Published on: **Mar 4, 2016**

- 1. POPULATION GENETICS AND THE HARDY-WEINBERG LAW ANSWERS TO SAMPLE QUESTIONSRemember the basic formulas:p2 + 2pq + q2 = 1 and p + q = 1p = frequency of the dominant allele in the populationq = frequency of the recessive allele in the populationp2 = percentage of homozygous dominant individualsq2 = percentage of homozygous recessive individuals2pq = percentage of heterozygous individuals 1. PROBLEM #1. A. Answer: 36%, as given in the problem itself. B. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%. C. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%. D. Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48). E. Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%. 2. PROBLEM #2. Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers). 3. PROBLEM #3. A. Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%). B. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%). C. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32). D.
- 2. 4. PROBLEM #4. Answers: The first thing youll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. A. that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. B. That would be p2 or (0.37)2 = 0.145. PROBLEM #5. A. Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we dont have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355. B. Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above). C. Answer: That would be 0.458 x 953 = about 436. D. Answer: Well, the "A" phenotype = 0.126 + 0.458 = 0.584 and the "a" phenotype = 0.416 (you already knew this from part A above). E. Answer: Simply put, The "A" phenotype = 0.584 x 1,245 = 727 tan-sided and the "a" phenotype = 0.416 x 1,245 = 518 red-sided ( or 1,245 - 727 = 518).6. PROBLEM #6. Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.7. PROBLEM #7. Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.8. PROBLEM #8. A. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3. B. Answer: This is a little harder to figure out. Try setting up a "Punnett square" type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:
- 3. MM (0.49) MN (0.42) NN (0.09) MM (0.49) 0.2401* 0.2058 0.0441 MN (0.42) 0.2058 0.1764* 0.0378 NN (0.09) 0.0441 0.0378 0.0081* Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three "unique" combinations. Thus, three of the possibilities must be doubled. MM x MM = 0.49 x 0.49 = 0.2401 MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116 MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882 MN x MN = 0.42 x 0.42 = 0.1764 MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756 NN x NN = 0.09 x 0.09 = 0.00819. PROBLEM #9. A. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%). B. Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%). C. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.10. PROBLEM #10. Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.11. PROBLEM #11. Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. You already know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 = 0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and 0.495 and these should equal 1.0 or very close to it. This type of problem may be on the exam.
- 4. 12. PROBLEM #12. (You will not have this type of problem on the exam) Answer: We need to solve for the following equation: q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or another way of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q, we can substitute 1 - q for p, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 - 4q. Subtracting q from both sides, and then 0=4-4q-q or 0 = 4-5q Subtract 4 also from both sides, we get -4 = -5q. We then divide through by -5 to get -4/-5 = q oranotherwards the answer is 0.8 =q.