The horizontal axis is labeled Year, and the vertical axis is labeled Resistant Isolates, by percent. Eight tick marks appear on the horizontal axis and are labeled, from left to right, one through 8, in increments of one. Five tick marks appear on the vertical axis and are labeled, from bottom to top, zero through 20, in increments of 5. On the graph, there are 8 data points connected by a curve drawn through the points. The first and last data points are the start and end of the curve, and the approximate coordinates of the data points are as follows: Point One: one comma 2. Point 2: two comma 2 point five. Point 3: three comma 2 Point 4: four comma 3. Point 5: five comma 4. Point 6: six comma 7. Point 7: seven comma 14. Point 8: eight comma 21. Over several years, bacteria were isolated from members of a human population and tested for antibiotic resistance. The percent of bacterial isolates that were found to be antibiotic resistant is presented in the graph above for each year of the study. Which of the following conclusions is best supported by the information presented in the graph?

The question refers to the following DNA strand and table of codons: Each triplet of DNA bases is numbered from one to seven. Triplet 1 is T, A, G, triplet 2 is T, T, C, triplet 3 is A, A, A, triplet 4 is C, C, G, triplet 5 is C, G, T, triplet 6 is A, A, C, triplet 7 is A, T, T. The left side of the table is labeled First Base in Codon, and labels the main rows, from top to bottom, U, C, A, G. The top side of the table is labeled Second Base in Codon, and labels the main columns, from left to right, U, C, A, G. The right side of the table is labeled, Third Base in Codon, and labels each of the main rows U C A G.The data in the table reads as follows: First Base U and Second Base U with Third Base U, results in U U U phenylalanine; with Third Base C results in U U C phenylalanine; with Third Base A, results in U U A leucine, and with Third Base G, results in U U G leucine. First Base C and Second Base U with Third Base U, results in C U U leucine; with Third Base C, results in C U C leucine; with: Third Base A, results in C U A leucine, and with Third Base G, results in C U A leucine. First Base A and Second Base U with Third Base U, results in A U U isoleucine; with Third Base C, results in A U C isoleucine; with Third Base A, results in A U A isoleucine; and with Third Base G, results in A U G methionine or start. First Base G and Second Base U with Third Base U, results in G U U valine; with Third Base C, results in G U C valine; with Third Base A, results in G U A valine, with Third Base G, results in G U G valine. First Base U and Second Base C with Third Base U, results in U C U serine; with Third Base C, results in U C C serine; with Third Base A, results in U C A serine; and with Third Base G, results in U C G serine. First Base C and Second Base C with Third Base U, results in C C U proline; with Third Base C, results in C C C proline; with Third Base A, results in C C A proline; and with Third Base G, results in C C G proline. First Base A and Second Base C with Third Base U, results in A C U threonine; with Third Base C, results in A C C threonine; with Third Base A, results in A C A threonine; and with Third Base G, results in A C G threonine. First Base G and Second Base C with Third Base U, results in G C U alanine; with Third Base C, results in G C C alanine; with Third Base A, results in G C A alanine; and with Third Base G, results in G C G alanine. First Base U and Second Base A with Third Base U, results in U A U tyrosine; with Third Base C, results in U A C tyrosine; with Third Base A, results in U A A stop; and with Third Base G, results in U A G stop. First Base C and Second Base A with Third Base U, results in C A U histidine; with Third Base C, results in C A C histidine; with Third Base A, results in C A A glutamine; and with Third Base G, results in C A G glutamine. First Base A and Second Base A with Third Base U, results in A A U asparagine; with Third Base C, results in A A C asparagine; with Third Base A, results in A A A lysine; and with Third Base G, results in A A G lysine. First Base G and Second Base A with Third Base U, results in G A U aspartate; with Third Base C, results in G A C aspartate; with Third Base A, results in G A A glutamate; and with Third Base G, results in GAG glutamate. First Base U and Second Base G with Third Base U, results in U G U cysteine; with Third Base C, results in U G C cysteine; with Third Base A, results in U G A stop; and with Third Base G, results in U G G tryptophan. First Base C and Second Base G with Third Base U, results in C G U arginine; with Third Base C, results in C G C arginine; with Third Base A, results in C G A arginine; and with Third Base G, results in C G G arginine. First Base A and Second Base G with Third Base U, results in A G U serine; with Third Base C, results in A G C serine; with Third Base A, results in A G A arginine; and with Third Base G, results in A G G arginine. First Base G and Second Base G with Third Base U, results in G G U glycine; with Third Base C, results in G G C glycine; with Third Base A, results in G G A glycine; and with Third Base G, results in G G G glycine. The mRNA transcribed from the DNA would read

