Supplemental Lecture (99/09/14 update) by Stephen T. Abedon (abedon.1@osu.edu)

  1. Chapter title: Hardy-Weinberg Problems
  2. Strategies for solving Hardy-Weinberg problems
    1. General strategies to employ when attempting Hardy-Weinberg problems
      1. Think of these problems as puzzles and then keep telling yourself that it's only a puzzle or a game; that is, just go with the flow; stressing yourself out at any juncture in life is counterproductive
      2. For introductory problems one can usually assume one locus, two allele, diploid genetics; if this is not the case, often one will be told so with the exception of questions that ask for speculation as to the number of loci, alleles, or the ploidy involved
      3. Always assume Hardy-Weinberg equilibrium unless information contradicts that assumption; you can always reject this hypothesis down the road.
    2. Work with decimals
      1. Decimals typically are easier to work with than percentages, fractions, or absolute numbers. Consequently:
        1. Always convert percentage information into decimal information (i.e., frequencies; 97% -- 0.97)
        2. Always convert fractions into decimals
        3. Always convert absolute information into decimal information
    3. Convert phenotypes to genotypes
      1. Always look for how you might convert phenotype information into genotype information, and then do so. Remember, solving Hardy-Weinberg problems is a game. Whenever you see phenotype frequencies, you should start looking forward to deducing genotype frequencies
      2. Remember, frequencies must add up to one
      3. When phenotypes map onto genotypes one-to-one (i.e., codominance), then genotype frequencies are equal to phenotype frequencies
      4. When phenotypes do not map onto genotypes one-to-one, then try to figure out the phenotype frequency of one homozygote, and then assume Hardy-Weinberg equilibrium (unless you have reason not to)
      5. Remember, again, frequencies must add up to one
    4. Convert genotypes to alleles
      1. When you have genotype frequencies, use that information to calculate allelic frequency. Remember, again that solving Hardy-Weinberg problems is only a game. Whenever you see phenotype frequencies, you should start looking forward to deducing genotype frequencies
      2. If you have Hardy-Weinberg equilibrium and have the frequency of one of a homozygote, then you can calculate at least one allele's frequency as the square root of the homozygote's frequency.
      3. If you don't know allelic frequencies but do know genotype frequencies then you can calculate allelic frequencies by multiplying the frequency of each genotype by a coefficient equal to the number of alleles of one type in each genotype, then divide the quantity by 2 (e.g., f(A) = (f(AA)*2 + f(Aa)*1 + f(aa)*0)/2)
      4. If you have not yet calculated genotype frequencies, but do know the absolute number of each genotype, then it is possible to skip a step and calculate allelic frequencies directly from absolute genotype numbers; just multiplying absolute numbers by the coefficient, as above, and then divide by twice the sum of the population size (i.e., the sum of the absolute numbers of genotypes)
      5. Remember, frequencies must add up to one
    5. Convert alleles to genotypes
      1. When you have allele frequencies, you can then calculate genotype frequencies using the Hardy-Weinberg equation, i.e., f(AA) = p2, f(Aa) = 2pq, and f(aa) = q2.
      2. Remember, frequencies must add up to one.
      3. Remember that you can generate genotype frequencies using the Hardy-Weinberg equation only if Hardy-Weinberg conditions apply.
    6. Incorporating selection
      1. Incorporating selection into Hardy-Weinberg problems complicates things somewhat
      2. Selection is doable, however, so long as you keep in mind that the effect of selection is to reduce absolute genotype/phenotype/allele numbers
      3. Operationally this may be accomplished by multiplying frequencies (typically genotype frequencies) by a selection coefficient
      4. Remember that following such a procedure the resulting "frequencies" will no longer add to one, but must be re-calculated such that genotype frequencies following selection are presented as decimals which do add up to one
    7. Practice
      1. As in much of life, the key to success is practice and dedication so as follows are a number of Hardy-Weinberg and related problems for you do solve
      2. Anything that you can do to make a difficult task an enjoyable one will always serve to make that task (and your life) easier; so lighten up and learn to enjoy doing population genetics
      3. Remember, it's only a game
  3. Practice questions
    1. Home-grown questions
      1. The study of allele and genotype frequencies within populations is called what? [PEEK]
