KEY H-W Practice Problems

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Feb 20, 2024

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BZ220 Introduction to Evolution KEY TO POPULATION GENETICS ( HARDY-WEINBERG) PRACTICE PROBLEMS OBJECTIVES: 1. Learn how to use Hardy-Weinberg as a null model/hypothesis for evolution. 2. Understand that in natural systems, many individual loci are in Hardy-Weinberg equilibrium, which implies that Hardy-Weinberg is useful in a practical context. I. HARDY-WEINBERG Hardy-Weinberg#1: Wing color polymorphisms occur regularly in butterflies. A population of butterflies in a butterfly garden consists mostly of red-spotted butterflies (Rr) with some black- winged butterflies (rr) and some red-winged butterflies (RR). The observed frequency of red- spotted butterflies is 0.40 and the observed frequency of black-winged butterflies is 0.20. Given this information, answer the following questions and determine whether the population is in Hardy-Weinberg equilibrium. a. What is the observed frequency of genotype Rr? 0.40 (determined from the problem statement) b. What is the observed frequency of genotype rr? 0.20 (determined from the problem statement) c. What is the observed frequency of genotype RR? 1- (0.40 + 0.20) = 0.40 (equation: f(RR) = 1 – [f(Rr)+f(rr)]) d. What is the observed frequency of allele R? R = 0.40 + (0.5)*(0.40) = 0.60 (equation: f(R) = p = f(RR) + ½f(Rr)) e. What is the observed frequency of allele r? r = 0.20 + (0.5)*(0.40) = 0.40 (equation: f(r) = q = f(rr) + ½f(Rr) … OR… 1 – p f. Assuming that the population is in Hardy-Weinberg, determine the frequency that you would predict for genotype rr. rr = (0.40) 2 = 0.16 (equation: f(rr) = q 2 ) g. Assuming that the population is in Hardy-Weinberg, determine the frequency that you would predict for genotype RR. RR = (0.60) 2 = 0.36 (equation: f(RR) = p 2 ) h. Assuming that the population is in Hardy-Weinberg, determine the frequency that you would predict for genotype Rr. Rr = 1 – 0.36 – 0.16 = 0.48 (equation: f(Rr) = 2pq …OR…1 -p 2 -q 2 ) i. Test the hypothesis that this butterfly population is in Hardy-Weinberg equilibrium by comparing your observations to the prediction of the hypothesis. Is the population in

Hardy-Weinberg equilibrium? (Hint: it may be helpful to organize your data in a table before reporting an answer.) No, because the observed and expected genotype frequencies do not match. f(RR) f(Rr) f(rr) Expected (H-W) 0.36 0.48 0.16 Observed 0.40 0.40 0.20 j. Which genotype/s are in excess (the observed frequency is greater than what you would predict) in this population? Genotype RR and rr are observed in excess. Hardy-Weinberg #2: Your professor counted the grasshoppers in her backyard, and found there were 300 of them. They came in three different sizes: short legged (genotype SS), medium legged (genotype SL), and long legged (genotype LL). She sorted the grasshoppers by leg size and found 75 short legged grasshoppers and 45 long legged grasshoppers before she became bored and stopped counting grasshoppers. All of the remaining uncounted grasshoppers were medium legged. a. What are the observed genotype frequencies (i.e., the frequency of the SS type, SL type, and LL type)? SS = 75/300 = 0.25 LL = 45/300 = 0.15 SL = 1-0.25-0.15 = 0.60 b. What are the observed allele frequencies (i.e., the frequency of the S allele and the frequency of the L allele)? S = 0.25 + 0.5*0.6 = 0.55 (equation: f(S) = p = f(SS) + ½f(SL) L = 1 – 0.55 = 0.45 f(r) = q = f(rr) + ½f(Rr) … OR… 1 – p) c. Determine the genotype frequencies that you would predict if the assumptions of Hardy- Weinberg were met. SS = 0.55 2 = 0.30 (equation: f(SS) = p 2 LL = 0.45 2 = 0.20 f(LL) = q 2 SL = 1-0.30-0.20 = 0.50 f(SL) = 2pq (or 1-p 2 -q 2 ) d. Test the hypothesis that this grasshopper population is in Hardy-Weinberg equilibrium by comparing your observations to the prediction of the hypothesis. Is the population in Hardy-Weinberg equilibrium? (Hint: it may be helpful to organize your data in a table before reporting an answer.) No, because the observed and expected genotype frequencies do not match. e. Which genotype/s are in excess (the observed frequency is greater than what you would predict) in this population? Genotype SL is observed in excess II. GENETIC DISORDERS

Genetic Disorder #1: The autosomal recessive disorder Phenylketonuria , if left untreated, can lead to problems with brain development and seizures. This genetic disorder affects approximately 1 in 12,000 people. Let “A” be the normal allele and let “a” be the allele for Phenylketonuria. Assuming Hardy-Weinberg conditions: a. What is the genotype of individuals that have Phenylketonuria? aa b. What is the frequency of affected individuals? aa = 1/12000 = 0.000083 c. What is the frequency of the disease allele? a = sqrt(0.000083) = 0.009 (equation: f(a) = q = √q 2 , where q 2 = f(aa)) Genetic Disorder #2: Non-classical congenital adrenal hyperplasia (NCCAH) is a recessive disorder that affects 1 in 100 individuals in New York City (no kidding!). Let “a” be the allele for NCCAH and “A” be the normal allele. Assuming Hardy -Weinberg conditions: a. What is q, the frequency of “a” in New York City? f(aa) = 1/100 = 0.01 f(a) = sqrt(0.01) = 0.10 (equation: f(a) = q = √q 2 , where q 2 = f(aa)) b. What is q t+100 , the frequency of “a” in New York City after one hundred generations? Hardy-Weinberg equilibrium means the population is at equilibrium and allele frequencies do not change, so after 100 generations the allele frequency is still 0.1 c. What is the frequency of individuals that are carriers (i.e., heterozygotes) for NCCAH in New York City? Carriers “carry” the disease allele, but do not exhibit disease symptoms. So, we are talking about the heterozygotes. Under Hardy-Weinberg assumptions, the frequency of heterozygotes is 2*(0.1)*(0.9) = 0.18 (equation: f(Aa) = 2pq)

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