The diagram below shows a simple food web.   At the bottom of the diagram is a box labeled “Grass.” From the grass box, three arrows point upward to three boxes arranged horizontally in the middle level: “Rabbits” on the left, “Mice” in the center, and “Squirrels” on the right. From each of these three boxes, arrows point upward and converge on a single box at the top labeled “Foxes.” The arrows indicate that grass is consumed by rabbits, mice, and squirrels, and that all three of these herbivores are consumed by foxes. Which group of organisms is expected to have the highest overall productivity?

Conservationists are concerned that many native terrestrial species have become rare because of recent human activities. Which activity is likely the most important cause today for the decline of native terrestrial species of plants and animals worldwide?

Staphylococcus aureus is a pathogenic bacterium that can infect a wide range of host species, including humans. S. aureus has a particular protein that binds with hemoglobin from the host organism. Hemoglobin is the iron-containing protein used to transport oxygen in the blood. Since iron is important for growth, S. aureus have evolved the ability to absorb the iron from the host’s hemoglobin. Different S. aureus strains preferentially infect different hosts and have different amino acid sequences at their hemoglobin-binding domains (Table 1; letters indicate different amino acids). In an experiment, different S. aureus strains were mixed with hemoglobin from macaque monkeys and their binding ability was measured (Figure 1). The differences in amino acid sequences contributed to the differential binding abilities observed. Table 1. Selected amino acid sequences and preferred host for four strains of S. aureus S. aureus Strain Amino Acid Sequence Host Species 1 Q Q F Y H Y A R S Species A 2 R Q A Y H Y A R T Species B 3 Q Q A Y H Y A R T Macaque 4 R Q A A H Y Q L T Species C The horizontal axis is labeled S. aureus Strain, and the numbers 1, 2, 3 and 4 are indicated. The vertical axis is labeled Percent Hemoglobin Binding, and the numbers 0 through 100, in increments of 10, are indicated. The data represented in the graph are as follows. Note that all values are approximate. S. aureus Strain 1, 25 percent hemoglobin binding. S. aureus Strain 2, 60 percent hemoglobin binding. S. aureus Strain 3, 97 percent hemoglobin binding. S. aureus Strain 4, 35 percent hemoglobin binding. Figure 1. Macaque hemoglobin binding ability of different strains of S. aureus Which of the following experiments would be most appropriate to determine whether populations of S. aureus are continuously adapting in order to obtain iron from hosts more effectively?

In community ecology, the “non-equilibrium model” is based on the idea that

When a marsh is sprayed with pesticides to control mosquitoes, trace amounts of DDT* are released. DDT can passively cross cell membranes, so it is able to accumulate in the cells of various small aquatic organisms. When feeders up the food chain, such as clams and fish, eat these organisms, they consume DDT. DDT accumulates in the cells of consumers at levels up to ten times greater than organisms at the previous stage.  This is an example of ______. *The US banned the use of DDT in 1972, but some countries still used the chemical to control mosquitoes that spread malaria.

A scientist is using an ampicillin-sensitive strain of bacteria that cannot use lactose because it has a nonfunctional gene in the lac operon. She has two plasmids. One contains a functional copy of the affected gene of the lac operon, and the other contains the gene for ampicillin resistance. Using restrictions enzymes and DNA ligase, she forms a recombinant plasmid containing both genes. She then adds a high concentration of the plasmid to a tube of the bacteria in a medium for bacterial growth that contains glucose as the only energy source. This tube (+) and a control tube (-) with similar bacteria but no plasmid are both incubated under the appropriate conditions for growth and plasmid uptake. The scientist then spreads a sample of each bacterial culture (+ and -) on each of the three types of plates indicated below. The columns show the different media on the plates starting with glucose medium, then glucose medium with ampicillin and then glucose medium with ampicillin and lactose. The rows are bacterial strain with added plasmid, also labeled positive, and bacterial strain with no plasmid, also labeled negative. The plates are labeled number one through number six, with one, two and three with positive bacteria on the media in order, and four, five and six with negative bacteria on the media in order. Plate one is bacteria with plasmid on glucose medium. Plate two is bacteria with plasmid on glucose medium with ampicillin. Plate three is bacteria with plasmid on glucose medium with ampicillin and lactose. Plate four is bacteria without plasmid on glucose medium. Plate five is bacteria without plasmid on glucose medium with ampicillin. Plate six is bacteria without plasmid on glucose medium with ampicillin and lactose If the scientist had forgotten to use DNA ligase during the preparation of the recombinant plasmid, bacterial growth would most likely have occurred on which of the following?

Water in a pond contaminated with the weed killer atrazine is suspected of inhibiting metamorphosis in northern leopard frogs. A team of scientists collected fertilized northern leopard frog eggs from a different pond that is not contaminated. Which of the following is the best experimental design to determine whether atrazine is responsible for inhibiting metamorphosis in northern leopard frogs?

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 1, 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 1. Figure 1 represents the universal genetic code. Table 1. Selected mammalian groups. Group Selected Members Biosynthesis of Ascorbic Acid Nonprimate mammals Elephant, mouse Yes Primate mammals Lemur Yes Orangutan, chimpanzee No Human No   Table 2. DNA sequence alignment* 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. Figure 1. Universal genetic code 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. Which of the following phylogenetic trees best illustrates (with the X) the point at which the mutation in the GULO gene most likely occurred during the evolutionary history of these organisms?

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?