      2. A plant population consists of two flower morphs, white and red, controlled by a single locus and two alleles. The white morph represents the recessive homozygote. The red morph consists of some combination of heterozygotes and dominant homozygotes. Assume that 50% of the population has red flowers. What are the frequencies of all three genotypes assuming Hardy-Weinberg equilibrium? [PEEK]
      3. A population of 40 guinea pigs undergoing natural selection is reduced to 20 individuals in only five generations. At this point genotype frequencies are 0.25, 0.5, and 0.25 for the genotypes AA, Aa, and aa, respectively. Assume that natural selection continues to act on this population. Is the population in Hardy-Weinberg equilibrium? [PEEK]
      4. Given a locus with three alleles and allele frequencies of 0.4, 0.4, and 0.2, what is the frequency of the least prevalent heterozygote? [PEEK]
        1. 0.06
        2. 0.16
        3. 0.26
        4. 0.36
        5. 0.10
        6. 0.08
      5. Given a population that up to now had been in Hardy-Weinberg equilibrium. Assume two alleles, one locus, p = 0.5, and distinctly different (and unambiguous) phenotypes associated with each genotype. Now assume internal fertilization and that all matings over one generation are 100% assortative with regard to the trait in question. What are the genotype frequencies before the round of assortative mating? What are the genotype frequencies in the generation that follows this round of assortative mating? Note: To answer this question you must know what assortative mating is and consequently understand the structural impact of assortative mating on the gene pool. [PEEK]
      6. What is a balanced polymorphism? [PEEK]
      7. Relative fitnesses that vary as a function of genotype frequency is an example of what? [PEEK]
      8. You have a population that, before selection, consists of 0.5 AA individuals, 0.25 Aa individuals, and 0.25 aa individuals. The relative fitnesses associated with each of these three genotypes are 0.3, 0.5, and 1.0, respectively. Define p, the frequency of A, (i) before selection and (ii) following one round of selection. [PEEK]
      9. Given a population with genotype frequencies of 0.2, 0.2, and 0.6 for genotypes AA, Aa, and aa, what should be the genotype frequencies assuming Hardy-Weinberg equilibrium? [PEEK]
      10. What is assortative mating and what does it do to a population's gene pool? [PEEK]
      11. You have a haploid population (hey, a bunch of T4 phage!) and one loci, two allele genetics. 50% of the A individuals die per round of replication but all of the a individuals survive. Survivors reproduce at the same rate, independent of genotype. You begin with a population in which the frequency of A (i.e., p) is equal to the frequency of a (i.e., q). What is the ratio of genotypes following one round of selection? [PEEK]
      12. The B allele is dominant to the b allele. The phenotype associated with the former is brown eyes, while blue eyes is the phenotype associated with the latter. The brown eye allele is present in the population at a frequency of 0.2. Given Hardy-Weinberg equilibrium and no differences in allele frequencies between genders, what is the probability that a blue-eyed woman will marry a brown-eyed man? [PEEK]
      13. Given a one locus, two allele system in which the relative fitness of AA is 0.5, Aa is 1.0, and aa is 0.0, this is an example of what? [PEEK]
      14. Draw a population of 10 giraffes following strong (i.e., highly effective) diversifying selection. Try to avoid being ambiguous in your drawings. [PEEK]
      15. The integration of Darwinism with Mendelian genetics is called the _________. [PEEK]
      16. What is the maximum number of alleles that a diploid individual can have at each locus? Consider only loci found on autosomal chromosomes. [PEEK]
      17. A population is in Hardy-Weinberg equilibrium. Consider only a single locus and two alleles found at that locus. If the frequency of the a allele is 0.4, what is the frequency of all of the possible genotypes at this locus? Call the other allele A. Assume that these organisms are diploid. [PEEK]
      18. A population consists of 200 aa individuals and is in Hardy-Weinberg equilibrium. Assuming one-locus, two-allele genetics, what is the frequency of the a allele if the population consists of a total of 1,000 individuals? What if the 200 aa individual population consists of a total of 10,000 individuals? [PEEK]
      19. A population is in Hardy-Weinberg equilibrium. The frequency of the dominant phenotype is 0.99. What fraction of individuals that carry at least one recessive allele are homozygous at this locus? Assume one-locus, two-allele genetics. [PEEK]
      20. Distinguish genetic bottleneck from founder effects. [PEEK]
      21. Inbreeding and assortative mating are both examples of what? [PEEK]
