Population Genetics Practice ProblemsPractice 1 In a population gene pool, the alleles A and a are at initial frequencies p and q, respectively. Prove that the gene frequencies and the zygotic frequencies do not change from generation to generation as long as the Hardy-Weinberg conditions are maintained. Solution 1 Zygotic frequencies generated by random mating are p2 (AA) + 2 pq(Aa) + q2(aa) =1 All of the gametes of AA individuals and half of the gametes of heterozygotes will bear the dominant allele A. Then the frequency of A in the gene pool of the next generation is p2 + pq = p2 + p (1 – p) = p2 + p – p2 = p Thus, each generation of random mating under Hardy-Weinberg conditions fails to change either the allelic or zygotic frequencies. Half the offspring from this mating are expected to be AA [(1/2) (4p3q) = 2 p3q]. and half are expected to be Aa (again with the frequency 2p3q). The matings AA × Aa occur with the frequency 4p3q. . Solution 2 There are six kinds of matings (ignoring male-female differences) that are easily generated in a mating table. Similar reasoning generates the frequencies of genotypes among the progeny shown in the following table.Practice 2 Prove the Hardy-Weinberg law by finding the frequencies of all possible kinds of matings and from these generating the frequencies of genotypes among the progeny using the symbols shown below. Practice 3 At what allelic frequency does the homozygous recessive genotype (aa) become twice as frequent as the heterozygous genotype Aa in a Hardy-Weinberg population? Solution 3 Let q = frequency of recessive allele. p = frequency of dominant allele. The frequency of homozygous recessives (q2) is twice as frequent at heterozygotes (2pq) when Therefore. or Proof: Practice 4 . either q = 0 (which is obviously an incorrect solution). 6 or 60%. the frequency of allele B is 0. each bearing one of the alleles at the locus under consideration). let us calculate the frequency of the CR allele. There are 144 roan individuals each carrying only one CR allele. calculate the estimated frequencies of the CR allele and the CW allele in the gene pool of the population. Suppose that a sample of 900 sheep of the Rambouillet breed in Idaho gave the following data: 891 white and 9 black. we can take the square root of that percentage of the population that is of the recessive genotype (phenotype) as our estimator for the frequency of the recessive allele. and CWCW is white. (b) If this population is completely panmictic. The fraction of all alleles in our sample of type CR becomes 360/600 = 0. what zygotic frequencies would be expected in the next generation? (c) How does the sample data in part (a) compare with the expectations for the next generation in part (b)? Is the population represented in part (a) in equilibrium? Solution 5 a. Since each individual is a diploid (possessing two sets of chromosomes. 48 white. Thus. 2 × 108 = 216 CR alleles. the total number of alleles represented in this sample is 300 × 2 = 600. There are 108 red individuals each carrying two CR alleles. We can arrive at this estimate for CW by following the same procedure as above. Practice 5 In Shorthorn cattle. There are 48 × 2 = 96 CW alleles represented in . The other 40%of the alleles in the gene pool must be of type CW. Since p + q = 1.0 If we assume the population is in equilibrium.White wool is dependent upon a dominant allele B and black wool upon its recessive allele b. and 144 roan animals were found in a sample of Shorthorns from the central valley of California. Estimate the allelic frequencies. the genotype CRCR is phenotypically red. First. 1 × 144 = 144 CR alleles. CRCW is roan (a mixture of red and white). Solution 4 p2 (BB) + 2pq (Bb) + q2 (bb) = 1. (a) If 108 red.9. the total number of CR alleles in our sample is 216 + 144 = 360. 6. the allele for long finger s2 is recessive.4) = 0. Solution 6 Since the dominance relationships are reversed in the two sexes.4)2 = 0. 96 + 144 = 240. Calculate the expected frequencies of long and short index fingers in females of this population. let us use all lowercase letters with superscripts to avoid confusion with either dominance or codominance symbolism. 0. Practice 7 .6) (0. A sample of the males in this population was found to contain 120 short and 210 long index fingers. Then according to the Hardy Weinberg law. and the frequency of the CW allele be represented by q = 0. we would expect as genotypic frequencies in the next generation p2 = (0.48(300) = 144 CRCW (roan).16CWCW c. The other 96% should possess long index fingers.6)2 = 0/36CRCR: 2pq = 2 (0. short index finger is recessive. q = frequency of s2 allele. and 0.16(300) = 48 CWCW (white). the original population must already be in equilibrium.04 or 4% of the females of this population will probably be short fingered.4 or 40% CW alleles. Recall that panmixis is synonymous with random mating. 