x.HmOne big problem... On a beautiful fall day, you are out examining a populati
ID: 4806 • Letter: X
Question
x.HmOne big problem...On a beautiful fall day, you are out examining a population of osageorange trees (Maclura pomifera - google "osage orange tree" - they'recool) in Forest Park in St. Louis. Osage oranges make large,grapefruit-sized fruits. Some fruits contain a particularly strongfragrance, while others are fragrance-free. Conveniently, the fragranceof Osage Orange Fruits is coded by a single locus with two alleles (A1,A2). There three genotypes A1A1, A1A2, and A2A2. The allele forfragrance (A1) is dominant over the allele for no fragrance (A2), so inheterozygotes the dominant allele (A1) masks the recessive allele (A2).What are the phenotypes associated with each genotype?
a. A1A1 fragrant, A1A2 not fragrant, A2A2 fragrant
b. A1A1 fragrant, A1A2 not fragrant, A2A2 not fragrant
c. A1A1 not fragrant, A1A2 not fragrant, A2A2 fragrant
d. A1A1 fragrant, A1A2 fragrant, A2A2 not fragrant
You examine 50 trees in the population, and observe that 37 makefragrant fruit, and 13 trees have fruits with no fragrance. After doingsome genetic work in the lab, you determine the genotype for each treeyou sampled. As it turns out, the number of genotypes in your sampleis: A1A1 = 23, A1A2 = 14, A2A2= 13. What are the genotype frequenciesyou observed in the population you sampled?
a. Frequency of A1A1 = 0.46, frequency of A1A2 = 0.28,frequency of A2A2 = 0.26
b. Frequency of A1A1 = 0.60, frequency of A1A2 = 0.24,frequency of A2A2 = 0.16
c. Frequency of A1A1 = 0.60, frequency of A1A2 = 0.24,frequency of A2A2 = 0.16
d. Frequency of A1A1 = 0.80, frequency of A1A2 = 0.12,frequency of A2A2 = 0.08
What are the allele frequencies in your population?
a. p = 0.60, q = 0.40
b. p = 0.75, q = 0.25
c. p = 0.80, q = 0.20
d. p = 0.70, q = 0.30
If you know the allele frequencies in your population, you can determinethe expected number of each genotype under Hardy-Weinberg Equilibrium.Use the allele frequencies you calculated in Question 3 to determinethe expected genotype frequencies under the Hardy-Weinberg principle.
a. Frequency of A1A1 = 0.36, frequency of A1A2 = 0.48,frequency of A2A2 = 0.16
b. Frequency of A1A1 = 0.46, frequency of A1A2 = 0.28,frequency of A2A2 = 0.26
c. Frequency of A1A1 = 0.32, frequency of A1A2 = 0.48,frequency of A2A2 = 0.20
d. Frequency of A1A1 = 0.18, frequency of A1A2 = 0.64,frequency of A2A2 = 0.18
Given the expected frequencies determined in Question 4, how manyfragrant trees and non-fragrant trees would you have expected in yourpopulation?
a. 30 fragrant trees, 20 non-fragranttrees
b. 50 fragrant trees, 0 non-fragrant trees
c. 42 fragrant trees, 8 non-fragrant trees
d. 37 fragrant trees, 13 non-fragrant trees
Based on these data, in this population you observed
a. more non-fragrant trees than expected
b. can’t tell from these data
c. more fragrant trees than expected
d. the same number of fragrant and non-fragrant trees asyou would expect
Based on the results calculated in previous questions, does thispopulation conform to the expectations of Hardy Weinberg?
a. yes
b. no
c. can't tell
After many happy and healthy years, you are now approaching the age of70. You decide that you want to repeat your collecting trip, and decideto visit the Osage Orange population in Forest Park again. You andyour husband/wife/partner/friend and the kids/grandkids/pets load up thecar and head back to Forest Park in St. Louis. The trees youoriginally censused in 2010 have died, but their offspring (the nextgeneration) are growing vigorously in the same spot. You examine 50randomly chosen new Osage Orange trees and discover that thedistribution of genotypes in the population is now: A1A1 = 8, A1A2 = 7,A2A2 = 35. What are the observed genotype frequencies in the OsageOrange population in 2060?