      22. In a hypothetical population of 2500 people, 2275 people have brown eyes and 225 people have blue eyes (the homozygous-recessive phenotype). If there are 4000 children produced by this generation, how many (i.e., what number) of the children would be expected to be heterozygous for eye color? Assume Hardy-Weinberg equilibrium. [PEEK]
      23. Describe, in terms of heritable variation, a population lacking in polymorphisms. [PEEK]
      24. How does heterozygous advantage contribute to the maintenance of polymorphisms, i.e., balanced polymorphisms? [PEEK]
      25. What characteristic of the two parent populations is typically a necessary prerequisite for the occurrence of hybrid vigor? [PEEK]
      26. In stabilizing selection, what categories of phenotypes is selection "editing out" of the population? [PEEK]
      27. In many human societies both genders are responsible for the care of offspring though with the duties typically separated such that the female is more responsible for meeting the physical needs of offspring while the male is more responsible for meeting the economic needs of offspring (obviously there exists a great deal of variation both within and between societies). As a consequence of these shared responsibilities toward raising offspring, human males, unlike those of many species, are often as picky about choosing their long-term mate as human females are about picking theirs. Perhaps consequently, both human genders, at sexual maturity, divert much time and energy away from survival and growth and instead toward self grooming, presumably to enhance their attractiveness to the other gender. Those humans most successful at such grooming display an edge in the procurement of the most desirable mates. One could even argue that such grooming represents a means of fooling prospective mates into believing that the groomer is healthier or otherwise more desirable as a mate than otherwise might be the case. In evolutionary terms, what form of natural selection is likely responsible for the evolution of such grooming behavior as exhibited by both human males and human females? [PEEK]
      28. In addition to no mutation, no natural selection, no genetic drift, and no migration, what further criteria must be met for a population to be in Hardy-Weinberg equilibrium? [PEEK]
      29. What is assortative mating? [PEEK]
      30. Considering eye color, if B is the brown-eye allele and b is the blue-eye allele, given a population consisting of the following, 323 BB, 23 Bb, and 45 bb, what are the frequencies of allele B and of allele b? Is this population in Hardy-Weinberg equilibrium? Quantitatively justify your answer (i.e., calculate the expected Hardy-Weinberg genotype frequencies and absolute genotype numbers, then compare these numbers to those observed in the populations). [PEEK]
    2. Questions from p. 259-261 of Campbell, 1996:
      1. In a population with two alleles for a particular locus, B and b, the allele frequency of B is 0.7. What would be the frequency of heterozygotes if the population is in Hardy-Weinberg equilibrium? [PEEK]
        1. 0.7
        2. 0.49
        3. 0.21
        4. 0.42
        5. 0.09
      2. In a population that is in Hardy-Weinberg equilibrium, 16% of the individuals show the recessive trait. What is the frequency of the dominant allele in the population? [PEEK]
        1. 0.84
        2. 0.36
        3. 0.6
        4. 0.4
        5. 0.48
      3. Levy and Levin (1975) used electrophoresis to study the phosphoglucose isomerase-2 locus in the evening primrose Oenothera biennis. . . They observed two alleles affecting electrophoretic mobility of the enzyme, PGI-2a and PGI-2b. In 57 strains, they observed 35 PGI-2a/PGI-2a, 19 PGI-2a/PGI-2b, and 3 PGI-2b/PGI-2b. . . Calculate the expected numbers of the three genotypes . . . assuming that the genotypes occur in Hardy-Weinberg proportions. [PEEK]
      4. Kelus (cited in Mourant et al., 1976) reports a study of 3100 Poles, of which 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N (and) the expected numbers of the three genotypic classes. [PEEK]
      5. Mourant et al. (1976) cite date on 400 Basques from Spain, of which 230 were Rh+ and 170 were Rh-. Calculate the allele frequencies of D (i.e., the allele which in homozyogous form results in the Rh+ phenotype) and d (the allele which in homozyogous form results in the Rh- phenotype; recall that the phenotype of Dd is Rh+). How many of the Rh+ individuals would be expected to be heterozygous? [PEEK]
      6. Phenylketonuria is a severe form of mental retardation due to a rare autosomal recessive. About one in 10,000 newborn Caucasians are affected with the disease. Calculate the frequency of the carriers (heterozygotes). [PEEK]
