*No in-class activities can be made up for unexcused absences. See syllabus.

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1 ICA 13 Key *No in-class activities can be made up for unexcused absences. See syllabus. Bluegill Q1. A large population of bluegill (a freshwater fish) was observed over ten consecutive summers. When traits were compared for the bluegill population, researchers noticed that about 6% of each generation showed small lips, which is believed to be a rare recessive trait. Use this information to answer the following questions: A. What is the estimated frequency of the small-lipped allele the population? The problem tells us that 6% of the populations is homozygous recessive. Put another way the frequency of the homozygous recessive phenotype is 0.06 Since we are told nothing else and are asked to estimate frequency we can use H-W 1 to solve for allele frequencies. There are two equations to remember: o p + q = 1 o p 2 +2pq = q 2 = 1 Remember: o p = allele frequency of dominate allele o q = allele frequency of recessive allele o And p 2 = homozygous dominant genotype frequency o 2pq = heterozygous genotype frequency o q 2 = homozygous recessive genotype frequency So since we need q and know the genotype frequency we can do the following q 2 = homozygous recessive genotype frequency q 2 = 0.06 q 2 = (0. 06) q = small lip allele frequency B. What proportion of the population is heterozygous for the trait? We need the heterozygous genotype frequency We know 2pq = heterozygous genotype frequency We don t know p but q we solved for above We know that p + q = 1 p = 1 p = Heterozygous genotype frequency = 2pq 2(0.755)(0.245) = 0.37 heterozygous fish in the population. C. What proportion of the population is homozygous dominant for the trait? We need the homozygous genotype frequency We know that p 2 = homozygous genotype frequency p 2 = (0.755) 2 = 0.57 homozygous dominant genotypes in population 1 Hardy-Weinberg (H-W) 1

2 C continued. Check your math p 2 +2pq = q 2 = =? It should be 1 and if so it appear the math is right. D. What must be true about small lips for it to remain stable after 10 generations? Random mating No evolutionary mechanisms (There are four could you list them on an exam?) Question 2. Blood Types Suppose in a hypothetical population, the blood type of 1000 random individuals was sampled and the following genotypes were determined: Blood Genotype Number of Individuals AA 100 AB 600 BB 300 A. What is the frequency of the A allele (dominant) in the population? Solution 1. Count the alleles: ( ) ( ) or (2000) = 0.4 A alleles in the population = 0.4 Solution 2. Calculate using the genotype frequencies: = 0.1 ; AB = = 0.6 ; BB = = 0.3 *Check your work: p 2 + 2pq + q 2 = = 1 o Ok so p = q 2 + ½(2pq) = Alleles from homozygous dominant individuals + Dominant allele from heterozygous individuals o So we get p = (0.1) + ½(0.6) = 0.4 A alleles in the population You can even use both methods to check your answers; if you have time on the test, but definitely on the homework. DANGER Because we gave you all the genotype frequencies you have to use one of the methods above to solve for the allele frequency. If you solved it like you did for question 1.A you would get a different answer to the p allele frequency than if you used all the data available. For example: 1000 = q2 = 0.01; then we would use q 2 = 0.01 = This is somewhat close but not close enough and subsequent answers will be wrong We only solve for p or q using this method when we don t know everything about the population. When you know more use it (HINT: we did not ask you to estimate, and we know the genotypes of individuals in the population sampled). We give you the genotype frequencies, use them because they give you more accurate calculations of the allele frequencies. 2