a. Frequency of A1A1 = 0.18, frequency of A1A2 = 0.64,frequency of A2A2 = 0.18
b. Frequency of A1A1 = 0.36, frequency of A1A2 = 0.48,frequency of A2A2 = 0.16
c. Frequency of A1A1 = 0.46, frequency of A1A2 = 0.28,frequency of A2A2 = 0.26
d. Frequency of A1A1 = 0.16, frequency of A1A2 = 0.14,frequency of A2A2 = 0.70
Based on the results of Question 8, what are the allele frequencies inthe 2060 population?
a. p = 0.60, q = 0.40
b. p = 0.23, q = 0.77
c. p = 0.45, q = 0.55
Explanation / Answer
n a beautiful fall day, you are out examining a population of osageorange trees (Maclura pomifera - google "osage orange tree" -they're cool) in Forest Park in St. Louis. Osage oranges makelarge, grapefruit-sized fruits. Some fruits contain a particularlystrong fragrance, while others are fragrance-free. Conveniently,the fragrance of Osage Orange Fruits is coded by a single locuswith two alleles (A1, A2). There three genotypes A1A1, A1A2, andA2A2. The allele for fragrance (A1) is dominant over the allele forno fragrance (A2), so in heterozygotes the dominant allele (A1)masks the recessive allele (A2). What are the phenotypes associatedwith each genotype? d. A1A1 fragrant, A1A2 fragrant, A2A2 not fragrant You examine 50 trees in the population, and observe that 37 makefragrant fruit, and 13 trees have fruits with no fragrance. Afterdoing some genetic work in the lab, you determine the genotype foreach tree you sampled. As it turns out, the number of genotypes inyour sample is: A1A1 = 23, A1A2 = 14, A2A2= 13. What are thegenotype frequencies you observed in the population yousampled? a. Frequency of A1A1 = 0.46, frequency ofA1A2 = 0.28, frequency of A2A2 = 0.26 What are the allele frequencies in your population? a. p = 0.60, q = 0.40 If you know the allele frequencies in your population, you candetermine the expected number of each genotype under Hardy-WeinbergEquilibrium. Use the allele frequencies you calculated in Question3 to determine the expected genotype frequencies under theHardy-Weinberg principle. d. Frequency of A1A1 = 0.18, frequency of A1A2 = 0.64, frequency ofA2A2 = 0.18 Given the expected frequencies determined in Question 4, how manyfragrant trees and non-fragrant trees would you have expected inyour population? c. 42 fragrant trees, 8 non-fragranttrees Based on these data, in this population you observed a. more non-fragrant trees thanexpected Based on the results calculated in previous questions, does thispopulation conform to the expectations of Hardy Weinberg? a. yes After many happy and healthy years, you are now approaching the ageof 70. You decide that you want to repeat your collecting trip, anddecide to visit the Osage Orange population in Forest Park again.You and your husband/wife/partner/friend and thekids/grandkids/pets load up the car and head back to Forest Park inSt. Louis. The trees you originally censused in 2010 have died, buttheir offspring (the next generation) are growing vigorously in thesame spot. You examine 50 randomly chosen new Osage Orange treesand discover that the distribution of genotypes in the populationis now: A1A1 = 8, A1A2 = 7, A2A2 = 35. What are the observedgenotype frequencies in the Osage Orange population in 2060? d. Frequency of A1A1 = 0.16, frequency of A1A2 = 0.14, frequency ofA2A2 = 0.70 Based on the results of Question 8, what are the allele frequenciesin the 2060 population? b. p = 0.23, q =0.77 What do these data tell you about the evolution of this populationof osage orange trees over the past 50 years? a. The population conforms to theexpectation of Hardy Weinberg; evolution ishappening
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