      7. The IA "allele" for the ABO blood groups actually consists of two subtypes, IA1 and IA2, either being considered "IA". In Caucasions, about 3/4 of the IA alelles are IA1 and 1/4 are IA2 (Cavalli-Sforza and Edwards, 1967). Among individuals of genotype IAIO, what fraction would be expected to be IA1IO? What fraction IA2IO? What would be the expected proportions of IA1IA1, IA1IA2, and IA2IA2 among IAIA individuals? [PEEK]
      8. If the frequency of the "green" form of red-green color blindness (due to an X-linked locus) is 5 percent among males, what fraction of females would be affected? What fraction of females would be heterozygous? [PEEK]
      9. Imagine an autosomal locus with four alleles, A1, A2, A3, and A4, at frequencies .1, .2, .3, and .4, respectively. Calculate the expected random mating frequencies of all possible genotypes. [PEEK]
      10. Consider a locus with two alleles (A1 and A2) and another locus with three alleles (B1, B2, B3). Let p1 = .3 be the allele frequency of A1, q1 = .2 be that of B1, and q2 = .3 be that of B2. Calculate the frequencies of all possible gametes, assuming that the loci are in linkage equilibrium. [PEEK]
      11. Suppose genotypes AA, Aa, and aa have frequencies in zygotes of 0.16, 0.48, and 0.36, respectively, and relative viabilities of w11 = 1.0, w12 = 0.8, and w22 = 0.6, respectively. Calculate the genotype frequencies in the zygotes in the next generation. [PEEK]
    3. Questions from pp. 253 of Sinnott et al., 1958:
      1. The frequencies in per cent of the blood group alleles in a Scottish population were computed to be IA = 20.62, IB = 7.56, and i = 71.83 (i.e., these are the ABO blood groupings where IA is frequency of the A allele, IA is the frequency of the B allele, and i is the frequency of the O allele,). What are the expected phenotype frequencies in this population, on the assumption of random mating? [PEEK]
      2. In a Chinese population, 99 per cent of all persons tested were Rh-positive (D+). What are the frequencies of the three genotypes DD, Dd, and dd expected on the assumption of random mating? What proportion of matings in this population would be subject to the risk of having a baby with erythroblastosis due to D incompatibility? [PEEK] (note: to answer this question you absolutely, positively must be aware of when and how erythroblastosis occurs)
      3. (If mating is at random and red-green color blindness (which is sex/X linked) does not affect survival or fertility, what should be the proportion of color-blind women in a population at (Hardy-Weinberg) equilibrium in which 8 per cent of the men are color blind? [PEEK]
    4. Questions from p. 484-485 of Hartl, 1983:
      1. Among a sample of 1000 Britishers, the number of individuals with each of the MN blood group phenotypes was as follows M: 298, MN: 489, N: 213. What is the M allele frequency? What is the N allele frequency? What number of each of the genotypes would be expected with random mating? [PEEK]
      2. A certain randomly mating population has a frequency of Rh- blood types of 16 percent. What is the frequency of the d allele (i.e., the Rh- allele)? What the frequency of the D allele (i.e., the Rh+ allele)? What are the expected genotype frequencies? [PEEK]
      3. The dry type of ear cerumen ("wax") is due to homozygosity for a simple Mendelian recessive. Among American Indians the frequency of dry-cerumen individual is 66 percent. What is the frequency of the recessive allele? What is the overall frequency of heterozygotes? Among individuals with the wet type of cerumen, what is the frequency of heterozygotes? (hint: to answer this question you must assume that the population is in Hardy-Weinberg equilibrium) [PEEK]
      4. In Zurich, Switzerland, the allele frequencies of IA, IB, and IO are 0.27, 0.06, and 0.67, respectively. What are the expected frequencies of blood types A, B, AB, and O? [PEEK]
  4. Practice question answers
    1. Home-grown questions
      1. population genetics
      2. the frequency of a is 0.707 (the square root of 0.5); the frequency of A is 0.293 (= 1 - 0.707); the frequency of aa is 0.5, the frequency of AA is 0.086; the frequency of Aa is 0.414 (note that I had to use three significant figures to avoid rounding error)
      3. No, it is being subjected to natural selection and does not have a large size
      4. (2) 2 * 0.4 * 0.2 = 0.16
      5. AA parents are 25% of population and produce only AA children; aa parents are 25% of the population and produce only aa children; Aa parents are 50% of the population and produce children who are 25% AA, 50% Aa, and 25% aa. Since these latter children make up only 50% of the population, the final frequencies are: AA = 0.25 + (0.25 * 0.5) = 0.375; aa = 0.375; and last, but surely not least, the frequency of the heterozygote is 0.5 * 0.5 = 0.25. 0.375 + 0.375 + 0.25 = 1.0. Note that assortative mating has decreased the frequency of heterozygotes in the population.
      6. a stably-existing polymorphism
