There is an arrow extending from Germany to the United Kingdom, and another arrow extending from Germany to Spain. Blackcap birds (Sylvia atricapilla) migrate out of Germany before wintertime. Prior to the 1960s, all members of a particular blackcap population flew to Spain, which had an abundant natural food source. Now, some members of the same blackcap population fly to the United Kingdom, where food placed in feeders by humans is abundant. The blackcaps return to the same forests in Germany to nest during the breeding season. Some blackcaps that migrate to the United Kingdom have become distinguishable by certain physical and behavioral traits from blackcaps that migrate to Spain. Which of the following best predicts the effect on the blackcap population if humans in the United Kingdom continue to place food in feeders during the winter?
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The column labels, from left to right; column 1, Diet Before…
The column labels, from left to right; column 1, Diet Before Change; column 2, Diet After Change. The row labels, from top to bottom; row 1, Principal diet; row 2, Main type of carbohydrate supplied by diet; row 3, Abundance of Streptococcus by diet; row 4, Abundance of Ruminococcus albus in the rumen; row 5, p H of rumen fluid; row 6, Clinical warning signs in cattle. From top to bottom, the data is as follows; Principal diet; Diet before change, Hay, diet after change, Grain. Main type of carbohydrate supplied by diet; Diet before change, Cellulose, diet after change, Starch. Abundance of Streptococcus by diet; Diet before change, Low, diet after change, High. Abundance of Ruminococcus albus in the rumen; Diet before change, 6 to 7, diet after change, < 5.6. Clinical warning signs in cattle; Diet before change, None, diet after change, Weakness. Researchers conducted a controlled experiment to investigate the effect of diet on the health of beef cattle. The initial hypothesis was that an abrupt change in diet will benefit beef cattle by reducing the sizes of bacterial populations living in the digestive systems of the cattle. In the experiment, the researchers determined the relative abundance of two bacterial species found in the rumen of cattle. The rumen is a part of the stomach that acts as a fermentation chamber in cattle and other ruminants. Shown in the table are the results from before and after an abrupt change in the cattle’s diet. Based on the results, which of the following best explains why the initial hypothesis should be revised?
In an investigation of interspecies competition, researchers…
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. Which of the following statements best justifies the use of the experimental results in an investigation of interspecies competition?
Two populations of a species of squirrel are geographically…
Two populations of a species of squirrel are geographically isolated from each other. Although they have the same population density, one population is significantly larger in number than the other. A new bacterial disease, which is easily spread and extremely virulent, affects both populations at the same time. Which of the following is the best prediction of how the new disease will affect the two populations?
A naturalist studying competitive interactions between flowe…
A naturalist studying competitive interactions between flower-visiting animals in a meadow observes that hummingbirds always prevent butterflies from feeding on blue flowers. What would most likely occur upon removal of hummingbirds from the meadow?
Which of the following statements concerning a gene is corre…
Which of the following statements concerning a gene is correct?
Sickle-cell anemia is an inherited blood disorder in which o…
Sickle-cell anemia is an inherited blood disorder in which one of the hemoglobin subunits is replaced with a different form of hemoglobin. Partial DNA sequences of the HBB gene for normal hemoglobin and for sickle-cell hemoglobin are shown in Figure 1. Partial sequence for normal hemoglobin: The D N A sequence, in nucleotide triplets is T G A, G G A, C T C, C T C, T T C. The R N A sequence, in nucleotide triplets is A C U, C C U, G A G, G A G, A A G. The amino acid sequence is Threonine, Proline, Glutamic Acid, Glutamic Acid, Lysine. A shaded box highlights the D N A triplet C T C, the R N A triplet G A G, and the amino acid Glutaminc Acid. Partial sequence for sickle-cell hemoglobin: The D N A sequence, in nucleotide triplets is T G A, G G A, C A C, C T C, T T C. The R N A sequence, in nucleotide triplets is A C U, C C U, G U G, G A G, A A G. The amino acid sequence is Threonine, Proline, Valine, Glutamic Acid, Lysine. A shaded box highlights the D N A triplet C A C, the R N A triplet G U G, and the amino acid Valine. Figure 1. Comparison of partial DNA sequences for normal hemoglobin and hemoglobin with a sickle-cell mutation Which of the following best describes the type of mutation shown in Figure 1 that leads to sickle-cell anemia?
Rhagoletis pomonella is a parasitic fly native to North Amer…
Rhagoletis pomonella is a parasitic fly native to North America that infests fruit trees. The female fly lays her eggs in the fruit. The larvae hatch and burrow through the developing fruit. The next year, the adult flies emerge. Prior to the European colonization of North America, the major host of Rhagoletis was a native species of hawthorn, Crataegus marshallii. The domestic apple tree, Malus domestica, is not native to North America, but was imported by European settlers in the late 1700s and early 1800s. When apple trees were first imported into North America, there was no evidence that Rhagoletis could use them as hosts. Apples set fruit earlier in the season and develop faster, where hawthorns set later and develop more slowly. Recent analysis of Rhagoletis populations has shown that two distinct populations of flies have evolved from the original ancestral population of flies that were parasitic on hawthorns. One population infests only apple trees, and the other infests only hawthorns. The life cycles of both fly populations are coordinated with those of their host trees. The flies of each population apparently can distinguish and select mates with similar host preferences and reject mates from the population specific to the other host tree. There is very little hybridization (only about 5 percent) between the two groups. The divergence between the two populations of Rhagoletis must have occurred very rapidly because
The graph below shows changes in a population of wild sheep…
The graph below shows changes in a population of wild sheep that were introduced to the island of Tasmania in the early 1800’s. The horizontal axis is labeled year from 1820 to 1940 with tick marks in increments of ten. The vertical axis is labeled number of sheep in thousands from zero to two thousand five hundred, with tick marks in increments of 500. The population curve starts at zero in 1820, increases slowly then increases rapidly to two point two five million by 1850. This increase is enclosed in a bracket. After 1850 the population curve fluctuates between two point two five million and one point two five million, with a horizontal dashed line through the middle of these population fluctuations. The dashed line on the graph represents the
The growth curve starts at the intersection of the x and y a…
The growth curve starts at the intersection of the x and y axes and increases exponentially then flattens out halfway up the y-axis. Figure two is a graph with x-axis labeled time and y-axis labeled algae population. The growth curve starts at the intersection of the x and y axes and increases exponentially then flattens out halfway up the y-axis. Phosphate is added after the growth curve flattens out, then increases exponentially again and flattens out at the top of the y-axis. Figure I shows the growth of an algal species in a flask of sterilized pond water. If phosphate is added as indicated, the growth curve changes as shown in Figure II. Which of the following is the best prediction of the algal growth if nitrate is added instead of phosphate?