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Answers%20to%20Population%20Genetics%202B%20F09%20Revised -...

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BIS 2B Fall 2009 _____________________________________________________________________________________ PART II: QUESTIONS FROM OLD EXAMS AND QUESTIONS ON POPULATION GENETICS 1. A population of peppered moths has the following genotypic frequencies: aa: 0.16, Aa: 0.04, AA: 0.80. The frequency of the 'a' allele in this population is… [show your work for full credit] The frequency of the ‘a’ allele in this population is: (a) = 0.18 You cannot assume that this population is in Hardy–Weinberg equilibrium, so you must count up the alleles to determine the frequency. Remember that a frequency is the number of times something occurs divided by the number of times it could have occurred. There are 2000 moths in the population, so there could be 4000 ‘a’ alleles (2000 x 2 alleles each; this is the denominator). Now we need to know how many times the ‘a’ allele did occur. It was present two times in aa moths that are 16% of the population (0.16 x 2000 x 2) and present only once in Aa moths that are 4% of the population (0.04 x 2000 x 1). We need to put both these numbers in the numerator: f(a) = (0.16 x 2000 x 2) + (0.04 x 2000 x 1) 2000 x 2 You could calculate this number, but before you do, notice that the population size (2000) is in every term, so it cancels out, and we can just work with the proportions given in the question: f(a) = (0.16 x 2) + (0.04 x 1) = 0.32 + 0.04 = 0.36 = 0.18 2 2 2 frequency of ‘a’ is _0.18_ Is this population in Hardy–Weinberg equilibrium? (Hint: calculate the expected genotype frequencies given the allele frequency you just calculated, then compare the observed and expected frequencies to test for equilibrium.) NO To find out if the population is in H–W equilibrium, you must compare the observed genotypic frequencies (given in the question) with the genotypic frequencies expected under H–W. If they match, then the population is in equilibrium. The expected genotypic frequencies depend on
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BIS 2B, Spring 2009 Page 2 the allele frequencies. You calculated the frequency of the ‘a’ allele in the first part of this question (f(a) = 0.18), so the f(A) = 1– f(a) or f(A) = 1– 0.18 = 0.82. You should check to make sure these two frequencies add up to one. If they don’t, you've made a math error. Now, to get to the expected genotype frequencies, we just use our formula: Expected f(aa) = f(a) x f(a) or q 2 . Here it is 0.18 x 0.18 = 0.03 Expected f(Aa) = 2(f(A) f(a)) or 2pq. Here it is 2(0.82 x 0.18) = 0.30 Expected f(AA) = f(A) x f(A) or p 2 . Here it is 0.82 x 0.82 = 0.67 Note: you can round these numbers to the nearest tenth of a decimal (that is 0.18 rounds to 0.2) to get approximate values here. The numbers will be close enough to use in the comparison below and then all the math can be done in your head. Now, we compare expected to observed frequencies. f(aa) = 0.03 0.16; not close, more observed than expected f(Aa) = 0.30 0.04; not close, fewer observed than expected f(AA) = 0.67 0.80; not close, more observed than expected There are big differences between H–W expectations and observed allelic frequencies, so the population is NOT in H–W equilibrium.
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Answers%20to%20Population%20Genetics%202B%20F09%20Revised -...

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