      7. frequency dependent selection
      8. p = 0.625 [(0.5*2 + 0.25*1)/2] prior to selection and [(0.5 * 0.3 * 2) + (0.25 * 0.5 * 1)] * 1 / [(0.5 * 0.3 * 2) + (0.25 * 0.5 * 2) + (0.25 * 1.0 * 2)] = (.3 + 0.125) / (0.3 + 0.25 + 0.5) = 0.425 / 1.05 = 0.405 after selection
      9. p = 0.3; q = 0.7; p2 = 0.09, 2pq = 0.42, q2 = 0.49
      10. It is non-random mating in which phenotypically similar individuals preferentially mate. It subdivides the population into more than one gene pool; alternatively, you might want to think of assortative mating's effect on the gene pool as one of subdividing by parent phenotype the vat into which we bestow our germ cells.
      11. p = 0.5; following selection the p = (0.5 * 0.5) / [(0.5 * 0.5) + 0.5] = 0.33; the ratio is either 0.33 or 3. If 70% of A died then p = (0.5 * 0.7) / [(0.5 * 0.7) + 0.5].
      12. p = 0.2; q = 0.8; p2 = 0.04; 2pq = 0.32; q2 = 0.64; the frequency of brown-eyed people (men or women) is 0.04 + 0.32 = 0.36; the frequency of blue-eyed people (men or women) is 0.64; the probability that a brown-eyed man will marry a blue eyed woman therefore is 0.36 * 0.64 = 0.23. That is, 23% of all marriages, assuming random marrying, will be between blue-eyed women and brown-eyed men (another 23% will be between blue-eyed men and brown-eyed woman, 41% will consist of all blue-eyed people, and 13% will be between all brown-eyed people.
      13. heterozygote advantage
      14. the giraffes should be distributed into two, preferably quantitative phenotypes such as with one set of giraffes having short necks and the other group having long necks
      15. Modern synthesis
      16. 2
      17. f(aa) = 0.16, f(Aa) = 0.48, f(AA) = 0.36; 0.16 + 0.48 + 0.36 = 1.0
      18. square-root of 0.2 in a population of 1000 = 0.45; square-root of 0.02 in a population of 10,000 = 0.14
      19. 1-0.99 = frequency of the recessive homozygote = 0.01; The frequency of the recessive allele is 0.1; the frequency of the heterozygote therefore is 2 * 0.1 * (1-0.1) = 0.18; the fraction of individuals that carry the recessive alleles that are homozygous for this allele therefore is 0.01 / (0.18 + 0.01) = 0.056 or approximately one in 18
      20. genetic bottlenecks result from a reduction in the size of a population or the maintenance of a population at a low population size; founder effects result when new populations are founded by relatively few individuals. Founder effects may be followed or preceded by genetic bottlenecks but are not necessarily so (i.e., the subpopulation that is doing the founding could be derived from a population that is relatively small or could fail to increase in number following the founding event thus resulting in an extended genetic bottleneck in addition to the founding effec.
      21. non-random mating
      22. 225 / 2500 = 0.09 = the frequency of the blue-eye allele individuals; (0.09)1/2 = 0.3 = the frequency of the blue-eye allele; 2 * (0.3) * (1 - 0.3) = 0.42 = the frequency of the heterozygote; 0.42 * 4000 = 1680 = number children of 4000 that would be expected to be heterozygous for eye color
      23. A population lacking heritable variation would also be lacking in polymorphisms, i.e., each locus would be associated with only a single allele
      24. with heterozygous advantage, two different alleles are required to display the phenotype/genotype with the greatest associated fitness. Thus, selection favors the maintenance of two alleles per that locus and the maintenance of two or more alleles per given locus is the definition of balanced polymorphism
      25. Inbreeding or, at least, the fixing of deleterious alleles not shared with the other parental population such that masking of deleterious alleles in the offspring may occur
      26. both extremes of the phenotype, i.e., in terms of quantitative , the low-end expression of the trait as well as the high-end expression
      27. sexual selection
      28. random mating between individuals
      29. choosing a mate on the basis of resemblance to that individual
      30. frequency of B is equal to [(323 * 2) + (23 * 1) + (45 * 0)] / [(323 * 2) + (23 * 2) + (45 * 2)] = 669 / 782 = 0.855; frequency of b is equal to 1 - 0.855 = 0.145; f(BB) = (0.855)2 = 0.73, f(Bb) = 0.25, f(bb) = 0.021, which corresponds to actual numbers of 391*0.73» 285, 391*0.25» 98, 391*0.021» 8 for BB, Bb, and bb, respectively
    2. Questions from p. 259-261 of Campbell, 1996:
      1. (4) 0.42 = 2 * 0.7 * 0.3 = 2 * 0.7 * (1 - 0.3) = 2 * p * q = 2 * p * (1 - p).
      2. (3) 0.6; the frequency of the recessive phenotype is given. This phenotype can only be achieved in the homozygote. Therefore you know what the frequency of the recessive homozygote is. This frequency is equal to q2 where q is the frequency of the recessive allele. The square root of 0.16 is 0.4 which is the value of q. The frequency of the dominant allele, p is simply 1 - q = 1 - 0.4 = 0.6.