Percent of Moths Recaptured in Two Different Environments Trial Moth Color and Environment Percent of Released Moths Recaptured I Light-colored moths were released in an unpolluted environment. 13% II Light-colored moths were released in a polluted environment. 12% III Dark-colored moths were released in an unpolluted environment. 7% IV Dark-colored moths were released in a polluted environment. 28% A researcher released large numbers of moths into different environments in an attempt to better understand a mechanism of evolution. The moths were released in four trials as described in the table above. Each of the released moths had a small mark on the underside of a wing for identification. After an appropriate amount of time, the researcher recaptured as many of the released moths as possible. Data from the experiment are included in the table above. Which of the following claims is best supported by the data?

Ascorbic acid (vitamin C) is an organic molecule necessary for the health of plants and animals. The majority of animals, including most mammals, synthesize ascorbic acid from organic precursors, but some primates are unable to synthesize ascorbic acid and must instead acquire it from dietary sources, such as certain fruits and vegetables. The L-gulonolactone oxidase (GULO) gene encodes an enzyme that catalyzes a required step in the biosynthesis of ascorbic acid. Most mammals carry a functional copy of the GULO gene, but some primates carry only a GULO pseudogene, which is a nonfunctional variant. A comparison of GULO genes and GULO pseudogenes from different animals can provide insight into the evolutionary relatedness of the animals. In Table I, selected members of some mammalian groups are listed, along with an indication of their ability to synthesize ascorbic acid. Table II shows an alignment of amino acid coding sequences from homologous regions of the GULO genes and GULO pseudogenes of the organisms listed in Table I. Figure 1 represents the universal genetic code. The title of the table is SELECTED MAMMALIAN GROUPS. The top row contains the column labels, from left to right: column one, Group; column two, Selected Members; column three, Biosynthesis of Ascorbic Acid. From top to bottom, the data is as follows: Row two: Group, Nonprimate mammals; Selected Members, Elephant, mouse; Biosynthesis of Ascorbic Acid, Yes. Row three: Group, Primate mammals; Selected Members, Lemur; Biosynthesis of Ascorbic Acid, Yes. Row four: Group, Primate mammals; Selected Members, Orangutan, chimpanzee; Biosynthesis of Ascorbic Acid, No. Row five: Group, Primate mammals; Selected Members, Human; Biosynthesis of Ascorbic Acid, No. It lists the relative positions of nucleotides in a non-template (coding) sequence. The table consists of six rows and twenty-seven columns. The row headers are as follows: elephant, mouse, lemur, orangutan, chimp, and human. The column headers run from 5 prime to 3 prime, displaying the positions from 1 (at 5 prime) through 27 (at 3 prime). The row-wise entries from the table are as follows. Row 1, Elephant. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: C (shaded), 6: C (shaded), 7: C, 8: A, 9: T, 10: C (shaded), 11: T (shaded), 12: G (shaded), 13: A, 14: A, 15: G, 16: A (shaded), 17: A (shaded), 18: G (shaded), 19: T, 20: C, 21: G, 22: G (shaded), 23: A (shaded), 24: A (shaded), 25: T, 26: A, 27 (3 prime): C. Row 2, Mouse. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: G (shaded), 6: C (shaded), 7: C, 8: A, 9: C, 10: C (shaded), 11: T (shaded), 12: G (shaded), 13: A, 14: A, 15: G, 16: A (shaded), 17: A (shaded), 18: G (shaded), 19: T, 20: C, 21: T, 22: G (shaded), 23: A (shaded), 24: G (shaded), 25: T, 26: A, 27 (3 prime): C. Row 3, Lemur. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: G (shaded), 6: C (shaded), 7: C, 8: A, 9: C, 10: C (shaded), 11: T (shaded), 12: G (shaded), 13: A, 14: A, 15: G, 16: A (shaded), 17: G (shaded), 18: G (shaded), 19: T, 20: C, 21: C, 22: G (shaded), 23: A (shaded), 24: G (shaded), 25: T, 26: A, 27 (3 prime): C. Row 4, Orangutan. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: G (shaded), 6: C (shaded), 7: en-dash, 8: A, 9: T, 10: T, 11: G (shaded), 12: G (shaded), 13: A (shaded), 14: A, 15: G, 16: A, 17: A (shaded), 18: A (shaded), 19: T (shaded), 20: C, 21: T, 22: G, 23: A (shaded), 24: G (shaded), 25: G (shaded), 26: A, 27 (3 prime): C. Row 5, Chimp. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: G (shaded), 6: C (shaded), 7: en-dash, 8: A, 9: T, 10: T, 11: G (shaded), 12: G (shaded), 13: A (shaded), 14: A, 15: G, 16: A, 17: A (shaded), 18: A (shaded), 19: T (shaded), 20: C, 21: T, 22: G, 23: A (shaded), 24: G (shaded), 25: G (shaded), 26: A, 27 (3 prime): C. Row 6, Human. 