Blog

Which of the following observations best represents a mutualistic relationship?

Which of the following describes the mechanism by which a plant stem grows toward light?

Tay-Sachs disease is a rare inherited disorder caused by an autosomal recessive allele of the HEXA gene. Affected individuals exhibit severe neurological symptoms and do not survive to reproductive age. Individuals who inherit one copy of the allele (Tay-Sachs carriers) typically show no symptoms of the disorder. The frequencies of Tay-Sachs carriers in the general population of North America and in three different subpopulations are presented in the table. FREQUENCY OF TAY-SACHS CARRIERS IN DIFFERENT POPULATIONS Population Frequency of Tay-Sachs Carriers General population 0.004 Subpopulation I 0.037 Subpopulation II 0.035 Subpopulation III 0.020 A researcher claims that Tay-Sachs carriers are protected against the infectious disease tuberculosis (TB). Which of the following observations about the annual incidence of tuberculosis in subpopulation II could best be used to support the researcher’s claim?

The lionfish is a venomous fish found primarily in the Red Sea and the Indian Ocean. In the 1990s, lionfish were accidentally released into the Atlantic Ocean, where they found abundant resources and favorable environmental conditions. Which of the following scenarios is most likely to result in the lionfish having a major impact on the communities into which they were introduced?

Barriers that prevent mating between species before fertilization are called ______.

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. Which of the following best explains the differences in the armor of the Lake Washington stickleback population summarized in Figure 1?

The most important human cause of the loss of biodiversity is attributable to which factor?

A model of a process involving nucleic acids is shown in Figure 1. The figure presents a diagram of a replication fork that is growing to the right. Strands that are complementary to the separated template strands on the left side of the fork have already been synthesized. The template strands on the right side remain single stranded. The left end of the upper template strand is labeled 5 prime, and the right end is labeled 3 prime. Two sequential arrows that point from right to left represent the process of strand synthesis with nucleotides that are complementary to the upper template strand. The left end of each arrow is labeled 3 prime, and the right end of each is labeled 5 prime. The left end of the lower template strand is labeled 3 prime, and the right end is labeled 5 prime. A single long arrow that points from left to right represents the process of strand synthesis with nucleotides that are complementary to the lower template strand. The left end of the arrow is labeled 5 prime, and the right end is labeled 3 prime. Figure 1. Model of a process involving nucleic acids Which of the following best explains what process is represented in Figure 1?

Antigens are foreign proteins that invade the systems of organisms. Vaccines function by stimulating an organism’s immune system to develop antibodies against a particular antigen. Developing a vaccine involves producing an antigen that can be introduced into the organism being vaccinated and which will trigger an immune response without causing the disease associated with the antigen. Certain strains of bacteria can be used to produce antigens used in vaccines. Which of the following best explains how bacteria can be genetically engineered to produce a desired antigen?

The question refers 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 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. The mRNA transcribed from the DNA would read