      3. Make p the frequency of PGI-2a and q the frequency of PGI-2b. The frequency of PGI-2a is equal to 35/57 + 0.5*19/57 = 0.78. Thus, p = 0.78 and q = 1 - 0.78 = 0.22. Given Hardy-Weinberg proportions, the expected frequency of PGI-2a/PGI-2a = p2 = 0.61, of PGI-2a/PGI-2b = 2pq = 0.34, and of PGI-2b/PGI-2b = q2 = 0.05. As a check, 0.61 + 0.34 + 0.5 do indeed equal 1.0. 0.61 * 57 = 35, 0.34 * 57 is a little greater than 19, and 0.05 * 57 is a little less than 3. Though they didn't ask, you probably would, to a first approximation (i.e., it's always good form to do the statistics even on the obvious), assume that this population is in Hardy-Weinberg equilibrium.
      4. This question is answered in the same manner above. That is, the frequency of allele M is equal to (1101 + 0.5 * 1496)/3100 = 0.60. Therefore the frequency of N is 0.40, of MM 0.36, of MN 0.48, and 0.16 for NN assuming Hardy-Weinberg equilibrium. Expected numbers are 0.36 * 3100 = 1116, 0.48 * 3100 = 1488, and 0.16 * 3100 = 496, respectively, which, of course, are pretty similar to the observed numbers.
      5. Since Rh- is the phenotype of the homozygous recessive, the d allele frequency is equal to the square root of 170/400 = 0.65. The expected frequency of heterozygotes is equal to 2 * 0.65 * (1 - 0.65) = 0.46 which is 182 individuals of 400.
      6. The frequency of the homozygous recessive is 1 in 10,000 = 0.0001 (= 10-4). Assuming simple genetics (e.g., all homozygotes are born and counted at the same rate as non-affected individuals), the frequency of the recessive allele is the square root of this, 0.01. From there the simple answer is about one in 50. Why, because when the recessive is sufficiently rare, the dominant allele is sufficiently close to 1 that the expression 2pq is essentially equal to 2q. Since q = 0.01, 2q = 0.02 = 1 / 50. Note that if you prefer to do things without taking short cuts, the answer would be 2 * 0.01 * (1 - 0.01) = 0.0198. Basically 0.02.
      7. This question is of a type that may be categorized as almost unfairly easy. That is because it is actually so simple that one wants to read far more into it than there actually is, and thus distracts oneself maximally, or at least reach unwanted levels of anxiety. The answers are: 3/4, 1/4, 9/16, 6/16, and 1/16. Why? First, the question was made rediculously easy simply by asking only for answers which are frequencies among individuals already carrying IA. Thus, among IAIO individuals, the fraction with the IA1 has to be the same (on average, of course) as its fraction in the entire population, which is 3/4. Similarly, the fraction of IA2 has to be equal to the frequency of this allele in the total population, which is 1/4. What is the fraction of the genotypes made up of only the IA1 and IA2 alleles, among IAIA individuals? Again, this is a far simpler question than what is the fraction of these genotypes among the total population? and is calculated simply using the familiar p2, 2pq, and q2 from the Hardy-Weinberg equation where p is the frequency of IA1 and q is the frequency of IA2, which are 3/4 and 1/4, respectively.
      8. The frequency of the phenotype among males is 0.05. Recall that males are haploid for the X chromosome. Therefore the rate at which they carry an X-linked allele is equal to the frequency of the allele in the population, which therefore is also 0.05. The probability that a female will carry one copy of this realtively rare allele is actually a little less than twice the allelic frequncy, or a little less than 0.10. Particularly, nearly twice the male probability because the female has two chances of carrying the allele, i.e., one chance for each X chromosome she carries, but, since the allele is relatively rare, a relatively low chance of picking up both alleles (the latter chance is part of the reason this value is a little less than twice the male rate, i.e., the odds of picking up one allele is some value less the odds of picking up two alleles). In fact, the more precise calculation of the frequency of the heterozygote in this case is simply 2pq or 2 * 0.05 * (1 - 0.05) which equals 0.095. The probability of female affliction is equal to the frequency of the homozygous recessive, q2 or 0.052 = 0.0025. Note, for the sake of checking these answers, that the rate at which females are neither afflicted nor carriers is p2 or (1 - 0.05)2 = 0.9025. These values should all add up to one and they do: 0.095 + 0.0025 + 0.9025 = 1.000.
      9. Answering this question is conceptionally easy, but a lot of work in practice. First figure out the possible genotypes, then multiply out the allele frequencies for each allele of a given genotype, then check yourself by making sure that frequences add up to 1.0 (did you remember to multiply the frequency of all of the heterozygotes by 2, i.e., as in 2pq?). Thus:

1.                                                                              A1A1, 0.1 * 0.1 = 0.01

2.                                                                              A1A2, 0.1 * 0.2 * 2 = 0.04

3.                                                                              A1A3, 0.1 * 0.3 * 2 = 0.06

4.                                                                              A1A4, 0.1 * 0.4 * 2= 0.08

5.                                                                              A2A2, 0.2 * 0.2 = 0.04

6.                                                                              A2A3, 0.2 * 0.3 * 2 = 0.12

7.                                                                              A2A4, 0.2 * 0.4 * 2 = 0.16

8.                                                                              A3A3, 0.3 * 0.3 = 0.09

9.                                                                              A3A4, 0.3 * 0.4 * 2 = 0.24

10.                                                                          A4A4, 0.4 * 0.4 = 0.16

      1. Stating that the loci are in linkage equilibrium simply means that we are assuming that there are no biases in allele combinations. It is very important in answering this question that you keep in mind that what we are looking for are gamete frequencies, not diploid frequencies! Here, then, the frequency of any given gamete is equal to the product of the frequencies of the constituting alleles (which also, by the way, need to be calculated to answer this question: p2 = 1 - p1 = 1 - 0.3 = 0.7 and q3 = 1 - q1 - q2 = 1 - 0.2 - 0.3 = 0.5). However, unlike above, we are not calculating heterozygote frequencies so avoid multiplying by 2! Finally, as usual, check yourself by making sure that all of the frequencies add up to 1.0. Thus:

