . Therefore, each gamete (sperm or egg) has a 40% chance of carrying allele TL and a 60% chance of carrying allele Ts , assuming mating is random.*