Which of the following correctly compares the two processes shown above? The first process is fertilization. A sperm fusing with an egg is shown. The second process is transduction. A five step process is shown. Step 1. Virus attaches to cell surface and inserts viral DNA into cell. Step 2. The cell’s circular DNA is broken into many pieces. Step 3. new viral particles are formed in the cell that include DNA from the host cell. Step 4. One of the new viral particles infects another cell. Step 5. The DNA from the virus is inserted into the circular chromosome of the shot cell.

Which of the following is NOT one of the ways an individual organism can respond to external environmental changes?

Which of the following observations best supports the hypothesis that a large object collided with Earth in a period of time associated with a mass extinction?

A current challenge for doctors involves the bacterial strain Clostridioides difficile, which no longer responds to traditional antibiotic treatments. Which of the following best explains why this particular strain of bacteria is resistant to antibiotic treatment?

Stickleback fish are found in both marine and freshwater habitats. The marine fish have no scales but have hardened, armorlike plates along their sides. The plates are thought to protect sticklebacks from certain predators. In the late 1980s, sticklebacks from a marine population colonized Loberg Lake, a freshwater lake in Alaska. Starting in 1990, researchers sampled fish from the lake every four years and recorded the armor-plate phenotypes of the male sticklebacks in each sample. The armor-plate phenotypes were categorized as either complete (plates extending from head to tail), partial (plates extending from head to abdomen), or low (a few plates near the head only). The results are shown in the table below. ARMOR-PLATE VARIATION IN THE STICKLEBACK POPULATION OF LOBERG LAKE Percent of Males in the Sample with Each Armor-Plate Phenotype Year Low Partial Complete 1990 1% 2% 97% 1994 45% 14% 41% 1998 58% 16% 26% 2002 76% 15% 9% 2006 90% 6% 4% Which of the following best explains the changes in the phenotype frequencies of the stickleback population in Loberg Lake?

All of the following are density-dependent factors that limit animal populations EXCEPT

African elephants, Loxodonta africana, are often hunted illegally for their tusks. Both male and female elephants have tusks, although the tusks are much larger in the males. Researchers have followed the elephant population in Gorongosa National Park in Mozambique for many years. Figure 1 shows the percent of tuskless female elephants expected in wild populations throughout the species’ range, as well as the percent of tuskless females that survived the hunting between 1977 and 1992 (those over 25 years old) and the percent of female elephants born after 1992 without tusks in the park. The data are based on 200 known female elephants in the park. There are 3 categories indicated along the horizontal axis: Wild Population, Females Older Than 25, and Females Born After 1992. The vertical axis is labeled Percent of Tuskless Females, and the numbers 0 through 100, in increments of 20, are indicated. The following data is represented by the graph. Wild Population, 4 percent of Tuskless Females. Females Older Than 25, 51 percent. Females Born After 1992, 32 percent. Figure 1. Comparison of the frequency of tuskless females by age cohort Which of the following best describes the process responsible for the change in the percent of tuskless female elephants in the Gorongosa National Park population shown in Figure 1?

In a hypothetical population of beetles, there is a wide variety of color, matching the range of coloration of the tree trunks on which the beetles hide from predators. The graphs below illustrate four possible changes to the beetle population as a result of a change in the environment due to pollution that darkened the tree trunks. All graphs have a y axis labeled population frequency, and an x axis labeled beetle color from light on the left to dark on the right. On graph one, the original population is a wide bell curve that begins a short distance to the right of light on the x axis and ends the bell curve at dark on the right side. The population after pollution is a bell curve that begins all the way at light on the left side and ends approximately halfway between light and dark. On graph two the original population is a wide bell curve that begins on the left side at light and ends on the right side at dark, with the peak in the center. The population after pollution is a curve with two peaks, the first peak is on the light side and the second peak is on the dark side, dropping in the middle in between both peaks.On graph three the original population is a wide bell curve that begins on the left side at light and ends on the right side at dark, with the peak in the center. The population after pollution is a narrow bell curve that fits under the original population bell curve, beginning to the right of left and ending to the left of dark.On graph four two the original population is a wide bell curve that begins on the left side at light and ends on the right side at dark, with the peak in the center. The population after pollution is bell cure that is narrower than the original population’s bell curve, beginning just before the middle of the graph and ending to the right of the original population on the dark side. Which of the following includes the most likely change in the coloration of the beetle population after pollution and a correct rationale for the change?

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?

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?