1 (5 prime): G, 2: A, 3: C, 4: A (shaded), 5: G (shaded), 6: C (shaded), 7: en-dash, 8: A, 9: T, 10: T, 11: G (shaded), 12: G (shaded), 13: A (shaded), 14: A, 15: G, 16: A, 17: A (shaded), 18: A (shaded), 19: T (shaded), 20: C, 21: T, 22: G, 23: A (shaded), 24: G (shaded), 25: G (shaded), 26: A, 27 (3 prime): C. A footnote below the table reads: For each D N A segment, the alternating shaded and unshaded nucleotides indicate the triplet codons of the open reading frame, shown from left (5 prime) to right (3 prime) as the non-template (coding) strand. An “en-dash” indicates the absence of a nucleotide. The left side of the table is 5 Prime First Base, and labels the main rows, from top to bottom, U, C, A, G. The top side of the table is labeled Second Base, and labels the main columns, from left to right, U, C, A, G. The right side of the table is labeled, 3 Prime Third Base, and labels each of the main rows U C A G. The data in the table reads as follows; First Base U and Second Base U with Third Base U, results in U U U phenylalanine; with Third Base C results in U U C phenylalanine; with Third Base A, results in U U A leucine, and with Third Base G, results in U U G leucine First Base C and Second Base U with Third Base U, results in C U U leucine; with Third Base C, results in C U C leucine; with Third Base A, results in C U A leucine, and with Third Base G, results in C U G leucine First Base A and Second Base U with Third Base U, results in A U U isoleucine; with Third Base C, results in A U C isoleucine; with Third Base A, results in A U A isoleucine; and with Third Base G, results in A U G methionine or start First Base G and Second Base U with Third Base U, results in G U U valine; with Third Base C, results in G U C valine; with Third Base A, results in G U A valine, with Third Base G, results in G U G valine First Base U and Second Base C with Third Base U, results in U C U serine; with Third Base C, results in U C C serine; with Third Base A, results in U C A serine; and with Third Base G, results in U C G serine First Base C and Second Base C with Third Base U, results in C C U proline; with Third Base C, results in C C C proline; with Third Base A, results in C C A proline; and with Third Base G, results in C C G proline First Base A and Second Base C with Third Base U, results in A C U threonine; with Third Base C, results in A C C threonine; with Third Base A, results in A C A threonine; and with Third Base G, results in A C G threonine First Base G and Second Base C with Third Base U, results in G C U alanine; with Third Base C, results in G C C alanine; with Third Base A, results in G C A alanine; and with Third Base G, results in G C G alanine First Base U and Second Base A with Third Base U, results in U A U tyrosine; with Third Base C, results in U A C tyrosine; with Third Base A, results in U A A stop; and with Third Base G, results in U A G stop First Base C and Second Base A with Third Base U, results in C A U histidine; with Third Base C, results in C A C histidine; with Third Base A, results in C A A glutamine; and with Third Base G, results in C A G glutamine First Base A and Second Base A with Third Base U, results in A A U asparagine; with Third Base C, results in A A C asparagine; with Third Base A, results in A A A lysine; and with Third Base G, results in A A G lysine First Base G and Second Base A with Third Base U, results in G A U aspartate; with Third Base C, results in G A C aspartate; with Third Base A, results in G A A glutamate; and with Third Base G, results in G A G glutamate First Base U and Second Base G with Third Base U, results in U G U cysteine; with Third Base C, results in U G C cysteine; with Third Base A, results in U G A stop; and with Third Base G, results in U G G tryptophan First Base C and Second Base G with Third Base U, results in C G U arginine; with Third Base C, results in C G C arginine; with Third Base A, results in C G A arginine; and with Third Base G, results in C G G arginine First Base A and Second Base G with Third Base U, results in A G U serine; with Third Base C, results in A G C serine; with Third Base A, results in A G A arginine; and with Third Base G, results in A G G arginine First Base G and Second Base G with Third Base U, results in G G U glycine; with Third Base C, results in G G C glycine; with Third Base A, results in G G A glycine; and with Third Base G, results in G G G glycine. Lemurs are primates that live on the island of Madagascar off the coast of Africa. Lemurs have a functional GULO gene and are able to produce ascorbic acid. However, primates that live in other places (e.g., humans, chimpanzees, and orangutans) have a GULO pseudogene and are unable to produce ascorbic acid. Which of the following best explains the genetic variation among primate species?