240/600 = 0. the homoygotes and 144 in the heterozygotes.2)2 = 0.36(300) = 108 CRCR (red).48CRCW: q2 = (0. We will let the frequency of the CR allele be represented by p = 0. Since the genotypic and gametic frequencies are not expected to change in the next generation. Then In females. Inmales. Note that these figures correspond exactly to those of our sample. p2 (s1s1) + 2pq (s1s2) + q2 (s2s2) = 1. Let p = frequency of s1 allele.0. b. Practice 6 In the human population.4. an index finger shorter than the ring finger is thought to be governed by a sex-influenced gene that appears to be dominant in males and recessive in females. Then p2 = (0. In a sample of size 300 we would expect 0. Three alleles. Presenting the solutions in a slightly different form. Let p = frequency of IA allele. B. 12% type B. and 3% type AB. q = frequency of IB allele. The expansion of (p + q + r)2 yields the zygotic ratio expected under random mating. B. Solving for the frequency of the recessive allele i. r = frequency of i allele. the frequencies of the ABO blood groups were found to be approximately 49% type O. Frequency of allele i . IA. and i. form the dominance hierarchy (IA = IB) > i. following the method for obtaining the frequency of the IA allele. what percentage of type A individuals are probably homozygous? Solution 7 a. What are the allelic frequencies in this population? (d) Given the population in part (c) above. (c) Among New York Caucasians. Let A. c. IB. respectively. (a) Determine the genotypic and phenotypic expectations for this blood group locus from a population in genetic equilibrium. and O. (b) Derive a formula for use in finding the allelic frequencies at the ABO blood group locus. 36% type A. Solving for the frequency of the IA allele.The ABO blood group system is governed by a multiple allelic system in which some codominant relationships exist. b. Or. and O represent the phenotypic frequencies of blood groups A. Solving for the frequency of the IB allele q = 1 – p – r. Since females possess two X chromosomes (hence two alleles).22 + 0. (b) What percentage of the females in this population would be expected to be white-eyed? Solution 8 a. Thus. b. (a) Estimate the frequency of the w+ allele and the w allele in the gene pool. 48/356 = 0.00 d. Frequency of IB allele Frequency of allelle IA Check: p + q + r = 0. A laboratory population of Drosophila was found to contain 170 red-eyed males and 30 white-eyed males. 30 of the 200 X chromosomes in this sample carry the recessive allele w.0 or 100% of the females .135 or 13.70 = 1. Practice 8 White eye color in Drosophila is governed by a sex-linked recessive gene w and wild-type (red) eye color is produced by its dominant allele w+.5% of all group A individuals in this population are expected to be homozygous.08 + 0. p2 (w+w+) + 2pq (w+w) + q2 (ww) = 1. Thus. their expectations may be calculated in the same manner as that used for autosomal genes. 893 = 0. Solution 9 The total number of CB alleles in this sample is 311 + 2(277) + 54 = 919. Therefore. we must calculate the allelic frequencies.The total number of alleles (X chromosomes) in this sample is 353 + 2(338) = 1029. CBCY female are tortoiseshell (blotches of yellow and black). q2 = (0. Are the genotypes in this sample conforming to the frequencies expected for a Hardy-Weinberg population within statistically acceptable limits? Solution 10 First. the frequency of the CB allele is 919/1029 = 0. 35 Hp1/Hp2. CYCY females or CYY males are yellow.107. Practice 9 The genetics of coat colors in cats was presented in Example 9.4: CBCB females or CBY males are black. The frequency of the CY allele would then be 1 – 0.15)2 = 0.893. A population of cats in London was found to consist of the following phenotypes: Determine the allelic frequencies using all of the available information. Let From these gene (allelic) frequencies we can determine the genotypic frequencies expected according to the Hardy-Weinberg equation. . A sample of 100 individuals has 10 Hp1/Hp1. and 55 Hp2/Hp2.0225 or 2:25% of all females in the population are expected to be white- eyed. Practice 10 A human serum protein called haptoglobin has two major electrophoretic variants produced by a pair of codominant alleles Hp1 and Hp2. This is not a significant x2 value. The frizzled phenotype is produced by the heterozygous genotype MNMF. and we may accept the hypothesis that this sample (and hence presumably the population from which it was drawn) is conforming to the equilibrium distribution of genotypes." The other homozygous genotype MNMN has normal plumage. Is this population in equilibrium? Solution 11 Let . we can do a chi-square test. that for "frizzled" feathers.Converting these genotypic frequencies to numbers based on a total sample size of 100. Practice 11 One of the "breeds" of poultry has been largely built on a single-gene locus. One homozygote MFMF produces extremely frizzled birds called "woolies. 150 normal. and 50 wooly birds. A sample of 1000 individuals of this "breed" in the United States contained 800 frizzled. The frizzled heterozygotes represent the "breed" type and are kept for show purposes as well as for breeding by bird fanciers. Much artificial selection (by people) is being practiced. The explanation for the large deviation from the equilibrium expectations is twofold. . and lay fewer eggs than do the normal birds. are slower to reach sexual maturity.Chi-square test for conformity to equilibrium expectations gives the following results: This highly significant chi-square value will not allow us to accept the hypothesis of conformity with equilibrium expectations. Such breeders dispose of (cull) many normal and wooly types. Natural selection is also operative on the wooly types because they tend to lose their feathers (loss of insulation) and eat more feed just to maintain themselves.
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