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The horizontal axis is labeled Species Present, and the vertical axis is labeled Nitrogen Influx in kilograms per year. On the horizontal axis, from left to right, there are evenly spaced tick marks labeled N, B, S, and S B. The vertical axis has evenly spaced tick marks labeled 300 through 600,in increments of 100. The pointover N is at about 375. Its error bar extends from about 350 to about 425. The pointover B is centered at about 375. Its error bar extendsfrom about 360 to about 405. The pointover S is centered at about 430. Its error bar extends from about 410 to about 470. The pointover S B is centered at about 510. Its error bar extendsfrom about 500 to about 530. The caption on the figure reads: Figure 1. Mean nitrogen influx plus or minus 2 times the standard error of the mean as a function of species present. N equals neither salmon nor bears present, B equals only bears present, S equals only salmon present, and S B equals both salmon and bears present. Figure 1. Mean nitrogen influx ±2SEẍ as a function of species present. N = neither salmon nor bears present, B = only bears present, S = only salmon present, and SB = both salmon and bears present. Pacific salmon and black bears have often been cited as examples of keystone species. Pacific salmon spawn in freshwater streams but spend most of their lives at sea. When mature salmon return to the freshwater streams to spawn, they are preyed upon by bears and other predators. When salmon migrate from their marine habitat to the freshwater streams, they bring nitrogen and other marine-derived nutrients that subsequently remain in the areas surrounding the streams—a process called nitrogen influx. In an investigation, the relationship between black bears, salmon, and influx of marine nitrogen into the area around a southwestern Alaskan stream was studied. The investigators established several test plots of the same size along the stream with the following species composition: no salmon or black bears (N), bears but not salmon (B), salmon but not bears (S), and a plot where salmon and bears interact (SB). Nitrogen influx in the different sampling areas was measured as a means of assessing the impact of the different species on the health of the ecosystem. The data are plotted in Figure 1. Which of the following most likely describes how the interaction between bears and salmon influences nitrogen dynamics in the environment?

The following questions refer 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 figure shows the universal codon table with 4 main data rows and 4 main data columns. 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. In which of the following would there NOT be a change in the amino acid sequence of the peptide coded for by this DNA?

A small number of lizards from a mainland population have been deposited on four isolated islands because of the effects of a rare strong storm. Which of the following best predicts the outcome of these lizards reproducing for many generations on the islands?

The population of Japanese sika deer in central Japan was determined each year from 2005 to 2014. The sika deer population underwent logistic growth starting at 26,000 deer in 2005 and stabilized at 58,000 deer between 2012 and 2014. Based on these data, the rmax for this population was determined to be 0.57. Central Japan contains a variety of habitats, including forests and grasslands. Sika deer benefit from the resources in grasslands more than forests; if deforestation occurred, producing more grasslands in the region, the carrying capacity for sika deer population would increase. What would the population size of sika deer be one year after the carrying capacity increased to 70,000 as a result of deforestation? Assume rmax does not change.

More than 90 percent of the nitrogenous waste that is processed and excreted by humans is derived from the breakdown of proteins. Most of the remaining nitrogenous waste material is derived from the breakdown of

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. 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. Evolution of a new trait typically takes many generations. Yet a dramatic shift in the extent of armor plating in the Lake Washington stickleback population occurred in the 50 years following the cleanup of the lake. Which of the following best describes the mechanism of the rapid evolution of the armor phenotype in the Lake Washington sticklebacks?

Simpsons Diversity Index is a way to quantify the diversity of a community.  The equation can be written as: D = N(N-1) / ∑ n ( n– 1)  Where: D = Diversity index N = Total number of individuals of all species n = Number of individuals of a specific species   Community A has 3 species (A, B, C). There are 5 individuals of each species. (N=15) Community B has 6 species (A, B, C, D, E). There are 3 individuals each of species A, B, and C. There are 2 individuals each of species D, E, and F. (N=15) Community C has 5 species (A, B, C, D, E). There are 3 individuals each of species. (N=15)   Which community has the highest species diversity?  

Commercial bananas are grown as a monoculture, with all banana plants cloned from one original banana plant. The commercial strains of bananas are seedless, so each new banana plant has to be manually planted from a cutting of an existing banana root. In the 1950s, the Gros Michel banana strain, the dominant export banana at that time, was destroyed by the fungus Fusarium oxysporum. A new Fusarium resistant variety, the Cavendish banana, was developed and is currently the banana strain grown for export. Recently, a Fusarium strain that successfully attacks the Cavendish strain has been documented. Which of the following best provides reasoning supporting a method that would help protect commercial banana crops from infection by pathogenic organisms such as Fusarium fungi?

The enzyme lactase aids in the digestion of lactose, a sugar found in the milk of most mammals. In most mammal species, adults do not produce lactase. Continuing to produce lactase into adulthood in people is called lactase persistence. A number of different alleles have been identified that result in lactase persistence. Figure 1 shows the percentage of people in different geographic areas parts of the Old World that exhibit lactase persistence. of Europe, most of Africa, and parts of Asia with distributions of 5 ranges of percent of lactase persistence across the regions. The ranges of percent are 81 to 100 percent, 61 to 80 percent, 41 to 60 percent, 21 to 40 percent, and 0 to 20 percent. The smallest portion of the map is in the 81 to 100 percent range and represents all of the United Kingdom, Denmark, small regions of southern Sweden, and small parts of western Norway. A small portion of the map is in the 61 to 80 percent range and represents most of northern Europe, including the rest of Sweden, Norway, regions of Finland, Estonia, Latvia, Lithuania, Northern Germany, Northern France, and also Senegal in Africa. A large portion of the map is in the 41 to 60 percent range and represents the rest of Western Europe, central Africa, Saudi Arabia, Yemen, Oman, the United Arab Emirates, Qatar, and Pakistan. The largest portion of the map is in the 21 to 40 percent range and represents northern and southern Africa, Turkey, Syria, Iraq, Jordan, Israel, Palestine, parts of Iran, Turkmenistan, Uzbekistan, and almost all of Russia. A small portion of the map is in the 0 to 20 percent range and represents a small region in western Russia, a large portion of Kazakhstan, and southern Spain. Figure 1. Distribution of lactase persistence in Europe, North Africa, and parts of Asia Which of the following best explains the distribution of lactase persistence in the areas shown in Figure 1 ?

A new mutation that arose in one copy of gene X in a somatic cell resulted in the formation of a tumor. Which of the following pieces of evidence best describes how the new mutation directly caused the tumor?