When DNA replicates, each strand of the original DNA molecule is used as a template for the synthesis of a second, complementary strand. Which of the following figures most accurately illustrates enzyme-mediated synthesis of new DNA at a replication fork?
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Hemoglobin is a highly conserved protein used by all mammals…
Hemoglobin is a highly conserved protein used by all mammals to transport oxygen. Table 1 was constructed by comparing the amino acid sequence in the hemoglobin molecules from five different mammal species. Table 1. Hemoglobin amino acid differences among five different mammal species Species 1 Species 2 Species 3 Species 4 Species 5 Species 1 — 21 11 13 5 Species 2 — 18 17 20 Species 3 — 3 10 Species 4 — 12 Species 5 — Which of the following best describes the importance of the number of amino acid differences indicated in Table 1?
Populations of organisms have the potential to increase at e…
Populations of organisms have the potential to increase at exponential rates, yet most population growth fits a logistic model of growth. Why is this?
Which of the following is NOT one of the ways an individual…
Which of the following is NOT one of the ways an individual organism can respond to external environmental changes?
A biologist spent many years researching the rate of evoluti…
A biologist spent many years researching the rate of evolutionary change in the finch populations of a group of islands. It was determined that the average beak size (both length and mass) of finches in a certain population increased dramatically during an intense drought between 1981 and 1987. During the drought, there was a reduction in the number of plants producing thin-walled seeds. Which of the following procedures was most likely followed to determine the change in beak size?
Which of the following statements best explains the structur…
Which of the following statements best explains the structure and importance of plasmids to prokaryotes?
The left side of the table is labeled First Base in Codon, a…
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. 5′- GTT TGT CTG TGG TAC CAC GTG GAC TGA – 3′ The DNA sequence is a small part of the coding (nontemplate) strand from the open reading frame of β-hemoglobin gene. Given the codon chart, what would be the effect of a mutation that deletes the G at the beginning of the DNA sequence?
There is a key of plants below the time progression, contain…
There is a key of plants below the time progression, containing grasses, annual plants, shrubs, oak trees and other trees. In the first time point the grass covers the ground and there are a few annual plants. The second time point has fewer grasses, 2 shrubs, and 3 oak trees. The third time point has no grass, 2 shrubs, 2 oak trees that have grown taller, and 3 other trees. The fourth time point has one oak tree that has grown even taller, and three other trees that have grown taller. There are no grasses, annual plants, nor shrubs. The fifth and final time point has one large oak tree and two large other trees.There are no grasses, annual plants, nor shrubs. The diagram above shows the progression of ecological events after a fire in a particular ecosystem. Based on the diagram, which of the following best explains why the oak trees are later replaced by other trees?
In an experiment to determine the effect of light availabili…
In an experiment to determine the effect of light availability on species richness in an environment, a grassland was divided into 26 plots that were assigned to one of two treatment groups: 1. Clipped – plots were mowed to simulate the grazing of herbivores that maintain the low height of grasses. 2. Enclosure – plots were enclosed in fences, and grasses were allowed to grow undisturbed to their maximum height. After eight years, data was collected on the amount of light penetration to the soil (bottom light density) and the number of species in the two treatment groups. Figure 1 shows the averaged light penetration measurements, and Figure 2 shows the averaged number of species per plot type. The horizontal axis is labeled Treatment, and the two bars indicated along it are labeled Enclosed and Clipped respectively. An error range is given for each bar. The vertical axis is labeled Bottom Light Density, in watts per meter squared, and the numbers 0 through 600, in increments of 200, are indicated. Each bar is described as follows. Note that all values are approximate. Enclosed. 500, plus or minus 25. Clipped. 170, plus or minus 25. Figure 1. Average bottom light intensity in clipped plots versus enclosed plots The horizontal axis is labeled Treatment, and the two bars indicated along it are labeled Enclosed and Clipped respectively. An error range is given for each bar. The vertical axis is labeled Average Number of Species per Plot, and the numbers 0 through 20, in increments of 2, are indicated. Each bar is described as follows. Note that all values are approximate. Enclosed. 11.6, plus or minus 0.3. Clipped. 17, plus or minus 1. Figure 2. Richness comparison of two treatments Which of the following statements is consistent with the data shown in the figures?
Why would a community in the early stages of secondary succe…
Why would a community in the early stages of secondary succession be less productive than a community in the later stages of secondary succession?