3 B. What is the frequency of the B allele (recessive) in this population? Solution 1. Count the alleles ( ) ( ) or (2000) 0.6 B alleles in the population Solution 2. Calculate using genotype frequencies = = 0.1 ; AB = = 0.6 ; BB = = 0.3 o Check your work: q 2 + 2pq + q 2 = 1; = 1 o q = q 2 + ½(2pq) = Alleles from homozygous dominant individuals + Dominant allele from heterozygous individuals o q = (0.3) + ½(0.6) = 0.6 B alleles in population o You now know p and q, so p + q = 1, thank goodness they do. Solution 3. Calculate using p + q = 1 o p = 1 o p = 0.6 B alleles in the population C. How many individuals of the next generation would you expect to have the AB genotype, if 4000 randomly chosen individuals were sampled? This assumes H-W, we can use this to calculate the heterozygous individual (AB) for the next generation using our newly calculated allele frequencies. o 2pq = heterozygous individuals o 2(0.4)(0.6) = 0.48 of the population is heterozygous (not done yet) o This is out of 4000 individuals x 0.48 = 1920 individuals are heterozygous. D. How many would be blood type O? Zero but why? They are not in the population. We are really going to be sticking with 2 alleles. Like just two. Again H-W is somewhat unrealistic because we learned earlier there are many polygenic traits. E. What is the expected # of AA, AB, and BB individuals in a population of 1000 individuals assuming population in Hardy-Weinberg equilibrium? Translation: Using your newly calculated allele frequencies from above (p = 0.4; q = 0.6) calculate the EXPECTED numbers of individuals in a population of 1000 people. o p 2 = (0.4) 2 = 0.16 o 2pq = 2(0.4)(0.6) = 0.48 o q 2 = (0.6) 2 = 0.36 These should total up to 1 (Do they?) We are not done. What is the expected NUMBER of individuals out of 1000 with these genotypes? o 1000 x 0.16 = 160 individuals homozygous dominant AA o 1000 x 0.48 = 480 individuals heterozygous AB o 1000 x 0.36 = 360 individuals homozygous BB F. Are the expected genotype frequencies in this population in similar to the observed values is the population in Hardy-Weinberg equilibrium? The observed values are those from the question. 100/1000, 600/1000, 300/1000 (see above) The expected values are those based on the allele frequencies you calculated (p = 0.4; q = 0.6) and are used to calculated the expected values if H-W was true. 3

4 You just calculated those in problem 2.E. AA AB BB Observed Expected Remember: If OBSERVED = EXPECTED than population is in H-W. If OBSERVED EXPECTED than population NOT in H-W Researchers use statistics to determine if they are different but in this class you can estimate it. Does 0.1 = 0.16; 0.6 = 0.48; and 0.3 = 0.36 look different to you? o Yes is an acceptable answer o No is an acceptable answer But it must be consistent with potential problems that follow G. What the possible explanations for this population in or out of Hardy-Weinberg equilibrium? If you answer yes above, this answer should be something like: There must be random mating, no migration, no drift, no natural selection, and no mutation. If you answer no above the answer above your answer should say something like: It is possible this population is undergoing some change in allele frequency from non-random mating, migration, drift, natural selection or mutation. Answers from 2.F and 2.G should be consistent. Q3 Sickle Cell A. What is the mode of selection acting on this population of people with malaria and sickle cell present? Stabilizing selection B. If 9% of an African population is born with a severe form of sickle-cell anemia (bb), what estimated frequency of the population will be more resistant to malaria because they are heterozygous (Bb) for the sickle-cell gene? You are just given the % of recessive individuals and it says estimate.so what do you do? Assume q 2 = 0.09 Solve for q; q 2 = 0.09 = q = 0.3 s alleles in the population C. What is the chance of 2 parents being carriers and having a baby with sickle-cell anemia? First what is the frequency of heterozygous individuals in the population? o q = 0.3, p = 0.7 because p + q =1. o 2pq = heterozygous genotype frequency o 2(0.3)(0.7) = 0.42 or 42% chance of being a carrier Second, what is the frequency of having an offspring with a recessive allele? o During meiosis for Bb individual it gets split into 4 daughter cells each with 2 dominant alleles and 2 recessive alleles. o B = 0.5 and b = 0.5 Third, you have to consider 2 parents so. o What is the chance the father carries it? (= 0.42) x chance of passing it on (= 0.5) x the chance the mother carries it (= 0.42) x chance passing it on (= 0.5) Or (0.42) x (0.5) x (0.42) x (0.5) = or 4.4% chance 4

5 Additional Notes: When given observed frequencies (Question 2 from ICA and Case Study 1 from text; page ) follow these steps to determine if in H-W equilibrium. 1. Estimate genotype frequencies from the observations. a. These are the observed values (genotype frequencies) b. Example i. /10000 = 0.1 ii. AB = 600/1000 = 0.6 iii. BB = 300/1000 = Calculate allele frequencies. a. Dominant allele = p i. = Homozygous dominant frequency + ½ heterozygous frequency ii. = p 2 + ½(2pq) b. Recessive = q i. = homozygous recessive frequency + ½ heterozygous frequency ii. = q 2 + ½(2pq) 3. Use observed allele frequencies (part 2) to calculate expected genotype frequencies if the population is in H-W equilibrium. p 2, 2pq, and q 2 4. Compare the observed and expected values. If the difference between the observed and expected genotype frequencies are small enough we can support that evolution and random mating are not occurring. This is essentially question 2 from the ICA. 5