1.                                                                              A1 B1, 0.3 * 0.2 = 0.06

2.                                                                              A1 B2, 0.3 * 0.3 = 0.09

3.                                                                              A1 B3, 0.3 * 0.5 = 0.15

4.                                                                              A2 B1, 0.7 * 0.2 = 0.14

5.                                                                              A2 B2, 0.7 * 0.3 = 0.21

6.                                                                              A2 B3, 0.7 * 0.5 = 0.35

      1. Don't let the notation throw you. w11 is just the relative fitness of AA, w12 the relative fitness of Aa, etc., though note that here relative fitness is being described solely in terms of survival. Regardless, multiply frequencies by relative viabilities: (0.16)(1.0) = 0.16, (0.48)(0.8) = 0.384, and (0.36)(0.6) = 0.216. These sum to 0.76 and these numbers thus translate to frequencies of 0.211, 0.505, and 0.284 for AA, Aa, and aa, respectively. The question specifically asked for zygote frequencies. These are just allelic frequencies. So 0.211 + 0.505/2 = 0.464 is the frequency of the A gamete and 0.536 the frequency of the a gamete. Assuming Hardy-Weinberg-like reestablishment of the next generation's zygotes (i.e., random mating, large population, no additional evolutionary change in allele frequency) this implies frequencies of AA, Aa, and aa of (0.464)2, 2(0.464)(0.536), and (0.536)2 or 0.215, 0.497, and 0.287, respectively.
    1. Questions from pp. 253 of Sinnott et al., 1958:
      1. The frequency of IAIA is 0.2062 * 0.2062 = 0.0425. The frequency of IBIB is 0.0756* 0.0756 = 0.0057. The frequency of II is 0.7183 * 0.7183 = 0.5160. The frequency of IAIB is 2 * 0.2062 * 0.0756 = 0.0312. The frequency of IAI is 2 * 0.2062 * 0.7183 = 0.2962. The frequency of IBI is 2 * 0.0756 * 0.7183 = 0.1086. 0.0425 + 0.0057 + 0.5160 + 0.0312 + 0.2962 + 0.1086 = 1.0002 which is close enough to one (assuming rounding error) to assume that I have not only listed all of the possible genotypes but have properly determined their frequencies assuming Hardy-Weinberg conditions. The frequency of the A phenotype is equal to the sum of the frequencies of the IAIA and IAI genotypes = 0.0425 + 0.2962 = 0.3387. The frequency of the B phenotype is equal to the sum of the frequencies of the IBIB and IBI genotypes = 0.0057 + 0.1086 = 0.1143. The frequency of the AB phenotype is equal to the frequency of the IAIB genotypes = 0.0312. The frequency of the O phenotype is equal to the frequency of the II genotypes = 0.5160. Once again, check to make sure that the frequencies add up to one. This will only happen with high likelihood if you have done all of the calculations correctly: 0.3387 + 0.1143 + 0.0312 + 0.5160 = 1.0002.
      2. THE FOLLOWING CALCULATIONS START OFF WITH AN ERROR AND SHOULD BE DISREGARDED UNTIL THIS ERROR IS CORRECTED; MY APPOLOGIES. The frequency of D is 0.99 and 0.01 for d. The frequencies of DD, Dd, and dd would be 0.9801, 0.0198, and 0.0001, respectively. Erythroblastosis only occurs if the baby is Rh-positive and the mother Rh-negative. If the mother is Rh-negative, then the odds of the baby being Rh-positive are equal to the odds of the father being DD plus the frequency of the father being Dd divided by 2 which is 0.9801 + 0.0198/2 = 0.9900. The odds that the mother in any given mating is Rh-negative are 0.0001. Thus, the odds that an Rh-positive baby will be carried by an Rh-negative mother are 0.9900 * 0.0001 = 0.000099 (or almost every baby born to an Rh-negative mother in this population). However, that is not what the question asked (and do you ever get the feeling that perhaps the authors of question such as these put twists like that in their questions perhaps inadvertently?). Instead it asked about what the odds of any women having a baby who is affected by erythroblastosis, and this condition only occurs if a second Rh-positive baby is carried by an Rh-negative woman. Thus, the probability is dependent on how many babies these women have. If we assume that they have two babies, no more, no less, then the probability of having a baby which is affected by erythroblastosis are equal to the odds of any woman being Rh-negative and carrying an Rh-positive child, times the odds that an Rh-negative woman's second child will also be Rh-positive (since the woman stays constant in this exercise, she doesn't get refactored into the equation; if you asked the question, what are the odds of two woman-child pairs being picked at random where the woman is Rh-negative and the child is Rh-positive in this population, then you would square 0.000099 instead of multiplying it by 0.99), i.e., 0.000099 * 0.99 = 0.000098. This isn't a very high incidence in the population, but note that it means that 98 out of every 100 Rh-negative women who mate randomly in this population (though not necessarily with more than one man) and has two children will face this problem.
      3. Because male's are haploid for the X chromosome, q = 0.08. Since females are diploid for the X chromosome, their rate of color blindness in this population would be q2 = 0.0064 or 0.64%.
    2. Questions from p. 484-485 of Hartl, 1983:
      1. Note that since the frequency of the three phenotypes equals 1.0 (i.e., 298 + 489 + 213 = 1000) that this implies that either this is a 1 locus, 2 allele system, or that this population is not in Hardy-Weinberg equilibrium (i.e., if there were a third, recessive allele, it apparently is not found in this population in the homozygous state). In addition, it would appear that the two alleles display codominance. Thus, the three genotypes associated with the three phenotypes, M, MN, and N, are MM, MN, and NN. The allelic frequency of M is (298 + 0.5 * 489) / 1000 = 0.5425. The allelic frequency of N is (213 + 0.5 * 489) / 1000 = 0.4575 = 1 - 0.5425. Given Hardy-Weinberg equilibrium, the expected genotype frequencies of MM, MN, and NN are 0.54252, 2 * 0.5425 * 0.4575, and 0.45752, respectively, which translates to 0.29, 0.50, and 0.21, respectively. These add up to 1.0. The associated numbers given a population size of 1000 are 290, 500, and 210, respectively, which his pretty close to the numbers observed.
      2. Given that this is a standard autosomal recessive allele, the frequency of d is the square root of 0.16 which is 0.4. Thus the frequency of D must be 0.6. The three possible genotypes are DD, Dd, dd and though the problem didn't state it, we will assume the null state and calculate the frequencies of these genotypes assuming Hardy-Weinberg equilibrium which are 0.36, 0.48, and, of course, 0.16 respectively. 0.36 + 0.48 + 0.16 = 1.0.
      3. The frequency of the recessive allele is simply the square root of 0.66 = 0.81. The frequency of heterozygotes must therefore be 2 * 0.81 * (1 - 0.81) = 0.30. 0.66 + 0.30 + (1 - 0.81)2 = 1.0. The last question is the tough one since it is not asking for the overall frequency but instead the frequency among a subgroup. Individuals with wet type cerumen include the homozygous dominant and the heterozygotes. Thus, the frequency of the heterozygotes among individual with wet type cerumen is equal to the frequency of heterozygotes divided by the sum of the frequency of heterozygotes and the frequency of homozygous dominants: 0.30 / (0.30 + 0.035) = 0.90, or 90 per cent.
      4. The frequency of type A blood = IA2 + 2 * IA * IO = 0.4347. The frequency of type B blood = IB2 + 2 * IB * IO = 0.084. The frequency of type AB blood = 2 * IA * IA = 0.0324. The frequency of type O blood = IO2 = 0.4489. 0.4347 + 0.084 + 0.0324 + 0.4489 = 1.000.
  1. References
    1. Campbell, N. A. (1996). Biology. Fourth Edition. Benjamin/Cummings Publishing, Menlo Park, California. pp. 434-435.
    2. Hartl, D. L. (1980). Principles of Population Genetics. Sinauer Ass., Inc., Sunderland, Massachusetts. pp. 78, 137-139.
    3. Hartl, D.L. (1983). Human Genetics. Harper & Row, Publishers, New York. pp. 484-485.
    4. Sinnott, E.W., Dunn, L.C., Dobzhansky, T. (1958). Principles of Genetics. Fifth Edition. McGraw-Hill Book Co., Inc. New York. p. 253.