The appearance of a fertile, polyploid individual within a population of diploid organisms is a possible source of a new species. If this individual is capable of reproducing to form a new population, scientists would consider this to be an example of

A(n) ______ can be described as an “energy budget.” It is the way plants and animals manage the tradeoffs between dedicating energy to survival and dedicating energy to reproduction.

Which of the following best describes an event during step 2 in the simplified model above?

A biologist named Hamilton first proposed the idea that animals might have an evolutionary incentive to behave altruistically in certain circumstances. His idea of inclusive fitness can be explained by the fact that 

In an investigation of interspecies competition, researchers grew the unicellular protozoan Paramecium aurelia in a 5 mL culture and Paramecium caudatum in a separate 5 mL culture. P. aurelia and P. caudatum were grown together in a third 5 mL culture. Each day a small sample of each culture was removed so the total number of individuals could be estimated, and the remainder of the population was transferred to fresh growth medium. The experimental results are represented in the graphs below. The horizontal axis is labeled “Time, in days,” and the numbers 1 through 25 are indicated. The vertical axis is labeled “Number of Individuals per 5 milliliters,” and the numbers O through 700, in increments of 100, are indicated. The data represented by the points on the graph are as follows. Point 1, 1 day, O individuals perO milliliters. Point 2, 2 days, 10 individuals per 5 milliliters. Point 3, 3 days, 20 individuals per 5 milliliters. Point 4, 4 days, 60 individuals per 5 milliliters. Point 5, 5 days, 90 individuals per 5 milliliters. Point 6, 6 days, 190 individuals per 5 milliliters. Point 7, 7 days, 260 individuals per 5 milliliters. Point 8, 8 days, 320 individuals per 5 milliliters. Point 9, 9 days, 410 individuals per 5 milliliters. Point 10, 10 days, 500 individuals per 5 milliliters. Point 11, 11 days, 570 individuals per 5 milliliters. Point 12, 12 days, 610 individuals per 5 milliliters. Point 13, 13 days, 510 individuals per 5 milliliters. Point 14, 14 days, 580 individuals per 5 milliliters. Point 15, 15 days, 550 individuals per 5 milliliters. Point 16, 16 days, 550 individuals per 5 milliliters. Point 17, 17 days, 510 individuals per 5 milliliters. Point 18, 18 days, 570 individuals per 5 milliliters. Point 19, 19 days, 510 individuals per 5 milliliters. The horizontal axis is labeled “Time, in days,” and the numbers 1 through 25 are indicated. The vertical axis is labeled “Number of Individuals per 5 milliliters,” and the numbers O through 250, in increments of 50, are indicated. The data represented by the points on the graph are as follows. Point 1, 1 day, 0 individuals per 5 milliliters. Point 2, 2 days, 10 individuals per 5 milliliters. Point 3, 3 days, 10 individuals per 5 milliliters. Point 4, 4 days, 10 individuals per 5 milliliters. Point 5, 5 days, 20 individuals per 5 