The table below describes the action of two genes involved in the regulation of nervous system development in the nematode C. elegans. C. elegans nervous system regulation genes Gene A Gene B Observation Pattern 1 Inactive Inactive No neurons develop Pattern 2 Inactive Active No neurons develop Pattern 3 Active Inactive Greater-than-normal number of neurons develop Pattern 4 Active Active Normal number of neurons develop Which of the following claims is best supported by the data?

The three-spined stickleback (Gasterosteus aculeatus) is a small fish found in both marine and freshwater environments. Marine stickleback populations consist mainly of individuals with armor-like plates covering most of their body surface (completely plated). Approximately 10,000 years ago, some marine sticklebacks colonized freshwater environments. After many generations in the freshwater environments, the freshwater stickleback populations lacked the armor plating (low plated) typical of marine stickleback populations. Over the period between 1957 and 2005, one freshwater population, in Lake Washington, a lake in a coastal region of the northwestern United States, changed from having a majority of individuals of the low-plated phenotype to having more individuals of the completely-plated phenotype than of the low-plated phenotype. Figure 1 shows the distribution of plated phenotypes in Lake Washington sticklebacks at four time points between 1957 and 2005. The figure shows a vertical bar graph title Figure one, Armor plating phenotypes in Lake Washington stickleback population. There are 5 tick marks along the horizontal axis. Centered between each tick mark, from left to right, are the numbers 1957, 1968, 1976, and 2005. The vertical axis is label Percentage of Fish, and the numbers appearing on it, from bottom to top, are zero,20, 40, 60, 80, and 100. The graph shows 11 bars and a key indicates black bars are completely plated, shaded bars are partially plated, and white bars are low plated. From left to right, the data reads as approximately:1957: completely plated,no bar; partially plated, 10; low plated, 90.1968: completely plated, 7; partially plated, 24; low plated 70. 1976: completely plated, 40; partially plated, 35; low plated 24. 2005: completely plated, 50; partially plated, 35; low plated 15. A single gene, ectodysplasin (EDA), is thought to be responsible for the variation in the number of armor plates in sticklebacks. Figure 2 shows a phylogenetic tree constructed by comparing DNA sequences of the EDA gene from a number of stickleback populations with low-plated or completely plated phenotypes. Figure 3 shows a phylogenetic tree constructed by comparing the sequences of 25 genes that were randomly selected from the same populations as shown in Figure 2. In both figures, shaded populations display the completely plated phenotype. The figures show two phylogenetic trees titled Figure 2, Phylogeny based on EDA gene only, and Figure 3, Phylogeny based on 25 random genes. Shaded populations indicated completely plated phenotypes. Figure 2 on the left divides Populations 1 through 8 as low plated, and Populations 9 through 15 as completely plated.A large branch connects all low plated phenotypes to all completely plated phenotypes. On the top branch, a tree connects Populations 1 and 2 only, and a branch then connects them to Population 3. A branch then connects Populations 1 through 3 to Population 4. A tree connects Populations 5 and 6 only, and a branch is then connected from Populations 5 and 6 to Populations 1 through 4. This tree is then connected to Population 7.On the bottom branch, a tree connects Populations 8 and 9, and a tree connects Populations 10 and 11. A branch then connects Populations 8 and 9 to Populations 10 and 11. This branch is then connected to Population 12. A tree connects Populations 8 through 12 to Population 13, a branch connects Population 14 to Populations 8 through 13, and a branch connects Population 15 to Populations 8 through 14. Figure 3 on the right has a tree that connects Population 15 to Populations one through 14. A tree connects Populations 4 and 6 and a single branch extends to the tree connecting Population 15 to Populations one through 14. A tree connects Populations 14 and 7, and a branch connects this set to Population 5. A branch then connects this set to Population 12, another branch connects this set to Population 13, and another branch connects this set to Population 8. A tree connects Populations 11 and 9, a branch connects this set to Population 10, another branch connects this set to Population 1, another branch connects this set to Population 3, and another branch connects this set to Population 2. A tree connects Populations 14, 7, 5, 12, 13, and 8 to Populations 11, 9, 10, 1, 3 and 2. Prior to 1960, Lake Washington was highly polluted and underwater visibility was limited to one or two meters. In the late 1960s, a large cleanup effort reduced pollution, resulting in visibility that increased to six to seven meters by 1976. Which of the following best explains how the change in underwater visibility affected armor plating in Lake Washington sticklebacks between 1957 and 1976?

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