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. Which of the following modifications of the DNA would produce the greatest change in the primary structure of the polypeptide chain?

Data regarding the presence (+) or absence (-) of five derived traits in several different species are shown in the table below. Presence (+) or absense (-) of five derived traits. Species Traits 1 2 3 4 5 V + + + – – W + + – – – X + – – + + Y – – – – – Z + – – – + Which of the following cladograms provides the simplest and most accurate representation of the data in the table?

In each diagram, the lac operon is represented by a rectangle that is divided into 6 segments that are labeled from left to right as follows: C A P Site, an unlabeled segment, Operator, lac Z, lac Y and lac A. In each diagram, an oval labeled R N A Polymerase is positioned under the right half of the C A P site, the unlabeled segment, and the left one fourth of the Operator. In the top diagram a rectangle with a triangular notch at the bottom is labeled Repressor, and it sits under and touches the right one third of the Operator, lac Z, and the first one third of lac Y. An X is marked on the right side of the R N A polymerase, and an arrow extends from the X to the repressor. In the bottom diagram a rectangle with a semicircular notch at the bottom is labeled Repressor, and it does not contact the lac operon. Instead a circle labeled allolactose fits into the semicircular notch of the repressor. An arrow extending from the R N A polymerase points to the right. Figure 1. Model of lac operon, comparing repressed and active states Which is a scientific claim that is consistent with the information provided and Figure 1?

_______ is the process in which RNA is produced by using a DNA template.

Living cells typically have biosynthetic pathways to synthesize at least some of the amino acids used in making proteins. Some strains of E. coli, a prokaryote, can synthesize the amino acid tryptophan, while other E. coli strains cannot. Similarly, some strains of the yeast S. cerevisiae, a eukaryote, can synthesize tryptophan, while other S. cerevisiae strains cannot. Which of the following describes the most likely source of genetic variation found in the tryptophan synthesis pathways of both species?

The following questions refer to the following information and graph. The data presented in the figure below are measurements of the rate of oxygen consumption at differing body masses in a species of fish. Each point represents measurements from a different fish. Measurements were taken at different temperatures. (○= 10°C,●= 15°C, □= 20°C, ■= 25°C.) The horizontal axis is labeled body mass by kilograms, and the horizontal axis is labeled rate of oxygen consumption by milligrams of oxygen per fish per hour. Five tick marks appear on the horizontal axis and are labeled, from left to right in even increments, zero point zero two, zero point zero five, zero point one, zero point two, zero point four. Six tick marks appear on the vertical axis in even increments, and are labeled, from bottom to top, one, two, four, eight, sixteen and thirty-two. On the graph, there are four sets of data taken at four different temperatures, ten degrees Celsius, fifteen degrees Celsius, twenty degrees Celsius and twenty-five degrees Celsius. Each set of data has a line of best fit. Each line has a positive slope, and the line for each temperature is above the line for the temperature below it. There are 20 data points for ten degrees Celsius, represented by open circles. The line of best fit starts at zero on the vertical axis and zero point zero three on the horizontal axis and increases with a straight line to eight milligrams of oxygen consumed per fish per hour at a body mass of zero point four kilograms. There are 22 data points for fifteen degrees Celsius, represented by shaded circles. The line of best fit starts below one on the vertical axis and zero on the horizontal axis and increases with a straight line to sixteen milligrams of oxygen consumed per fish per hour at a body mass of zero point four kilograms. There are 21 data points for twenty degrees Celsius, represented by open squares. The line of best fit starts between one and two on the vertical axis and zero on the horizontal axis and increases with a straight line to between sixteen and thirty-two milligrams of oxygen consumed per fish per hour at a body mass of zero point four kilograms. There are 20 data points for twenty-five degrees Celsius, represented by shaded squares. The line of best fit starts at two on the vertical axis and zero on the horizontal axis and increases with a straight line thirty-two milligrams of oxygen consumed per fish per hour at a body mass of zero point four kilograms. The best explanation for the fact that not all points lie on any given line is that

In the “RNA world” model for the origin of life, the first protocells (probionts) relied on RNA, not DNA, for information storage and transmission. Which of the following could best be considered evidence in support of the RNA world model?

Nondisjunction during meiosis can negatively affect gamete formation. A model showing a possible nondisjunction event and its impact on gamete formation is shown in Figure 1. The first part of the figure shows a cell that has 5 pairs of chromosomes lined up at the equator of the cell. Each chromosome is composed of 2 chromatids. An arrow points to the next part of the figure that shows the same cell with the chromosomes starting to separate. 4 chromosomes and an additional unseparated pair of chromosomes are shown on the left side of the cell. 4 chromosomes are shown on the right side of the cell. Arrows point from the cell to the next part of the figure that shows 2 cells. One of the 2 cells has 6 chromosomes, each composed of a pair of chromatids, all in a single line along the equator of the cell. The other cell has 4 chromosomes, each composed of a pair of chromatids, all in a single line along the equator of the cell. Arrows point from each cell to 2 more cells in the final part of the figure. The 2 cells derived from the cell with 6 chromosomes each have 6 chromosomes. The 2 cells derived from the cell with 4 chromosomes each have 4 chromosomes. These four cells are labeled Gametes. Figure 1. Model of a nondisjunction event Which of the following best describes the most likely impact on an individual produced from fertilization between one of the daughter cells shown and a normal gamete?

Scientists want to determine whether the rapid economic growth in China between 1950 and 2000 caused a mass extinction of animal species during that time period. A mass extinction event is considered to occur when the rate of species extinction far exceeds the background extinction rate. The scientists collected data on the number of extinctions of selected vertebrate species in China between 1950 and 2000, as shown in Table 1. Extinction rates were ascertained by measuring the average percent of species lost (PSL). Table 1. The average percent of extinctions of 252 protected vertebrate species in 2,365 counties in China from 1950 to 2000. Mammals Birds Amphibians and Reptiles Background Rate Average PSL ±1 standard deviation 47.4±33.59 19.75±17.70 28.80±28.86 27.21±16.20 Which of the following is the null hypothesis most relevant for evaluating the data in Table 1?

The graph below illustrates a logistic growth curve for a population of Euglena. with x-axis labeled time in days from zero to twenty, and y-axis labeled population size in organisms per milliliter. The growth curve starts with a population just above zero at zero days and increases exponentially to three-hundred organisms per milliliter at day ten, then flattening out and remaining at three-hundred. There is a point labeled Q on the curve at day twenty. The population would be most likely to remain at the level shown at Q under which of the following conditions?