milliliters. Point 6, 6 days, 60 individuals per 5 milliliters. Point 7, 7 days, 110 individuals per 5 milliliters. Point 8, 8 days, 140 individuals per 5 milliliters. Point 9, 9 days, 165 individuals per 5 milliliters. Point 10, 10 days, 190 individuals per 5 milliliters. Point 11, 11 days, 220 individuals per 5 milliliters. Point 12, 12 days, 200 individuals per 5 milliliters. Point 13, 13 days, 200 individuals per 5 milliliters. Point 14, 14 days, 180 individuals per 5 milliliters. Point 15, 15 days, 190 individuals per 5 milliliters. Point 16, 16 days, 180 individuals per 5 milliliters. Point 17, 17 days, 190 individuals per 5 milliliters. Point 18, 18 days, 205 individuals per 5 milliliters. Point 19, 19 days, 208 individuals per 5 milliliters. The horizontal axis is labeled “Time, in days,” and the numbers 1 through 25 are indicated, The vertical axis is labeled “Number of Individuals per 5 milliliters,” and the numbers O through 450, in increments of 50, are indicated. A key indicates that one line represents P aurelia, and the other line represents P caudatum. Both lines begin at 1 day, and 0 individuals per 5 milliliters. The line representing P caudatum spikes momentarily above the line representing P aurelia after 2 days, but then falls back down toward the horizontal axis and remains below the line representing P Aurelia until both graphs end. The data represented by the points on each line are as follows. Point 1, 1 day. P aurelia, 0. P caudatum, 0. Point 2, 2 days. P aurelia, 10. P caudatum, 145. Point 3, 3 days. P aurelia, 25. P caudatum, 10. Point 4, 4 days. P aurelia, 55. P caudatum, 30. Point 5, 5 days. P aurelia, 95. P caudatum, 50. Point 6, 6 days. P aurelia, 200. p caudatum, 90. Point 7, 7 days. P aurelia, iss. p caudatum, 110. Point 8, 8 days. P aurelia, 220. p caudatum, 125. Point 9, 9 days. P aurelia, 295. p caudatum, 100. Point 10, 10 days. P aurelia, 240. P caudatum, 90. Point 11, 11 days. P aurelia, 300. P caudatum, 70. Point 12, 12 days. P aurelia, 300. P caudatum, 90. Point 13, 13 days. P aurelia, 340. P caudatum, 60. Point 14, 14 days. P aurelia, 390. P caudatum, 70. Point 15, 15 days. P aurelia, 340. P caudatum, 55. Point 16, 16 days. P aurelia, 360. P caudatum, 56. Point 17, 17 days. P aurelia, 335. P caudatum, 48. Point 18, 18 days. P aurelia, 360. P caudatum, 50. Point 19, 19 days. P aurelia, 305. P caudatum, 50. Point 20, 20 days. P aurelia, 350. P caudatum, 50. Point 21, 21 days. P aurelia, 325. P caudatum, 48. Point 22, 22 days. P aurelia, 350. P caudatum, 20. Point 23, 23 days. P aurelia, 350. P caudatum, 20. Point 24, 24 days. P aurelia, 325. P caudatum, 40. Point 25, 25 days. P aurelia, 350. P caudatum, 25. Based on the experimental results, which of the following statements best describes the relationship of the two populations that were studied in the investigation?

Nine percent of a population is homozygous recessive (aa) at a certain locus. Assuming that the population is in Hardy-Weinberg equilibrium, which of the following is closest to the frequency of the recessive allele (a)?