Name __________________________ BIO 2110 – Exam 3 5 November 2014 10 questions – 10 points/question see last page for potentially useful formulae show your work on all problemsName __________________________Name __________________________ 1. Shown below is an alignment of five different random sequences taken from a population. From this alignment, estimate the effective population size (Ne) based on both the nucleotide diversity (!) and the number of segregating sites (k). [µ = 10-8 mutations/generation/nucleotide. Positions with nucleotide variation shaded.] 1 GATCAGTCAT ACGGATCCTA GACTACGACT AGCCTAGCAT CCGAAATCAA 2 GATCAGTCAT ACCGATCCTA GACTACGACT AGCCTAGCAT CCGAAATCAA 3 GATCAGTCAT ACGGATCCTA GACTACGACT AACCTAGCAT CCGAAATCAA 4 GATCAGTCAT ACGGATCCTA GACTACGACT AACCTAGCAT CCGAAATCAA 5 GATCAGTCAT ACGGATCCTA GTCTACGACT AGCCTAGCAT CCGAAATCAAName __________________________ 2. The frequency at which humming fish hum is determined by allelic variation at a single gene (see below). This variation has selective consequences, lowered pitched hums can be used to communicate over greater distances than higher pitched hums. Given this selective regime and an initial frequency of the ‘A’ allele of p = 0.2, what allelic and genotypic frequencies would you expect after two generations of selection? genotype phenotype relative fitness AA 60 Hz WAA = 1.0 Aa 60 Hz WAa = 1.0 aa 470 Hz Waa = 0.3Name __________________________ 3. In two populations of giraffes, average neck lengths are six feet long. In crosses between these giraffe populations, F1 progeny have six foot long necks. Neck lengths in F2 progeny vary from two feet long to ten feet long. Out of 1,792 F2 giraffes, seven have necks that are two feet long. i. Why does phenotype variation appear in the F2 generation when it is absent in the F1 and P0 generations? ii. How many genes are involved in neck length variation in these giraffes? iii. What other neck lengths would be observed in the F2 generation? iv. Out of 1,792 giraffes, how many would be expected in each phenotypic class?Name __________________________ 4. In populations of medieval flying rabbits, one locus (A) on chromosome 3 has been found responsible for the carnivorous diet phenotype. Geneticists want to determine if an adjacent locus on chromosome 3 is in linkage equilibrium with locus A. It appears that the carnivorous diet phenotype is greatly associated with the long fang phenotype locus (B), recombinants are rare (i.e. long-fanged, vegans not so prominent in the population). Determine the relationship between these loci using the sample population given below. Are these two loci in linkage equilibrium? If not, then what is the disequilibrium value for this population? [variant positions high-lighted] A B ⇩ ⇩ Ind. A: GACTA CCCTA GGTAC TAACT TTTGG locus A: carnivorous diet Ind. B: GAGTA CCCTA GGTAC AAACT TTTGG A1 = C Ind. C: GAGTA CCCTA GGTAC AAACT TTTGG A2 = G Ind. D: GAGTA CCCTA GGTAC AAACT TTTGG Ind. E: GAGTA CCCTA GGTAC AAACT TTTGG Ind. F: GACTA CCCTA GGTAC AAACT TTTGG Ind. G: GACTA CCCTA GGTAC AAACT TTTGG locus B: long fang Ind. H: GACTA CCCTA GGTAC AAACT TTTGG B1 = A Ind. I: GAGTA CCCTA GGTAC TAACT TTTGG B2 = T Ind. J: GACTA CCCTA GGTAC AAACT TTTGG Ind. K: GACTA CCCTA GGTAC AAACT TTTGG Ind. L: GAGTA CCCTA GGTAC TAACT TTTGG Ind. M: GACTA CCCTA GGTAC AAACT TTTGG Ind. N: GACTA CCCTA GGTAC TAACT TTTGG Ind. O: GAGTA CCCTA GGTAC TAACT TTTGG Ind. P: GAGTA CCCTA GGTAC TAACT TTTGG Ind. Q: GACTA CCCTA GGTAC AAACT TTTGG Ind. R: GAGTA CCCTA GGTAC TAACT TTTGG Ind. S: GACTA CCCTA GGTAC AAACT TTTGG Ind. T: GAGTA CCCTA GGTAC TAACT TTTGGName __________________________ 5. In a population of barbary cats, vocalizations are determined by allelic variation at the voc gene. vocM homozygotes meow, vocR homozygotes roar and vocM/vocR heterozygotes bark. A census of vocalization sounds in 1,000 barbary cats is shown below. Based on these data, what is the inbreeding coefficient (FIS) for this barbary cat population? vocalization number meow 553 bark 294 roar 153Name __________________________ 6. The average per nucleotide divergence between two species of grackles is 0.0156. The mutation rate in these species is 3 x 10-8 mutations per nucleotide per generation. Assuming neutrality, how many generations has it been since these grackle species diverged?Name __________________________ 7. Wing length in pterodactyls is a quantitative character. From a QTL study, QTL and no QTL probabilities for three markers are shown below. Based on these probabilities, which of these markers would be linked to a QTL involved in the regulation of wing length? Use a Lod score of 3.0 as your cut-off for significance. Show your work. probabilities marker QTL no QTL A1-3 0.9900 0.0100 C2-6 0.9993 0.0007 E6-9 0.8692 0.1308Name __________________________ 8. In Dendrobates leucomelas, variation in toxin secretion results from allelic variation in a single gene. The presence of toxins is dominant. In population A, 84% of frogs secrete toxins (pA 0 = 0.6). In population B, 36% of frogs secrete toxins (pB 0 = 0.2). Frogs migrate from population A to population B (mAB = 0.1) but not from population B to population A (mBA = 0.0). How would allelic and genotypic frequencies change in population B over two generations of migration?Name __________________________ 9. Hump number in Dactrianedary camels varies from 1 to 3. This variation results from allelic variation in a single gene. Dominance relationships among the three alleles of this gene are shown below. In one population, there are 510 three-humped camels, 130 two-humped camels and 360 one-humped camels. Based on the frequencies of these phenotypes, what are the frequencies of each allele in this population? genotype number of humps HH Hh Hη three hh hη two ηη oneName __________________________ 10. In a population of 1,531 zebras, a recessive mutation arises that results in the absence of stripes. What is the initial frequency of this mutation? Assuming selective neutrality, what are the initial probabilities of fixation and extinction for this allele?Name __________________________Name __________________________ p2 + q2 + r2 + 2pq + 2pr + 2qr = 1 FIS = 1 - (fAa/2pq) pn+1 = (WAA)(p2) + (WAa)(pq) 1/2N (2N -1)/2N W p + q + r = 1 PA1 = G1 + G2